octonions and albert algebras over commutative rings · 2019-10-29 · holger p. petersson and...

Holger P. Petersson and Michel L. Racine

Octonions and Albert algebrasover commutative ringsPreliminary version

October 23, 2019

Springer

Contents

1 Prologue: the ancient protagonists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 The Graves-Cayley octonions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Cartan-Schouten bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Unital subalgebras of O and their Z-structures . . . . . . . . . . . . . . . . . . 114 Maximal quaternionic and octonionic Z-structures . . . . . . . . . . . . . . . 165 The euclidean Albert algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 Z-structures of unital real Jordan algebras . . . . . . . . . . . . . . . . . . . . . . 317 The Leech lattice. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2 Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398 The language of non-associative algebras . . . . . . . . . . . . . . . . . . . . . . . 399 Unital algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4210 Scalar extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4611 Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5612 Quadratic maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6013 Polynomial laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3 Alternative algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8514 Basic identities and invertibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8515 Strongly associative subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8916 Homotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9117 Isotopy involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9618 Finite-dimensional alternative algebras over fields. . . . . . . . . . . . . . . . 101

4 Composition algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10319 Conic algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10320 Conic alternative algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11021 The Cayley-Dickson construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11322 Basic properties of composition algebras. . . . . . . . . . . . . . . . . . . . . . . 12023 Hermitian forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13324 Ternary hermitian spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

v

vi Contents

25 Reduced composition algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14626 Norm equivalences and isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 15727 Affine schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16728 Faithfully flat etale splittings of composition algebras. . . . . . . . . . . . . 184

5 Jordan algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20529 Linear Jordan algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20530 Para-quadratic algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20931 Jordan algebras and basic identities. . . . . . . . . . . . . . . . . . . . . . . . . . . . 21832 Power identities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23133 Inverses, isotopes and the structure group. . . . . . . . . . . . . . . . . . . . . . . 23734 The Peirce decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

6 Cubic Jordan algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26335 From cubic norm structures to cubic Jordan algebras . . . . . . . . . . . . . 26336 Building up cubic norm structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28537 Elementary idempotents and cubic Jordan matrix algebras . . . . . . . . 295

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459

Chapter 1Prologue: the ancient protagonists

c.PROLOGUE

Prominent specimens belonging to the protagonists of the present volume will beintroduced here not only over the field R of real numbers but also over the ring Z ofrational integers. The results we obtain or at least sketch along the way will serveas a motivation for the systematic study we intend to carry out in the subsequentchapters of the book.

1. The Graves-Cayley octonions

s.GRACAAt practically the same time, William R. Hamilton (1843), John T. Graves (1843)and Arthur Cayley (1845) discovered what are arguably the most important non-commutative, or even non-associative, real division algebras: the Hamiltonian quater-nions and the Graves-Cayley octonions. It is particularly the latter, with their abilityto co-ordinatize the euclidean Albert algebra (cf. 5.5 below), that deserve our atten-tion. We begin with a brief digression into elementary linear algebra.

ss.CROPRO1.1. The cross product in 3-space. We regard complex column 3-space C3 as aright vector space over the field C of complex numbers in the natural way. It carriesthe canonical hermitian inner product

C3×C3 −→ C, (u,v) 7−→ utv,

with respect to which the unit vectors ei ∈C3, 1≤ i≤ 3, form an orthonormal basis;we write u 7→ ‖u‖ for the corresponding hermitian norm on C3. Recall that the cross(or vector) product on C3 can be defined by the formula

(u× v)tw = det(u,v,w) (u,v,w ∈ C3). (1) DECRO

The cross product is complex linear in each variable and alternating; more precisely,u× v = 0 if and only if u and v are linearly dependent; in particular u× u = 0.Moreover, (1) implies that the expression (u× v)tw remains invariant under cyclicpermutations of its arguments. The cross product of the unit vectors is determinedby the formula

ei× e j = el for (i jl) ∈ (123),(231),(312). (2) UNCRO

1

2 1 Prologue: the ancient protagonists

The specification of the indices i, j, l in this formula will henceforth be expressedby saying that it holds for all cyclic permutations (i jl) of (123). Finally, the crossproduct satisfies the Grassmann identity

(u× v)×w = vwtu−uwtv = w× (v×u) (3) GRAS

for all u,v,w ∈ C3, which immediately implies the Jacobi identity

(u× v)×w+(v×w)×u+(w×u)× v = 0. (4) JAC

ss.REALG1.2. Real algebras. A detailed dictionary of non-associative algebras will be pre-sented in 8.1 below. For the time being, it suffices to define a real algebra as avector space A over the field R of real numbers together with an R-bilinear product(x,y) 7→ xy from A×A to A that need neither be associative nor commutative.Norwill A in general admit an identity element, but if it does, it is said to be unital.Asub-vector space of A stable under multiplication is called a subalgebra; it is a realalgebra in its own right. We speak of a unital subalgebra B of A if B is a subalge-bra containing the identity element of A.For X ,Y ⊆ A, we denote by XY the additivesubgroup of A generated by all products xy, x∈ X , y∈Y ; we always write X2 := XXand Xy := Xy, xY := xY for all x ∈ X , y ∈ Y . If A and B are real algebras, wedefine a homomorphism from A to B as a linear map h : A→ B preserving products:h(xy) = h(x)h(y) for all x,y ∈ A. A real algebra A is said to be a division algebra ifA 6= 0 and for all u,v ∈ A, u 6= 0, the equations ux = v and yu = v can be solveduniquely in A; if A has finite dimension (as a vector space over R), this is equivalentto the absence of zero divisors: for all a,b ∈ A, the equation ab = 0 implies a = 0 orb = 0.

Now suppose A is a finite-dimensional real algebra and let (ei)1≤i≤n be a basis ofA over R. Then there is a unique family (γi jl)1≤i, j,l≤n of real numbers such that

e jel =n

∑i=1

γi jlei (1≤ j, l ≤ n). (1) RESTR

The γi jl are called the structure constants of A relative to the basis chosen; theydetermine the multiplication of A uniquely. But note that different bases of the samealgebra may have vastly different structure constants. Given a finite-dimensionalreal algebra, the objective must be to find a basis with a particularly simple set ofstructure constants. If A and B are two real algebras of the same finite dimension,with bases (ei)1≤i≤n and (di)1≤i≤n, respectively, then the linear bijection h : A→ Bsending ei to di for 1≤ i≤ n is easily seen to be an isomorphism of real algebras ifand only if the family of structure constants of A relative to (ei) is the same as thefamily of structure constants of B relative to (di).

ss.REQUAM1.3. Real quadratic maps. A map Q : V →W between finite-dimensional realvector spaces V,W is said to be quadratic if it is homogeneous of degree 2, soQ(αv) = α2Q(v) for all α ∈ R, v ∈V , and the map

1 The Graves-Cayley octonions 3

DQ : V ×V −→W, (v1,v2) 7−→ Q(v1 + v2)−Q(v1)−Q(v2), (1) DERQUAM

is (symmetric) bilinear. In this case, we call DQ the bilinearization or polar map ofQ. We mostly write Q(v1,v2) := DQ(v1,v2) if there is no danger of confusion.

For example, given a finite-dimensional real algebra A, the squaring

sq : A−→ A, x 7−→ x2, (2) SQUAM

is a quadratic map with bilinearization given by

sq(x,y) = Dsq(x,y) = xy+ yx (3) DERSQUAM

for all x,y ∈ A.Recall that a real quadratic form, i.e., a quadratic map q : V → R, is positive

(resp. negative) definite if q(v)> 0 (resp. < 0) for all non-zero elements v ∈V .ss.DEFGC

1.4. Defining the Graves-Cayley octonions. By restricting scalars from C to R,

O := C⊕C3

may be viewed as a real vector space (of dimension 8 whose elements will be writtenas a⊕u, a∈C, u∈C3. Following Zorn [129, p. 401], we make O into a real algebraby the multiplication

(a⊕u)(b⊕ v) := (ab− utv)⊕ (va+ub+ u× v) (a,b ∈ C, u,v ∈ C3). (1) PROGC

This algebra, also denoted by O, is called the algebra of Graves-Cayley octonions.It has an identity element furnished by

1O := 1⊕0 (2) UNGC

but is neither commutative nor associative. For example, an easy computation shows((0⊕ e1)(0⊕ e2)

)(0⊕ e3i) = (−i)⊕0 =−(0⊕ e1)

((0⊕ e2)(0⊕ e3i)

).

ss.NOTR1.5. Norm, trace and conjugation. The map nO : O→ R defined by

nO(a⊕u) := aa+ utu = |a|2 + utu (a ∈ C, u ∈ C3) (1) NOGC

is called the norm of O. It is a positive definite quadratic form whose bilinearizationmay be written as

nO(x,y) = ab+ ba+ utv+ vtu = 2Re(ab+ utv) (2) BLINOGC

for x = a⊕u, y = b⊕v, a,b ∈C, u,v ∈C3 and hence gives rise to a euclidean scalarproduct on O defined by


〈x,y〉 :=12

nO(x,y) = Re(ab+ utv), (3) EUSGC

making O into a real euclidean vector space with corresponding euclidean norm

‖x‖ := 〈x,x〉=√

nO(x) =√|a|2 +‖u‖2. (4) EUNGC

Definition (3) is the standard way of introducing a euclidean scalar product in thepresent set-up. On the other hand, expression of (2) is important in a more generalcontext when working over commutative rings where 1

2 is not available.The norm of O as defined in (1) canonically induces the trace of O, i.e., the linear

form tO : O→ R defined by

tO(x) := nO(1O,x) = 2Re(a) = a+ a, (5) TRAGC

and the conjugation of O, i.e., the linear map

ιO : O−→O, x 7−→ x := tO(x)1O− x, (6) COGC

which obviously satisfies

a⊕u = a⊕ (−u) (a ∈ C, u ∈ C3), (7) CONGC

leaves the norm invariant:

nO(x) = nO(x) (x ∈O), (8) NOCOGC

and has period 2: ¯x = x for all x ∈O. We view

O0 := Ker(tO) = x ∈O | x =−x= (iR)⊕C3. (9) PUGC

as a euclidean subspace of O. For a ∈ C, u ∈ C3, we use (1.4.1) to compute

(a⊕u)2 = (a2− utu)⊕(2Re(a)u

)= 2Re(a)(a⊕u)− (|a|2 +‖u‖2)(1⊕0),

so by (1), (5) every element x ∈O satisfies the quadratic equation

x2− tO(x)x+nO(x)1O = 0. (10) QUAGC

Combining this with (6), we conclude

xx = nO(x)1O = xx, x+ x = tO(x)1O. (11) CONOTR

Moreover, replacing x by x+ y, expanding and collecting mixed terms in (10), weconclude (see also (1.3.3))

x y := xy+ yx = tO(x)y+ tO(y)x−nO(x,y)1O. (12) BLAGC

1 The Graves-Cayley octonions 5

The process of passing from (10) to (12) will be encountered quite frequently in thepresent work and is called linearization. We also note

nO(x, y) = tO(xy) = nO(x,y) (13) TRANO

for x = a⊕u, y = b⊕ v, a,b ∈ C, u,v ∈O since (1.4.1), (5), 2 yield

tO(xy) = 2Re(ab− utv) = nO(a⊕ (−u),b⊕ v) = nO(x,y),

hence the second equation of (13), while the first now follows from (8) linearized.ss.ALTGC

1.6. Alternativity. As we have seen in 1.4, the algebra of Graves-Cayley octonionsis not associative. On the other hand, it follows from Exc. 1 below that it is alterna-tive: the associator [x,y,z] := (xy)z−x(yz) is an alternating (trilinear) function of itsarguments x,y,z ∈O. In particular, the identities

x2y = x(xy), (xy)x = x(yx), (yx)x = yx2

hold in all of O. Alternative algebras are an important generalization of associativealgebras and will be studied more systematically, under a much broader perspective,in later portions of the book, see particularly Chap. 3 below.

t.DIGC1.7. Theorem. The Graves-Cayley octonions form an eight-dimensional properlyalternative real division algebra, and the norm of O permits composition: nO(xy) =nO(x)nO(y) for all x,y ∈O.

Proof. As we have seen, the real algebra O is alternative (1.6) but not associative(1.4). It remains to show that the positive definite quadratic form nO permits compo-sition since this immediately implies that O has no zero divisors, hence is a divisionalgebra (1.2). Accordingly, let x = a⊕u, y = b⊕v, a,b ∈C, u,v ∈C3. Then (1.4.1),(1.5.1) yield

nO(xy) = nO((ab− utv)⊕ (va+ub+ u× v)

)=(ab− vtu

)(ab− utv)+(vta+ ut b+(u× v)t)(va+ub+ u× v

)= aabb− abutv− vtuab+ vtuutv+ vtvaa+ vtuab+ vt(u× v)+ utvab

+ utubb+ ut(u× v)+(u× v)tva+(u× v)tub+(u× v)t(u× v)

Observing (1.1.1), we obtain vt(u× v) = ut(u× v) = (u×v)tv = (u×v)tu = 0, while(1.1.1) combined with the Grassmann identity (1.1.3) implies

(u× v)t(u× v) = vt((u× v)× u)= vtvutu− vtuutv.

Hence the preceding displayed formula reduces to

nO(xy) = aabb+ aavtv+ utubb+ utuvtv = nO(x)nO(y),

as claimed.


r.LICO

1.8. Remark. The composition formula nO(xy) = nO(x)nO(y) is bi-quadratic inx,y ∈O and by (repeated) linearization yields

nO(x1y,x2y) = nO(x1,x2)nO(y),

nO(xy1,xy2) = nO(x)nO(y1,y2),

nO(x1y1,x2y2)+nO(x1y2,x2y1) = nO(x1,x2)nO(y1,y2)

for all x,y,x1,x2,y1,y2 ∈O.ss.FOIN

1.9. A formula for the inverse. Given x 6= 0 in O, Thm. 1.7 yields unique elementsy,z ∈O such that xy = zx = 1O. In fact, by (1.5.11), we necessarily have

y = z = x−1 :=1

nO(x)x.

We call x−1 the inverse of x in O.ss.HAQU

1.10. The Hamiltonian quaternions. The subspace H := R⊕R3 of O is actuallya subalgebra since

(α⊕u)(β ⊕ v) = (αβ −utv)⊕ (αv+βu+u× v). (α,β ∈ R, u,v ∈ R3) (1) PROH

This algebra is called the algebra of Hamiltonian quaternions. It contains an identityelement since 1H := 1O ∈H. Note that the Hamiltonian quaternions can be defineddirectly, without recourse to the Graves-Cayley octonions, by the product (1) on thereal vector space R⊕R3. Moreover, the vectors

1H := 1O, i := 0⊕ e1, j := 0⊕ e2, k := 0⊕ e3 (2) IJK

form a basis of H over R in which 1H acts as a two-sided identity element and, by(1) combined with (1.1.2),

i2 = j2 = k2 =−1H, ij = k =−ji, jk = i =−kj, ki = j =−ik. (3) MUIJK

Thus we have obtained a basis of the Hamiltonian quaternions with a particularlysimple family of structure constants (cf. 1.2).

Restricting the norm, trace, conjugation of O to H, we obtain what we call thenorm, trace, conjugation of H, denoted by nH, tH, ιH, respectively, which enjoy thesame algebraic properties we have derived for the Graves-Cayley octonions in 1.5.By the same token, H may also be viewed canonically as a real euclidean vectorspace. Consulting Thm. 1.7, we now obtain the following result.

c.ASDIH1.11. Corollary. The Hamiltonian quaternions H form an associative but not com-mutative real division algebra of dimension four, and the norm of H permits com-position: nH(xy) = nH(x)nH(y) for all x,y ∈H.

2 Cartan-Schouten bases 7

Proof. The only statement demanding a proof is the assertion that H is associative.But this can either be verified directly, or follows immediately from Exc. 1 combinedwith the Jacobi identity (1.1.4).

Exercises.

pr.ASSGC 1. The associator of O. The deviation of a real algebra A from being associative is measured by itsassociator [x,y,z] := (xy)z− x(yz) for x,y,z ∈ A. Prove for i = 1,2,3 and xi := ai⊕ui ∈O, ai ∈ C,ui ∈ C3, that

[x1,x2,x3] =(det(u1,u2,u3)−det(u1,u2,u3)

)⊕(∑((ui×u j)× uk +(ui× u j)(ak− ak)

)),

where the sum on the right is extended over all cyclic permutations (i jk) of (123). Conclude thatthe associator on O is an alternating (trilinear) function of its arguments.

pr.DIVO 2. Prove directly, i.e., without recourse to properties of the norm, that O is a division algebra.

pr.INVAS 3. Norm, trace and conjugation under the product of O.(a) Prove that the conjugation of O is an algebra involution, so not only ¯x = x but also xy = yx

for all x,y ∈O.(b) Show that trace and norm of O satisfy

tO(xy) = tO(yx), tO((xy)z

)= tO

(x(yz)

), nO(x,zy) = nO(xy,z) = nO(y, xz)

for all x,y,z ∈O.(c) Prove x(xy) = nO(x)y = (yx)x and xyx := (xy)x = x(yx) = nO(x, y)x−nO(x)y for all x,y∈O.

pr.MIMOU 4. The Moufang identities. We will see later (cf. 14.3) that the Moufang identities

x(y(xz)

)= (xyx)z, (xy)(zx) = x(yz)x,

((zx)y

)x = z(xyx)

(cf. Exc. 3 (c) for a parantheses-saving notation)) hold in arbitrary alternative algebras. Give adirect proof for the alternative algebra O by reducing to the case y,z ∈ 0⊕C3 and using as wellas proving the cross product identity

(u× v)wt +(v×w)ut +(w×u)vt = det(u,v,w)13 (1) CROMA

for all u,v,w ∈ C3.

2. Cartan-Schouten bases

s.CASHIn 1.10 we have exhibited a basis for the Hamiltonian quaternions with a particularlysimple family of structure constants. In this section, we will pursue the same objec-tive for the Graves-Cayley octonions. Due to their higher dimension and the lackof associativity, however, matters will be not quite as simple as in the quaternioniccase.


ss.DECS

2.1. Defining Cartan-Schouten bases (Cartan-Schouten [18]). We define a Cartan-Schouten basis of O as a basis consisting of the identity element 1O and additionalvectors u1, . . . ,u7 ∈ O such that the following two conditions are fulfilled, for allr = 1, . . . ,7:

u2r = −1O, (1) SQCS

ur+iur+3i = ur =−ur+3iur+i (i = 1,2,4), (2) MICS

where the indices in (2) are to be reduced mod7. We then put u0 := 1O, allow-ing us to write Cartan-Schouten bases as (ur)0≤r≤7. By (1) and (1.5.10), we havetO(ur) = 0, hence ur ∈ O0 for 1 ≤ r ≤ 7. The conscientious reader may wonderwhether equations (1), (2) really define a multiplication table for the basis chosen,i.e., whether all possible products between the basis vectors are well defined anduniquely determined by the preceding conditions. That this is indeed the case willbe settled affirmatively in Exc. 5 below. We also note by Exc. 6 that Cartan-Schoutenbases are orthonormal.

p.EXCS

2.2. Proposition. Cartan-Schouten bases of O exist.

Proof. Let (u1,u2) be a pair of ortho-normal vectors in the seven-dimensional eu-clidean vector space O0. Then (1.5.13) yields

tO(u1u2) = nO(u1,u2) =−nO(u1,u2) = 0,

hence u1u2 ∈ O0. Now let u3 ∈ O0 be an orthonormal vector that is perpendicularto u1, u2, and u1u2. Then Exc. 6 shows that u0 = 1O,u1,u2,u3,ur := ur−3ur−2 (4≤r ≤ 7) make up a Cartan-Schouten basis of O.

There is a remarkable interplay between Cartan-Schouten bases and projectiveplanes that provides a first glimpse at the profound connection between non-associative algebras and geometry; for more on this fascinating topic, we refer thereader to Faulkner [30]. Here we only sketch some details.

ss.INGE2.3. Incidence geometries. An incidence geometry G consists of

(i) two disjoint sets P = P(G ), whose elements are called points, and L =L (G ), whose elements are called lines,

(ii) a relation between points and lines, i.e., a subset I ⊆P×L .

If P ∈P , ` ∈L satisfy (P, `) ∈ I, we write P|` or `|P and say P is incident to ` (orP lies on `, or ` passes through P). If several points all lie on a single line, they aresaid to be collinear.

ss.PROP

2.4. Projective planes. An incidence geometry G as in 2.3 is called a projectiveplane if

(i) any two distinct points lie on a unique line,

2 Cartan-Schouten bases 9

(ii) any two distinct lines pass through a unique point,(iii) there are four points no three of which are collinear.

Important examples are provided by P2(F), the projective plane of a field F : itspoints (resp. lines) are defined as the one-(resp. two-)dimensional subspaces of(three-dimensional column space) F3 over F , and a point P is said to be incidentwith a line ` if P⊆ `. Working with the canonical scalar product (x,y) 7→ xty on F3,it is clear that P (resp. `) ⊆ F3 is a point (resp. a line) if and only if P⊥ (resp. `⊥)⊆ F3 is a line (resp. a point), and P is incident to ` if and only if `⊥ is incident toP⊥. In particular, over a finite field F , there are as many points as there are lines inP2(F).

ss.FACA2.5. The Fano plane and Cartan-Schouten bases. The Fano plane is the pro-jective plane P2(F2), where F2 stands for the field with two elements. The pointsof this geometry have the form 0,x with 0 6= x ∈ F3

2, hence identify canonicallywith the seven elements of F3

2 \0, while the lines of this geometry have the form0,x,y,x+ y, where x,y ∈ F3

2 \ 0 are distinct points. Hence each line, of whichthere are seven by what we have seen in 2.4, consists of three points (besides 0) thatare permuted cyclically under addition.

In the standard visualization of the Fano plane (see Fig. 1 below), its seven linesare represented by (i) the three sides, (ii) the three medians, and (iii) the inner circleof an isoscale triangle, while its seven points are located and numbered as shown.The entire picture fits into a directed graph whose nodes agree with the points of theFano plane and give rise to subdivisions of the seven lines, yielding fifteen edgesdirected in the way indicated. The key feature of this construction is that it allows to

u3 u2 u5

u6

u4 u1

u7

Fig. 1

recover the Graves-Cayley octonions on the free real vector space generated by thenodes ur, 1 ≤ r ≤ 7, and an additional element u0. In order to accomplish this, itsuffices to define a multiplication on the basis vectors that enjoys the characteristicproperties (2.1.1), (2.1.2) of Cartan-Schouten bases. We do so by setting u0ur :=ur =: uru0 for 0 ≤ r ≤ 7, u2

r = −u0 for 1 ≤ r ≤ 7, and by defining urus for 1 ≤


r,s ≤ 7, r 6= s, in the following way: let ut , 1 ≤ t ≤ 7, be the third point on the linecontaining ur and us. If, after an appropriate cyclic permutation of the indices r,s, t,the orientation of the edge joining ur and us leads from ur to us (resp. from us to ur),we put urus := ut (resp. urus :=−ut ). Then (2.1.1) holds by definition, while (2.1.2)can be verified in a straightforward manner..

ss.SYGC2.6. Symmetries of the Graves-Cayley octonions. One of the most important fea-tures of the Graves-Cayley octonions is the fact that, in spite of their non-associativecharacter, they have lots of symmetries. More specifically, we consider their auto-morphism group, denoted by Aut(O) and defined as the set of bijective linear mapsϕ : O→O satisfying ϕ(xy) = ϕ(x)ϕ(y) for all x,y ∈O; it is obviously a subgroupof GL(O), the full linear group of the real vector space O. We know from Exc. 7below that Aut(O) canonically embeds as a closed subgroup into the orthogonalgroup O(O0) ∼= O7(R) of the euclidean vector space O0, a compact real Lie groupof dimension 21, and hence is a compact real Lie group in its own right [19, Cor. ofProp. IV.XIV.2]. Our claim that O has lots of symmetries will now be corroboratedby the formula

dim(

Aut(O))= 14. (1) DIAU

At this stage, we will not be able to give a rigorous proof of this formula; instead,we will confine ourselves to a naive dimension count making the formula intuitivelyplausible.

First of all, the automorphism group of O acts simply transitively on the set ofCartan-Schouten bases of O. Each Cartan-Schouten basis, in turn, by Exc. 6 below,is completely determined by an element of

X0 := (u1,u2,u3) ∈ X | 〈u1u2,u3〉= 0, (2) PARCS

where X stands for the set of orthonormal systems of length 3 in the euclidean vec-tor space O0 ∼= R7. Thus we have dim(Aut(O)) = dim(X0). On the other hand,the orthogonal group O7(R) acts transitively on X , and for the first three unit vec-tors e1,e2,e3 ∈ R7, the isotropy group of (e1,e2,e3) ∈ X identifies canonically withO4(R). Hence

dim(X) = dim(

O7(R))−dim

(O4(R)

)=

7 ·62− 4 ·3

2= 21−6 = 15.

But (2) shows that X0 is a “hyper-surface” in X , which gives

dim(

Aut(O))= dim(X0) = dim(X)−1 = 14,

as claimed in (1).

Exercises.

3 Unital subalgebras of O and their Z-structures 11

pr.MUTA 5. Put M := r ∈ Z | 1≤ r ≤ 7×1,2,4 and show for s, t ∈ Z, 1≤ s, t ≤ 7, that s 6= t if and onlyif either there is a unique element (r, i) ∈ M satisfying r + i ≡ s mod 7 and r + 3i ≡ t mod 7, orthere is a unique element (r, i) ∈M satisfying r+3i≡ s mod 7 and r+ i≡ t mod 7.

pr.CHACS 6. Characterization of Cartan-Schouten bases. Prove for a family (ur)0≤r≤7 of elements in O thatthe following conditions are equivalent.

(i) (ur)0≤r≤7 is a Cartan-Schouten basis of O.(ii) u0 = 1O and

u2r =−1O = (urur+1)ur+3 = ur(ur+1ur+3) (1≤ r ≤ 7, indices mod 7). (3) ELGRO

(iii) u0 = 1O, and (ur)1≤r≤7 is a basis of O0 such that ‖ur‖= 1 for 1≤ r ≤ 7 and

ur+iur+3i = ur (1≤ r ≤ 7, i = 1,2,4, indices mod 7). (4) HAMI

(iv) u0 = 1O, and (u1,u2,u3) is an orthonormal system in the euclidean vector space O0 suchthat

nO(u1u2,u3) = 0, ur = ur−3ur−2 (4≤ r ≤ 7). (5) EXMI

In this case, (ur)1≤r≤7 is an orthonormal basis of O0.

pr.OPLU 7. The algebra O+. The real vector space O becomes a commutative real algebra O+ under themultiplication

x · y :=12

x y =12

tO(x)y+12

tO(y)x−12

nO(x,y)1O (x,y ∈O)

with identity element 1O+ := 1O. Show that Aut(O)⊆ Aut(O+) is a closed subgroup and that theassignment ϕ 7→ ϕ|O0 determines a topological isomorphism from Aut(O+) onto O(O0).

3. Unital subalgebras of O and their Z-structures

s.SUBGCZThe Graves-Cayley octonions, and the Hamiltonian quaternions as well, derive aconsiderable amount of their significance from the profound connections with seem-ingly unrelated topics in other areas of mathematics and physics. One of these con-nections pertains to the arithmetic theory of quadratic forms. Without striving forcompleteness or maximum generality, it will be briefly touched upon in the nexttwo sections. Our results, incomplete as they are, underscore the need for an under-standing of quaternion and octonion algebras not just over fields but, in fact, overarbitrary commutative rings.

ss.COGR3.1. The general set-up. (a) Throughout this section, we fix a real vector space Vof finite dimension n and assume most of the time, but not always, that V is equippedwith a positive definite quadratic form q : V → R. Speaking of (V,q) as a positivedefinite real quadratic space under these circumstances, we may and always willregard V as a euclidean vector space with the scalar product 〈x,y〉 := 1

2 q(x,y) anddenote the associated euclidean norm by ‖x‖=

√〈x,x〉=

√q(x), for all x,y ∈V .


(b) We also denote by D one of the unital subalgebra R,C,H,O of O. Via restric-tion, D inherits from O the data norm, trace, bilinearized norm and conjugation,denoted by nD, tD, DnD, ιD, respectively, that mutatis mutandis enjoy the variousproperties assembled in 1.5 for the Graves-Cayley octonions. These properties willbe freely used here without further ado. In particular, (D,nD) is a positive definitereal quadratic space as in (a), allowing us to regard D as a euclidean subspace of Oin a natural way.

ss.ZAL3.2. Algebras over Z and Q. There is nothing special about the base field R inthe definition of an algebra. It may be replaced by any commutative associative ringof scalars, a point of view adopted systematically in the remainder of the book. Forexample, we may pass to the field Q of rational numbers or the ring Z of rationalintegers: by a Q-algebra (resp. a Z-algebra) we mean Q-vector space (resp. an ad-ditive abelian group) A together with a Q-bilinear (resp. a bi-additive) map fromA×A to A. The conventions of 1.2 carry over to this modified setting virtually with-out change. Any real algebra may be viewed as a Q-algebra (resp. a Z-algebra) byrestriction of scalars.

ss.MIPOA

3.3. Power-associative algebras and the minimum polynomial. Let A be a finite-dimensional unital algebra over K, where either K = R or K = Q. Powers of x ∈ Awith integer coefficients ≥ 0 are defined inductively by x0 = 1A, xn+1 = xxn for n ∈N. We say A is power-associative if xm+n = xmxn for all x∈A, m,n∈N, equivalently,if

K[x] := ∑n∈N

Kxn ⊆ A (1) MONOG

is a unital commutative associative subalgebra, for all x ∈ A. Hence it makes senseto talk about the minimum polynomial of x (over K) in its capacity as an element ofK[x]. It will be denoted by µx, or µK

x to indicate dependence on K, and is the uniquemonic polynomial in K[t] that generates the ideal of all polynomials in K[t] killingx.

The preceding considerations apply in particular to K = R and the real algebraD; indeed, D is power-associative since, for x ∈ D, we may invoke (1.5.10) to con-clude that R[x] = R1D+Rx ⊆ D is a unital commutative associative subalgebra ofdimension at most 2. Moreover, again by (1.5.10),

µx = t2− tD(x)t+nD(x) ⇐⇒ x /∈ R1D, (2) MIPOT

while, of course, µα1D = t−α for α ∈ R.

ss.INTEL3.4. Integral elements. Let A be a finite-dimensioal unital power-associative alge-bra over K = R or K = Q. An element x ∈ A is said to be integral if f (x) = 0 forsome monic polynomial f ∈ Z[t]. For example, x is integral if its minimal polyno-mial has integral coefficients. The converse of this also holds in important specialcases.


p.CHAIN

3.5. Proposition. Let A be a finite-dimensional unital power-associative algebraover Q and x ∈ A. Then the following conditions are equivalent.

(i) x is integral.(ii) Z[x] := ∑n∈NZxn is a finitely generated abelian group.(iii) µx ∈ Z[t].

Proof. (i)⇔ (ii). Apply Bourbaki [11, V, §1, Thm. 1] to A := Z, R := Z[x].(iii)⇒ (i). Clear.(i) ⇒ (iii). Apply [11, V, §1, Cor. of Prop. 11] to A := Z, K := Q, K′ := Q[x]

to conclude that the coefficients of µx ∈ Q[t] are integral over Z. Hence (iii) holdssince Z is integrally closed.

ss.ILAS3.6. Integral lattices and Z-structures. (a) A subset L⊆V is called a lattice in Vif there exists a basis (e1, . . . ,en) of V (as a real vector space) which is associatedwith L in the sense that

L = Ze1⊕·· ·⊕Zen.

Then L is a free abelian group of rank n, and QL := Qe1 ⊕ ·· · ⊕Qen is an n-dimensional vector space over Q.(b) By an integral quadratic lattice in (V,q) (or simply in V , the quadratic formq being understood) we mean a lattice L ⊆ V such that q(L) ⊆ Z, which after lin-earization implies q(x,y) ∈ Z for all x,y ∈ L.(c) Let A be a finite-dimensional unital R-algebra. A lattice L in A is said to be uni-tal if 1A ∈ L. Note in particular that a unital integral quadratic lattice L⊆D satisfiestD(L)⊆Z and is stable under conjugation; in particular, the minimum polynomial ofx ∈ L by (3.3.2) has integer coefficients, so L consists entirely of integral elements.(d) Let A be a finite-dimensional unital R-algebra. By a Z-structure of A we meana unital lattice M ⊆ A that is closed under multiplication (M2 ⊆ M). Z-structuresof A are, in particular, unital Z-algebras, more precisely, Z-subalgebras of A. Notefurther that QM is a unital Q-subalgebra of A.

l.LIND3.7. Lemma. Let L ⊆ V be a lattice. A family of elements in QL that is linearlyindependent of Q is so over R.

Proof. Let (e1, . . . ,en) be a basis of V associated with L. If x1, . . . ,xm ∈QL are lin-early independent over Q, they can be extended to a Q-basis (x1, . . . ,xn) of QL. Thisimplies x j = ∑

ni=1 αi jei for 1≤ j ≤ n and some matrix (αi j) ∈ GLn(Q)⊆ GLn(R).

Thus (x1, . . . ,xn) is an R-basis of V , forcing x1, . . . ,xm to be linearly independentover R.

p.MINIQUR

3.8. Proposition. Let A ba finite-dimensional unital power-associative real alge-bra, M a Z-structure of A and x ∈ QM ⊆ A. Then µR

x = µQx , so the minimum poly-

nomial of x over R agrees with the minimum polynomial of x over Q. In particular,it belongs to Q[t].


Proof. By definition, µRx divides µQ

x over R. But both are monic polynomials inR[t] of the same degree, by Lemma 3.7. Hence µR

x = µQx .

p.STRULA3.9. Corollary. Every Z-structure M of D is a unital integral quadratic lattice.

Proof. Unitality being obvious, it remains to show nD(x) ∈ Z for all x ∈ M. Firstof all, QM is a finite-dimensional unital power-associative Q-algebra containing x,and µx as given in 3.3 by Prop. 3.8 is the minimum polynomial of x over Q. Onthe other hand, let Z[x] = ∑n≥0Zxn be the unital subalgebra of M generated by x.Since M is finitely generated as a Z-module, so is Z[x]⊆M. Thus x is integral overZ (Prop. 33.2). By Prop. 33.2 again and (3.3.2), this implies nD(x) ∈ Z, as claimed.

ss.BATR

3.10. Basis transitions. If E = (e1, . . . ,en) is a(n ordered) basis of V , then so isES = (e′1, . . . ,e

′n), e′j := ∑

ni=1 si jei, 1≤ j≤ n, for any invertible real quadratic matrix

S = (si j) of size n. In this way, GLn(R) acts on the set of bases of V from the right ina simply transitive manner, so any two bases E,E ′ of V allow a unique S ∈ GLn(R)such that ES = E ′. We call S the transition matrix from E to E ′.

Let E := (e1, . . . ,en) be an R-basis of V . Then we write

Dq(E) :=(q(ei,e j)

)1≤i, j≤n ∈Matn(R) (1) MABI

for the matrix of the symmetric bilinear form Dq : V ×V → R relative to the basisE. Along with Dq, the matrix Dq(E) is also positive definite. Given S ∈ GLn(R), itis well known and easily checked that

Dq(ES) = StDq(E)S. (2) TRAFO

If L is an integral quadratic lattice in V and the basis E is associated with L, then thepositive definite matrix Dq(E) has integral coefficients and its diagonal entries areeven.

p.DISL

3.11. Proposition. Let L⊆V be an integral quadratic lattice and E an R-basis ofV that is associated with L. Then

det(L) := det(Dq(E)

)is a positive integer that only depends on L and not on the basis chosen.

Proof. Since Dq(E) belongs to Matn(Z), its determinant is an integer, which mustbe positive since Dq(E) is positive definite. The transition matrix from E to anotherbasis E ′ of V associated with L not only has integral coefficients but is also uni-modular, i.e., has determinant ±1. Now (3.10.2) shows det(Dq(E ′)) = det(Dq(E)).

ss.DISC

3.12. The discriminant. Let L be an integral quadratic lattice of V . The quantitydet(L) exhibited in Prop. 3.13 is called the determinant of L; it is a positive integer.


By contrast, the non-zero (possibly negative) integer

disc(L) := (−1)bn2 c det(L)

is called the discriminant of L. By a unimodular integral quadratic lattice of V wemean an integral quadratic lattice of discriminant ±1.

p.CODIS3.13. Proposition. Let L′⊆ L be integral quadratic lattices of V . Then L/L′ is finiteand disc(L′) = [L : L′]2 disc(L).

Proof. Let E (resp. E ′) be an R-basis of V associated with L (resp. L′) and S thetransition matrix from E to E ′. Then S ∈Matn(Z)∩GLn(R), and the ElementaryDivisor Theorem [46, Thm. 3.8] implies that there exist P,Q ∈ GLn(Z) and a chainof successive non-zero integral divisors d1| · · · |dn satisfying S=Pdiag(d1, . . . ,dn)Q.Replacing E ′ by E ′T , T := Q−1, and E by EP, we may assume S = diag(d1, . . . ,dn).With E = (e1, . . . ,en) this implies E ′ = ES = (d1e1, . . . ,dnen), and

L/L′ ∼= (Z/d1Z)⊕·· ·⊕ (Z/dnZ)

is finite of order ∏ni=1 |di|. On the other hand, applying Prop. 3.11 and (3.10.2), we

conclude

disc(L′) = (−1)bn2 c det

(Dq(ES)

)= (−1)b

n2 c( n

∏i=1

di)2 det

(Dq(E ′)

)= [L : L′]2 disc(L),

as claimed. c.MAXIN

3.14. Corollary. A unimodular integral quadratic lattice of V is maximal in thesense that it is not contained in any other integral quadratic lattice of V .

r.SPEZ3.15. Remark. The preceding observations apply in particular to Z-structures of A.For example, a Z-structure of A that is unimodular (as an integral quadratic lattice)is maximal as an integral quadratic lattice by Cor. 3.14, hence as a Z-structure aswell.

e.GAI

3.16. Examples. Let M be a Z-structure and E an orthonormal basis of A (asa euclidean real vector space). If E is associated with M, then (3.10.1) showsDnA(E) = 2 ·1r, r := dimR(A), and we conclude

disc(M) = (−1)br2 c2r. (1) DIOR

We now consider a number of specific cases.(a) By Exc. 10 below, M := Z is the unique Z-structure of A := R, and we havedisc(Z) = 2.(b) The Gaussian integers Ga(C) := Z[i] = Z⊕Zi are a Z-structure of A := C, and


we have disc(Ga(C)) = −22 = −4. The Gaussian integers are the integral closureof Z in C (Neukirch [89, Thm. 1.5]), hence by Cor. 3.9 are a maximal Z-structure.(c) In (1.10.1) we have exhibited an orthonormal basis (1H, i, j,k) of A := H, withstructure constants by (1.10.3) equal to ±1. Thus

Ga(H) := Z1H⊕Zi⊕Zj⊕Zk (2) GAHA

is a Z-structure of H having discriminant disc(Ga(H)) = 24 = 16, called the Gaus-sian integers of H.(d) Let E = (ur)0≤r≤7 be a Cartan-Schouten basis of A :=O. From Exc. 6 we deducethat E is an orthonormal basis of O, while (2.1.1), (2.1.2) show that the correspond-ing structure constants are ±1. Hence

Ga(O) := GaE(O) :=7⊕

r=0

Zur (3) GAGC

is a Z-structure of O having discriminant disc(Ga(O)) = 28 = 256, called the Gaus-sian integers of O relative to E.

Exercises.

pr.CHALAT 8. Prove for an additive subgroup L⊆V that the following conditions are equivalent.

(i) L is a lattice.(ii) L spans V as a real vector space and is a free abelian group of rank at most dimR(V ).(iii) There are lattices L0,L1 ⊆V such that L0 ⊆ L⊆ L1.

pr.LASTRU 9. Let M be a Z-structure of D. Show that there exists an additive subgroup of D properly contain-ing M that is free of finite rank and closed under multiplication but not a Z-structure of D.

pr.REDONE 10. Let M be a Z-structure of D. Prove M ∩R = Z. (Hint: Note that M is a discrete additivesubgroup of D under the natural topology and that the non-zero discrete additive subgroups of Rhave the form Zv for some non-zero element v ∈ R.)

4. Maximal quaternionic and octonionic Z-structures

s.MAQOContrary to the Gaussian integers in C, their simple-minded analogues in H andO (cf. 3.16 (c),(d)) are not maximal Z-structures. In fact, they will be enlarged tomaximal ones in the present section.

ss.HUZH4.1. Towards the Hurwitz Z-structure of H. Starting out from the orthonormalbasis (1H, i, j,k) of H exhibited in (1.10.2), we put

h :=12(1H+ i+ j+k) ∈H (1) HIJK

and note

4 Maximal quaternionic and octonionic Z-structures 17

nH(h) = tH(h) = nH(i,h) = nH(j,h) = nH(k,h) = 1. (2) ENTH

After an obvious identification, we can realize the complex numbers via

C := R[i] = R1H⊕Ri (3) IDCH

as a subalgebra of H, which satisfies the relation

H= C⊕Ch (4) DECHC

as a direct sum of subspaces.. To see this, it suffices to show that the sum on theright of (4) is direct, so let u,v∈C and suppose u = vh. If v 6= 0, then h = v−1u∈C,a contradiction. Thus u = v = 0, as desired. Note that

Ga(C) = Z[i] = Z1H⊕Zi (5) GACEM

after the identification carried out in (3).

4.2. Theorem (Hurwitz [40]). With the notation and assumptions of 4.1,t.HURO

Hur(H) := Ga(C)⊕Ga(C)h (1) HURHA

is a Z-structure and a maximal integral quadratic lattice of H, called its Z-structureof (or simply the) Hurwitz quaternions. Hur(H) contains the Gaussian integers of Has a sub-Z-structure and has discriminant 4:

Ga(H)⊆ Hur(H), disc(

Hur(H))= 4. (2) DISCH

Moreover, (1H, i,h, ih) and (1H, i, j,h) are R-bases of H that are both associatedwithHur(H).

Proof. Since H is a division algebra by Cor. 1.11, the map H→H, x 7→ xh, is bijec-tive. Combining (4.1.4), (4.1.5) with (1), we therefore conclude that M :=Hur(H)⊆H is a unital lattice and that (1H, i,h, ih) is an R-basis of H associated withM.

Next we prove that M is closed under multiplication. Let u ∈ Ga(C). Then(1.5.12) implies uh+ hu = tH(u)h+ tH(h)u− nH(u,h)1H, where the coefficientsof the linear combination on the right by (4.1.2) are all integers. Hence M =Ga(C)⊕hGa(C). Combining this with (1.5.10), (4.1.2), we obtain

Ga(C)hGa(C)h⊆ Ga(C)Ga(C)+Ga(C)Ga(C)h2

= Ga(C)+Ga(C)(h−1H)⊆ Ga(C)⊕Ga(C)h = M.

Thus M is indeed multiplicatively closed and hence a Z-structure of H. Since

ih =12(i−1H+k− j) =−(1H+ j−h),


we see that E := (1H, i, j,h) is another R-basis of H associated withM. In particular,Ga(H)⊆M (since k = 2h−1H− i− j) and

DnH(E) =

2 0 0 10 2 0 10 0 2 11 1 1 2

.

Subtracting the arithmetic mean of the first two rows from the fourth, we conclude

disc(M) = det

2 0 0 10 2 0 10 0 2 10 0 1 1

= 4 · (2−1) = 4.

It remains to show that M is a maximal integral quadratic lattice of H, which weleave as an exercise (Exc. 12).

r.LIHU4.3. Remark. The Hurwitz quaternions are endowed with a rich arithmetic struc-ture that has been investigated extensively in the literature. For example, it is possi-ble to derive Jacobi’s famous formula [???] for the number of ways a positive integercan be expressed as a sum of four (integral) squares in a purely arithmetic fashionby appealing to properties of the Hurwitz quaternions [40, p. 335]. The reader mayconsult Brandt [14] or Conway-Smith [20, Ch. 5] for further details on the subject.There are also profound connections with the theory of automorphic forms, cf. Krieg[64] for a systematic and essentially self-contained study of this topic.

In the second part of the present section, we will imitate our approach to the Hurwitzquaternions on the octonionic level. This will lead us to a Z-structure of O that isnot only maximal but even, remarkably, a unimodular integral quadratic lattice.

ss.COZO4.4. Towards the Coxeter octonions. In 1.10, we have defined the Hamiltonianquaternions H explicitly as a unital subalgebra of O. For our subsequent consider-ations, it will be important to select a different realization of this kind, dependingon the choice of a Cartan-Schouten basis E = (ur)0≤r≤7 of O that will remain fixedthroughout the rest of this section. Recall from Exc. 6 (c) that E is an orthonormalbasis of O obeying the multiplication rules (2.1.2) which may be conveniently readoff from Fig. 1 in 2.5 above.

In particular, we have the relations u1u2 = u4, u2u4 = u1, u4u1 = u2, which showthat the Hamiltonian quaternions may also be identified via

H= R1O⊕Ru1⊕Ru2⊕Ru4 (1) HACS

as a unital subalgebra of O under the matching 1H = 1O, i = u1, j = u2, k = u4. Thisobviously implies


Ga(H) = Z1O⊕Zu1⊕Zu2⊕Zu4 (2) GACS

for the Gaussian integers of H. We now put

p :=12(1O+u1 +u2 +u3) ∈O (3) BFEL

and note

u1p =12(−1O+u1 +u4 +u7), (4) ONEL

u2p =12(−1O+u2−u4 +u5), (5) TWOL

u3p =12(−1O+u3−u5−u7), (6) THRL

u4p =12(−u1 +u2 +u4−u6) (7) FOUL

as well as

nO(p) = tO(p) = nO(u1,p) = nO(u2,p) = nO(u3,p) = 1, nO(u4,p) = 0. (8) ENTO

We also claim

O=H⊕Hp, (9) DECOH

which will follow once we have shown that the sum on the right is direct. Indeed,suppose u = vp 6= 0 for some u,v ∈ H. Then Exc. 3 (c) yields nO(v)p = v(vp) =vu ∈H, hence p ∈H, a contradiction to (1), (3).

4.5. Theorem (Coxeter [21]). With the notation and assumptions of 4.4,t.COGC

Cox(O) := CoxE(O) := Ga(H)⊕Ga(H)p (1) COXGC

is a Z-structure and a unimodular integral quadratic lattice of O, called its Z-structure of (or simply the) Coxeter octonions (relative to E). Cox(O) contains theGaussian integers of O as a sub-Z-structure, and

E ′ := (1O,u1,u2,u4,p,u1p,u2p,u4p) (2) BECO

is a an R-basis of O associated with Cox(O).

Proof. From (4.4.9) and (1) we deduce that M := Cox(O) is a unital lattice in Oand E ′ as defined in (2) is an R-basis of O associated withM.

Now we show that the lattice M is integral quadratic. Indeed, from (4.4.8) weobtain nO(Ga(H),p)⊆ Z. But then, given u,v ∈Ga(H), we may apply Exc. 3 (c) aswell as Thm. 1.7 and (4.4.3) to conclude

nO(u+ vp) = nO(u)+nO(u,vp)+nO(vp) = nO(u)+nO(vu,p)+nO(v) ∈ Z,


as claimed. Moreover, consulting (4.4.3)−(4.4.7), we see that Ga(O) ⊆ M, so Mcontains the Gaussian integers of O.

Our next aim is to prove that M is a Z-structure of O, which will follow once wehave shown that it is closed under multiplication, equivalently, that

Ga(H)(

Ga(H)p)⊆M, (3) GAGAL(

Ga(H)p)

Ga(H)⊆M, (4) GALGA(Ga(H)p

)(Ga(H)p

)⊆M. (5) GALGAL

Noting that M is a unital integral quadratic lattice, we first let u ∈ Ga(H) and apply(1.5.12) to obtain

up+pu = tO(u)p+ tO(p)u−nO(u,p)1O =(u−nO(u,p)1O

)+ tO(u)p,

which in view of (1.5.6) implies

up+pu≡ tO(u)p mod Ga(H), (6) USYL

pu≡ up mod Ga(H). (7) LUBAR

Now let u,v ∈ H. Linearizing the first alternative law in 1.6 and using (6), (7) aswell as Exc. 3 (a), we obtain

u(pv)≡ (up+pu)v−p(uv)≡ tO(u)pv−p(uv)

≡ p(uv)≡ (vu)p mod Ga(H),

hence u(pv) ∈M. But then (3) follows since u(vp)≡ u(pv) mod Ga(H) by (7). Onthe other hand,

(vp)u = u(vp)+(vp)u−u(vp)= tO(u)vp+ tO(vp)u−nO(u,vp)1O−u(vp),

which by (3) proves (4). And finally, by (7), (3), Exc. 4 and Exc. 3 (c),

(up)(vp)≡ (pu)(vp)≡ p(uv)p≡ nO(p, vu)p−nO(p)vu≡ 0 mod M,

which proves (5).It remains to show that M is unimodular as an integral quadratic lattice. To this

end, we compute the matrix DnO(E ′). Applying 1.8, we obtain nO(urp,usp) =nO(ur,us) = 2 ·δrs, so DnO(E ′) is the block matrix(

2 ·14 TT t 2 ·14

),

where T := (nO(ur,usp))r,s∈0,1,2,4 by (4.4.3)–(4.4.7) has the form


T =

1 −1 −1 01 1 0 −11 0 1 10 1 −1 1

∈Mat4(Z).

One checks immediately that the columns of T all have euclidean length√

3 and aremutually orthogonal. Hence Exc. 13 below implies det(DnO(E ′)) = (22−3)4 = 1,and the proof of the theorem is complete.

ss.VIPO4.6. Vista: even positive definite inner product spaces. One of the key notionsdominating the arithmetic theory of quadratic forms is that of a unimodular integralquadratic lattice as defined in 3.6 (b), which is studied, e.g., in Milnor-Husemoller[86] (see also Serre [117] or Kneser [57]), under the name “even positive definiteinner product space over Z”. A prominent and particularly important example isfurnished by the integral quadratic lattice underlying the Coxeter octonions. In orderto appreciate the significance of this example, we record the following facts.

(a) Every even positive definite inner product space over Z splits uniquely intothe orthogonal sum of indecomposable subspaces [86, (6.4)].

(b) The rank of an even positive definite inner product space over Z is divisibleby 8 [86, (5.1)].

(c) There exists an even positive definite inner product space of rank 8 over Z [86,(6.1)] which is unique up to isomorphism [???].

The integral quadratic lattice in (c), discovered independently by Smith [???] andKorkine-Zolotarev [63], is closely connected with the root system E8 and is there-fore called the E8-lattice. We will come back to this connection in ??? below. By(a), (b) above, the E8-lattice is indecomposable, while uniqueness in (c) shows thatit is isomorphic to the integral quadratic lattice underlying the Coxeter octonions.

r.RECOX4.7. Remarks. (a) Since the automorphism group of O acts simply transitively onthe set of CartanSchouten bases of O, the Coxeter octonions up to isomorphism donot depend on the Cartan-Schouten basis chosen.(b) The similarity of our approach to the Hurwitz quaternions on the one hand andto the Coxeter octonions on the other is not accidental: in Chap. ??? below, we willdevelop a purely algebraic formalism that contains both of these constructions asspecial cases.(c) The reader is referred to Van der Blij-Springer [125] and Conway-Smith [20,Chap. 9−11] for a detailed study of arithmetic properties of the Coxeter octonions.Concerning a purely arithmetic approach to a Jacobi-type formula for the numberof ways a positive integer may be expressed as a sum of eight squares using theCoxeter octonions, the reader may consult Rehm [111].

Exercises.

pr.HURCON 11. Characterization of the Hurwitz quaternions by congruence conditions. Prove


Hur(H) = m1H+ni+ pj+qk | m,n, p,q ∈ Z or m,n, p,q ∈ 12+Z.

pr.MAXHUR 12. Complete the proof of Thm. 4.2 by showing that the Hurwitz quaternions form a maximalintegral quadratic lattice in H: if L ⊆ H is an integral quadratic lattice containing Hur(H), thenHur(H) = L.

pr.DETFO 13. A determinant formula. Let p,q be positive integers and r,s be positive real numbers. Prove formatrices T1 ∈Matp,q(R), T2 ∈Matq,p(R) that the matrix

S :=(

r ·1p T1T2 s ·1q

)∈Matp+q(R)

has determinant det(S) = rp−q charT2T1 (rs), where “char” stands for the characteristic polynomialof a square matrix. Conclude in the special case T1 = T , T2 = T t , where all the columns of T ∈Matp,q(R) are assumed to have euclidean length

√a for some a > 0 and to be mutually orthogonal,

that det(S) = rp−q(rs− a)q and that S is positive definite if a < rs. What happens if we drop theassumption of the columns of T all having the same euclidean length but retain the one that theybe mutually orthogonal?

pr.UNIT 14. Units. An element u of an integral quadratic lattice L in a positive definite real quadratic space(V,q) is called a unit if q(u) = 1. The set of units in L (possibly empty) will be denoted by L×.(a) Let M be a Z-structure of D. Show 1D ∈M× and that M× is closed under multiplication. Showfurther for u ∈M that the following conditions are equivalent.

(i) u ∈M×.(ii) uv = 1D for some v ∈M.(iii) vu = 1D for some v ∈ D.

In this case, v in (ii) (resp. (iii)) is unique and v = u.(b) Determine the units of the Gaussian integers in C,H,O and of the Hurwitz quaternions.

pr.HURDE 15. (Loos) Alternate description of the Hurwitz quaternions. Show with

ε1 :=12(j−k), ε2 :=−1

2(j+k), ε3 :=

12(1H+ i), ε4 :=−1

2(1H− i)

that (εi)1≤i≤4 is an orthonormal basis of H relative to the euclidean scalar product 〈x,y〉 and that

Hur(H) = Z(ε1− ε2)⊕Z(ε2− ε3)⊕Z(ε3− ε4)⊕Z(ε3 + ε4).

Conclude

Hur(H) = 4

∑i=1

ξiεi | ξi ∈ Z (1≤ i≤ 4),4

∑i=1

ξi ∈ 2Z

andHur(H)× = ±εi ± ε j | 1≤ i < j ≤ 4.

Remark. The last two equations show that Hur(H) is the root lattice of the root system D4, whileHur(H)× consists precisely of the roots of that system. See ??? for more details.

pr.ROOC 16. (Loos) Alternate description of the Coxeter octonions and their units. Let E = (ur)0≤r≤7 be aCartan-Schouten basis of O.(a) Show with

ε1 :=12(−u0 +u2), ε2 :=

12(u0 +u2), ε3 :=−1

2(u1 +u3), ε4 :=

12(u1−u3),

ε5 :=12(−u4 +u5), ε6 :=

12(u4 +u5), ε7 :=

12(u6−u7), ε8 :=

12(u6 +u7)


that (εi)1≤i≤8 is an orthonormal basis of O relative to the euclidean scalar product 〈x,y〉.(b) Conclude that Cox(O) is the additive subgroup of O generated by the expressions

±εi ± ε j,12

8

∑i=1

siεi, (1) EEIGHT

where 1 ≤ i < j ≤ 8 for the first type of (1), while (si)1≤i≤8 in the second type of (1) varies overthe elements of ±18 such that the number of indices i = 1, . . . ,8 having si =−1 is even.(c) Show further that

Cox(O) = 8

∑i=1

ξiεi | ξi ∈ R, 2ξi,ξi−ξ j ∈ Z (1≤ i, j ≤ 8),8

∑i=1

ξi ∈ 2Z. (2) COXCON

(d) Deduce from (c) that the units of O are precisely the elements listed in (1) and that there areexactly 240 of them.(Hint: In order to derive (b) (resp. (c)), show that the additive subgroup of O generated by theelements in (1) (resp. by the right-hand side of (2)) is an integral quadratic lattice of O containingCox(O).)

Remark. The elements of (1) are precisely the roots of the root system E8 and, therefore, Cox(O)is the corresponding root lattice. See ??? for more details.

pr.KILA 17. The Kirmse lattice. (Kirmse [55]) Let E := (ur)0≤r≤7 be a Cartan-Schouten basis of O. Showthat

Kir(O) := KirE(O) := Ga(O)+4

∑i=1

Zvi

with

v1 :=12(1O+u1 +u2 +u4), v2 :=

12(u3 +u5−u6−u7),

v3 :=12(1O+u1 +u3−u7), v4 :=

12(1O+u2 +u3 +u5)

is a unital unimodular integral quadratic lattice in O which, however, contrary to what has beenclaimed in [55, p. 70], is not a Z-structure of O. (Hint. Consider the product v1v3.)Remark. It follows from 4.7 that Kir(O) is isomorphic to the Coxeter octonions as an integralquadratic lattice, hence, by what has been shown in Exc. 16 (d), must have exactly 240 units, inagreement with [55, p. 76].

The following exercise may be viewed as a corrected version of Kirmse’s approach to theconstruction of “integer octonions”.

pr.ALCO 18. An alternate model of the Coxeter octonions. Let (ui)0≤i≤7 be a Cartan-Shouten basis of O andput

v1 :=12(1O+u1 +u2 +u4), v2 :=

12(1O+u1 +u5 +u6),

v3 :=12(1O+u1 +u3 +u7), v4 :=

12(u1 +u2 +u3 +u5).

Then show that

R := Ga(O)+4

∑i=1

Zvi ⊆O

is a Z-structure isomorphic to the Coxeter octonions.


pr.HUCO 19. Show that there is an embedding of the Hurwitz quaternions into the Coxeter octonions asZ-algebras.

5. The euclidean Albert algebra

s.EUALThe euclidean Albert algebra made its first appearance in the work of Jordan-vonNeumann-Wigner [51], who suspected it to be what would later be called an excep-tional simple formally real Jordan algebra. But while the authors were able to showthat it is simple and formally real, the remaining two properties eluded them. Thisgap was subsequently closed by Albert [2] in an immediate follow-up to [51].

In the present section, we define the euclidean Albert algebra and derive someof its most basic properties. In doing so, we rely heavily on the Graves-Cayleyoctonions but also on rudiments from the theory of real Jordan algebras, which willbe developed here from scratch.

ss.STASU5.1. The standard subalgebras of O. Throughout this section, we fix a positiveinteger n and, as in 3.1, we write D for one of the four unital subalgebras R,C,H,Oof O. We also put d := dimR(D). As explained more fully in 3.1, the algebra Dinherits its unit element, norm, trace, conjugation from O via restriction. We identifythe base field R canonically inside D via R=R1D =R1O. Note by (1.5.7) that onlythe elements of R remain fixed under the conjugation of D.

ss.HERMAD5.2. Hermitian matrices over D. We denote by Matn(D) the real vector space ofn× n matrices with entries in D. It becomes a real algebra under ordinary matrixmultiplication, with identity element given by 1n, the n×n unit matrix. This algebrais associative for D = R,C,H, but highly non-associative for D = O. One checkseasily that the map

Matn(D)−→Matn(D), x 7−→ xt , (1) COTRAP

is an involution, i.e., it is R-linear of period 2 and satisfies the relation

xyt = yt xt (x,y ∈Matn(D)). (2) ANTI

We speak of the conjugate transpose involution in this context and denote by

Hern(D) :=

x ∈Matn(D) | x = xt (3) HERCO

the set of elements in Matn(D) that are hermitian in the sense that they remain fixedunder the conjugate transpose involution. Note by the conventions of 5.1 that thediagonal entries of x ∈ Hern(D) are all scalars, and so Hern(D)⊆Matn(D) is a realsubspace of dimension

5 The euclidean Albert algebra 25

dimR(

Hern(D))= n+

n(n−1)2

d. (4) DIMHER

The ordinary matrix units ei j, 1≤ i, j≤ n, of Matn(D) canonically induce hermitianmatrix units in Hern(D) according to the rules

v[i j] := vei j + ve ji (v ∈ D, 1≤ i, j ≤ n, i 6= j). (5) HERUNT

ss.SYMPRO5.3. The symmetric matrix product. The conjugate transpose involution by (5.2.2)does not preserve multiplication and hence the subspace Hern(D)⊆Matn(D) is nota subalgebra. In order to remedy this deficiency, we pass from ordinary matrix mul-tiplication to the symmetric matrix product

x• y :=12(xy+ yx) (x,y ∈Matn(D)). (1) SYMPRO

The ensuing real algebra, denoted by Matn(D)+, continues to be unital, with identityelement 1n, is obviously commutative, but fails to be associative even if D is.

On the positive side, the conjugate transpose involution, again by (5.2.2), doespreserve multiplication of Matn(D)+, allowing us to conclude that Hern(D) becomesa unital commutative real algebra under the symmetric matrix product and is, infact, a unital subalgebra of Matn(D)+. Note that, thanks to the factor 1

2 on the right-hand side of (1), squares of elements in Hern(D), computed in Matn(D) and inHern(D), respectively, coincide, though cubes in general do not.

ss.CASEN5.4. The case n = 3. The formalism described in 5.2, 5.3 will become particularlyimportant for n = 3. The unital subalgebra Her3(D) ⊆ Mat3(D)+ by (5.2.4) hasdimension

dimR(

Her3(D))= 3(d +1) (1) DIMTH

and consists of the elements

x =

α1 u3 u2u3 α2 u1u2 u1 α3,

= ∑(αieii +ui[ jl]

)(αi ∈ R, ui ∈O, 1≤ i≤ 3), (2) ELTH

where the sum on the very right is extended over all cyclic permutations (i jl) of(123). As will be seen in due course, the structure of this algebra becomes exceed-ingly delicate for D=O.

ss.EEU5.5. Enter the euclidean Albert algebra. The real algebra

Her3(O)


of 3× 3 hermitian matrices with entries in the Graves-Cayley octonions under thesymmetric matrix product is called the euclidean Albert algebra1. It is commutative,contains an identity element and by (5.4.1) has dimension 27.

The euclidean Albert algebra derives its importance from profound connectionswith various branches of mathematics and physics. Suffice it to mention at this stageconnections with

• Exceptional Lie groups of type E6,F4,D4, see ??? for more details,• Self-dual homogeneous convex cones (Koecher [62]),• Bounded symmetric domains (Loos [68]),• Theta functions (Dorfmeister-Walcher [25]),• Elementary particles (Dray-Manogue [26]).

In order to obtain a proper understanding of the algebras Her3(D), in particular ofthe euclidean Albert algebra, it will be crucial to connect them with the theory ofJordan algebras. We will do so in the remainder of this section.

ss.REJO5.6. Real Jordan algebras and euclidicity. By a real Jordan algebra we mean areal algebra J (the term “real” being understood if there is no danger of confusion)satisfying the following identities, for all x,y ∈ J.

xy = yx (commutative law), (1) RECOM

x(x2y) = x2(xy) (Jordan identity). (2) REJO

A real Jordan algebra J is said to be euclidean if, for all positive integers m, theequation ∑

mr=1 x2

r = 0 has only the trivial solution in J:

∀x1, . . . ,xm ∈ J :( m

∑r=1

x2r = 0 =⇒ x1 = · · ·= xm = 0

)(3) REUC

ss.SPEXRE5.7. Special and exceptional real Jordan algebras. (a) Standard examples of realJordan algebras are easy to construct: let A be an associative algebra with multiplica-tion (x,y) 7→ xy. Then it is readily checked that, in analogy to (5.3.1), the symmetricproduct

x• y :=12(xy+ yx) (1) SYMRE

converts A into a Jordan algebra, which we denote by A+. It follows that every subal-gebra of A+, which may or may not be one of A, is a Jordan algebra. Jordan algebraswhich are isomorphic to a subalgebra of A+, for some associative algebra A, are saidto be special, while non-special Jordan algebras are called exceptional.(b) The real algebras D= R,C,H are all associative. Hence so is Matn(D), and wededuce from 5.3 that Hern(D) is a special Jordan algebra. By contrast, the precedingargument breaks down for D=O since O is not associative. In fact, while Her3(O),

1 The prefix “euclidean” will be explained in 5.6 below.


as a delicate argument in Thm. 5.10 below will show, continues to be a Jordan alge-bra, it will be an exceptional one. Moreover, Hern(O) for n > 3 is not even a Jordanalgebra anymore (Exc. 21).(c) Our principal aim in the present section will be to show in a unified fashion thatHer3(D), for any D, including D=O, is a euclidean Jordan algebra, thereby justify-ing the terminology of 5.5. We do so not by following Albert’s original approach [2]but by adopting an idea of Springer’s [???], according to which some fundamentalproperties of ordinary 3×3-matrices over a field survive in the algebra Her3(D).

ss.NOTRAD5.8. Norm, trace and adjoint of Her3(D). As in (5.4.2), the elements of J :=Her3(D) will be written systematically as

x = ∑(αieii +ui[ jl]

), y = ∑

(βieii + vi[ jl]

), (1) ELXY

where αi,βi ∈ R, ui,vi ∈ D for 1 ≤ i ≤ 3, and unadorned summations will alwaysbe taken over the cyclic permutations (i jl) of (123). Then we define the norm N :=NJ : J → R, the trace T := TJ : J → R and the adjoint ] : J → J, x 7→ x], by theformulas

N(x) := α1α2α3−∑αinO(ui)+ tO(u1u2u3), (2) NORDE

T (x) := ∑αi, (3) TRADE

x] := ∑

((α jαl−nO(ui)

)eii +

(−αiui +u jul

)[ jl]). (4) ADDE

By Exc. 3 (b), the final term on the right of (2) is unambiguous. Moreover, the normN is a cubic form, i.e., after choosing a basis of Her3(D) as a real vector space, N isrepresented by a homogeneous polynomial of degree 3 in d variables. Similarly, thetrace T is a linear form, while the adjoint is a real quadratic map in the sense of 1.3.In particular, the expression x×y := (x+y)]−x]−y] is (symmetric) bilinear in x,y.More precisely,

x× y = ∑

((α jβl +β jαl−nO(ui,vi)

)eii +

(−αivi−βiui +u jvl + v jul

)[ jl])

(5) BADDE

Finally, we put

S(x) := T (x]) = ∑(α jαl−nO(ui)

). (6) QUADE

and note that S := SJ : J→ R is a quadratic form with bilinearization given by

S(x,y) := DS(x,y) = T (x× y) = ∑(α jβl +β jαl−nO(ui,vi)

). (7) BQUADE

ss.FUIDRE5.9. Fundamental identities. (a) For finite-dimensional real vector spaces V,W ,equipped with the natural topology, a non-empty open subset U ⊆ V and a smoothmap F : U →W (e.g., a polynomial map), we write DF(u)(x) for the directionalderivative of F at u ∈ U in the direction x ∈ V . Thus DF(u)(x) ∈W agrees withthe factor of t ∈ R, |t| sufficiently small, in the Taylor expansion of F(u+ tx). For


example, given a quadratic map Q : V →W , we have DQ(u)(x) = DQ(u,x) foru,x ∈ J, where the right-hand side is to be understood in the sense of 1.3.(b) For arbitrary elements

x = ∑(αieii +ui[ jl]

), y = ∑

(βieii + vi[ jl]

)(1) ELXYZ

and z of J := Her3(D), with αi,βi ∈ R, ui,vi ∈O, 1≤ i≤ 3, we denote by

x3 := x• x2 =12(xx2 + x2x) (2) CUBER

the cube of x in J (which in general is not the same as its cube in Mat3(O)) andclaim that the following relations hold.

x2 = ∑

((α

2i +nO(u j)+nO(ul)

)eii +

((α j +αl)ui +u jul

)[ jl]), (3) SQUARE

x• y = ∑

((αiβi +

12

nO(u j,v j)+12

nO(ul ,vl))eii (4) SYMRE

+12((α j +αl)vi +(β j +βl)ui +u jvl + v jul

)[ jl]),

T (x• y) = ∑(αiβi +nO(ui,vi)

), (5) TRSYRE

T((x• y)• z

)= T

(x• (y• z)

), (6) TASRE

S(x,y) = T (x)T (y)−T (x• y), (7) BST

x] = x2−T (x)x+S(x)13, (8) SHASQ

DN(x)(y) = T (x] • y) = T (x2 • y)−T (x)T (x• y)+S(x)T (y), (9) DIREN

x3 = T (x)x2−S(x)x+N(x)13, (10) CUBRE

x2 • y = 2T (x)x• y+T (y)x2−S(x)y (11) DICURE

−S(x,y)x+DN(x)(y)13−2x• (x• y).

The verification of these identities is either straightforward or part of Exc. 20 below.t.HEUJO

5.10. Theorem. Her3(D) is a unital euclidean Jordan algebra.

Proof. We know from 5.3 that the algebra J := Her3(D) is unital and commutative,so our first aim must be to establish the Jordan identity (5.6.2). To this end, wecombine (5.9.11) with (5.9.7), (5.9.9) and obtain

x2 • y = 2T (x)x• y+T (y)x2−2x• (x• y)−S(x)y−(T (x)T (y)−T (x• y)

)x (1) XSQY

+(T (x2 • y)−T (x)T (x• y)+S(x)T (y)

)13.

Multiplying this equation by x, we conclude


x• (x2 • y) = 2T (x)x• (x• y)+T (y)x3−2x•(x• (x• y)

)(2) XXSQY

−S(x)x• y−(T (x)T (y)−T (x• y)

)x2

+(T (x2 • y)−T (x)T (x• y)+S(x)T (y)

)x.

On the other hand, replacing y by x• y in (1) and applying (5.9.6), we deduce

x2 • (x• y) = 2T (x)x• (x• y)+T (x• y)x2−2x•(x• (x• y)

)−S(x)x• y−

(T (x)T (x• y)−T

(x• (x• y)

))x

+(

T(x2 • (x• y)

)−T (x)T

(x• (x• y)

)+S(x)T (x• y)

)13

= 2T (x)x• (x• y)+T (x• y)x2−2x•(x• (x• y)

)−S(x)x• y−

(T (x)T (x• y)−T (x2 • y)

)x

+(T (x3 • y)−T (x)T (x2 • y)+S(x)T (x• y)

)13

Subtracting this from (2) and pplying (5.9.10) twice now yields

x• (x2 • y)− x2 • (x• y) = T (y)x3−T (x)T (y)x2 +S(x)T (y)x

−T((

x3−T (x)x2 +S(x)x)• y)

13

= T (y)(x3−T (x)x2 +S(x)x−N(x)13

)= 0,

and the Jordan identity holds. Thus J is a real Jordan algebra. To establish its eu-clidicity is now anti-climactic: let m be a positive integer, xr = ∑(αireii+uir[ jl])∈ Jwith αir ∈ R, uir ∈O for 1≤ r ≤ m, 1≤ i≤ 3, and suppose ∑r x2

r = 0. Since someof the diagonal entries on the right-hand side of (5.9.3) are strictly positive unlessx = 0, we conclude x1 = · · ·= xm = 0, as claimed.

ss.CEJOMA5.11. Cubic euclidean Jordan matrix algebras. The algebras Her3(D) are calledcubic euclidean Jordan matrix algebras. This terminology is justified by Thm. 5.10combined with (5.9.10).

c.POAS5.12. Corollary. Let x ∈ Her3(D). Then

R[x] := R13 +Rx+Rx2 ⊆ Her3(D)

is a unital commutative associative subalgebra. In particular, Her3(D) is power-associative.

Proof. Since x • x2 = x3 by (5.9.2), we have x •R[x] ⊆ R[x]. Moreover, the Jordanidentity and (5.9.10) yield

(x2)2 = x2 • (x• x) = x• (x2 • x) = x• x3 ∈ x•R[x]⊆ R[x]. (1) SQSQ


Thus R[x] ⊆ J := Her3(D) is a unital subalgebra. While commutativity is inheritedfrom J, associativity may be checked on the spanning set 13,x,x2, where it is eitherobvious or a consequence of (1). The final assertion follows immediately from thedefinition 3.3.

ss.MIPOD

5.13. The minimum polynomial. (a) As in 3.3, we denote by µx ∈ R[t] the mini-mum polynomial of x in the finite-dimensional unital power-associative real algebraJ := Her3(D). From (5.9.10) we deduce

µx = t3−T (x)t2 +S(x)t−N(x) ⇐⇒ 13∧ x∧ x2 6= 0 in∧3

(J). (1) MIPOX

Note that, since powers of x in J are well behaved, J by euclidicity (cf. Thm. 5.10)contains no nilpotent elements other than zero.

c.MIDEC5.14. Corollary. The minimum polynomial of x∈Her3(D) splits into distinct linearfactors over R. We have

1≤ m := deg(µx) = dimR(R[x]

)≤ 3,

and there exists a basis (cr)1≤r≤m of R[x] as a real vector space that up to order isuniquely determined by the conditions

cr • cs = δrscr (1≤ r,s≤ m) andm

∑r=1

cr = 13. (1) MIDEM

Proof. Write µ for the product of the distinct irreducible factors of µx. Then µxdivides µn for some positive integer n, which implies µ(x)n = 0. But J := Her3(D)does not contain non-zero nilpotent elements (5.13). Hence µ(x) = 0, and we con-clude that µx = µ has only simple irreducible factors over R. Suppose one of thesehas degree 2. Then C, by the Chinese Remainder Theorem [65, II, Cor. 2.2], be-comes a (possibly non-unital) subalgebra of R[x]. As such it inherits euclidicityfrom J, contradicting the equation 12 + i2 = 0 in C. Thus µx splits into distinct lin-ear factors, and applying the Chinese Remainder Theorem again yields quantities cr,1 ≤ r ≤ m, with the desired properties. It remains to prove uniqueness up to order.Let (dr) be another basis of R[x] satisfying (1) after the obvious notational adjust-ments. Then, with indices always varying over 1,2,3, we have dr = ∑s αrscs forsome u := (αrs) ∈ GL3(R), and d2

r = dr yields αrs ∈ 0,1. On the other hand,∑r dr = 13 amounts to ∑r αrs = 1. Hence, given s, there is a unique index π(s) suchthat αrs = δrπ(s). Assuming π(s) = π(s′) for some s 6= s′ would therefore imply thecontradiction that two distinct columns of u ∈ GL3(R) are presented by the samevector in R3. Thus π ∈S3 and cr = dπ(r) for all r.

Exercises.

pr.FORMU 20. Verify the identities (5.9.3)−(5.9.11).

6 Z-structures of unital real Jordan algebras 31

pr.HEMAO 21. Show for a positive integer n that Hern(O) under the symmetric matrix product is a Jordanalgebra if and only if n≤ 3. (Hint. Prove by repeated linearization, equivalently, by repeatedly tak-ing directional derivatives, that a real Jordan algebra J satisfies the fully linearized Jordan identityu((vw)x)+ v((wu)x)+w((uv)x) = (uv)(wx)+(vw)(ux)+(wu)(vx) for all u,v,w,x ∈ J.)

pr.INVRE 22. Invertibility in Her3(D). Let x ∈ J := Her3(D) and denote by L0x : R[x]→ R[x] the linear map

given by L0xy = x • y for all y ∈ R[x]. We say x is invertible in J if it is invertible in the unital

commutative associative subalgebra R[x]⊆ J. Prove:

(a) If 13∧ x∧ x2 6= 0 in∧3(J), then

µx = det(t1R[x]−L0

x), N(x) = det(L0x), T (x) = tr(L0

x).

(b) N(u• v) = N(u)N(v) for all u,v ∈ R[x] but not for all u,v ∈ J..(c) x is invertible in J if and only if N(x) 6= 0. In this case,

x−1 = N(x)−1x].

(d) x]] = N(x)x (adjoint identity) and N(x]) = N(x)2.

(Hint. For (b) and (d), use Zariski densitiy arguments [???].)

pr.AURE 23. Automorphisms of Her3(D). Prove with J := Her3(D) that a linear bijection ϕ : J → J is anautomorphism of J if and only if it preserves units and norms: ϕ(13) = 13, N ϕ = N. Concludefor 0 6= u ∈O that ϕ : J→ J defined by

ϕ

(∑(αieii +ui[ jl]

)):= ∑αieii +(u−1u1)[23]+ (u2u−1)[31]+ (uu3u)[12]

for αi ∈ R, ui ∈O, 1≤ i≤ 3, is an automorphism of J if and only if nO(u) = 1.

6. Z-structures of unital real Jordan algebras

s.ZUNREJExtending the notion of a Z-structure from the subalgebras D = R,C,H,O of theGraves-Cayley octonions to real Jordan algebras, most notably the cubic euclideanJordan matrix algebras Her3(D), turns out to be a remarkably delicate task. In partic-ular, the idea of copying verbatim the formal definition 3.6 (d) in the Jordan setting,leading to the notion of a linear Z-structure in the process, is practically uselesssince, as will be seen in due course, linear Z-structures are marred by a number ofserious deficiencies, the lack of natural examples being one of the most notorious.

In order to overcome this difficulty, we take up an idea of Knebusch [56] bydefining what we call quadratic Z-structures of finite-dimensional unital real Jor-dan algebras. In contradistinction to their linear counterpart, quadratic Z-structuresare based not on the bilinear product xy but on the cubic operation Uxy provided bythe U-operator (see 6.4 below) of the ambient Jordan algebra. This not only yieldsan abundant variety of natural examples but also a first glimpse at how Jordan alge-bras should be treated over commutative rings in which 1

2 is not available: throughMcCrimmon’s theory [75] of quadratic Jordan algebras.

Throughout this section, we let J be a finite-dimensional unital real Jordan alge-bra. As before, D stands for any of the unital subalgebras R,C,H,O of O. We begin


by naively rephrasing the definition of a Z-structure, with a slight terminologicaltwist, in the Jordan setting.

ss.LIZE6.1. Linear Z-structures. By a linear Z-structure of J we mean a lattice Λ ⊆ Jwhich is unital in the sense that 1J ∈ Λ and which is closed under multiplication:xy ∈Λ for all x,y ∈Λ .

One is tempted to regard the preceding definition as a very natural one because,for instance, any linear Z-structure in J may canonically be regarded as a Jordanalgebra over the integers in its own right. And yet it comes equipped with seriousdeficiencies which already come to the fore when trying to construct examples.

e.JODE6.2. Examples: the Jordan algebras D+. Either by definition (5.7) or by Exc. 25,J := D+ is a special unital real Jordan algebra, and it would be perfectly natural toexpect any Z-structure of D to be a linear one of J. But this, though valid in someisolated cases (see Exc. 26 below) is not true in general. In fact, it fails to be validin the following important examples.(a) (Knebusch [56, p. 175]) Let D := H and M := Hur(H) ⊆ H be the Z-structureof Hurwitz quaternions. Then i,h ∈M by Thm. 4.2, while (1.5.12) and (4.1.2) yield

i•h =12(ih+hi) =

12(tH(i)h+ tH(h)i−nH(i,h)1H

)=

12(−1H+ i),

which by Exc. 11 does not belong to M. Thus Hur(H) ⊆ H+ is not a linear Z-structure.(b) Similarly, let D := O and M := CoxE(O) ⊆ O be the Z-structure of Coxeteroctonions relative to a Cartan-Shouten basis E = (ui)0≤i≤7 of O. Then u1,p ∈M byThm. 4.5, while (1.5.12) and (4.4.8) yield

u1 •p=12(u1p+pu1) =

12(tO(u1)p+tO(p)u1−nO(u1,p)1O

)=

12(−1O+u1)∈H.

If this were an element of M, then (4.5.1) would imply u1 • p ∈ CoxE(O)∩H =Ga(H), a contradiction. Thus u1 •p /∈ M, and we conclude that CoxE(O) is not alinear Z-structure of O+.

e.CUBEU6.3. Examples: cubic euclidean Jordan matrix algebras. Again it would be per-fectly natural to expect any Z-structure of D giving rise, via 3×3-hermitian matri-ces, to a linear Z-structure of J := Her3(D). But this does never hold. Indeed, let Mbe a Z-structure of D. Then M is stable under conjugation, so

Λ := Her3(M) := x ∈Mat3(M) | x = xt

makes sense and is a unital lattice in J. However, Λ fails to be closed under multi-plication since, e.g., 1[23] and 1[31] both belong to Λ while 1[23]•1[31] = 1

2 [21] =12 [12] obviously does not.


In view of the deficiencies highlighted in the preceding examples, we will aban-don linear Z-structures of real Jordan algebras and replace them by quadratic ones,which are based on the following concept.

ss.UOPRE6.4. The U-operator. For x ∈ J, the linear map

Ux : J −→ J, y 7−→Uxy := 2x(xy)− x2y, (1) UOPREL

is called the U-operator of x. The map

U : J −→ J, x 7−→Ux (2) UOPRE

is called the U-operator of J. Note that the U-operator of J is a quadratic map whosebilinearization yields the Jordan triple product

xyz :=Ux,zy = (Ux+z−Ux−Uz)y = 2(x(zy)+ z(xy)− (xz)y

)(x,y,z ∈ J). (3) JOTRE

The U-operator is of the utmost importance for a deeper understanding of Jordanalgebras. Its fundamental algebraic properties will be addressed in ???? below. Forthe time being, it will be enough to observe a few elementary facts assembled inExercises 24, 25.

ss.QUAZE6.5. Quadratic Z-structures (cf. Knebusch [56, p. 175]). By a quadratic Z-structure of J we mean a unital lattice Λ ⊆ J such that Uxy ∈ Λ for all x,y ∈ Λ .This implies that Λ is closed under the Jordan triple product (6.4.3) and, in par-ticular, 2xy = xy1J ∈ Λ for all x,y ∈ Λ . However, while an inspection of (6.4.1)shows that linear Z-structures are always quadratic ones, the converse does not hold,as may be seen from the following examples.

e.QUADE6.6. Examples. Let M be a Z-structure of D. Then 1D+ = 1D ∈ M, and Exc. 25yields Uxy = xyx ∈ M for all x,y ∈ M. Thus M is a quadratic Z-structure of D+;as such it will be denoted by M+. Now it follows from 6.2 that both Hur(H)+ andCoxE(O)+ are quadratic Z-structures of H+ and O+, respectively, but not linearones.

We have observed in 6.3 that

Λ := Her3(M) = x ∈Mat3(M) | x = xt

is never a linear Z-structure of J := Her3(D). But we claim that it is always aquadratic one. Indeed, by Cor. 3.9, trace and norm of D take on integral valueson M. Hence, given

x = ∑(αieii +ui[ jl]), y = ∑(βieii + vi[ jl]) ∈Λ ,

we have αi,βi ∈M∩R= Z by Exc. 10, ui,vi ∈M for 1≤ i≤ 3, and (5.9.5), (5.8.4)yield T (x• y) ∈ Z, x] ∈Λ . Linearizing ad applying Exc. 24 (b), we therefore obtainUxy = T (x• y)x− x]× y ∈Λ , and the proof is complete.


Our next aim will be to show that, in analogy to Cor. 3.9, the linear form T , thequadratic form S, and the cubic form N of J := Her3(D) all take on integral valueson any quadratic Z-structure of J. Actually, we will be able to establish this resultunder slightly less restrictive conditions. The following lemma paves the way.

l.TSNQ6.7. Lemma. Let Λ ⊆ J := Her3(D) be a unital lattice such that QΛ ⊆ J is asubalgebra over Q. Then T (x),S(x),N(x) ∈Q for all x ∈QΛ .

Proof. From Prop. 3.8 we deduce µx := µRx = µQ

x ∈ Q[t], so m := deg(µx) is atmost 3. If m = 3, then (5.13.1) implies the assertion. At the other extreme, if m = 1,then x = α1J for some α ∈ Q, which again implies the assertion. We are left withthe case m = 2, so µx = t2−α1t+α2 for some α1,α2 ∈Q, and a short computationgives

x3 = (α21 −α2)x−α1α213.

Similarly, invoking (5.9.10), we obtain

x3 =(α1T (x)−S(x)

)x−(α2T (x)−N(x)

)13,

and comparing coefficients, we conclude

α1T (x)−S(x), α2T (x)−N(x) ∈Q. (1) ALTSN

On the other hand, by Zariski density, we can find elements y ∈ QΛ and α ∈ Q×such that

deg(µRy ) = deg(µQ

y ) = deg(µRαx+y) = deg(µQ

αx+y) = 3.

This implies T (y) ∈ Q and αT (x)+T (y) = T (αx+ y) ∈ Q, hence T (x) ∈ Q. Butnow (1) yields S(x),N(x) ∈Q as well.

r.SUAL6.8. Remark. Let B be a Q-subspace of J that is closed under squaring. Then B isa Q-subalgebra since xy = 1

2 ((x+ y)2− x2− y2) ∈ B for all x,y ∈ B.

p.TSNZ6.9. Proposition. Let Λ ⊆ J := Her3(D) be a lattice that is stable under takingpowers with arbitrary integer exponents ≥ 0. Then every x ∈ Λ is integral andT (x),S(x),N(x) ∈ Z.

Proof. For x ∈ Λ we have 1J = x0 ∈ Λ (so Λ is unital) and x2 ∈ Λ . This propertycarries over to QΛ which, by Remark 6.8, is a unital subalgebra of J. Given x ∈Λ ,Lemma 6.7 therefore implies T (x),S(x),N(x) ∈Q. Moreover, Λ being stable underpowers, Z[x]⊆Λ is an additive subgroup which, along with Λ , is finitely generated.Thus Propositions 33.2 and 3.8 show that x is integral and µx := µR

x = µQx ∈ Z[t].

Since the remaining assertions of the proposition will follow, either for obviousreasons or from Exc. 10 and 5.13, if m := deg(µx) ∈ 1,3, we may assume m = 2.Then Cor. 5.14 yields non-zero elements c1,c2 ∈ R[x] and scalars α1,α2 ∈ R suchthat


c1 + c2 = 13, c21 = c1, c2

2 = c2, c1 • c2 = 0, (1) CONET

x = α1c1 +α2c2. (2) XAC

Note that R[x]⊆ J is a unital commutative associative subalgebra and write R for theintegral closure of Z in R[x] (Bourbaki [11, V, §1, Def. 3]). From what we have justseen we deduce x∈ R, while (1) implies cr ∈ R for r = 1,2. Hence αrcr = xcr ∈ R by(1), (2), and we conclude that αr ∈ R is integral. On the other hand, Exc. 27 belowimplies, or allows us to assume,

N(c1) = N(c2) = 0, T (c1) = 2, S(c1) = 1, T (c2) = 1, S(c2) = 0, (3) ENCE

which by (5.9.7) implies

c]1 = c1−2c1 +13 = c2, c]2 = c2− c2 = 0. (4) CESH

Combining (3), (4) with (5.9.7) and Exc. 28, it follows that the rational numbers

T (x) = 2α1 +α2,

S(x) = α21 +α1α2S(c1,c2) = α

21 +α1α2

(T (c1)T (c2)−T (c1 • c2)

)= α

21 +2α1α2,

N(x) = α31 N(c1)+α

21 α2T (c]1 • c2)+α1α

22 T (c1 • c]2)+α

32 N(c2) = α

21 α2

are all integral, hence belong to Z.

Our next objective will be to derive a useful criterion for a unital lattice in J to be aquadratic Z-structure. Before doing so, we need a lemma.

l.EXQU6.10. Lemma. Let X ⊆Q be closed under squaring (ξ ∈X⇒ ξ 2 ∈X) and supposethe additive subgroup of Q generated by X is finitely generated. Then X ⊆ Z.

Proof. Let dξ > 0 be the exact denominator of ξ ∈ X . By hypothesis, dξ | ξ ∈ Xis a bounded set of positive integers. Pick η ∈ X such that dη is maximal. Then d2

η

is the exact denominator of η2 ∈ X , which implies d2η ≤ dη by maximality, hence

dη = 1 and therefore dξ = 1 for all ξ ∈ X . Thus X ⊆ Z.

t.CHAQUA6.11. Theorem (cf. Racine [109, Prop. IV.1]). For a unital lattice Λ ⊆ J :=Her3(D) the following conditions are equivalent.

(i) Λ is a quadratic Z-structure in J.(ii) x2 ∈Λ for all x ∈Λ .(iii) x] ∈Λ for all x ∈Λ .

Proof. (i)⇒ (ii). Let x ∈Λ . Then (6.4.1) shows x2 =Ux13 ∈Λ .Before we can deal with the remaining implications, it will be important to note

that


N(Λ) generates a finitely generated additive subgroup of R. (1) FIGE

In order to see this, let (e1, . . . ,en) be an R-basis of J associated with Λ . Since N(Λ)by Exc. 28 is contained in the additive subgroup of R generated by the quantitiesN(ei) (1 ≤ i ≤ n), T (e]i ,e j) (1 ≤ i, j ≤ n, i 6= j), T (ei× e j,el) (1 ≤ i < j < l ≤ n),it generates a finitely generated additive subgroup of R on its own, as claimed. Wealso note

N(13± x) = 1±T (x)+S(x)±N(x) (2) ONEX

for all x ∈ J.(ii) ⇒ (iii). Since Λ is closed under squaring, so is QΛ , which therefore is a

rational subalgebra of J (Remark 6.8). Now Lemma 6.7 implies T (x),S(x),N(x)∈Qfor all x ∈Λ . On the other hand, Exc. 22 (b) and (1) imply that X := N(Λ) satisfiesthe hypotheses of Lemma 6.10. Hence N(Λ) ⊆ Z. Combining (5.9.7) with (2) forx ∈Λ , we therefore conclude

T (x)2−T (x2) = S(x,x) = 2S(x) = N(13 + x)+N(13− x)−2 ∈ Z. (3) TSNX

But from (2) we also deduce 2T (x) = N(13 + x)−N(13− x)− 2N(x) ∈ Z, whichcombined with (3) shows that X :=Z+T (Λ) satisfies the conditions of Lemma 6.10.Thus T (x) ∈ Z for all x ∈Λ , and (2) gives S(x) ∈ Z. Now (5.9.8) shows x] ∈Λ andwe have (iii).

(iii)⇒ (i). Since Λ is stable under the adjoint map, so is QΛ . For 0 6= x ∈ QΛ ,we apply Exc. 22 (d) and obtain N(x)x = x]] ∈QΛ . Extending x to a Q-basis of QΛ

yields an R-basis of J (Lemma 3.7), and we conclude N(x) ∈ Q. Now Exc. 22 (b)and (1) show that X := N(Λ) staisfies the hypotheses of Lemma 6.10. Thus N(Λ)⊆Z. On the other hand, by (5.9.8) linearized, (iii) implies T (x)13− x = 13× x ∈ Λ ,hence T (x)13 ∈ Λ and then T (x)3 = N(T (x)13) ∈ Z. Hence T (x) ∈ Q is integral,which implies T (x) ∈ Z, S(x) = T (x]) ∈ Z. Now (5.9.7) shows T (x • y) ∈ Z for allx,y∈Λ , and from (6.4.1) combined with Exc. 24 (b) we conclude Uxy = T (x•y)x−x]× y ∈Λ . Thus Λ ⊆ J is a quadratic Z-structure.

Exercises.

pr.UORE 24. Properties of the U-operator. Let J be a real Jordan algebra. Prove:

(a) If J is a subalgebra of A+ for some associative real algebra A, then Uxy = xyx for all x,y ∈ Jin terms of the associative product (x,y) 7→ xy of A.

(b) If J = Her3(D), thenUxy = T (x• y)x− x]× y

for all x,y ∈ J.

pr.OPLSP 25. Show that a linear map between commutative real algebras is a homomorphism if and only ifit preserves squares. Conclude that the algebra O+ of Exc. 7 is a unital special real Jordan algebraalgebra such that Uxy = xyx for all x,y ∈O. Is O+ euclidean?

pr.GAULI 26. Let (ei)1≤i≤n be an orthonormal basis of D as a euclidean real vector space such that e0 = 1D.Prove that

7 The Leech lattice. 37

Λ :=n

∑i=1

Zei

is a linear Z-structure of the special real Jordan algebra D+ (cf. Exc. 25). Conclude that the Gaus-sian integers Ga(D) form linear Z-structure of D, denoted by Ga(D)+.

pr.IDREJO 27. Idempotents. Let c ∈ J := Her3(D) be an idempotent in the sense that c2 = c, and assume0 6= c 6= 13. Prove N(c) = 0 and either T (c) = 1, S(c) = 0 or T (c) = 2, S(c) = 1.

pr.EXPNO 28. ProveN(x+ y) = N(x)+T (x],y)+T (x,y])+N(y)

for all x,y ∈ J := Her3(D).

7. The Leech lattice.

s.LELA

Chapter 2Foundations

c.FOUND

Our main purpose in this chapter will be to introduce a number of important con-cepts and terminological prerequisites in a degree of generality that will be requiredin the subsequent development of the book. Throughout we let k be an arbitrary com-mutative ring. All k-modules are supposed to be unital left k-modules. Unadornedtensor products are always to be taken over k.

The possibility of k = 0 being the zero ring is expressly allowed. The onlymodule over k = 0 is the zero module M = 0. It is free of rank one with basis0.

8. The language of non-associative algebras

s.LANIn this section, we give a quick introduction to the language of non-associative al-gebras. Without striving for completeness or maximum generality, we confine our-selves to what is indispensable for the subsequent development.

ss.CONAS8.1. The concept of a non-associative algebra. A non-associative algebra or justan algebra over k (or a k-algebra for short) is a k-module A together with a bilinearmap A×A→ A, called the multiplication (or product) of A and usually indicatedby juxtaposition: (x,y) 7→ xy. Thus k-algebras satisfy both distributive laws and arecompatible with scalar multiplication but may fail to be associative or commutativeor to contain a unit element. Nevertheless, the standard vocabulary of ring theory(ideals, homomorphisms, quotients, direct sums, direct products, . . . ) easily extendsto this more general setting and will be used here mostly without further comment.To mention just two examples, a subalgebra of A is a submodule closed under mul-tiplication, and a homomorphism h : A→ B of k-algebras is a linear map preservingthe product: h(xy) = h(x)h(y) for all x,y ∈ A. Examples of k-algebras for k = R ork = Z have been discussed in the preceding chapter.

For the rest of this section we fix an algebra A over k.ss.LRMUL

8.2. Left and right multiplication. For x ∈ A, the linear map

Lx : A−→ A, y 7−→ Lxy := xy,

is called the left multiplication by x in A. Similarly, the right multiplication by x inA will be denoted by

Rx : A−→ A, y 7−→ Rxy := yx .

39

40 2 Foundations

The linear map

L : A−→ Endk(A),x 7−→ Lx, (resp. R : A−→ Endk(A),x 7−→ Rx,)

is called the left (resp. right) multiplication of A.ss.GEN

8.3. Generators. For arbitrary subsets X ,Y ⊆ A, we write XY as in 26.11 for theadditive subgroup of A spanned by all products xy, x∈ X , y∈Y . Similar conventionsapply to other multi-linear mappings in place of the product of A. We abbreviateX2 := XX . A submodule B⊆ A is a subalgebra if and only if B2 ⊆ B.

Let X ⊆ A be an arbitrary subset. Then the smallest subalgebra of A containing Xis called the subalgebra generated by X. Roughly speaking, it consists of all linearcombinations of finite products, bracketed arbitrarily, of elements in X . To make thisa bit more precise, we introduce the following definition.

ss.MON8.4. Monomials over a subset of an algebra. For a subset X ⊆ A, we define sub-sets Monm(X) ⊆ A, m ∈ Z, m > 0, recursively by setting Mon1(X) = X and byrequiring that Monm(X), m ∈ Z, m > 1, consist of all products yz, y ∈ Monn(X),z ∈Monp(X), n, p ∈ Z, n, p > 0, n+ p = m. The elements of

Mon(X) :=⋃

m∈Z,m>0

Monm(X)

are called monomials over X . With these definitions it is clear that the subalgebra ofA generated by X agrees with k Mon(X), i.e., with the submodule of A spanned bythe monomials over X .

ss.ACA

8.5. Associators and commutators. The trilinear map

A×A×A−→ A, (x,y,z) 7−→ [x,y,z] := (xy)z− x(yz),

is called the associator of A, which we have described in Exc. 1 for the real algebraof Graves-Cayley octonions. Similarly, the bilinear map

A×A−→ A, (x,y) 7−→ [x,y] := xy− yx ,

is called the commutator of A. It is straightforward to check that they satisfy therelations

[xy,z]− x[y,z]− [x,z]y = [x,y,z]− [x,z,y]+ [z,x,y] , (1) ss.ACA.1

[xy,z,w]− [x,yz,w]+ [x,y,zw] = x[y,z,w]+ [x,y,z]w (2) ss.ACA.2

for all x,y,z,w ∈ A. Observe that the commutator is alternating, so [x,x] = 0 for allx ∈ A, while the associator in general is not. This gives rise to the important conceptof an alternative algebra that we encountered already in the study of Graves-Cayleyoctonions (1.6) and that will be discussed more systematically in the next chapter.

8 The language of non-associative algebras 41

ss.COMASS8.6. Commutative and associative algebras. The k-algebra A is commutative if itsatisfies the commutative law xy = yx, equivalently, if its commutator is the zeromap. Similarly, A is associative if it satisfies the associative law (xy)z = x(yz),equivalently, if its associator is the zero map. Note that A is commutative if andonly if its left and right multiplications are the same, and that the following condi-tions are equivalent.

(i) A is associative.(ii) The left multiplication L : A→ Endk(A) is an algebra homomorphism: Lxy =

LxLy for all x,y ∈ A.(iii) The right multiplication R : A→ Endk(A) is an algebra anti-homomorphism:

Rxy = RyRx for all x,y ∈ A.ss.POW

8.7. Powers. We define the powers with base x ∈ A and exponent n ∈ Z, n > 0,recursively by the rule

x1 = x, xn+1 = xxn

and write k1[x] = ∑n≥1 kxn for the submodule of A spanned by all powers of x withpositive integral exponents. It consists of all “polynomials” in x with coefficientsin k and zero constant term. A is said to be power-associative if xmxn = xm+n forall x ∈ A and all m,n ∈ Z, m,n > 0. This is equivalent to k1[x] ⊆ A, x ∈ A, being acommutative associative subalgebra; in fact, it is then the subalgebra of A generatedby x.

ss.IDEMP8.8. Idempotents. An element c∈ A is called an idempotent if c2 = c. In particular,we count 0 as an idempotent. Two idempotents c,d ∈ A are said to be orthogonalif cd = dc = 0; in this case, c+ d is also an idempotent. By an orthogonal systemof idempotents in A we mean a family (ci)i∈I consisting of mutually orthogonalidempotents: cic j = δi jci for all i, j ∈ I.

ss.ASBILIFO8.9. Associative bilinear and linear forms. A bilinear form σ : A×A→ k is saidto be associative if it is symmetric and satisfies the relation σ(xy,z) = σ(x,yz) forall x,y,z∈ A. If in this case I ⊆ A is an ideal, then so is I⊥ = x ∈ A | σ(x, I) = 0,its orthogonal complement relative to σ . To every linear form t : A→ k correspondscanonically a bilinear form σt : A× A → k via σt(x,y) := t(xy) for all x,y ∈ A.We say that t is associative if σt is, equivalently, if t vanishes on all commutatorsand associators of A. For example, it follows immediately from (1.4.1) and (1.5.5)combined with Exc. 3 (a) that the trace form of the Graves-Cayley octonions isassociative, even though the Graves-Cayley octonions themselves are not.

ss.STRUCO8.10. Structure constants. Generalizing the set-up described in 1.2, suppose A isfree as a k-module, with basis (ei)i∈I . Then there exists a unique family (γi jl)i, j,l∈Iof scalars in k such that, for all j, l ∈ I,

42 2 Foundations

γi jl = 0 (for almost all i ∈ I), (1) FISU

e jel = ∑i∈I

γi jlei. (2) MUTA

The γi jl , i, j, l ∈ I, are called the structure constants of A relative to the basis (ei).Conversely, let M be a free k-module with basis (ei)i∈I and γi jl a family of scalarsin k satisfying (1), then there is a unique algebra structure A on M making the γi jlthe structure constants of A relative to (ei): just define the multiplication of A on thebasis vectors by (2) and extend it bilinearly to all of M.

Exercises.

pr.HOSI 29. Let A,B be k-algebras and f : A→ B a k-linear map such that, for all x,y ∈ A, we have f (xy) =± f (x) f (y). Show that f or − f is a homomorphism from A to B.

pr.NILRAD 30. The nil radical. (Behrens [9]) Let A be a k-algebra. An element x ∈ A is said to be nilpotent if0∈Mon(x) is a monomial over x (8.4). This concept of nilpotency is the usual one for associativeor, more generally, for power-associative algebras. A is said to be a nil algebra if it consists entirelyof nilpotent elements, ditto for a nil ideal. Prove for any ideal I ⊆ A that A is a nil algebra if andonly if I is a nil ideal and A/I is a nil algebra. Conclude that the sum of all nil ideals in A is a nilideal, called the nil radical of A and denoted by Nil(A).

pr.ID 31. Lifting Idempotents. Let A be a power-associative k-algebra.(a) Suppose x ∈ A satisfies a monic polynomial with invertible least coefficient, i.e., there existintegers n > d > 0, and scalars αd , . . . ,αn−1 ∈ k with αd ∈ k× and αdxd + · · ·+αn−1xn−1 +xn = 0.Given r ∈ Z, r > 0, write kr[x] for the k-submodule of A spanned by the powers xn, n ∈ Z, n ≥ r.Then show that there is a unique element c ∈ kd [x] satisfying cxd = xd . Conclude that c ∈ A is anidempotent. (Hint. Use the fact that a surjective linear map from a finitely generated k-module toitself is bijective [116, 38.15].)(b) Let ϕ : A→ A′ be a surjective homomorphism of unital power-associative algebras over k andsuppose Ker(ϕ) ⊆ A is a nil ideal (Exc. 30). Conclude from (a) that every idempotent c′ ∈ A′ canbe lifted to A, i.e., there exists an idempotent c ∈ A satisfying ϕ(c) = c′.

9. Unital algebras

s.UNALLet A be a k-algebra.

ss.UNELT9.1. The unit element. As usual, e ∈ A is said to be a unit (or identity) element ofA if ex = xe = x for all x ∈ A. A unit element may not exist, but if it does it is uniqueand called the unit element of A, written as 1A. In this case, A is said to be unital. Aunital subalgebra of a unital algebra is a subalgebra containing the unit element. IfA is unital, the unital subalgebra of A generated by X ⊆ A is defined as the smallestunital subalgebra of A containing X . A unital homomorphism of unital algebras is ahomomorphism of algebras preserving unit elements.

ss.UNHU9.2. The unital hull. Every k-algebra A may be embedded into its unital hull, givenon the k-module A = k⊕A by the multiplication

9 Unital algebras 43

(α⊕ x)(β ⊕ y) := αβ ⊕ (αy+βx+ xy) (α,β ∈ k, x,y ∈ A).

Indeed, A is unital with unit element 1A = 1⊕ 0, and A embeds into A throughthe second summand via x 7→ 0⊕ x. Identifying A ⊆ A accordingly, we obtain A =k1A +A as a direct sum of submodules, and A ⊆ A is an ideal. The unital hull maybe characterized by the universal property that every homomorphism from A to aunital k-algebra B extends uniquely to a unital homomorphism from A to B.

ss.POID9.3. Powers and idempotents revisited. Suppose A is unital.(a) For x ∈ A, we define x0 := 1A and, combining with 8.7, obtain powers xn forall n ∈ N. The submodule of A spanned by these powers will be denoted by k[x],so we have k[x] = k1A + k1[x]. If A is power-associative, then k[x] ⊆ A is a unitalcommutative associative subalgebra and, in fact, agrees with the unital subalgebraof A generated by x.(b) An orthogonal system F = (ci)i∈I of idempotents in A is said to be complete ifci = 0 for almost all i ∈ I and ∑i∈I ci = 1A. Any orthogonal system F = (ci)i∈I ofidempotents in A having ci = 0 for almost all i ∈ I can be enlarged to a completeone: with an additional index i0 /∈ I and I0 := i0∪ I put F := (ci)i∈I0 where ci0 :=1A−∑i∈I ci.

ss.MULALG9.4. The multiplication algebra. The unital subalgebra of Endk(A) generated byall left and right multiplications in A is called the multiplication algebra of A andwill be denoted by Mult(A). We may view A as a (left) Mult(A)-module in a naturalway. The Mult(A)-submodules of A are precisely its (two-sided) ideals.

ss.CENT9.5. The centre. Let A be unital. An element a ∈ A is said to be central if [a,x] =[a,x,y] = [x,a,y] = [x,y,a] = 0 for all x,y ∈ A. The totality of central elements in A,written as Cent(A), is called the centre of A. By (8.5.1), (8.5.2), Cent(A) is a unitalcommutative associative subalgebra of A. The assignment α 7→ α1A gives a unitalhomomorphism from k to Cent(A). If this homomorphism is an isomorphism, A issaid to be central. By restricting the product of A to Cent(A)×A, we obtain a scalarmultiplication on A by elements of the centre, that converts A into a central algebraover Cent(A), denoted by Acent and called the centralization of A.

ss.NUC9.6. The nucleus. Slightly less important than the centre but still useful is the nu-cleus of a unital k-algebra A, defined by

Nuc(A) :=

a ∈ A | [a,A,A] = [A,a,A] = [A,A,a] = 0.

Following (8.5.2), the nucleus is a unital associative subalgebra of A, but will fail ingeneral to be commutative. A subalgebra B ⊆ A is said to be nuclear if it is unital(as a subalgebra) and is contained in the nucleus of A.

ss.SIMAL9.7. Simplicity and division algebras. A is said to be simple if it has non-trivialmultiplication, so A2 6= 0, and 0 and A are the only (two-sided) ideals of A.

44 2 Foundations

Examples are fields or, more generally, division algebras, where A is called a di-vision algebra if it is non-zero and, for all u,v ∈ A, u 6= 0, the equations ux = v,yu = v can be solved uniquely in A. Division algebras have no zero divisors: if A isa division algebra and x,y ∈ A satisfy xy = 0, then x = 0 or y = 0. Examples of di-vision algebras are provided by arbitrary field extensions but also, for k =R, by theGraves-Cayley octonions (1.7) and the Hamiltonian quaternions (1.11). Note thatan algebra A over k with non-trivial multiplication is simple if and only if it is anirreducible Mult(A)-module.

ss.MAT9.8. Matrices. For n ∈ Z, n > 0, we write Matn(A) for the k-module of n× n-matrices over A; it becomes a k-algebra under ordinary matrix multiplication. More-over, if A is unital, then so is Matn(A), with identity element given by the n×n unitmatrix 1n = (δi j1A)1≤i, j≤n in terms of the Kronecker-delta, and the usual matrixunits ei j, 1≤ i, j ≤ n, make sense in Matn(A). More generally, in obvious notation,

Matn(A) =n⊕

i, j=1

Aei j

as a direct sum of k-modules (where each summand on the right identifies canoni-cally with A as a k-module), and the expressions aei j, a ∈ A, 1≤ i, j ≤ n, satisfy themultiplication rules

(aei j)(belm) = δ jl(ab)eim (a,b ∈ A, 1≤ i, j, l,m≤ n). (1) MAMU

p.IDMA9.9. Proposition. Let n ∈ Z, n > 0, and suppose A is unital. Then the assignmenta 7→Matn(a) gives an inclusion preserving bijection from the set of ideals in A tothe set of ideals in Matn(A).

Proof. It suffices to show that any ideal in Matn(A) has the form Matn(a) for someideal a in A, so let I ⊆Matn(A) be an ideal and put a := a ∈ A | ae11 ∈ I. From(9.8.1) we deduce

(ab)e11 = (ae11)(be11), ae1 j = (ae11)e1 j, ae j1 = e j1(ae11), aei j = ei1(ae1 j),

for a,b∈ A, 1≤ i, j≤ n, which implies that a⊆ A is an ideal satisfying Matn(a)⊆ I.Conversely, let x = ∑i j ai jei j ∈ I, ai j ∈ A. For 1 ≤ l,m ≤ n, the matrix alme11 =

∑i, j(e1l(ai jei j))em1 =(e1lx)em1 is contained in I, forcing alm ∈ a, hence x∈Matn(a),and the proposition is proved.

c.SIMA9.10. Corollary. If A is a simple unital k-algebra, then so is Matn(A), for anyinteger n > 0.

Exercises.

pr.IDDIRSUM 32. Ideals of direct sums. Let (A j)1≤ j≤n be a finite family of unital k-algebras. Show that the idealsof the direct sum A = A1⊕·· ·⊕An (under the componentwise multiplication) are precisely of theform I1⊕ ·· · ⊕ In where I j are ideals of A j for 1 ≤ j ≤ n. Does this conclusion also hold if the

9 Unital algebras 45

A j,1 ≤ j ≤ n, are not assumed to be unital? Conclude that the decomposition of a unital algebrainto the direct sum of finitely many simple ideals, if at all possible, is unique up to order.

pr.ALGEL 33. Algebraic elements in power-associative algebras. Let A be a unital power-associative algebraover a field F . An element x ∈ A is said to be algebraic (over F) if the subalgebra F [x] ⊆ A isfinite-dimensional. In this case, the unique monic polynomial of least degree in F [t] killing x iscalled the minimum polynomial of x (over F) and is denoted by µx; note that µx generates the idealof all polynomials in F [t] killing x. We say that x is split algebraic (over F) if it is algebraic andits minimum polynomial decomposes into linear factors over F . The algebra A is called algebraic(resp. split algebraic) if every element of A has this property.

Now let x be a split algebraic element of A and write

µx =r

∏i=1

(t−αi)ni (1) SPLIMI

with positive integers r,n1, . . . ,nr and α1, . . . ,αr ∈ F distinct. Then prove:

(a) Setting µi := µx(t−αi)

ni ∈ F [t] for 1 ≤ i ≤ r, there are polynomial f1, . . . , fr ∈ F [t] such that∑

ri=1 µi fi = 1. Conclude that (c1, . . . ,cr) with ci := µi(x) fi(x) for 1 ≤ i ≤ r is a complete

orthogonal system of idempotents in F [x], and that there exists an element v∈F [x] satisfying

x =r

∑i=1

αici + v, vn = 0 (n := max1≤i≤r ni). (2) MIDEC

(b) c∈ F [x] is an idempotent if and only if there exists a subset I ⊆1, . . . ,r such that c = cI :=∑

ri=1 ci.

(c) The following conditions are equivalent.

(i) Nil(F [x]) = 0.(ii) n1 = · · ·= nr = 1.(iii) x = ∑

ri=1 αici.

In this case, x is called split semi-simple.(d) x is invertible in F [x] if and only if αi 6= 0 for all i = 1, . . . ,r.(e) Assume char(F) 6= 2 and αi ∈F×2 for 1≤ i≤ r. Then there exists a y∈F [x] such that y2 = x.

(Hint. Reduce to the case α1 = · · ·= αr = 1.)

pr.CENIDSPLIDE 34. Central idempotents and direct sums of ideals. Let A be a unital k-algebra. Idempotents of Abelonging to the centre are said to be central. Show for a positive integer n that the assignment

(e j)1≤ j≤n 7−→ (Ae j)1≤ j≤n

yields a bijection from the set of complete orthogonal systems of n central idempotents in A ontothe set of decompositions of A into the direct sum of n complementary ideals.

pr.NUCMAT 35. Nucleus and centre of matrix algebras. Let A be a unital k-algebra and n a positive integer.Show

Nuc(

Matn(A))= Matn

(Nuc(A)

), Cent

(Matn(A)

)= Cent(A)1n.

pr.ASLIFO 36. Associative linear forms and unitality. Let A be a k-algebra. Show that A is unital provided itadmits an associative linear form whose corresponding (symmetric) bilinear form is non-singular.

pr.FIDIA 37. Let F be an algebraically closed field. Show that every finite-dimensional non-associative di-vision algebra over F is isomorphic to F .

46 2 Foundations

10. Scalar extensions

s.SCEXScalar extensions belong to the most useful techniques in the study of modules andnon-associative algebras over commutative rings. In this section, we briefly recallthe main ingredients of this technique, remind the reader of some standard factsabout projective modules and give a few application to scalar extensions of simplealgebras.

ss.SCEM10.1. Scalar extensions of modules. We denote by k-alg the category of unitalcommutative associative k-algebras. The objects of this category are commutativeassociative k-algebras containing an identity element, while its morphism are k-algebra homomorphisms taking 1 into 1. Similarly, we write k-mod for the categoryof k-modules, its objects being (left) k-modules and its morphisms being k-linearmaps.

Now let M ∈ k-mod and R ∈ k-alg. Then the k-module M⊗R may be convertedinto an R-module by the scalar multiplication

s(x⊗ r) = x⊗ (rs) (x ∈M, r,s ∈ R). (1) RMUL

This R-module, denoted by MR, will be called the scalar extension or base changeof M from k to R. If f : M→ N is a k-linear map between k-modules, then

fR := f ⊗1R : MR −→ NR, x⊗ r 7−→ f (x)⊗ r (2) RLEX

is an R-linear map between R-modules. Summing up, we thus obtain a functor fromk-mod to R-mod. We also have a natural map

can := canM := canM,R : M −→MR, x 7−→ xR := x⊗1R, (3) KTOR

which is k-linear but in general neither injective nor surjective. Moreover, givenan R-module M′ and a k-linear map g : M → M′, there is a unique R-linear mapg′ : MR→M′ such that g′ i = g, namely the one given by

g′(x⊗ r) = rg(x) (x ∈M, r ∈ R). (4) REX

ss.REDID10.2. Reduction modulo an ideal. Let M be a k-module and a⊆ k an ideal. If wewrite α 7→ α (resp. x 7→ x) for the canonical map from k to k := k/a (resp., from Mto M := M/aM), then M becomes a k-module under the well defined natural action

(α, x) 7−→ αx (α ∈ k, x ∈M)

from k× M to M. We have k ∈ k-alg and a natural identification Mk = M as k-modules such that

x⊗ α = αx (1) ALPHEX

10 Scalar extensions 47

for x ∈M and α ∈ k.ss.ITSCA

10.3. Iterated scalar extensions. Iterated scalar extensions collapse to simpleones: for R ∈ k-alg, we have R-alg ⊆ k-alg canonically, and, given a k-module M,any S ∈ R-alg yields a natural identification (MR)S = MS as S-modules via

(x⊗ r)⊗R s = x⊗ (rs), x⊗ s = (x⊗1R)⊗R s = xR⊗R s (1) ITBA

for all x ∈M, r ∈ R, s ∈ S. This identification is functorial in M, so we have (xR)S =xS for all x ∈ M and (ϕR)S = ϕS for all k-linear maps ϕ : M → N of k-modulesM,N. Moreover, the unit homomorphism ϑ : R→ S, r 7→ r · 1S, is a morphism ink-alg (even in R-alg) such that

(1M⊗ϑ)(x) = xS (2) TENTH

for all x ∈MR.ss.LOPROM

10.4. Localizations. Particularly important instances of scalar extensions are pro-vided by localizations at prime ideals. Here are some of the relevant facts. Unlessexplicitly saying otherwise, we refer the reader to Bourbaki [11, II] for details andfurther reading.

We denote by Spec(k) the prime spectrum of k, i.e., the totality of prime idealsin k, endowed with the Zariski topology. Recall that a basis for this topology isprovided by the principal open sets

D( f ) := p ∈ Spec(k) | f /∈ p ⊆ Spec(k) ( f ∈ k). (1) PROP

We also put

V (Z) := p ∈ Spec(k) | Z ⊆ p (Z ⊆ k), (2) ZACLO

V ( f ) :=V ( f) = Spec(k)\D( f ) for f ∈ k and have V (Z) =V (kZ) for Z ⊆ k. Theopen (resp. closed) subsets of Spec(k) relative to the Zariski topology are calledZariski-open (resp. Zariski-closed).

Let ϑ : k→ k′ be a homomorphism of commutative rings, i.e., a morphism inZ-alg. Then

Spec(ϑ) : Spec(k′)→ Spec(k), p′ 7→ Spec(ϑ)(p′) := ϑ−1(p′), (3) SPEPH

is a well defined continuous map. In this way, Spec becomes a contra-variant functorfrom the category of commutative rings to the category of topological spaces.

The localization of k at a prime ideal p⊆ k will be denoted by kp; it is a local ringwhose residue field, written as κ(p), agrees with the quotient field of k/p. Now letM be a k-module. We write Mp := M⊗ kp (resp. M(p) := M⊗κ(p) = Mp⊗kp κ(p)for the base change of M from k to kp (resp. to κ(p)), and obtain natural maps

48 2 Foundations

canp := canM,kp : M −→Mp, x 7−→ xp := xkp , (4) CANPE

can(p) := canM,κ(p) : M→M(p), x 7−→ x(p) := xκ(p). (5) CANOP

Moreover, for any k-linear map ϕ : M → N of k-modules M,N, we put ϕp :=ϕkp : Mp→ Np and ϕ(p) := ϕκ(p) : M(p)→ N(p).

Given a ring homomorphism ϑ : k→ k′ as above, let p′ ∈ Spec(k′) and put p :=Spec(ϑ)(p′) = ϑ−1(p′). Then we obtain a commutative diagram

k ϑ //

canp

k′

canp′

kp

ϑ ′ //

canp(p)

k′p′

canp′ (p′)

κ(p)

ϑ

// κ(p′),

(6) KAYPE

where ϑ ′ : kp → k′p′ is the local homomorphism canonically induced by ϑ and inturn determines the field homomorphism ϑ : κ(p)→ κ(p′), making κ(p′) a fieldextension of κ(p). Now let M be a k-module, regard k′ (resp. k′p′) as an element ofk-alg (resp. kp-alg) by means of ϑ (resp. ϑ ′) and put M′ := Mk′ as k′-modules. Then(10.3.1) and 10.2 yield natural identifications

M′p′ = Mp⊗kp k′p′ (7) MPRLO

as k′p′ -modules that are functorial in M. Similarly,

M′(p′) = M(p)⊗κ(p) κ(p′) (8) MPRFI

as vector spaces over κ(p′).ss.PROPSP

10.5. Principal open sets as spectra. For f ∈ k, we denote by

k f := α/ f n | α ∈ k, n ∈ N (1) KAF

the ring of fractions associated with the multiplicative subset f n | n ∈ N ⊆ k andby

can f : k −→ k f , α 7−→ α/1, (2) CANF

the canonical homomorphism, making k f a k-algebra. The continuous map

Spec(can f ) : Spec(k f )−→ Spec(k)

induces canonically a homeomorphism


Φ f : Spec(k f )∼−→ D( f )⊆ Spec(k)

such that

Φ f (q) = α ∈ k | α/1 ∈ q (q ∈ Spec(k f )), (3) FIEF

Φ−1f (p) = p f := α/ f n | α ∈ k, n ∈ N (p ∈ D( f )). (4) FIIN

Moreover, for g ∈ k, n ∈ N, the principal open set D(g/ f n)⊆ Spec(k f ) satisfies

Φ f(D(g/ f n)

)= D( f )∩D(g) = D( f g). (5) DEGF

Finally, for a k-linear map ϕ : M → N of k-modules M,N, we denote ny ϕ f :=ϕk f : Mk f =: M f → N f := Nk f its k f -linear extension.

ss.PROPID10.6. Principal open sets of idempotents. Let ε ∈ k be an idempotent and M ak-module. We put ε+ := ε , ε− := 1− ε , k± = ε±k, M± := ε±M = k±M and havek = k+⊕k− as a direct sum of ideals, M = M+⊕M− as a direct sum of submodules.The projections

π± : k −→ k±, α 7−→ α± := ε±α, (1) PIPM

make k± elements of k-alg. Similarly, we have the projections M→M±, x 7→ x± :=ε±x, and obtain natural identifications

Mk± = M⊗ k± = M± (2) EMPM

as k±-modules such that

x⊗α± = α±x = α±x± (α± ∈ k±, x ∈M). (3) EXTEA

Moreover, any k-linear map ϕ : M→ N of k-modules M,N induces k±-linear mapsϕ± : M±→ N± via restriction and ϕ± = ϕk± is the base change of ϕ from k to k±.

Specializing f to ε in 10.5, we obtain

kε = α/ε | α ∈ k+= α/1 | α ∈ k+= α/1 | α ∈ k. (4) KEPS

There is a unique morphism π+ε : kε → k+ in k-alg that makes the diagram

kπ+ //

canε

k+

kε ,

∼=π+ε

88

(5) CANPI

commutative and is actually an isomorphism. By 10.5, therefore,

Spec(π+) : Spec(k+)→ Spec(k)

50 2 Foundations

induces canonically a homeomorphism Spec(k+)∼→ D(ε)⊆ Spec(k). Note that

Spec(π+)(p+) = p+⊕ k− ∈ D(ε) (6) SPEPI

for all p+ ∈ Spec(k+).ss.PROMOD

10.7. Projective modules. Recall that a k-module M is projective if it is a directsummand of a free k-module. This property is stable under base change, i.e., ifM is projective over k, then so is MR = M⊗R over R, for all R ∈ k-alg. Given aprojective k-module M, a theorem of Kaplansky [53] (see also Scheja-Storch [116,§ 88, Aufg. 45]) shows that Mp is a free kp-module, possibly of infinite rank, for allprime ideals p⊆ k. Hence to every projective k-module M we can associate its rankfunction

Spec(k)−→ N∪∞, p 7−→ rkp(M) := rkkp(Mp) = dimκ(p)

(M(p)

).

By [11, II, Theorem 1] a k-module M is finitely generated projective if and only if,for all p ∈ Spec(k), the kp-module Mp is free of finite rank and the rank functionof M is locally constant relative to the Zariski topology of Spec(k). For a projectivek-module M and r ∈ N, we say M has rank r if rkp(M) = r for all p ∈ Spec(k).This condition forces M to be finitely generated and determines r uniquely, unlessk = 0, in which case M = 0 has rank r for any r ∈ N. If M is projective but notnecessarily finitely generated, the preceding terminology makes sense and will beused also for r = ∞.

Finally, we deviate slightly from the terminology of Knus [58, I §9, p. 52] bycalling a k-module faithfully projective if it is projective and faithful.

ss.LIBU10.8. Line bundles. As usual, a line bundle over k is defined as a finitely gener-ated projective k-module of rank 1. This means that the following two equivalentconditions are fulfilled.

(i) L is a finitely generated k-module and, for all p ∈ Spec(k), the kp-module Lp

is free of rank 1.(ii) There are finitely many elements f1, . . . , fm ∈ k such that ∑k fi = k and, for

each i = 1, . . . ,m, the k fi -module L fi is free of rank 1.

For lines bundles L,L′ over k, we recall the following elementary but fundamentalfacts.

(a) L⊗L′ and L∗ = Homk(L,k) are line bundles over k.(b) Pic(k) := [L]|L is a line bundle over k, where [L] stands for the isomorphism

class of L, is an abelian group under the operation [L][L′] := [L⊗L′], with unitelement and inverse of [L] given by [k] and[L∗], respectively. Pic(k) is calledthe Picard group of k.

(c) For R ∈ k-alg, LR = L⊗R is a line bundle over R and L 7→ LR gives a grouphomomorphism Pic(k)→ Pic(R). Thus we obtain a (covariant) functor Picfrom k-algebras to abelian groups.


(d) There is a natural isomorphism L∗⊗ L′ ∼→ Homk(L,L′) sending x∗⊗ x′, forx∗ ∈ L∗, x′ ∈ L′, to the linear map L→ L′, x 7→ 〈x,x∗〉x′.

(e) If ϕ : L→ L′ is an epimorphisms, it is, in fact, an isomorphism.ss.UMOEL

10.9. Faithful and unimodular elements. Fixing a k-module M, we write M∗ =Homk(M,k) for the dual module of M and (x∗,x) 7→ 〈x∗,x〉 for the canonical pairingM∗×M→ k. Given x ∈M,

Ann(x) := α ∈ k | αx = 0, 〈M∗,x〉 := 〈x∗,x〉 | x∗ ∈M∗ (1) ANX

are ideals in k, the former being called the annihilator of x, and we have

Ann(x)〈M∗,x〉= 0. (2) ANM

We say x is faithful if Ann(x) = 0, equivalently, if the map k→ M, α 7→ αx, isinjective. A much stronger condition is provided by the notion of unimodularity:x is said to be unimodular if kx ⊆ M is a free submodule (of rank 1) and a directsummand of M at the same time. This obviously happens if and only if 〈x∗,x〉 = 1for some x∗ ∈M∗, i.e., 〈M∗,x〉= k; in this case, the k-module kx has x as a basis.Note that for an element of a projective (hence free) module M over a local ring k tobe unimodular it is necessary and sufficient that it can be extended to a basis of M.

Unimodular elements are faithful but not conversely since unimodularity is stableunder base change while faithfulness is not. Therefore we call x ∈M strictly faithfulif xR ∈ MR is faithful for all R ∈ k-alg. Given a unital k-algebra A, the identityelement 1A ∈ A is faithful if and only if A is faithful as a k-module: αA = 0implies α = 0, for all α ∈ k.

ss.NOAM10.10. A notational ambiguity. Let M be a k -module, x∗ ∈ M∗ and R ∈ k-alg.Then the symbol x∗R := (x∗)R can be interpreted in two ways: on the one hand,as x∗R = x∗ ⊗ 1R via (10.1.3), which belongs to (M∗)R, on the other as x∗R =x∗⊗ 1R : MR → kR = R via (10.1.6), which belongs to (MR)

∗. In general, these in-terpretations lead to completely different objects, but, as the following result shows,in an important special case they may be identified under a canonical isomorphism.In fact, this isomorphism survives as a homomorphism in full generality as follows:

For a k-module M and any R ∈ k-alg as above, the assignment x∗ 7→ x∗ ⊗ 1Rdetermines a k-linear map M∗→ (MR)

∗, which by (10.1.4) gives rise to an R-linearmap ϕ : (M∗)R→ (MR)

∗ satisfying ϕ(x∗⊗ r) = r(x∗⊗1R) for all x∗ ∈M∗, r ∈ R.l.PRODU

10.11. Lemma. Let M be a finitely generated projective k-module and R ∈ k-alg.Then the natural map

ϕ : (M∗)R∼−→ (MR)

∗, x∗⊗ r 7−→ r(x∗⊗1R)

is an isomorphism of R-modules. Identifying (M∗)R = (MR)∗ =: M∗R by means of this

isomorphism, we have 〈x∗,x〉R = 〈x∗R,xR〉 for all x ∈M, x∗ ∈M∗, in other words, the

52 2 Foundations

canonical pairing M∗R×MR→ R is the R-bilinear extension of the canonical pairingM∗×M→ k.

Proof. We must show that ϕ is bijective. Localizing if necessary, we may assumethat M is free of finite rank, with basis (ei)1≤i≤n. Let (e∗i )1≤i≤n be the correspondingdual basis of M∗. Then (eiR), (e∗i ⊗ 1R) are R-bases of MR, (M∗)R, respectively,while the family (e∗i ⊗ 1R) of elements of (MR)

∗ is dual to (eiR), hence forms thecorresponding dual basis of (MR)

∗. But ϕ(e∗i ⊗1R) = e∗i ⊗1R for 1≤ i≤ n, forcing ϕ

to be an isomorphism. Finally, using ϕ to identify (M∗)R = (MR)∗ and letting x∈M,

x∗ ∈M∗, we conclude 〈x∗R,xR〉 = 〈x∗⊗ 1R,xR〉 = 〈x∗⊗ 1R,x⊗ 1R〉 = 〈x∗,x〉⊗ 1R =〈x∗,x〉R.

l.UNIMOD10.12. Lemma. (Loos [71]) Assume M is a finitely generated projective k-moduleand let x ∈M. If p⊆ k is a prime ideal, then x(p) = 0 if and only if 〈M∗,x〉 ⊆ p.

Proof. M(p) being a vector space over κ(p), we have x(p) = 0 if and onlyif 〈y∗,x(p)〉 = 0 for all y∗ ∈ M(p)∗. Identifying M(p)∗ = M∗(p) by means ofLemma 10.11, we therefore obtain

x(p) = 0⇐⇒ 〈x∗(p),x(p)〉= 0 for all x∗ ∈M∗

⇐⇒ 〈x∗,x〉(p) = 0 for all x∗ ∈M∗

⇐⇒ 〈x∗,x〉 ∈ p for all x∗ ∈M∗

⇐⇒ 〈M∗,x〉 ⊆ p.

l.FAITH

10.13. Lemma. Consider the following conditions, for a k-module M and x ∈M.

(i) x is unimodular.(ii) x is strictly faithful.(iii) xR 6= 0 for all R ∈ k-alg with R 6= 0.(iv) xK 6= 0 for all fields K ∈ k-alg.(v) x(p) 6= 0 for all p ∈ Spec(k).

Then the implications

(i) =⇒ (ii) =⇒ (iii)⇐⇒ (iv)⇐⇒ (v)

hold, and if M is finitely generated projective, then all five conditions are equivalent.

Proof. Since the implications (i)⇒ (ii)⇒ (iii)⇒ (iv)⇒ (v) are obvious, while (v)⇒ (i) follows immediately from Lemma 10.12 if M is finitely generated projective,it remains to prove (v)⇒ (iv)⇒ (iii).

(v)⇒ (iv). If K ∈ k-alg is a field, the kernel of the natural map k→ K is a primeideal p⊆ k, making K/κ(p) a field extension, and xK = x(p)K 6= 0 by (v).

(iv) ⇒ (iii). Let 0 6= R ∈ k-alg and m ⊆ R be a maximal ideal. Then K :=R/m ∈ k-alg is a field and (xR)K = xK 6= 0 by (iv), forcing xR 6= 0.


ss.TEPRAL10.14. Tensor products of algebras. Given k-algebras A,B, the k-module A⊗B isagain a k-algebra under the multiplication

(x1⊗ y1)(x2⊗ y2) = (x1x2)⊗ (y1y2) (xi ∈ A,yi ∈ B, i = 1,2). (1) TEPRO

Moreover, if A and B are both unital, so is A⊗B, with unit element 1A⊗B = 1A⊗1B.For example, given any unital k-algebra A and a positive integer n, we have a naturalidentification Matn(k)⊗A = Matn(A) such that x⊗a = xa = (ξi ja) for x = (ξi j) ∈Matn(k), a ∈ A.

Now let A be a k-algebra and R∈ k-alg. Then the R-module structure of AR (10.1)is compatible with the k-algebra structure of A⊗R as defined in (1). In other words,AR is canonically an R-algebra, and the observations made in 10.1−10.9 carry overmutatis mutandis from modules to algebras. In particular, the assignment A 7→ ARgives a functor from k-algebras to R-algebras.

We now give applications of the preceding set-up to simple algebras.ss.p-CENTMAL

10.15. Proposition. Let A be a unital k-algebra. Then the right multiplication of Ainduces an isomorphism

R : Cent(A) ∼−→ EndMult(A)(A)

of k-algebras.

Proof. For a ∈ Cent(A), Ra clearly belongs to EndMult(A)(A), so we obtain a uni-tal homomorphism R : Cent(A)→ EndMult(A)(A), which is obviously injective. Toshow that it is also surjective, let d ∈ EndMult(A)(A) act on A from the right byjuxtaposition. Then (xy)d = x(yd) = (xd)y for all x,y ∈ A. Putting y = 1A, weconclude d = Ra, where a := 1Ad. Hence the preceding relation is equivalent to(xy)a = x(ya) = (xa)y, which in turn is easily seen to imply that a belongs to thecentre of A.

ss.c-CENTSIM1

10.16. Corollary. The centre of a unital simple algebra is a field.

Proof. Since Mult(A) acts irreducibly on A, it suffices to combine Prop. 10.15 withSchur’s lemma [47, p. 118].

For the remainder of this section, we use Prop. 10.15 to identify Cent(A) =EndMult(A)(A) for any unital k-algebra A.

ss.c-CENTSIM210.17. Corollary. Let A be a unital finite-dimensional algebra over a field k andsuppose A is simple. Then EndCent(A)(A) = Mult(A).

Proof. Since Mult(A) acts faithfully and irreducibly on A, it is a primitive Ar-tinian k-algebra [47, Def. 4.1]. Moreover, by Prop. 10.15, its centralizer in Endk(A)is Cent(A). Thus the assertion follows from the double centralizer theorem [47,Thm. 4.10].

54 2 Foundations

Algebras that are both central and simple are called central simple.ss.c-CENTSIM3

10.18. Corollary. A unital finite-dimensional algebra A over a field k is centralsimple if and only if it is non-zero and Mult(A) = Endk(A).

Proof. If A is central simple, the assertion follows from Cor. 10.17. Conversely,suppose A 6= 0 and Mult(A) = Endk(A). Then A is simple, and Cor. 10.17 impliesthat Cent(A) belongs to the centre of Endk(A). Hence A is central.

ss.c-CENTSIM410.19. Corollary. For a unital finite-dimensional algebra A over a field k, the fol-lowing conditions are equivalent.

(i) A is central simple.(ii) Every base field extension of A is central simple.(iii) A⊗ k is simple, where k denotes the algebraic closure of k.

Proof. (i)⇒ (ii). Let k′ be an extension field of k and put A′ = A⊗k′ as a k′-algebra.After identifying Endk′(A′)=Endk(A)⊗k′ canonically, a moment’s reflection showsMult(A′) = Mult(A)⊗ k′. Hence the assertion follows from Cor. 10.18.(ii)⇒ (iii). Obvious.(iii) ⇒ (i). If I ⊆ A is a non-trivial ideal, then so is I⊗ k ⊆ A⊗ k, a contradiction.Hence A is simple. By Exc. 44 (a), Cent(A)⊗ k agrees with the centre of A⊗ k andtherefore is a finite algebraic field extension of k. As such, it has degree 1, forcingCent(A) = k1A, and A is central.

Exercises.

pr.CEBA 38. Centre and nucleus under base change. Show that the centre (resp. the nucleus) of a unitalk-algebra A in general does not commute with base change, even if one assumes that the centre(resp. the nucleus) be free of finite rank and a direct summand of A as a k-module. (Hint. Let Abe the unital Z-algebra given on a free Z-module of rank 3 with basis 1A,x,y by the multiplicationtable x2 = 2 ·1A, xy = yx = y2 = 0 and consider its base change to R := Z/2Z ∈ Z-alg.)

pr.FIB 39. Fibers of prime spectra. Let ϕ : k→ k′ be a homomorphism of commutative rings, let p ∈Spec(k) and view k′ as a k-algebra by means of ϕ . Consider the ring homomorphism

ψ : k′ −→ k′⊗κ(p), α′ 7−→ ψ(α ′) := α

′⊗1κ(p)

and show that the continuous map

Spec(ψ) : Spec(k′⊗κ(p)

)−→ Spec(k′)

induces canonically a homeomorphism

Spec(k′⊗κ(p)

) ∼−→ Spec(ϕ)−1(p),

where the right-hand side carries the topology induced from the Zariski topology of Spec(k′).

pr.LOCHOM 40. Put X := Spec(k).

(a) Let ψ,χ : M→ N be k-linear maps of k-modules and suppose M is finitely generated. Showthat

U := p ∈ X | ψp = χpis a Zariski-open subset of X .


(b) Let ϕ : A→ B be a k-linear map of k-algebras and suppose A is finitely generated as ak-module. Prove that

V := p ∈ X | ϕp : Ap→ Bp is an algebra homomorphism

is a Zariski-open subset of X .

pr.IDEPAR 41. Idempotents and partitions. Consider X = Spec(k) under the Zariski topology as in 10.4.(a) Let ε ∈ k be an idempotent. Prove

D(ε) = p ∈ X | εp = 1p= p ∈ X | εp 6= 0=V (1− ε).

Conclude for an orthogonal system (εi)i∈I of idempotents in k that⋃

D(εi) =D(∑εi) and the unionon the left is disjoint.(b) Prove that the assignment (εi)i∈I 7→ (D(εi))i∈I yields a bijection from the set of complete or-thogonal systems of idempotents in k in the sense of 9.3 (b) onto the set of decompositions of Xinto the disjoint union of open subsets almost all of which are empty. (Hint. To prove surjectivity,you may either argue with the structure sheaf of the geometric (= locally ringed) space attached tok (Grothendieck [36, Chap. 1, § 1], Demazure-Gabriel [22, I, § 1, no. 1,2, particularly Prop. 2.6],Hartshorne [39, II, § 2]) or imitate the proof of Bourbaki [11, II, § 4, Prop. 15], but watch out inthe latter case for a sticky point on lines 2,3, p. 104.)

pr.RANDEC 42. Rank decomposition. Let M be a finitely generated projective k-module. Prove:

(a) The setRk(M) := rkp(M) | p ∈ Spec(k)

is finite.(b) There exists a unique complete orthogonal system (εi)i∈N of idempotents in k such that, with

the induced decompositions

k =⊕i∈N

ki, ki = kεi (i ∈ N), (1) KKEI

M =⊕i∈N

Mi, Mi = M⊗ ki (i ∈ N) (2) MMEI

as direct sums of ideals (resp. of additive subgroups), the ki-modules Mi are finitely generatedprojective of rank i, for all i ∈ N. Show further that εi = 0 for all i ∈ N\Rk(M).

Remark. Equation (2) is called the rank decomposition of M. It remains virtually unchanged byignoring the components belonging to some or all indices i ∈ N \Rk(M). Note also that, if Mcarries an algebra structure, (2) is a direct sum of ideals.

pr.LOCSIALG 43. Residually simple algebras. (Cf. Knus [58, III, (5.1.8)]) Let A, A′ be unital k-algebras that arefinitely generated projective of the same rank r ∈ N as k-modules. Suppose A is residually simple,so A(p) is a simple algebra over the field κ(p), for all p ∈ Spec(k). Prove that every unital algebrahomomorphism from A to A′ is an isomorphism.

pr.GRANTO 44. Groups of antomorphisms1. By an algebra with a group of antomorphisms over k we mean apair (A,G) consisting of a k-algebra A and a group G of automorphisms or anti-automorphismsof A. An ideal of (A,G) is an ideal of A which is stabilized by every element of G. We say that(A,G) is simple if A2 6= 0 and there are no ideals of (A,G) other than 0 and A. If A containsan identity element, in which case we also say that (A,G) is unital, we define the centre of (A,G)as the set of elements in the centre of A that remain fixed under G:

1 This is not a misprint.

56 2 Foundations

Cent(A,G) = a ∈ Cent(A) | g(a) = a for all g ∈ G.

If the natural map from k to Cent(A,G) is an isomorphism of k-algebras, then (A,G) is said tobe central. Unital algebras with a group of antomorphisms which are both central and simple arecalled central simple. By the multiplication algebra of (A,G), written as Mult(A,G), we meanthe unital subalgebra of Endk(A) generated by G and all left and right multiplications affectedby arbitrary elements of A. Base change of algebras with a group of antomorphisms is definedcanonically. Note that ordinary k-algebras are algebras with a group of antomorphisms in a naturalway, as are algebras with involution.

Now let k be a field and (A,G) a k-algebra with a group of antomorphisms. Then prove:(a) Cent((A,G)⊗R) = (Cent(A,G))⊗R as R-algebras, for any R ∈ k-alg.(b) The right multiplication of A induces an isomorphism

R : Cent(A,G)∼−→ EndMult(A,G)(A)

of k-algebras.(c) The centre of a unital simple algebra with a group of antomorphisms is a field.(d) If (A,G) is finite-dimensional, then EndCent(A,G)(A) = Mult(A,G), and for (A,G) to be centralsimple it is necessary and sufficient that A 6= 0 and Mult(A,G) = Endk(A).

pr.GRANTOCS 45. Central simplicity of algebras with groups of antomorphisms. Let (A,G) be a finite-dimensionalunital algebra with a group of antomorphisms over a field k. Then show that the following condi-tions are equivalent.

(i) (A,G) is central simple.(ii) Every base field extension of (A,G) is central simple.(iii) (A,G)⊗ k is simple, where k denotes the algebraic closure of k.

pr.TPGRANTO 46. Tensor products of algebras with groups of antomorphisms. Let (A,G),(A′,G′) be unital k-algebras with a group of antomorphisms and suppose (i) A is commutative or G′ ⊆ Aut(A′), and(ii) A′ is commutative or G ⊆ Aut(A). Define (A,G)⊗ (A′,G′) = (A⊗A′,G⊗G′), where G⊗G′

is supposed to consist of all g⊗ g′, g ∈ G,g′ ∈ G′ having the same parity in the sense that theyare both either automorphisms or anti-automorphisms of A,A′, respectively. Show that if (A,G) iscentral simple and (A′,G′) is simple, then (A,G)⊗ (A′,G′) is simple.

11. Involutions

s.INVInvolutions of associative algebras are a profound concept with important connec-tions to other branches of algebra and arithmetic. An in-depth account of this con-cept over fields may be found in Knus-Merkurjev-Rost-Tignol [59]. Over arbitrarycommutative rings, the reader may consult Knus [58, III, § 8] for less complete butstill useful results on the subject. In the present section, elementary properties ofinvolutions will be studied for unital non-associative algebras.

ss.ALIN11.1. Algebras with involution. An involution of a unital k-algebra B is a k-linearmap τ : B→ B satisfying the following conditions.

(i) τ is involutorial, i.e., τ2 = 1B.(ii) τ is an anti-homomorphism, i.e., τ(xy) = τ(y)τ(x) for all x,y ∈ B.

11 Involutions 57

In particular, τ is bijective, hence by (ii) may be viewed as an isomorphism τ : B→Bop of k-algebras, where the opposite algebra Bop lives on the same k-module asB under the new multiplication x · y := yx for x,y ∈ B. Examples of involutions areprovided by the conjugation of the Graves-Cayley octonions or the Hamiltonionquaternions (Exc. 3 (a)).

ss.HOID11.2. Homomorphisms, base change and ideals of algebras with involution. Bya k-algebra with involution we mean a pair (B,τ) consisting of a unital k-algebraB and an involution τ of B. A homomorphism h : (B,τ)→ (B′,τ ′) of k-algebraswith involution is a unital homomorphism h : B→ B′ of k-algebras that respectsthe involutions, i.e., τ ′ h = h τ . In this way, we obtain the category of k-algebraswith involution. If (B,τ) is a k-algebra with involution, then (B,τ)R := (BR,τR) forR∈ k-alg is an R-algebra with involution, called the scalar extension or base changeof (B,τ) from k to R. By an ideal of (B,τ) we mean an ideal I ⊆ B that is stabilizedby τ: τ(I) = I. In this case, (B,τ)/I := (B, τ), with B := B/I and τ : B→ B beingthe k-linear map canonically induced by τ , is a k-algebra with involution makingthe canonical projection B→ B a homomorphism (B,τ)→ (B, τ) of k-algebras withinvolution.

ss.CEIN

11.3. The centre of an algebra with involution. If (B,τ) is a k-algebra with invo-lution, then τ stabilizes the centre of B, and via restriction we obtain an involutionof Cent(B), i.e., an automorphism of period 2. We call

Cent(B,τ) := a ∈ Cent(B) | τ(a) = a

the centre of (B,τ). It is a unital (commutative associative) subalgebra of Cent(B).ss.SWINV

11.4. Simplicity and the exchange involution Let (B,τ) be a k-algebra with invo-lution. We say that (B,τ) is simple (as an algebra with involution) if B has non-trivialmultiplication and there are no ideals of (B,τ) other than 0 and B. If B is simple,so obviously is (B,τ). The converse, however, does not hold. To see this, we con-sider the following class of examples.

Let A be a unital k-algebra. Then a straightforward verification shows that themap

εA : Aop⊕A−→ Aop⊕A, x⊕ y 7−→ εA(x⊕ y) := y⊕ x,

is an involution, called the exchange involution (or switch) of Aop⊕A.p.SIMINV

11.5. Proposition. Let (B,τ) be a k-algebra with involution. For (B,τ) to be simpleas an algebra with involution it is necessary and sufficient that B be simple or thereexist a simple unital k-algebra A such that (B,τ)∼= (Aop⊕A,εA).

Proof. By Exc. 32, the condition is clearly sufficient. Conversely, suppose (B,τ) issimple but B is not. Let 0 6= I ⊂ B be any proper ideal of B. Then τ(I)+ I,τ(I)∩I are both ideals of (B,τ), the former being different from 0, the latter beingdifferent from B. Since (B,τ) is simple, we conclude B = τ(I)⊕ I as a direct sum

58 2 Foundations

of ideals. Regarding A = I as a unital k-algebra in its own right, any non-zero idealJ of A is a proper ideal of B, so by what we have just shown, B = τ(J)⊕ J, whichimplies J = A, and A is simple. One now checks easily that the map

(Aop⊕A,εA)∼−→ (B,τ), x⊕ y 7−→ τ(x)+ y,

is an isomorphism of algebras with involution.

ss.SYSK11.6. Symmetric and skew elements. Let (B,τ) be a k-algebra with involution.Following the notational conventions of [59, I, § 2], we put

H(B,τ) := Sym(B,τ) := x ∈ B | τ(x) = x, (1) SYM

Symd(B,τ) := y+ τ(y) | y ∈ B, (2) SYMD

Skew(B,τ) := x ∈ B | τ(x) =−x, (3) SKEW

Alt(B,τ) := y− τ(y) | y ∈ B, (4) ALT

which are all submodules of B, the first one among these containing the identityof B. The elements of Sym(B,τ) (resp. Skew(B,τ)) are called τ-symmetric (resp.τ-skew). If k contains 1

2 , then Sym(B,τ) = Symd(B,τ), Skew(B,τ) = Alt(B,τ), andB = Sym(B,τ)⊕Skew(B,τ) as a direct sum of submodules. At the other extreme,if 2 = 0 in k, then

Symd(B,τ) = Alt(B,τ)⊆ Sym(B,τ) = Skew(B,τ),

and in general this containment is proper.ss.AMSU

11.7. Ample submodules. Let (B,τ) be an associative k-algebra with involution.Then Symd(B,τ) is a τ-ample submodule of B in the sense that xSymd(B,τ)τ(x)⊆Symd(B,τ) for all x ∈ B. This follows from (11.6.2) and x(z+τ(z))τ(x) = xzτ(x)+τ(xzτ(x)) for all z ∈ B.

ss.CONTRA11.8. The conjugate transpose involution. Let B be a unital k-algebra and τ : B→B, x 7→ x, an involution of B. Given n ∈ Z, n > 0, it is readily checked that the map

Matn(τ) : Matn(B)−→Matn(B), x 7−→ xt ,

which sends an n×n matrix over B into its conjugate transpose, is an involution ofMatn(B), called the conjugate transpose involution induced by τ . We put

Matn(B,τ) :=(

Matn(B),Matn(τ)), Symn(B,τ) := Sym

(Matn(B),Matn(τ)

).

In the special case B = k,τ = 1k, we obtain

τort : Matn(k)−→Matn(k), S 7−→ τ(S) := St ,

called the split orthogonal involution of degree n over k. We put

11 Involutions 59

Symn(k) := Symn(k,1k) = S ∈Matn(k) | S = St, (1) SYMA

Skewn(k) := Skew(

Matn(k),τort)= S ∈Matn(k) | St =−S. (2) SKMA

In particular, Symn(k) is a free k-module of rank 12 n(n+1).

ss.TWINV11.9. Twisting involutions. Let (B,τ) be a k-algebra with involution and q ∈Nuc(B)× an invertible element of the unital associative subalgebra Nuc(B) ⊆ B(9.6). Then the map B→ B, x 7→ q−1xq is an unambiguously defined automorphism,forcing

τq : B−→ B, x 7−→ τ

q(x) := q−1τ(x)q (1) TWINV

to be an anti-automorphism of B. Moreover τ(q) is an involution provided τ(q) =±q, in which case we sometimes call τq the q-twist of τ .

ss.ALMA11.10. Alternating matrices. For n ∈ Z, n > 0, a matrix S ∈Matn(k) is said to bealternating if it satisfies one (hence all) of the following equivalent conditions.

(i) S is skew-symmetric and its diagonal entries are zero.(ii) S = T −T t for some T ∈Matn(k).(iii) xtSx = 0 for all x ∈ kn.

The submodule of Matn(k) consisting of all alternating n× n-matrices over kwill be denoted by Altn(k). By 11.8 and (11.6.4), we obviously have Altn(k) =Alt(Matn(k),τort), which is a free k-module of rank n(n−1)

2 .

ss.SPSPL11.11. The split symplectic involution of degree n. Let n be a positive integer.We view the elements of Mat2n(k) as 2× 2-blocks of n× n-matrices over k. Withthis in mind, we put

I := In :=(

0 1n−1n 0

)∈ GL2n(k) (1) SYMPTWI

and have

I2 =−12n, I−1 = It =−I. (2) TRINV

Hence we may form the I-twist (11.9) of the split orthogonal involution of degree2n over k (11.8), which we call the split symplectic involution of degree n over k,denoted by

τspl : Mat2n(k)−→Mat2n(k), S 7−→ τspl(S) := ISt I−1 = ISt It = I−1St I. (3) TAUSY

A verification shows

τspl(S) =(

dt −bt

−ct at

)(for S =

(a bc d

)∈Mat2n(k)), (4) SYEX

and we conclude

Sym(

Mat2n(k),τspl)=(a b

c at

)| a ∈Matn(k), b,c ∈ Skewn(k)

. (5) SYSP

60 2 Foundations

On the other hand, combining 11.10 (ii) with (11.6.4) and (4), we deduce that

Symd(

Mat2n(k),τspl)=(a b

c at

)| a ∈Matn(k), b,c ∈ Altn(k)

(6) SYSD

is a free k-module of rank 2n2−n.

12. Quadratic maps

s.QUAMAIn this section, we collect a few elementary facts about quadratic maps that in thisgenerality are somewhat hard to find in the literature. For a brief appearance of thisconcept over the reals, see 1.3.

ss.COQUA12.1. The concept of a quadratic map. Let M,N be k-modules. A map Q : M→Nis said to be quadratic if it satisfies the following two conditions.

(i) Q is homogeneous of degree 2: Q(αx) = α2Q(x) for all α ∈ k and all x ∈M.(ii) The map

DQ : M×M −→ N, (x,y) 7−→ (DQ)(x,y) := Q(x+ y)−Q(x)−Q(y)

is (symmetric) bilinear.

We sometimes call DQ the bilinearization orpolar map belonging to Q. Condi-tions (i), (ii) imply (DQ)(x,x) = 2Q(x) for all x∈M, so the quadratic map Q may berecovered from its polar map DQ if 1

2 ∈ k but not in general. At the other extreme, if2 = 0 in k, then DQ is alternating. In the general set-up, we often relax the notationand simply write Q(x,y) instead of (DQ)(x,y) for x,y ∈M if there is no danger ofconfusion. Quadratic forms are quadratic maps taking values in the base ring, hencearise in the special case N = k. The polar map of a quadratic form is of course calledits polar form. Examples are provided by the positive definite real quadratic formsof 3.1 and, more specifically, by the norm of the Graves-Cayley octonions (1.5).

Elementary manipulations of quadratic maps, like scalar multiples, compositionwith linear maps and direct sums, are defined in the obvious manner; we omit thedetails. Less obvious are scalar extensions, which we address here in a slightly moregeneral set-up. We begin with two preparations.

ss.TEBI12.2. Tensor products of bilinear maps. If b : M×M→N and b′ : M′×M′→N′

are bilinear maps between k-modules, then so is

b⊗b′ : (M⊗M′)× (M⊗M′)→ N⊗N′, (x⊗ x′,y⊗ y′) 7−→ b(x,y)⊗b′(x′,y′).

The same simple-minded approach does not work for quadratic maps, see Milnor-Husemoller [86, p. 111] for comments.

ss.RAQUA12.3. Radicals of bilinear and quadratic maps. Let M,N be k-modules andb : M×M → N a symmetric or skew-symmetric bilinear map. Then the submod-ule

12 Quadratic maps 61

Rad(b) := x ∈M | b(x,y) = 0 for all y ∈M ⊆M (1) RABI

is called the radical of b. Now suppose we are given a k-module M1 and a surjectivelinear map π : M→M1 such that Ker(π)⊆Rad(b). Then b factors uniquely throughπ×π to a symmetric or skew-symmetric bilinear map b1 : M1×M1→ N satisfyingRad(b1) = Rad(b)/Ker(π).

Along similar lines, let Q : M→ N be a quadratic map. Then

Rad(Q) := x ∈ Rad(DQ) | Q(x) = 0 (2) RAQUA

= x ∈M | Q(x) = (DQ)(x,y) = 0 for all y ∈M

is a submodule of M, called the radical of Q. If, for M1,π as above, we as-sume Ker(π) ⊆ Rad(Q), then Q factors uniquely through π to a quadratic mapQ1 : M1→ N (well) defined by Q1(π(x)) := Q(x) for x ∈M. Moreover, Rad(Q1) =Rad(Q)/Ker(π) and D(Q1) = (DQ)1.

p.TEQUA12.4. Proposition. Let M,N,V,W be k-modules, Q : M→ N a quadratic map andb : V ×V →W a symmetric bilinear map. Then there exists a unique quadratic map

Q⊗b : M⊗V −→ N⊗W

such that

(Q⊗b)(x⊗ v) = Q(x)⊗b(v,v) (x ∈M, v ∈V ), (1) QUATE

D(Q⊗b) = (DQ)⊗b. (2) BILTE

Proof. Uniqueness is clear, so we only have to show existence. We first consider thecase that the k-module M is free, with basis (ei)i∈I where we may assume the indexset I to be totally ordered. Then every element z ∈M⊗V can be written uniquely asz = ∑i ei⊗ vi, vi ∈V , and setting

(Q⊗b)(z) := ∑i

Q(ei)⊗b(vi,vi)+∑i< j

Q(ei,e j)⊗b(vi,v j),

we obtain a quadratic map which, since b is symmetric, has the desired properties.Now let M be arbitrary. Then there exists a short exact sequence

0 // L i // F π // M // 0

of k-modules, where F is free. Since the functor −′ := −⊗V is right exact [12, II,§ 3, Prop. 5], we obtain an induced exact sequence

L′ i′ // F ′ π ′ // M′ // 0,

and the special case just treated yields a quadratic map Q := (Q π)⊗ b : F ′ →N⊗W satisfying (1), (2). Let z = ∑u j⊗ v j ∈ L′, u j ∈ L, v j ∈ V , and x ∈ F , v ∈ V .

62 2 Foundations

Then

Q(i′(z)

)=((Qπ)⊗b

)(i′(∑u j⊗ v j)

)=((Qπ)⊗b

)(∑ i(u j)⊗ v j)

= ∑(Qπ i)(u j)⊗b(v j,v j)+∑j<l

(Qπ i)(u j,ul)⊗b(u j,ul) = 0,

Q(i′(z),x⊗ v

)=((Qπ)⊗b

)(∑ i(u j)⊗ v j,x⊗ v)

= ∑(Qπ i)(u j,x)⊗b(v j,v) = 0,

hence i′(z) ∈ Rad(Q). We have thus shown Ker(π ′) = Im(i′) ⊆ Rad(Q), and 12.3yields a quadratic map Q⊗b : M′→ N⊗W such that (Qπ)⊗b = (Q⊗b)π ′ andD((Qπ)⊗b) = [D(Q⊗b)] (π ′×π ′). Hence (1), (2) hold for Q⊗b.

c.BAQUA12.5. Corollary. Let Q : M→N be a k-quadratic map of k-modules and R∈ k-alg.Then there exists a unique R-quadratic map QR : MR→ NR of R-modules making acommutative diagram

MQ //

can

N

can

MR QR

// NR.

(1) BAQUA

In particular, D(QR) = (DQ)R is the R-bilinear extension of DQ. We call QR thescalar extension or base change of Q from k to R.

Proof. Applying Prop. 12.4 to Q and the symmetric bilinear map b : R×R→ Rgiven by the multiplication of R, we obtain a k-quadratic map QR : MR→NR, whichis in fact a quadratic map over R satisfying the condition of the corollary.

We now proceed to list a few elementary properties of bilinear and quadratic forms.ss.NOBIQ

12.6. Notation. Let M be a k-module. Then we write Bil(M) for the k-moduleof bilinear forms on M. The submodules of Bil(M) consisting of all symmetric,skew-symmetric, alternating bilinear forms on M will be denoted respectively bySbil(M), Skil(M), Abil(M). Recall that a bilinear form σ : M×M→ k is alternatingif σ(x,x) = 0 for all x ∈M. We then have Abil(M)⊆ Skil(M) with equality if 1

2 ∈ k.Finally, we write Quadk(M) := Quad(M) for the k-module of quadratic forms on M.The assignment q 7→ Dq defines a linear map from Quad(M) to Sbil(M).

ss.MAFO12.7. Connecting matrices with bilinear and quadratic forms. Let n be a posi-tive integer. For S ∈Matn(k), the map

〈S〉 : kn× kn −→ k, (x,y) 7−→ 〈S〉(x,y) := xtSy (1) BILS

is a bilinear form on kn, and the assignment S 7→ 〈S〉 gives a linear bijectionfrom Matn(k) onto Bil(kn) matching Symn(k), Skewn(k), Altn(k) respectively withSbil(kn), Skil(kn), Abil(kn). If S = diag(α1, . . . ,αn) ∈Matn(k) is a diagonal matrix,


we put 〈α1, . . . ,αn〉 := 〈S〉 as a symmetric bilinear form on kn; it sends

(ξ1...

ξn

,

η1...

ηn

) ∈ kn× kn ton

∑i=1

αiξiηi.

On the other hand, the map

〈S〉quad : kn −→ k, x 7−→ 〈S〉quad(x) := xtSx (2) QUAS

is a quadratic form on kn such that D〈S〉quad = 〈S+ St〉, and the assignment S 7→〈S〉quad gives rise to a short exact sequence

0 // Altn(k) // Matn(k) // Quad(kn) // 0 (3) MAQUA

of k-modules. Thus, roughly speaking, quadratic forms on kn are basically the sameas arbitrary n×n-matrices modulo alternating n×n-matrices over k. Again, If S =diag(α1, . . . ,αn) ∈Matn(k) is a diagonal matrix, we put 〈α1, . . . ,αn〉quad := 〈S〉quadas a quadratic form form on kn; it sendsξ1

...ξn

∈ kn ton

∑i=1

αiξ2i .

ss.FOFR12.8. Bilinear and quadratic forms on free modules. Let M be a free k-moduleof finite rank n > 0 and (ei)1≤i≤n a basis of M over k. Given a bilinear form σ : M×M→ k, we call S := (σ(ei,e j))1≤i, j≤n ∈Matn(k) the matrix of σ with respect to thebasis (ei), and identifying M = kn by means of this basis matches σ with the bilinearform〈S〉 on kn (12.7). On the other hand, given a quadratic form q : M→ k, we finea matrix S = (si j) ∈Matn(k) by

si j :=

q(ei,e j) (1≤ i < j ≤ n),q(ei) (1≤ i = j ≤ n),0 (1≤ j < i≤ n).

(1) QUAMA

We call S the matrix of q with respect to the basis (ei) and, again, identifying M = kn

by means of this basis matches q with the quadratic form 〈S〉quad on kn.

ss.REBI12.9. Regularity conditions on bilinear forms. Let σ : M×M→ k be a symmet-ric or skew-symmetric bilinear form on a k-module M. For a submodule N ⊆M, thesubmodule

N⊥ := N⊥σ := x ∈M | σ(x,y) = 0 for all y ∈ N ⊆M (1) ORCO

64 2 Foundations

is called the orthogonal complement of N in M. In particular, M⊥ = Rad(σ) (12.3)is the kernel of the natural map

σ : M −→M∗, x 7−→ σ(x,−). (2) TIBI

σ is said to be non-singular if

(i) M is finitely generated projective as a k-module,(ii) σ : M ∼→M∗ is an isomorphism.

This implies Rad(σ) = 0 but not conversely. Since passing to the dual of a finitelygenerated projective module is compatible with base change (Lemma 10.11), so isthe property of a (skew-)symmetric bilinear form to be non-singular. If M = kn

is free of finite rank n and S ∈ Matn(k) is (skew-)symmetric, then the (skew-)symmetric bilinear form〈S〉 : kn×kn→ k is non-singular if and only if S ∈GLn(k).

l.NOSU12.10. Lemma. Let σ : M×M → k be a symmetric or skew-symmetric bilinearform over k and suppose N ⊆M is a submodule on which σ is non-singular. ThenM = N⊕N⊥.

Proof. We write σ ′ for the restriction of σ to N×N. Since N∩N⊥=Rad(σ ′)= 0,it remains to show M = N+N⊥, so let x∈M. Then σ(x,−) restricts to a linear formon N, which by non-singularity of σ ′ has the form σ ′(u,−) for some u ∈ N. Thusσ(x,y) = σ(u,y) for all y ∈ N, and we conclude x = u+ v, v := x−u ∈ N⊥.

ss.REQUA12.11. Regularity conditions on quadratic forms. A quadratic form q : M→ kis said to be non-degenerate if Rad(q) = 0. Unfortunately, this property is notinvariant under base change, even if we allow M to be finitely generated projective.Following Loos [70, 3.2] in a slightly more general vein, we therefore call q sep-arable if M is projective, possibly of infinite rank, and the scalar extension qK isnon-degenerate for all fields K ∈ k-alg. Indeed, the property of a quadratic form tobe separable is clearly stable under base change. As a simple example, we note thatthe one-dimensional quadratic form 〈α〉quad, α ∈ k×, is always separable but be-comes degenerate if 2 = 0 in k and k is not reduced, i.e., contains non-zero nilpotentelements.

A regularity condition even stronger than separability is provided by the conceptof non-singularity: q is said to be non-singular if its induced symmetric bilinearform Dq has this property. In this case, M is finitely generated projective by 12.9 (ii),and if 2 = 0 in k, then Dq is an alternating non-singular bilinear form, forcingrkp(M) to be even for all p ∈ Spec(k). For a finitely generated free module M = kn

over any commutative ring k and S ∈Matn(k), the quadratic form 〈S〉quad : kn→ kis non-singular if and only if S+St ∈ GLn(k).

Finally, q is said to be weakly non-singular if for all u∈M, the relation q(u,v)= 0for all v ∈ M implies u = 0. This is equivalent to Dq : M → M∗ being injective.Note that weak non-singularity is not stable under base change, making this notiondistinctly less interesting than the previous one.


ss.QUAMO12.12. Quadratic modules and spaces. It is sometimes linguistically convenientto think of a quadratic form q : M→ k as a quadratic module (M,q). Given quadraticmodules (M,q) and (M′,q′) over k, a homomorphism h : (M,q)→ (M′,q′) is a k-linear map h : M→ M′ satisfying q′ h = q. In this way one obtains the categoryof quadratic modules over k. Isomorphisms between quadratic modules over k arecalled isometries. A quadratic module (M,q) over k is said to be a quadratic space ifq is non-singular. We write (M,q)⊥ (M′,q′) for the orthogonal sum of the quadraticmodules (M,q), (M′,q′) defined on the direct sum M⊕M′ by the quadratic form

q⊥ q′ : M⊕M′ −→ k, x⊕ x′ 7−→ q(x)+q′(x′).

We sometimes use the alternate notation (M,q)⊕ (M′,q′) := (M,q) ⊥ (M′,q′) andq⊕q′ := q⊥ q′.

ss.POQUAM12.13. Pointed quadratic modules. By a pointed quadratic module over k wemean a triple (M,q,e) consisting of a quadratic module (M,q) and a distinguishedelement e∈M, called the base point, such that q(e)= 1. Homomorphisms of pointedquadratic modules are defined as homomorphism of the underlying quadratic mod-ules preserving base points in the obvious sense. We speak of (M,q,e) as a pointedquadratic space if (M,q) is a quadratic space.

Let (M,q,e) be a pointed quadratic module over k. We call q its norm, the linearform t := (Dq)(e,−) on M its trace and the linear map

ι : M −→M, x 7−→ x := t(x)e− x, (1) COPOQUA

its conjugation. Obviously, the trace satisfies t(e) = 2, while the conjugation hasperiod 2 and preserves base point, norm and trace.

The following lemma, which is obvious but quite useful, has been communicatedto me by O. Loos and E. Neher.

l.UNBA12.14. Lemma. Let (M,q,e) be a pointed quadratic module over k and supposethere exists a bilinear form β : M×M→ k, possibly not symmetric, such that q(x) =β (x,x) for all x ∈ M (which holds automatically if M is projective as a k-module(Exc. 50)), then the base point e ∈M is unimodular.

Proof. The linear form λ := β (e,−) on M has λ (e) = β (e,e) = q(e) = 1.

ss.NADI12.15. The naive discriminant. The multiplicative group k×2 acts on the additivegroup k by multiplication form the right, allowing us to form the quotient k/k×2.Now suppose we are given a quadratic module (M,q) over k such that the underlyingk-module M is free of finite rank. In analogy to 3.10–3.12 we let E = (e1, . . . ,en) beany k-basis of M, put

Dq(E) :=(q(ei,e j)

)1≤i, j≤n ∈Matn(k) (1) NAMABI

and have

66 2 Foundations

Dq(ES) = StDq(E)S(S = (si j)1≤i, j≤n ∈ GLn(k)

), (2) NATRAF

where ES = (e′1, . . . ,e′n) is the k-basis of M defined by e′j := ∑i si jei for 1 ≤ j ≤ n.

Hence the expression

disc((M,q)

):= (−1)[

n2 ] det

((M,q)

)mod k/k×2

is independent of the basis chosen and called the discriminant of (M,q). SinceZ×2 = 1, this definition generalizes the one of the discriminant of an integralquadratic lattice as given in 3.12. For a more sophisticated definition of the discrim-inant that makes sens for arbitrary quadratic spaces, we refer the reader to Knus [58,Chap. IV, §4].

ss.ISVE12.16. Isotropic vectors. Let (M,q) be a quadratic space over k. An element u∈Mis said to be isotropic (relative to q or in (M,q)) if u is unimodular and satisfies therelation q(u) = 0. As in the case of quadratic forms over fields, isotropic vectors mayalways be completed to hyperbolic pairs: given u ∈M isotropic, there exists v ∈Msuch that (u,v) is a hyperbolic pair, i.e., q(u) = q(v) = 0, q(u,v) = 1. Indeed, since qis non-singular, unimodularity of u yields an element w ∈M satisfying q(u,w) = 1,and one checks that (u,v), v :=−q(w)u+w, is a hyperbolic pair.

The quadratic space (M,q) (or q) will be called isotropic if M contains anisotropic vector relative to q. A submodule N ⊆ M is said to be totally isotropic(relative to q) if it is a direct summand of M and satisfies q(N) = 0.

ss.HYSP12.17. Hyperbolic spaces. Let M be a finitely generated projective k-module. Then

hM : M∗⊕M −→ k, v∗⊕ v 7−→ hM(v∗⊕ v) := 〈v∗,v〉, (1) HYPA

is a non-singular quadratic form since M∗∗ identifies canonically with M and(DhM)∼ : M∗⊕M→M⊕M∗ (cf. (12.9.2)) is the switch v∗⊕ v 7→ v⊕ v∗. We alsowrite

hM := (M∗⊕M,hM)

as a quadratic space and call this the hyperbolic space associated with M or simply ahyperbolic space. Observe that both M∗ and M may be viewed canonically as totallyisotropic submodules of M∗ ⊕M relative to hM . The functor assigning to M itsassociated hyperbolic space is additive, so for another finitely generated projectivek-module N we obtain a natural isomorphism

hM⊕N ∼= hM⊕hN .

If M ∼= kn is free of rank n ∈N, we speak of the split hyperbolic space of rank n. ForL a line bundle in the sense of 10.8, hL will be referred to as a hyperbolic plane.,which is said to be split if L is free of rank 1.

A quadratic space is said to be hyperbolic (resp. a (split) hyperbolic plane) if itis isometric to some hyperbolic space (resp. (the split) hyperbolic plane).


Exercises.

pr.SPLHYPLA 47. A splitness criterion for hyperbolic planes. Let L be a line bundle over k. Show that the follow-ing conditions are equivalent.

(i) The hyperbolic plane hL is split.(ii) L∗⊕L contains a hyperbolic pair relative to hL.(iii) L∼= k is free.

(Hint. To prove (ii)⇒ (iii) , find a unimodular vector in L.)

pr.HYPPA 48. Let h be the split hyperbolic plane over k and (e1,e2) a hyperbolic pair in h. Show for u1,u2 ∈ hthat the following conditions are equivalent.

(i) (u1,u2) is a hyperbolic pair in h.(ii) There exists a decomposition k = k+⊕ k− of k as a direct sum of ideals such that in the

induced decompositions

h = h+⊕h−, e j = e j+⊕ e j−, u j = u j+⊕u j− ( j = 1,2), (2) HYPM

where h± = hk± is the split hyperbolic plane over k± with the corresponding hyperbolic pair(e1±,e2±) = (e1,e2)k± , the quantities u j± = (u j)k± ∈ h± ( j = 1,2) satisfy the relations

u1+ = γ+e1+, u2+ = γ−1+ e2+, (3) GAHY

u1− = γ−e2−, u2− = γ−1− e1− (4) GOPHY

for some γ± ∈ k×± .

pr.MULQUA 49. Multi-quadratic maps. For a positive integer n and k-modules M1, . . . ,Mn,M, a map

F : M1×·· ·×Mn −→M

is called k-multi-quadratic or k-n-quadratic if for all i = 1, . . . ,n and for all u j ∈ M j , 1 ≤ j ≤ n,j 6= i,

Mi −→M, ui 7−→ F(u1, . . . ,ui−1,ui,ui+1, . . . ,un)

is a quadratic map over k. Show for R ∈ k-alg and a k-n-quadratic map F : M1×·· ·×Mn→M thatthere exists a unique R-n-quadratic map FR : M1R×·· ·×MnR→MR rendering the diagram

M1×·· ·×MnF //

can

M

can

M1R×·· ·×MnR FR

// MR.

(5) MULQUA

commutative, where can: M1×·· ·×Mn→M1R×·· ·×MnR is defined by

can(u1, . . . ,un) := (u1R, . . . ,unR)

for all ui ∈ Mi, 1 ≤ i ≤ n. Writing Quad(M1, . . . ,Mn;M) for the k-module of k-n-quadratic mapsfrom M1× ·· ·×Mn to M, show further that the assignment F 7→ FR defines a k-linear map fromQuad(M1, . . . ,Mn;M) to Quad(M1R, . . . ,MnR;MR). We call FR the R-n-quadratic extension of F .

pr.QUABIL 50. Let (M,q) be a quadratic module over k and suppose M is projective. Show that there exists abilinear form β : M×M→ k, possibly not symmetric, such that q(x) = β (x,x) for all x ∈M.

pr.SEQUAFI 51. Separable quadratic forms over fields. Let q : V → F be a quadratic form over a field F andK/F an algebraically closed extension field. Show that the following conditions are equivalent.

68 2 Foundations

(i) q is separable.(ii) q is non-degenerate and the radical of Dq has dimension at most 1 over F .(iii) The extended quadratic form qK : VK → K over K is non-degenerate.

pr.LIQUA 52. Let (V,q) be a quadratic space of dimension n over the field F .

(a) If n ≥ 3, show that every element in V can be decomposed into the sum of at most twoanisotropic elements in V .

(b) Assume that (V,q) has Witt index at least 3. Show for anisotropic vectors x1,x2 ∈ V thatq(x1,x2) 6= 0 or there exists an anisotropic vector y ∈V satisfying q(x1,y) 6= 0 6= q(y,x2).

Can the extra conditions on (V,q) in (a), (b) be avoided?

13. Polynomial laws

s.POLAThe differential calculus for polynomial (or rational) maps between finite-dimensionalvector spaces, as explained in Braun-Koecher [15] or Jacobson [43], for exam-ple, belongs to the most useful techniques from the toolbox of elementary non-associative algebra. It elevates the linearization procedures that we have encounteredso far, and that will become much more involved in subsequent portions of the book(see, e.g., the identities in 31.3 below), to a new level of systematic conciseness.While it works most smoothly over infinite base fields, it can be slightly adjusted towork over finite ones as well. There is also a variant due to McCrimmon [76] thatallows infinite-dimensional vector spaces.

Extending this formalism to arbitrary modules over arbitrary commutative rings,however, requires a radically new approach. The one adopted in the present volume,due to Roby [112], is based on the concept of a polynomial law. In the presentsection, we explore this concept in detail and show, in particular, how the differentialcalculus for polynomial maps over infinite fields can be naturally extended to thismore general setting. For a different approach, see Faulkner [29].

Throughout, we let k be an arbitrary commutative ring and M,N,M1, . . . ,Mn (n∈Z, n > 0) be arbitrary k-modules.

ss.POMA13.1. Reminder: polynomial maps. For the time being, let V,W be finite-dimensi-onal vector spaces over an infinite field F and (vi)1≤i≤m, (w j)1≤ j≤n be bases ofV,W , respectively, over F . By a polynomial map from V to W we mean a set mapf : V →W such that there exist polynomials p1, . . . , pn ∈ F [t1, . . . , tm] satisfying theequations

f (m

∑i=1

αivi) =n

∑j=1

p j(α1, . . . ,αm)w j (α1, . . . ,αm ∈ F). (1) DEPOMA

Since F is infinite, (1) determines the polynomials p j, 1 ≤ j ≤ n, uniquely. More-over, the concept of a polynomial map is obviously independent of the bases chosen.In the special case W = F , we speak of a polynomial function on V . The totality ofall polynomial functions on V forms a unital commutative associative F-algebra,

13 Polynomial laws 69

denoted by F [V ] and canonically isomorphic to the polynomial ring F [t1, . . . , tm]over F .

Returning to our polynomial map f : V →W as above, we generalize (1) bydefining a family of set maps fR : VR→WR, one for each R ∈ F-alg, given by

fR(m

∑i=1

riviR) :=n

∑j=1

p j(r1, . . . ,rm)w jR (r1, . . . ,rm ∈ R). (2) BAPOMA

The key property of this family may be expressed by a coherence condition, sayingthat its constituents fR vary functorially with R ∈ k-alg: every morphism ϕ : R→ Sin k-alg yields a commutative diagram

VRfR //

1V⊗ϕ

WR

1W⊗ϕ

VS fS

// WS.

(3) FUPOMA

as shown. This coherence condition will now be isolated in the formal definition ofa polynomial law.

ss.COPOLA13.2. The concept of a polynomial law. With the k-module M we associate a(covariant) functor Ma : k-alg → set (where set stands for the category of sets)by setting Ma(R) = MR as sets for R ∈ k-alg and Ma(ϕ) = 1M ⊗ϕ : MR → MS asset maps for morphisms ϕ : R→ S in k-alg. Analogously, we obtain the functorNa : k-alg→ set associated with the k-module N. We then define a polynomial lawf from M to N (over k) as a natural transformation f : Ma→ Na. In explicit terms,this means that, for all R ∈ k-alg, we are given set maps fR : MR→ NR varying func-torially with R, so whenever ϕ : R→ S is a unital homomorphism of k-algebras, thediagram

MRfR //

1M⊗ϕ

NR

1N⊗ϕ

MS fS

// NS.

(1) FUPOLA

commutes. A polynomial law from M to N will be symbolized by f : M → N, inspite of the fact that it is not a map from M to N in the usual sense. But it induces one,namely fk : M→ N, which, however, does not determine f uniquely. If ϕ : R→ Sis a morphism in k-alg, then S belongs to R-alg via ϕ , and identifying (MR)S = MS,(NR)S = NS canonically by means of (10.3.1), we conclude from (10.3.2) that (1) isequivalent to

fR(x)S = fS(xS) (x ∈MR). (2) FUBA

70 2 Foundations

The totality of polynomial laws from M to N will be denoted by Pol(M,N), or byPolk(M,N) to indicate dependence on k. It is a k-module in a natural way, the sum off ,g ∈ Pol(M,N) being given by ( f +g)R = fR+gR for all R ∈ k-alg, ditto for scalarmultiplication. The multiplication of polynomial laws M→N by scalar polynomiallaws M→ k is defined analogously. In particular, Pol(M,k) is a unital commutativeassociative k-algebra. If f : M→ N and g : N → P are polynomial laws over k, soobviously is g f : M→ P, given by (g f )R = gR fR,R ∈ k-alg. Every polynomiallaw f : M → N over k gives rise to its scalar extension or base change f ⊗ R :MR→ NR, a polynomial law over R determined by the condition ( f ⊗R)S := fS forall S ∈ R-alg ⊆ k-alg, where (MR)S = MS, NR.S = NS are canonically identified asbefore. We mostly write fR, or simply f , for f ⊗R if there is no danger of confusion.

ss.COMUIN13.3. Convention: multi-indices. For R ∈ k-alg it is often convenient to usethe multi-index notation rν := rν1

1 · · ·rνnn , where r = (r1, . . . ,rn) ∈ Rn and ν =

(ν1, . . . ,νn) ∈ Nn are sequences of length n ≥ 1 in R,N, respectively. Also, we put|ν | := ∑

ni=1 νi.

For example, given a finite chain T = (t1, . . . , tn) of indeterminates, the elementsof the polynomial algebra k[T] = k[t1, . . . , tn] may be written as ∑ν∈Nn αν Tν , where(αν)ν∈Nn is a family of finite support in k. Moreover, observing the identificationsof 31.2, we obtain

Mk[T] =⊕

ν∈Nn

(Tν M) (1) EMKATE

as a direct sum of k-modules. For a polynomial law f : M→ N over k, these identi-fications (replacing k by R and f by f ⊗R) yield

fR[T](x) = fR(x) (R ∈ k-alg, x ∈MR) (2) EFERTE

since fR[T](x) = ( f ⊗R)R[T](xR[T]) = ( f ⊗R)R(x)R[T] = fR(x) by 13.2, particularlyby (13.2.2). Similarly, given a positive integer q and writing k[ε] for the free k-algebra on a single generator ε , subject to the relation εq = 0 (so k[ε] = k[t1]/(tq

1),ε = t1 +(tq

1)), we have

Mk[ε] =q−1⊕j=0

(ε jM.) (3) EMKAEP

ss.STANID13.4. A standard identification. For all R ∈ k-alg, we will systematically adoptthe canonical identification

(M1×·· ·×Mn)R = M1R×·· ·×MnR

as R-modules such that


(v1⊕·· ·⊕ vn)⊗ r = (v1⊗ r)⊕·· ·⊕ (vn⊗ r), (1) VEONVE

(v1⊗ r1)⊕·· ·(vn⊗ rn) =n

∑i=1

(0⊕·· ·⊕0⊕ vi⊕0⊕·· ·⊕0)⊗ ri

for all vi ∈M, ri ∈ R, 1≤ i≤ n. Given a morphism ϕ : R→ S in k-alg, this identifi-cation implies

(1M1 ⊗ϕ)×·· ·× (1Mn ⊗ϕ) = 1M1×···×Mn ⊗ϕ. (2) ONEMON

ss.HOPOLA13.5. Homogeneous polynomial laws. A polynomial law f : M → N is said tobe homogeneous of degree d ∈ N if fR(rx) = rd fR(x) for all R ∈ k-alg, r ∈ R, x ∈MR. More generally, a polynomial law f : M1× ·· ·×Mn → N is said to be multi-homogeneous of multi-degree d = (d1, . . . ,dn) ∈ Nn if

fR(r1x1, . . . ,rnxn) = rd11 · · ·r

dnn fR(x1, . . . ,xn)

for all R ∈ k-alg, ri ∈ R, xi ∈ MiR, i = 1, . . . ,n. Here and in the sequel, we alwaysidentify (M1 × ·· · ×Mn)R = M1R × ·· · ×MnR canonically by means of (13.4.1).Thanks to Exc. 56, multi-homogeneous (resp. homogeneous) polynomial laws ofmulti-degree 1 = (1, . . . ,1) (resp. of degree 2) identify canonically with multi-linear(resp. quadratic) maps in the usual sense (resp. in the sense of 12.1). Notice alsothat a multi-homogeneous polynomial law of multi-degree d ∈ Nn is homogeneousof degree |d|. Scalar homogeneous polynomial laws are called forms. We speak oflinear, quadratic, cubic, quartic, ... forms instead of forms of degree d = 1,2,3,4, . . . .

ss.POFIN13.6. Local finiteness. A family ( fi)i∈I of polynomial laws fi : M→ N (i ∈ I) issaid to be locally finite if, for all R ∈ k-alg and all x ∈MR, the family ( fiR(x))i∈I ofelements in NR has finite support. In this case, we obtain a well defined polynomiallaw

∑i∈I

fi : M −→ N

over k by setting

(∑i∈I

fi)R(x) := ∑i∈I

fiR(x) (R ∈ k-alg, x ∈MR).

Note that this definition mutatis mutandis makes sense also if each fi, i∈ I, is merelyassumed to be a family of set maps fiR : MR → NR, R ∈ k-alg, which may or maynot vary functorially with R.

t.BILEXP13.7. Theorem. Let f : M→N be a polynomial law over k and n a positive integer.Then there exists a unique locally finite family of polynomial laws

Πν f : Mn −→ N (ν ∈ Nn)

such that

72 2 Foundations

fR(n

∑i=1

rixi) = ∑ν∈Nn

rν(Π ν f )R(x) (1) EFERSU

for all R ∈ k-alg, r = (r1, . . . ,rn) ∈ Rn, x = (x1, . . . ,xn) ∈ (MR)n. In particular,

fR[T](n

∑i=1

tixi) = ∑ν∈Nn

Tν(Π ν f )R(x), (2) EFERBT

and this condition alone determines the family (Π ν f )ν∈Nn uniquely.

Proof. We begin by showing uniqueness, so suppose (Π ν f )ν∈Nn is a family of thedesired kind. Replacing R by R[T] and ri by ti but keeping xi ∈MR for 1≤ i≤ n in(1), the identification (13.3.1) implies

fR[T](n

∑i=1

tixi) = ∑ν∈Nn

Tν(Π ν f )R[T](x),

where the right-hand side by (13.3.2) agrees with the one of (2). Hence (2) holds, andin view of (13.3.1) again, (2) determines the quantities (Π ν f )R(x) ∈ NR uniquely.Thus the family (Π ν f )ν∈Nn is unique. In order to establish its existence, we let R ∈k-alg and x = (x1, . . . ,xn) ∈Mn

R. Then fR[T](∑ni=1 tixi) ∈ NR[T], and (13.3.1) yields a

unique family ((Π ν f )R(x))ν∈Nn of elements in NR having finite support such that(2) holds. We have thus obtained a locally finite family

(Π ν f )ν∈Nn =((

(Π ν f )R)

R∈k-alg

)ν∈Nn

of set maps (Π ν f )R : MnR→ NR, ν ∈ Nn, R ∈ k-alg, satisfying (2).

Next we show (1), so let R ∈ k-alg, r = (r1, . . . ,rn) ∈ Rn, x = (x1, . . . ,xn) ∈MnR

and ϕ : R[T]→ R the R-algebra homomorphism having ti 7→ ri for 1≤ i≤ n. Usingthe identification (10.3.1), we conclude (1M ⊗ϕ)(Tν y) = rν y for ν ∈ Nn, y ∈Mn

R,ditto form N in place of M. Hence (13.2.1) and (2) imply

fR(n

∑i=1

rixi) = fR (1M⊗ϕ)(n

∑i=1

tixi) = (1N⊗ϕ) fR[T](n

∑i=1

tixi)

= ∑ν∈Nn

(1N⊗ϕ)(Tν(Π ν f )R(x)

)= ∑

ν∈Nnrν(Π ν f )R(x),

which completes the proof of (1). It remains to show that the (Π ν f )R, ν ∈ Nn

vary functorially with R ∈ k-alg, so let ϕ : R → S be any morphism in k-algand ψ : R[T] → S[T] its natural extension fixing ti for 1 ≤ i ≤ n. One checks(1M ⊗ψ)(Tν y) = Tν(1M ⊗ϕ)(y) for ν ∈ Nn, y ∈ MR, ditto for N in place of M,and (13.4.2) yields (1M⊗ϕ)n = 1Mn ⊗ϕ . Hence


∑ν∈Nn

Tν(1N⊗ϕ) (Π ν f )R(x) = (1N⊗ψ)(

∑ν∈Nn

Tν(Π ν f )R(x))

= (1N⊗ψ) fR[T](n

∑i=1

tixi)

= fR[T] (1M⊗ψ)(n

∑i=1

tixi)

= fR[T]( n

∑i=1

ti(1M⊗ϕ)(xi))

= ∑ν∈Nn

Tν(Π ν f )R((1M⊗ϕ)(x1), . . . ,(1M⊗ϕ)(xn)

)= ∑

ν∈NnTν((Π ν f )R (1Mn ⊗ϕ)

)(x),

and comparing coefficients, the assertion follows.

ss.LIPOLA13.8. Linearizations. The polynomial laws Π ν f for ν ∈ Nn, n ∈ Z, n > 0 as con-structed in Thm. 13.7 are called the linearizations or polarizations of the polynomiallaw f : M→ N over k. We list a few elementary properties.(a) For ν ∈ Nn, the polynomial law Π ν f : Mn→ N is multi-homogeneous of multi-degree ν . Indeed, replacing xi by rixi (ri ∈ R, 1 ≤ i ≤ n) on the left-hand side of(13.7.2) amounts to the same as carrying out the substitution ti 7→ riti (1 ≤ i ≤ n).Hence the assertion follows from (13.7.1), (13.3.2).(b) Writing π.ν := (νπ−1(1), . . . ,νπ−1(n)) for ν = (ν1, . . . ,νn) ∈ Nn and π ∈ Sn, wehave

(Π ν f )R(xπ(1), . . . ,xπ(n)) = (Π π.ν f )R(x1, . . . ,xn) (1) PERLIN

for all R ∈ k-alg, x1, . . . ,xn ∈MR, which follows immediately from the fact that theleft-hand side of (13.7.2) remains unaffected by any permutation of the summands.(c) If f is homogeneous of degree d ∈N, then Π ν f = 0 for all ν ∈Nn, n ∈ Z, n > 0,unless |ν | = d. In order to see this, let s be a new variable, replace ti by sti for1≤ i≤ n in (13.7.2) and compare coefficients of s|ν |.

In particular, assume d > 0, let n = d and ν := 1 := (1, . . . ,1) ∈ Nd . Then (a)shows that Π 1 f : Md→N is multi-homogeneous of multi-degree 1, i.e., it is a multi-linear map (Exc. 56 (a)), while we conclude from (b) that it is totally symmetric.Putting x1 = · · ·= xd =: y ∈MR in (13.7.2) and comparing coefficients of t1 · · · td in

( d

∑i1,...,in=1

ti1 · · · tin)

fR(y) = (d

∑i=1

ti)d fR(y) = ∑

ν∈Nd

Tν(Π ν f )R(y, . . . ,y),

we conclude

(Π 1 f )R(y, . . . ,y) = d! fR(y) (2) TOTLIN

74 2 Foundations

for all R ∈ k-alg, y ∈MR. We call Π 1 f the total linearization of f .c.VANCRI

13.9. Corollary. A polynomial law over k is zero if and only if all its linearizationsvanish identically over k: for all polynomial laws f : M→ N over k, we have

f = 0 ⇐⇒ ∀n ∈ Z, n > 0, ∀ν ∈ Nn : (Π ν f )k = 0.

Proof. The implication from left to right is obvious. Conversely, suppose (Π ν f )k =0 for all n ∈ Z, n > 0, ν ∈ Nn. We must show fR = 0 for all R ∈ k-alg. Everyx ∈MR can be written as x = ∑

ni=1 riviR for some n ∈ Z, n > 0, r = (r1, . . . ,rn) ∈ Rn,

v = (v1, . . . ,vn) ∈Mn. From (13.7.1) and (13.2.2) we therefore deduce

fR(x) = ∑ν∈Nn

rν(Π ν f )R(vR) = ∑ν∈Nn

rν(Π ν f )k(v)R = 0,

as desired. c.FREPOLA

13.10. Corollary. Assume the k-modules M and N are free of finite rank, with bases(v1, . . . ,vm) and (w1, . . . ,wn), respectively. A family of set maps fR : MR→ NR, R ∈k-alg, is a polynomial law over k if and only if there exist polynomials p1, . . . , pn ∈k[t1, . . . , tm] such that

fR(m

∑i=1

riviR) =n

∑j=1

p j(r1, . . . ,rm)w jR (1) FREPOLA

for all R ∈ k-alg, r1, . . . ,rm ∈ R.

Proof. If such polynomials exist, it is readily checked that the fR vary functoriallywith R ∈ k-alg, hence give rise to a polynomial law f : M → N. Conversely, letf : M→ N be a polynomial law over k. For R ∈ k-alg and r = (r1, . . . ,rm) ∈ Rm, weput v := (v1, . . . ,vm) and apply (13.7.1) to conclude

fR(m

∑i=1

riviR) = ∑ν∈Nm

rν(Π ν f )R(vR) = ∑ν∈Nm

rν(Π ν f )k(v)R.

Here (Π ν f )k(v) ∈ N, ν ∈ Nm, may be written as

(Π ν f )k(v) =n

∑j=1

β jν w j

with unique coefficients β jν ∈ k, 1≤ j ≤ n. Since the family (Π ν f )ν∈Nm is locallyfinite, we can form the polynomials

p j := ∑ν∈Nm

β jν Tν ∈ k[T] (T := (t1, . . . , tm), 1≤ j ≤ n)

which obviously satisfy (1).


c.POMALA13.11. Corollary. Let V,W be finite-dimensional vector spaces over an infinite fieldF. Then the assignment f 7→ fF defines a linear bijection from PolF(V,W ) onto thevector space of polynomial maps from V to W. For bases (v1, . . . ,vm), (w1, . . . ,wn)of V,W, respectively, over F, the inverse of this bijection assigns to every polynomialmap f : V →W the polynomial law given by the family of set maps fR : VR→WR,R ∈ k-alg, as defined in (13.1.2).

Proof. This follows from Cor. 13.10 combined with 13.1.

ss.BILI13.12. Binary linearizations. It is sometimes useful to rewrite the formalism ofThm. 13.7 for n = 2, so let f : M→ N be any polynomial law over k. With indepen-dent variables s, t, we then apply (13.7.2) and obtain

fR[s,t](sx+ ty) = ∑m≥0,n≥0

smtn(Π (m,n) f )R(x,y) (1) BILI

for all R ∈ k-alg, x,y ∈MR, and (13.8.1) yields

(Π (m,n) f )R(x,y) = (Π (n,m) f )R(y,x). (2) BIPERLI

If f is homogeneous of degree d ∈ N, we obtain Π (m,n) f = 0 for m,n ≥ 0 unlessm+n = d (13.8 (c)), so (1) collapses to

fR[s,t](sx+ ty) =d

∑n=0

sd−ntn(Π (d−n,n) f )R(x,y)

for all R ∈ k-alg, x,y ∈MR.ss.TODE

13.13. Total derivatives. Let f : M → N be any polynomial law over k. ByThm. 13.7, the family (Π (m,n) f )m,n≥0 of polynomial laws M×M → N is locallyfinite, allowing us to define, for all n ∈ N, a polynomial law

Dn f := ∑m≥0

Π(m,n) f : M×M −→ N, (1) ENDER

called the n-th (total) derivative of f . By 13.8 (a), it is homogeneous of degree n inthe second variable, i.e.,

(Dn f )R(x,ry) = rn(Dn f )R(x,y). (R ∈ k-alg, r ∈ R, x,y ∈MR)

In particular, Exc. 53 and (13.7.1) imply

(D0 f )R(x,y) = ∑m≥0

(Π (m,0) f )R(x,y) = ∑m≥0

(Π m f )R(x) = fR(x), (2) ZERDER

hence D0 f = f as polynomial laws over k. Moreover, abbreviating D f := D1 f ,

76 2 Foundations

(D f )R(x) : MR −→ NR,y 7−→ (D f )R(x)(y) := (D f )R(x,y),

by Exc. 56 is an R-linear map for all x ∈MR,R ∈ k-alg. Similarly,

(D2 f )R(x) : MR −→ NR, y 7−→ (D2 fR)(x,y),

is an R-quadratic map in the usual sense. Specializing s 7→ 1 in (13.12.1), we obtainthe relation

fR[t](x+ ty) = ∑n≥0

tn(Dn f )R(x,y), (3) TAYL

called the Taylor expansion of f at x. For q ∈ Z, q > 0, it is sometimes convenientto replace R[t] by the algebra R[ε],εq+1 = 0, and to apply (13.7.1). The ensuingrelation

fR[ε](x+ εy) =q

∑n=0

εn(Dn f )R(x,y) (4) TAYLEPS

determines the first q derivatives of f uniquely. In particular, for q = 1, R[ε] is thealgebra of dual numbers and

fR[ε](x+ εy) = fR(x)+ ε(D f )R(x,y). (5) TAYLDUAL

ss.TODEHO13.14. Total derivatives of homogeneous polynomial laws Let f : M→ N be ahomogeneous polynomial law of degree d ≥ 0 over k. By 13.8 (c), the terms Π (m,n) fin the sum on the right of (13.13.1) vanish unless m+n = d. Thus Dn f = 0 for n > dand

Dn f = Π(d−n,n) f (0≤ n≤ d). (1) DEEFPI

Comparing this with (13.12.2), we conclude

(Dn f )R(x,y) = (Dd−n f )R(y,x) (R ∈ k-alg, 0≤ n≤ d, x,y ∈MR). (2) HOSYMDIF

We also obtain Euler’s differential equation

(D f )R(x,x) = d fR(x) (3) EULDIFEQ

by setting y = x in (13.13.5), which gives fR(x)+ ε(D f )(x,x) = (1+ ε)d fR(x) =fR(x)+ εd fR(x), as claimed.

ss.DICA13.15. Differential calculus The standard rules of differentiation are valid for ar-bitrary polynomial laws and will henceforth be used without further comment. Forconvenience, we mention just a few examples.

Let f : M→ N, g : N→ P, fi : M→ Ni (i = 1,2) be polynomial laws over k andsuppose N1×N2→ N is a k-bilinear map written multiplicatively. Furthermore, let


λ : M→ N (resp. Q : M→ N) be a linear (resp. quadratic) map. Then, dropping thesubscript “R” for simplicity (e.g., by writing f (x) instead of fR(x), we have

(Dλ )(x,y) = λ (y), (1) DILA

(DQ)(x,y) = Q(x,y), (2) DIQU

(D2Q)(x,y) = Q(y), (3) DISQ(D(g f )

)(x,y) = (Dg)

(f (x),(D f )(x,y)

), (4) DICH(

D(gλ ))(x,y) = (Dg)

(λ (x),λ (y)

)(5) DICL(

D2(g f ))(x,y) = (Dg)

(f (x),(D2 f )(x,y)

)(6) SEDICH

+(D2g)(

f (x),(D f )(x,y)),(

D( f1 f2))(x,y) = (D f1)(x,y) f2(x)+ f1(x)(D f2)(x,y), (7) DIPR(

D2( f1 f2))(x,y) = (D2 f1)(x,y) f2(x)+(D f1)(x,y)(D f2)(x,y) (8) SEDIPR

+ f1(x)(D2 f2)(x,y)

for all R ∈ k-alg and all x,y ∈MR. Note that (4) (resp. (6)) is the first (resp. second)order chain rule, while (7) (resp. (8)) is the first (resp. second) order product rule ofthe differential calculus. Moreover, (5) by (1) is a special case of (4)

ss.DIDE13.16. Directional derivatives Let y∈M. For any polynomial law f : M→N overk, the rule

(∂y f )R(x) := (D f )R(x,yR) ∈ NR (R ∈ k-alg, x ∈MR)

defines a new polynomial law ∂y f : M→N, called the derivative of f in the directiony, and the map

∂y : Polk(M,N)−→ Polk(M,N)

is obviously k-linear.Our next aim will be to show that the operators ∂y,y ∈ M, commute and that

their iterations relate to the total linearization of a homogeneous polynomial law. Toaccomplish this, we need some preparation.

l.TODEFI13.17. Lemma. Let f : M→ N be a polynomial law over k. Then

(Dp f )R[T](n

∑i=1

tixi,xn+1) = ∑ν∈Nn

Tν(Π (ν ,p) f )R(x1, . . . ,xn,xn+1)

for all R ∈ k-alg, p,n ∈ N, n > 0, x1, . . . ,xn,xn+1 ∈MR.

Proof. With an additional indeterminate tn+1, we apply (13.7.2) and obtain

fR[T,tn+1](n+1

∑i=1

tixi) = ∑ν∈Nn,p≥0

Tν tpn+1(Π

(ν ,p) f )R(x1, . . . ,xn,xn+1).

78 2 Foundations

Invoking the Taylor expansion (13.13.3) and comparing coefficients of tpn+1, the

lemma follows. l.LIDIDE

13.18. Lemma. Let f : M→ N be a polynomial law over k and y ∈M. Then

(Π ν∂y f )R(x1, . . . ,xn) = (Π (ν ,1) f )R(x1, . . . ,xn,yR)

for all R ∈ k-alg, n ∈ Z, n > 0, ν ∈ Nn0, x1, . . . ,xn ∈MR.

Proof. Applying (13.7.2) to ∂y f , we obtain

(∂y f )R[T](n

∑i=1

tixi) = ∑ν∈Nn

Tν(Π ν∂y f )R(x1, . . . ,xn).

On the other hand, treating the left-hand side to the definition of ∂y (13.16) andapplying Lemma 13.17 for p = 1, we conclude

(∂y f )R[T](n

∑i=1

tixi) = ∑ν∈Nn

Tν(Π (ν ,1) f )R(x1, . . . ,xn,yR),

as desired. p.ITDIDE

13.19. Proposition. Let f : M → N be a polynomial law over k, p ∈ N, andy1, . . . ,yp ∈M. Then

(∂y1 · · ·∂yp f )R(x) = ∑i≥0

(Π (i,1,...,1) f )R(x,y1R, . . . ,ypR) (1) ITDIDE

for all R ∈ k-alg, x ∈MR. In particular, the operators ∂y,y ∈M, commute.

Proof. To establish the first part, we argue by induction on p. For p= 0, the assertionis just (13.7.1), while the induction step follows immediately from Lemma 13.18.The second part now derives from the first for p = 2 and (13.8.1).

c.DIDETOLI13.20. Corollary. Let f : M → N be a homogeneous polynomial law of degreed > 0. Then the total linearization of f relates to its directional derivatives by theformula

(Π 1 f )k(y1, . . . ,yd−1,yd) = (∂y1 · · ·∂yd−1 f )k(yd) (1) DIDETOLI

for all y1, . . . ,yd−1,yd ∈M. In particular, the right-hand side is totally symmetric iny1, . . . ,yd .

Proof. . Since f is homogeneous of degree d, the expression on the right of (13.19.1)by 13.8 (c) collapses to the single term (Π (d−p,1,...,1) f )R(x,y1R, . . . ,ypR), and (1)follows for p = d−1. The rest is clear.

Our next aim in this section will be to show that a particularly simple and usefulZariski density argument easily extends from the setting of finite-dimensional vector


spaces over infinite fields to arbitrary modules over commutative rings. The key tothis extension is the following concept, which is modeled after standard notions inalgebraic geometry (cf. Demazure-Gabriel [22, I, § 1, 1.2] or Jantzen [50, 1.5], seealso 27.15 below).

ss.SUFU13.21. Subfunctors Let A be a unital associative k-algebra, possibly not commu-tative, and g : M→ A a polynomial law over k. For R ∈ k-alg, we put

D(g)(R) := x ∈MR |gR(x) ∈ A×R ,

where A×R on the right as usual stands for the group of invertible elements in AR.Clearly, D(g) is a subfunctor of Ma, i.e., D(g)(R)⊆Ma(R) = MR is a subset for allR ∈ k-alg and (1M⊗ϕ)(D(g)(R))⊆ D(g)(S) for all morphisms ϕ : R→ S in k-alg.

p.ZADE13.22. Proposition. Let A be a unital associative k-algebra and g : M→ A a poly-nomial law over k such that D(g)(k) 6= /0. If f : M→N is a homogeneous polynomiallaw over k that vanishes on D(g), i.e., fR|D(g)(R) = 0 for all R ∈ k-alg, then f = 0.

For most applications, this proposition is totally adequate. But sometimes the fol-lowing more detailed, but also more technical, version will be needed.

l.ZADE13.23. Lemma. With A and g as in Prop. 13.22, let f : M→ N be a homogeneouspolynomial law over k and R∈ k-alg such that fS|D(g)(S) = 0 for all S∈R-alg⊆ k-algwhich are free of positive rank as R-modules. Then fR = 0.

Proof. After replacing g by g⊗R and f by f ⊗R if necessary, we may assume R= k.Pick e ∈D(g)(k), so e ∈M satisfies u := gk(e) ∈ A×. Substituting Lu−1 g for g, wemay assume gk(e) = 1A. Now write d for the degree of f and consider the k-algebraS := k[ε], εd+1 = 0, which is free as a k-module of rank d + 1. Picking any x ∈Mand using the identifications of (13.3.3, the Taylor expansion (13.13.4) implies that

gS(e+ εx) = 1A +d

∑n=1

εn(Dng)k(e,x)

is invertible in AS. Hence eS + εx ∈ D(g)(S) and, by hypothesis,

0 = fS(e+ εx) =d

∑n=0

εn(Dn f )k(e,x).

Comparing coefficients of εd , we conclude from (13.13.2), (13.14.2) that

fk(x) = (D0 f )k(x,e) = (Dd f )k(e,x) = 0.

For our subsequent applications, it will sometimes be necessary to consider a slighttwist of polynomial laws that relates to ordinary polynomial laws as semi-linear

80 2 Foundations

maps relate to linear ones. The general set-up of this twist may be described asfollows.

ss.TWIMOD

13.24. Twisting modules and linear maps. Let σ : K′ → K be a morphism ink-alg, i.e., a homomorphism of k-algebras taking 1 into 1. For a K-module M, wedenote by Mσ the K′-module which has the same additive group structure as M (inparticular, Mσ = M as sets), and whose scalar multiplication is the one of M twistedby σ :

a′x := a′.σ x := σ(a′)x (a′ ∈ K′, x ∈M). (1) TWISCA

Note that the identity map of M may be viewed as a σ -semi-linear bijection

σM : Mσ −→M. (2) SIEM

For x ∈M, we put xσ := (σM)−1(x) ∈Mσ and have

σM(xσ ) = x, a′xσ =(σ(a′)x

)σ(a′ ∈ K′, x ∈M). (3) APEX

Given another morphism τ : K′′ → K′ in k-alg, and abbreviating στ := σ τ , weclearly have (Mσ )τ = Mστ and (στ)M = σM τMσ , hence (xσ )τ = xστ for all x∈M.If ϕ : M → N is a K-linear map of K-modules, then ϕσ : Mσ → Nσ defined byσN ϕσ = ϕ σM , equivalently,

ϕσ (xσ ) = ϕ(x)σ (x ∈M), (4) PHISIG

is a K′-linear map of K′-modules. Note that ϕσ = ϕ as set maps and (ϕσ )τ = ϕστ .In this way we obtain a functor from K-mod to K′-mod that restricts canonically toa functor from K-alg to K′-alg.

ss.SEMPOL13.25. The concept of a semi-polynomial law. Let σ : K′→ K be a morphism ink-alg, M a K-module and M′ a K′-module. Then 13.2 yields a functor Ma : K-alg→set. Similarly, utilizing the formalism of 13.24, we obtain a functor σ∗M′a : K-alg→set by setting (σ∗M′a)(R) := M′⊗K′ Rσ as sets and

(σ∗M′a)(ϕ) := 1M′ ⊗K′ ϕσ : M′⊗K′ R

σ −→M′⊗K′ Sσ

as set maps for morphisms ϕ : R→ S in K-alg. A σ -semi-polynomial law from M′

to M over k is now defined as a natural transformation f : σ∗M′a→Ma. This meansthat we are given a family of set maps

fR : M′Rσ = M′⊗K′ Rσ −→M⊗K R = MR,

one for each R ∈ K-alg, which varies functorially with R, so for every morphismϕ : R→ S in K-alg, the diagram


M′Rσ

fR //

1M′⊗K′ϕσ

MR

1M⊗ϕ

M′Sσ

fS// MS.

(1) SIGPOL

commutes.

Exercises.

pr.ZERLIN 53. Let f : M→ N be a polynomial law over k, R ∈ k-alg, n, p be positive integers and x ∈ MnR,

y ∈MpR . Prove for ν ∈ Nn that

(Π (ν ,0,...,0) f )R(x,y) = (Π ν f )R(x).

pr.HOCO 54. (Roby [112]) Let f : M→ N be a polynomial law over k. Show that there is a unique family( fd)d≥0 of polynomial laws M→ N over k such that the following conditions hold: (i) the family( fd)d≥0 is locally finite, (ii) f = ∑ fd , (iii) fd is homogeneous of degree d for all d ≥ 0. Give anexample where fd 6= 0 for all d ≥ 0.

pr.COPOLA 55. Constant polynomial laws (Roby [112]). Show that the homogeneous polynomial laws M→Nof degree 0 are precisely of the form w,w ∈ N, where wR : MR → NR for R ∈ k-alg is given bywR(x) := wR,x ∈MR.

pr.MULI 56. (Roby [112]) (a) Let µ : M1 × ·· · ×Mn → N be a multi-linear map. Show that µ : M1 ×·· ·×Mn → N given by µR := µ ⊗R (the R-multi-linear extension of µ) for R ∈ k-alg is a multi-homogeneous polynomial law of multi-degree 1 = (1, . . . ,1). Show that, conversely, every multi-homogeneous polynomial law M1×·· ·×Mn→ N of multi-degree 1 uniquely arises in this way.(b) If Q : M→ N is a quadratic map, show that Q : M→ N given by QR := Q⊗R (the base changeof Q from k to R in the sense of 12.5) for R ∈ k-alg is a homogeneous polynomial law of degree2 over k. Show that, conversely, every homogeneous polynomial law M→ N of degree 2 over kuniquely arises in this way. Identifying quadratic maps and homogeneous polynomial laws of de-gree 2 accordingly, prove that the total first derivative DQ : M×M→ N of Q as a polynomial lawis the same as the bilinearization of Q as quadratic map.

pr.HIDIDE 57. (Roby [112]) Let f : M→ N be a polynomial law over k and p ∈ N.(a) Show for n ∈ Z,n > p,ν1, . . . ,νn ∈ N,x,x1, . . . ,xp ∈MR,R ∈ k-alg that

(Π (ν1,...,νp,νp+1,...,νn) f )R(x1, . . . ,xp,x, . . . ,x)

=(νp+1 + · · ·+νn)!

νp+1! · · ·νn!(Π (ν1,...,νp,νp+1+···+νn) f )R(x1, . . . ,xp,x)

(b) Show for y ∈M that ∂[p]y f : M→ N given by (∂

[p]y f )R(x) = (Dp f )R(x,yR) (R ∈ k-alg, x ∈MR)

is a polynomial law over k. Moreover, the map ∂[p]y : Polk(M,N)→ Polk(M,N) satisfies (∂y)

p =

p!∂ [p]y .

(c) Show that ∂[2]y+z = ∂

[2]y +∂y∂z +∂

[2]z for y,z ∈M.

pr.EXOCUB 58. Exotic cubic forms. Even when working over a field, homogeneous polynomial laws of degree≥ 3 are no longer determined by the set maps they induce over the base ring. Examples for thisphenomenon will be discussed in the present exercise.(a) Let f : M → N be a homogeneous polynomial law of degree 3 over k. Simplify notation bywriting f (x) = fR(x) for R ∈ k-alg, x ∈MR (ditto for other polynomial laws), and put

82 2 Foundations

f (x,y) := (D f )(x,y), f (x,y,z) := (Π (1,1,1) f )(x,y,z) (2) EFEXY

for x,y,z ∈MR. Then prove

f (x+ ty) = f (x)+ t f (x,y)+ t2 f (y,x)+ t3 f (y), (3) EFEXTE

f (x+ y,z) = f (x,z)+ f (x,y,z)+ f (y,z), (4) EFEXPLY

f (x,y,z) = f (x+ y+ z)− f (x+ y)− f (y+ z)− f (z+ x) (5) EFEXYZ

+ f (x)+ f (y)+ f (z),

f (n

∑i=1

rixi) =n

∑i=1

r3i f (xi)+ ∑

1≤i, j≤n,i6= jr2

i r j f (xi,x j) (6) EFLICO

+ ∑1≤i< j<l≤n

rir jrl f (xi,x j,xl)

for all R ∈ k-alg, n ∈ Z, n > 0, a variable t and elements x,y,z,x1, . . . ,xn ∈MR, r1, . . . ,rn ∈ R.(b) Let F be a field, n ∈ Z, n > 0, Fn n-dimensional column space over F with the canonical basis(ei)1≤i≤n and g : Fn → F a cubic form. View Fn as a unital commutative associative F-algebraunder the component-wise multiplication and prove that the following conditions are equivalent.

(i) The set map gF : Fn→ F is zero but g itself is not.(ii) F ∼= F2 consists of two elements,

gF (ei) = gF (x,x) = gF (x,y,z) = 0 (1≤ i≤ n, x,y,z ∈ Fn), (7) BASGEF

and there are x0,y0 ∈ Fn such that gF (x0,y0) 6= 0.(iii) F ∼= F2 consists of two elements and there exists a non-zero alternating matrix S ∈Matn(F)

such thatgR(x) = xt SRx2

for all R ∈ F-alg and all x ∈ (Fn)R = Rn.

pr.CUMA 59. Cubic maps. In this exercise, we compare Faulkner’s approach [29] to homogeneous polyno-mial maps, for simplicity restricted here to the special case of degree 3, with the formalism ofpolynomial laws.(a) Show that the totality of polynomial laws over k can be canonically converted into a category,denoted by k-polaw, and regard the homogeneous polynomial laws of degree 3 as a full subcate-gory, denoted by k3-holaw, of k-polaw.(b) Let M and N be k-modules. Following Faulkner [29], define a cubic map from M to N as a pair( f ,g) of set maps f : M→M and g : M×M→ N satisfying the following conditions.

(i) f is homogeneous of degree 3: f (αx) = α3 f (x) for all α ∈ k and all x ∈M.(ii) g is quadratic-linear: For all x ∈M,

g(x,−) : M −→ N, y 7−→ g(x,y),

is a linear map, and for all y ∈M,

g(−,y) : M −→ N, x 7−→ g(x,y),

is a quadratic map.(iii) ( f ,g) satisfies the expansion

f (x+ y) = f (x)+g(x,y)+g(y,x)+ f (y)

for all x,y ∈M.


(iv) Euler’s differential equation holds:

g(x,x) = 3 f (x)

for all x ∈M.

Cubic maps from M to N as above, symbolized by ( f ,g) : M→ N, form a k-module in the obviousway. Now prove:

(c) If ( f ,g) : M→ N is a cubic map, then the assignment

(x,y,z) 7−→ g(x,y,z) := g(x+ y,z)−g(x,z)−g(y,z) (8) TRICU

defines a trilinear map M×M×M→ N which is totally symmetric. Moreover,

Rad( f ,g) := x ∈M | ∀y ∈M : f (x) = g(x,y) = g(y,x) = 0, (9) CURAD

called the radical of ( f ,g), is a submodule of M such that

g(

Rad( f ,g),M,M)= 0, (10) TRIRAD

and for any linear surjection π : M→M1 of k-modules having Ker(π)⊆ Rad( f ,g), there isa unique cubic map ( f1,g1) : M1→ N such that f1 π = f and g1 (π×π) = g.

(d) The totality of cubic maps between k-modules can be canonically converted into a cate-gory, denoted by k-cumap. Furthermore, the assignment f 7→ ( fk,(D f )k) on objects and theidentity on morphisms yields a well defined isomorphism of categories from k3-holaw ontok-cumap, the inverse of that isomorphism on objects being denoted by ( f ,g) 7→ f ∗g.

pr.NONSC 60. Matching non-scalar polynomial laws with scalar ones. (Loos [67, 18.5]) Prove for k-modulesM,N that there exists a unique k-linear map

Ψ : N⊗Pol(M,k)−→ Pol(M,N)

satisfying (Ψ(v⊗ f )

)R(x) = v⊗ fR(x) (11) PSIVE

for all v ∈ N, f ∈ Pol(M,k), R ∈ k-alg, x ∈ MR. Moreover, Ψ is an isomorphism if N is finitelygenerated projective.

pr.SEPOLA 61. Let σ : K′→ K, τ : K′′→ K′ be morphisms in k-alg, M a K-module, M′ a K′-module and M′′

a K′′-module

(a) Let ϕ : M′→M be a σ -semi-linear map. Show for R ∈ k-alg that there is a unique σ -semi-linear map ϕR : M′Rσ →MR satisfying

ϕR(x′⊗K′ rσ ) = ϕ(x′)⊗K r (12) PHIAR

for all x′ ∈M′, r∈R. Moreover, ϕR is σR-semi-linear, and the family of set maps ϕR : M′Rσ →MR, R ∈ k-alg, is a σ -semi-polynomial law ϕ : M′→M over k.

(b) Let f : M′→M (resp. g : M′′→M′) be a σ - (resp. τ-) semi-polynomial law over k. Showthat the family of set maps

( f g)R := fR gRσ : M′′Rστ = M′′(Rσ )τ −→MR (R ∈ k-alg)

defines a στ-semi-polynomial law f g : M′′→M over k, called the composition of f andg. Conclude that the composition ( f ,g) 7→ f g is associative in the obvious sense.

84 2 Foundations

(c) Let ϕ : M′→M (resp. ψ : M′′→M′) be σ - (resp. τ-) semi-linear maps. Prove (ϕ ψ)∼ =ϕ ψ as an equality of semi-polynomial laws.

(d) Assume σ : K′ ∼→ K is an isomorphism in k-alg. Prove that f : M′ → M is a σ -semi-polynomial law over k if and only if there exists an ordinary polynomial law g : M′ →Mσ

over K′ satisfying f = (σM)∼ g, and that, in this case, g is unique.

pr.POSEPO 62. Extend all concepts and results of the present section from ordinary polynomial laws to semi-polynomial laws.

Chapter 3Alternative algebras

c.ALT

Alternative algebras belong to the most important building blocks of Albert alge-bras. They will be presented here with this connection always in mind. We beginby deriving a number of important identities, proceed to establish a fairly generalversion of Artin’s associativity theorem and investigate homotopes of alternative al-gebras, which in turn give rise to the study of isotopy involutions in the final sectionof this chapter.

14. Basic identities and invertibility

s.BASIDSTRASSHaving gained some proficiency in the language of arbitrary non-associative alge-bras, we are now adequately prepared to deal with the more specific class of alter-native algebras. After giving the formal definition, we derive the Moufang identitiesand introduce the notion of invertibility. Throughout, we let k be an arbitrary com-mutative ring.

ss.CONALAL14.1. The concept of an alternative algebra. A k-algebra A is said to be alterna-tive if its associator (cf. 8.5), i.e., the trilinear map (x,y,z) 7→ [x,y,z] = (xy)z−x(yz)from A×A×A to A, is alternating. This means that the following identities hold inA:

x(xy) = x2y, (left alternative law) (1) LALT

(yx)x = yx2, (right alternative law) (2) RALT

(xy)x = x(yx). (flexible law) (3) FLEX

In fact, since the symmetric group on three letters is generated by any two of thetranspositions (1,2), (2,3), (3,1), any two of the above equations imply the third,hence force A to be alternative. An alternative algebra and its opposite have thesame associator, so Aop is alternative if and only if A is. It is sometimes convenientto express the alternative laws in operator form, i.e.,in terms of left and right multi-plications as follows:

L2x = Lx2 , (4) LLALT

R2x = Rx2 , (5) RRALT

LxRx = RxLx. (6) RLEX

ss.LIN

85

86 3 Alternative algebras

14.2. Linearizations. For the rest of this section, we fix an alternative algebra Aover k. Then, thanks to flexibility, the expression xyx is unambiguous in A. Also,since (14.1.1)−(14.1.3) are quadratic in x, they may be linearized to yield new iden-tities valid in A. For example, replacing x by x+z in (14.1.1), collecting mixed termsand interchanging y with z, we obtain

x(yz)+ y(xz) = (xy+ yx)z. (1) LINLALT

Similarly,

(xy)z+(xz)y = x(yz+ zy), (2) LINRALT

(xy)z+(zy)x = x(yz)+ z(yx). (3) LINFLEX

Again these relations can be expressed in terms of left and right multiplications;details are left to the reader. Also, it is now straightforward to check that the propertyof an algebra to be alternative remains stable under base change: AR is alternative,for any R ∈ k-alg.

ss.MOUF14.3. The Moufang identities. Less obvious is the fact that the Moufang identities

(xyx)z = x(y(xz)

), (left Moufang identity) (1) LMOUF

z(xyx) =((zx)y

)x, (right Moufang identity) (2) RMOUF

(xy)(zx) = x(yz)x (middle Moufang identity) (3) MMOUF

hold in any alternative algebra. Since the associator is alternating, (1) follows from(14.1.1), (14.2.2) and

(xyx)z− x(y(xz)

)= [xy,x,z]+ [x,y,xz] =−[x,xy,z]− [x,xz,y]

= − (x2y)z− (x2z)y+ x((xy)z+(xz)y

)= − x2(yz+ zy)+ x

(x(yz+ zy)

)= 0.

Reading the left Moufang identity in the opposite algebra Aop gives the right Mo-ufang identity. Finally, again using the fact that the associator is alternating, but also(1), we obtain

(xy)(zx)− x(yz)x = [x,y,zx]− x[y,z,x] =−[x,zx,y]− x[z,x,y]

= − (xzx)y+ x((zx)y

)− x((zx)y

)+ x(z(xy)

)= 0,

and the middle Moufang identity is proved.Viewing (1), (2) as linear maps in z and (3) as a bilinear one in y,z, the Moufang

identities may also be expressed as

14 Basic identities and invertibility 87

Lxyx = LxLyLx, (4) LLMOUF

Rxyx = RxRyRx, (5) RRMOUF

(Lxy)(Rxz) = LxRx(yz). (6) RLMOUF

The linearization process is useful also in the present context. For example, lineari-zing the right Moufang identity (2) we deduce

z((wy)x

)+ z((xy)w

)=((zw)y

)x+((zx)y

)w = z

(w(yx)

)+ z(x(yw)

). (7) LRMOUF

ss.INVALT14.4. Inverses. There is no useful concept of invertibility in arbitrary non-associa-tive algebras. Fortunately, however, the standard notion for associative algebras car-ries over to the alternative case without change. If A is unital (with identity element1A), an element x ∈ A is said to be invertible if there exists an element y ∈ A, calledan inverse of x in A, that satisfies the relations xy = 1A = yx. Indeed, the concept ofinvertibility enjoys the usual properties. Before proving this, we need a preparation.

ss.UOPALT14.5. The U-operator. The U-operator of A (no longer assumed to be unital) isdefined as the quadratic map

U : A−→ Endk(A), x 7−→Ux := LxRx = RxLx, (1) UOPALT

which acts on individual elements as

Uxy = xyx. (2) UEL

Note that the U-operator does not change when passing to the opposite algebra.Moreover, it may be used to rewrite the middle Moufang identity in the form

Ux(yz) = (Lxy)(Rxz). (3) UMMOUF

Finally, the U-operator satisfies the following important relations:

Uxy = LxUyRx = RyUxLy (4) ULR

for all x,y ∈ A. Indeed, using (14.3.7), (14.3.3), (14.3.2), (14.1.1), we obtain

Uxyz =((xy)z

)(xy) = x

((yz)(xy)+ [(xy)z]y

)−([x(xy)]z

)y

= x(y(zx)y+ x(yzy)

)− x2(yzy) = x

(y(zx)y

)= LxUyRxz

for all z ∈ A, giving the first part of (4). The second one follows by passing to Aop.p.INVALT

14.6. Proposition. Let A be unital and x ∈ A. Then the following conditions areequivalent.

(i) x is invertible.


(ii) Lx is bijective.(iii) Rx is bijective.(iv) Ux is bijective.(v) Lx and Rx are surjective.(vi) Ux is surjective.(vii) 1A ∈ Im(Lx)∩ Im(Rx).(viii) 1A ∈ Im(Ux).

In this case, x has a unique inverse in A, denoted by x−1, and

x−1 = L−1x 1A = R−1

x 1A =U−1x x, (1) INVALT

Lx−1 = L−1x , Rx−1 = R−1

x , Ux−1 =U−1x . (2) OPINVALT

Proof. (i)⇔ (ii). If x is invertible with inverse y, we obtain xy2x = 1A by (14.3.3),hence LxLy2Lx = 1A by (14.3.4), and (ii) follows. Conversely, suppose Lx is bijective.Then there is a unique y ∈ A satisfying xy = 1A, and the relation x(yx) = (xy)x =1Ax = x1A combined with (ii) shows yx = 1A, hence (i).

(i)⇔ (iii). This is just (i)⇔ (ii) in Cop.(iii)⇒ (iv). Obvious since (iii) implies (ii).(iv)⇒ (v). Obvious since Lx and Rx commute by flexibility (14.1.6).(v)⇒ (vi)⇒ (vii). Obvious, again by flexibility.(vii) ⇒ (viii). We find elements y,z ∈ A satisfying xy = zx = 1A, and (14.3.3)

gives x(yz)x = 1A.(viii)⇒ (ii). We find an element w ∈ A satisfying xwx = 1A, and (14.3.1) implies

LxLwLx = 1A, forcing Lx to be bijective.Uniqueness of the inverse and (1) now follow from (ii), (iii), (iv). Furthermore,

xx−1x = x and the (14.3.4) implies LxLx−1Lx = Lx, so (ii) gives Lx−1 = L−1x . Reading

this in Aop yields Rx−1 = R−1x , hence Ux−1 =U−1

x , and the proof of (2) is complete.

ss.SINVALT

14.7. The set of invertible elements. If A is unital, then the set of its invertibleelements will be denoted by A×. Clearly, 1A ∈ A×, and x ∈ A× implies x−1 ∈ A×

with (x−1)−1 = x. Moreover, A× is closed under multiplication. More precisely, ifx,y ∈ A are invertible, so is xy and

(xy)−1 = y−1x−1. (1) MINVALT

Indeed, setting z = y−1x−1, the Moufang identities and (14.6.2) imply ((xy)z)y =x(y(y−1x−1)y) = x(x−1y) = y, hence (xy)z = 1A by Prop. 14.6 (iii). Replacing x byy−1 and y by x−1 yields z(xy) = 1A, and (1) follows.

r.ALDI14.8. Remark. Prop. 14.6 shows that a unital alternative algebra is a division alge-bra in the sense of 9.7 if and only if 1A 6= 0 and every non-zero element is invertible.

15 Strongly associative subsets 89

15. Strongly associative subsets

s.STRASOur aim in this section will be to prove Artin’s associativity theorem, which saysthat every alternative algebra on two generators is associative. Actually, we willderive a somewhat more general result by adopting the approach of Bourbaki [12,III, Appendix, § 1] (see also Braun-Koecher [15, VII, § 1]). Throughout we let k baa commutative ring and A an alternative k-algebra.

ss.COSTRO15.1. The concept of a strongly associative subset. A subset X ⊆ A is said to bestrongly associative if [x,y,z] = 0 provided at least two of the elements x,y,z ∈ Abelong to X . Since the associator is alternating, this is equivalent to the condition[X ,X ,A] = 0. Hence, if X ⊆A is strongly associative, so is the submodule spannedby X , and X ∪1A provided A is unital. Examples of strongly associative subsetsare X = x, for any x ∈ A (by (14.1.1)), and X = x,x−1, for A unital and anyx ∈ A× (by (14.1.1), (14.6.2)).

l.STABSUBMOD15.2. Lemma. Let X ⊆ A be a set of generators for A as a k-algebra. SupposeM ⊆ A is a k-submodule that contains X and satisfies XM+MX ⊆M. Then M = A.

Proof. We begin by showing that A′, the set of elements x ∈ A satisfying xM +Mx⊆M, is closed under multiplication, so let x,y ∈ A′. Then, for all z ∈M, (xy)z =[x,y,z]+ x(yz) = x(yz)− [x,z,y] = x(yz)− (xz)y+ x(zy) ∈M and, similarly, z(xy) ∈M, forcing xy∈ A′, as claimed. It follows that A′, being a subalgebra of A containingX , agrees with A, which implies AM ⊆ M. But then M must be a subalgebra of Acontaining X , and we conclude M = A.

p.STRASSALG15.3. Proposition. If X is a strongly associative subset of A, then so is the subal-gebra of A generated by X.

Proof. By 8.4, it suffices to show that Mon(X) =⋃

m>0 Monm(X), the set of mono-mials over X , is a strongly associative subset of A. To this end, we only need toprove

[Monm(X),Monn(X),A] = 0 (m,n ∈ Z,m,n > 0), (1) STRASSALG

and we do so by induction on p = m+ n ≥ 2. The case p = 2 being obvious byhypothesis, let us assume p > 2. Since (1) is symmetric in m,n by alternativity, wemay even assume m> 1. Hence, given u∈Monm(X),v∈Monn(X),a∈A, we obtainu = u1u2, ui ∈Monmi(X), mi ∈ Z, mi > 0, i = 1,2, m1 +m2 = m, and (8.5.2) yields

−[u,v,a] = [u1u2,a,v] =−u1[u2,v,a]+ [u1,u2,a]v− [u1,v,u2a]− [u1,u2,av],

where all terms on the right vanish by the induction hypothesis. Hence [u,v,a] = 0,which completes the induction.

p.UNSTRASSUB15.4. Proposition. Let X ,Y be strongly associative subsets of A. Then the subalge-bra of A generated by X and Y is associative.


Proof. We may not only assume that A itself is generated by X and Y but also, byProp. 15.3, that X ,Y ⊆ A are subalgebras. Then the set B of elements z ∈ A suchthat [X ,Y,z] = 0 is a submodule of A containing X and Y since they are bothstrongly associative. The proof will be complete once we have shown B = A sincethis implies [X ∪Y,X ∪Y,A] = 0, forcing X ∪Y to be a strongly associative subsetof A; this property is inherited by the subalgebra generated by X and Y , which showsthat A is associative, as desired. In order to prove B = A, we apply Lemma 15.2 andhence must show

XB+BX +Y B+BY ⊆ B. (1) UNSTRASSUB

Accordingly, let x,x′ ∈ X ,y ∈ Y,z ∈ B. Then (8.5.2) yields

[x′x,z,y]− [x′,xz,y]+ [x′,x,zy] = x′[x,z,y]+ [x′,x,z]y = 0

since z ∈ B and X is strongly associative, which implies [x′,x,zy] = 0 as well. But Xis also a subalgebra of B, whence [x′x,z,y] = 0. Altogether, we conclude [X ,Xz,Y ] =0, forcing Xz⊆ B. Interchanging the roles of X and Y and passing to the oppositealgebra, we obtain (1), and the proof is complete.

c.ARTIN15.5. Corollary. (Artin’s theorem) Let x,y∈A. Then the subalgebra of A generatedby x and y is associative. If A is unital, the same conclusion holds for the unitalsubalgebra of A generated by x,y and (if they exist) their inverses.

Proof. In the non-unital case, put X = x,Y = y. In the unital case, put X =x,1A if x is not invertible, X = x,1A,x−1 otherwise, ditto for Y .

c.POWASSALT15.6. Corollary. Alternative algebras are power-associative. Moreover, if A is uni-tal, then xmxn = xm+n and (xm)n = xmn for all x ∈ A×, m,n ∈ Z.

Exercises.

pr.COMALT 63. Show that an alternative algebra A over k satisfies the relation 3[A,A,A] ⊆ [A,A] +A[A,A] +[A,A]A. Conclude that commutative alternative algebras without 3-torsion are associative.

pr.UNALT 64. A characterization of unital alternative algebras (McCrimmon [79]). Let A be an alternativek-algebra. Show that the following conditions are equivalent.

(i) A is unital.(ii) For some x ∈ A, both Lx and Rx are bijective.(iii) For some x ∈ A, both Lx and Rx are surjective.(iv) For some x ∈ A, Ux is surjective.(v) For some x,y ∈ A, both Lx and Ry are bijective.

pr.ONESINVALT 65. One-sided inverses (McCrimmon [79]). Let A be a unital alternative k-algebra and x,y ∈ A.Show that the following conditions are equivalent.

(i) xy = 1A.(ii) LxLy = 1A.(iii) RyRx = 1A.

16 Homotopes 91

(Hint. For the implication (i)⇒ (ii) prove that E = LxLy, F = RyRx are projections of the k-moduleA satisfying EF = 1A.)If these conditions are fulfilled, x (resp. y) is said to be right-invertible (resp. left-invertible) withright (resp. left) inverse y (resp. x) (in general not unique).

pr.INVPRODALT 66. (McCrimmon [79]) Let A be a unital alternative algebra over k and x,y ∈ A such that xy isinvertible. Show in the language of Exc. 65 that x is right-invertible with y(xy)−1 as a right inverseand y is left-invertible with (xy)−1x as a left inverse. (Hint. For the first assertion, use (14.5.4) toshow that Ly is injective.)

pr.PEIRCEALT 67. The singular Peirce decomposition of alternative algebras (cf. Schafer [114]). Let A be a unitalalternative algebra over k and c ∈ A an arbitrary idempotent. Put c1 = c,c2 = 1A− c to prove thatthe maps Lci Rc j : A→ A (i, j = 1,2) form a complete orthogonal system of projections satisfying(cix)c j = ci(xc j) =:= cixc j for all x ∈ A. (Hint. Expand Uc1+c2 .) Conclude

A = A11⊕A12⊕A21⊕A22 (1) PALTDEC

as a direct sum of submodules

Ai j := Ai j(c) := x ∈ A | cixc j = x= x ∈ A | cix = x = xc j (2) PALTC

= x ∈C | cx = (2− i)x, xc = (2− j)x ⊆ A (i, j = 1,2)

that satisfy the multiplication rules

Ai jA jl ⊆ Ail , (3) PALTU

AiiA jl = Al jAii = 0, (i 6= j) (4) PALTD

A2i j ⊆ A ji (i 6= j) (5) PALTT

for all i, j, l = 1,2. Prove x2 = 0 for all x ∈ Ai j , i 6= j, and that (5) can be sharpened to A2i j = 0

(for i 6= j) if A is associative.

16. Homotopes

s.HOMOTALTHomotopes have been established a long time ago as an extremely versatile toolin the theory of Jordan algebras and can look back to a long, respectable history.Though they entered the scene of alternative algebras only much later, without everreceiving the amount of attention they were accustomed to from the Jordan setting, itwill be shown in the present book that, also in these new surroundings, they providea useful and convenient formalism for many problems related to Albert algebras.

In the present section, the conceptual foundations for homotopes and isotopes ofalternative algebras will be laid down following McCrimmon [79]. Throughout, welet k be an arbitrary commutative ring and A an alternative algebra over k. We beginwith an easy special case of the general concept.

ss.HOMOTASS16.1. Digression: associative algebras For the time being, let B be an associa-tive k-algebra and p ∈ B. Define a new k-algebra B(p) on the k-module B by themultiplication


x .p y := xpy (x,y ∈ B).

We call B(p) the p-homotope of B, which is obviously associative; moreover, themaps Lp,Rp : B(p) → B is an algebra homomorphism. In particular, if B is unitaland p is invertible, B(p) ∼= B under Lp or Rp. This explains why homotopes do notplay a significant role in (associative) ring theory.

ss.CONHOMOTALT16.2. The concept of a homotope Returning to our original alternative algebra A,let p,q ∈ A. On the k-module A we define a new k-algebra A(p,q) by the multiplica-tion

x .p,q y := (xp)(qy) (x,y ∈ A). (1) HOMOTALT

We call A(p,q) the p,q-homotope (or just a homotope) of A. We obviously have

(A(p,q))op = (Aop)(q,p), A(α p,α−1q) = A(p,q) (α ∈ k×). (2) HOMOP

A k-algebra is said to be homotopic to A if it is isomorphic to some of its homotopes.If A is associative, (1) collapses to x .p,q y = xpqy, so A(p,q) = A(pq) as in 16.1.

Our next aim will to show that homotopes of alternative algebras are alternative,and that homotopes of homotopes are homotopes. More precisely, we obtain thefollowing proposition.

p.ITHOMOTALT16.3. Proposition. Let p,q, p′,q′ ∈ A. Then A(p,q) is an alternative k-algebra and

(A(p,q))(p′,q′) = A(p′′,q′′), p′′ := p(qp′)p, q′′ := q(q′p)q. (1) ITHOMOTALT

Proof. For the first part, it suffices to verify the left alternative law (14.1.1) sinceby passing to the opposite algebra and invoking (16.2.2), we will obtain the rightalternative law as well. So let x,y ∈ A. Then (14.5.4) and the Moufang identities(14.3.1), (14.3.2) imply

x . p,q (x .p,q y) = (xp)(q[(xp)(qy)]

)= (Uxpq)(qy) = (LxUpRxq)(qy)

=(x[p(qx)p]

)(qy) =

([(xp)(qx)]p

)(qy) = (x .p,q x) .p,q y,

hence (14.1.1) for A(p,q). The verification of the second part is straightforward andleft to the reader.

ss.FULT

16.4. Functoriality. We denote by k-alt the category of (possibly non-unital) alter-native k-algebras, morphisms being ordinary k-algebra homomorphisms (8.1). Bycontrast, the category of weakly 2-pointed alternative k-algebras will be denotedby k-twalt. Its objects are triples (A, p,q) consisting of an alternative k-algebra Aand a pair of elements p,q ∈ A, while its morphisms have the form h : (A, p,q)→(A′, p′,q′) with weakly two-pointed alternative k-algebras (A, p,q), (A′, p′,q′) and

16 Homotopes 93

an algebra homomorphism h : A→A′ satisfying h(p)=α p′, h(q)=α−1q′ for someα ∈ k×. It is then clear that the assignment (A, p,q) 7→ A(p,q) gives rise to a (covari-ant) functor from k-twalt to k-alt which is the identity on morphisms.

ss.UNHOMOTALT16.5. The connection with unital algebras We are particularly interested in ho-motopes containing an identity element. In order to find necessary and sufficientconditions for this to happen, we consider the U-operator U (p,q) of A(p,q) (p,q ∈ A),so U (p,q)

x = L(p,q)x R(p,q)

x by (14.5.1), where L(p,q),R(p,q) stand for the left, right mul-tiplication of A(p,q). We claim

U (p,q)x =UxUpq (x ∈ A). (1) UOPHOMOTALT

To prove this, we let x,y ∈ A and compute, using flexibility in A(p,q), the Moufangidentities and (14.5.2), (14.5.4)

U (p,q)x y =

([(xp)(qy)]p

)(qx) =

(x[p(qy)p]

)(qx) = x(RqUpLqy)x =UxUpqy,

as desired.p.UNITHOMOTALT

16.6. Proposition. For p,q ∈ A, the p,q-homotope A(p,q) has a unit element if andonly if A has a unit element and pq is invertible in A. In this case, 1(p,q)

A := (pq)−1

is the unit element of A(p,q).

Proof. If e ∈ A is a unit element for A(p,q), then (16.5.1) implies 1A = U (p,q)e =

UeUpq, so Ue is surjective. But then A is unital (Exc. 64), and e must be invertible inA (Prop. 14.6). Hence so is pq with U−1

pq =Ue and (pq)−1 =U−1pq (pq) =Ue(pq) =

(ep)(qe) = e .p,q e = e since e is a unit element for A(p,q). Conversely, let A be unitaland suppose pq ∈ A is invertible with inverse e := (pq)−1. Then Exc. 66 implies(ep)q = 1A, forcing LepLq = 1A by Exc. 65, and we conclude e .p,q x = (ep)(qx) = xfor all x ∈ A, so e is a left unit element for A(p,q). Passing to the opposite algebraof A will show that e is also a right unit for A(p,q), forcing A(p,q) to be unital withidentity element e.

ss.ISOTALT16.7. Isotopes If A is unital and p,q ∈ A are both invertible (which is stronger thanjust requiring pq to be invertible), then A(p,q) is called the p,q-isotope (or simply anisotope) of A. By Props. 16.3, 16.6, A(p,q) is a unital alternative algebra in this case,with unit element

1(p,q)A = (pq)−1 = q−1 p−1, (1) UNITISOTALT

and (16.3.1) gives

(A(p,q))(p′,q′) = A (p′ := q−1 p−2,q′ := q−2 p−1). (2) ITISOTALT


In particular, calling a k-algebra isotopic to A if it is isomorphic to some of itsisotopes, isotopy is an equivalence relation on unital alternative algebras.

By (1), the algebra A(p,q), for A unital and p,q ∈ A×, determines the product pquniquely. But it is important to note that the factors of this product are not uniquelydetermined by A(p,q). Indeed, bringing in the nucleus of A (cf. 9.6), we have

p.NUCISOTALT16.8. Proposition. Let A be unital and p,q, p′,q′ ∈ A×. Then A(p,q) = A(p′,q′) if andonly if p′ = pu,q′ = u−1q for some u ∈ Nuc(A)×.

Proof. p′ = pu,q′ = u−1q for some u ∈ Nuc(C)× clearly implies A(p,q) = A(p′,q′).Conversely, assume A(p,q) = A(p′,q′). Then

(xp)(qy) = (xp′)(q′y) (x,y ∈ A). (1) NUCISOTALT

Setting u := p−1 p′, we obtain p′ = pu, and (1) for x = p′−1,y = 1A yields q′ = u−1q,hence u= qq′−1. It remains to prove u∈Nuc(A) and, since A is alternative, it sufficesto show [A,u,A] = 0. To this end, we put y = q′−1 (resp. x = p′−1) in (1) to obtain(xp)u = xp′ (resp. u−1(qy) = q′y). Hence (1) reads (xp)(qy) = ((xp)u)(u−1(qy)),and since p,q are invertible, this amounts to xy = (xu)(u−1y) for all x,y ∈ A. Re-placing y by uy and invoking (14.6.2), we conclude [A,u,A] = 0, as desired.

ss.UNTISOTALT16.9. Unital isotopes. If A is unital, an isotope of A is said to be unital if it has thesame identity element as A. By (16.7.1), unital isotopes have the form

Ap := A(p−1,p) (p ∈ A×). (1) UNTISOTALT

By Exc. 68, invertibility and inverses in A and Ap (p∈ A×) are the same, while from(16.2.2), (16.3.1) we conclude

(Ap)op = (Aop)p−1, (Ap)q = Apq, Aα p = Ap (p,q ∈ A×, α ∈ k×). (2) ITUNTISOTALT

Moreover, Prop. 16.8 implies

Ap = Aq⇐⇒ p = uq for some u ∈ Nuc(A)× (p,q ∈ A×). (3) EQIS

In particular, Ap = A for all p ∈ A× if A is associative.

Unital isotopes are important for various reasons: for example, they play a usefulrole in the two Tits constructions of cubic norm structures. Moreover, arbitrary iso-topes are always isomorphic to appropriate unital ones (cf. Exc. 73 below). Andfinally, working in unital isotopes turns out to be computationally smooth, as maybe seen from the following lemma.

l.UNIPO16.10. Lemma. Let A be unital and p ∈ A×. Then A and Ap have the same U-operators as well as the same powers xm for x ∈ A, m ∈N (resp. for x ∈ A×, m ∈ Z).

16 Homotopes 95

Proof. The equality of U-operators follows from (16.5.1). Since Uxmxn = (Ux)mxn =

x2m+n for x ∈ A, m,n ∈N (resp. x ∈ A×, m,n ∈ Z), the remaining assertions can nowbe derived by induction.

ss.FULTONE16.11. Functoriality. Adjusting the terminology of 16.4 to the unital case, we de-note by k-alt1 the category of unital alternative k-algebras, morphisms being unitalk-algebra homomorphisms (9.1). By contrast, the category of pointed alternative k-algebras will be denoted by k-palt. Its objects are pairs (A, p) consisting of a unitalalternative k-algebra A and an invertible element p ∈ A, while its morphisms havethe form h : (A, p)→ (A′, p′) with pointed alternative k-algebras (A, p), (A′, p′) anda unital algebra homomorphism h : A→ A′ satisfying h(p) = α p′ for some α ∈ k×.Again, it is then clear that the assignment (A, p) 7→ Ap gives rise to a (covariant)functor from k-palt to k-alt1 which is the identity on morphisms.

ss.ISOTISOMALT16.12. Isotopy versus isomorphism. Recall that unital alternative k-algebras A,Bare isotopic if A ∼= B(p,q) for some invertible elements p,q ∈ B, equivalently byExc. 73 (a), if A ∼= Bp for some invertible element p ∈ B. It is then a natural ques-tion to ask whether isotopic unital alternative algebra are always isomorphic. Thisquestion has a trivial affirmative answer in the associative case (cf. 16.1), but a lesstrivial negative one in general, cf. McCrimmon [79, p. 259] for a counter example.It is not known, however, whether isotopic unital alternative k-algebras exist that arefinitely generated projective as k-modules (or even finite-dimensional over a field)but not isomorphic.

Exercises.

pr.INVISOTALT 68. Inverses in isotopes. Let A be a unital alternative k-algebra and p,q ∈ A×. Show that the invert-ible elements of A and A(p,q) are the same and that, for x ∈ A× = A(p,q)×,

x(−1,p,q) :=U−1pq x−1

is the inverse of x in A(p,q).

pr.ALBISOT 69. Albert isotopies (Albert [3]). Given non-associative k-algebras A,B, an Albert isotopy from Ato B is a triple ( f ,g,h) of k-linear bijections f ,g,h : A→ B such that f (xy) = g(x)h(y) for allx,y ∈ A. Albert isotopies from A to itself are called Albert autotopies of A. They form a subgroupof GL(A)×GL(A)×GL(A), denoted by Atp(A). Now prove Schafer’s isotopy theorem (Schafer[113], McCrimmon [79]). If A,B are k-algebras, with A alternative and B unital, and if ( f ,g,h) isan Albert isotopy from A to B, then A,B are both unital alternative, g−1(1B),h−1(1B) are invertiblein A, and f : A(p,q)→ B, p := h−1(1A)

−1, q := g−1(1A)−1, is an isomorphism.

pr.STRGRALT 70. The structure group of an alternative algebra (Petersson [99]). Let A be a unital alternativek-algebra and Str(A) the set of all triples (p,q,g) composed of elements p,q ∈ A× and an isomor-phism g : A ∼→ A(p,q). Show that Str(A) is a group under the multiplication

(p,q,g)(p′,q′,g′) :=(

p[qg(p′)]p,q[g(q′)p]q,gg′)

(1) STRGRALT

for (p,q,g),(p′,q′,g′) ∈ Str(A) by performing the following steps.(a) Str(A) is closed under the operation (1).(b) For p,q ∈ A×, g ∈ GL(A), the following conditions are equivalent.


(i) (p,q,g) ∈ Str(A).(ii) g(xy)p = g(x)[(pq)

(g(y)p

)] for all x,y ∈ A.

(iii) qg(xy) = [(qg(x)

)(pq)]g(y) for all x,y ∈ A.

(c) The assignment ( f ,g,h) 7→ (h(1A)−1,g(1A)

−1, f ) determines a well defined bijection fromAtp(A), the group of Albert autotopies of A (Exc. 69), onto Str(A that is compatible with mul-tiplications and whose inverse is given by (p,q,g) 7→ (g,Rpg,Lqg).

What is the unit element of Str(A)? What is the inverse of (p,q,g) ∈ Str(A)?

pr.EXTLRMULT 71. Extended left and right multiplications (Petersson [99]). Let A be a unital alternative algebraand u ∈ A×. Prove that

Lu := (u,u−2,Lu), Ru := (u−2,u,Ru)

belong to the structure group of A (Exc. 70). Show further for u,v∈ A× that the following identitieshold in Str(A):

Luvu = LuLvLu, (Lu)−1 = Lu−1 ,

Ruvu = RuRvRu, (Ru)−1 = Ru−1 ,

LuRv = (v−2u,u−1vu−1,LuRv),

RvLu = (v−1u−1v−1,vu−2,RvLu),

LuRu = RuLu.

pr.UNITSTRGRALT 72. The unital structure group (Petersson [99]). Let A be a unital alternative k-algebra. Show thatthe set Str1(A) of all elements (p,q,g) ∈ Str(A) such that g(1A) = 1A (equivalently, p = q−1) is asubgroup of Str(A), called the unital structure group of A. More precisely, show:

(a) Abbreviating the elements of Str1(A) as (p,g) := (p−1, p,g) (so for (p,g), p ∈ A×, g ∈GL(A),to belong to the unital structure group of A it is necessary and sufficient that g : A ∼→ Ap be anisomorphism), its group structure is determined by 1Str1(A) = (1A,1A) and

(p,g)(p′,g′) =(

pg(p′),gg′)

for (p,g),(p′,g′) ∈ Str1(A).(b) Int(p) := (p−3,LpRp−1 ) belongs to the unital structure group of A. Why can Int(p) be viewed

as the alternative version of the inner automorphism affected by an invertible element of a unitalassociative algebra?

pr.UNITARBISOTALT 73. Let A be a unital alternative algebra over k.

(a) Show that arbitrary isotopes of A are canonically isomorphic to unital ones. (Hint. For p,q ∈A×, consider the extended right multiplication by pq (Exc. 71).)

(b) Computing the iterated unital isotope ((Ap)q)r for p,q,r ∈ A× in two different ways seemsto imply (pq)r = up(qr) for some invertible element u ∈ Nuc(A). What’s wrong with thisargument and with this conclusion?

pr.NUCISOTALT 74. Nucleus and centre of an isotope (Petersson [95]). Let A be a unital alternative k-algebra andp,q ∈ A×. Prove Nuc(A(p,q)) = Nuc(A)(pq)−1 and Cent(A(p,q)) = Cent(A)(pq)−1. (Hint. Reduceto the case of unital isotopes.)

17. Isotopy involutions

s.ISINVWe have learned in 11.9 how to twist involutions of unital non-associative algebrasby symmetric or skew-symmetric invertible elements in the nucleus. This procedure

17 Isotopy involutions 97

is useful when the nucleus is big, e.g., for associative algebras, but distinctly less sowhen it is small, e.g., when it agrees with the scalar multiples of the identity element,as happens, for example, in the case of octonions (Exc. ??? below), arguably themost important class of properly alternative algebras. In the present section, wedescribe a way out of this impasse by introducing the notion of isotopy involution foralternative algebras that allows twisting by arbitrary symmetric or skew-symmetricinvertible elements.

Throughout, we let k be an arbitrary commutative ring. We begin with a simplebut useful preparation.

l.TAUPO17.1. Lemma. Let B be a unital alternative k-algebra, p ∈ B× and τ : B→ Bp aunital homomorphism or anti-homomorphism. Then τ preserves U-operators andarbitrary powers: τ Ux = Uτ(x) τ for all x ∈ B and τ(xn) = τ(x)n for all x ∈ B,n ∈ N (resp. x ∈ B×, n ∈ Z).

Proof. By Lemma 16.10, U-operators and powers do not change when passing tothe opposite or a unital isotope of B.

ss.COISINV17.2. The concept of an isotopy involution. Let B be a unital alternative algebraover k and ε = ±. By an isotopy involution of type ε of B we mean a pair (τ, p)satisfying the following conditions.

(i) p ∈ B× and τ(p) = ε p.(ii) τ : B→ Bp is an anti-isomorphism.(iii) τ2 = 1B.

We often speak of isotopy involutions if the type is clear from the context. In explicitterms, condition (ii) is equivalent to τ being a linear bijection satisfying

τ(xy) =(τ(y)p−1)(pτ(x)

)(x,y ∈ B). (1) ISINV

Note that the preceding conditions are compatible in the following sense: supposethe pair (τ, p) satisfies (i), (ii). Then Lemma 17.1 implies τ(p−1) = ε p−1, we obtainisomorphisms τ : B→ (Bp)op, τ : Bop→ Bp, and functoriality 16.11 applied to thelatter yields the second arrow in

Bτ// (Bp)op = (Bop)p−1

τ// (Bp)τ(p−1) = (Bp)p−1

= B.

Thus, regardless of (iii), it follows from (i), (ii) alone that τ2 : B→ B is an isomor-phism, so in the presence of (i), (ii), condition (iii) makes sense.

Trivial examples of isotopy involutions are provided by the observation that theisotopy involutions of an associative algebra B have the form (τ, p), where τ : B→B is an ordinary involution and p ∈ B× is symmetric or skew-symmetric relative toτ . Less trivial examples may be found by consulting the following lemma, whichgives a first indication of twisting in the setting of alternative algebras.

l.TWIALT


17.3. Lemma. Let (B,τ) be an alternative k-algebra with involution and supposeq ∈ B× satisfies τ(q) = εq, for some ε =±. Then (τq,q3) with

τq : B−→ Bq3

, x 7−→ τq(x) := q−1

τ(x)q

is an isotopy involution of type ε of B.

Proof. By Cor. 15.5, the definition of τq is unambiguous, and we have (τq)2 = 1B,τq(q3) = εq3. From Exc. 72 we deduce that Int(q−1) = Lq−1Rq : B→ Bq3

is an

isomorphism, forcing τq := Int(q−1) τ : B→ Bq3to be an anti-isomorphism.

ss.HOIDIS17.4. Homomorphisms and base change. Fix ε =±. By an alternative k-algebrawith isotopy involution of type ε we mean a triple (B,τ, p) consisting of a unitalalternative k-algebra B and an isotopy involution (τ, p) of type ε of B. We also speakof an alternative algebra with isotopy involution of positive (resp. negative) type ifε = + (resp. ε = −). A homomorphism h : (B,τ, p)→ (B′,τ ′, p′) of alternative k-algebras with isotopy involution of type ε is a unital homomorphism h : B→ B′ ofk-algebras that respects the isotopy involutions: τ ′ h = h τ and satisfies h(p) =α p′ for some α ∈ k×. In this way, we obtain the category of alternative k-algebraswith isotopy involution of type ε . If (B,τ, p) is an alternative k-algebra with isotopyinvolution of type ε over k, then (B,τ, p)R := (BR,τR, pR) is one over R, for anyR ∈ k-alg, called the scalar extension or base change of (B,τ, p) from k to R. Notethat (B,τ,1B) is an alternative algebra with isotopy involution if and only if (B,τ)is an alternative algebra with involution.

We now proceed to assemble a few elementary properties of isotopy involutions.For the time being, we fix an alternative algebra (B,τ, p) with isotopy involution oftype ε =± over k.

l.IDTAUIS17.5. Lemma. The following identities hold for all x,y ∈ B and all n ∈ Z.

τ(xpn) = εn pn

τ(x), τ(pnx) = εnτ(x)pn, (1) POP

τ(τ(x)(py)

)= ετ(y)(px), τ

((τ(x)p−1)y)= ε

(τ(y)p−1)x, (2) TAUTAU

τ((xy)p

)= ε pτ(xy) = ε

(pτ(y)

)τ(x), (3) TAUP

τ(

p−1(xy))= ετ(xy)p−1 = ετ(y)

(τ(x)p−1). (4) TAUPI

Proof. Since τ preserves powers by Lemma 17.1, applying (17.2.1) to y = pn (resp.x = pn) yields (1). Using this for n = 1 and (17.2.1) again, we deduce τ(τ(x)(py)) =(τ(py)p−1)(px) = ε((τ(y)p)p−1)(px) = ετ(y)(px), hence the first relation of (2);the second one follows analogously. To derive the first relation in (3), one applies (1)for n = 1, while the second one is a consequence of the Moufang identities (14.3.1),(14.3.3): pτ(xy) = p((τ(y)p−1)(pτ(x))) = [p(τ(y)p−1)p]τ(x) = (pτ(y))τ(x). Re-lation (4) follows in a similar way.

ss.SYSP

17 Isotopy involutions 99

17.6. Symmetric elements. For ε ′ =±, we put

Symε ′(B,τ, p) := x ∈ B | τ(x) = ε′x, (1) SYSEP

and more specifically, as in 11.6,

Sym(B,τ, p) := Sym+(B,τ, p) = x ∈ B | τ(x) = x, (2) SYSP

Skew(B,τ, p) := Sym−(B,τ, p) = x ∈ B | τ(x) =−x. (3) SKESP

They are both submodules of B, and by 17.2 (i) we have p∈ Symε(B,τ, p). Applying(17.5.2) for y = x yields

τ(x)(px),(τ(x)p−1)x ∈ Symε(B,τ, p) (x ∈ B). (4) ELSYSP

By contrast,(τ(x)p)x or τ(x)(p−1x) will in general not belong to Symε(B,τ, p) (seeExc. 76 below).

The twisting of isotopy involutions, which we now proceed to discuss, takes on aslightly different form from what we have seen in the case of ordinary involutions(Lemma 17.3).

p.TWISINV17.7. Proposition. Let ε ′ =± and q ∈ Symε ′(B,τ, p) be invertible in B. Then

(B,τ, p)q := (Bq,τq, pq) (1) TWISINV

with

τq(x) := ε

′q−1τ(qx), pq = pq (x ∈ B) (2) TWIEL

is an alternative k-algebra with isotopy involution of type εε ′, called the q-isotopeof (B,τ, p), such that

τq(xq) = ε

′τ(x)q, (3) ISRQ

Symε ′′(Bq,τq, pq) = q−1 Symε ′ε ′′(B,τ,q) = Symε ′ε ′′(B,τ, p)q, (4) SYSQ

((B,τ, p)q)q′ = (B,τ, p)qq′ (5) TWISIT

for all x ∈ B, ε ′′ =± and all q′ ∈ Symε ′′(Bq,τq, pq) that are invertible in B.

Proof. Starting with (3), we apply Lemma 17.1 to obtain τq(xq) = ε ′q−1τ(qxq) =ε ′q−1τ(Uqx) = ε ′q−1Uτ(q)τ(x) = ε ′q−1[qτ(x)q], and (3) follows. Setting x = p, thisimplies τq(pq) = ε ′τ(p)q = εε ′pq, while applying τq to (3) yields (τq)2(xq) =ε ′τq(τ(x)q) = τ2(x)q = xq, hence (τq)2 = 1Bq . In order to establish (1) as an al-ternative k-algebra with isotopy involution of type εε ′, it therefore suffices to showthat τ : Bq→ (Bq)pq

= Bqpq is an anti-isomorphism, equivalently, that(τ

q(y)(qpq)−1)((qpq)τq(x))= τ

q((xq−1)(qy))

(x,y ∈ B). (6) TAUQAN


In order to do so, we replace y by yq and apply (2), (3), (17.5.3), the Moufangidentities and (17.2.1) to compute(

τq(yq)(qpq)−1)((qpq)τq(x)

)=[((

τ(y)q)q−1)

p−1]q−1 · q

[p(

q(q−1

τ(qx)))]

=(τ(y)p−1)q−1 · q

(pτ(qx)

)= ε

′(τ(y)p−1)q−1 · q

(pτ(x)

)q

= ε′((

τ(y)p−1)(pτ(x)))

q

= ε′τ(xy)q = τ

q((xy)q)

= τq((

xq−1)(q(yq)))

,

and (6) holds. It remains to establish (4), (5).While (4) follows immediately from(2), (3), we note in (5) that (Bq)q′ = Bqq′ by (16.9.2). Also, with p′ := pq it must beborne in mind that the expression (p′)q′ has to be computet not in B but in Bq, so(p′)q′ = ((pq)q−1)(qq′) = p(qq′) = pqq′ . Moreover, by (4), τ(qq′) = ε ′ε ′′qq′, so theright-hand side of (5) makes sense. And finally, (3) gives

(τq)q′(x(qq′))= (τq)q′

(((xq)q−1)(qq′)

)= ε

′′(τ

q(xq)q−1)(qq′)

= ε′ε′′τ(x)(qq′) = τ

qq′(x(qq′))

for all x ∈ B, and the proof of (5) is complete.

c.ENISINV17.8. Corollary. Up to isomorphism, the alternative k-algebras with isotopy invo-lution of type ε =± are precisely of the form

(B,τ,1B)q = (Bq,τq,q),

where (B,τ) is an alternative k-algebra with involution and q ∈ B is invertible sat-isfying τ(q) = εq. Moreover,

τq(x) = q−1

τ(x)q (x ∈ B). (1) NOFOIS

Proof. Let (B′,τ ′, p′) be an alternative k-algebra with isotopy involution and putq := p′. Then Prop. 17.7 implies (B′,τ ′, p′) = (B,τ,1B)

q, with B := (B′)q−1, τ :=

(τ ′)q−1. In particular, (B,τ) is an alternative k-algebra with involution, which com-

bines with (17.7.2) to yield τ ′(x)= τq(x)= εq−1τ(qx)= εq−1τ(x)τ(q)= q−1τ(x)q,hence (1).

Remark. Equation (17.8.1) shows that the notational conventions of Prop. 17.7 andCor. 17.8 are compatible with the ones of Lemma 17.3.

Exercises.

18 Finite-dimensional alternative algebras over fields. 101

pr.CAPI 75. Let k be a commutative ring and ε =±. Show that the category of alternative k-algebras withisotopy involution of type ε as defined in 11.2 is isomorphic to the category of ε-pointed (orsimply pointed) alternative k-algebras with involution, which we define as follows: its objectsare the pairs ((B,τ),q) where (B,τ) is an alternative k-algebra with involution and q ∈ B× (thebase point) satisfies τ(q) = εq, while its morphisms are homomorphisms of alternative algebraswith involution preserving base points up to an invertible scalar factor: for ε-pointed alternativek-algebras ((B,τ), p), ((B′,τ ′), p′) with involution, a morphism h : ((B,τ), p)→ ((B′,τ ′), p′) is ahomomorphism h : (B,τ)→ (B′,τ ′) of alternative k-algebras with involution such that h(p) = α p′

for some α ∈ k×.

pr.SWISINV 76. Let A be a unital alternative k-algebra and q ∈ A×. Show that (Aop⊕Aq,εA, p), where εA is theswitch (11.4) and p := q−1⊕ q−1, is an alternative k-algebra with isotopy involution of type +.Show further that (εA(x)p)x ∈ Sym(Aop⊕Aq,εA, p) if and only if q2 ∈ Nuc(A).

18. Finite-dimensional alternative algebras over fields.

s.FIDALOur sole purpose in this short section will be to recall the basic results of the struc-ture theory for finite-dimensional alternative algebras over an arbitrary field. Sincethis theory, mostly due to Zorn [128]–[130] is well documented in book form (see,e.g., Schafer [114, Chap. III]), proofs in the present account will be mostly omitted.

Throughout we let k be a commutative ring and F a field of arbitrary characte-ristic.

ss.NILPA

18.1. Nilpotent algebras. Let A be a non-associative algebra over k. For a positiveinteger n, we write

An := ZMonn(A) (1) ENPO

for the additive subgroup of A generated by all products, in some association, of nfactors in A. By induction on n, one sees easily that

A = A1 ⊇ A2 ⊇ ·· · ⊇ An ⊇ An+1 ⊇ ·· · (2) ENDES

forms a descending chain of ideals in A. The algebra A is said to be nilpotent ifAn = 0 for some integer n ≥ 1, equivalently, if some positive integer n has theproperty that all products of n factors in A, no matter how associated, are equal tozero.

ss.NIRA

18.2. The nil radical. Let A be an alternative algebra over k. Since A is power-associative (Cor. 15.6), an element x ∈ A is nilpotent in the sense of Exc. 30 if andonly if xn = 0 for some integer n ≥ 0. Recall from loc. cit. that the nil radical of A,denoted by Nil(A) and defined as the sum of all nil ideals in A, is a nil ideal itself,hence the biggest nil ideal in A. Moreover, the nil radical of A/Nil(A) is zero.

ss.SEMS18.3. Semi-simplicity. Let A be a finite-dimensional alternative algebra over thefield F . We say that A is semi-simple if Nil(A) = 0.


18.4. Theorem (Schafer [114, Thm. 3.2]). The nil radical of a finite-dimensionalt.NILNPalternative algebra over F is nilpotent.

18.5. Theorem (Schafer [114, Thm. 3.10]). A finite-dimensional semi-simple al-t.SSIDternative F-algebra contains an identity element.

18.6. Theorem (Schafer [114, Thm. 3.12]). A finite-dimensional alternative alge-t.SSIMPbra A over F is semi-simple if and only if there is a decomposition

A = A1⊕·· ·⊕Ar

of A as a finite direct sum of simple ideals. In this case, the decomposition is uniqueup to isomorphism and up to the order of the summands.

18.7. Theorem (Schafer [114, Thm. 3.17]). A finite-dimensional alternative alge-t.CESIMbra over F is central simple if and only if it is either isomorphic to Matn(D) forsome positive integer n and some finite-dimensional central associative division al-gebra D over F, or it is an octonion algebra over F as defined in 22.20 below. Inthe former case, n is unique and D is unique up to isomorphism.

ss.PRONI18.8. Properly nilpotent elements. Along the way towards proving the preced-ing structure theorems, it turns out to be convenient to take advantage of properlynilpotent elements: given an alternative algebra A over k, an element x ∈ A is saidto be properly nilpotent if xu ∈ A is nilpotent for all u ∈ A. This is equivalent toux ∈ A being nilpotent for all u ∈ A since (ux)n+1 = u(xu)nx for all positive integersn, which follows from Artin’s theorem (Cor. 15.5). The most important technicalresults pertaining to the notion of properly nilpotent elements are as follows.

l.ALPRI18.9. Lemma. Let A be a unital alternative algebra of finite dimension over F andsuppose there are no idempotents in A other than 0 and 1A. Then every element ofA is either invertible or nilpotent. Moreover, the set of nilpotent elements in A is anideal.

Proof. The first assertion holds by [114, Lemma 3.5], which als shows that the setof properly nilpotent elements in A is an ideal. Hence it suffices to show that everynilpotent element is properly nilpotent. Otherwise, there exist x,y ∈ A such that x isnilpotent and xy is invertible. Let n be the least positive integer satisfying xn = 0.Then xn−1(xy) = xny = 0 implies xn−1 = 0, a contradiction.

18.10. Theorem (Schafer [114, Thm. 3.7]). The nil radical of a finite-dimensionalt.NIRAPROalternative algebra over a field agrees with the set of its properly nilpotent elements.

18.11. Corollary (Schafer [114, Cor. 3.8]). Let A be a finite-dimensional alterna-c.NIRAIDtive algebra over F and c ∈ A an idempotent. Then Nil(Aii(c)) = (Nil(A))∩Aii(c)for i = 1,2.

Chapter 4Composition algebras

c.COMAL

We have seen in Thm. 1.7 that the algebra of Graves-Cayley octonions as definedin 1.4 carries a positive definite real quadratic form that permits composition. In thepresent chapter we will study a class of non-associative algebras over an arbitrarycommutative ring for which such a property is characteristic. Many results we de-rived for the Graves-Cayley octonions in the first chapter will resurface here underfar more general circumstances, and with much more natural proofs attached.

19. Conic algebras

s.CONALBy (1.5 .10), every element x in the algebra of Graves-Cayley octonions satisfies aquadratic equation which is universal in the sense that its coefficients depend “al-gebraically” on x. This innocuous but useful property gives rise to the notion of aconic algebra that will be studied in the present section.

The term “conic algebra” made its first appearance in a paper by Loos [72] andderives its justification from the fact that conic algebras, just like ordinary conics ascurves in the (affine or projective) plane, are intimately tied up with quadratic equa-tions; for a more sophisticated motivation of this term, see 19.3 below. In derivingthe main properties of conic algebras, we adhere rather closely to the treatment ofMcCrimmon [82], who calls them degree 2 algebras, while other authors speak ofquadratic algebras in this context. By contrast, the term “quadratic algebra” will beused here in a much more restrictive sense, as in Knus [58, I, (1.3.6)].

Throughout we let k be an arbitrary commutative ring.ss.COCO

19.1. The concept of a conic algebra. By a conic algebra over k we mean a unitalk-algebra C together with a quadratic form nC : C→ k such that

nC(1C) = 1, x2−nC(1C,x)x+nC(x)1C = 0 (x ∈C). (1) COCO

We call nC the norm of C. Most of the time, we will just speak of a conic algebra Cover k, its norm nC being understood1.

Let C be a conic algebra over k. Viewing C merely as a k-module, (C,nC,1C) is apointed quadratic module over k with norm nC, base point 1C and trace tC given bytC(x) = nC(1C,x) for all x ∈C, in the present context called the trace of C.

1 It follows from Exc. 86 below that the algebra C and condition(1) do not determine the quadraticform nC uniquely. In spite of this, we feel justified in phrasing our definition, as well as similarones later on, in this slightly informal manner because it is convenient and there is no danger ofconfusion.

103

104 4 Composition algebras

The linear map tC : C→ k defined by tC(x) = nC(1C,x) for x ∈ C is called the(linear) trace of C, while we refer to ιC : C→ C, x 7→ x := tC(x)1C − x (which islinear as well and of period two) as the conjugation of C. We also denote by tcthe bilinear trace of C, i.e., the bilinear form C×C → k, (x,y) 7→ tC(xy), whichin general is not symmetric. Given a submodule M ⊆ C, we always write M⊥ =x ∈C | nC(x,M) = 0 for the orthogonal complement of M in C relative to thepolarized norm.

Let C′ be another conic algebra over k. By a homomorphism h : C→C′ of conicalgebras we mean a unital k-algebra homomorphism which preserves norms in thesense that nC′ h = nC. It is clear that homomorphisms of conic algebras also pre-serve (linear as well as bilinear) traces and conjugations. If B ⊆ C is a unital sub-algebra, it may and always will be regarded as a conic algebra in its own right bydefining its norm nB := nC|B as the restriction of the norm of C to B; in this way, theinclusion B →C becomes a homomorphism of conic algebras.

Conic algebras are clearly invariant under base change. If C is a conic algebraover k, then so is Cop, with the same norm, linear trace and conjugation as C, whilethe bilinear trace changes in the obvious way to tCop(x,y) := tC(y,x).

ss.MOTEX19.2. Examples of conic algebras. Examples of conic algebras exist in abundance.Aside from the Graves-Cayley octonions, the Hamiltonian quaternions and, moregenerally, any unital subalgebra of O, the following examples seem to be particu-larly noteworthy.(a) The base ring k itself, with norm nk : k→ k given by the squaring: nk(α) = α2

for α ∈ k. We have tk(α) = 2α , and the conjugation of k is the identity.(b) R= k⊕k (as a direct sum of ideals), with norm nR : R→ k given by nR(α⊕β ) =αβ for α,β ∈ k. We have tR(α ⊕ β ) = α + β , and the conjugation of R is the“switch”: α⊕β = β ⊕α . In particular, the quadratic module (R,nR) is the splithyperbolic plane (12.17).(c) R= k[ε],ε2 = 0, the k-algebra of dual numbers, with norm, trace, conjugation re-spectively given by nR(α1R+βε)=α2, tR(α1R+βε)= 2α , α1R +βε =α1R−βε

for α,β ∈ k.(d) R = K, where k is a field and K/k is a quadratic field extension. Then R is aconic k-algebra, with nR = NK/k (resp. tR = TK/k) the field norm (resp. the fieldtrace) of K/k. Moreover, ιR is the non-trivial Galois automorphism of K/k if K/k isseparable, and the identity otherwise.

ss.MOT19.3. Motovation. Let k be a field and X ⊆ P2

k a smooth conic in the plane, givenby a separable quadratic form in three variables. Then one checks easily that thescheme-theoretic intersection (cf. [36, I , 4.4]) of X with appropriate lines in P2

k hasthe form Spec(R) (viewed as an affine scheme), where R is one of the algebras listedin 19.2 (b),(c),(d) above. This gives another motivation of the term “conic algebra”.

19 Conic algebras 105

19.4. Quadratic algebras.2 In analogy to Knus [58, I, (1.3.6)], an algebra R over kss.QUALGis said to be quadratic if it contains an identity element and is projective of rank 2 asa k-module. In this case, we claim that 1R ∈ R is unimodular, R is commutative asso-ciative and its conjugation is an involution, i.e., a k-automorphism of period two. Asto the first part, we note that R(p), for any p ∈ Spec(k), is a unital two-dimensionalκ(p)-algebra, forcing 1R(p) 6= 0. Hence 1R is unimodular by Lemma 10.13. Sincethe remaining assertions are local on k, we may assume that k is a local ring, makingR a free k-module of rank 2. Extending 1R to a basis of R, which we are allowed todo by unimodularity, the commutative and the associative law for the multiplicationof R trivially hold, as does the property of the conjugation to be a k-automorphismof period two. Thus our assertion is proved.

It now follows from the Hamilton-Cayley theorem that a quadratic k-algebra Ris conic, with norm and trace given by

nR(x) = det(Lx), tR(x) = tr(Lx) (x ∈ R). (1) QUAMU

ss.BASID19.5. Basic identities. Let C be a conic algebra over k. The following identitieseither hold by definition or are straightforward to check.

x2 = tC(x)x−nC(x)1C, (1) CAQUAD

tC(x) = nC(1C,x), (2) CATN

nC(1C) = 1, tC(1C) = 2, (3) CAA1

x = tC(x)1C− x, 1C = 1C, x = x, (4) CACONJ

x y : = xy+ yx = tC(x)y+ tC(y)x−nC(x,y)1C, (5) CAQUADL

xx = nC(x)1C = xx, x+ x = tC(x)1C, (6) CACONJC

nC(x) = nC(x), tC(x) = tC(x), (7) CACONJO

tC(x2) = tC(x)2−2nC(x), (8) CATQUAD

tC(x y) = tC(x,y)+ tC(y,x) = 2[tC(x)tC(y)−nC(x,y)], (9) CABILT

nC(x, y) = tC(x)tC(y)−nC(x,y), (10) CANOCO

xy− yx =(tC(x,y)−nC(x, y)

)1C. (11) CACONJM

By (1) and 8.7, k[x] agrees with the submodule of C spanned by 1C and x; it is a com-mutative associative subalgebra. In particular, conic algebras are power-associative.

Before we can proceed, we require two short digressions.ss.UMORE

19.6. Unimodular elements revisited. For several of our subsequent results it willbe important to know that 1C ∈ C is unimodular. While this is not always true, itdoes hold automatically if the linear trace of C is surjective, since this is equivalentto tC(x) = 1 for some x ∈C and so the linear form λ : C→ k, y 7→ tC(xy), satisfiesλ (1C) = 1.

2 I am indebted to Erhard Neher, who discovered a faulty argument in the original presentation ofthis subsection.


For example, if 12 ∈ k, then tC(x) = 1 for x = 1

2 ·1C and 1C is unimodular. More-over, we have

C = k1C⊕C0 (1) TRAZE

as a direct sum of submodules, where

C0 = Ker(tC) = x ∈C | x =−x, (2) PURCO

and

H(C, ιC) = k1C. (3) HACE

Another important condition that ensures unimodularity of 1C will be stated sepa-rately.

p.PROUNIM19.7. Proposition. Let C be a conic algebra over k which is projective as a k-module. Then 1C ∈C is unimodular and Cp 6= 0 for all prime ideals p⊆ k.

Proof. The first part follows immediately from Lemma 12.14 (applied to the pointedquadratic module (C,nC,1C)), which in turn implies the second, by Lemma 10.13.

ss.TRICO

19.8. Trivial conjugation. A conic algebra C over k is said to have trivial con-jugation if ιC = 1C. Having trivial conjugation is a rather exotic phenomenon. Forexample, if 1

2 ∈ k, then (19.6.109), (19.6.2) show that C does not have trivial con-jugation unless C ∼= k. More generally, the same conclusion holds if tC is surjective(Exc. 85). On the other hand, if 2 = 0 in k and tC = 0 (e.g., if k is a field and C aninseparable quadratic field extension of k), then C does have trivial conjugation.

In subsequent applications, conic algebras which are flexible (i.e., satisfy the iden-tity (xy)x = x(yx)) or whose conjugation is an (algebra) involution play an importantrole. These two properties will now be related to one another in various ways.

p.CHAFL19.9. Proposition. Let C be a conic algebra over k.(a) The conjugation of C is an involution if and only if

tC(x,y)−nC(x, y) ∈ Ann(C)

for all x,y ∈C.(b) C is flexible if and only if(

tC(x,y)−nC(x, y))x =

(nC(x,xy)− tC(y)nC(x)

)1C (1) CONFL

for all x,y ∈C.

Proof. (a) follows immediately from (19.5.11). In (b) one simply notes (xy)x−x(yx) = (xy) x− x(x y) and expands the right-hand side using (19.5.5), (19.5.1).


c.INVFL

19.10. Corollary. Let C be a conic k-algebra whose conjugation is an involution.Then C is flexible if and only if nC(x,xy)− tC(y)nC(x) ∈ Ann(C) for all x,y ∈C.

p.NORAS19.11. Proposition. The following (collections of) identities are equivalent in anyconic algebra C over k.

nC(x,yx) = tC(y)nC(x), (1) WRBILCOMP

nC(x,xy) = tC(y)nC(x), (2) WLBILCOMP

nC(xy,z) = nC(x,zy), (3) RASSNOR

nC(xy,z) = nC(y, xz), (4) LASSNOR

tC(x,y) = nC(x, y), tC(xy,z) = tC(x,yz). (5) ASSBILT

Proof. Since (2) (resp. (4)) is (1) (resp. (3)) in Cop, the equivaelnce of (1)–(5) willfollow once we have established the following implications.

(1)⇔ (2). Combine (19.5.2) with (19.5.5) to conclude nC(x,xy) = 2tC(y)nC(x).(2)⇒ (3). Linearize (2) and apply (19.5.4).(4)⇒ (2). Put z = x in (4) and apply (19.5.6).(3)⇒ (5). Put z = 1C in (3) and apply (19.5.2) to deduce the first two equations

of (5). But then, by Prop. 19.9 (a), ιC is an involution of C, yielding the final state-ment of the proposition, while the last equation of (5) follows by a straightforwardcomputation.

(5)⇒ (1). By Prop. 19.9 (a) and the first equation of (5), ιC is an involution of C,which implies nC(x,yx) = tC(x, xy) = tC(xx, y) by (5), and (1) follows from (19.5.6),(19.5.7).

ss.NORAS19.12. Norm-associative conic algebras. A conic algebra satisfying one (henceall) of the identities (19.11.1) – (19.11.5) is said to be norm-associative. If C is anorm-associative conic algebra, then (19.11.5) combined with (19.5.10) shows thatthe bilinear trace of C is an associative bilinear form, so

tC(xy) = tC(yx), tC((xy)z

)= tC

(x(yz)

)=: tC(xyz). (1) ADMASSBILT

Norm-associativity is clearly invariant under scalar extensions. Also, if C is a norm-associative conic algebra, then so is Cop.

p.NAFL19.13. Proposition. Let C be a norm-associative conic algebra over k. Then Cis flexible and its conjugation is an involution, called the canonical involution ofC. Conversely, suppose C is a flexible conic algebra and projective as a k-module.Then C is norm-associative.

Proof. If C is a norm-associative conic k-algebra, then by (19.11.5) and (19.11.2)combined with Prop. 19.9 (a), C is flexible and its conjugation is an involution.Conversely, suppose C is flexible and projective as a k-module. Given y ∈ C, it


suffices to show

tC(x,y) = nC(x, y) (x ∈C) (1) TECNEC

since (19.9.1) and unimodularity of 1C (Prop. 19.7) then imply (19.11.2). To estab-lish (1), we may assume that k is a local ring and extend e0 := 1C to a basis (ei)of the k-module C. Then (1) holds for x = e0 by (19.5.7) and for x = ei, i 6= 0, by(19.9.1), hence on all of C by linearity.

c.CHORAS19.14. Corollary. For a faithful conic algebra over k to be norm-associative it isnecessary and sufficient that it be flexible and its conjugation be an involution.

Proof. By Prop. 19.13, the condition is clearly necessary. Conversely, suppose Cis a flexible faithful conic k-algebra whose conjugation is an involution. CombiningProp 19.9 with faithfulness, we see that (19.11.2) holds, so C is norm-associative.

p.UNINOR19.15. Proposition. Let C be a conic algebra over k. If C is projective as a k-module, then the norm of C is uniquely determined by the algebra structure of C.

Proof. Let n : C→ k be any quadratic form making C a conic algebra and write tfor the corresponding linear trace. Then λ := tC− t (resp. q := nC− n) is a linear(resp. a quadratic) form on C, and (19.5.1) yields

λ (x)x = q(x)1C, (1) DITRNO

for all x∈C. We have to prove λ = q = 0. Since 1C ∈C is unimodular by Porp. 19.7,it suffices to show λ = 0. This can be checked locally, so we may assume that k isa local ring, allowing us to extend e0 := 1C to a basis (ei) of C as a k-module. Butλ (e0) = 0 by (19.5.3). Hence it remains to show λ (ei) = 0 for all i 6= 0, whichfollows immediately from (1).

Exercises.

pr.NILRADCON 77. Let C be a conic algebra over k. Prove

(a) nC(yz) = nC(y)nC(z) for all y,z ∈ k[x].(b) x is invertible in (the commutative associative k-algebra) k[x] if and only if nC(x) is a unit in

k, in which casex−1 = n(x)−1x.

(c) x is a nilpotent element of C (Exc. 30) if and only if tC(x) and nC(x) are nilpotent elementsof k.

pr.CONALGHOMNOR 78. Let C,C′ be conic k-algebras and suppose C is projective as k-modules. Prove that every in-jective homomorphism ϕ : C→ C′ of unital k-algebras is, in fact, one of conic algebras, i.e., itpreserves norms (hence traces and conjugations as well). Does this conclusion continue to holdwithout the hypothesis of ϕ being injective?

pr.CONMO 79. Co-ordinates for conic algebras (Loos [72]). Let k be a commutative ring, X a k-module, e∈ Xa unimodular element and λ : X → k a linear form such that λ (e) = 1. Putting Mλ := Ker(λ ), wethen have X = ke⊕Mλ as a direct sum of submodules.


(a) Let (T,B,K) be a conic co-ordinates system relative to (X ,e,λ ) in the sense that T : Mλ → kis a linear form, B : Mλ ×Mλ → k is a (possibly non-symmetric) bilinear form and

K : Mλ ×Mλ −→Mλ , (x,y) 7−→ x× y,

is an alternating bilinear map. Define a k-algebra structure C :=Con(T,B,K)=CX ,e,λ (T,B,K)on X by the multiplication

(αe+ x)(βe+ y) :=(αβ −B(x,y)

)e+((α +T (x))y+βx− x× y

)for α,β ∈ k, x,y ∈Mλ . Show that C is a conic k-algebra with identity element e and normnC : C→ k given by

nC(αe+ x) := α2 +αT (x)+B(x,x)

for α ∈ k and x ∈Mλ .(b) Conversely, suppose C is a conic k-algebra with underlying k-module X and identity element

1C = e. Show that (TC,BC,KC), where T = TC : Mλ → k, B = BC : Mλ ×Mλ → k and K =KC : Mλ ×Mλ →Mλ , (x,y) 7→ x× y are defined by

T (x) := tC(x), B(x,y) :=−λ (xy), x× y := tC(x)y− xy+λ (xy) (x,y ∈Mλ ),

is a conic co-ordinate system for (X ,e,λ ). Show further that the assignments (T,B,K) 7→Con(T,B,K) and C 7→ (TC,BC,KC) define inverse bijections between the set of conic co-ordinate system for (X ,e,λ ) and the set of conic k-algebras with underlying k-module X andidentity element e.

Remarks. (a) It is sometimes convenient to define conic co-ordinate systems on a k-module thatis independent of the choice of λ , namely, on X/ke. This point of view, systematically adopted byLoos [72], turns out to be particularly useful when analyzing the question of how conic co-ordinatesystems change with λ .(b) Let C be a conic k-algebra, X = C as a k-module and e := 1C . If 2 ∈ k is a unit, then conicco-ordinate systems may be taken relative to (X ,e,λ ) with λ := 1

2 tC , in which case the conicco-ordinate system corresponding to C is already in Osborn [91, Thm. 1].

pr.CONFIELD 80. (Dickson [23]) Let k be a field of characteristic not 2 and C a unital algebra over k. Show thatC is conic if and only if 1C,x,x2 are linearly dependent over k for all x ∈C. (Hint. If this conditionis fulfilled, linearize the expression 1C ∧x∧x2 ∈

∧3 C to show that 0 and the elements u ∈C−k1Csatisfying u2 ∈ k1C form a vector subspace of C.)

pr.DICON 81. The Dickson condition Generalize Exc. 80 in the following way: let C be a unital algebra overk which is either free or finitely generated projective as a k-module and whose identity element isunimodular. Show that there exists a quadratic form nC : C→ k making C a conic k-algebra if andonly if C satisfies the Dickson condition: For all R ∈ k-alg and all x ∈CR, the element x2 ∈CR isan R-linear combination of x and 1CR .

pr.IDEMP 82. Elementary idempotents. (a) Let C be a conic k-algebra. Show that, if k 6= 0 is connected(so 0,1 are the only idempotents of k, equivalently by Exc. 41, the topological space Spec(k) isconnected), an element c ∈ C is an idempotent 6= 0,1C if and only if nC(c) = 0 and tC(c) = 1.Conclude that, for any commutative ring k and any element c ∈ C, the following conditions areequivalent.

(i) c is an idempotent satisfying cR 6= 0,1CR for all R ∈ k-alg, R 6= 0.(ii) c is an idempotent satisfying cp 6= 0,1Cp for all prime ideals p⊆ k.(iii) nC(c) = 0, tC(c) = 1.(iv) c is an idempotent and the elements c,1C− c are unimodular.


If these conditions are fulfilled, we call c an elementary idempotent of C.

pr.LILID 83. Let C be a conic k-algebra and suppose I ⊆ k is a nil ideal. Write α 7→ α , x 7→ x for thecanonical projection k→ k := k/I,

C −→ C :=C⊗ k =C/IC,

respectively, and show that any elementary idempotent c′ ∈ C can be lifted to an elementary idem-potent c ∈C satisfying c = c′.

pr.DECID 84. Let C be a conic algebra over k. Prove for c ∈C that the following conditions are equivalent.

(i) c is an idempotent in C.(ii) There exists a complete orthogonal system (ε(0),ε(1),ε(2)) of idempotents in k, giving rise

to decompositions

k = k(0)⊕ k(1)⊕ k(2), C =C(0)⊕C(1)⊕C(2)

as direct sums of ideals, where k(i) = kε(i) and C(i) = ε(i)C = Ck(i) as a conic algebra overk(i) for i = 0,1,2, such that

c = 0⊕ c(1)⊕1C(2) ,

where c(1) is an elementary idempotent of C(1).

In this case, the idempotents ε(i), i = 0,1,2, in (ii) are unique and given by

ε(0) :=

(1−nC(c)

)(1− tC(c)

), ε

(1) :=(1−nC(c)

)tC(c), ε

(2) := nC(c). (2) VEPI

pr.TRICO 85. Let C be a conic k-algebra that is faithful as a k-module and whose linear trace is surjective.Show that C has trivial conjugation if and only if C ∼= k.

20. Conic alternative algebras

s.COALTIn 5.3, we have defined the euclidean Albert algebra as a commutative non-associativereal algebra that lives on the 3× 3 hermitian matrices with entries in the Graves-Cayley octonions. As we will show in due course, this important construction canbe generalized to arbitrary conic alternative algebras over commutative rings once apeculiar additional hypothesis has been inserted. The elementary properties of conicalternative algebras needed to carry out this generalization will be assembled in thepresent section. Throughout we let k be an arbitrary commutative ring.

We begin by identifying the “peculiar additional hypothesis” alluded to above asfollows.

ss.MULCO

20.1. Multiplicative conic algebras. A conic algebra C over k is said to be multi-plicative if its norm permits composition:

nC(xy) = nC(x)nC(y) (x,y ∈C) (1) MUPL

Linearizing this identity repeatedly, we conclude that multiplicative conic algebrasalso satisfy the relations

20 Conic alternative algebras 111

nC(x1y,x2y) = nC(x1,x2)nC(y), (2) MUPLL

nC(xy1,xy2) = nC(x)nC(y1,y2), (3) MUPLR

nC(x1y1,x2y2)+nC(x1y2,x2y1) = nC(x1,x2)nC(y1,y2) (4) MUPLT

for all x,x1,x2,y,y1,y2 ∈C. We conclude from this that multiplicative conic algebrasare stable under base change and that, if C is a multiplicative conic algebra, then sois Cop.

Putting y1 := 1C, y2 := y in (3), we see that multiplicative conic algebras satisfythe identity (19.11.2). Therefore Prop. 19.13 immediately implies the first part ofthe following observation, while the second part is a consequence of (19.4.1) .

p.MULAD20.2. Proposition. (a) Multiplicative conic algebras are norm-associative. In par-ticular, they are flexible, and their conjugation is an involution.(b) Quadratic algebras are multiplicative.

ss.IDCO20.3. Identities in conic alternative algebras. Let C be a conic alternative algebraover k. Combining the left and right alternative laws, x(xy) = x2y and (yx)x = yx2,with (19.5.1), (19.5.4), we deduce Kirmse’s identities [55, p. 67]

x(xy) = nC(x)y = (yx)x. (1) KID

We can also derive a formula for the U-operator,

Uxy = xyx = nC(x, y)x−nC(x)y, (2) QAUOP

which follows by using (19.5.1), (19.5.5), (19.5.4) to manipulate the expressionxyx = (x y)x− yx2, see Exc. 3 (c) in the special case k = R, C = O. Applying thenorm to the right-hand side of (2) and expanding, we conclude

nC(Uxy) = nC(xyx) = nC(x)2nC(y). (3) NOUP

In view of this, one might be tempted to conjecture that conic alternative algebras aremultiplicative. But, according to Exc. 86 below, this is not so. Fortunately, however,(3) is strong enough to characterize invertibility in conic alternative algebras.

p.CHIN20.4. Proposition. Let C be a conic alternative algebra over k. An element x∈C isinvertible in C if and only if nC(x) ∈ k is invertible in k. In this case, x−1 = nC(x)−1xand nC(x−1) = nC(x)−1.

Proof. If nC(x) ∈ k×, then Kirmse’s identities (20.3.1) show that y := nC(x)−1x sa-tisfies xy = 1C = yx, forcing x ∈C× and y = x−1 by Prop. 14.6. Conversely, supposex is invertible in C. Then Uxx−2 = 1C, and (20.3.3) yields 1 = nC(x)2nC(x−2), hencenC(x) ∈ k×. The final formula follows from the fact that the conjugation of C leavesits norm invariant.


While it is not true general that conic alternative algebras are multiplicative, thisimplication does hold under natural conditions on the module structure.

p.ALMU20.5. Proposition. Let C be a conic alternative algebra over k which is projectiveas a k-module. Then C is multiplicative.

Proof. As an alternative algebra, C is flexible, whence Prop. 19.13 implies that Cis norm-associative. In particular, the conjugation of C is an involution. Let x,y ∈C. By Artin’s Theorem (Cor. 15.5), the unital subalgebra of C generated by x,y isassociative. Moreover, by (19.5.4), it contains x and y. Hence (19.5.6) yields

nC(xy)1C = xyxy = xyyx = nC(y)xx = nC(x)nC(y)1C,

and since 1C is unimodular, C is indeed multiplicative.

Exercises.

pr.NONUNNOR 86. (McCrimmon [82]) The following exercise will produce an example of an algebra that supportsmany different conic algebra structures, some of which are multiplicative (resp. norm-associative),while others are not.

Let k = k0[ε], ε2 = 0, be the algebra of dual numbers over a commutative ring k0, and view k0as a k-algebra via the natural map k→ k0, ε 7→ 0, the corresponding module action k× k0 → k0being indicated by (α,α0) 7→ α.α0. In particular, (column) 3-space k3

0 becomes a k-module in thisway. In fact, it becomes a k-algebra under the multiplicationα1

α2α3

β1β2β3

:=

00

α1β2

(αi,βi ∈ k0, 1≤ i≤ 3). (1) ETHR

Furthermore, let

z =

δ1δ2δ3

∈ k30

be arbitrary. Now consider the k-algebra C defined on the k-module k⊕ k30 by the multiplication

(α1⊕u1)(α2⊕u2) := α1α2⊕(α1.u2 +α2.u1 +[u1,u2]

)for αi ∈ k, ui ∈ k3

0 , i = 1,2, where [u1,u2] is the commutator belonging to the k-algebra structureof k3

0 just defined, and let nz : C→ k be given by

nz(α⊕u) := α2 +(α.(zt u)

)ε

for α ∈ k, u ∈ k30 . Then show that C is an associative conic k-algebra with norm nC := nz, and that

the following conditions are equivalent.

(i) C is multiplicative.(ii) C is norm-associative.(iii) The conjugation of C is an involution.(iv) δ3 = 0.

pr.NIRA 87. Let C be a multiplicative conic algebra over k. Show that its nil radical has the form

Nil(C) = x ∈C|nC(x),nC(x,y) ∈ Nil(k) for all y ∈C.

21 The Cayley-Dickson construction 113

pr.ARTCONALG 88. Artin’s theorem for conic alternative algebras. Let C be a conic alternative k-algebra that isunitally generated by two elements x,y ∈C. Show that C is spanned by 1C,x,y,xy as a k-module.Conclude without recourse to Artin’s theorem (Cor. 15.5) that C is associative.

pr.NORMA 89. Norms of associators. (Garibaldi-Petersson [34]) Let C be a multiplicative conic alternativealgebra over k. Prove

nC([x1,x2,x3]) = 4nC(x1)nC(x2)nC(x3)−∑ tC(xi)2nC(x j)nC(xl)

+∑ tC(xix j)nC(xi,x j)nC(xl)− tC(x1x2)tC(x2x3)tC(x3x1)

+ tC(x1x2x3)tC(x2x1x3)

for all x1,x2,x3 ∈C, where both summations on the right are to be taken over the cyclic permuta-tions (i jl) of (123).

pr.ISTQALT 90. Isotopes of conic alternative algebras. Let C be a conic alternative k-algebra. Prove for p,q ∈C× that C(p,q) is again a conic alternative k-algebra with norm nC(p,q) = nC(pq)nC . Moreover, traceand conjugation of C(p,q) are given by

tC(p,q) (x) = nC(pq,x), ιC(p,q) (x) = x(p,q) = nC(pq)−1 pq x pq (x ∈C).

Show further that, if in addition, C is multiplicative, then so is C(p,q).

pr.CONIS 91. Let C be a multiplicative conic alternative algebra over k and suppose q ∈C× has trace zero.Show that (C,τ,q) with τ : C→ C defined by τ(x) := q−1xq for x ∈ C is an alternative algebrawith isotopy involution of negative type over k and describe the pointed alternative k-algebra withinvolution corresponding to (C,τ,q) via Exc. 75.

21. The Cayley-Dickson construction

s.CADIThe only example we have encountered so far of an alternative algebra which isnot associative is the real algebra of Graves Cayley octonions. The Cayley Dicksonconstruction will provide us with a tool to accomplish the same over any commuta-tive ring. Moreover, it will play a crucial role in the structure theory of compositionalgebras over fields later on.

Throughout this section, we fix an arbitrary commutative ring k. Our first aimwill be to present what might be called an internal version of the Cayley-Dicksonconstruction.

21.1. Proposition. (Internal Cayley-Dickson construction) Let C be a multiplica-p.INTCDtive conic alternative algebra over k and B ⊆ C a unital subalgebra. If l ∈ C isperpendicular to B relative to DnC, then B+Bl ⊆C is the subalgebra of C gener-ated by B and l. Moreover, setting µ :=−nC(l), the identities


u(vl) = (vu)l, (1) UVEL

(vl)u = (vu)l, (2) VELU

(ul)(vl) = µ vu, (3) ULVEL

nC(u+ vl) = nC(u)−µnC(v), (4) NOEL

tC(u+ vl = tC(u), (5) TREL

u+ vl = u− vl (6) COEL

hold for all u,v ∈ B, and

Bl = lB⊆ B⊥. (7) BEEL

Proof. The first assertion follows from (1)–(3), so it will be enough to establish (1)–(7). Since l is orthogonal to B and, in particular, has trace zero, (19.5.5) implies

u l = tC(u)l, (8) VELT

hence lu = tC(u)l − ul, and we have (2) in the special case v := 1C. Combin-ing this with the linearized left alternative law (14.2.1) and (8), we concludeu(vl) = u(lv) = (u l)v− l(uv) = tC(u)lv− l(uv) = l(u v) = (vu)l, since ιC is aninvolution by Prop. 20.2 (a). Thus (1) holds. Reading (1) in Cop and invoking (8)once more gives (2) in full generality. In order to establish (3), we combine themiddle Moufang identity (14.3.3) with (20.3.2) to conclude (ul)(vl) = (lu)(vl) =l(uv)l = nC(l, vu)l−nC(l)vu = µ vu, as claimed. Turning to (4), we use (19.11.4) toexpand the left-hand side and obtain nC(u+ vl) = nC(u)+nC(u,vl)+nC(v)nC(l) =nC(u)+ nC(vu, l)− µnC(v) = nC(u)− µnC(v), and the proof of (4) is complete. Itis now straightforward to verify (5), (6). Finally, turning to (7), we have Bl = lB by(2) for v = 1B and note that C is norm-associative by Prop. 20.2 (a). Hence (19.11.4)yields nC(u,vl) = nC(vu, l) = 0 for all u,v∈B, which implies Bl ⊆B⊥ and (7) holds.

Remark. The above sum B+Bl of k-submodules of C need not be direct. For ex-ample, it could happen that 0 6= l ∈ Rad(DnB)⊆ B.

ss.EXTCD21.2. The external Cayley-Dickson construction. We will now recast the pre-ceding considerations on a more abstract, but also more general, level. Let B be anyconic algebra over k and µ ∈ k an arbitrary scalar (playing the role of −nC(l) inProp. 21.1). We define a k-algebra C on the direct sum B⊕B j of two copies of B asa k-module by the multiplication

(u1 + v1 j)(u2 + v2 j) := (u1u2 +µv2v1)+(v2u1 + v1u2) j,

for ui,vi ∈ B, i = 1,2, and a quadratic form nC : C→ k by

nC(u+ v j) := nB(u)−µnB(v) (u,v ∈ B).


C together with nC is said to arise from B,µ by means of the Cayley-Dickson con-struction, and is written as Cay(B,µ) in order to indicate dependence on the pa-rameters involved. Note that 1C = 1B + 0 · j is an identity element for C and thatthe assignment u 7→ u+0 · j gives an embedding, i.e., an injective homomorphism,B →C of unital k-algebras, allowing us to identify B⊆C as a unital subalgebra.

p.CDCON21.3. Proposition. Let B be a conic k-algebra and µ ∈ k an arbitrary scalar. ThenC = Cay(B,µ) as defined in 21.2 is a conic k-algebra whose algebra structure,unit element, norm, polarized norm, trace, conjugation relate to the correspondingobjects belonging to B by the formulas

(u1 + v1 j)(u2 + v2 j) = (u1u2 +µv2v1)+(v2u1 + v1u2) j, (1) CDMULT

1C = 1B, (2) CDUNT

nC(u+ v j) = nB(u)−µnB(v), (3) CDNOR

nC(u1 + v1 j,u2 + v2 j) = nB(u1,u2)−µnB(v1,v2), (4) CDNORL

tC(u+ v j) = tB(u), (5) CDTR

ιC(u+ v j) = u+ v j = u− v j (6) CDCONJ

for all u,ui,v,vi ∈ B, i = 1,2. In particular,

B j = jB⊆ B⊥. (7) CDPERP

Proof. (1)–(3) are simply repetitions of things stated in 21.2 and immediately imply(4). Hence nC(1C,x) = tB(u) for x = u+v j, u,v∈ B, and from (1)), (19.5.4), (19.5.1)we conclude

x2 =(u2 +µnB(v)1B

)+ tB(u)v j = tB(u)(u+ v j)−

(nB(u)−µnB(v)

)j

= nC(1C,x)x−nC(x)1C.

Thus C is a conic k-algebra. (1)–(6) are now clear, while (7) follows directly from(1) and (4).

r.NOCD21.4. Remark. In the situation of Prop. 21.3, the norm of the Cayley-Dickson con-struction Cay(B,µ) may be written more concisely as

nCay(B,µ) = nB⊕ (−µ)nB = 〈1,−µ〉⊗nB.

ss.ITCD21.5. The Cayley Dickson process. Since the Cayley-Dickson construction stabi-lizes the category of conic algebras, it can be iterated: given a conic algebra B overk and scalars µ1, . . . ,µn ∈ k, we say that

Cay(B; µ1,µ2, . . . ,µn) := Cay(. . .(Cay

(Cay(B,µ1),µ2

)). . . ,µn

).

arises from B and µ1, . . . ,µn by means of the Cayley-Dickson process.


Confronting the external Cayley-Dickson construction with the internal one as de-scribed in Prop. 21.1, one obtains the following useful result.

21.6. Proposition. (Universal property of the Cayley-Dickson construction) Letp.CDUNVg : B→ C be a homomorphism of conic k-algebras and suppose in addition thatC is multiplicative alternative. Given l ∈ g(B)⊥ ⊆C and setting µ :=−nC(l), thereexists a unique extension of g to a homomorphism h : Cay(B,µ)→C of conic alge-bras over k sending j to h( j) = l. The image of h is the subalgebra of C generatedby g(B) and l, while its kernel has the form

Ker(h) = u+ v j | u,v ∈ B,g(u) =−g(v)l.

Proof. Any such h has the form h(u+ v j) = g(u)+ g(v)l for u,v ∈ B. Conversely,defining h in this manner, we obtain h( j) = l and deduce from (21.1.1)–(21.1.4)combined with (21.3.1 that h is a homomorphism of conic algebras. The remainingassertions are clear.

Before we can proceed, we will have to insert an easy technicality.ss.ZEDIMO

21.7. Zero divisors of modules. Let M be a k-module. In accordance with Bour-baki [11, I, 8.1], an element µ ∈ k is called a zero divisor of M if the homothetyx 7→ µx from M to M is not injective. We claim that, if M is projective and con-tains unimodular elements, the zero divisors of M are precisely the zero divisors ofk. Indeed, since the injectivity of linear maps can be tested locally, we may assumethat k is a local ring, in which case M 6= 0 is free as a k-module, and the assertionfollows.

c.CDEMB

21.8. Corollary. Under the hypotheses of Prop. 21.6, assume that g is injective, Bis projective as a k-module, nB is weakly non-singular and µ is not a zero divisor ofk. Then h is an isomorphism from Cay(B,µ) onto the subalgebra of C generated byB and l.

Proof. We need only show that h is injective, so let u,v ∈ B satisfy u+v j ∈Ker(h).By Prop. 21.6 and (21.1.7), we have g(u) =−g(v)l ∈ g(B)∩g(B)⊥, forcing g(u) =g(v)l = 0 by weak non-singularity of nB, and (21.1.3) yields g(µv) = µg(v) =(g(v)l)l = 0, hence µv = 0. But µ , not being a zero divisor of k, neither is oneof B by 21.7. Thus v = 0, as claimed.

e.CDR21.9. Examples. Considering the conic algebras of Chap. 1 over the field k =R ofreal numbers, we note that i∈C belongs to (R1C)

⊥ and has norm 1. Hence Cor. 21.8yields a canonical identification C= Cay(R,−1). Similarly, identifying C=R[i] asa subalgebra of H, the Hamiltonian quaternions, j ∈H belongs to C⊥ and again hasnorm 1, which implies H= Cay(C,−1) = Cay(R;−1,−1). And finally, viewing Hvia 1.10 as a subalgebra of O, the Graves-Cayley octonions, and consulting (1.5.1),(1.5.2), we conclude that j := 0⊕ (ie1) ∈ H⊥ ⊆ O has norm 1. Therefore O =Cay(H,−1) = Cay(C;−1,−1) = Cay(R;−1,−1,−1).


Useful properties of conic algebras preserved by the Cayley-Dickson constructionare in short supply. For instance, the property to be projective as a k-module triviallycarries over from a conic algebra B to any Cayley-Dickson construction Cay(B,µ),µ ∈ k. Other examples are provided by the following result.

p.CDNORAS21.10. Proposition. Let B be a conic algebra over k and µ ∈ k. If the conjugationof B is an involution, then so is the conjugation of C := Cay(B,µ). If B is norm-associative, then so is C.

Proof. The first part follows by a straightforward computation. As to the second, itsuffices to verify (19.12.2), so we must show

nC(u1 + v1 j,(u1 + v1 j)(u2 + v2 j)

)= tC(u2 + v2 j)nC(u1 + v1 j)

for ui,vi ∈ B, i = 1,2. To this end, one expands the left-hand side using (21.3.1),(21.3.4) and observes that (19.12.4) holds for B. Details are left to the reader.

On the other hand, preserving flexibility under the Cayley-Dickson construction isonly possible with a caveat.

p.CDFL21.11. Proposition. For a conic k-algebra B and µ ∈ k, the Cayley-Dickson con-struction C = Cay(B,µ) is flexible if and only if B is flexible and the conjugation ofB is an involution.

Proof. Suppose first that C is flexible. Then B is flexible and (19.9.1) holds for allx,y ∈C. In particular, for u1,u2,v1 ∈ B, we set x := u1 + v1 j, y := u2 and concludethat (

tC(x,y)−nC(x, y))x =

(tC((u1 + v1 j)u2)−nC(u1 + v1 j, u2)

)(u1 + v1 j)

=(

tB(u1u2 +(v1u2) j

)−nB(u1, u2)

)(u1 + v1 j)

=(tB(u1,u2)−nB(u1, u2)

)(u1 + v1 j),

belongs to k1B ⊆ B. Comparing B j-components, we obtain tB(u1,u2)−nB(u1, u2) ∈Ann(B), whence the conjugation of B is an involution (Prop. 19.9 (a)).

Conversely, suppose B is flexible and its conjugation is an involution. Then, byProp. 21.10, the conjugation of C is an involution, and we deduce from Cor. 19.10that it suffices to show

nC(x,xy)− tC(y)nC(x) ∈ Ann(C)

for all x,y ∈C. By linearity (in y), assuming x = u1+v1 j with u1,u2 ∈ B, we are leftwith the following cases.Case 1. y = u2 ∈ B. Then


nC(x,xy)− tC(y)nC(x) = nC(u1 + v1 j,u1u2 +(v1u2) j

)− tB(u2)nC(u1 + v1 j)

= nB(u1,u1u2)−µnB(v1,v1u2)

− tB(u2)nB(u1)+µtB(u2)nB(v1)

=(nB(u1,u1u2)− tB(u2)nB(u1)

)−µ

(nB(v1,v1u2)− tB(u2)nB(v1)

),

where both summands on the right by the hypotheses on B and Cor. 19.10 belong toAnn(B) = Ann(C). Hence so does nC(x,xy)− tC(y)nC(x).Case 2. y = v2 j, v2 ∈ B. Then

nC(x,xy)− tC(y)nC(x) = nC(u1 + v1 j,µ v2v1 +(v2u1) j

)= µ

(nB(v2v1,u1)−nB(v1,v2u1)

).

It therefore remains to show nB(u,vw)− nB(vu,w) ∈ Ann(B) for all u,v,w ∈ B.Since Bop is a flexible conic algebra whose conjugation is an involution, Cor. 19.10implies nB(u,vu)− tB(v)nB(u) ∈ Ann(B), so after linearization, Ann(B) containsnB(u,vw)+nB(w,vu)− tB(v)nB(u,w) = nB(u,vw)−nB(vu,w), as desired.

ss.CDCOMASS21.12. Commutators and associators. Let B be a conic algebra over k. For µ ∈k, we wish to find conditions that are necessary and sufficient for the the Cayley-Dickson construction C = Cay(B,µ) to be commutative, associative, alternative,respectively. To this end, we will describe the commutator and the associator ofC in terms of B and µ under the assumption that, in case of the associator, theconjugation of B be an involution. Then, letting ui,vi ∈ B, i = 1,2,3 and keeping thenotation of 21.2, a lengthy but straightforward computation yields

[u1 + v1 j,u2 + v2 j] = u+ v j,

[u1 + v1 j,u2 + v2 j,u3 + v3 j] = u+ v j,

where

u = [u1,u2]+µ(v2v1− v1v2), (1) CDCOMONE

v = v2(u1−u1)− v1(u2−u2), (2) CDCOMTWO

u = [u1,u2,u3]+µ((v2v1)u3− (u3v2)v1+ (3) CDASSONE

v3(v2u1)+ v3(v1u2)−u1(v3v2)− (u2 v3)v1),

v = v3(u1u2)− (v3u2)u1 +(v2u1)u3 +(v1u2)u3 (4) CDASSTWO

− (v2u3)u1− v1(u3 u2)+µ(v3(v2v1)− v1(v2v3)

).

With the aid of these identities, we can now prove the following important result.t.CDID

21.13. Theorem. For a conic k-algebra B, an arbitrary scalar µ ∈ k and the cor-responding Cayley-Dickson construction C := Cay(B,µ), the following statementshold.


(a) C is commutative if and only if B is commutative and has trivial conjugation.(b) C is associative if and only if B is commutative associative and its conjugation

is an involution.(c) C is alternative if and only if B is associative and its conjugation is an involu-

tion.

Proof. (a) If C is commutative, then so is B and (21.12.2) for v2 = 1B,v1 = 0 impliesιB = 1B. Conversely, let B be commutative and suppose ιB = 1B. Then an inspectionof (21.12.1), (21.12.2) shows that C is commutative as well.

(b) If C is associative, then so is B, its conjugation is an involution by Prop. 21.11,and (21.12.4) for u3 = v1 = v2 = 0,v3 = 1C shows that B is commutative as well.Conversely, if B is commutative associative and its conjugation is an involution, aninspection of (21.12.3), (21.12.4) shows that C is associative.

(c) If C is alternative, then the conjugation of B by Prop. 21.11 is an involution.Moreover, v = 0 for u1 = u2, v1 = v2, v3 = 0 and (21.12.4) combined with (19.5.4)yield

0 = (v1u1)u3 +(v1u1)u3− (v1u3)u1− v1(u3 u1)

= tB(u1)v1u3− (v1u3)u1− tB(u1)v1u3 + v1(u3u1)

= − [v1,u3,u1].

Hence B is associative. Conversely, let this be so and suppose ιB is an involution. Set-ting u1 = u2,v1 = v2 in (21.12.3), (21.12.4), Kirmse’s identities (20.3.1) and(19.5.4),(19.5.6) imply

u = µ(nB(v1)u3−nB(v1)u3 + tB(u1)v3v1− tB(u1)v3v1

)= 0,

v = tB(u1)v1u3− tB(u1)v1u3 +µ(nB(v1)v3−nB(v1)v3

)= 0,

forcing C to be alternative.

c.CDID21.14. Corollary. In addition to the above, assume that B is projective as a k-module. Then

(a) C is associative if and only if B is commutative associative.(b) C is alternative if and only if B is associative.

Proof. All the algebras of Thm. 21.13 are flexible. Hence in each case Prop. 19.13shows that the hypothesis of the conjugation of B being an involution is automatic.

e.CDAL

21.15. Examples. (a) Let R be a quadratic k-algebra whose conjugation is non-trivial. For any µ1 ∈ k, Cor. 21.14 (a) combined with Thm. 21.13 (a) shows thatthe conic algebra B := Cay(R,µ1) is associative but not commutative. ApplyingCor. 21.14 again, we therefore conclude for any µ2 ∈ k that the conic algebra C :=Cay(B,µ2) = Cay(R; µ1,µ2) is alternative but not associative. In view of Ex. 21.9,these results generalize what we have found in Exc. 1 and Cor. 1.11.


(b) At the other extreme, assume 2 = 0 in k. If B is a commutative associative conick-algebra with trivial conjugation (e.g., B = k), then by Thm. 21.13 and (21.3.6) sois the Cayley-Dickson process C := Cay(B; µ1, . . . ,µn), for any positive integer nand any µ1, . . . ,µn ∈ k.

Exercises.

pr.CDMUL 92. Let B be a multiplicative conic algebra over k and µ ∈ k. Show that the Cayley-Dickson con-struction Cay(B,µ) is multiplicative if and only if [B,B,B]⊆ Rad(DnB).

pr.CDSCAPAR 93. Let B be a multiplicative conic k-algebra, and let µ ∈ k, a ∈ Nuc(B). Show that the map

ϕ : Cay(B,nB(a)µ

)−→ Cay(B,µ)

defined by ϕ(u+ v j) := u+(av) j for u,v ∈ B is a homomorphism of conic algebras. Moreover, itis an isomorphism if and only if a is invertible in Nuc(B).

pr.ZERDIV 94. Zero divisors of algebras. Let A be a k-algebra. An element x ∈ A is called a right (resp. left)zero divisor of A if the left (resp. right) multiplication Lx : A→ A (resp. Rx : A→ A) of x in A is notinjective. We say A has left (resp. right) zero divisors if it contains left (resp. right) zero divisorsother than zero.

Now let C be a conic alternative k-algebra which is projective as a k-module. Prove for x ∈Cthat the following conditions are equivalent.

(i) x is a right zero divisor of C.(ii) x is a left zero divisor of C.(iii) nC(x) is a zero divisor of k.

(Hint. For the implication (i)⇒ (iii), argue indirectly and pass to the base change C f =C⊗k f , f =nC(x).)

pr.VARCD 95. A variant of the Cayley-Dickson construction. Let B be a conic k-algebra and µ ∈ k. On thedirect sum B⊕ j′B of two copies of B as a k-module we define a k-algebra structure Cay′(B,µ) bythe formula

(u1 + j′v1)(u2 + j′v2) := (u1u2 +µv2v1)+ j′(u1v2 +u2v1)

for u1,u2,v1,v2 ∈ B. Show that there is a natural isomorphism

Cay(Bop,µ)∼= Cay′(B,µ)op

and conclude that Cay′(B,µ) is a conic k-algebra with norm, trace and conjugation canonically iso-morphic to the corresponding objects attached to Cay(B,µ). Show further that, if the conjugationof B is an involution, then

Cay(B,µ) ∼−→ Cay′(B,µ), u+ v j 7−→ u+ j′v,

is an isomorphism of conic algebras.

22. Basic properties of composition algebras.

s.BAPROBefore being able to deal with the main topic of this section, it will be necessaryto introduce an auxiliary notion that can hardly stand on its own but turns out to betechnically useful. Throughout, we let k be an arbitrary commutative ring.

22 Basic properties of composition algebras. 121

ss.PRECO22.1. Pre-composition algebras. By a pre-composition algebra over k we mean ak-algebra C satisfying the following conditions.

(i) C is unital.(ii) C is projective as a k-module.(iii) There exists a non-degenerate quadratic form n : C→ k that permits compo-

sition :

n(1C) = 1, (1) PERUN

n(xy) = n(x)n(y) (x,y ∈C). (2) PERCOM

In this case, (2) may be linearized (repeatedly) and yields the relations

n(xy,xz) = n(x)n(y,z), (3) LPERCOMR

n(xy,zy) = n(x,z)n(y), (4) LPERCOML

n(xy,wz)+n(wy,xz) = n(x,w)n(y,z) (5) LPERCOMT

for all x,y,z,w ∈C.In view of Prop. 20.5, conic alternative algebras which are projective as k-

modules are pre-composition algebras provided their norm is a non-degeneratequadratic form. Remarkably, the converse of this is also true.

p.PRECOAL22.2. Proposition. Let C be a pre-composition algebra over k and n : C→ k anynon-degenerate quadratic form that permits composition. Then C is a conic alter-native k-algebra with unique norm nC = n. Moreover,

C⊥ := Rad(DnC) = x ∈C | nC(x,y) = 0 for all y ∈C ⊆ Cent(C)

is a central ideal of C satisfying

2C⊥ = nC(x,y)C⊥ = 0

for all x,y ∈C.

Proof. For the first part, it suffices to show that C is a conic alternative algebrawith norm nC := n since uniqueness follow from Prop. 19.15. By (22.1.1), we haven(1C) = 1. Now we put z = 1C in (22.1.3). Then

n(xy,x) = n(x)t(y), (1) PCBILCOMP

where t(y) := n(1C,y). Setting y = x,w = z = 1C in (22.1.5), we obtain t(x2) +n(x,x) = t(x)2, hence

t(x2) = t(x)2−2n(x). (2) PCTQUAD

Furthermore, setting z = x,w = 1C in (22.1.5), we also obtain n(xy,x)+ n(x2,y) =t(x)n(x,y), so by (1),


n(x2− t(x)x+n(x)1C,y) = 0.

Similarly, one can show n(x2− t(x)x+n(x)1C) = 0 by expanding the left-hand sideand using (1) (for y = x) as well as (2). Since n is non-degenerate, C is there-fore a conic k-algebra with norm nC = n. Moreover, (1) shows that C is norm-associative, hence flexible by Prop. 19.13. Thus alternativity will follow once wehave established the left alternative law, equivalently, the first of Kirmse’s identities(20.3.1). To this end, we combine (19.11.4) with (22.1.3)and the multiplicativityof n to obtain n(x(xy),z)) = n(xy,xz) = n(n(x)y,z). A similar computation yieldsn(x(xy)− n(x)y) = 0, and non-degeneracy of n implies the first Kirmse identityx(xy) = n(x)y, as claimed.

We now turn to C⊥. By non-degeneracy of n, n : C⊥ → k is an embedding ofadditive groups. In particular, n(2x) = 2n(x,x) = 0 for x ∈C⊥ implies 2C⊥ = 0.Moreover, (19.11.3) and (19.11.4) show that C⊥ ⊆ C is an ideal. We claim thatthis ideal belongs to the centre of C. Indeed, for x,y ∈ C,z ∈ C⊥, evaluating n at(xy)z,x(yz) ∈ C⊥ yields the same value n(x)n(y)n(z), which implies [x,y,z] = 0.Similarly, [y,z] = 0, and the assertion follows. It remains to prove n(x,y)z = 0. Sincethe conjugation of C is the identity on C⊥, this follows from (19.11.5) and n(x,y)z =t(xy)z = (xy)z+(yx)z = (zx)y+ yzx = n(zx,y)1C = 0.

Pre-composition algebras suffer the serious disadvantage of not being stable underbase change. This may be seen from the following example.

e.PURINS22.3. Example. Let K/k be a purely inseparable field extension of characteristic 2and exponent at most 1, so K2 ⊆ k (we allow the degree of K/k to be infinite). ThenK is a pre-composition algebra over k whose norm, given by the squaring K →k,x 7→ x2, is an anisotropic quadratic form with zero bilinearization. In particular,for [K : k] > 1, nK becomes isotropic, hence degenerate, when extending scalars tothe algebraic closure, so the corresponding scalar extension of the k-algebra K isnot a pre-composition algebra anymore.

Our next objective in this section will be to define the concept of a compositionalgebra in such a way that it remains stable under base change and always, evenwhen 2 is not a unit, includes the base ring as an, albeit trivial, example. This isensured by insisting on suitable separability conditions on the constituents enteringinto the definition of a composition algebra. We then proceed to derive a numberof important properties that are all related in one way or another to the Cayley-Dickson construction. We show in particular that composition algebras over semi-local rings always contain quadratic etale subalgebras, which in turn may be used torecover the ambient composition algebra by means of the Cayley-Dickson process.It follows that, over any commutative ring, composition algebras (if they have a rankat all) exist only in ranks 1,2,4,8. The section concludes with a brief discussion ofquaternion and octonion algebras.

ss.CONCOMALG


22.4. The concept of a composition algebra. Basically following a suggestion ofO. Loos, we define a composition algebra over k as a k-algebra C satisfying thefollowing conditions.

(i) C is unital.(ii) C is projective as a k-module.(iii) The rank function p 7→ rkp(C) from Spec(k) to N∪∞ is locally constant

with respect to the Zariski topology of Spec(k).(iv) There exists a separable quadratic form n : C→ k that permits composition:

n(1C) = 1, (1) COUN

n(xy) = n(x)n(y) (x,y ∈C). (2) COAL

If even a non-singular quadratic form (rather than just a separable one) exists on Cpermitting composition, we speak of a non-singular composition algebra. Note thatcondition (iii) above holds automatically if C is not only projective but also finitelygenerated as a k-module (Bourbaki [11, II, Thm. 1]).

Comparing conditions (i)−(v) above with the corresponding ones of 22.1, we seethat the main difference between pre-composition algebras and (non-singular) com-position algebras lies in the respective regularity condition imposed on the quadraticform permitting composition. In view of this difference, non-singular compositionalgebras are always pre-composition algebras. On the other hand, ordinary compo-sition algebras are pre-composition algebras if the base ring k is a field, but not ingeneral. For example, k is always a composition algebra over itself, while it is apre-composition algebra if and only if, for all α ∈ k, the relations α2 = 2α = 0 im-ply α = 0, which fails to be the case if, e.g., k contains non-zero nilpotent elementsand 2 = 0 in k. At the other extreme, if 1

2 ∈ k, then k will always be a non-singularcomposition algebra.

The most important feature of composition algebras (as opposed to pre-compositionalgebras, cf. Ex. 22.3 above) is that they are stable under base change. This consti-tutes the main reason why the former are to be preferred to the latter. Our next aimwill be to examine the relationship between (non-singular) composition algebrasand pre-composition algebras more closely. We first note that if C is a (non-singular,resp. pre-) composition algebra over k, so is Cop. We also need to characterize a classof composition algebras sitting inside conic algebras as “small” unital subalgebras.

p.SMASU22.5. Proposition. Let C be a conic algebra over k and u ∈C. Then D := k[u]⊆Cis a unital commutative associative subalgebra and the following conditions areequivalent.

(i) D is a non-singular composition algebra of rank 2.(ii) D is free of rank 2 as a k-module with basis (1C,u), and the quadratic form

nC is non-singular on D.(iii) tC(u)2−4nC(u) ∈ k×.

In this case,


disc((D,nD)

)=(tC(u)2−4nC(u)

)mod k×2 (1) DISCU

is the discriminant of the quadratic space (D,nD).

Proof. Since conic algebras are power-associative by 19.5, the first part is clear.(i)⇔ (ii). Suppose (i) holds. Since D is a finitely generated projective k-module

of rank 2, the natural surjection k2 → D determined by the elements 1C,u must bea bijection, giving the first part of (ii). As to the second, D is a pre-composition al-gebra, hence a conic one, so nC by Prop. 22.2 restricts to the unique non-degenerate(actually, non-singular) quadratic form on D permitting composition. Conversely,(i) is a consequence of (ii) by Exc. 77 (a).

(ii) ⇔ (iii). Since D is flexible, (ii) combined with Prop. 19.13 implies that itis norm-associative, so by (19.11.5) the bilinear trace tD = tC|D×D is non-singularalong with nD = nC|D×D. Computing the determinant of DnD relative to the basis(1D,u) of D, we obtain

det(

nC(1C,1c) nC(1C,u)nC(u,1C) nC(u,u)

)= det

(2 tC(u)

tC(u) 2nC(u)

)(2) DISC

= 4nC(u)− tC(u)2,

hence (iii), and the formula for the discriminant. Conversely, if (iii) holds, it sufficesto show that 1C,u are linearly independent over k, so suppose α,β ∈ k satisfy therelation α1C +βu = 0. Then αu+βu2 = 0, and taking traces we conclude(

2 tC(u)tC(u) tC(u)2−2nC(u)

)(α

β

)=

(tC(1C) tC(u)tC(u) tC(u2)

)(α

β

)= 0.

But by (iii) the matrix on the very left is invertible, forcing α = β = 0, as desired.

22.6. Proposition (Kaplansky [52]). If k is a field, a k-algebra C is a pre-compositionp.PRECOMFIELDalgebra if and only if it is either a (finite-dimensional) non-singular composition al-gebra or a purely inseparable field extension of characteristic 2 and exponent atmost 1.

Proof. In view of the Ex. 22.3, we only have to prove that a pre-composition al-gebra C over k has the form indicated in the proposition. Adopting the notationof Prop. 22.2, we first assume C⊥ 6= 0. Since 2C⊥ = 0 and C⊥ is a centralideal of C whose non-zero elements, by non-degeneracy, are anisotropic relative tonC, hence invertible in C (Prop. 20.4), we conclude that k has characteristic 2 andK := C = C⊥ = Cent(C) is an extension field of k whose trace (in its capacity asa conic algebra) vanishes identically. Thus K/k is purely inseparable of exponent≤ 1. We are left with the case C⊥ = 0, so nC is weakly non-singular. It sufficesto show that C is finite-dimensional. Our first aim will be to exhibit a unital subal-gebra D ⊆C of dimension at most 2 on which nC is non-singular. For char(k) 6= 2,D := k1C will do, so suppose char(k) = 2. Then weak non-singularity of nC pro-duces an element u ∈ C of trace 1, and Prop. 22.5 shows that D := k[u] ⊆ C is a


subalgebra of the desired kind. Now let B⊂C be any proper unital subalgebra of fi-nite dimension on which nC is non-singular. Then C = B⊕B⊥ by Lemma 12.10, andsince nC is weakly non-singular on C, we find an anisotropic vector l ∈ B⊥. But Cis a conic alternative k-algebra, so Cor. 21.8 leads to an embedding Cay(B,µ) →C,µ =−nC(l)∈ k×, whose image continues to be a unital subalgebra of C on which nCis non-degenerate. Assuming C were infinite-dimensional and starting from D, wecould repeat this procedure indefinitely but, by Cor. 21.14, after at most four steps,would arrive at a subalgebra of C that is no longer alternative. This contradiction toProp. 22.2 shows that C is indeed finite-dimensional.

Remark. The preceding proof actually yields more than is claimed in the proposi-tion. But since the situation described here will soon be re-examined in the moregeneral set-up of semi-local rings, we see no point at this stage to state this addi-tional piece of information explicitly.

c.COMALGLOC22.7. Corollary. Let C be a composition algebra over k. Then C is finitely gene-rated projective as a k-module, and if it has rank r, then C is either non-singular, orr = 1 (i.e., C ∼= k) and 1

2 /∈ k.

Proof. In view of 22.4 (iii), the assertion is local on k, so we may assume that kis a local ring with maximal ideal m. The C(m) is a composition algebra, hencealso a pre-composition algebra, over the field κ(m) anf thus, by Prop. 22.6 com-bined with Ex. 22.3, is either finite-dimensional non-singular or one-dimensional ofcharacteristic 2.

t.COMALT

22.8. Theorem. A k-algebra C is a composition algebra (resp. a non-singular com-position algebra) if and only if it is a conic alternative algebra which is finitely gen-erated projective as a k-module and has a separable (resp. non-singular) norm. Inthis case, the norm nC (cf. Prop. 19.15) is the only separable quadratic form on Cpermitting composition.

Proof. A conic alternative algebra which is finitely generated projective as a k-module and has a separable (resp. non-singular) norm is a composition algebra(resp. a non-singular composition algebra) by Prop. 20.5. Conversely, let C be acomposition algebra and n : C→ k a separable quadratic form permitting composi-tion.By Cor. 22.7, we need only show that C is conic alternative with norm nC = n.By Prop. 22.2, we are done if C is a pre-composition algebra. Otherwise, C is asingular composition algebra, and Cor. 22.7 implies C = k and n = ε〈1〉q for someidempotent ε ∈ k such that ε(p) = 1 in κ(p) for all p ∈ Spec(k) since n is separable.Hence ε = 1.

ss.EXCOMALG

22.9. Examples. (a) We know from 19.2 that D = k⊕k is a quadratic commutativeassociative k-algebra whose norm is the hyperbolic plane on D given by nD(α ⊕β ) = αβ for α,β ∈ k. Hence D is a non-singular composition algebra, called (forreasons that will become apparent later) the split composition algebra of rank 2.(b) The algebra Mat2(k) of 2-by-2 matrices over k is a non-singular associativecomposition algebra whose norm is the determinant det : Mat2(k)→ k. Indeed, det


is a quadratic form that permits composition and is isometric to 2h, the direct sumof two copies of the hyperbolic plane, hence non-singular.

In order to gain a more detailed understanding of composition algebras, they willnow be exposed to the Cayley-Dickson construction.

t.COALCD22.10. Theorem. Let B be a conic k-algebra and µ ∈ k an arbitrary scalar. ThenC = Cay(B,µ) is a composition algebra if and only if B is a non-singular associa-tive composition algebra and µ is invertible in k. In this case, C is a non-singularcomposition algebra as well.

Proof. Assume first that B is a non-singular associative composition algebra andµ ∈ k is a unit. Then B is conic associative by Thm. 22.8, forcing C to be conicalternative by Cor. 21.14 and nC ∼= nB⊕ (−µ)nB (by (21.3.3)) to be a non-singularquadratic form. By Cor. 22.7 and Thm. 22.8, therefore, C is a non-singular com-position algebra. Conversely, assume that C is a composition algebra. Since C isconic alternative by Thm. 22.8, B is conic associative by Cor. 21.14. Furthermore,for every field K ∈ k-alg, the decomposition (nC)K ∼= (nB)K ⊕ (−µK)(nB)K com-bines with the non-degeneracy of (nC)K to show that (nB)K is non-degenerate aswell and µK 6= 0. Hence B is a composition algebra, and specializing K = κ(p) forall p ∈ Speck implies µ ∈ k×. It remains to prove that B is non-singular. Otherwise,Bp would be a singular composition algebra over kp, for some p ∈ Speck, forcingCp to be a singular composition algebra as well. But then Cor. 22.7 implies that Cp

has rank 1, a contradiction.

Before exploiting the Cayley-Dickson construction still further, it is advisable toinsert a technicality.

l.COMTRI22.11. Lemma. The linear trace of a non-singular composition algebra is surjec-tive. Up to isomorphism, the only composition algebra over k having trivial conju-gation is k itself.

Proof. For the first part, we combine the non-singularity of the bilinear trace with theunimodularity of the identity to find an element in C having trace 1. For the secondpart, we let C be a composition algebra over k with trivial conjugation. Localizingif necessary, we may assume by Cor. 22.7 that C is non-singular. Then the assertionfollows from the first part and Exc. 85.

Our next aim will be to show that, under suitable conditions on k, all compositionalgebras arise from composition algebras of rank 2 by means of the Cayley-Dicksonprocess. Within the framework of the present section, it would be enough to doso for k being a local ring. But with an eye on subsequent applications, it will beimportant to relax this condition in a way that is tied up with the following concept.

ss.SEMLOC22.12. Semi-local rings. Our base ring k is said to be semi-local if it satisfies thefollowing two equivalent conditions (Bourbaki [11, II, §3, Prop. 16]):

(i) The set of maximal ideals in k is finite.


(ii) The quotient k/Jac(k) is a finite direct product of fields.

Here Jac(k) denotes the Jacobson radical of k (Bourbaki [13, VII, §6, Def. 3]. Whileevery local ring is semi-local, the zero ring k = 0 is semi-local without being local.Less trivial examples arise as follows (Bourbaki [11, IV, §2, Cor. 3 of Prop. 9]):Suppose k0 is a (commutative) Noetherian semi-local ring and k ∈ k0-alg is finitelygenerated as a k0-module. Then k is Noetherian and semi-local as well. But notethat k need not be local even if k0 is.

We proceed by collecting a few useful properties of semi-local rings, addressedto projective modules and quadratic forms.

22.13. Proposition (Bourbaki [11, II, §5, Prop. 5]). Let k be a semi-local ring. Thenp.PROSEMevery finitely generated projective k-module of rank r ≥ 0 is free.

Remark. Since semi-local rings need not be connected, the condition on the rankfunction cannot be avoided.

l.QUAFORUNT

22.14. Lemma. Let k be a semi-local ring. Then every quadratic space (M,q) overk satisfying Supp(M) = Spec(k) represents some unit in k.

Proof. The assertion clearly holds over a finite direct product of fields since it isvalid over each individual factor. Now let k be arbitrary and write π for the canonicalmap from k to R := k/Jac(k). By the special case just treated, applied to (M,q)⊗R,some x ∈M satisfies π(q(x)) ∈ R×, hence q(x) ∈ k×.

Returning after this brief detour to the setting of compositions algebras, we can nowestablish the following basic result.

t.COMSUBALG22.15. Theorem. Let k be a semi-local ring and C a composition algebra of rankr over k.

(a) If B ⊆ C is a non-singular composition subalgebra of rank s < r, there ex-ists a unit µ ∈ k× such that the inclusion B → C extends to an embeddingCay(B,µ)→C.

(b) If r > 1, then C contains a composition subalgebra of rank 2 having the formD = k[u] for some u ∈C of trace 1.

Proof. (a) Since nC is non-singular on B, Lemma 12.10 yields an orthogonal split-ting C = B⊕B⊥, and the assumption s < r implies that (B⊥,nC|B⊥) is a quadraticspace over k with Supp(B⊥) = Spec(k). Thus, by Lemma 22.14, there exists anelement l ∈ B⊥ satisfying µ :=−nC(l) ∈ k×. Now Cor. 21.8 implies (a).

(b) We consider three cases.Case 1. k is a field. Suppose first char(k) 6= 2. Then (a) yields a unital embedding

of D := Cay(k,µ) = k⊕ k j into C for some µ ∈ k×, and D is a composition algebraof rank 2 (Prop. 22.5). Moreover, D= k[u] with u= 1

2 + j ∈D having trace 1. We areleft with the case char(k) = 2. Since C is non-singular, we find an element u∈C withtC(u) = 1. Then tC(u)2−4nC(u)∈ k×, so D := k[u]⊆C is a composition subalgebraof rank 2 by Prop. 22.5.


Case 2. k is a finite product of fields. Then k = k1⊕·· ·⊕ km as a direct sum ofideals, with m ∈ N and fields ki, 1 ≤ i ≤ m. This implies C = C1⊕·· ·⊕Cm, whereCi = C⊗ ki (1 ≤ i ≤ m) is a composition algebra of dimension > 1 over ki. ThusCase 1 yields a composition ki-subalgebra Di⊆Ci of rank 2 generated by an elementof trace 1, forcing D := D1⊕·· ·⊕Dm ⊆C to have the same properties over k.

Case 3. The general case. Then Case 2 applies to the base change C′ := C⊗Rover R := k/Jac(k), and we find an element v ∈C such that v′ := vR ∈C′ has trace1 and generates a rank 2 composition subalgebra of C′. Invoking Prop. 22.5, weconclude that v generates a rank 2 composition subalgebra of C and tC(v) ∈ 1+Jac(k)⊆ k×, so u := tC(v)−1v ∈C is an element of the kind we are looking for.

c.ENUM22.16. Corollary. Every composition algebra of rank > 1 over a semi-local ringarises from a composition algebra of rank 2, and even from the base ring itself if 2is a unit, by an application of the Cayley-Dickson process.

The preceding theorem has important consequences also in the case when the basering is arbitrary.

22.17. Corollary (cf. Legrand [66]). Let C be a composition algebra of rank r overc.RKCOMALGk and assume k 6= 0. Then r = 1,2,4 or 8 and the following statements hold.

(a) C is non-singular unless r = 1 and 12 /∈ k.

(b) If r = 2, then C is commutative associative and has non-trivial conjugation.(c) If r = 4, then C is associative but not commutative.(d) If r = 8, then C is alternative but not associative.

Proof. We conclude from Cor. 22.16 that r = 2s is a power of 2. Localizing when-ever necessary, (a)−(d) now follow by a straightforward combined application ofCor. 22.7, Lemma 22.11 and Thms. 22.10, 22.15, 21.13. Finally, Thm. 22.8 andCor. 21.14 show that s > 3 is impossible.

We conclude this section by taking a closer look at the remaining cases.ss.QUAET

22.18. Quadratic etale algebras. Let D be a unital commutative associative k-algebra that is finitely generated projective as a k-module. Then the following con-ditions are easily seen (and well known) to be equivalent.

(i) For all maximal ideals m ⊆ k, the algebra D(m) over the field κ(m) = k/mmay be written as a (finite) direct product of (finite) separable field exten-sions.

(ii) For all prime ideals p⊆ k, the κ(p)-algebra D(p) may be written as a (finite)direct product of (finite) separable field extensions.

(iii) The bilinear trace D×D→ k,(x,y) 7→ tr(Lxy), L being the left multiplicationof D, is a non-singular symmetric bilinear form.

If these conditions are fulfilled, D is said to be finite etale (or separable) over k. If,in addition, D has rank 2 as a finitely generated projective k-module, we speak of a


quadratic etale k-algebra. Comparing (iii) with (19.11.5), 19.4 and Cor. 22.17, wesee that being a quadratic etale k-algebra and a composition algebra over k of rank2 are equivalent notions; therefore these terms will henceforth be used interchange-ably.

Typical examples of quadratic etale k-algebras arise as follows.

(iv) Let λ ∈ k and D := k[t]/(t2 − t + λ ) = k[u], where u, the canonical im-age of t in D, has trace 1 and norm λ . Then D is free of rank 2 as a k-module, with basis 1D,u, hence a quadratic k-algebra whose norm satisfiesnD(α1D +βu) = α2 +αβ +λβ 2 for all α,β ∈ k. By Prop. 22.5, the algebraD is quadratic etale if and only if 1−4λ ∈ k×.

(v) If k contains 12 , the Cayley-Dickson construction D := Cay(k,µ), µ ∈ k,

yields a quadratic algebra over k with norm nD ∼= 〈1,−µ〉 in the sense of12.7. Moreover, D is etale if and only if µ ∈ k×.

ss.QUATALG22.19. Quaternion algebras. Quaternion algebras over k are defined as compo-sition algebras of rank 4. By Cor. 22.17, they are associative but not commutative.Typical examples arise from the Cayley-Dickson construction as follows:

(i) B = Cay(D,µ), D a quadratic etale k-algebra and µ ∈ k×, by Thm. 22.10 is aquaternion algebra over k with norm

nB = nD⊕ (−µ)nD = 〈1,−µ〉⊗nD.

(ii) More specifically, if 12 ∈ k, then B = Cay(k; µ1,µ2), µ1,µ2 ∈ k×, is a quater-

nion algebra over k with norm

nB = 〈1,−µ1,−µ2,µ1µ2〉quad (1) QUATNOR

The most prominent examples of quaternion algebras are provided by

(iv) the algebra Mat2(k) of 2-by-2 matrices over k, cf. Example 22.9 (c), and(v) the Hamiltonian quaternions over the field R of real numbers as defined in

1.10.

But note that the Hurwitz quaternions of Thm. 4.2, though conic associative andfree of rank 4 as a Z-module, are not a quaternion algebra over the ring of rationalintegers since they have discriminant 4. Instead, they belong to a wider class investi-gated by Loos [72] under the term quaternionic algebras: associative conic algebrasover a commutative ring that are finitely generated projective of rank 4 as modules.

ss.OCTALG22.20. Octonion algebras. Octonion algebras over k are defined as compositionalgebras of rank 8. By Cor. 22.17, they are alternative but not associative. Typicalexamples arise again from the Cayley-Dickson construction:

(i) C = Cay(B,µ), B a quaternion algebra over k, µ ∈ k×, by Thm. 22.10 is anoctonion algebra over k with norm

nC ∼= nB⊕ (−µ)nB ∼= 〈1,−µ〉⊗nB.


(ii) C = Cay(D; µ1,µ2), D a quadratic etale k-algebra and µ1,µ2 ∈ k×, is an oc-tonion algebra with norm

nC = nD⊕ (−µ1)nD⊕ (−µ2)nD⊕ (µ1µ2)nD = 〈1,−µ1,−µ2,µ1µ2〉⊗nD.

(iii) If 12 ∈ k, then C = Cay(k; µ1,µ2,µ3), with µ1,µ2,µ3 ∈ k×, is an octonion

algebra over k, whose norm is given by norm

nC = 〈1,−µ1,−µ2,µ1µ2,−µ3,µ1µ3,µ2µ3,−µ1µ2µ3〉quad (1) OCTNOR

The most prominent example of an octonion algebra is provided by

(iv) the Graves-Cayley octonions O over the field R of real numbers as defined in1.4.

In a more arithmetic vein,

(v) the Coxeter octonions of Thm. 4.5 form an octonion algebra over the ringof rational integers. Since they are indecomposable as an integral quadraticlattice (cf. 4.6), they provide an example of an octonion algebra that cannotbe obtained from the Cayley-Dickson construction. For other examples of thisremarkable phenomenon, see Knus-Parimala-Sridharan [61] or Thakur [122].

We are not ready yet to define the notion of a split octonion algebra; this task has tobe postponed to the next section.

Exercises.

pr.COMPUN 96. Let C be a unital k-algebra and n : C→ k a quadratic form such that all the conditions of 22.1hold, with the possible exception of (22.1.1). Show that the following conditions are equivalent.

(i) C is a pre-composition algebra.(ii) 1C ∈C is unimodular in the sense of 10.9.(iii) 1C ∈C is faithful in the sense of 10.9.

pr.CONIL 97. Let F be a field. Show that a two-dimensional unital F-algebra is precisely one of the following.

(i) a separable quadratic extension field of F ,(ii) split quadratic etale,(iii) an inseparable quadratic extension field of F ,(iv) isomorphic to the F-algebra of dual numbers.

Conclude that, if F is perfect of characteristic 2 and C is a conic F-algebra without nilpotentelements other than 0, then C has dimension at most 2.

pr.COSIM 98. Let F be a field and C a conic F-algebra whose conjugation is an involution and whose normis a non-degenerate quadratic form. Prove that (C, ιC) is simple as an algebra with involution andconclude that C is either simple or split quadratic etale.

pr.CDDIVALG 99. For a conic algebra over a field to be a division algebra it is necessary that its norm beanisotropic. Show that the converse of this statement does not hold, even in the finite-dimensionalcase, by proving Brown’s theorem (Brown [16, Thm. 3]): given an octonion algebra B over a fieldof characteristic not 2 and a non-zero scalar µ , the Cayley-Dickson construction C = Cay(B,µ) isa division algebra if and only if µ is not the norm of an element in B and −µ is not the norm of atrace zero element in B. In order to do so, let F be a field of arbitrary characteristic and with B,µas above perform the following steps.


(a) Suppose µ /∈ nB(B×). Show for 0 6= xi = ui + vi j ∈C, ui,vi ∈ B, i = 1,2 that x1x2 = 0 if andonly if ui 6= 0 6= vi for i = 1,2 and the following relations hold.

(i) nB(u1) =−µnB(v1),(ii) (u1u2)v1 =−u1(u2v1),(iii) v2 =−(v1u2)u−1

1 .

(b) Conclude from (a) that if F has characteristic 2, then C is a division algebra if and only if itsnorm is anisotropic.

(c) Suppose char(F) 6= 2 and B is a division algebra. Then non-zero elements x,y,z ∈ B satisfythe relation (xy)z=−x(yz) if and only if there is a quaternion subalgebra A⊆B with x,y∈A,x y = 0, z ∈ A⊥.

(d) Now prove Brown’s theorem.(e) Conclude from (d) that the norm of the real sedenions

S := Cay(O;−1)

is anisotropic but the algebra itself fails to be a division algebra.

Remark. The final statement of (e) follows also from the Bott-Milnor-Kervaire theorem [10, 54]according to which finite-dimensional real division algebras exist only in dimensions 1,2,4,8. Fora more precise statement about the zero divisors of S, see Exc. 132 below.

pr.FROBTHE 100. Frobenius’s theorem for alternative real division algebras. Prove that a finite-dimensionalalternative real division algebra is isomorphic to R,C,H, or O. (Hint. Exc. 64, Exc. 80.)

pr.CDISOT 101. Isotopes and the Cayley-Dickson construction. Let B be a multiplicative conic associativealgebra over k, µ ∈ k and p ∈ B×. Prove that the assignment u+ v j 7→ p−1up+ v j determinesan isomorphism from Cay(B,µ) onto the unital isotope Cay(B,µ)p. Conclude for an octonionalgebra C over k and p,q ∈C× that the algebras C and C(p,q) are isomorphic provided the elementpq2 belongs to a quaternion subalgebra of C.

pr.AZUQUAD 102. Centre and nucleus of quaternion and octonion algebras.

(a) Show for a quadratic k-algebra R and u∈ R that u− u is invertible if and only if R is etale andgenerated by u. Show further that a quadratic etale k-algebra D contains an element u withu− u ∈ D× provided k is a semi-local ring. Conclude for k arbitrary that H(D, ιD) = k1D.

(b) Conclude from (a) and Cor. 22.16 that a quaternion algebra over any commutative ring k iscentral, hence an Azumaya algebra (cf. Knus-Ojanguren [60, III, §5]). Prove similarly thatan octonion algebra C over k satisfies

Nuc(C) = x ∈C | xy = yx for all y ∈C= k1C.

pr.ETAUT 103. Automorphisms of quadratic etale algebras. Let D be a quadratic etale k-algebra. Prove: ak-linear map ϕ : D→ D is an automorphism of D if and only if there exists a decompositionk = k+⊕ k− of k as a direct sum of ideals such that the induced decompositions

D = D+⊕D−, D± := Dk± , ϕ = ϕ+⊕ϕ−, ϕ± := ϕk±

satisfy ϕ+ = 1D+ , ϕ− = ιD− .

pr.IDEALSCOAL 104. Ideals in composition algebras (Petersson [93]) Let C be a composition algebra over k andview k ⊆C as a subalgebra in the natural way. Show that the assignments

a 7−→ aC, I 7−→ k∩ I

give inclusion preserving inverse bijections between the ideals of k and the ideals of (C, ιC) asan algebra with involution. Show further that, if C has rank r 6= 2 as a projective k-module, the


same assignments yield inclusion preserving bijections between the ideals of k and those of C.(Hint. Show first that an ideal I of (C, ιC) (resp. of C) (depending on the hypotheses stated) withI∩ k = 0 is zero.)

pr.PEIRCEELID 105. Peirce decomposition relative to elementary idempotents. Let C be a multiplicative conicalternative algebra over k. If c ∈ C is an elementary idempotent (cf. Exc. 82), put c1 := c, c2 :=1C− c, Ci j :=Ci j(c) for i, j = 1,2 and show

Cii = kci (i = 1,2). (2) ONEONEPEIR

Show further for elements

x = α1c1 + x12 + x21 +α2c2, y = β1c1 + y12 + y21 +β2c2, (3) PEXY

in C, where αi,βi ∈ k, xi j,yi j ∈Ci j for i, j = 1,2, i 6= j (Excs. 67,82), that

nC(x) = α1α2 +nC(x12,x21), (4) NORPEIR

nC(x,y) = α1β2 +α2β1 +nC(x12,y21)+n(x21,y12), (5) NORPEIRLIN

tC(x) = α1 +α2, (6) TRPEIR

x = α2c1− x12− x21 +α1c2. (7) CONJPEIR

Moreover, if C is a composition algebra, then the k-modules C12,C21 are in duality to each otherunder DnC . Finally, for i, j = 1,2, the trilinear form

C3i j −→ k, (x,y,z) 7−→ tC(xyz),

(cf. (19.12.1)) is alternating.

pr.MISPLET 106. Minimal splitting of quadratic etale algebras. Let D be a quadratic etale k-algebra.

(a) Prove that the map ϕ : D⊗D ∼→ D⊕D defined by

ϕ(x⊗d) := (xd)⊕ (xd)

for x,d ∈D is an isomorphism of D-algebras. (Hint. For u ∈D compute ϕ(u⊗1D−1D⊗ u)and ϕ(1D⊗u−u⊗1D).)

(b) Conclude from (a) that

σ := 1D⊗ ιD : D⊗D−→ D⊗D,

σ′ := ϕ σ ϕ

−1 : D⊕D−→ D⊕D

are ιD-semi-linear involutions of D⊗D, D⊕D, respectively, with

σ′(a⊕b) = b⊕ a

for all a,b ∈ D.

pr.QUACOMQUA 107. Quadratic forms permitting composition on quadratic algebras. Let R be a quadratic algebraover k.

(a) Prove for an idempotent e ∈ R with nR(e) = 0 that

tR Le : R−→ k, x 7−→ tR(ex)

is a (possibly non-unital) algebra homomorphism.

23 Hermitian forms 133

(b) Conclude from (a) that for idempotents ε ∈ k, e ∈ R with εe = 0, nR(e) = 0, the quadraticform

q : R−→ k, x 7−→ tR(ex2)+ εnR(x)

permits composition.(c) Let D = k⊕ k be the split quadratic etale k-algebra. Show that a quadratic form q : D→ k

permits composition if and only if there exists an orthogonal system (ε1,ε2,ε3) of idempo-tents in k (possibly incomplete) such that

q(α⊕β ) = ε1α2 + ε2αβ + ε3β

2 (8) MAQU

for all α,β ,γ ∈ k(d) Finally, show that if D is etale, all quadratic forms on D permitting composition have the

form described in (b). (Hint. Reduce to the split case D = k⊕ k by using (c) and Exc. 106.)

pr.QUACOMCOM 108. Quadratic forms on composition algebras permitting composition. Given any quadratic space(M,Q) over k and writing Alt(M,Q) for the k-module of elements f ∈ Endk(M) satisfyingQ( f (x),x) = 0 for all x ∈M, make use of the short exact sequence

0−→ Alt(M,Q)−→ Endk(M)−→ Quad(M)−→ 0

to show that a quadratic form q on a composition algebra C of rank r 6= 2 over k permits compositionif and only if q = εnC for some idempotent ε ∈ k.

pr.COVQUATALG 109. Embeddings into quaternion subalgebras. Let C be an octonion algebra over a semi-local ringk and R ⊆C a quadratic subalgebra that is a direct summand of C as a k-module. Show that thereexists a quaternion subalgebra of C containing R. (Hint. Show more precisely that there exists anelement b ∈C making the subalgebra of C generated by R and b a quaternion algebra and reducethis assertion to the case that k is a field by arguing as in the proof of Thm. 22.15 (b).)

Remark. Over fields, the preceding result amounts to [121, Prop. 1.6.4]: every element of anoctonion algebra over field can be embedded into a quaternion subalgebra.

pr.RINV 110. Reflections and involutions of composition algebra. (cf. Jacobson [41] and Racine-Zel’manov[110]) Let k be a commutative ring containing 1

2 and C a composition algebra of rank r > 1 over k.

(a) Let σ be a reflection of C, i.e., an automorphism of order 2. Show that its fixed algebraFix(σ) := x ∈ C | σ(x) = x ⊆ C is a composition subalgebra of rank r

2 and that the as-signment σ 7→ Fix(σ) determines a bijection from the set of reflections of C onto the setof composition subalgebras of C having rank r

2 . Show further that two reflections of C areconjugate under Aut(C) if and only if their fixed algebras are.

(b) Show that an involution of C commutes with its conjugation. Use this and (a) to set up abijective correspondence between the set of involutions of C that are distinct from its conju-gation (resp. the set of their isomorphism classes) and the set of composition subalgebras ofC having rank r

2 (resp. the set of their conjugacy classes under Aut(C)). Finally, show for aninvolution τ 6= ιC of C that H(C,τ)⊆C is a finitely generated projective submodule of rankr2 +1.

23. Hermitian forms

s.HERFORBefore being able to proceed with the study of composition algebras, it will benecessary to insert a few basic facts about hermitian forms over commutative ringsthat will be useful not only in the present context but also for cubic Jordan algebras


later on. A systematic account of the subject may be found in Knus [58] or Hahn-O’Meary [38].

Throughout we let k be an arbitrary commutative ring and (B,τ) an associativealgebra with involution over k in the sense of § 11. In particular, B contains anidentity element. We also write x := τ(x) for x ∈ B.

ss.LERI23.1. Passing from left to right modules and conversely. Any left B-module Mmay be converted into a right B-module by defining xa := ax for x ∈M and a ∈ B.We denote this right module by Mτ . A B-linear map f : M→ N, x 7→ (x) f , of leftB-modules may be viewed as a B-linear map f τ : Mτ → Nτ of right B-modules, sowe have f τ(x) = (x) f for all x ∈Mτ . The preceding conventions make equally goodsense with left and right modules interchanged. We then have Mττ = M for any left(right) module M over B and f ττ = f for any B-linear map f : M→ N of left (right)B-modules.

Writing BB (resp. BB) for B viewed as a left (resp. right) B-module, τ : (BB)τ →BB is a linear bijection of right B-modules. Hence, if a left B-module is (finitelygenerated) projective (resp. free), then so is Mτ as a right B-module, and conversely.

ss.TWIDU23.2. The twisted dual of a module. Let M be a right B-module. Then the addi-tive group M• := HomB(M,B) canonically becomes a left B-module if one definesax• : M → B by (ax•)(y) := ax•(y) for a ∈ B, x• ∈ M• and y ∈ M. Using the for-malism of 23.1, we may then convert the left B-module M• into the right B-moduleM∗ := M•τ , which we call the τ-twisted dual, or simply the twisted dual, of M. Notethat for B commutative and τ = 1B, the terms “right B-module” and “left B-module”may be used interchangeably, and the twisted dual of M agrees with the ordinarydual as defined in 10.9. On the other hand, if (B still being commutative) τ is notthe identity, the twisted dual and the ordinary dual are different notions. However, itwill always be clear from the context which one of the two we have in mind.

For any right B-module M, we have its canonical pairing

capM : M∗×M −→ B, (x∗,y) 7−→ 〈x∗,y〉 := x∗(y), (1) PATWI

which is anti-linear in the first variable and linear in the second:

〈x∗a,yb〉= a〈x∗,y〉b (a,b ∈ B, x∗ ∈M∗, y ∈M). (2) LIPA

If f : M→ N is a linear map of right B-modules, then the assignment y∗ 7→ y∗ fdefines a linear map f ∗ : N∗→M∗, called the adjoint of f because it is characterizedby the relation

〈 f ∗(y∗),x〉= 〈y∗, f (x)〉 (y∗ ∈ N∗, x ∈M). (3) ADLI

In this way, we obtain a contravariant additive functor from the category of rightB-modules to itself.

ss.SEFO

23 Hermitian forms 135

23.3. Sesquilinear forms. By a sesquilinear form over B we mean a bi-additivemap h : M×N→ B, where M,N are right B-modules such that h(xa,yb) = ah(x,y)bfor all a,b∈ B and all x∈M, y∈N. In this case, we define the adjoint map or simplythe adjoint of h as the B-linear map

ϕh : M→ N∗, x 7−→ ϕh(x) := h(x,−).

Conversely, given a B-linear map ϕ : M→ N∗, we obtain a sesquilinear form hϕ :M×N→ B via hϕ(x,y) := 〈ϕ(x),y〉 for x ∈M, y ∈ N, and the two constructions areinverse to each other.

As an example, let M be any right B-module. Then the canonical pairing of(23.2.1) by (23.2.2) is a sesquilinear form over B whose adjoint map M∗ → M∗

is the identity on M∗.ss.SESBA

23.4. Base change. For R∈ k-alg, we can form the base change (B,τ)R = (BR,τR),which is an associative algebra with involution over R, and for a right B-module M,the R-module MR = M⊗R becomes a right BR-module in a natural way. If f : M→N is a homomorphism of right B-modules, then its R-linear extension fR : MR→NRis in fact one of right BR-modules.

For x∗ ∈M∗, we clearly have x∗⊗1R ∈ (MR)∗, and the assignment x∗ 7→ x∗⊗1R

gives a k-linear map M∗→ (MR)∗, which in turn induces an R-linear map

φ : (M∗)R→ (MR)∗, x∗⊗ r 7−→ r(x∗⊗1R).

This map is, in fact, a homomorphism of right BR-modules and will henceforth bereferred to as the canonical homomorphism from (M∗)R to (MR)

∗.Any sesquilinear form h : M×N → B over B yields canonically an extended

sesquilinear form hR : MR×MR → BR over BR, called the base change or scalarextension of h from k to R, such that the diagram

MR(ϕh)R //

ϕ(hR) ''

(N∗)R

(NR)

∗

(1) ADBA

commutes, where the vertical arrow refers to the canonical homomorphism from(N∗)R to (NR)

∗.ss.SEMO

23.5. Sesquilinear modules. By a sesquilinear module over B, we mean a pair(M,h) consisting of a right B-module M and a sesquilinear form h : M×M → B.Given two sesquilinear modules (M,h) and (M′,h′) over B, a homomorphism from(M,h) to M′,h′) (or from h to h′) is a B-linear map f : M→M′ preserving sesquilin-ear forms in the sense that h′ ( f × f ) = h. In this way one obtains the category ofsesquilinear modules over D. Isomorphims in this category are called isometries.

ss.SESMAT


23.6. Sesquilinear forms and matrices. Let p,q be positive integers. ViewingM := Bp, N := Bq as free right B-modules in the natural way, every matrix T ∈Matp,q(B) determines a sesquilinear form

〈T 〉sesq : Bp×Bq −→ B, (x,y) 7−→ xtTy,

and every sesquilinear form on Bp×Bq can be written uniquely in this way. Identify-ing a vector x ∈ Bq with the linear form Bq→ B,y 7→ xty, we obtain an identificationBq = Bq∗. The adjoint of 〈T 〉sesq then agrees with the linear map T t : Bp→ Bq.

ss.DODU23.7. The double dual. Given a right B-module M, we obtain a natural B-linearmap canM : M→M∗∗ determined by the condition

canM(x)(y∗) := 〈y∗,x〉 (x ∈M,y∗ ∈M∗).

In important cases, canM is an isomorphism. For example, if M =Bn, then Bn∗∗=Bn

and canM = 1Bn under the identifications of 23.6. Now let h : M×M → B be asesquilinear form. Then so is h∗ : M×M → B defined by h∗(x,y) := h(y,x) forx,y ∈ M, and the adjoints of h,h∗ are related to one another by the commutativediagram

Mϕh∗ //

canM

M∗

M∗∗.(ϕh)

∗

<<

Hence if canM is bijective, allowing us to identify M = M∗∗ accordingly, the adjointof h∗ agrees with the adjoint of the adjoint of h.

ss.HEFO23.8. Hermitian forms. By a hermitian form on a right B-module M we mean asesquilinear form h : M×M→ B satisfying h(y,x) = h(x,y) for all x,y ∈M; this isequivalent to h = h∗. For T ∈Matn(B), we obtain (〈T 〉sesq)

∗ = 〈T t〉sesq, so 〈T 〉sesqis a hermitian form if and only if T = T t is a hermitian matrix. By a hermitian mod-ule we mean a pair (M,h) consisting of a right B-module M and a hermitian formh : M×M→ B. We view hermitian modules as a full subcategory of sesquilinearmodules.

24. Ternary hermitian spaces

s.TERHERThe Cayley-Dickson construction discussed in some of the preceding sections isnot an appropriate tool when dealing with octonion algebras that fail to contain anyquaternion subalgebras. In the sequel, we therefore propose a different method ofconstructing composition algebras which is due to Thakur [122] over rings contain-ing 1

2 and has been sketched by the author [100] over fields of arbitrary character-

24 Ternary hermitian spaces 137

istic3 The main constituents of this construction are quadratic etale algebras andternary hermitian spaces. They always lead to octonion algebras and, conversely,every octonion algebra containing a quadratic etale subalgebra arises in this man-ner; in particular, this holds true for the Graves-Cayley octonions over the reals asdefined in 1.4 and for arbitrary octonion algebras over any commutative ring thatcontains 2 in its Jacobson radical (cf. Prop. 24.16 below). A slight generalization ofour method leads to a similar construction of quaternion algebras due to to Pumplun[107] provided 1

2 belongs to the base ring. Both constructions, the octonionic as wellas the quaternionic one, will be treated here in a unified fashion.

In this section, we fix a composition algebra D of rank at most 2 over an arbitrarycommutative ring k. By Cor. 22.17, D is commutative associative and is endowedwith its canonical involution, which we abbreviate as ι := ιD, a 7→ a. The set-upof the preceding section may thus be specialized to (B,τ) := (D, ι). By Exc. 102,we can identify k = H(D, ι) as a unital subalgebra of D, and this identification iscompatible with base change.

ss.UFID24.1. Some useful identifications. Fix R ∈ k-alg and let M be a right D-module.Then we obtain a natural identification

MR = M⊗R = M⊗D (D⊗R) = MDR

as right DR-modules such that

x⊗ r = x⊗D (1D⊗ r), x⊗D (a⊗ r) = (xa)⊗ r (x ∈M, r ∈ R, a ∈ D), (1) ARDE

ditto for left D-modules. It follows that, if M is finitely generated projective over D,then so is MR over DR.

Now suppose that M is a left D-module. Since Mι =M as k-modules, we obtain acanonical identification (Mι)R = (MR)

ιR , and (1) yields an identification (Mι)DR =(MDR)

ιDR as right DR-modules matching x⊗D (a⊗ r) in (Mι)DR with x⊗D (a⊗ r) in(MDR)

ιDR , for x ∈M, a ∈ D, r ∈ R.Combining the previous identifications with 23.4 and Lemma 10.11, we obtain

l.PROTWI24.2. Lemma. Let M be a finitely generated projective right D-module and R ∈k-alg. Then the canonical homomorphism

(M∗)R −→ (MR)∗, x∗⊗ r 7−→ r(x∗⊗1R),

is an isomorphism of right DR-modules. Identifying (M∗)R = (MR)∗=: M∗R by means

of this isomorphism, we have 〈x∗,x〉R = 〈x∗R,xR〉 for all x ∈ M, x∗ ∈ M∗, in otherwords, the canonical pairing M∗R ×MR → DR is the R-bilinear extension of thecanonical pairing M∗×M→ D.

3 If I remember correctly, Harald Holmann from the University of Fribourg (Switzerland), in hisFernstudienkurs “Aufbau des Zahlsystems” written in the early 1980ies for the Department ofMathematics of Fernuniversitat, used this method to construct the Graves-Cayley octonions overthe reals. This should be checked, and the appropriate historical context should be investigated.


ss.NONSES24.3. Non-singular sesquilinear forms. A sesquilinear form h : M×N→ D overD with right D-modules M,N is said to be non-singular if M and N are both finitelygenerated projective and the adjoint ϕh : M→N∗ is an isomorphism. Note that for hto be non-singular it is necessary that M and N be finitely generated projective rightD-modules having the same rank function. Combining (23.4.1) with Lemma 24.2 wesee that the property of a sesquilinear form to be non-singular is stable under basechange. By a sesquilinear (resp. hermitian) space over D we mean a sesquilinear(resp. hermitian) module (M,h) such that the sesquilinear (resp. hermitian) formh : M×M→ D is non-singular. Given a positive integer n, we speak of a hermitianspace of rank n if the underlying module has rank n as a finitely generated rightD-module.

ss.EXPO

24.4. Exterior powers. Let M,N be right D-modules and h : M × N → D asesquilinear form. Given a positive integer n, we may pass to the n-th exterior power∧n

h :∧n

M×∧n

N −→ D

(well) defined by

(∧n

h)(x1∧·· ·∧ xn,y1∧·· ·∧ yn) = det((

h(xi,y j))

1≤i, j≤n

)(1) EXSE

for xi ∈ M, y j ∈ N, 1 ≤ i, j ≤ n. We call∧n h, which is again a sesquilinear form

over D, the n-th exterior power of h. Passing to the adjoint maps, we conclude thatthe diagram

∧n M∧n

ϕh //

ϕ∧nh ((

∧n(N∗)

ϕ∧n capN

(∧n N)∗

(2) EXAD

commutes. Since exterior powers of right D-modules commute with base change(Bourbaki [12, III, § 7, Prop. 8]), so do exterior powers of sesquilinear forms.

l.EXBA24.5. Lemma. If h : M×N→ D is a non-singular sesquilinear form over D, thenso is its n-th exterior power, for any positive integer n.

Proof. it suffices to show that the vertical arrow in (24.4.2) is an isomorphism. SinceN is finitely generated projective over D, we may assume that it is free of finite rankp≥ n. Let (ei)1≤i≤p be a k-basis of N and (e∗i )1≤i≤p the corresponding dual basis ofN∗. Then

(e j1 ∧·· ·∧ e jn)1≤ j1<···< jn≤p (1) BASEX

is a k-basis of∧n N, while


(e∗i1 ∧·· ·∧ e∗in)1≤i1<···<in≤p (2) SPAEX

is a spanning family of∧n(N∗) as a right D-module such that(

ϕ∧n capN(e∗i1 ∧·· ·∧ e∗in)

)(e j1 ∧·· ·∧ e jn) = det

((〈e∗iλ ,e jµ 〉)1≤λ ,µ≤n

),

which is 1 for (i1, . . . , in) = ( j1, . . . , jn) and 0 otherwise. Thus ϕ∧n capNmaps the

spanning family (2) of∧n(N∗) onto the dual of the basis (1) of

∧n N, hence must bean isomorphism.

r.DUEXT24.6. Remark. Let M be a finitely generated projective right D-module and n apositive integer. Then the preceding result (or its proof) yields an identification∧n(M∗) = (

∧n M)∗ such that the canonical pairing∧n(M∗)×

∧nM = (

∧nM)∗×

∧nM −→ D

is given by〈x∗1∧·· ·∧ x∗n,y1∧·· ·∧ yn〉= det

((〈x∗i ,y j〉)1≤i, j≤n

)for x∗1, . . . ,x

∗n ∈M∗, y1, . . . ,yn ∈M.

ss.DET24.7. Determinants. Let (M,h) be a hermitian space of rank n over D. An isomor-phism ∆ :

∧n V ∼→ D may not exist but if it does, we follow Loos-Petersson-Racine[73, 4.3] and call it an orientation of M; it is unique up to an invertible factor inD. Given an orientation ∆ :

∧n V ∼→ D, there exists a unique element det∆ (h) ∈ k×,called the ∆ -determinant of h, such that ∆ :

∧n(M,h) ∼→ (D,〈det∆ (h)〉sesq) is anisometry. The ∆ -determinant changes with ∆ according to the rule

deta∆ (h) = nD(a)−1 det∆ (h) (a ∈ D×). (1) TRANSDH

If M is free of rank n as a right D-module with basis (ei) and T = (h(ei,e j)) ∈GLn(D) stands for the corresponding hermitian matrix, then det∆ (h) = det(T ),where ∆ :

∧n M ∼→ D is the orientation normalized by ∆(e1∧·· ·∧ en) = 1.ss.HERVE

24.8. Ternary hermitian spaces and the hermitian vector product. Let (M,h)be a hermitian space over D which is ternary in the sense that it has rank n = 3 andsuppose ∆ :

∧3 M ∼→ D is an orientation of M. By non-singularity of h, there existsa unique map M×M→M,(x,y) 7→ x×h,∆ y, such that

h(x×h,∆ y,z) = ∆(x∧ y∧ z). (x,y,z ∈M)

We call ×∆ ,h the hermitian vector product induced by h and ∆ . It is obviously bi-additive, alternating and anti-linear in both arguments. Moreover, the expressionh(x×∆ ,h y,z) remains unaffected by a cyclic change of variables and vanishes if twoof them coincide.

e.HERVE


24.9. Example. In keeping with the previous conventions, we regard D3 as a freeright D-module of rank 3 that is equipped with the canonical basis (e1,e2,e3) ofordinary unit vectors, and denote by x× y the usual vector product of x,y ∈ D3, asdefined 1.1 for the special case k = R, D = C. It satisfies the Grassmann identity(1.1.3), i.e.,

(x× y)× z = y(ztx)− x(zty), (1) GRARE

but also

(Sx)× (Sy) = (S])t(x× y) (2) MAVE

for all x,y,z ∈ D3 and all S ∈ Mat3(D), where S] ∈ Mat3(D) stands for the usualadjoint of S in the sense of linear algebra.

Now suppose T ∈ GL3(D) is a hermitian matrix and consider the ternary hermi-tian space (D3,〈T 〉sesq). Writing ∆0 :

∧3 D3 ∼→ D for the ordinary determinant, i.e.,for the orientation normalized by ∆0(e1∧e2∧e3) = 1, any other volume element onD3 has the form ∆ = a∆0 for some a ∈ D×, and it is easily checked that the hermi-tian vector product induced by 〈T 〉sesq and ∆ relates to the ordinary one accordingto the formula

x×〈T 〉sesq,∆ y = T−1(x× y)a = (T x×Ty)det(T )−1a. (3) HEROR

p.HERGRAS24.10. Proposition. Let (M,h) be a ternary hermitian space over D and suppose∆ :∧3 M ∼→D is a volume element of M. Then the hermitian vector product induced

by h and ∆ satisfies the hermitian Grassmann identity

(x×h,∆ y)×h,∆ z =(yh(z,x)− xh(z,y)

)det∆ (h)−1 (x,y,z ∈M). (1) HERGRAS

Proof. The question is local on D, so we may assume M = D3, h = 〈T 〉sesq, ∆ = a∆0as in Example 24.9. Then the assertion follows from (24.7.1), (24.9.1), (24.9.3) bya straightforward computation.

We will now be able to derive the first main result of this section. It turns out to bea direct generalization of the construction leading to the Graves-Cayley octonions(1.4) and to Thm. 1.7.

24.11. Theorem (Thakur [122]). Let D be a non-singular composition algebra oft.TERHERrank r≤ 2 over k, (M,h) a ternary hermitian space over D and suppose ∆ :

∧3 M ∼→D is an orientation of M satisfying det∆ (h) = 1. Then the k-module D⊕M becomesa composition algebra over k under the multiplication

(a⊕ x)(b⊕ y) :=(ab−h(x,y)

)⊕ (ya+ xb+ x×h,∆ y) (1) TEMU

for a,b ∈D, x,y ∈M. Identifying k⊆D canonically, this composition algebra, writ-ten as


C = Ter(D;M,h,∆),

has unit element, norm, linearized norm, trace, conjugation given by

1C = 1D⊕0, (2) TEUN

nC(a⊕ x) = nD(a)+h(x,x), (3) TENO

nC(a⊕ x,b⊕ y) = nD(a,b)+ tD(h(x,y)

), (4) TELINO

tC(a⊕ x) = tD(a), (5) TETR

a⊕ x = a⊕ (−x) (6) TECONJ

for all a,b ∈ D, x,y ∈M.

Proof. C is clearly a k-algebra with unit element given by (2), and it follows fromExc. 111 below that the quadratic form nC : C→ k as defined in (3), which triviallysatisfies (4), is non-singular. Therefore the theorem will follow once we have shownthat nC permits composition relative to the multiplication (1). Accordingly, usingstandard properties of the hermitian vector product (24.8), we expand

nC((a⊕ x)(b⊕ y)

)= nC

((ab−h(x,y))⊕ (ya+ xb+ x×h,∆ y)

)= nD(ab)−nD

(ab,h(x,y)

)+nD

(h(x,y)

)+h(ya+ xb+ x×h,∆ y,ya+ xb+ x×h,∆ y)

= nD(a)nD(b)−nD(ab,h(x,y)

)+nD

(h(x,y)

)+nD(a)h(y,y)+abh(y,x)+ a bh(x,y)

+nD(b)h(x,x)+h(x×h,∆ y,x×h,∆ y).

Here abh(y,x)+ a bh(x,y) = nD(ab,h(x,y)) by (19.5.4), (19.11.5), and the hermitianGrassmann identity (24.10.1)yields

h(x×h,∆ y,x×h,∆ y) = h(y,(x×h,∆ y)×h,∆ x

)= h(y,yh(x,x)− xh(x,y)

)= h(x,x)h(y,y)−nD

(h(x,y)

).

Hence nC does indeed permit composition relative to (1) and the proof is complete.

ss.TERHER

24.12. The ternary hermitian construction. The composition algebra

C = Ter(D;M,h,∆)

obtained in Thm. 24.11 is said to arise from the parameters involved by means of theternary hermitian construction. Note that C has rank 4r. After the canonical identi-fication a = a⊕0 for a ∈D, the composition algebra C contains D as a compositionsubalgebra. The entire construction will now be reversed by showing that any com-


position algebra containing D as a composition subalgebra arises from D by meansof the ternary hermitian construction.

24.13. Theorem (Thakur [122]). Let C be a composition algebra of rank 4r over kt.HERTERcontaining D as a non-singular composition subalgebra of rank r ≤ 2. Then thereexist a ternary hermitian space (M,h) over D and an orientation ∆ :

∧3 M ∼→ Dsatisfying det∆ (h) = 1 such that the inclusion D → C extends to an isomorphismfrom Ter(D;M,h,∆) onto C.

Proof. Identifying k = k1C ⊆C throughout, we proceed in several steps.10. Since nC is non-singular on D, Lemma 12.10 yields a decomposition

C = D⊕M, M = D⊥ (1) DEPER

as a direct sum of k-submodules. Associativity of the norm ((19.11.3),(19.11.4))implies MD ⊆ M and we claim that M becomes a right D-module in this way.Localizing if necessary we may assume that D is generated by a single element(Thm. 22.15 (b)), in which case the assertion follows immediately from the alterna-tive law. We also observe M ⊆ Ker(tC) and

ax = xa (x ∈M,a ∈ D) (2) COTWI

since ax+ xa = tD(a)x by (19.5.5) and (1).20. Next we define k-bilinear maps h : V ×V → D, ×D : M×M→M by

xy =−h(x,y)+ x×D y, h(x,y) ∈ D, x×D y ∈M, (3) HERFORVEC

for all x,y ∈M. From x2 =−nC(x) we conclude

nC(x) = h(x,x) (4) NORHER

and that the map ×D is alternating. Moreover, xy = yx = (−y)(−x) = yx yields

h(x,y) = h(y,x), (5) HERCONJ

hence

nC(x,y) = tD(h(x,y)

). (6) NORTR

In particular, h(x,y) = 0 for all y ∈M implies x = 0.30. We claim that (M,h) is a hermitian module over D. By (5), it suffices to showthat h is linear in the second variable. Using (19.11.3),(19.11.4) repeatedly and ob-serving that M is a right D-module by 10, we obtain nC(x(yb),a) = nC((xy)b,a) forall x,y ∈ M, a,b ∈ D, and (3) in conjunction with non-singularity of nD gives theassertion.40. Consider the k-trilinear map δ : M3 → D defined by δ (x,y,z) = h(x×D y,z),which by 30 is D-linear in z. We clearly have δ (x,x,z) = 0, but also δ (x,y,y) = 0


since (2), (3) imply (x×D y)y = h(x,y)y+ xy2 = yh(x,y)−nC(y)x ∈M. Hence δ isalternating, forcing it to be in fact D-trilinear, and we obtain a unique D-linear map∆ :

∧3 M→ D satisfying

∆(x∧ y∧ z) = h(x×D y,z). (x,y,z ∈M). (7) VOLEL

50. We now show that (M,h) is a ternary hermitian space over D, ∆ :∧3 M ∼→D is

an orientation of M satisfying det∆ (h) = 1 and ×D is the hermitian vector productinduced by h and ∆ . The final statement follows immediately from (7) as soon asthe preceding ones have been established. To do so, we may assume that k is alocal ring, forcing C to arise from D by a twofold application of the Cayley-Dicksonconstruction (Cor. 22.16): there are units µ1,µ2 ∈ k satisfying

C = Cay(D; µ1, µ2) = D⊕D j1⊕D j2⊕D j3,

where j1 ∈D⊥=M satisfies nC( j1)=−µ1, j2 ∈ (D⊕D j1)⊥ satisfies nC( j2)=−µ2,and j3 = j1 j2. Hence, by (2),

M = j1D⊕ j2D⊕ j3D.

More precisely, (3), (4) and the relations j1 j2 = j3 ∈M, j1 j3 = µ1 j2 ∈M, j3 j2 =µ2 j1 ∈ M show that ( j1, j2, j3) is a basis of M over D with respect to which thematrix of h has the form diag(−µ1,−µ2,µ1µ2)∈GL3(D). Hence (M,h) is a ternaryhermitian space over D; moreover, j1×D j2 = j3 by (3). Thus (7) gives ∆( j1∧ j2∧j3) = h( j3, j3) = µ1µ2 ∈ k×, so ∆ is indeed an orientation of M. The remainingassertion det∆ (h) = 1 is now straightforward to check.60. In view of 50 we can form the composition algebra C′ := Ter(D;V,h,∆) andour construction yields a natural identification C = C′ matching D with the firstsummand of C′.

24.14. Corollary (Pumplun [107]). For 12 ∈ k, every quaternion algebra over k hasc.HERQUAT

the form Ter(k;M,β ,∆), where (M,β ) is a ternary symmetric bilinear space overk and ∆ is a volume element of M satisfying det∆ (β ) = 1. Conversely, every suchalgebra is a quaternion algebra.

e.GCRE24.15. Example. Working over the field k = R of real numbers, we have

O= Ter(C;C3,〈13〉sesq,∆0), H= Ter(R;R3,〈13〉,∆0),

where ∆0 is the normalized orientation of 24.9.

The Coxeter octonions and the examples of Knus-Parimala-Sridharan [61] (see alsoThakur [122]) show that there are octonion algebras to which the ternary hermitianconstruction does not apply since they do not contain any quadratic etale subalge-bras. On the other hand, if 2 ∈ k is sufficiently far removed from being a unit, thenquadratic etale subalgebras always exist.


p.ETSU24.16. Proposition. Let C be a composition algebra of rank r > 1 over k and as-sume 2 ∈ k is contained in the Jacobson radical of k (e.g., 2 = 0 in k). Then Ccontains quadratic etale subalgebras.

Proof. By Lemma 22.11, C contains an element u of trace 1. Then tC(u)2−4nC(u)=1−4nC(u) ∈ k× by hypothesis, so k[u]⊆C is a quadratic etale subalgebra by Prop.22.5.

ss.ZOVE

24.17. Zorn vector matrices. Let D := k⊕k be the split quadratic etale k-algebra.Then ι = ιD is the exchange involution, and the free right D-module D3 = k3⊕ k3

is endowed with the normalized orientation ∆ :∧3 D3 ∼→ D as in 24.7 and with the

unit hermitian form h := 〈13〉sesq which satisfies det∆ (h) = 1. Hence we may formthe octonion algebra

C := Ter(D;D3,h,∆)

over k, which may now be described more explicitly as follows.Writing elements a,b ∈ D, x,y ∈ D3 as

a = α1⊕α2, b = β1⊕β2, x = u1⊕u2, y = v1⊕ v2 (1) RECO

with αi,βi ∈ k, ui,vi ∈ k3, i = 1,2, we have

h(x,y) = u1⊕u2t(v1⊕ v2) = (u2⊕u1)

t(v1⊕ v2),

hence

h(x,y) = (ut2v1)⊕ (ut

1v2), (2) HARE

and applying (24.9.3), we obtain

x×h,∆ y = x× y = (u2× v2)⊕ (u1× v1). (3) VERE

Visualizing the elements of C = D⊕D3 in matrix form as

a⊕ x =(

α1 u2u1 α2

), b⊕ y =

(β1 v2v1 β2

),

we deduce from (24.11.1), (2), (3) that(α1 u2u1 α2

)(β1 v2v1 β2

)=

(α1β1−ut

2v1 α1v2 +β2u2 +u1× v1β1u1 +α2v1 +u2× v2 α2β2−ut

1v2

). (4) ZOMU

Consulting Thm. 24.11, we therefore conclude that the multiplication rule (4) con-verts the k-module

Zor(k) :=(

k k3

k3 k

)(5) ZOMA


into an octonion algebra, called the octonion algebra of Zorn vector matrices overk. Norm, trace and conjugation of this octonion algebra are given by

nZor(k)

((α1 u2u1 α2

))= α1α2 +ut

1u2, (6) ZONO

tZor(k)

((α1 u2u1 α2

))= α1 +α2, (7) ZOTR(

α1 u2u1 α2

)=

(α2 −u2−u1 α1

). (8) ZOCONJ

ss.SPLICO24.18. Split composition algebras. Let C be a composition algebra over k. We saythat C is

(a) split of rank 1 if C ∼= k,(b) split of rank 2 or split quadratic etale if C ∼= k⊕ k (as a direct sum of ideals).(c) split of rank 4 or a split quaternion algebra if C ∼= Mat2(k),.(d) split of rank 8 or a split octonion algebra if C ∼= Zor(k).

If C is an arbitrary composition algebra over k, we consider its rank decomposition(§10, Exercise 42), which in the special case at hand, thanks to Corollary 22.17,at-tains the form

k = k0⊕ k1⊕ k2⊕ k3, ki := kεi (0≤ i≤ 3), (1) KKEIF

C =C0⊕C1⊕C2⊕C3, Ci :=C⊗ ki (0≤ i≤ 3) (2) CKEIF

as direct sums of ideals induced by a complete orthogonal system (εi)0≤i≤3 of idem-potents in k uniquely determined by the condition that Ci is a composition algebraof rank 2i over ki for 0 ≤ i ≤ 3. Then C is said to be split if Ci is split of rank 2i

for all i = 0,1,2,3. Note that the property of a composition algebra to be split (resp.split of rank r ∈ 1,2,4,8) is stable under base change.

For the split composition algebras of rank r = 1,2,4,8 over k exhibited in (a)−(d)above, it is sometimes helpful to introduce a unified notation. We put

C0r(k) :=

k for r = 1,k⊕ k for r = 2,Mat2(k) for r = 4,Zor(k) for r = 8

(3) NOTASP

and haveC0r(k)R =C0r(R)

for all R ∈ k-alg. We call C0r(k) the standard split composition algebra of rank rover k.

Exercises.


pr.INQUA 111. Let D be a quadratic etale k-algebra, use Exc. 102 to identify k = H(D, ιD)⊆ D canonically,and let (M,h) be a hermitian space of rank n over D.

(a) If k is a local ring, show that h can be diagonalized: there exist a basis (ei)1≤i≤n of M as aright D-module and scalars α1, . . . ,αn ∈ k× such that h(ei,e j) = δi jαi for 1≤ i, j ≤ n.

(b) Deduce from (a) that (M,q), where M is viewed canonically as a k-module and q : M→ kis defined by q(x) := h(x,x) for x ∈M, is a quadratic space of rank 2n over k.

pr.HERISOT 112. Isotopes of the ternary hermitian construction. Let D be a quadratic etale k-algebra, (M,h) aternary hermitian space over D and ∆ :

∧3 M ∼→D an orientation of M satisfying det∆ (h) = 1. PutC = Ter(D;M,h,∆). For p ∈ D× ⊆C×, we refer to the concept of unital p-isotopes as defined in16.9 and have D = Dp ⊆ Cp, so Thm. 24.13 yields a ternary hermitian space (M,h)p = (Mp, hp)over D and an orientation ∆ p of Mp such that Cp = Ter(D;Mp,hp,∆ p). Describe Mp,hp,∆ p ex-plicitly. What does this description mean for the octonion algebras of Zorn vector matrices?

pr.TERFU 113. (Thakur [122]) For i = 1,2, let (Mi,hi) be ternary hermitian spaces over a quadratic etalek-algebra D and ∆i :

∧3 Mi∼→ D an orientation of Mi satisfying det∆i (hi) = 1. Put

Ci := Ter(D;Mi,hi,∆i) = D⊕Mi (i = 1,2)

and prove for any map χ : M1→M2 that the following conditions are equivalent.

(i) χ : (M1,h1,∆1)∼→ (M2,h2,∆2) is an isomorphism, i.e., χ : (M1,h1)

∼→ (M2,h2) is an isom-etry satisfying ∆2 (

∧3χ) = ∆1.

(ii) χ : (M1,h1)∼→ (M2,h2) is an isometry satisfying

χ(x×h1,∆1 y) = χ(x)×h2,∆2 χ(y). (x,y ∈M1).

(iii) 1D⊕χ : C1∼→C2 is an isomorphism.

pr.PRESPLI 114. A presentation of the split octonions. Show that C := Zor(k) is the free unital alternativek-algebra on three generators E,X1, ,X2 satisfying the relations

E2 = E, X21 = X2

2 = 1C, X1X2X1 =−X2, (4) PRESPLI

X1EX1 = X2EX2 =−(X1X2)E(X1X2) = E := 1C−E,

i.e., that C has three generators E,X1,X2 satisfying (4) and, conversely, given any unital alterna-tive k-algebra A and three element e,x1,x2 ∈ A such that (4) holds (after the obvious notationaladjustments), there exists a unique homomorphism C→ A of unital k-algebras sending E,X1,X2respectively to e,x1,x2. Conclude that, for some ideal a ⊆ k, the subalgebra of A generated bye,x1,x2 is isomorphic to Zor(k/a) as a k-algebra.

25. Reduced composition algebras

s.REDCOIn the classical theory of finite-dimensional linear non-associative algebras overfields, the standard way to define the notion of a reduced algebra consists in re-quiring the existence of a complete orthogonal system of absolutely primitive idem-potents. This approach will be adapted here to the setting of composition algebrasover arbitrary commutative rings. We then proceed to explore more accurately thestructure of reduced composition algebras and compare it with the more restrictivenotion of splitness as defined in 24.18.

Throughout we let k be an arbitrary commutative ring.

25 Reduced composition algebras 147

ss.PRIAB25.1. Primitive and absolutely primitive idempotents. Recall from, e.g., Braun-Koecher [15, p. 52] or Schafer [114, pp. 39,56] that an idempotent of an algebra overa field is said to be primitive if it is non-zero and it cannot be decomposed into thesum of two non-zero orthogonal idempotents; we speak of an absolutely primitiveidempotent if it remains primitive in every base field extension. These definitionshave natural extensions to algebras over arbitrary commutative rings as follows.

Let A be a k-algebra. An idempotent c ∈ A is called primitive if c 6= 0 and forall orthogonal idempotents c1,c2 ∈ A satisfying c = c1 + c2, there exists a completeorthogonal system (ε1,ε2) of idempotents in k such that ci = εic for i = 1,2; thisnotion obviously reduces to the previous one if k is a field (or, more generally,a connected commutative ring). c is said to be absolutely primitive if it remainsprimitive in every scalar extension AR, R ∈ k-alg, R 6= 0.

Note that for an alternative algebra A over a field, an idempotent c∈A is primitiveif and only if the subalgebra A11(c)⊆ A contains no idempotents other than 0 and c.

p.ABUN25.2. Proposition. Let A be an algebra over k and suppose A is finitely gener-ated projective as a k-module. Then every absolutely primitive idempotent of A is aunimodular element.

Proof. Let c ∈ A be an absolutely primitive idempotent. Then, for all p ∈ Spec(k),the idempotent c(p)∈ A(p) is primitive, hence non-zero. The assertion follows fromLemma 10.13.

ss.SEPALTAL

25.3. Separable alternative algebras over rings. Recall from Schafer [114, p. 58]that a finite-dimensional alternative algebra A over a field F is said to be separableif, for every field extension K/F , the extended alternative algebra AK over K is semi-simple in the sense of 18.3. Using the theory of Azumaya algebras (Knus-Ojanguren[60, III, Thm. 5.1]) or that of separable Jordan pairs (Loos [69]) as a guide, thisnotion will now be extended to algebras over commutative rings as follows.

A unital alternative algebra C over k is said to be separable if

(i) C is finitely generated projective as a k-module.(ii) C(p) is a (finite-dimensional) separable (alternative) algebra over the field

κ(p), for all p ∈ Spec(k).

We claim that composition algebras are separable alternative. Indeed, alternativityand condition (i) follow from Thm. 22.8 while, in order to establish condition (ii),it suffices to show that a composition algebra C over a field F is semi-simple. As-suming (as we may) dimF(C) > 1, the norm of C is a non-singular quadratic form(Cor. 22.17), forcing Nil(C) = 0 by Exc. 77.

For arbitrary separable alternative algebras, the property of an idempotent tobe absolutely primitive can be characterized by means of its Peirce decomposition(Exc. 67).

p.ABPRI


25.4. Proposition. Let A be a separable alternative algebra over k. For an idem-potent c ∈ A to be absolutely primitive it is necessary and sufficient that A11(c) befree of rank 1 as a k-module. In this case, c is a basis of A11(c).

Proof. Note first that by Exc. 67 the Peirce components of A relative to c are com-patible with base change, i.e., Ai j(c)R ∼= (AR)i j(cR) canonically, for all i, j = 1,2,R ∈ k-alg. In order to prove sufficiency, suppose A11(c) is free of rank 1 as a k-module. Then c is a basis of A11(c) and obviously a primitive idempotent, hencean absolutely primitive one since the property of A11(c) to be free of rank 1 re-mains stable under base change. Conversely, in order to prove necessity, suppose cis absolutely primitive. The Peirce decomposition shows that A11(c) is finitely gen-erated projective as a k-module. For any prime ideal p ⊆ k, let K ∈ k-alg be thealgebraic closure of the residue field κ(p). Then cK ∈CK is a primitive idempotentand, combining Lemma 18.9 with Cor. 18.11, we see that A11(c)K ∼= (AK)11(cK) is afinite-dimensional alternative division algebra over the algebraically closed field K.By Exc. 37, this implies rkp(A11(c)) = dimK((AK)11(cK)) = 1, so A11(c) is a linebundle. But c ∈ A11(c) is a unimodular vector by Prop. 25.2, forcing A11(c) to befree of rank 1 with basis c.

ss.REDCOAL

25.5. Reduced composition algebras A composition algebra over k is said to bereduced if it contains an absolutely primitive idempotent. The base ring k itself isclearly a reduced composition algebra of rank 1. In order to describe the reducedcomposition algebras of rank > 1, we require a few preparations. The first one ofthese connects absolutely promitive idempotents with elementary ones as defined inExc. 82.

p.ABPRIELID

25.6. Proposition. Elementary idempotents in conic algebras are absolutely prim-itive. Conversely, let C be a composition algebra of rank r > 1 over k. The everyabsolutely primitive idempotent in C is elementary.

Proof. Let C be a conic k-algebra and suppose c ∈C is an elementary idempotent.Since the property of an idempotent to be elementary is stable under base change, tsuffices to show for the first part of the proposition that c is primitive. Accordingly,assume c = c1 +c2 with orthogonal idempotents ci ∈C, i = 1,2. Then 2ci = cci =tC(c)ci + tC(ci)c−nC(c,ci)1C and we conclude ci = tC(ci)c−nC(c,ci)1C. Multiply-ing this by c, we obtain ci = εic for some εi ∈ k, and since c is unimodular, (ε1,ε2) isa complete orthogonal system of idempotents in k. Thus c is primitive, as claimed.

For the second part of the proposition, assume C is a composition algebra of rankr > 1 and let c ∈ C be an absolutely primitive idempotent. Then so is cR, for anyR ∈ k-alg, R 6= 0. This gives cR 6= 0 by definition, but also cR 6= 1CR since r > 1and C11(c) is free of rank 1 as a k-module by Prop. 25.4. Hence c is elementary.

Remark. The proof of the first part of the preceding proposition shows, more gen-erally,

kc = x ∈C | cx = x = xc

for any elementary idempotent in a conic algebra C over k.


p.CHAREDCOAL25.7. Proposition. Let C be a composition algebra of rank r > 1 over k. Then thefollowing conditions are equivalent.

(i) C is reduced.(ii) C contains a split quadratic etale subalgebra.(iii) The norm of C is isotropic.(iv) The norm of C is hyperbolic.

If these conditions are fulfilled and c∈C is an absolutely primitive idempotent, then

nC ∼= hkc⊕C12(c). (1) NORHYP

Proof. (i)⇒ (ii). By definition and Prop. 25.6, C contains an elementary idempotentc, whence Exc. 82 shows that kc⊕ kc⊆C is a split quadratic etale subalgebra.

(ii)⇒ (iii). Let D ⊆C be a split quadratic etale subalgebra. Since nD = nC|D isisotropic (Ex 22.9 (a)), so is nC.

(iii)⇒ (i). Choose an isotropic vector u ∈C relative to nC. Since u is unimodularand nC is non-singular, some v ∈ C satisfies tC(uv) = nC(u, v) = 1. But nC(uv) =nC(u)nC(v) = 0, so (by Exc. 82) c = uv ∈ C is an elementary, hence absolutelyprimitive, idempotent.

(i)⇒ (iv). Let c ∈C be an elementary idempotent with Peirce components Ci j =Ci j(c), i, j = 1,2. By (4), (5) of Exc. 105, the norm nC determines a duality betweenthe k-modules kc⊕C12 and kc⊕C21, on which it is identically zero. Hence (1) and(iv) follow.

(iv) ⇒ (iii). If (iv) holds, C = M1 ⊕M2 may be written as the direct sum oftwo totally isotropic submodules Mi ⊆C relative to nC, i = 1,2. In particular, 1C =e1 + e2, ei ∈Mi, i = 1,2, which implies nC(ei) = 0, 1 = nC(1C) = nC(e1,e2). Thus(e1,e2) is a hyperbolic pair in (C,nC).

c.RECOT25.8. Corollary. The reduced composition algebras of rank 2 are precisely the splitquadratic etale k-algebras.

ss.TWIMA25.9. Twisted 2×2-matrices. Let L be a line bundle over k. Then there is a naturalidentification

Endk(k⊕L) =(

k L∗

L k

)as k-algebras, where multiplication on the right is the usual matrix product, takingadvantage of the canonical pairing L∗×L→ k at the appropriate place. We claimthat C := Endk(k⊕L) is a quaternion algebra over k, with norm, trace, conjugationrespectively given by


nC

((α1 u∗

v α2

))= α1α2−〈u∗,v〉, (1) TWINO

tC((

α1 u∗

v α2

))= α1 +α2, (2) TWITR(

α1 u∗

v α2

)=

(α2 −u∗

−v α1

). (3) TWICONJ

for α1,α2 ∈ k, u∗ ∈ L∗ and v ∈ L. Since passing to the dual of a finitely generatedprojective module by Lemma 10.11 commutes with base change, so does the con-struction of C. Hence our claim may be checked locally, in which case L is a freek-module of rank 1 and C becomes isomorphic to the split quaternion algebra of2× 2-matrices, with (1)–(3) converted into the formulas for the ordinary determi-nant, trace, conjugation, respectively, of 2× 2-matrices. We call C = Endk(k⊕ L)the quaternion algebra of L-twisted 2×2-matrices over k. Note that

e :=(

1 00 0

), e′ :=

(0 00 1

)(4) DIACON

form a complete orthogonal system of elementary idempotents in C with off-diagonal Peirce components

C12(e) =(

0 L∗

0 0

), C21(e) =

(0 0L 0

). (5) OFFPE

We can now characterize reduced quaternion algebras in the following way.p.RECOF

25.10. Proposition. Up to isomorphism, the reduced quaternion algebras over kare are precisely the quaternion algebras of L-twisted 2×2-matrices, for some linebundle L over k. More precisely, let C be a quaternion algebra over k and c ∈ Can absolutely primitive idempotent. Then there exist a line bundle L over k and anisomorphism

φ : C ∼−→ Endk(k⊕L)

such that

φ(c) =(

1 00 0

). (1) COABF

Proof. Let L be a line bundle over k. By (25.9.1), the norm of C := Endk(k⊕ L)is hyperbolic, and Prop. 25.7 shows that C is reduced. Conversely, suppose C isa reduced quaternion algebra over k and c1 := c ∈ C is an absolutely primitiveidempotent. Then c is an elementary idempotent by Proposition 25.6, so the cor-responding Peirce decomposition of C takes on the form C = kc1⊕C12⊕C21⊕kc2,with c2 := 1C − c1 and finitely generated projective k-modules C12, C21 which,by §22, Exercise 105, are in duality under the bilinearized norm. Counting ranks,we conclude that L := C21 is a line bundle over k and C12 = L∗ under the natural


identification induced by the bilinearized norm. Since C is associative, §15, Exer-cise 67, shows C2

i j = 0. By the same token, given x21 ∈C21, y∗21 ∈C12, we obtainy∗21x21 = αc1 for some α ∈ k, and taking traces in conjunction with (7) of §22, Ex-ercise 105, yields

α = tC(y∗21,x21) = nC(y∗21, x21) =−nC(y∗21,x21) =−〈y∗21,x21〉.

Thus y∗21x21 =−〈y∗21,x21〉c1. But then

x21y∗21 = (−x21)(−y∗21) = x21y∗21 = y∗21x21 =−〈y∗21,x21〉c2.

Now one checks easily that

φ : C ∼−→ End(k⊕L), α1c1⊕ x∗21⊕ x21 +α2c2 7−→(

α1 −x∗21x21 α2

)is an isomorphism of quaternion algebras having the desired property.

ss.TWIZO25.11. Twisted Zorn vector matrices. Let M be a finitely generated projective k-module of rank 3 and θ an orientation of M, i.e., a k-linear bijection θ :

∧3 M ∼→ k.Dualizing by means of the identification 24.6, we obtain a k-linear bijection θ ∗ : k =k∗ ∼→ (

∧3 M)∗ =∧3(M∗), hence an induced orientation θ ∗−1 on M∗ uniquely deter-

mined by the condition

θ(v1∧ v2∧ v3)θ∗−1(v∗1∧ v∗2∧ v∗3) = det

((〈v∗i ,v j〉)1≤i, j≤3

)(1) DUVO

for v j ∈M, v∗i ∈M∗, 1 ≤ i, j ≤ 3. These two orientations in turn give rise to alter-nating bilinear maps

M×M −→M∗, (v,w) 7−→ v×θ w,

M∗×M∗ −→M, (v∗,w∗) 7−→ v∗×θ w∗,

called the associated vector products, according to the rules

θ(u∧ v∧w) = 〈u×θ v,w〉, θ∗−1(u∗∧ v∗∧w∗) = 〈w∗,u∗×θ v∗〉 (2) TRIVE

for u,v,w ∈M,u∗,v∗,w∗ ∈M∗. Now the k-module

Zor(M,θ) =

(k M∗

M k

)becomes a unital k-algebra under the multiplication(

α1 v∗

v α2

)(β1 w∗

w β2

)=

(α1β1−〈v∗,w〉 α1w∗+β2v∗+ v×θ w

β1v+α2w+ v∗×θ w∗ −〈w∗,v〉+α2β2

)(3) TWIZOMU

for αi,βi ∈ k,v,w∈M,v∗,w∗ ∈M∗, i= 1,2, whose unit element is the identity matrix12. We call Zor(M,θ) the algebra of (M,θ)-twisted Zorn vector matrices over k.


We claim that C := Zor(M,θ) is an octonion algebra over k, with norm, trace,conjugation given by

nC

((α1 u∗

v α2

))= α1α2−〈u∗,v〉, (4) NOTWI

tC((

α1 u∗

v α2

))= α1 +α2, (5) TRTWI(

α1 u∗

v α2

)=

(α2 −u∗

−v α1

). (6) CONJTWI

for α1,α2 ∈ k, u∗ ∈M∗, v ∈M. Since the construction of C, as in the quaternioniccase, commutes with base change, this may be checked locally, so we may assumethat M is a free k-module of rank 3. Hence there exist elements u1,u2,u3 ∈M satis-fying the following equivalent conditions.

(i) B := (u1,u2,u3) is a basis of M.(ii) u1∧u2∧u3 ∈

∧3(M) is unimodular.(iii) θ(u1∧u2∧u3) ∈ k×.

Replacing, e.g., u3 by an appropriate scalar multiple, we may therefore assume B

to be θ -balanced in the sense that θ(u1 ∧ u2 ∧ u3) = 1. Hence we find a naturalidentification M = k3 such that B = (e1,e2,e3) is the basis of ordinary unit vectorsand θ = det :

∧3(k3)→ k is given by the ordinary determinant. This in turn yieldsa natural identification M∗ = k3 such that the canonical pairing M∗×M→ k agreeswith the map k3× k3 → k, (u,v) 7→ utv. Thus the basis (ei)1≤i≤3 is selfdual, andapplying (1) we conclude θ ∗−1 = det as well. Hence both vector products×det agreewith the ordinary vector product × on k3 and Zor(k3,det) = Zor(k) is the same asthe algebra of ordinary Zorn vector matrices over k. The assertion follows. As in thecase of reduced quaternion algebras,

e :=(

1 00 0

), e′ :=

(0 00 1

)(7) DIACONO

form a complete orthogonal system of elemenatary idempotents in C with off diag-onal Peirce components

C12(e) =(

0 M∗

0 0

), C21(e) =

(0 0M 0

)(8) OFFPEO

Remark. In [98], the algebras of (M,θ)-twisted Zorn vector matrices have beencalled split octonion algebras. K. McCrimmon (oral communication) has rightlyobjected that splitness for composition algebras should be defined in such a way asto allow, up to isomorphism, precisely one example in each rank 1,2,4,8. We haveadopted his point of view in the definition 24.18 above.


25.12. Theorem (Petersson [98]). Up to isomorphism, the reduced octonion alge-t.REDCOEbras over k are precisely the algebras of (M,θ)-twisted Zorn vector matrices, forsome finitely generated projective k-module M of rank 3 and some orientation θ ofM. More precisely, given an octonion algebra C over k and an absolutely primitiveidempotent c∈C, there exist M,θ as above and an isomorphism φ : C ∼→ Zor(M,θ)such that

φ(c) =(

1 00 0

). (1) COABE

Proof. Let M be a finitely generated projective k-module of rank 3 and θ an orien-tation of M. Then (25.11.4) shows that the norm of the octonion algebra Zor(M,θ)is hyperbolic, forcing the algebra itself to be reduced (Prop. 25.7). Conversely, sup-pose C is a reduced octonion algebra over k and c1 := c∈C is an absolutely primitiveidempotent. Then c is an elementary one by Proposition 25.6, so the correspondingPeirce decomposition takes on the form C = kc1⊕C12⊕C21⊕kc2, with c2 := 1C−c1and finitely generated projective k-modules C12, C21 in duality under the bilinearizednorm (§22, Exercise 105). Counting ranks, we conclude that M := C21 is a finitelygenerated projective k-module of rank 3 and C12 = M∗ under the natural identifi-cation induced by the bilinearized norm. Given v ∈ M = C21, u∗ ∈ M∗ = C12, weclaim

u∗v =−〈u∗,v〉c1, vu∗ =−〈u∗,v〉c2. (2) COVO

Indeed, the Peirce rule (3) of §15, Exercise 67, yield u∗v = αc1 for some α ∈ k, andtaking traces in conjunction with (7) of §22, Exercise 105, we conclude

α = tC(u∗v) = nC(u∗, v) =−nC(u∗,v) =−〈u∗,v〉.

Hence the first equation of (2) holds; the second one follows analogously. Now com-bine (19.11.5) with §22, Exercise 105, and observe that the expression tC(uvw) =nC(uv, w) = −nC(uv,w) is alternating trilinear in u,v,w ∈ M. Since M2 ⊆ M∗

by the Peirce rules (§15, Exercise 67), we therefore obtain a unique linear mapθ :

∧3(M)→ k such that

θ(u∧ v∧w) =−tC(uvw) = nC(uv,w) = 〈uv,w〉 (u,v,w ∈M). (3) COVE

We claim that θ is an isomorphism. In order to see this, we may assume that kis a local ring and first treat the case that k is a field. Then we must show θ 6= 0.Otherwise, 〈uv,w〉 = nC(uv,w) = 0 for all u,v,w ∈ M, hence M2 = 0. But then,given u,v ∈ M, w∗ ∈ M∗, linearized right alternativity yields (uv)w∗ + (uw∗)v =u(vw∗)+u(w∗v), which combines with (2) to imply 〈w∗,u〉c2v= 〈w∗,v〉(uc2+uc1),hence the contradiction 〈w∗,u〉v = 〈w∗,v〉u. We are left with the case that k is a localring. Given any basis (u1,u2,u3) of M, the case just treated implies that θ(u1∧u2∧u3) does not vanish after passing to the residue field of k. Hence θ(u1 ∧u2∧u3) isa unit in k, and we conclude that θ is indeed an isomorphism. Moreover, combining


(3) with (25.11.2) we obtain u×θ v = uv for the associated vector product of u,v ∈M. Our next aim is to prove

θ∗−1(u∗∧ v∗∧w∗) = nC(u∗v∗,w∗) = 〈w∗,u∗v∗〉 (u∗,v∗,w∗ ∈M∗). (4) COVEST

Localizing if necessary, we may assume that M is a free k-module, with basis(ei)1≤i≤3 chosen in such a way that θ(e1 ∧ e2 ∧ e3) = 1. If we write (e∗i )1≤i≤3 forthe corresponding dual basis of M∗, then (25.11.1) shows θ ∗−1(e∗1 ∧ e∗2 ∧ e∗3) = 1,and since both sides of (4) are alternating trilinear in u∗,v∗,w∗, the proof will becomplete once we have shown e∗1e∗2 = e3. Consulting (3), and making use of its al-ternating character, we obtain e1e2 = e∗3, whence a cyclic change of variables alsoyields e2e3 = e∗1, e3e1 = e∗2. Since ei = −ei, e∗i = −e∗i for i = 1,2,3 by (7) of §22,Exercise 105, the middle Moufang identity (14.3.3) combined with (20.3.2) yields

e∗1e∗2 = e∗1e∗2 = (e3e2)(e1e3) = e3(e2e1)e3 =−e3(e1e2)e3 =−e3e∗3e3

= −nC(e3, e∗3)e3 +nC(e3)e∗3 = 〈e∗3,e3〉e3 = e3,

and (4) is proved. Consequently, u∗×θ v∗ = u∗v∗ ∈ M is the corresponding vectorproduct of u∗,v∗ ∈M∗. Now one checks easily that φ : C ∼→ Zor(M,θ) defined by

φ(α1c1⊕ v∗⊕u⊕α2c2) :=(

α1 v∗

u α2

)for α1,α2 ∈ k, v∗ ∈M∗ =C12, u ∈M =C21 is an isomorphism of octonion algebrassatisfying (1).

c.SPLIRE25.13. Corollary. Split composition algebras over any commutative ring are re-duced. Conversely, if k is a local ring, then every reduced composition algebra overk is split.

Proof. The first part follows by comparing the list 24.18 with Prop. 25.7. Con-versely, let k be a local ring and C a reduced composition algebra of rank r = 1,2,4,8over k. While the case r = 1 is trivial, and the case r = 2 follows immediatelyfrom Prop. 25.7, the remaining cases r = 4,8 are a consequence of Prop. 25.10and Thm. 25.12 since finitely generated projectives over k are free.

c.SPLIDI25.14. Corollary. For a composition algebra C of dimension > 1 over a field F,the following conditions are equivalent.

(i) C is split.(ii) The norm of C is isotropic.(iii) The norm of C is hyperbolic.(iv) C is not a division algebra.

Proof. By Cor. 25.13, C is split if and only if it is reduced. Thus (i)–(iii) are equiv-alent (Prop. 25.7). Moreover, the implication (i)⇒(iv) is obvious, while (iv)⇒(ii)follows from Prop. 20.4.


p.CORED25.15. Proposition. Let C be a composition algebra over k and D⊆C a quadraticetale subalgebra. Then CD, the base change of C from k to D, is reduced.

Proof. By Lemma 12.10, we have C = D⊕D⊥ as a direct sum of k-modules, whichimplies CD = DD⊕ (D⊥)D as a direct sum of D-modules. But DD is split quadraticetale (Exc. 106 (a)), forcing C to be reduced by Prop. 25.7.

c.COSPLI25.16. Corollary. Let C be a composition algebra of rank r > 1 over the local ringk. Then every quadratic etale subalgebra of C splits C.

Proof. This follows immediately by combining Prop. 25.15 with Cor. 25.13.

Remark. Quadratic etale subalgebras of C as In Cor. 25.16 exist by Thm. 22.15 (b).

Exercises.

pr.DECAPID 115. Let C be a composition algebra over k and c an idempotent in C. Use Exc. 84 to show that cis absolutely primitive if and only if there exist a decomposition

k = k(1)⊕ k(2) (1) DECAPBAS

as a direct sum of ideals, a composition algebra C(1) over k(1) and an elementary idempotent c(1) ∈C(1) such that

C =C(1)⊕ k(2) (2) DECAPAL

as composition algebras over k andc = c(1)⊕1k(2) .

Conclude that C is reduced if and only if it allows decompositions as in (1), (2) such that C(1) is acomposition algebra over k(1) whose rank (as a function Spec(k)→ Z) is nowhere equal to 1.

pr.ELIDREDQUAT 116. Elementary idempotents in reduced quaternion algebras. Let L0 be a line bundle over k andwrite

B = Endk(k⊕L0) =

(k L∗0

L0 k

)for the corresponding reduced quaternion algebra over k. Prove:

(a) For c ∈ B, the following conditions are equivalent.

(i) c is an elementary idempotent.(ii) Viewing c as a linear map k⊕L0→ k⊕L0,

Lc := Im(c)⊆ k⊕L0

is a line bundle over k.(iii) There exist a line bundle L over k and an isomorphism

Φ : B ∼−→ Endk(k⊕L) =(

k L∗

L k

)such that

Φ(c) =(

1 00 0

).


In this case c ∈ B is an elementary idempotent as well and

k⊕L0 = Lc⊕Lc, L0 ∼= Lc⊗Lc. (3) ELCE

Moreover, L in (iii) is unique up to isomorphism and

L∼= B21(c)∼= L0⊗L∗⊗2c , B12(c)∼= L∗0⊗L⊗2

c . (4) BETCE

(b) Two elementary idempotents c,d ∈ B are conjugate under inner automorphisms of B if andonly if Lc ∼= Ld . They are conjugate under arbitrary automorphisms of B if and only ifL⊗2

c∼= L⊗2

d .(c) If L is any line bundle over k, then L∼= Lc for some elementary idempotent c ∈ B if and only

if L is (isomorphic to) a direct summand of k⊕L0.

pr.LITWO 117. Line bundles on two generators. Let L be a line bundle over k.

(a) Let n be a positive integer and suppose there are elements f1, . . . , fn ∈ k such that k = ∑k fiand L fi is free (of rank one) over k fi for 1≤ i≤ n. Show that L is generated by n elements.

(b) Show that the following conditions are equivalent.

(i) L is generated by two elements.(ii) L⊕L∗ is a free k-module of rank two.(iii) There exists an elementary idempotent c ∈ B := Mat2(k) such that L∼= Lc.(iv) There exist elements f1, f2 ∈ k such that k f1 + k f2 = k and L fi is free (of rank one)

over k fi for i = 1,2.

Conclude that if L satisfies one (hence all) conditions (i)−(iv) above, then so does L⊗n,for all n ∈ Z. In particular, there exists an elementary idempotent c(n) ∈ Mat2(k), uniqueup to conjugation by inner automorphisms of B, satisfying L⊗n ∼= Lc(n) . Describe such anidempotent as explicitly as possible.

Remark. Let k be a Dedekind domain. Then there is a canonical identification of the class groupof k with its Picard group, and every fractional ideal of k is generated by two elements (O’Meara[90, 22:5b]). Hence the preceding two exercises yield a bijective correspondence between the setof conjugacy classes of elementary idempotents in Mat2(k) under inner automorphisms and theclass group of k. In particular, for k the ring of integers of an algebraic number field K := Quot(k),the number of these conjugacy classes agrees with the class number of K.

pr.ISOREDQUAT 118. Isomorphisms of reduced quaternion algebras. Let L0,L′0 be line bundles over k and B =Endk(k⊕L0), B′ = Endk(k⊕L′0) the corresponding reduced quaternion algebras. Use Exc. 116 toshow that the following conditions are equivalent.

(i) B∼= B′.(ii) Every line bundle L over k that is a direct summand of k⊕L0 admits a line bundle L′ over

k that is a direct summand of k⊕L′0 and satisfies

L0⊗L′⊗2 ∼= L′0⊗L⊗2. (5) ISOCOND

(iii) There exist line bundles L,L′ over k that are direct summands of k⊕L0, k⊕L′0, respectively,and satisfy (5).

pr.FQUATSPL 119. Let L be a line bundle over k. Prove that the reduced quaternion algebra Endk(k⊕L) is splitif and only if L ∼= L′⊗2 for some line bundle L′ on two generators over k. Use this to constructexamples of reduced quaternion algebras over k that are free as k-modules but not split.

pr.FAIWITCAN 120. Failure of Witt cancellation. Let L be a line bundle on two generators over k that is not ofperiod 2 in Pic(k). Put L′ := L⊗2 and show h⊥ h∼= h⊥ hL′ even though hL′ is not split. (Hint. UseExc. 119 to prove that the reduced quaternion algebra Endk(k⊕L′) is split.)

26 Norm equivalences and isomorphisms 157

pr.GRASSSPL 121. Let M be a finitely generated projective k-module of rank 3 and θ an orientation of M. Show

(u×θ v)×θ w∗ = 〈w∗,u〉v−〈w∗,v〉u, (u∗×θ v∗)×θ w = 〈u∗,w〉v∗−〈v∗,w〉u∗

for all u,v,w ∈M, u∗,v∗,w∗ ∈M∗.

pr.TEREM 122. Work out the connection between the ternary hermitian construction of octonion algebras(24.12) and twisted Zorn vector matrices (25.11). Use this to give a new proof of Thm. 25.12 byreducing it to Thm. 24.13.

pr.OCTDEDDOM 123. Prove that every reduced octonion algebra over a Dedekind domain is split. (Hint. Use thewell known structure of finitely generated projective modules over Dedekind domains, cf. Lang[65, III, Exc. 11 – 13].)

pr.DEDOC 124. (Petersson [93]) Let k be an Dedekind domain and C an octonion algebra over k. Show thatI ⊆C is a one-sided ideal if and only if I = aC for some ideal a ⊆ k. (Hint. Assume first that k isa field, then, that it is a discrete valuation ring, before using Exc. 104 to treat the problem in fullgenerality.)

Remark. The preceding result generalizes the Mahler [74]-Van der Blij-Springer [125] theorem,which reaches the same conclusion for the Coxeter octonions over k = Z. Whether it remains validover arbitrary commutative rings seems to be an open question. It certainly fails for quaternionalgebras, the split quaternions over any ring serving as a particularly easy counter example.

26. Norm equivalences and isomorphisms

s.NOREQOne of the most important results in the classical theory of composition algebrasis the norm equivalence theorem. It says that composition algebras over fields areclassified by their norms, in other words, two composition algebras over a fieldare isomorphic if and only if their norms are equivalent, i.e., isometric. Our firstaim in this section will be to establish this result in the more general setting ofsemi-local rings. As in the case of fields, the key ingredient of the proof is Wittcancellation of non-singular quadratic forms, which fails over arbitrary commutativerings (see Exc. 120 below) but is valid over semi-local ones. As an application, weclassify composition algebras over special fields, like the reals and the complexes,finite fields, the p-adics and algebraic number fields. The section also contains a fewgeneral comments on norm equivalence over arbitrary commutative rings.

Let k be a commutative ring. We begin by introducing the concept of norm equi-valence and some useful modifications.

ss.NORSIM

26.1. Norm similarities. Let C, C′ be conic algebras over k. A norm similarityfrom C to C′ is a bijective k-linear map f : C→ C′ such that there exists a scalarµ f ∈ k× satisfying nC′ f = µ f nC; in this case, µ f = nC′( f (1C)) is unique and calledthe multiplier of f . Norm similarities with multiplier 1 are called norm isometries ornorm equivalences. We say that C and C′ are norm similar (resp. norm equivalent)if there exists a norm similarity (resp. a norm equivalence) from C to C′. Norm sim-ilarities f : C→C′ preserving units (so f (1C) = 1C′ ) automatically have multiplier1; we speak of unital norm equivalences in this context.


It is clear that the preceding notions are stable under base change. Generallyspeaking, the principal objective of the present section is to understand the connec-tion between unital norm equivalences and isomorphisms (resp. anti-isomorphisms)of composition algebras.

p.NORSIM26.2. Proposition. Let C, C′ be conic alternative k-algebras.

(a) If C is multiplicative, then for any a ∈C×, the maps La,Ra : C→C are normsimilarities, with multipliers µLa = µRa = nC(a).

(b) Let f : C→C′ be a norm similarity. Then f (C×) =C′×. Moreover, if f is evena unital norm equivalence, then f preserves norms, traces, conjugations andinverses.

Proof. (a) follows from the property of nC permitting composition combined withProp. 14.6. In order to establish (b), it suffices to combine Prop. 20.4 with (19.5.2),(19.5.4).

c.NORSIM26.3. Corollary. Multiplicative conic alternative algebras C,C′ over k are normsimilar if and only if there exists a unital norm equivalence from C to C′.

Proof. Let f : C→ C′ be a norm similarity. By Prop. 26.2 (b), the element a′ :=f (1C) is invertible in C′×, and La′−1 f : C→C′ is a unital norm equivalence.

p.ISNEQ26.4. Proposition. (a) Let C, C′ be conic algebras over k that are projective ask-modules. Then every isomorphism or anti-isomorphism from C to C′ is a unitalnorm equivalence.(b) A unital norm equivalence from one quadratic k-algebra onto another is anisomorphism.

Proof. (a) follows immediately from Exc. 78 below.(b) Let R, R′ be quadratic k-algebras and f : R→ R′ a unital norm equivalence. In

order to show that f is an isomorphism, we may assume that k is a local ring, whichimplies R= k[u] for some u∈R such that 1R,u is a basis of R over k. Then R′ = k[u′],u′ := f (u), and since f preserves units, norms and traces by Prop. 26.2 (b), we havef (u2) = u′2 = f (u)2 and f is an isomorphism.

We proceed by briefly recalling Witt extension and Witt cancellation of quadraticforms over semi-local rings.

t.WITEXT26.5. Theorem (Baeza [8, III, 4.1]). (Witt extension theorem) Let (M,Q) be aquadratic space over the semi-local ring k and suppose the submodule N ⊆ M isa direct summand such that the set Q(N⊥) := Q(x) | x ∈ N⊥ generates the ring k.Then every monomorphism from (N,Q|N) to (M,Q) whose image is a direct sum-mand of M can be extended to an isometry from (M,Q) onto itself.

c.WITCAN26.6. Corollary. (Witt cancellation theorem) Let (M1,q1), (M2,q2), (M′2,q

′2) be

quadratic spaces over the semi-local ring k and suppose


(M1,q1)⊥ (M2,q2)∼= (M1,q1)⊥ (M′2,q′2).

Then (M2,q2)∼= (M′2,q′2).

Proof. We may assume that M2 and M′2 both have rank r > 0 . By Lemma 22.14 thisimplies

q2(M2)∩ k× 6= /0 6= q′2(M′2)∩ k×. (1) QINV

Now put

(M,Q) := (M1,q1)⊥ (M2,q2), (M′,Q′) := (M1,q1)⊥ (M′2,q′2)

and let i1 : (M1,q1)→ (M,Q), i′1 : (M1,q1)→ (M′,Q′) be the natural embeddings.Then N := i1(M1) ⊆ M, N′ := i′1(M1) ⊆ M′ are submodules giving rise to non-singular subspaces (N,Q|N)⊆ (M,Q), (N′,Q′|N′)⊆ (M′,Q′) such that

(N⊥,Q|N⊥)∼= (M2,q2), (N′⊥,Q′|N′⊥)∼= (M′2,q′2), (2) ENQU

and given any isometryΨ : (M′,Q′) ∼−→ (M,Q),

we obtain a monomorphism α : (N,Q|N)→ (M,Q) determined by

α(i1(x1)

)=Ψ

(i′1(x1)

)(x1 ∈M1).

The image of α is Ψ(N′), hence a direct summand of M by Lemma 12.10. By (1),(2) and Thm. 26.5, therefore, α can be extended to an isometry Λ from (M,Q) ontoitself. Then

Φ :=Ψ−1 Λ : (M,Q)

∼−→ (M′,Q′)

is an isometry that maps N onto N′, hence N⊥ onto N′⊥. By (2), this means(M2,q2)∼= (M′2,q

′2).

26.7. Norm Equivalence Theorem (cf. Jacobson [41], Van der Blij-Springer [125]).Let k be a semi-local ring and C, C′ be composition algebras over k. Then the fol-t.NOREQlowing conditions are equivalent.

(i) C and C′ are isomorphic.(ii) C and C′ are norm equivalent.(iii) C and C′ are norm similar.

Proof. (i)⇒ (ii). Prop. 26.4 (a).(ii)⇒ (iii). Obvious.(iii)⇒ (i). We may assume that C,C′ both have rank r > 0. There is nothing to

prove for r = 1, so let us assume r ≥ 2. By Cor. 26.3, there exists a unital normequivalence f : C→C′, and Thm. 22.15 (a) yields a quadratic etale subalgebra D =k[u]⊆C, for some u∈C having trace 1. Since f preserves units, norms and traces byProp. 26.2 (b), we conclude that D′ := f (D) = k[u′], u′ := f (u), is a quadratic etale


subalgebra of C′ (Prop. 22.5) and f : D→ D′ is an isomorphism (Prop. 26.4 (b)).Now suppose B⊆C, B′ ⊆C′ are any non-singular composition subalgebras of ranks < r that are isomorphic. Then B,B′ are associative (Cor. 22.17), and combiningLemma 22.14 with the Witt cancellation theorem (Cor. 26.6), we find invertibleelements l ∈ B⊥ ⊆ C, l′ ∈ B′⊥ ⊆ C′ such that nC(l) = nC′(l′) ∈ k×. Therefore B, l(resp. B′, l′) generate non-singular composition subalgebras B1 ⊆C (resp. B′1 ⊆C′)of rank 2s that are isomorphic (Cor. 21.8). Continuing in this manner, we eventuallyobtain an isomorphism from C to C′ (actually, we do so after at most two steps).

c.NORCIS26.8. Corollary. Let B be a non-singular associative composition algebra over thesemi-local ring k and µ,µ ′ ∈ k×. Then

Cay(B,µ)∼= Cay(B,µ ′) ⇐⇒ µ ≡ µ′ mod nB(B×).

Proof. Put C := Cay(B,µ), C′ := Cay(B,µ ′). By Thm. 26.7, C ∼= C′ impliesnC ∼= nC′ , and Remark 21.4 combines with Witt cancellation (Cor. 26.6) to yieldan isometry ϕ : µnB

∼→ µ ′nB. But then µ = µ ′nB(u), u = ϕ(1B) ∈ B×. Conversely,if µ = µ ′nB(u) for some u ∈ B×, then Lu : µnB

∼→ µ ′nB is an isometry, so nC, nC′ byRemark 21.4 are isometric, forcing C and C′to be isomorphic by the norm equiva-lence theorem 26.7.

Remark. Cor. 26.8 fails if B is singular. For example, let k be a field of characteristic2 and µ,µ ′ ∈ k. Then one checks easily that Cay(k,µ) and Cay(k,µ ′) are isomorphicif and only if µ ′ = µ +α2 for some α ∈ k.

NORCOM26.9. Comments. Apart from its potential for a great many important applications,some of which will be dealt with in the remainder of this section, the norm equiva-lence theorem enjoys a few remarkable properties of a different kind. Here are someexamples.(a) The norm equivalence theorem does not claim that every unital norm equivalencebetween composition algebras over a field (or a semi-local ring) is an isomorphismor an anti-isomorphism. While this is certainly true for quadratic etale algebras overany ring (Prop. 26.4 (b)) and, as will be seen later in ??? below (see also Knus [58,V, (4.3.2)] or Gille [35, Thm. 2.4]), is basically true for quaternion algebras, againover any ring, it fails miserably in the octonionic case. The reader may either con-sult Exc. 130 below to see this or turn to the octonion algebra O over k = R: in thelatter case, the automorphisms or anti-automorphisms of O form a closed subset ofGL(O) that may be written as the union of two 14-dimensional pieces (2.6), whilethe unital norm equivalences of O identify canonically with the Lie group O7, whichhas dimension 21.(b) It is a natural question to ask whether the norm equivalence theorem holdsover an arbitrary commutative ring. By (a), the answer is yes for composition al-gebras of rank 2 or 4. On the other hand, it has recently been shown by Gille [35,Thm. 3.3] that the answer is no in rank 8, so there exist commutative rings k andnon-isomorphic octonion algebras over k whose norms are isometric. More surpris-


ingly still, there are non-split octonion algebras over an appropriate commutativering whose norms are split hyperbolic [35, p. 308](c) The norm equivalence theorem was anticipated already by Zorn [129, p. 399],whereone finds the following remarkable statement:

Das System wurde allein aus der quadratischen Form xx gewonnen, die Aquiva-lenz der Formen ist also mit der Aquivalenz der Systeme gleichbedeutend.4

By “das System” (resp. the expression “x x”), Zorn is referring to an octonionalgebra (resp. to its norm). However, the reason given by him for the validity ofthe norm equivalence theorem is not convincing since, e.g., it would imply that thetheorem would hold over any commutative ring, contrary to Gille’s result mentionedin (b), and that any unital norm equivalence would be an isomorphism.

One reason for the importance of the norm equivalence theorem is that we know a lotabout quadratic forms over special fields, and this knowledge may be used to classifycomposition algebras over these fields. Recall from Cor. 25.14 that compositionalgebras over any field are either split or division algebras.

ss.ALGCLOFIE26.10. Algebraically closed fields. Quadratic spaces over an algebraically closedfield k are classified by their dimension. Hence the norm equivalence theorem im-plies that, up to isomorphism, in each dimension 1,2,4,8 there is exactly one com-position algebra over k, namely, the split one.

ss.REAL26.11. The reals. Thanks to Sylvester’s inertia theorem, quadratic spaces over R(or, more generally, over any real closed field) are classified by their dimension andtheir signature. In particular, in each dimension there are precisely two isometryclasses of anisotropic quadratic forms, a positive definite and a negative definiteone. Hence each dimension r = 1,2,4,8 allows exactly one isomorphism class ofcomposition division algebras over R: uniqueness follows from the norm equiva-lence theorem 26.7, while existence is provided by the classical examples R, C, H,O of § 1.

ss.FINFIE26.12. Finite fields. The key fact to be used here is Chevalley’s theorem (Lang [65,IV, Ex. 7 (a)]): every form (= homogeneous polynomial) of degree d in n > d inde-terminates over a finite field k = Fq, q a prime power, has a non-trivial zero in Fn

q. Itfollows that every quadratic form in at least three variables over Fq is isotropic. Inparticular, quaternion and octonion algebras over Fq are all split (Cor. 25.14), whilethe only non-split composition algebra of dimension 2 over Fq up to isomorphismis the unique quadratic field extension Fq2/Fq.

ss.PAD26.13. p-adics and their finite extensions. Let p be a prime. The completion ofQ relative to its p-adic valuation is Qp, the field of p-adic numbers. Given a finitealgebraic extension K of Qp, the following classical results on quadratic forms over

4 Emphasis in the original.


K are due to Hasse (see Scharlau [115, Chap. 5, Thm. 6.3] in the case K =Qp andO’Meara [90, 63:11b, 63:17, 63:19] in general for details).

(i) Up to isometry, there is a unique anisotropic quadratic form of dimension 4over K, and this form is universal in the sense that it represents every elementof K.

(ii) Every quadratic form in at least five variables over K is isotropic.(iii) Up to isomorphism, there is a unique quaternion division algebra over K.

By (ii), all octonion algebras over K are split. To complete the picture, it remainsto classify composition algebras of dimension 2 over K, which are parametrized bythe group K×/K×2. Describing this group is pretty straightforward if p is odd, butrequires some care for p = 2; again we refer to Scharlau [115, Chap. 5, § 6] andO’Meara [90, § 63 A, particularly 63:9] for details.

ss.ALNUM26.14. Algebraic number fields. In this subsection, we assume some familiaritywith the foundations of algebraic number theory. Let K be an algebraic numberfield, i.e., a finite algebraic extension of the field Q of rational numbers. We denoteby Ω the set of places of K, including the infinite ones, and by Kv the completion ofK at the place v ∈ Ω ; the natural embedding λv : K → Kv makes the image λv(K)a dense subfield of Kv relative to the v-adic topology. The set of real places of Kwill be denoted by S, so we have a natural identification Kv = R for v ∈ S; in fact,the assignment v 7→ λv determines a bijective correspondence between S and the setof embeddings K → R. Given an algebra A (resp. a quadratic form Q) over K, weabbreviate Av = A⊗K Kv (resp. Qv = Q⊗K Kv) for v ∈Ω .

The key to understand composition algebras over K is provided by the Hasse-Minkowski theory of quadratic forms. Referring to O’Meara [90, 66:1, 66:4], Schar-lau [115, Chap. 5, 7.2, 7.3] for details, one of its most fundamental results reads asfollows.

26.15. Theorem (Hasse-Minkowski). With the notation of 26.14, the followingt.HASMINstatements hold.

(a) A quadratic form Q over K is isotropic if and only if Qv is isotropic over Kvfor all v ∈Ω .

(b) Non-singular quadratic forms Q, Q′ over K are isometric if and only if Qv, Q′vare isometric over Kv for all v ∈Ω .

Since a non-singular quadratic form Q over a field represents a scalar α if and onlyif the form 〈α〉quad ⊥ Q is isotropic, the first part of the Hasse-Minkowski theoremimmediately implies:

c.HASMIN26.16. Corollary. A non-singular quadratic form Q over K represents an elementα ∈ K if and only if Qv represents λv(α) ∈ Kv for all v ∈Ω .

Combining the second part of Thm. 26.15 with the norm equivalence theorem 26.7,it follows that composition algebras over number fields are completely determinedby their local behavior:


c.LOCOM26.17. Corollary. Composition algebras C and C′ over K are isomorphic if andonly if Cv and C′v are isomorphic over Kv for all v ∈Ω .

Cor. 26.17 may be regarded as the first step towards the classification of compositionalgebras over number fields. In order to complete this classification, it seems natu-ral to invoke another fundamental fact from Hasse-Minkowski theory: every “localfamily” of non-singular quadratic forms Q(v) over Kv, v ∈Ω , has a “global realiza-tion” by a non-singular quadratic form Q over K satisfying Qv ∼= Q(v) for all v ∈Ω

if and only if a certain obstruction, based on Hilbert’s reciprocity law and involvingHasse symbols, vanishes. But since there is no a priori guarantee that the propertyof being the norm of a composition algebra descends from the totality of forms Qv,v ∈Ω , to the form Q, one has to argue in a different manner.

For simplicity, we confine ourselves to quaternion and octonion algebras. Theclassification of the former is accomplished by the following result from class fieldtheory, which we record (again) without proof.

26.18. Theorem (O’Meara [90, 71:19]). With the notation of 26.14, let T ⊆Ω be at.QUANUMfinite set consisting of an even number of real or finite places of K. Then there existsa quaternion algebra B over K, unique up to isomorphism, such that Bv is split forall v ∈Ω \T and a division algebra for all v ∈ T .

Remark. Recall from 26.11 and 26.13 (iii) that Kv, for v ∈ T , admits precisely onequaternion division algebra, so uniqueness of B follows from Cor. 26.17.

Surprisingly, the classification of octonion algebras over K turns out to be muchsimpler. In fact, it can be deduced quite easily from the results assembled so far.

26.19. Theorem (Albert-Jacobson [7]). With the notation of 26.14, write B :=t.OCTNUCay(K;−1,−1) as a quaternion algebra over K. Then every octonion algebra overK is isomorphic to Cay(B,µ) for some µ ∈ K×. Moreover, given µ,µ ′ ∈ K×, thefollowing conditions are equivalent.

(i) Cay(B,µ)∼= Cay(B,µ ′).(ii) λv(µµ ′)> 0 for all v ∈ S.

Proof. For the first part, it will be enough to prove that every octonion algebra C overK contains a unital subalgebra isomorphic to B. To this end, we claim that every non-singular sub-form of nC having dimension at least 5 represents the element 1 ∈ K.

Indeed, let Q be such a sub-form. By Cor. 26.16, it suffices to show for all v ∈Ω

that Qv represents the element 1 ∈ Kv. If Cv is split, nCv has maximal Witt index,and a dimension argument shows that every maximal totally isotropic subspace ofCv relative to nCv intersects Qv non-trivially. Thus Qv is isotropic, hence universaland so represents 1. On the other hand, if Cv is a division algebra, v ∈ S must bereal (26.10, 26.13), forcing nCv to be positive definite. But then so is Qv, whichtherefore again represents 1. This proves our claim. We first apply the claim ton0

C, the restriction of nC to the pure octonions C0 = Ker(tC), and find an elementj1 ∈C satisfying tC( j1) = 0, nC( j1) = 1. By Cor. 21.8 and Prop. 22.5, D := K[ j1]∼=


Cay(K,−1) is a quadratic etale subalgebra of C. Applying our claim once more, thistime to the restriction of nC to D⊥, yields an element j2 ∈D⊥ satisfying nC( j2) = 1,and invoking Cor. 21.8 again, we see that the subalgebra of C generated by D andj2 is isomorphic to B, giving the first part of the theorem. As to the second, weconclude Cv ∼= Cay(H,λv(µ)), C′v ∼= Cay(H,λv(µ

′)) for all v ∈ S, and since Cv,C′vare both split for v ∈ Ω \ S (26.10, 26.13), we can apply Corollaries 26.17, 26.8 toobtain the following chain of equivalent conditions.

C ∼=C′ ⇐⇒ ∀v ∈ S : Cv ∼=C′v⇐⇒ ∀v ∈ S : Cay

(H,λv(µ)

)∼= Cay(H,λv(µ

′))

⇐⇒ ∀v ∈ S : λv(µµ′) ∈ nH(H×) = R×+.

26.20. Corollary (Zorn [129]). With the notation of 26.14, there are precisely 2|S|c.NUMOCisomorphism classes of octonion algebras over K.

Proof. 5 Adopting the notation of Thm. 26.19, let T ⊆ S. We apply the weak ap-proximation theorem (O’Meara [90, 11:8]) and find an element µT ∈ K satisfying

|λv(µT )+1|< 1 (v ∈ T ), |λv(µT )−1|< 1 (v ∈ S\T ).

Up to isomorphism, the octonion algebra CT := Cay(B,µT ) does not depend on thechoice of µT (Thm. 26.19), so the assignment T 7→ CT gives a well defined mapfrom 2S to the set of isomorphism classes of octonion algebras over K. Conversely,let C be an octonion algebra over K. Then TC := v ∈ S |Cv ∼= O is a subset of S,and C 7→ TC gives a map in the opposite direction. Since the two maps thus definedare easily seen to be inverse to one another, the assertion follows.

Remark. Given a family of octonion algebras C(v) over Kv, v∈Ω , Cor. 26.20 showsthat there always exists an octonion algebra C over K, unique up to isomorphism,which satisfies Cv ∼= C(v) for all v ∈ Ω , so the Hilbert reciprocity law yields noobstructions when dealing with octonion norms.

Exercises.

pr.SKONOCO 125. The Skolem-Noether theorem for composition algebras. Let k be a semi-local ring, C a com-position algebra of rank r over k and B,B′ ⊆C two composition subalgebras of the same rank s≤ r.Show that every isomorphism from B to B′ can be extended to an automorphism of C. Concludethat, up to conjugation by automorphisms of O (the Graves-Cayley octonions), the only non-zerosubalgebras of O are R,C,H and O.

pr.COMPRI 126. (cf. Van der Blij-Springer [125, (3.4)]) Let k be a principal ideal domain and C a compositionalgebra of rank > 1 over k that has (left or right) zero divisors (cf. Ex. 94). Show that C is split.

pr.UNINOR 127. Unital norm equivalences of conic alternative algebras. (a la Jacobson-Rickart [49]) Let C,C′ be conic alternative k-algebras and suppose f : C→C′ is a unital norm equivalence.

5 According to loc. cit., p. 400, there should be a reference to the work of H. Brandt.


(a) Show

f (Uxy) =U f (x) f (y), (1) COMUOP(f (xy)− f (x) f (y)

)(f (xy)− f (y) f (x)

)= [ f (x), f (xy), f (y)] (2) ISAN

for all x,y ∈C. In particular, the left-hand side of (2) is symmetric in x,y.(b) Conclude from (1) and its linearizations that if c ∈ C is an elementary idempotent, so is

c′ := f (c) ∈C′ andf(C12(c)+C21(c)

)=C′12(c

′)+C′21(c′).

pr.UNORQUA 128. Unital norm equivalences of quaternion algebras (Petersson [93]). (a) Show for quaternionalgebras B,B′ over k: if f : B→ B′ is a unital norm equivalence, then there exists a decompositionk = k+⊕ k− of k as a direct sum of ideals such that, with the corresponding decompositions

B = B+⊕B−, B± = Bk± , B′ = B′+⊕B′−, B′± = B′k± , f = f+⊕ f−, f± = fk± , (3) QUATPM

f+ : B+ → B′+ is an isomorphism of quaternion algebras over k+, and f− : B− → B′− is a anti-isomorphism of quaternion algebras over k−. (Hint. Reduce to the case that k is a local ring byapplying § 10, Exercise 41. Then imitate the beginning of the proof of the implication (iii)⇒ (i) inthe norm equivalence theorem.)(b) Use (a) to prove a slightly weakened version of Knus’s theorem [58, V, (4.3.2)] (see also Gille[35, 2.4]): for quaternion algebra B,B′ over k, the following conditions are equivalent.

(i) B and B′ are isomorphic.(ii) B and B′ are norm equivalent.(iii) B and B′ are norm similar.

pr.ELIDNO 129. Elementary idempotents and unital norm equivalences. Write ei j (i, j = 1,2) for the usualmatrix units in the split quaternion algebra B = Mat2(k) and prove for any element c ∈ B that thereexists an automorphism of B sending e11 to c if and only if there exists a unital norm equivalenceof B sending e11 to c.

pr.OCTNOREQU 130. Octonionic norm equivalences. Let C be an octonion algebra over k and p ∈C×. Show thatLpRp−1 is

(a) always a unital norm equivalence of C,(b) an automorphism of C if and only if p3 ∈ k1C ,(c) never an anti-automorphism of C.

(Hint. Use Ex. 72 (c))

pr.ISTIS 131. Isotopes of composition algebras. Show that isotopes of a composition algebra C over k arecomposition algebras which are non-singular if C is, and which are isomorphic to C if k is a semi-local ring..

pr.ZEPA 132. Zero divisor pairs of the real sedenions. Recall that the real sedenions S = Cay(O,−1) =O⊕O j as defined in Exc. 99 form a conic real algebra with norm nS canonically isometric tonO ⊥ nO. By a zero divisor pair of S we mean a pair (x,y) of non-zero elements in S such thatxy = 0. Note that if (x,y) is a zero divisor pair of S, then so is (αx,βy) for all α,β ∈ R×. Thusthe study of arbitrary zero-divisor pairs in S is equivalent to the one of zero divisor pairs withpre-assigned norm. With this in mind, we define

Zer(S) := (x,y) ∈ S×S | xy = 0, nS(x) = nS(y) = 2 (4) ZEPA

and let G := Aut(O) act diagonally first on S and then on Zer(S) via

G×Zer(S)−→ Zer(S),(σ ,(x,y)

)7−→

(σ(x),σ(y)

).


Use Exc. 99 to show that Zer(S) becomes a principal homogeneous G-space in this way , i.e., theaction of G on Zer(S) is simply transitive.

Remark. This result is a refined version of Moreno’s theorem [87, Cor. 2.14], which says thatZer(O)⊆ S×S is a closed subset homeomorphic to Aut(O).

The following sequence of exercises is designed to put the units of the Coxeter octonions (Exc. 16)in a broader perspective.

pr.ZOFI 133. Let q be a prime power and C := Zor(Fq) the unique octonion algebra over the field with qelements. Show

|C×|= q3(q−1)(q4−1), |x ∈C | nC(x) = 1|= q3(q4−1),

pr.MIVE 134. Positive definite integral quadratic modules and minimal vectors. By a positive definite in-tegral quadratic module we mean a quadratic module (M,Q) over Z such that M is a finitelygenerated free abelian group and the quadratic form Q : M→ Z is positive definite. Then M is anintegral quadratic lattice in the positive definite real quadratic space (MR,QR) in the sense of 3.6and, conversely, every integral quadratic lattice in a positive definite real quadratic space becomesa positive definite integral quadratic module in a natural way.The discriminant of a positive def-inite integral quadratic module may be defined as in 3.12. A positive definite integral quadraticmodule is called indecomposable if it cannot be written as the sum of two orthogonal non-zerosubmodules.

Now let (M,Q) be a positive definite integral quadratic module over Z. A vector x ∈M is saidto be minimal if cannot be written as the sum of two vectors of strictly shorter length: x = y+z withy,z ∈ M and Q(y) < Q(x), Q(z) < Q(x) is impossible. The set of minimal vectors in (M,Q) willbe denoted by Min(M,Q). Two minimal vectors x,y ∈M are said to be equivalent if there existsa finite sequence x = x0,x1, . . . ,xk−1,xk = y of minimal vectors in M such that Q(xi−1,xi) 6= 0 for1≤ i≤ k.

(a) Show that Min(M,Q) generates the additive group M.(b) Show that equivalence of minimal vectors defines an equivalence relation on Min(M,Q).(c) Denoting by [x] the equivalence class of x ∈ Min(M,Q) with respect to the equivalence

relation defined in (b), and setting M[x] := ∑y∈[x]Zy, prove Eichler’s theorem[???]: (M,Q)splits into the orthogonal sum of indecomposable integral quadratic submodules M[x], where[x] varies over the equivalence classes of Min(M,Q).

pr.ODDI 135. Let C be a multiplicative conic alternative algebra over Z such that the following conditionsare fulfilled.

(a) C is free of rank 8 as a Z-module.(b) The norm of C is positive definite.(c) The discriminant of the integral quadratic module (C,nC) is odd.

Prove that |C×| ≤ 240, and that if |C×|= 240, then (C,nC) is an indecomposable positive definiteintegral quadratic module. (Hint. Reduce C mod 2 to obtain an octonion algebra C† over F2 andshow that the fibers of the natural map C×→C†× consist of two elements. Then apply Excs. 133–134.)

pr.QEZE 136. Show that a quadratic etale algebra over the integers is split. Conclude that the algebra of Hur-witz quaternions cannot be obtained from the Cayley-Dickson construction: there are no quadraticZ-algebra R and no scalar µ ∈ Z such that Hur(H)∼= Cay(R,µ).

Remark. It can also be shown that a quaternion algebra over the integers is split, but the proofrequires more work.

27 Affine schemes 167

pr.OCZE 137. Classification of octonion algebras over the integers (Van der Blij-Springer [125]). Show thatan octonion algebra over Z is either split or isomorphic to the Coxeter octonions. In order to do so,let C be a non-split octonion algebra over Z with product x · y, write xy for the product in Cox(O)and perform the following steps.

(a) Reduce to the case that C = Cox(O) as additive groups, 1C = 1O and nC = nCox(O) (Hint.Use 4.6 and Exc. 126).

(b) Show that there is an isomorphism ψ : C⊗Z F2∼→ Cox(O)⊗Z F2 and use the fact that the

orthogonal group of nC ⊗Z F2 is generated by orthogonal transvections (Dieudonne [24,Prop. 14]) to lift ψ to an orthogonal transformation ϕ : C→ Cox(O) such that ϕ(x · y) =±ϕ(x)ϕ(y) for all x,y ∈C×. (Hint. Use the fact, known from the solution to Exc. 135, thatthe map C×→ (C⊗ZF2)

× induced by the natural surjection C→C⊗ZF2 is itself surjectiveand that each fiber consists of two elements.)

(c) Now prove that ϕ or −ϕ is an isomorphism.

pr.SPLIFICO 138. Splitting fields of composition algebras. Let C be a composition algebra over a field F . A fieldextension K/F is said to be a splitting field of C if the extended composition algebra CK over K issplit. Prove:

(a) (Ferrar [31, Lemma 5], Petersson-Racine [92, Proposition 4.3]). For a quadratic field exten-sion K/F to be a splitting field of C it is necessary and sufficient that one of the followinghold.

(i) C ∼= F .(ii) C is split quadratic etale.(iii) There exists an F-embedding K →C.

(b) The separable closure of F is a splitting field of C.

27. Affine schemes

s.AFFA treatment of octonions and Albert algebras over commutative rings would beincomplete without taking advantage of, and giving applications to, the theory ofaffine group schemes. Our aim in the present section will be to carry out a firstthrust into this important topic. We do so by explaining the most elementary andbasic notions from scheme theory and by illustrating them through a number ofsimple examples. We always focus attention on what is absolutely essential for theintended applications. Though a considerable amount of what we will be doing hereretains its validity under much more general circumstances (e.g., in the setting ofcategory theory), generalizations of this sort will be completely ignored.

Throughout we let k be an arbitrary commutative ring. In our treatment of affinek-schemes, we follow Demazure-Gabriel [22] by adopting the functorial point ofview. However, in order to keep matters simple, the delicate formalism developedby loc. cit. in order to stay within a consistent framework of set theory will beavoided; instead, we favor a more stream-lined “naive” approach as given, e.g., byJantzen [50]. Our notation will combine that of Jantzen [50] and Loos [71].

ss.KAFU27.1. k-functors, subfunctors and direct products. By a k-functor we mean afunctor X from the category of unital commutative associative k-algebras to the


category of sets:X : k-alg−→ set.

Morphisms of k-functors are defined as natural transformations, so if X,X′ are k-functors, a morphism f : X→X′ is a family of set maps f (R) : X(R)→X′(R), onefor each R ∈ k-alg, such that, for each morphism ρ : R→ R′ in k-alg, the diagram

X(R)f (R) //

X(ρ)

X′(R)

X′(ρ)

X(R′)f (R′)// X′(R′)

(1) MOKAFU

commutes. The category of k-functors will be denoted by k-fct. Given k-functorsX,X′, we will denote by Mor(X,X′) the totality of morphisms from X to X′.Note that a morphism f : X→ X′ is an isomorphism if and only if the set mapsf (R) : X(R)→ X′(R) are bijective for all R ∈ k-alg.

By a subfunctor of a k-functor X we mean a k-functor Y such that Y(R)⊆ X(R)for all R∈ k-alg and the inclusion maps i(R) : Y(R) →X(R) give rise to a morphismi : Y→ X of k-functors; in other words, for any morphism ρ : R→ R′ in k-alg, wehave X(ρ)(Y(R)) ⊆ Y(R′), and the set map Y(ρ) : Y(R)→ Y(R′) is induced byX(ρ) via restriction. We sometimes write Y⊆ X for Y being a subfunctor of X.

Let X1,X2 be k-functors. Then we define a k-functor X1×X2, called their di-rect product, by setting (X1 ×X2)(R) := X1(R)×X2(R) for all R ∈ k-alg and(X1×X2)(ρ) := X1(ρ)×X2(ρ) for all morphisms ρ : R→ R′ in k-alg. The di-rect product comes equipped with two projection morphisms pi : X1×X2→ Xi fori = 1,2 such that pi(R) is the projection from X1(R)×X2(R) onto the i-th factor,for each R ∈ k-alg. It then follows that X1×X2 together with p1, p2 is an honest-to-goodness direct product in the category k-fct by satisfying the correspondinguniversal property.

ss.COASCH27.2. The concept of an affine k-scheme. Let R ∈ k-alg. We define

Spec(R) := Homk-alg(R,−) : k-alg−→ set, (1) BFSP

so Spec(R) is the k-functor given by

Spec(R)(S) = Homk-alg(R,S) (2) SPOB

for all S ∈ k-alg and

Spec(R)(σ) : Homk-alg(R,S)−→ Homk-alg(R,S′), (3) SPMO

Homk-alg(R,S) 3 ϕ 7−→ Spec(R)(σ)(ϕ) := σ ϕ ∈ Homk-alg(R,S′)


for all morphisms σ : S→ S′ in k-alg. By an affine k-scheme or an affine schemeover k we mean a k-functor that is isomorphic to Spec(R), for some R ∈ k-alg. Weview affine k-schemes as a full subcategory of k-fct, denoted by k-aff.

Let ϕ : R′→ R be a morphism in k-alg. Then

Spec(ϕ) := Homk-alg(−,ϕ) : Spec(R)−→ Spec(R′) (4) SPEPH

is a morphism of k-functors, explicitly given by

Spec(ϕ)(S) : Spec(R)(S)−→ Spec(R′)(S), (5) SPEPS

Homk-alg(R,S) 3 ψ 7−→ Spec(ϕ)(S)(ψ) := ψ ϕ ∈ Homk-alg(R′,S)

for all S ∈ k-alg. If ϕ ′ : R′′→ R′ is another morphism in k-alg, we have

Spec(1R) = 1Spec(R), Spec(ϕ ϕ′) = Spec(ϕ ′)Spec(ϕ). (6) SPECO

Summing up we conclude that the data presented in (1), (4) define a contra-variantfunctor

Spec : k-alg−→ k-fct, (7) BFSPEC

which may also be regarded as a contra-variant functor

Spec : k-alg−→ k-aff. (8) SPAFF

In the latter capacity, it will be seen in due course to induce an anti-equivalence ofcategories.

ss.AFSP27.3. Affine n-space. Let n be a positive integer. We define a k-functor An

k , calledaffine n-space, by setting An

k(R) := Rn as a set and

Ank(ρ) := ρ

n : Rn −→ R′n, Rn 3 (r1, . . . ,rn) 7−→(ρ(r1), . . . ,ρ(rn)

)∈ R′n (1) AFMO

for a morphism ρ : R→ R′ in k-alg as a set map. With independent indeterminatest1, . . . , tn, we claim that

Ank∼= Spec(k[t1, . . . , tn]) (2) AFSCH

and, in particular, that Ank is an affine scheme over k. Indeed, letting R ∈ k-alg and

r1, . . . ,rn ∈ R, write

εn(R)(r1, . . . ,rn) : k[t1, . . . , tn]−→ R

for the morphism in k-alg given by

εn(R)(r1, . . . ,rn)(g) := g(r1, . . . ,rn) (g ∈ k[t1, . . . , tn]). (3) EPSUB


Note that εn(R)(r1, . . . ,rn) is uniquely determined by the condition of sending ti tori for 1≤ i≤ n. In this way we obtain a bijective set map

εn(R) : Rn ∼−→ Homk-alg

(k[t1, . . . , tn],R) = Spec

(k[t1, . . . , tn]

)(R)

varying functorially with R. Hence

εn : An

k∼−→ Spec

(k[t1, . . . , tn]

)is an isomorphism of k-functors.

27.4. Proposition (Yoneda Lemma). Let X be a k-functor and R ∈ k-alg. Then thep.YONassignment

Mor(Spec(R),X

)3 f 7−→ΦR,X( f ) := f (R)(1R) ∈ X(R) (1) YOEL

defines a bijection

ΦR,X : Mor(Spec(R),X

) ∼−→ X(R), (2) YOMA

and we have

Φ−1R,X(x)(S)(ϕ) = X(ϕ)(x) (3) YOIN

for all x ∈ X(R), S ∈ k-alg, ϕ ∈ Homk-alg(R,S).

Proof. Given x ∈ X(R), S ∈ k-alg, we define a set map Θ(x)(S) : Spec(R)(S)→X(S) by

Θ(x)(S)(ϕ) := X(ϕ)(x)(ϕ ∈ Homk-alg(R,S)

). (4) THEX

For a morphism σ : S→ S′ in k-alg it follows easily from (27.2.3) that the diagram

Spec(R)(S)Θ(x)(S) //

Spec(R)(σ)

X(S)

X(σ)

Spec(R)(S′)

Θ(x)(S′)// X(S′)

commutes. Thus Θ(x)∈Mor(Spec(R),X), and we have obtained a map Θ : X(R)→Mor(Spec(R),X). It is now straightforward to verify that Θ ΦR,X is the identity onMor(Spec(R),X) and ΦR,X Θ is the identity on X(R). Hence ΦR,X is bijective withinverse Θ .

c.YONAF

27.5. Corollary. Let R,R′ ∈ k-alg.(a) The map


Mor(Spec(R),Spec(R′)

) ∼−→ Homk-alg(R′,R), (1) YAFL

f 7−→ΦR,Spec(R′)( f ) = f (R)(1R)

is a bijection with inverse

Homk-alg(R′,R)∼−→Mor

(Spec(R),Spec(R′)

), ϕ 7−→ Spec(ϕ). (2) YAFIN

(b) ϕ ∈ Homk-alg(R′,R) is an isomorphism if and only if

Spec(ϕ) ∈Mor(Spec(R),Spec(R′))

is an isomorphism, and in this case Spec(ϕ)−1 = Spec(ϕ−1).

Proof. Specializing Proposition 27.4 to X := Spec(R′) and applying (27.2.3),(27.2.5), we obtain (a). It remains to establish (b). If ϕ is an isomorphism, then so isSpec(ϕ), by (27.2.6), with Spec(ϕ)−1 = Spec(ϕ−1). Conversely, suppose Spec(ϕ)is an isomorphism. Then there exists a morphism g : Spec(R′) → Spec(R) suchthat gSpec(ϕ) = 1Spec(R), Spec(ϕ)g = 1Spec(R′). Here (a) implies g = Spec(ϕ ′)for some morphism ϕ ′ : R → R′ in k-alg. From (27.2.6) we therefore deduceSpec(ϕ ϕ ′) = 1Spec(R), Spec(ϕ ′ ϕ) = 1Spec(R′), and (a) again shows that ϕ isan isomorphism with inverse ϕ ′.

ss.REFU27.6. Regular functions. With affine 1-space A1

k , also called the affine line, weconsider the contra-variant functor

k[−] := Mor(−,A1k) : k-fct−→ set, (1) KABL

so we have

k[X] := Mor(X,A1k) (2) KABLO

for all k-functors X and

k[ f ] : k[X′]−→ k[X], (3) KABLMO

Mor(X′,A1k) 3 f ′ 7−→ k[ f ]( f ′) := f ′ f ∈Mor(X ,A1

k)

for all morphisms f : X→ X′ of k-functors. Since A1k(R) = R for all R ∈ k-alg, the

set k[X], for any k-functor X, carries the structure of a unital commutative associativek-algebra by defining the scalar multiple α f ∈ k[X], the sum f1 + f2 ∈ k[X] and theproduct f1 f2 ∈ k[X] according to the rules

(α f )(R)(x) := α f (R)(x),

( f1 + f2)(R)(x) := f1(R)(x)+ f2(R)(x), (4) KABALG

( f1 f2)(R)(x) := f1(R)(x) f2(R)(x)

for α ∈ k, f , f1, f2 ∈ k[X], R ∈ k-alg and x ∈ X(R). Thus k[X] ∈ k-alg, with


1k[X](R)(x) = 1R(R ∈ k-alg, x ∈ X(R)

). (5) UNKAB

We call k[X] the k-algebra of regular functions on X. Note for R ∈ k-alg that thebijection

ΦR,A1k

: k[Spec(R)] ∼−→ R (6) KABSP

of Proposition 27.4 is an isomorphism of k-algebras.If f : X→ X′ is a morphism of k-functors, then one checks easily that

k[ f ] : k[X′]−→ k[X]

is a morphism of unital commutative associative k-algebras. Thus the functor (1)may actually be viewed as a contra-variant functor

k[−] = Mor(−,A1k) : k-fct−→ k-alg, (7) KABLALG

from which we recover (1) by composing (7) with the forgetful functor k-alg→ set.On the other hand, restricting the functor (7) to the category of affine k-schemes, weobtain a contra-variant functor

k[−] = Mor(−,A1k) : k-aff−→ k-alg. (8) KABLAF

The fact that all these functors are denoted by the same symbol will not cause anyconfusion.

e.REFUAF

27.7. Example. Let n be a positive integer. For g ∈ k[t1, . . . , tn], the set maps

g(R) : Rn −→ R, (r1, . . . ,rn) 7−→ g(R)(r1, . . . ,rn) := g(r1, . . . ,rn) (1) TILGE

vary functorially with R ∈ k-alg, hence give rise to an eelement g ∈Mor(Ank ,A

1k) =

k[Ank ]. On the other hand, g determines a unique morphism g∗ : k[t]→ k[t1, . . . , tn]

in k-alg given by

g∗(h) := h(g) (h ∈ k[t]), (2) GEAST

and one checks that the diagram

Rn g(R) //

εn(R) ∼=

R

∼= ε1(R)

Spec(k[t1, . . . , tn])(R) Spec(g∗)(R)// Spec(k[t])(R)

(3) TIGVEP

commutes. Hence, by Corollary 27.5 (a), since every morphism k[t]→ k[t1, . . . , tn]in k-alg has form g∗ for a unique g ∈ k[t1, . . . , tn],


k[t1, . . . , tn]∼−→ k[An

k ], g 7−→ g (4) GTIG

is an isomorphism of k-algebras. We usually identify

k[t1, . . . , tn] = k[Ank ], k[t1, . . . , tn] 3 g = g ∈ k[An

k ] (5) GIDTIG

accordingly.p.ADJSPE

27.8. Proposition. The contra-variant functors

Spec : k-alg−→ k-fct, k[−] : k-fct−→ k-alg

are adjoint to one another in the sense that, for all k-functors X and all R ∈ k-alg,there exists a natural bijection

ΨX,R : Mor(X,Spec(R)

) ∼−→ Homk-alg(R,k[X])

given by

ΨX,R( f )(r)(S)(x) = f (S)(x)(r) (1) ADJEL

for all f ∈Mor(X,Spec(R)), r ∈ R, S ∈ k-alg, x ∈ X(S). Moreover,

Ψ−1

X,R(g)(S)(x)(r) = g(r)(S)(x) (2) ADJIN

for all g ∈ Homk-alg(R,k[X]), S ∈ k-alg, x ∈ X(S), r ∈ R.

Proof. We define Ψ :=ΨX,R by (1). Since f (S)(x) : R→ S is a morphism in k-alg,so will be Ψ( f ) : R→ k[X] once we have shown that the left-hand side of (1) variesfunctorially with S. Thus, fixing r ∈ R and a morphism σ : S→ S′ in k-alg, we mustshow with g :=Ψ( f ) that the diagram

X(S)g(r)(S) //

X(σ)

S

σ

X(S′)

g(r)(S′)// S′

(3) ADJFU

commutes, which follows by a straightforward computation from (1) and the com-mutativity of

X(S)f (S) //

X(σ)

Spec(R)(S)

Spec(R)(σ)

X(S′)

f (S′)// Spec(R)(S′).

Conversely, define Ψ ′ : Homk-alg(R,k[X])→Mor(X,Spec(R)) by


Ψ′(g)(S)(x)(r) := g(r)(S)(x) (4) ADJPR

for g ∈ Homk-alg(R,k[X]), S ∈ k-alg, x ∈ X(S), r ∈ R. Again we must show that theset map Ψ ′(g)(S) : X(S)→ Spec(R)(S) varies functorially with S, i.e., that for amorphism σ : S→ S′ in k-alg the diagram

X(S)Ψ ′(g)(S) //

X(σ)

Spec(R)(S)

Spec(R)(σ)

X(S′)

Ψ ′(g)(S′)// Spec(R)(S′)

commutes, which follows from the commutativity of (3) for all r ∈ R. Finally, thedefinitions (1), (4) show that the maps Ψ , Ψ ′ are inverse to one another.

e.ENDOB27.9. Example. The affine k-scheme Spec(k) satisfies

Spec(k)(R) = Homk-alg(k,R) = ϑR (1) SPEK

for all R ∈ k-alg, where

ϑR : k −→ R, α 7−→ (ϑR)(α) := α1R. (2) BUR

is the unit morphism in k-alg corresponding to R. Hence

Spec(k)(ϕ)(ϑR) = ϑS (3) SPEPHI

for all morphisms ϕ : R→ S in k-alg.Now let X be any k-functor. By Proposition 27.8, there is a unique morphism

σX : X→ Spec(k), called the structure morphism of X, and (1), (3) imply for allR ∈ k-alg that

σX(R) : X(R)−→ Spec(k)(R) = ϑR

is the constant mapX(R) 3 x 7−→ ϑR ∈ Spec(k)(R).

Moreover, every morphism f : X→ X′ of k-functors is one over Spec(k) in thesense that the triangle

Xf

//

σX ##

X′

σX′Spec(k)

commutes. Finally, if X = Spec(R) is an affine k-scheme, then

σX = Spec(ϑR). (4) SIGMAX

p.CRIAF


27.10. Proposition. (a) Let X be a k-functor. With the notation of Proposition 27.8,

fX :=Ψ−1

X,k[X](1k[X]) : X−→ Spec(k[X]) (1) KAXIN

is a morphism of k-functors such that

fX(R)(x)(g) = g(R)(x) (2) KAXINEL

for all R ∈ k-alg, x ∈ X(R), g ∈ k[X], and if h : X→ X′ is any morphism of k-functors, then the diagram

XfX //

h

Spec(k[X])

Spec(k[h])

X′fX′// Spec(k[X′])

(3) KAXINDI

commutes.(b) With the notation of Proposition 27.4,

fSpec(R) = Spec(ΦR,A1k) (4) EFSPE

(c) A k-functor X is an affine k-scheme if and only if fX is an isomorphism.

Proof. (a) That the k-functor fX satisfies (2) follows immediately from (27.8.2).Using (2), it is now straightforward to verify the commutativity of (3).

(b) Put X := Spec(R). We have noted in (27.6.6) that ΦR,A1k

: k[X]→ R is anisomorphism of k-algebras. By Corollary 27.5, therefore,

Spec(ΦR,A1k) : X ∼−→ Spec(k[X]) (5) EXISOSP

is an isomorphism of k-functors. In order to establish (4), let S ∈ k-alg, x ∈ X(S),g ∈ k[X]. With Φ := ΦR,A1

kwe must show Spec(Φ)(S)(x)(g) = fX(S)(x)(g), which

follows easily by direct computation since the diagram

Spec(R)(R)g(R) //

Spec(R)(x)

R

x

Spec(R)(S)

g(S)// S

commutes.(c) follows immediately from (3), (4) and (5).

c.ANTEQ27.11. Corollary. The contra-variant functors


k-algSpec // k-affk[−]

oo

define an anti-equivalence of categories.

Proof. Let X be an affine k-scheme. Then fX : X ∼→ Spec(k[X]) is an isomorphismby Proposition 27.10 (c), and (27.10.3) shows that the family fX, X ∈ k-aff, de-termines an isomorphism of functors from 1k-aff to Spec k[−]. Conversely, letR ∈ k-alg. Then the bijective map

Φ = ΦR,A1k

: k[Spec(R)] ∼−→ R

of Proposition 27.4 by (27.6.6) is an isomorphism of k-algebras. Now let ρ : R→ R′

be a morphism of k-algebras. Since for all f ∈ k[Spec(R)] the diagram

Spec(R)(R)f (R) //

Spec(R)(ρ)

R

ρ

Spec(R)(R′)

f (R′)// R′

commutes, one checks that so does

k[Spec(R)]ΦR,A1

k

∼= //

k[Spec(ρ)]

R

ρ

k[Spec(R′)]

ΦR′,A1k

∼= // R′.

Thus the family ΦR,A1k, R ∈ k-alg, determines an isomorphism of functors from the

composition k[−]Spec to 1k-alg.

ss.NOCON27.12. Notational conventions. (a) Let f : X→ X′ be a morphism of k-functors.If there is no danger of confusion, the set map f (R) : X(R)→ X′(R) for R ∈ k-algwill simply be written as f :

f : X(R)−→ X′(R) (R ∈ k-alg). (1) SINOT

For a morphism ρ : R→ R′ in k-alg and x ∈ X(R), we therefore have

f(X(ρ)(x)

)= X′(ρ) f (x). (2) SIFU

(b) Let X be an affine k-scheme. Then we identify


X = Spec(k[X]) (3) IDSPEK

by means of the isomorphism fX of Proposition 27.10. Given R ∈ k-alg, x ∈ X(R)and f ∈ k[X], an application of (27.10.2) yields x( f ) = fX(R)(x)( f ) = f (R)(x),hence

x( f ) = f (x). (4) IDEV

(c) Let f : X→ X′ be a morphism of affine k-schemes. By Corollary 27.11, there isa unique morphism f ∗ : k[X′]→ k[X] in k-alg having f = Spec( f ∗). We call f ∗ theco-morphism of f .

ss.PRAFF27.13. Direct products of affine schemes. Let Xi for i = 1,2 be affine k-schemesThen the k-functor X1×X2 of 27.1 is an affine k-scheme as well. More precisely,one checks easily that

Π : X1×X2∼−→ Spec(k[X1]⊗ k[X2])

defined byΠ(x1,x2)( f1⊗ f2) := x1( f1)x2( f2) = f1(x1) f2(x2)

for R∈ k-alg, xi ∈Xi(R), fi ∈ k[Xi], i = 1,2 is an isomorphism. Note that the projec-tions pi : X1×X2→Xi correspond to the co-morphisms p∗i : k[Xi]→ k[X1]⊗k[X2]given by p∗1( f1) = f1⊗1k[X2] and p∗2( f2) = 1k[X1]⊗ f2.

ss.CLOSU27.14. Closed subfunctors. Let X be an affine k-scheme. For any subset I ⊆ k[X],we use (27.12.4) to define a subfunctor V(I) of X by setting

V(I)(R) :=

x ∈ X(R) | x(I) = 0=

x ∈ X(R) | ∀ f ∈ I : f (x) = 0

(1) CLOVE

for all R ∈ k-alg. We call V(I) the closed subfunctor of X determined by I or simplya closed subfunctor of X. It clearly depends only on the ideal generated by I. On theother hand, if I ⊆ k[X] is an ideal, with canonical projection π : k[X]→ k[X]/I, thenSpec(π) : Spec(k[X]/I)→ Spec(k[X]) may be regarded as an isomorphism

Spec(π) : Spec(k[X]/I) ∼−→ V(I). (2) CLAFF

In particular, closed subfunctors of X are affine k-schemes. If I′ is another ideal ink[X], with canonical projection π ′ : k[X]→ k[X]/I′, then we claim

I ⊆ I′ ⇐⇒ V(I′)⊆ V(I). (3) REVER

The implication from left to right being obvious, let us assume V(I′)⊆ V(I). Sincethese are both affine k-schemes, the inclusion V(I′) → V(I) of closed subfunctorsof X has the form Spec(ρ) for some morphism ρ : R/I→ R/I′ in k-alg satisfyingρ π = π ′. Thus I ⊆ Ker(ρ π) = Ker(π ′) = I′.

ss.OSUP


27.15. Open subfunctors. Let X be an affine k-scheme. For any subset I ⊆ k[X],we define a subfunctor D(I) of X by setting

D(I)(R) :=

x ∈ X(R) | Rx(I) = R=

x ∈ X(R) | ∑f∈I

R f (x) = R

(1) OPEDE

for all R ∈ k-alg. This is indeed a subfunctor of X since, for all morphismsρ : R→ R′ in k-alg and all x ∈ X(R), we find finitely many ri ∈ R, fi ∈ I such that∑ri fi(x) = 1R, which by (27.12.2) implies ∑ρ(ri) fi(X(ρ)(x)) = ∑ρ(ri)ρ( fi(x)) =ρ(∑ri fi(x)) = 1R′ , and we conclude X(ρ)(x) ∈ D(I)(R′).

Subfunctors of X having the form D(I) for some I ⊆ k[X] are said to be open.Of particular importance is the case of a principal open subfunctor, defined by theproperty that I = f, f ∈ k[X], is a singleton. We put X f := D( f ) := D( f) andhave

X f (R) =

x ∈ X(R) | f (x) ∈ R×=

x ∈ X(R) | x( f ) ∈ R×. (2) PRIOP

The elements x ∈ X(R) = Homk-alg(k[X],R) having x( f ) ∈ R× can be characterizedby the property that they factor through the localization k[X] f , in which case theydo so uniquely. Thus there is a natural identification

X f = Spec(k[X] f

)(3) PRIAF

such that the co-morphism of the inclusion X f → X is the canonical map k[X]→k[X] f . In particular, principal open subfunctors of affine k-schemes are affine. Note,however, that an arbitrary open subfunctor of an affine k-scheme will in general notbe affine.

ss.GROSCH

27.16. k-group schemes. By a k-group functor we mean a functor from the cate-gory k-alg to the category of groups:

G : k-alg−→ grp.

By composing G with the forgetful functor grp→ set, we obtain a k-functor, alsodenoted by G, to which the formalism of the preceding subsections applies. By asubgroup functor of G we mean a subfunctor H of G (viewed as a k-functor) suchthat H(R) is a subgroup of G(R) for all R ∈ k-alg. Then H may be regarded as ak-group functor in its own right. More generally, morphisms of k-group functors aredefined as natural transformations. Thus, given k-group functors G,G′, a morphismf : G→ G′ is nothing else than a morphism of k-functors making f (R) : G(R)→G′(R) a group homomorphism for all R ∈ k-alg.

By an affine k-group scheme, we mean a k-group functor that, regarded as a k-functor, is isomorphic to an affine k-scheme, so we have G ∼= Homk-alg(R,−) ask-functors, for some R ∈ k-alg. From now on, we will drop the prefix “affine” andjust talk about k-group schemes to mean affine k-group schemes.

e.EXGAM


27.17. Examples. (a) The additive group of k is defined as the k-group functor Gagiven by Ga(R) = R (viewed as an additive group) for all R ∈ k-alg and Ga(ϕ) = ϕ

for all morphisms ϕ : R→ S in k-alg. Thus Ga = A1k as k-functors and k[Ga]∼= k[t]

canonically.(b) The multiplicative group of k is defined as the k-group functor Gm given byGm(R)=R× (viewed as a multiplicative group) for all R∈ k-alg and Gm(ϕ) : R×→S× induced by a morphism ϕ : R→ S in k-alg via restriction. Thus Gm = (A1

k)t ask-functors and k[Gm]∼= k[t, t−1] canonically.

e.EXMA27.18. Example. Let M be a k-module. Generalizing the additive group of k as inExample 27.17 (a), we obtain a k-group functor Ma by setting Ma(R) :=MR =M⊗R(viewed as an additive group) for all R ∈ k-alg and Ma(ϕ) := 1M ⊗ϕ : Ma(R)→Ma(S) (viewed as an additive group homomorphism) for all morphisms ϕ : R→ Sin k-alg; regarded just as a k-functor. Ma has made its appearance before, in ourtreatment of polynomial laws (§13). Writing S(M∗) for the symmetric algebra of thedual of M (Bourbaki [12, III, §6]), we claim that if M is finitely generated projective,then Ma is a k-group scheme with k[Ma]∼= S(M∗) canonically.

Indeed, for any R ∈ k-alg, the universal property of the symmetric algebra com-bined with 10.1 and Lemma 10.11 yield the following chain of canonical isomor-phisms.

Homk-alg(S(M∗),R

) ∼= Homk(M∗,R)∼= HomR(M∗⊗R,R)∼= (M∗⊗R)∗ ∼= M∗∗⊗R∼= M⊗R = Ma(R).

Keeping track of how these isomorphisms act on individual elements, we obtain anidentification

Ma(R) = Homk-alg(S(M∗),R

)such that u⊗ r for u ∈ M, r ∈ R is the unique unital k-algebra homomorphismS(M∗)→ R satisfying

(u⊗ r)(v∗) = 〈v∗,u〉r (1) MAAFF

for all v∗ ∈M∗. It is now easily checked that, for any morphism ϕ : R→ S in k-alg,this identification matches Ma(ϕ) with Homk-alg(S(M∗),ϕ), which completes theproof.

ss.SCAPORE27.19. Scalar polynomial laws revisited Since A1

k , the affine line, is nothing elsethan the forgetful functor k-alg→ set, it follows for any k-module M that k[Ma] =Mor(Ma,A1

k) agrees with the k-algebra of scalar polynomial laws on M as definedin 13.2. Thus Example 27.18 combined with Corollary 27.11 implies Polk(M,k) ∼=S(M∗) as k-algebras provided M is finitely generated projective.

e.GEEL27.20. Example. Let A be a unital associative k-algebra which is finitely gener-ated projective as a k-module. We claim that GL1(A) defined by GL1(A)(R) := A×R(viewed as a multiplicative group) for all R ∈ k-alg and GL1(A)(ϕ) : A×R → A×S


induced from a morphism ϕ : R→ S in k-alg via restriction of 1A⊗ϕ : AR→ AS isa group scheme over k. Indeed, we know from Example 27.18 that Aa is a groupscheme over k. Let f ∈ k[Aa] be any regular function on Aa such that x ∈ AR,R ∈ k-alg, is invertible if and only if f (x) is a unit in R (for instance, one couldchoose f = detL, where L : A → Endk(A) is the left multiplication of A anddet : Endk(A)→ k is the determinant, which are both compatible with base change).Then GL1(A) = (Aa) f as k-functors and (27.15.3) shows that GL1(A) is a k-groupscheme satisfying

k[GL1(A)]∼= k[Aa] f . (1) REGEEL

The preceding construction specializes to Gm = GL1(k) but also to

GLn := GL1(

Matn(k)), (2) GEELEN

where we have

k[GLn]∼= k[ti j | 1≤ i, j ≤ n]det. (3) REGLEN

ss.SOMID27.21. Some identifications. We fix two k-modules M,N and let R ∈ k-alg. By10.1, the k-linear map

Homk(M,N)−→ HomR(MR,NR), η 7−→ ηR

extends uniquely to an R-linear map

Φk,R : Homk(M,N)R −→ HomR(MR,NR)

such that

Φk,R(η⊗ r)(u⊗ r′) = η(u)⊗ rr′ (1) EXLI

for η ∈ Homk(M,N), u ∈ M, r,r′ ∈ R. Now suppose ϕ : R→ S is a morphism ink-alg, so we have S ∈ R-alg via ϕ , identify MS = (MR)S, NS = (NR)S by means of(10.3.1) and let

Ψ : HomR(MR,NR)−→ HomS((MR)S,(NR)S

)= HomS(MS,NS)

be the R-linear map given by

Ψ(ζ ) := ζS = ζ ⊗R 1S(ζ ∈ HomR(MR,NR)

). (2) PSIZE

Then it is straightforward to check that the diagram


Homk(M,N)a(R)Φk,R //

Homk(M,N)a(ϕ))

HomR(MR,NR)

Ψ

Homk(M,N)a(S)

Φk,S

// HomS(MS,NS)

(3) PHIKAR

commutes.Finally, let us assume that M,N are both finitely generated projective. Then Φk,R

is an isomorphism (Bourbaki [12, II, §5, Prop. 7]), allowing us to identify

Homk(M,N)R = HomR(MR,NR) (4) EXHOM

by means of Φk,R, ditto for S in place of R, and since (3) commutes, we conclude

ζS = Homk(M,N)a(ϕ)(ζ ) (5) ZETHOM

for all ζ ∈ Homk(M,N)R.e.GEELM

27.22. Example. Let M be a finitely generated projective k-module and R ∈ k-alg.The identifications of 27.21 show Endk(M)R = EndR(MR) as R-algebras. Thus

GL(M) := GL1(

Endk(M))

(1) GEELM

is a k-group scheme having GL(M)(R) = GL(MR) for all R ∈ k-alg and

GL(M)(ϕ) : GL(MR)−→ GL(MS), GL(MR) 3 η 7−→ ηS ∈ GL(MS) (2) GEELPH

for any morphism ϕ : R→ S in k-alg. Moreover, (27.20.1) implies

k[GL(M)] = k[Endk(M)a]det. (3) KAGEEL

e.AUTNAS27.23. Example. Let A be a non-associative k-algebra. We define a k-group functorAut(A) by setting Aut(A)(R) := Aut(AR) for all R ∈ k-alg and

Aut(A)(ϕ) : Aut(AR)−→ Aut(AS), Aut(AR) 3 η 7−→ ηS ∈ Aut(AS)

for all morphisms ϕ : R→ S in k-alg, where we regard S as an R-algebra via ϕ anduse (10.3.1) to identify AS = (AR)S as S-algebras. We claim:

(∗) if A is finitely generated projective as a k-module, then Aut(A) is a closedsubfunctor of GL(A) and hence, in particular, a k-group scheme, called theautomorphism group scheme of A.

In order to see this, let u,v ∈ A and w∗ ∈ A∗. For R ∈ k-alg we define a set map

fu,v,w∗(R) : GL(AR)−→ R

by


fu,v,w∗(η) := fu,v,w∗(R)(η) :=⟨w∗R,η(uRvR)−η(uR)η(vR)

⟩for η ∈GL(AR). It follows immediately from Lemma 10.11 and (27.22.2) that theseset maps vary functorially with R. Thus fu,v,w∗ ∈ k[GL(A)]. Moreover, since thecanonical pairing A∗×A→ k is non-singular, η ∈GL(AR) belongs to Aut(AR) if andonly if fu,v,w∗(η) = 0 for all u,v ∈ A and all w∗ ∈ A∗. Hence we deduce from 27.14that Aut(A) is the closed subfunctor of GL(A) determined by the ideal I⊆ k[GL(A)]generated by the quantities fu,v,w∗ , u,v ∈ A, w∗ ∈ A∗. In particular, assertion (∗)follows. Note that, since A and A∗ are both finitely generated as k-modules, so is Ias an ideal in k[GL(A)].

e.ORGR27.24. Example. In a similar vein, let Q := (M,q) be a quadratic module over k.Then

O(Q) := O(M,q) := η ∈ GL(M) | qη = q (1) ORGR

is a subgroup of GL(M), called the orthogonal group of Q. Moreover, this groupmay be converted into a k-group functor by using Corollary 12.5 to define O(Q)(R) :=O(QR), QR := (MR,qR), for all R ∈ k-alg, and

O(Q)(ϕ) : O(Q)(R)−→O(Q)(S), O(QR) 3 η 7−→ ηS ∈ O(QS) (2) SCOR

for all morphisms ϕ : R→ S in k-alg,where we employ the same identifications asin 27.23.. If M is finitely generated projective, we let (wi)1≤i≤n be a finite family ofgenerators for M and define set maps

fi(R, fi j(R) : GL(MR)−→ R

for 1≤ i, j ≤ n and all R ∈ k-alg by setting

fi(R)(η) := qR(η(wiR)

)−qR(wiR),

fi j(R)(η) := qR(η(wiR),η(w jR)

)−qR(wiR,w jR)

for η ∈ GL(MR). These set maps vary functorially with R, hence define elementsfi, fi j ∈ k[GL(M)], and writing I ⊆ k[GL(M)] for the ideal they generate, O(Q) isclearly the closed subfunctor of GL(M) determined by I. In particular, O(Q) is ak-group scheme, called the orthogonal group scheme of Q.

ss.BACHFU27.25. Base change. Let k′ be a fixed commutative associative k-algebra with 1.Under restriction of scalars, every k′-algebra becomes a k-algebra, and every ho-momorphism of k′-algebras becomes one of k-algebras. In particular, k′-alg may beviewed canonically as a subcategory, though not a full one, of k-alg:

k′-alg⊆ k-alg. (1) PKSUB

Restricting a k-functor X as defined in 27.1 to k′-alg, we obtain a k′-functor, denotedby Xk′ and called the base change or scalar extension of X from k to k′. Similarly,


restricting a morphism f : X→ Y of k-functors to k′-alg, we obtain a k′-functorfk′ : Xk′ → Yk′ , called the base change or scalar extension of f from k to k′.

Most of our preceding constructions commute with base change. For instance,we have

(Ank)k′ = An

k′ (2) PAFFEN

for any positive integer n. If M is a k-module, then

(Ma)k′ = (Mk′)a (3) PEMA

under the identification (10.3.1) since (Ma)k′(R′) = M ⊗ R′ = (M ⊗ k′)⊗k′ R′ =(Mk′)R′ = (Mk′)a(R′) for all R′ ∈ k′-alg, similarly for morphisms in k′-alg.

Let R ∈ k-alg. For R′ ∈ k′-alg, (10.1.4) yields a bijection

canR(R′) : Homk-alg(R,R′)∼−→ Homk′-alg(Rk′ ,R

′) (4) CANARP

given by

canR(R′)(ρ)(r⊗α′) = α

′ρ(r) (5) CANARL

for all ρ ∈Homk-alg(R,R′), r ∈ R, α ′ ∈ k′, and this bijection depends functorially onR′. Thus we obtain an isomorphism

canR :(Spec(R)

)k′∼−→ Spec(Rk′). (6) BASPE

Moreover, for a morphism ϕ : R→ S in k-alg, one checks that the diagram(Spec(S)

)k′

∼=canS

//(Spec(ϕ)

)k′

Spec(Sk′)

Spec(ϕk′ )

(Spec(R)

)k′

∼=canR

// Spec(Rk′)

(7) CANES

commutes.ss.REFUBA

27.26. Regular functions under base change. We continue the discussion begunin 27.25. Let X be a k-functor. An element f ∈ k[X] by (27.6.2) is a morphismf : X→ A1

k of k-functors, hence gives rise to a morphism fk′ : Xk′ → (A1k)k′ = A1

k′of k′-functors, and we conclude fk′ ∈ k′[Xk′ ]. By (27.6.4), the map

k[X]−→ k′[Xk′ ], f 7−→ fk′ , (1) EXREF

is a morphism in k-alg and thus gives rise to a morphism

canX : k[X]k′ −→ k′[Xk′ ] (2) CANEX


in k′-alg given by

canX( f ⊗α′) = α

′ fk′ (3) CANEXEX

for f ∈ k[X] and α ′ ∈ k′-alg. Consulting the morphisms fX (resp. fXk′ ) describedin (27.10.2) it is now easily checked, using (27.25.5), (27.25.6) and (3), that thediagram

Xk′ fXk′

//

( fX)k′

Spec(k′[Xk′ ])

Spec(canX)

(Spec(k[X])

)k′

∼=cank[X]

// Spec(k[X]k′)

(4) EFEXP

commutes. Now suppose X is an affine k-scheme. Since ( fX)k′ and fXk′ are bothisomorphisms of k′-functors, by Proposition 27.10 (c), so is Spec(canX) by (4), andwe conclude from Corollary 27.5 (b) that

canX : k[X]k′∼−→ k′[Xk′ ]

is an isomorphism of k′-algebras.

28. Faithfully flat etale splittings of composition algebras.

s.FPPFCAGiven a commutative ring k, remaining fixed throughout this section, and a primeideal p ⊆ k, it follows immediately from §26, Exercise 138 (b), that a compositionalgebra over k becomes split after extending scalars to the separable closure of κ(p).Unfortunately, this observation is as obvious as it is useless. For example, the basechange from k to any field in k-alg trivializes the linear algebra of k and thus destroysall the relevant information one could possibly have about this important ingredient.

In the present section, this deficiency will be overcome by dealing with scalarextensions relative to faithfully flat etale k-algebras. After having derived (resp. re-called) a number of basic properties pertaining to this important concept, we willindeed be able to show that every composition algebra C over k becomes split aftera faithfully flat etale base change. The proof is not at all obvious and, following[73], will be based on a number of scheme-theoretic concepts and results due toGrothendieck [37] and Demazure-Gabriel [22] which will be quoted here in duecourse. As an iomportant by-producr of our approach we also show that Aut(C) asdefined in Example 27.23 is a smooth group scheme in the sense of 28.16 below.

We begin our preparations for the main results of this section by recalling a fewelementary facts from Bourbaki [12], Knus-Ojanguren [60] and Waterhouse [126],sometimes with proofs, sometimes without.

ss.FLAFA

28 Faithfully flat etale splittings of composition algebras. 185

28.1. Flat and faithfully flat modules. Let N be a k-module. By [12, II, §3, Corol-lary of Proposition 5], the functor −⊗N : k-mod→ k-mod is right exact, so when-ever we are given an exact sequence

M′ϕ// M

ψ// M′′ // 0 ,

of k-modules, the induced sequence

M′⊗Nϕ⊗1N

// M⊗Nψ⊗1N

// M′′⊗N // 0

is also exact. N is said to be flat if the functor −⊗N is exact in the sense that itpreserves exact sequences, equivalently, if for every injective k-linear map ϕ : M′→M of k-modules, the induced linear map ϕ⊗1N : M′⊗N→M⊗N is also injective.We say N is faithfully flat provided a sequence of k-modules is exact if and onlyif it becomes so after tensoring with N. A free k-module is always flat, while it isfaithfully flat if and only if it is non-zero.

ss.FFBAS28.2. (Faithful) flatness under base change. Let k′ ∈ k-alg and N be a k-module.(a) Generalizing iterated scalars extensions as in 10.3, we consider a k′-module M′

and let k′ act on M′⊗N = M′⊗k N through the first factor, making M′⊗N a moduleover k′. We then have a natural identification

M′⊗k N = M′⊗k′ Nk′ (1) EMPREN

of k′-modules such that

x′⊗k y = x′⊗k′ yk′ , x′⊗k′ (y⊗α′) = (α ′x′)⊗k y (2) ELPREN

for x′ ∈M′, y∈N, α ′ ∈ k′. Moreover, for a k′-linear map ϕ ′ : M′→M′1 of k′-modulesand a k-linear map ψ : N→ N1 of k-modules, we obtain

ϕ′⊗k ψ = ϕ

′⊗k′ ψk′ (3) FIPRPS

under this identification.(b) Let

M′1ϕ ′1

// M′2ϕ ′2

// M′3 (4) PEMPFI

be a sequence of k′-modules. Since 1Nk′ = (1N)k′ , we may apply (a) to obtain acommutative diagram


M′1⊗k′ Nk′ϕ ′1⊗k′1Nk′

//

1 ∼=

M′2⊗k′ Nk′ϕ ′2⊗k′1Nk′

//

1 ∼=

M′3⊗k′ Nk′

1 ∼=

M′1⊗k Nϕ ′1⊗k1N

// M′2⊗k Nϕ ′2⊗k1N

// M′3⊗N.

(5) PEMPEN

Now suppose N is flat over k and assume (4) is exact. Then so is the bottom row of(5), hence also its top row, and we conclude that Nk′ is flat over k′. Moreover, if N isfaithfully flat over k and the top row of (5) is exact, so is its bottom row, hence also(4). Thus Nk′ is faithfully flat over k′.In summary:(faithful) flatness is stable under base change.

ss.FLAFALG28.3. Flat and faithfully flat algebras. By a (faithfully) flatk-algebra we mean aunital commutative associative algebra over k, i.e., an object of k-alg, that is (faith-fully) flat as a k-module.

Let R be a flat k-algebra. If M is a k-module and N ⊆M is a k-submodule, thenthe inclusion i : N → M gives rise to an R-linear injection iR : NR → MR, whichmay and always will be used to identify NR ⊆ MR as an R-submodule. Extendingthe short exact sequence 0→ N→M→M/N→ 0 from k to R, we obtain

(M/N)R = MR/NR. (1) MMODN

Similarly, given any k-linear map f : M → M′ of k-modules and doing the samewith

0 // Ker( f ) // Mf// M′ // Coker( f ) // 0

yields

Ker( fR) = Ker( f )R, Coker( fR) = Coker( f )R. (2) KERR

Finally, the factorization

Mf

//

g ""

M′

Im( f )-

<<

""0,

where g : M → Im( f ) is the unique k-linear map induced by f , gives rise to thecommutative diagram


MR fR//

gR ""

M′R

Im( f )R

-

<<

##0

of R-modules, which shows

Im( fR) = Im( f )R. (3) IMF

p.CHAFF28.4. Proposition. Let R be a flat k-algebra. Then the following conditions areequivalent.

(i) R is faithfully flat.(ii) For all k-modules M, the linear map canM := canM,R : M→MR, x 7→ xR, is

injective.(iii) MR = 0 implies M = 0, for all k-modules M.

Proof. (i)⇒ (ii). Put ϕ := canM . By faithful flatness, it suffices to show that

ϕR : M⊗R−→ (M⊗R)⊗R

is injective. But since it is easily checked that the k-linear map

ψ : (M⊗R)⊗R−→M⊗R, (x⊗ r1)⊗ r2 7−→ x⊗ (r1r2),

satisfies ψ ϕR = 1M⊗R, the assertion follows.(ii)⇒ (iii). Obvious.(iii)⇒ (i). Let

M′ϕ//

Mψ//

M′′

M′R ϕR

// MR ψR// M′′R

(1) PEMPHIM

be a commutative diagram, with the top row being given as any sequence ink-mod, and the bottom row assumed to be exact. We must show that the top rowis exact as well. The k-linear map f := ψ ϕ : M′ → M′′ satisfies fR = 0, henceIm( f )R = Im( fR) (by (28.3.3)) = 0 and then f = 0 by (iii). Thus Im(ϕ) ⊆Ker(ψ). On the other hand, (28.3.2)), (28.3.3)) and exactness of the bottom rowimply Im(ϕ)R = Im(ϕR) = Ker(ψR) = Ker(ψR). From (28.3.1) we therefore deduce(Ker(ψ)/ Im(ϕ))R = 0, and (iii) again shows Im(ϕ) = Ker(ψ), as claimed.

ss.EQU


28.5. Equalizers. Let C be a category. By an equalizer of morphisms f ,g : X→Yin C we mean a morphism e : E → X in C such that f e = g e and, for anymorphism u : U→X in C such that f u= gu, there is a unique morphism h : U→E in C such that the diagram

E e// X

f //g// Y

U

∃!h

OO

u

??

commutes. Clearly, if an equalizer exists, it is unique up to a unique isomorphism.For more details on this concept, see D. Pumplun [106, 4.2].

It is sometimes important to realize the linear map canM in condition (ii) ofProposition 28.4 as an equalizer. To this end, we consider an arbitrary R ∈ k-algand two morphisms

ρi = ρRi : R−→ S := R⊗R (i = 1,2) (1) RHOI

in k-alg defined by

ρ1(r) := r⊗1R, ρ2(r) := 1R⊗ r (r ∈ R). (2) RHOEL

Givena k-module M, we also put

ρMi := ρ

R,Mi := 1M⊗ρi : MR −→MS. (3) RHOIM

p.RHOIQ28.6. Proposition. Let M be a k-module and suppose R ∈ k-alg is faithfully flat.With the notation of 28.5, the sequence

0 // M canM// MR

ρM1 //

ρM2

// MS (1) RHOIQ

is exact. In particular, the natural map canM : M→MR is an equalizer of ρM1 , ρM

2in the category k-mod.

Proof. The final statement follows immediately from the exactness of (1). By faith-ful flatness of R, it therefore suffices to show that

0 // M⊗RcanM⊗1R

// MR⊗Rρ⊗1R

// MS⊗R (2) RHOIQAR

is exact, where ρ := ρM1 − ρM

2 . Since canM is injective by Proposition 28.4, so iscanM⊗1R by flatness of R, and we have exactness of (2) at M⊗R. Since, obvi-ously, ρ canM = 0, we conclude (ρ ⊗ 1R) (canM⊗1R) = (ρ canM)⊗ 1R = 0,hence Im(canM⊗1R) ⊆ Ker(ρ ⊗ 1R), and exactness of (2) at MR⊗R will followonce we have shown the reverse inclusion. For this purpose, we define a k-linear


map ϕ : MS⊗R→MR⊗R by

ϕ

((x⊗ (r1⊗ r2)

)⊗ r3

):= (x⊗ r1)⊗ (r2r3)

for all x ∈M and all r1,r2,r3 ∈ R. A straightforward verification shows

ϕ (ρM1 ⊗1R) = 1MR⊗R, Im

(ϕ (ρM

2 ⊗1R))⊆ Im(canM⊗1R).

For z ∈ Ker(ρ ⊗ 1R), we therefore conclude z = ϕ((ρM1 ⊗ 1R)(z)) = ϕ((ρM

2 ⊗1R)(z)) ∈ Im(canM⊗1R), which completes the proof.

p.CHARFIB28.7. Proposition. Let R ∈ k-alg and p ∈ Spec(R). Writing ϑR : k→ R for the unithomomorphism corresponding to R as in 27.9, κ(p) for the algebraic closure of thefield κ(p) and setting X := Spec(R) as an affine k-scheme, we consider the followingconditions on R and p.

(i) Spec(ϑR)−1(p) 6= /0.

(ii) R⊗κ(p) 6= 0.(iii) X

(κ(p)

)6= /0.

Then the implications

(i) ⇐⇒ (ii) ⇐= (iii) (1) CHARFIB

hold. Moreover, if R is finitely generated as a k-algebra, then all three conditionsare equivalent.

Proof. Put K := κ(p).(i)⇔ (ii). §10, Exercise 39.(iii) ⇒ (ii). If X(K) 6= /0, then there exists a morphism R→ K in k-alg, which

in turn induces a unital homomorphism R⊗κ(p)→ K of κ(p)-algebras. Hence (ii)holds.

We have thus shown (1). Assuming that R is finitely generated as a k-algebra,it remains to verify the implication (ii) ⇒ (iii). If (ii) holds, then R⊗ κ(p) is anon-zero finitely generated κ(p)-algebra. By [11, V, §3, Proposition 1], therefore,we find a morphism R⊗ κ(p)→ K in κ(p)-alg. Composing with the natural mapR→ R⊗κ(p) yields an element of X(K). Hence (iii) holds.

p.FLAFAITH28.8. Proposition. For a flat k-algebra R, the following conditions are equivalent.

(i) R is faithfully flat.(ii) R⊗κ(p) 6= 0 for all p ∈ Spec(k).(iii) The natural map Spec(R)→ Spec(k) induced by the unit homomorphism k→

R is surjective.(iv) mR 6= R for all maximal ideals m⊆ k.

Proof. (i)⇒ (ii). Apply Proposition 28.4.(ii)⇒ (iii). Apply Proposition 28.7.


(iii)⇒ (iv). By (ii), some prime ideal q⊆ R lies above m. Hence mR⊆ q⊂ R.(iv) ⇒ (i). By Proposition 28.4 we have to show MR 6= 0 for all k-modules

M 6= 0. Let 0 6= x ∈ M and I := Ann(x) ⊂ k. The kx ∼= k/I as k-modules, so wehave an exact sequence 0→ k/I→M. Since R is flat, this implies that the sequence0→ (k/I)⊗ R = R/IR→ MR is also exact. Let m ⊆ k be a maximal ideal in kcontaining I. Then IR⊆mR⊂ R by (iv), which implies R/IR 6= 0 and then MR 6=0.

The two preceding results are related to a concept that turns out to be useful lateron.

ss.FAIAFF

28.9. Faithful affine k-schemes. An affine k-scheme X is said to be faithful ifX(K) 6= /0 for all algebraically closed fields K ∈ k-alg. In this case, we also say thatX has non-empty geometric fibers.

ss.CONVEN28.10. Convention. It is sometimes convenient to use notions originally defined forunital commutative associative k-algebras also for affine k-schemes and vice versa.This convention is justified by the anti-equivalence of these categories establishedin Corollary 27.11. For example, an affine k-scheme X is faithfully flat if and onlyif k[X] ∈ k-alg has this property.

c.FAFFF28.11. Corollary. Consider the following conditions, for any affine k-scheme X.

(i) X is faithful and flat.(ii) X is faithfully flat.

Then (i) implies (ii), and both conditions are equivalent if k[X] is finitely generatedas a k-algebra.

Proof. If condition (i) holds, so does condition (ii) of Proposition 28.7, for R := k[X]and any p∈ Spec(k). Hence that proposition combined with Proposition 28.8 showsthat X is faithfully flat. Conversely, suppose X is faithfully flat and R is finitelygenerated as a k-algebra. If K ∈ k-alg is an algebraically closed field, the kernel ofthe unit homomorphism ϑK : k→ K is some prime ideal p ∈ Spec(k), making K aκ(p)-algebra in a natural way. Hence the unit homomorphism ϑK factors uniquelythrough the unit homomorphism

ϑL : k −→ L, L := κ(p).

On the other hand, since Spec(ϑR) : Spec(R)→ Spec(k) is surjective by Proposi-tion 28.8, we have X(L) 6= /0 by Proposition 28.7. Therefore X(K) 6= /0, and X isfaithful.

The property of an algebra to be finitely generated, which shows up as an importantingredient of Proposition 28.7, is sometimes not enough for the intended applica-tions and has to be replaced by the following refinement.


ss.FIPRAL28.12. Finitely presented k-algebras. By a presentation of a k-algebra R ∈ k-alg,we mean a short exact sequence

0 // Ii// k[T]

π// R // 0, (1) FIPRAL

where T = (t1, . . . , tn) is a finite chain of independent indeterminates, π is a mor-phism in k-alg and I ⊆ k[T] is an ideal. For a presentation of R to exist it is necessaryand sufficient that R be finitely generated as a k-algebra. The presentation (1) of R issaid to be finite if the ideal I ⊆ k[T] is finitely generated. We say a k-algebra R (thecondition R ∈ k-alg being understood) is finitely presented if a finite presentation ofR exists.

Tensoring (1) with any k′ ∈ k-alg, we obtain a commutative diagram

Ik′ ik′//

jk′

k′[T]πk′// Rk′

// 0

0 // I′i′

==

0,

where I′ := ik′(Ik′) ⊆ k′[T] and jk′ is induced by ik′ . Since this diagram is every-where exact and I′ ⊆ k′[T] is obviously a finitely generated ideal if I ⊆ k[T] is, itfollows that the property of a k-algebra to be finitely presented is stable under basechange. We wish to show, among other things, that this property is also stable underfaithfully flat descent. We require a preparation.

p.APREFI28.13. Proposition. Every presentation of a finitely presented k-algebra is finite.

Proof. Let R ∈ k-alg be finitely presented and let

0 // I // k[T]π// R // 0 (1) ARPRE

be any presentation of R as in (28.12.1). By hypothesis, there exists a finite presen-tation

0 // J // k[S]µ// R // 0 (2) FIPRE

of R, so J ⊆ k[S] is a finitely generated ideal. We must shoiw that I ⊆ k[T] is afinitely generated ideal as well. Writing S = (s1, . . . ,sm), T = (t1, . . . , tn), the quan-tities π(t j) ∈ R by (2) have a lift under µ to polynomials g j ∈ k[S]. Thus

µ(g j) = π(t j) (1≤ j ≤ n). (3) MUG


The morphism

ϕ : k[S,T] = k(S)⊗ k[T]µ⊗π

// R⊗RmultR

// R (4) FIES

is surjective satisfying, in obvious notation,

ϕ(S) = µ(S), ϕ(T) = π(T). (5) FIEST

We now claim

Ker(ϕ) = J+n

∑j=1

k[S,T](t j−g j). (6) KERFI

Consulting (2), (5), we see that the right-hand side is contained in the left. Con-versely, let f ∈Ker(ϕ), write g := (g1, . . . ,gn) ∈ k[S]n and regard f as a polynomialh(T) ∈ k[S][T]. Then (5), (3), (2) imply

h(g(S)

)= f(S,g(S)

)∈ J, (7) HAGES

while the Taylor expansion (cf. (13.13.3) below) yields

h(T) = h(g(S)+T−g(S)

)= h(g(S)

)+ ∑

r≥1(Drh)

(g(S),T−g(S)

),

where the first summand on the right by (7) belongs to J. On the other hand,(Drh)(g(S),T) for r ≥ 1 is homogeneous of degree r in T and thus belongs to theideal in k[S,T] generated by t1, . . . , tn. We therefore conclude

(Drh)(g(S),T−g(S)

)∈

n

∑j=1

k[S,T](t j−g j),

which completes the proof of (6). Now let hi ∈ k[T] for 1 ≤ i ≤ m be lifts of µ(si)under π , so

π(hi) = µ(si) (1≤ i≤ m). (8) PIHI

Setting h := (h1, . . . ,hm) ∈ k[T]m, we consider the surjective homomorphism

ψ : k[S,T]→ k[T]

of unital k-algebras given by

ψ(S) = h, ψ(T) = T. (9) PSIST

Since Ker(ϕ)⊆ k[S,T] is a finitely generated ideal by (6), the proof will be completeonce we have shown


ψ(

Ker(ϕ))= I. (10) PSIKER

By (2), (3), (6), (8), (9), the left-hand side is clearly contained in the right. Con-versely, let f (T) ∈ I. Then (1), (3) show f (g(S)) ∈ J, hence

f (T) = f(g(S)+T−g(S)

)= f(g(S)

)+ ∑

r≥1(Dr f )

(g(S),T−g(S)

)∈ J+ ∑

r≥1k[S,T](t j−g j).

Now (6) implies f (T) ∈Ker(ϕ)∩k[T], and from (9) we deduce f (T) = ψ( f (T)) ∈ψ(Ker(ϕ)), which completes the proof of (10).

c.FAFLAD28.14. Corollary. Let R,k′ ∈ k-alg and suppose k′ is faithfully flat. If Rk′ is finitelygenerated (resp. finitely presented) over k′, then so is R over k.

Proof. Assume first that Rk′ is finitely generated over k′. Then there exists an exactsequence

k′[T′]π ′// Rk′

// 0 (1) KAYPR

in k′-alg with T′ = (t′1, . . . , t′n), and the quantities π ′(t′j) ∈ Rk′ for 1≤ j ≤ n may be

written as

π′(t′j) =

m

∑i=1

ri j⊗α′i j, ri j ∈ R, ,α ′i j ∈ k′ (1≤ i≤ m, 1≤ j ≤ n). (2) PIPR

Let T = (ti j)1≤i≤m,1≤ j≤n be a chain of independent variables and π : k[T]→ R bethe morphism in k-alg given by π(ti j) = ri j for 1 ≤ i ≤ m, 1 ≤ j ≤ n. Since theπ ′(t′j), 1≤ j≤ n, generate Rk′ as a k′-algebra, so do the (ri j)k′ , 1≤ i≤m, 1≤ j≤ n,by (2). Thus the sequence

k[T]π// R // 0 (3) KAYBFT

in k-alg becomes exact after tensoring with k′, hence must have been so all alongsince k′ is faithfully flat over k. Thus it follows from the exactness of (3) that R is afinitely generated k-algebra.

Now suppose Rk′ is finitely presented over k′. By what we have just seen, R isfinitely generated over k, so we have a presentation

0 // I // k[T]π// R // 0 (4) IKET

of R as in (28.12.1). By faithful flatness of k′, the extended sequence

0 // I⊗ k′ // k′[T]πk′// Rk′

// 0


continues to be exact. By Proposition 28.13, therefore, I⊗ k′ ⊆ k′[T] is a finitelygenerated ideal. On the other hand, from (28.2.1) we deduce I⊗ k′ = I⊗k[T] k′[T]as k′[T]-modules. Applying Exercise 139 (a), we conclude that I ⊆ k[T] is a finitelygenerated ideal. Hence R is finitely presented as a k-algebra.

ss.ETAL28.15. Etale k-algebras. The notion of a finite etale algebra as defined in 22.18will now be generalized as follows. A k-algebra R ∈ k-alg is said to be etale if itis finitely presented and satisfies one of the following equivalent conditions ([36,(17.1.1), (17.3.1), (17.6.2)], applied to the structure morphism Spec(R)→ Spec(k))of 27.9, and § 10, Exercise 42.

(i) For all k′ ∈ k-alg and all ideals I′ ⊆ k′ satisfying I′2 = 0, the set map

Homk-alg(R,k′) Homk-alg(R,π)// Homk-alg(R,k′/I′)

induced by the projection π : k′→ k′/I′ is bijective.(ii) R is flat over k, and for all p ∈ Spec(k), the extended algebra R(p) = R⊗κ(p)

over the field κ(p) is a possibly infinite direct sum of finite separable extensionfields of κ(p).

ss.SMOAFF28.16. Smooth affine k-schemes. Since the original definition of smoothness asgiven in [22, I, §4, 4.1] is rather technical, although more akin to what one expectsfrom the study of classical algebraic varieties or differentiable manifolds, we preferto recall the one of [37, (17.1.1), (17.3.1)] (see also the characterization in [22, I,$4, 4.6]) because it is more easily accessible in the present context. Accordingly, anaffine k-scheme X is said to be smooth if X is finitely presented and, for all R∈ k-algand all ideals I ⊆ R having I2 = 0, the set map X(R)→ X(R/I) induced by theprojection from R to R/I is surjective.

ss.THREEFF28.17. Three fundamental facts. Let X be an affine k-scheme.

(i) ([37, (17.16.2)], 28.11) If X is faithful, flat and finitely presented, abbreviatedfppf (“fidelement plat et de presentation finis”), then there exists an fppf R ∈k-alg such that X(R) 6= /0.

(ii) ([37, (17.16.3), (ii)], 28.8) If X is faithful and smooth, then there exists anetale fppf R ∈ k-alg such that X(R) 6= /0.

(iii) ([37, (17.7.3), (ii)], 27.26) If R ∈ k-alg is faithfully flat, then for X to besmooth over k it is necessary and sufficient that the base change

XR ∼= Spec(k[X]R

)∈ R-aff

be smooth over R.ss.TORS

28.18. Torsors. Let X be an affine k-scheme and G a k-group scheme acting on Xfrom the right, so we have a morphism X×G→ X of k-functors such that, for allR ∈ k-alg,


X(R)×G(R)−→ X(R), (x,g) 7−→ xg,

is a group action in the usual sense depending functorially on R. Note that X(R) maywell be empty! We say X is a torsor in the flat topology with structure group G if

(i) the action of G on X is simply transitive, i.e., for all R ∈ k-alg and all x,y ∈X(R), there is a unique g ∈G(R) satisfying y = xg.

(ii) There exists an fppf S ∈ k-alg such that X(S) 6= /0

In this case, fixing S as in (ii), we may apply (i) to obtain an isomorphism XS∼→GS

of affine S-schemes, allowing us to conclude from 28.17 (iii) that, e.g., X is smoothif and only if so is G.

Finally, if instead of(ii), we even have

(iii) there exists an etale fppf S ∈ k-alg having X(S) 6= /0,

then X is called a torsor (with structure group G) in the etale topology.ss.SETUP

28.19. The set-up. After the preparations we have presented in the preceding partsof this section, we will finally be able to describe the set-up for the applications wehave in mind.

Unless other arrangements have been made, we fix a composition algebra C ofrank r > 1 over k. By Corollaries 22.17 and 22.7, r ∈ 2,4,8 and C is non-singular.Moreover, given an elementary idempotent e∈C, the Peirce components C12(e) andC21(e) by §22, Exercise 105 are in duality to each other under the bilinearized normof C. Hence they are finitely generated projective k-modules of rank m = r

2 −1.

ss.SPLIDAT28.20. The concept of a splitting datum. In order to define splitting data for C,we discuss the cases r = 2,4,8 separately.(a) For r = 2, a splitting datum for C by definition has the form ∆ = (e)∈C1, wheree ∈C is an elementary idempotent.(b) For r = 4, a splitting datum for C by definition has the form ∆ = (e,x,y) ∈C3,where e ∈ C is an elementary idempotent and x ∈ C21(e), y ∈ C12(e) satisfy thefollowing conditions, with e′ := 1C− e.

xy = e′, tC(xy) = 1, yx = e. (1) SPLIDAQ

Actually, one checks easily that these equations are mutually equivalent.(c) For r = 8, a splitting datum for C by definition has the form ∆ = (e,x1,x2,x3) ∈C4, where e ∈C is an elementary idempotent and x1,x2,x3 ∈C21(e) satisfy the fol-lowing conditions.

(x1x2)x3 =−e, tC(x1x2x3) =−1, (xix j)xl =−e (2) SPLIDAO

for all cyclic permutations (i jl) of (123). Again one checks that these equations aremutually equivalent.

In summary, splitting data of C belong to Cn, where n = 1,2,4 for r = 2,4,8,respectively. Moreover, they are preserved by isomorphisms: if η : C → C′ is an


isomorphism of composition algebras of rank r over k, and ∆ is a splitting datumfor C, then the linear bijection ηn : Cn → C′n maps ∆ ⊆ Cn to the splitting datumη(∆) := ηn(∆)⊆C′n of C′. Finally, splitting data are stable under base change, soif ∆ ⊆Cn is a splitting datum for C, then ∆R ⊆Cn

R is one for CR, for all R ∈ k-alg.ss.AFSPLI

28.21. The affine scheme of splitting data. Again we treat the cases r = 2,4,8separately and let R ∈ k-alg be arbitrary.(a) Let r = 2. Then (e) ∈C1

R by §19, Exercise 82 is a splitting datum for CR if andonly if

nC(e) = 0, tC(e) = 1. (1) AFSPLIT

(b) Let r = 4. Then (e,x,y) ∈C3R is a splitting datum for CR if and only if

nC(e) = 0, tC(e) = 1,〈u∗R,ex〉= 〈u∗R,xe− x〉= 〈u∗R,ye〉= 〈u∗R,ey− y〉= 0, (2) AFSPLIF

tC(xy) = 1

for all u∗ ∈C∗.(c) Let r = 8. Then (e,x1,x2,x3) ∈C4

R is a splitting datum for CR if and only if

nC(e) = 0, tC(e) = 1,〈u∗R,exi〉= 〈u∗R,xie− xi〉= 0, (3) AFSPLIE

tC(x1x2x3) =−1

for all u∗ ∈C∗ and all i = 1,2,3.Setting n = 1,3,4 for r = 2,4,8, respectively, we therefore conclude that equations(1), (2), (3), respectively, define a closed subscheme of Cn

a := (Cn)a = (Ca)n in the

sense of 27.14, denoted by Splid(C) and called the affine scheme of splitting datafor C. By definition we have

Splid(C)(R) := Splid(CR) := ∆ | ∆ is a splitting datum for CR (4) SPLIDO

for all R ∈ k-alg and, in view of (10.3.2),

,Splid(C)(ϕ) : Splid(C)(R)−→ Splid(C)(S), (5) SPLIDM

Splid(CR) 3 ∆ 7−→ ∆S = (1Cn ⊗ϕ)(∆) ∈ Splid(CS),

for all morphisms ϕ : R→ S in k-alg. Passing from C to its affine scheme of splittingdata is obviously compatible with base change.

As will be seen in due course, for a splitting datum to exist it is necessary andsufficient that the ambient composition algebra be split. In fact, a much more precisestatement will be derived in Proposition 28.25 below. Before proceeding to thisresult, we discuss a few examples.


ss.STANEX28.22. Standard examples of splitting data. Here we present examples of split-ting data for the standard split composition algebras C0 :=C0r(k) of rank r > 1 overk as described in 24.18 (b)−(d). Again we treat the cases r = 2,4,8 separately.(a) r = 2. Then C = k⊕ k is the direct sum of two copies of k as ideals and

∆0 := ∆02(k) := (E), E := 1⊕0 ∈C (1) STANEXT

is a splitting datum for C.(b) r = 4. Then C0 = Mat2(k) is the algebra of 2×2-matrices with entries in k and

∆0 := ∆04(k) := (E,X ,Y ), (2) STANEXF

E := E11 =

(1 00 0

), X := E21 =

(0 01 0

), Y := E12 =

(0 10 0

)is a splitting datum for C0.(c) r = 8. The C0 = Zor(k) is the algebra of Zorn vector matrices over k and, writing(ei)1≤i≤3 for the canonical basis of k3 over k,

∆0 := ∆08(k) := (E,X1,X2,X3), E =

(1 00 0

), Xi =

(0 0ei 0

)(1≤ i≤ 3) (3) STANEXE

is a splitting datum for C since (24.17.4) implies

(X1X2)X3 =

(0 e1× e20 0

)(0 0e3 0

)(4) SPLIDZ

= − (e1× e2)te3

(1 00 0

)=−det(e1,e2,e3)E =−E.

The splitting datum ∆0r(k) exhibited in (1), (2), (3) above will henceforth be referredto as the standard splitting datum for C0r(k) (r = 2,4,8). We clearly have ∆0r(k)R =∆0r(R) for all R ∈ k-alg.

p.SPLISMO28.23. Proposition. The affine k-scheme of splitting data for C is smooth.

Proof. Consulting (28.21.1)−(28.21.3) we see that X := Splid(C) is defined byfinitely many equations as a closed subscheme of Cn

a . By Exercise 141 (b) and Exer-cise 142, therefore, X is finitely presented. Hence, by 28.16, it suffices to show thatthe set map X(R)→X(R/I) induced by the projection R→ R/I is surjective, for allideals I ⊆ R satisfying I2 = 0. In order to do so, we may assume R = k and writeα 7→ α , x 7→ x for the projection k→ k := k/I, C −→ C := C⊗ k = C/IC, respec-tively. We must show that every splitting datum ∆ ′ of C can be lifted to a splittingdatum ∆ of C satisfying ∆ = ∆ ′. In order to do so, we treat the cases r = 2,4,8separately.


(a) r = 2. A splitting datum for C has the form ∆ ′ = (e′) for some elementaryidempotent e′ ∈ C. By §19, Exercise 83, e′ can be lifted to an elementary idempotente ∈C. Thus ∆ := (e) is a splitting datum for C such that ∆ = ∆ ′.

(b) r = 4. A splitting datum for C has the form ∆ ′ = (e′,x′,y′) for some elemen-tary idempotent e′ ∈ C, where x′ ∈ C21(e′), y′ ∈ C12(e′) satisfy tC(x

′y′) = 1k. As in(a), we find an elementary idempotent e ∈C satisfying e = e′. The canonical projec-tion C→ C induces surjections Ci j(e)→ Ci j(e′) for i, j= 1,2. Hence x′,y′ canbe lifted to elements u ∈C21(e), v ∈C12(e), respectively, satisfying u = x′, v = y′.Hence

tC(uv) = tC(x′y′) = 1k,

and we conclude tC(uv) = 1+α for some α ∈ I. This implies 1+α ∈ k× withinverse 1−α since I2 = 0. Setting x := u ∈ C21(e), y := (1−α)v ∈ C12(e), wetherefore deduce not only x = x′, y = y′ but also tC(xy) = 1, so ∆ := (e,x,y) is asplitting datum for C such that ∆ = ∆ ′.

(c) r = 8. A splitting datum for C has the form ∆ ′ = (e′,x′1,x′2,x′3) for some

elementary idempotent e′ ∈ C and some x′1,x′2,x′3 ∈ C21(e′) satisfying tC(x

′1x′2x′3) =

−1k. Again e′ can be lifted to an elementary idempotent e ∈ C satisfying e = e′,and again x′i can be lifted to an element ui ∈C21(e) satisfying ui = x′i for 1≤ i≤ 3.Hence

tC(u1u2u3) = tC(x′1x′2x′3) =−1k,

and we conclude tC(u1u2u3) = −1 + α for some α ∈ I. This implies −1 + α ∈k× with inverse −(1+α). Setting xi := ui for i = 1,2 and x3 := −(1+α)u3, wetherefore deduce not only xi ∈C21(e) and xi = x′i for 1≤ i≤ 3 but aoso tC(x1x2x3) =−1. Thus ∆ := (e,x1,x2,x3) is a splitting datum for C such that ∆ = ∆ ′.

ss.FUNCIS28.24. The k-functor of isomorphisms. Let A,B be non-associative k-algebras.(a) In partial generalization of 27.23, we consider the set of (k-algebra) isomor-phisms from A to B:

Isomk(A,B) := Isom(A,B) := η | η : A ∼−→ B is a k-isomorphism. (1) ISOMAB

This set will in general be empty but it gives rise to to a k-functor

Isomk(A,B) := Isom(A,B) : k-alg−→ set

by defining

Isom(A,B)(R) := IsomR(AR,BR) (2) FUNCIS

for all R ∈ k-alg and

Isom(A,B)(ϕ) : Isom(A,B)(R)−→ Isom(A,B)(S), (3) ISOMPH

Isom(AR,BR) 3 η 7−→ ηS ∈ IsomS(AS,BS)


for all morphisms ϕ : R→ S in k-alg, where we view S as an R-algebra via ϕ andidentify AS = (AR)S, BS = (BR)S as S-algebras via (10.3.1).(b) The k-group functor Aut(A) of 27.23 acts canonically on Isom(A,B) from theright via

IsomR(AR,BR)×Aut(AR)−→ Isom(AR,BR), (η ,ζ ) 7−→ η ζ ,

and this action is simply transitive.(c) Returning to our composition algebra C of rank r > 1 over k, we define a splittingof C as an isomorphism from C0r(k) onto C, where C0r(k) is the split compositionalgebra over k described in (24.24.18). Thus Isom(C0r(k),C) is the set of splittingsof C(d) The terminology introduced in (a)−(c) above applies equally well to other al-gebraic structures, like quadratic forms (Exercise ???), associative algebras with orwithout involution (Exercise ???) or para-quadratic algebras (§??? below).

p.CASPLI28.25. Proposition. Let C be a composition algebra of rank r > 1 over k and de-note by ∆0r(k) the standard splitting datum for C0r(k) as defined in 28.22. Then theassignment

η 7−→ η(∆0r(k)

)defines a bijection from the set of splittings of C onto the set of splitting data of C:

Φ := Φ(k) : Isom(C0r(k),C

) ∼−→ Splid(C).

Proof. Since ∆0 := ∆0r(k) is a splitting datum for C0 :=C0r(k), its image under anisomorphism η : C0

∼→C is a splitting datum for C. Thus the map Φ is well defined,and it remains to show that it is bijective.

We begin with injectivity, which will follow once we have shown that ∆0 gener-ates C0 as a unital k-algebra. By 28.22, this is trivial for r = 2 and obvious for r = 4,while for r = 8 it suffices to note XiX j =

(0 el0 0

)for all cyclic permutations (i jl) of

(123), which follows immediately from (24.17.4) and (28.22.3).In order to show that Φ is surjective, we pick any splitting datum ∆ of C and have

to find an isomorphism η : C0∼→C sending ∆0 to ∆ . We do so again by treating the

cases r = 2,4,8 separately.r = 2. Then ∆ = (e) for some elementary idempotent e ∈ C. From §19, Exer-

cise 82, we deduce that e, e are unimodular and ke+ ke is a quadratic etale subalge-bra of C. Hence C = ke⊕ ke since C has rank 2 as a k-module, and

η : C0 −→C, α⊕β 7−→ αe+β e

is an isomorphism sending ∆0 to ∆ .r = 4. Then ∆ = (e,x,y), with e ∈C an elementary idempotent and x,y ∈C21(e)

satisfying (28.20.1). By Proposition 25.10, we may assume


C = Endk(k⊕L) =(

k L∗

L k

), e =

(1 00 0

),

for some line bundle L over k, and (25.9.5) yields elements u ∈ L, v∗ ∈ L∗ such thatx =

(0 0u 0

), y =

(0 v∗0 0

). Now (28.20.1) implies 〈v∗,u〉= 1, forcing L,L∗ to be free k-

modules of rank 1 with dual basis vectors u,v∗, respectively. Hence the assignment(α β

γ δ

)7−→

(α βv∗

γu δ

)(α,β ,γ,δ ∈ k)

defines an isomorphism η : C0∼→C sending ∆0 to ∆ .

r = 8. Then ∆ = (e,x1,x2,x3), where e ∈ C is an elementary idempotent andx1,x2,x3 ∈C21(e) satisfy (28.20.2). By Theorem 25.12, we may assume

C = Zor(M,θ) =

(k M∗

M k

), e =

(1 00 0

),

for some finitely generated projective k-module M of rank 3 and some orientationθ of M, where (25.11.8) yields elements u1,u2,u3 ∈ M satisfying xi =

( 0 0ui 0)

fori = 1,2,3. Combining (28.20.2) with (25.11.3), (25.11.2), we conclude

−e = (x1x2)x3 =

(0 u1×u20 0

)(0 0u3 0

)=

(−〈u1×θ u2,u3〉 0

0 0

)= −θ(u1∧u2∧u3)e,

hence θ(u1∧u2∧u3) = 1. Thus (u1,u2,u3) is a θ -balanced basis of M in the senseof 25.11, which also implies that there is an identification C = Zor(k) =C08(k) =C0matching ui with ei for i = 1,2,3. But this means we have found an isomorphismfrom C0 to C sending ∆0 to ∆ .

28.26. Theorem (Loos-Petersson-Racine [73, 4.10]). Let C be a composition alge-t.ISCTORbra of rank r ∈ 1,2,4,8 over k. Then the k-functor

Isom(C0r(k),C)

is an affine smooth torsor in the etale topology with structure group G=Aut(C0r(k)).

Proof. The assertion is trivial for r = 1 since in this case, applying Exercise 143below, we obtain a natural identification of Isom(C01(k),C) with the closed sub-scheme 1C ⊆ Ca. Hence we may assume r > 1. Putting X := Isom(C0r(k),C), theset maps

Φ(R) : X(R) ∼−→ Splid(C)(R)

given by Proposition 28.25 for any R ∈ k-alg are bijective and one checks that theyare compatible with base change, hence give rise to an isomorphism

Φ : X ∼−→ Splid(C)


of k-functors. By Proposition 28.23, therefore, X is a smooth affine k-scheme actedupon by G from the right in a simply transitive manner (28.24 (b)). In view of28.9, 28.17 (ii), the proof of the theorem will be complete once we have shownX(K) 6= /0 for all algebraically closed fields K ∈ k-alg. But this follows immediatelyfrom 26.10.

The preceding theorem has two corollaries which, up to a point, will be provedsimultaneously.

28.27. Corollary (Loos-Petersson-Racine [73, 4.11]). For any k-algebra C, thec.SPLICFPPFfollowing conditions are equivalent.

(i) C is a composition algebra over k.(ii) There exists a faithfully flat k-algebra R ∈ k-alg making CR a composition

algebra over R.(iii) There exists a faithfully flat k-algebra R ∈ k-alg making CR a composition

algebra over R that is split in the sense of 24.18.(iv) There exists a fppf etale k-algebra R ∈ k-alg making CR a split composition

algebra over R.

28.28. Corollary (Loos-Petersson-Racine [73, 4.12]). Let C be a composition al-c.SPLICSMgebra over k. Then Aut(C) is a smooth k-group scheme.

Proof of 28.27, 28.28. We first note that in 28.27, the implications (iv)⇒ (iii)⇒ (ii)are obvious, while the implication (ii)⇒ (i) follows from the fact that C is a conicalgebra (by Exercise 146) whose norm, nC, is separable (by Exercise 144, since(nC)R = nCR is and R is faithfully flat) and permits composition (since (nC)R doesand the natural map C→CR by Proposition 28.4 is injective). In 28.27, therefore, itremains to prove the implication (i)⇒ (iv).

Next we reduce both corollaries to the case that

C has rank r ∈ 1,2,4,8 as a k-module. (1) REDRAN

This is accomplished by considering the rank decomposition of C, which, by 24.18,attains the form

k = k0⊕ k1⊕ k2⊕ k3, C =C0⊕C1⊕C2⊕C3 (2) RADECO

as direct sums of ideals, where C j :=Ck j is a composition algebra of rank 2 j over k jfor 0 ≤ j ≤ 3. Hence, assuming the implication (i)⇒ (ii) of 28.27 if (1) holds. wefind fppf etale k j-algebras R j for 0 ≤ j ≤ 3 such that the composition algebra C jR j

over R j is split of rank 2 j. It is now straightforward to check that R := R0⊕R1⊕R2⊕R3 is an fppf etale k-algebra making CR = C0R0 ⊕C1R1 ⊕C2R2 ⊕C3R3 a splitcomposition algebra over k. This completes the reduction for 28.27.

Assuming 28.28 if (1) holds, let R ∈ k-alg and I ⊆ R be an ideal having I2 = 0.Using §9, Exercise 32, and arguing as before, we have


R = R0⊕R1⊕R2⊕R3,

I = I0⊕ I1⊕ I2⊕ I3, (3) ARZERI

CR =C0R0 ⊕C1R1 ⊕C2R2 ⊕C3R3 ,

CR/I =C0,R0/I0 ⊕C1,R1/I1 ⊕C2,R2/I2 ⊕C3,R3/I3 .

Since C j,R j = 1k jCR, we conclude that Aut(CR) stabilizes C j,R j for 0≤ j ≤ 3. Thus

Aut(CR) = Aut(C0R0)×Aut(C1R1)×Aut(C2R2)×Aut(C3R3),

and, similarly,

Aut(CR/I) = Aut(C0,R0/I0)×Aut(C1,R1/I1)×Aut(C2,R2/I2)×Aut(C3,R3/I3).

Since C jR j has rank 2 j over R j, the k j-group scheme Aut(C j) is smooth, forcing it tobe finitely presented and the natural map Aut(C jR j)→Aut(C j,R j/I j) to be surjectivefor 0≤ j ≤ 3. Hence so is the natural map Aut(CR)→ Aut(CR/I), which completesthe reduction also for 28.28.

For the remainder of the proof, we may therefore assume that (1) holds. HenceTheorem 28.26 implies that X := Isom(C0r(k),C) is an affine smooth torsor in theetale topology, with structure group G = Aut(C0r(k)). Thus condition (iii) of 28.18implies X(R) 6= /0 for some etale fppf R ∈ k-alg. Hence CR ∼=C0k(R) is split of rankr over R.This completes the proof of 28.27. Moreover, the chain of isomorphisms

Aut(C)R ∼= Aut(CR)∼= Aut(C0r(R)

)∼= GR ∼= XR

shows that Aut(C)R is smooth over R. But then, by 28.17 (iii), Aut(C) must besmooth over k.

ss.CONREM

28.29. Concluding remarks. The three fundamental facts from algebraic geome-try assembled in 28.17 form an extremely versatile tool to derive non-trivial resultsabout non-associative algebras over commutative rings, having first been appliedby Loos [69] in the setting of Jordan pairs more than forty years ago. Further ap-plication in many different directions will be given in subsequent portions of thiswork.

Exercises.

pr.MOFF 139. Modules and faithful flatness. Let M be a k-module.

(a) If k′ ∈ k-alg is faithfully flat and Mk′ is finitely generated as a k′-module, show that so is Mas a k-module.

(b) If M is projective of rank r ∈ N, show that there exists a faithfully flat k-algebra k′ makingMk′ a free k′-module of rank r.

pr.TRAFIPR 140. Let R be a finitely presented k-algebra.

(a) Prove¡ that a finitely presented R-algebra is finitely presented over k.(b) Conclude from (a) that R f is a finitely presented k-algebra, for any f ∈ R.


pr.IDFIPR 141. Let ϕ : R→ R′ be a surjective morphism in k-alg. Prove:

(a) (cf. [22, I, §3, 1.3 (b)]) If R is finitely generated and R′ is finitely presented, then Ker(ϕ)⊆ Ris a finitely generated ideal.

(b) If R is finitely presented and Ker(ϕ) ⊆ R is a finitely generated ideal, then R′ is finitelypresented.

pr.EMMA 142. Let M be a finitely generated projective k-module. Show that the affine k-scheme Ma of 27.18is finitely presented.

pr.SGLT 143. Let M be a k-module and w ∈ M. Show that w : k-alg→ set define by w(R) = wR ⊆ MRfor all R ∈ k-alg is a subfunctor of Ma, and even a finitely presented closed affine subscheme if Mis finitely generated projective.

pr.FFSEP 144. Let M be a projective k-nodule, q : M→ k a quadratic form and R a faithfully flat k-algebra.Show that if qR : MR→ R is separable over R in the sense of 12.11, then so is q over k.

pr.FFPOLA 145. Faithfully flat descent of polynomial laws. Let R be a faithfully flat k-algebra. As in 28.5,consider the two morphisms

ρi : R−→ S := R⊗R (i = 1,2) (1) RHOIX

in k-alg defined by

ρ1(r) := r⊗1R, ρ2(r) := 1R⊗ r (r ∈ R). (2) RHOELX

Pulling back scalar multiplication from S to R by means of ρi converts any T ∈ S-alg into someTi ∈ R-alg such that T1 = T2 = T as k-algebras. Now let M,N be k-modules and suppose g : MR→NR is a polynomial law over R. Then prove:(a) For i = 1,2, there are unique polynomial laws gi : MS→ NS over S such that giT := (gi)T = gTias set maps from (MS)T = MT = (MR)Ti to (NS)T = NT = (NR)Ti for all T ∈ S-alg.(b) There exists a polynomial law f : M→ N over k satisfying f ⊗R = g if and only if g1 = g2. Inthis case, f is unique, and for all k′ ∈ k-alg, the morphism

ψ : k′ −→ Rk′ , α′ 7−→ 1R⊗α

′,

in k-alg makes the diagram

Mk′ 1Mk′

//

1M⊗ψ

Mk′ fk′//

canMk′ ,Rk′

Nk′ 1Nk′

//

canNk′ ,Rk′

Nk′

1N⊗ψ

MRk′ 1MRk′

// (Mk′ )Rk′ = (MR)Rk′ gRk′// (NR)Rk′ = (Nk′ )Rk′ 1NRk′

// NRk′

(3) CHARF

commutative.

pr.FFCON 146. Faithfully flat descent of conic algebras. Let C be a non-associative k-algebra und supposeR ∈ k-alg is faithfully flat. Prove:

(a) If CR is unital, then so is C.(b) If C is projective as a k-module and CR carries a quadratic form over R making it a conic

R-algebra, then C carries a quadratic form over k making it a conic k-algebra such that thebase change from k to R of C as a conic k-algebra is CR as a conic R-algebra.


Remark. This exercise is a routine application of the previous one. Applications of this kind willoccur quite frequently in the present volume. Rather than always carrying out the details, we fromnow on refer to these applications by saying that the corresponding results are obtained by faithfullyflat descent.

pr.FFSPLIQ 147. Let Q := (M,q) be a quadratic space of rank 2n, n ∈N, over k. By a hyperbolic basis of Q wemean a family (wi)1≤i≤2n of elements in M such that

q(wi) = q(wn+i) = q(wi,w j) = q(wn+i,wn+ j) = 0, q(wi,wn+ j) = δi j (4) HYPB

for 1≤ i, j ≤ n. The set of hyperbolic bases of Q will be denoted by Hyp(Q)⊆M2n.

(a) Show that the k-functorHyp(Q) : k-alg−→ set

given byHyp(Q)(R) := Hyp(QR), QR := (MR,qR),

for all R ∈ k-alg and

Hyp(Q)(ϕ) : Hyp(Q)(R)−→Hyp(Q)(S),

Hyp(QR) 3 (wi)1≤i≤2n 7−→ (wiS)1≤i≤2n ∈ Hyp(QS)

for all morphisms ϕ : R→ S in k-alg is a smooth closed k-subscheme of M2na .

(b) Coclude from (a) that

(i) there exists a fppf etale k-algebra R making QR a split hyperbolic quadratic space overR in the sense of 12.17.

(ii) O(Q) is a smooth group scheme.

(Hint. Imitate the arguments of 28.24−28.28.)

Chapter 5Jordan algebras

c.JORAL

In the preceding chapter, we investigated some fundamental properties of composi-tion algebras, particularly of octonions, over arbitrary commutative rings. Our nextmain objective will be to accomplish the same for what we call cubic Jordan al-gebras, among which Albert algebras are arguably the most important. In order toachieve this objective, a few prerequisites from the general theory of Jordan alge-bras are indispensable. Rather than striving for maximum generality, we confineourselves to what is absolutely necessary for the intended applications.

29. Linear Jordan algebras

s.LIJOLinear Jordan algebras have been introduced by Albert, who in three fundamen-tal papers [4, 5, 6] developed a virtually complete structure theory of the finite-dimensional ones over arbitrary fields of characteristic not 2. While the main focusof the present volume is primarily on (quadratic) Jordan algebras, linear Jordan al-gebras are still of some interest since, e.g., they motivate the study of quadratic onesand provide useful illustrations of why certain weird phenomena can only occur incharacteristic 2. In this section, the most elementary properties of linear Jordan al-gebras will be discussed. Since linear Jordan algebras have been extensively treatedin book form (Braun-Koecher [15], Jacobson [43], Zhevlakov et al [127], and par-ticularly McCrimmon [83]), proofs will often be omitted.

Throughout we let k be a commutative ring containing 12 . We begin by repeating

the definitions 5.6, 5.7.ss.LIJO

29.1. The concept of a linear Jordan algebra. By a linear Jordan algebra overk we mean a (non-associative) k-algebra J satisfying the following identities, for allx,y ∈ J.

xy = yx (commutative law), (1) LIJCOM

x(x2y) = x2(xy) (Jordan identity). (2) LIJO

ss.SPEEXLI29.2. Special and exceptional linear Jordan algebras. (a) Let A be a k-algebrawith multiplication (x,y) 7→ xy. Then the symmetric product

x• y :=12(xy+ yx)

converts A into a commutative algebra over k, denoted by A+. Moreover, one checkseasily that, if A is associative, then A+ is a linear Jordan algebra. Linear Jordan

205

206 5 Jordan algebras

algebras that are isomorphic to a subalgebra of A+, for some associative algebra A,are said to be special. Non-special linear Jordan algebras are called exceptional.(b) For example, let B be a unital k-algebra and τ : B → B an involution. ThenH(B,τ) = x ∈ B | τ(x) = x is a unital subalgebra of B+. In particular, if B isassociative, then H(B,τ) is a unital special Jordan algebra.

ss.ELEIDE29.3. Elementary identities. The Jordan identity (29.1.2) can be expressed interms of left multiplication operators as

[Lx,Lx2 ] = 0, (1) LELIJO

so a commutative algebra is linear Jordan if and only if (1) holds, i.e., if and only ifthe left multiplication operators Lx and Lx2 commute.

Now let J be a linear Jordan algebra over k. Replacing x by αx+ y in (1) forx,y ∈ J, α ∈ k, expanding and comparing mixed terms by using the fact that J iscommutative and k contains 1

2 , we conclude

2[Lx,Lxy]+ [Ly,Lx2 ] = 0. (2) ONELE

Repeating this procedure and dividing by 2 yields

[Lx,Lyz]+ [Ly,Lzx]+ [Lz,Lxy] = 0, (3) TWOLE

which, when applied to any w ∈ J, amounts to

x((yz)w

)+ y((zx)w

)+ z((xy)w

)= (yz)(xw)+(zx)(yw)+(xy)(zw). (4) TOLE

We call (3) or (4) the fully linearized Jordan identity. Viewing (4) as a linear operatorin x, we deduce, after an obvious change of notation,

L(xy)z = LxyLz +LyzLx +LzxLy−LxLzLy−LyLzLx. (5) ELTRI

Here the sum of the first three terms on the right is symmetric in x,y,z, hence remainsunaffected by interchanging x and z. Since [x,y,z] = (xy)z− (zy)x = z(xy)− x(zy),this implies

L[x,y,z] = L[Lz,Lx]y = [[Lz,Lx],Ly]. (6) LIETR

For any integer m > 1, we may put y := xm−1, z := x in (5) and obtain the identity

Lxm+1 = 2LxmLx +Lx2Lxm−1 −L2xLxm−1 −Lxm−1L2

x , (7) LIPOM

which for m = 2 reduces to

Lx3 = 3LxLx2 −2L3x . (8) LIPOTR

p.LIJOBA

29 Linear Jordan algebras 207

29.4. Proposition. If J is a linear Jordan algebra over k, then JR is a linear Jordanalgebra over R, for any R ∈ k-alg.

Proof. It suffices to check the Jordan identity (29.1.2) for JR, which is straightfor-ward, using the identities derived in 29.3. See[83, p. 149] for details.

p.LIJOP29.5. Proposition. Let J be a linear Jordan algebra over k and x ∈ J.(a) For u ∈ k1[x] (resp. u ∈ k[x] if J is unital), Lu is a polynomial in Lx and Lx2 .(b) For u,v ∈ k1[x] (resp. u,v ∈ k[x] if J is unital), Lu and Lv commute: [Lu,Lv] = 0.

Proof. (a) We may assume u = xm for some m ∈ Z, m > 0 (resp. m ∈ N). Then theassertion follows from (29.3.7) by induction on m.

(b) Since Lx,Lx2 commute by (29.3.1), so do Lu,Lv by (a).

c.LIJOPA29.6. Corollary. Linear Jordan algebras over k are power-associative.

Proof. This follows from a straightforward application of Prop. 29.5 (b). Details areleft to the reader.

r.LIPOSI

29.7. Remark. In view of Prop. 29.5, equation (29.3.7) simplifies to

Lxm+1 = 2(Lxm −Lxm−1Lx)Lx +Lx2Lxm−1 . (1) LIPOSI

We have encountered examples of linear Jordan algebras in 29.2 (special Jordanalgebras) and in Thm. 5.10 (cubic euclidean Jordan matrix algebras, in particularthe euclidean Albert algebra). Other examples, seemingly of a completely differentnature, will now be introduced.

ss.LIJPOQ29.8. The linear Jordan algebra of a pointed quadratic module. Let (M,q,e)be a pointed quadratic module over k, with trace t and conjugation ι , cf. 12.13for details. Then the k-module M becomes a k-algebra J = J(M,q,e) under themultiplication

xy =12(t(x)y+ t(y)x−q(x,y)e

)(x,y ∈M). (1) LIMPOQ

J is obviously commutative and unital, with identity element 1J = e, and we have

x2− t(x)x+q(x)1J = 0 (2) SQLIPO

for all x ∈ J. Thus Lx2 , being a linear combination of Lx and 1J , commutes withLx, and we conclude that J is a unital linear Jordan algebra., called the linear Jor-dan algebra of the pointed quadratic module (M,q,e). Extending the proof of [43,Thm. VII.1] from fields of characteristic not 2 to commutative rings containing 1

2 inthe obvious way, it follows that J(M,q,e) is a special Jordan algebra.

Since q(e) = 1, we conclude from (2) that J = J(M,q,e) is a commutative conick-algebra in the sense of 19.1 whose norm, trace, conjugation agree with the corre-sponding data attached to (M,q,e). Conversely, consider any conic algebra C over k.


As we have noted before, (C,nC,1C) is a pointed quadratic module, and comparing(1) with (19.5.5) divided by 2, we obtain

J(C,nC,1C) =C+. (3) LIJOCO

We now extend the definition of the U-operator as given in 6.4 to the present moregeneral context.

ss.UOPLIJ29.9. The U-operator of a linear Jordan algebra. Let J be a linear Jordan algebraover k. For x ∈ J, the linear map

Ux : J −→ J, y 7−→Uxy := 2x(xy)− x2y, (1) UOPLIX

is called the U-operator of x. The quadratic map

U : J −→ Endk(J), x 7−→Ux = 2L2x−Lx2 , (2) UOPLIJ

is called the U-operator of J. Its bilinearization gives rise to the tri-linear Jordantriple product

xyz :=Ux,zy = (Ux+z−Ux−Uz)y = 2(x(zy)+ z(xy)− (xz)y

). (3) TRILIJ

Viewing this as a map acting on z, we obtain the linear operators

Vx,y := 2(Lxy +[Lx,Ly]

), (4) VELIJ

uniquely determined by the condition

Vx,yz = xyz. (5) VEELIJ

In particular, we have

Vx :=Vx,1J =V1J ,x = 2Lx, (6) VELELI

so up to the factor 2, the operator Vx agrees with the left multiplication by x in J.e.UOLIEX

29.10. Examples. In addition to the U-operator of cubic Jordan matrix algebras(Exc. 24 (b)), we now discuss the following cases.(a) Let A be an associative algebra over k. As in Exc. 24 (a), the U-operator of helinear Jordan algebra A+ is given by the formula

Uxy = xyx (x,y ∈ A) (1) UOAS

in terms of the associative product in A. In particular, any subalgebra of A+ is closedunder the binary operation (x,y) 7→ xyx.(b) Let (M,q,e) be a pointed quadratic module over k, with trace t and conjugationι , x 7→ x. Then the U-operator of the linear Jordan algebra J(M,q,e) is given by the

30 Para-quadratic algebras. 209

formula

Uxy = q(x, y)x−q(x)y (x,y ∈ J) (2) UOPOL

since (29.8.1), (29.8.2), (29.9.1) and (12.13.1) yield

Uxy = 2x(xy)− x2y = t(x)xy+ t(y)x2−q(x,y)x− t(x)xy+q(x)y

= q(x,e)t(y)x−q(x)t(y)e−q(x,y)x+q(x)y

= q(x, t(y)e− y

)x−q(x)

(t(y)e− y

)= q(x, y)x−q(x)y,

as claimed.

The U-operator and its variants described in 29.9 are of the utmost importance for aproper understanding of (linear) Jordan algebras. This is primarily due to a numberof fundamental identities satisfied by these operators.

ss.ADVID29.11. Advanced identities. Let J be a linear Jordan algebra over k. Then thefollowing identities hold for all x,y,z,u,v ∈ J, where for the validity of the first,J is required to be unital.

U1J = 1J , (1) UONEL

LyUx +UxLy =Uxy,x, (2) LYUX

LyUx,z +Ux,zLy =Uxy,z +Uzy,x, (3) LYUXZ

xyz= zyx, (4) TRISYM

[Vx,y,Vu,v] =Vxyu,v−Vu,yxv, (5) FILI

Vx,yUx =UxVy,x =UUxy,x, (6) VXYU

VUxy,y =Vx,Uyx, (7) VUXY

UUxy =UxUyUx. (8) FUFOL

A verification of these identities, which (with the exception of (1) and (4)) are highlynon-trivial, will be omitted because they are logically not strictly necessary for thesubsequent applications to cubic Jordan algebras. The interested reader is referredto McCrimmon [83, p. 202] or Meyberg [85].

30. Para-quadratic algebras.

s.PAQUAJust as linear Jordan algebras fit naturally into the more general framework of arbi-trary non-associative algebras, i.e., of modules (over a commutative ring) equippedwith a binary operation that is linear in each variable, (quadratic) Jordan algebras tobe investigated below fit naturally into the more general framework of what we callpara-quadratic algebras−modules equipped with a binary operation that is quadraticin the first variable and linear in the second. It is the purpose of the present section


to extend the language of (linear) non-associative algebras as explained in § 8 to thismodified setting.

Throughout we let k be an arbitrary commutative ring. With an eye on subsequentapplications, we discuss the notion of a para-quadratic algebra only in the presenceof a base point which serves as a “weak identity element”.

ss.COPAQU30.1. The concept of a para-quadratic algebra. By a para-quadratic algebraover k we mean a k-module J together with a quadratic map

U : J −→ Endk(J), x 7−→Ux, (1) UPAQ

the U-operator, and a distinguished element 1J ∈ J, the base point, such that

U1J = 1J . (2) UUNPAQ

We then write

xyz :=Ux,zy = (Ux+z−Ux−Uz)y (x,y,z ∈ J) (3) TRIPAQ

for the associated trilinear triple product, and

x y := x1Jy (x,y ∈ J) (4) CIRPAQ

for the associated bilinear circle product. Note that the triple product (3) is sym-metric in the outer variables, while the circle product (4) is commutative. Moreover,xyx = 2Uxy for all x,y ∈ J. Viewing (3) (resp. (4)) as a linear operator in z, weobtain linear maps

Vx,y : J −→ J, z 7−→ xyz, (5) VEPAQ

Vx :=Vx,1J : J→ J, y 7−→ x y = x1Jy. (6) VUNPAQ

that depend bilinearly on x,y (resp. linearly on x). Most of the time,we simply writeJ for a para-quadratic algebra, its U-operator, base point, triple and circle productbeing understood. In keeping with our introductory promise, 1J because of (2), (4)may be regarded as a weak identity element for J.

If J and J′ are para-quadratic algebras over k, a homomorphism from J to J′ isdefined as a k-linear map ϕ : J→ J′ preserving U-operators and base points in thesense that

ϕ(Uxy) =Uϕ(x)ϕ(y), ϕ(1J) = 1J′ (7) HOMPAQ

for all x,y ∈ J. In this case, ϕ also preserves triple and circle products, so we have

ϕ(xyz) = ϕ(x)ϕ(y)ϕ(z), ϕ(x y) = ϕ(x)ϕ(y) (x,y ∈ J). (8) HOMTRI

Summing up we obtain the category k-paquad of para-quadratic k-algebras.ss.UNPA


30.2. Unital para-quadratic algebras. A para-quadratic algebra J over k is saidto be unital if xy = 1Jxy for all x,y ∈ J. Since the triple product is symmetric inthe outer variables, and because of (30.12.4), we then have

x y = 1Jxy= x1Jy= xy1J (1) UNPA

for all x,y ∈ J and, in particular, 1J x = 2U1J x = 2x. If J is unital, the weak identityelement 1J is called the unit or identity element of J.

For the rest of this section, we fix a para-quadratic algebra J over k.ss.PASUB

30.3. Subalgebras. For X ,Y,Z ⊆ J we denote by UXY (resp. XY Z) the additivesubgroup of J generated by the expressions Uxy (resp. xyz) for x∈ X , y∈Y , z∈ Z.We say that J′ is a subalgebra of J if J′ ⊆ J is a k-submodule satisfying 1J ∈ J′ andUJ′J′ ⊆ J′. Then J′J′J′+ J′ J′ ⊆ J′ and there is a unique way of viewing J′

as a para-quadratic k-algebra in its own right such that the inclusion J′ → J is ahomomorphism. This implies not only 1J′ = 1J but also that the triple (resp. circle)product of J′ is obtained from the triple (resp. circle) product of J via restriction.

e.UNFL30.4. Example. Let A be a flexible unital k-algebra, so we have (xy)x = x(yx) =:xyx for all x,y ∈ A. Then the U-operator defined by

Uxy := xyx (x,y ∈ A) (1) UOFL

and the unit element of A convert A into a para-quadratic k-algebra denoted by A(+).Triple and circle product of A(+) are given by

xyz= (xy)z+(zy)x = x(yz)+ z(yx), x y = xy+ yx (x,y,z ∈ A). (2) TRIFL

In particular, the para-quadratic algebra A(+) is unital.ss.PAID

30.5. Ideals. We say I is an ideal in J if it is a k-submodule satisfying the inclusionrelations

UIJ+UJI +JJI ⊆ I. (1) PAID

In this case, there is a unique way of making the k-module J′ := J/I into a para-quadratic k-algebra such that the canonical map from J to J′ is a homomorphism.Conversely, the kernel of any homomorphism of para-quadratic algebras is an ideal.Moreover, if I1 and I2 are ideals in J, then so is I1+ I2, and the standard isomorphismtheorems of abstract algebra continue to hold in this modified setting.

ss.PAIN30.6. Inner and outer ideals. There is a vague analogy between one-sided idealsin ring theory and the following notions for para-quadratic algebras. A k-submoduleI ⊆ J is said to be an inner (resp. an outer) ideal if


UIJ ⊆ I (resp. UJI +JJI ⊆ I). (1) PAIN

Thus a submodule of J is an ideal if and only if it is an inner and an outer ideal. Butthe analogy to one-sided ideals goes only so far: for example, let I ⊆ J be an outerideal and x∈ I, y∈ J. The 2Uxy = xyx ∈ JJI ⊆ I, and we conclude that, if 1

2 ∈ k,then outer ideals of J are ideals.

ss.DISUID30.7. Direct sums of ideals. Let J1, . . . ,Jr be para-quadratic algebras over k. Then

J := J1⊕·· ·⊕ Jr,

their direct sum as a k-module, becomes a para-quadratic k-algebra, with U-operatorand base point respectively given by

Ux1⊕···xr(y1⊕·· ·yr) = (Ux1y1)⊕·· ·⊕ (Uxr yr), 1J = 1J1 ⊕·· ·⊕1Jr

for xi,yi ∈ Ji, 1 ≤ i ≤ r. It follows immediately from the definition that also thetriple and circle product of J are carried out component-wise. In particular, J isunital if and only if Ji is unital, for each i = 1, . . . ,r. Identifying Ji ⊆ J canonicallyfor 1≤ i≤ r, we clearly have UJiJ j = 0 for 1≤ i, j ≤ r, i 6= j and JiJ jJl= 0for 1≤ i, j, l ≤ r unless i = j = l.

Conversely, let J be a any para-quadratic algebra over k and suppose I1, . . . , Ir ⊆ Jare ideals such that J = I1⊕·· ·⊕ Ir as a direct sum of submodules (i.e., of ideals).For all i, j, l = 1, . . . ,r, this implies UIi I j = 0 unless i= j and IiI jIl= 0 unlessi = j = l. It follows that I1, . . . , Ir are para-quadratic k-algebras in their own right,and J identifies canonically with their direct sum as para-quadratic algebras.

ss.PAOW30.8. Powers. Let x ∈ J. We define the powers xn ∈ J for n ∈ N inductively by

x0 = 1J , x1 = x, xn =Uxxn−2 (n ∈ N n≥ 2) (1) PAOW

and write

k[x] := ∑n∈N

kxn (2) PAMOPO

for the submodule of J spanned by the powers of x. More generally, we define

kr[x] := ∑n≥r

kxn (3) PAMOPOR

for r ∈ N as a submodule of k[x]. We say J is power-associative at x if

Uxmxn = x2m+n, xmxnxp= 2xm+n+p (4) PAOWAS

for all m,n, p ∈ N. This is easily seen to imply

(xm)n = xmn (5) PAIT


for all m,n ∈ N, and that k[x]⊆ J is a para-quadratic subalgebra.For example, let A be a unital flexible k-algebra as in 30.4.. Then the powers of

x ∈ A are the same in A and A(+). In particular, if A is power-associative, then so isA(+) and conversely.

ss.PAIM30.9. Idempotents. An element c ∈ J is called an idempotent if c3 = c2 = c. Thisimplies cn = c for all positive integers n, hence k[c] = k1J + kc; in particular, J ispower-associative at c. There are always the trivial idempotents 0 and 1J but possiblyno others. Two idempotents c,d ∈ J are said to be orthogonal, written as c⊥ d, if

Ucd =Udc = ccd= ddc= cd = 0. (1) PAORID

Orthogonality of idempotents is obviously a symmetric relation. Moreover, c⊥ d iseasily seen to imply that c+d ∈ J is an idempotent. If c ∈ J is an idempotents, thenso is 1J− c, and the idempotents c,1J− c are orthogonal.

Let A be a unital flexible k-algebra. Then the idempotents of A and A(+) arethe same. Moreover, for two idempotents c,d in A to be orthogonal in A(+) it isnecessary and sufficient that they be orthogonal in A, i.e., cd = dc = 0.

ss.PAMU30.10. The multiplication algebra. The subalgebra of Endk(J) generated by thelinear operators Ux,Vx,y for x,y ∈ J is called the multiplication algebra of J, denotedmy Mult(J). Note by (30.1.2) that Mult(J) is a unital subalgebra of Endk(J), so 1J ∈Mult(J). We may view J canonically as a Mult(J)-left module. Then the Mult(J)-submodules of J are precisely the outer ideals of J.

ss.PASIM30.11. Simplicity and division algebras. J is said to be simple if it is non-zero andhas only the trivial ideals 0 and J. We say J is outer-simple if it is non-zero andthe only outer ideals are 0 and J. Note by 30.6 that simplicity and outer simplicityare equivalent notions if 1

2 ∈ k but not in general, see ??? below for examples. Notefurther by 30.10 that outer simplicity (and not simplicity) is equivalent to J being anirreducible Mult(J)-module.

And finally, J is said to be a division algebra if it is non-zero and Ux : J→ J isbijective for all nonzero elements x∈ J. For example, if A is a unital flexible divisionalgebra, then A(+) is a para-quadratic division algebra.

ss.PASCA30.12. Scalar extensions. For an R ∈ k-alg, the assignment ϕ 7→ ϕR = ϕ ⊗ 1Rdetermines a unital homomorphism Endk(J)→ EndR(JR) of k-algebras, which inturn yields canonically a unital R-algebra homomorphism

ω :(

Endk(J))

R −→ EndR(JR), ϕ⊗ r 7−→ rϕR = ϕ⊗ (r1R). (1) PAOM

Composing this with the R-quadratic extension of the U-operators of J (Cor. 12.5),we obtain an R-quadratic map U : JR→ EndR(JR)


JR UR

//

U

""

(Endk(J)

)R

ω

EndR(JR)

(2) PASCA

which, together with 1JR := (1J)R, makes JR a para-quadratic algebra over R, calledthe scalar extension or base change of J from k to R. The para-quadratic algebrastructure of JR is characterized by the conditions that

Ux⊗r(y⊗ s) = (Uxy)⊗ (r2s) (x,y ∈ J, r,s ∈ R) (3) PASCAU

and that the triple (resp. circle) prdoduct of JR is the R-trilinear (resp. R-bilinear)extension of the triple (resp. circle) product of J. The standard properties enjoyedby the scalar extensions of k-modules or linear non-associative algebras over k(cf. 10.1) now carry over to this modified setting without change.

ss.PACE30.13. The centroid. Due to the non-linear character of para-quadratic algebras, itseems impossible to define a meaningful analogue of the centre inside the algebrasthemselves. Instead, just as in the case of linear non-associative algebras without aunit (e.g., of Lie algebras, cf. Jacobson [42, Chap. X, § 1]), one has to work insidetheir endomorphism algebras.

Accordingly, we define the centroid of J, denoted by Cent(J), as the set of all el-ements a∈ Endk(J) such that, writing ax := a(x) (a∈ Endk(J), x∈ J) for simplicity,

Uax = a2Ux, Uax,y = aUx,y, aUx =Uxa, (x,y ∈ J). (1) PACE

The difficulty with this definition is that one of the conditions imposed on the ele-ments of the centroid is no longer linear in a, and hence it is not at all clear whetherCent(J)⊆ Endk(J) is a submodule, let alone a commutative subalgebra. Before dis-cussing this question any further, let us observe for a ∈ Cent(J) that

aUx,y =Uax,y =Ux,ay =Ux,ya, aVx,y =Vax,y =Vx,ay =Vx,ya (x,y ∈ J). (2) AUV

These relations either follow by linearizing (1) or by straightforward verification,e.g., aVx,yz = aUx,zy = Uax,zy = Vax,yz for z ∈ J. Note that, in view of (1), (2), theelements of the centroid behave “just like scalars” not only with respect to the U-operator but also with respect to the triple and circle product:

axyz= (ax)yz= x(ay)z= xy(az), (3) PACTRI

a(x y) = (ax) y = x (ay) (4) PACCI

for all a ∈ Cent(J), x,y,z ∈ J.


Our next aim will be to exhibit conditions under which the centroid becomes acommutative subalgebra (resp. a subfield) of the endomorphism algebra of J.

30.14. Proposition (cf. McCrimmon [78, Thm. 2]). The following conditions arep.PACOSUequivalent.

(i) Cent(J) is a commutative (unital) subalgebra of Endk(J).(ii) Cent(J) is a (unital) subalgebra of Endk(J).(iii) Cent(J) is an additive subgroup of Endk(J).(iv) The elements of Cent(J) commute by pairs: [a,b] = 0 for all a,b ∈ Cent(J).

Proof. In (i), (ii), unitality is automatic since 1J ∈ Cent(J).(i)⇒ (ii)⇒ (iii). Obvious.Before tackling the remaining implications (iii)⇒(iv)⇒(i), we claim, for all

a,b ∈ Cent(J),

a+b ∈ Cent(J) ⇐⇒ ∀x ∈ J : U(a+b)x = (a+b)2Ux ⇐⇒ [a,b] = 0. (1) SUMCEN

Indeed, since the last two conditions of (30.13.1) are linear in a, the first equivalencein (1) is obvious. As to the second, we use (30.13.1), (30.13.2) and compute

U(a+b)x− (a+b)2Ux =Uax +Uax,bx +Ubx−a2Ux− (ab+ba)Ux−b2Ux

= (2ab−ab−ba)Ux = [a,b]Ux,

which by (30.1.2) is zero for all x ∈ J if and only if [a,b] = 0. This completes theproof of (1).

(iii)⇒ (iv). This follows immediately from (1).(iv) ⇒ (i). By (1) and (iv), we need only show that Cent(J) is closed under

multiplication, so let a,b ∈ Cent(J). Then

U(ab)x = a2Ubx = a2b2Ux = (ab)2Ux

since a and b by (iv) commute. Thus ab ∈ Cent(J).

ss.CEPA30.15. Central para-quadratic algebras. J is said to be central if the linear mapk→ Endk(J), α 7→ α · 1J , is injective with image Cent(J). In this case, Cent(J) ⊆Endk(J) is a unital subalgebra isomorphic to k. Conversely, suppose Cent(J) ⊆Endk(J) is a subalgebra, automatically unital and commutative by Prop. 30.14. Thenthe natural action of Cent(J) on J converts J into a central para-quadratic algebraover Cent(J), denoted by Jcent and called the centralization of J. See 9.5 for theanalogous, but more elementary, concept in the context of linear non-associativealgebras.

ss.EXRA30.16. The extreme radical. We wish to show that, under some mild extra condi-tion, the centroid of a simple para-quadratic algebra is a field. This extra conditionis best understood in terms of the extreme radical of J, which is defined by


Rex(J) := z ∈ J |Uz =Uz,x = 0 for all x ∈ J (1) EXRA

and obviously a submodule of J. In fact, the extreme radical of J agrees with theradical of the quadratic map x 7→Ux as defined in 12.3.

30.17. Theorem (Schur’s lemma) (cf. McCrimmon [78, Thm. 3]). The centroid oft.PACENSIa simple para-quadratic algebra with zero extreme radical is a field.

Proof. Let a,b ∈ Cent(J) and x,y ∈ J. Then (30.13.1), (30.13.2) yield

U[a,b]x =Uabx−bax =Uabx−Uabx,bax +Ubax = a2Ubx−aUbx,axb+Uaxb2

= a2Uxb2−2a2Uxb2 +a2Uxb2 = 0,U[a,b]x,y =Uabx−bax,y = aUx,yb−aUx,yb = 0.

Hence [a,b]x∈Z(J)= 0, and we deduce from Prop. 30.14 that Cent(J)⊆Endk(J)is a commutative unital subalgebra. It remains to show that its non-zero elements areinvertible, so let a ∈ Cent(J) be non-zero. For x,y,z ∈ J we have

Uaxy = a2Uxy ∈ Im(a), Uyax = aUyx ∈ Im(a), yz(ax)= ayzx ∈ Im(a),

whence Im(a) ⊆ J is a non-zero ideal. Thus Im(a) = J by simplicity, and a is sur-jective. On the other hand, let z ∈ I := Ker(a). Then axyz = xy(az) = 0 for allx,y ∈ J, forcing JJI ⊆ I. Similarly, UJI ⊆ I. And finally, since a is surjective,x = aw for some w ∈ J, which implies aUzx = aUzaw = a2Uzw =Uazw = 0, henceUzx ∈ I. Thus I ⊂ J is an ideal, and we conclude I = 0. Summing up, we haveshown that a : J→ J is bijective.

Exercises.

pr.PARNIL 148. Monomials and the nil radical in para-quadratic algebras. Let J be a para-quadratic alge-bra over k. For X ⊆ J and m ∈ N we define subsets Monm(X) ⊆ J by setting Mon0(X) := 1J,Mon1(X) := X and by requiring that Monm(X) for m > 1 consist of all elements Uyz, y ∈Monn(X ,z ∈Monp(X), n, p ∈ N, n > 0, m = 2n+ p. The elements of

Mon(X) :⋃

m∈NMonm(X)

are called monomials (in J) over X .

(a) Prove Monm(x) = xm for all x ∈ J and all m ∈ N if J is power-associative.(b) An element x ∈ J is said to be nilpotent if 0 ∈Mon(x) is a monomial over x. We say

I ⊆ J is a nil ideal if it is an ideal consisting entirely of nilpotent elements. Prove that theimage of a nilpotent element under a homomorphism of para-quadratic algebras is nilpotent,and that for ideals I′ ⊆ I in J, the ideal I is nil if and only if I′ is nil and I/I′ is a nil idealin J/I′. Conclude that the sum of all nil ideals in J is a nil ideal, called the nil radical of J,denoted by Nil(J).

(c) Prove Nil(k)J ⊆ Nil(J).

pr.EVALPOL 149. Para-quadratic evaluation. Let J be a para-quadratic algebra over k and x ∈ J such that J ispower-associative at x. We define the evaluation at x ∈ J as the linear map εx : k[t]→ J determinedby εx(tn) = xn for all n ∈ N and put f (x) := εx( f ) for all f ∈ k[t].


(a) Show that εx : k[t](+) → J is a homomorphism of para-quadratic k-algebras and concludethat

I := Ix := Ker(εx) := f ∈ k[t] | f (x) = 0is an ideal in k[t](+). Show further that

I0 := I0x := Ker0(εx) = f ∈ k[t] | f (x) = (t f )(x) = 0

is the unique largest ideal in k[t] contained in I. Moreover, both f 2 and 2 f belong to I0 forall f ∈ I. It follows that R := k[t]/I0 is a unital commutative associative k-algebra, and withthe canonical projection ε0

x : k[t]→ R, there is a unique homomorphism π : R(+) → J ofpara-quadratic algebras making the diagram

k[t](+)

ε0x

||

εx

!!R(+)

π

// J

commutative. In particular, k[x] = Im(π) ⊆ J is a para-quadratic subalgebra of J, and a2 =2a = 0 for all a ∈ Ker(π).

(b) Suppose 2 = 0 in k and let n≥ 2 be an integer. Show that

In := ktn + tn+2k[t]

is an ideal in k[t](+) but not in k[t]. Conclude from the relations

xn+1 6= 0 = xn = xn+2 = xn+3 = · · ·

for the image of t under the canonical projection k[t](+) → Jn := k[t](+)/In that Jn has nolinear structure, i.e., there is no unital flexible algebra A over k satisfying Jn ∼= A(+).

pr.PAPONI 150. Let J be a para-quadratic algebra over k and x ∈ J. Show that, if J is power-associative at x,the following conditions are equivalent.

(i) x is nilpotent in the sense of Exc. 148 (b).(ii) xm for some positive integer m.(iii) There exists a positive integer m such that xn = 0 for all integers n≥ m.

Show further for the para-quadratic subalgebra k[x]⊆ J (cf. Exc. 149 (a)) that

Nil(k[x]) = v ∈ k[x] | v is nilpotent.

pr.PARALIFT 151. Para-quadratic lifting of idempotents. Let J be a para-quadratic algebra over k and x ∈ J suchthat J is power-associative at x.

(a) For v ∈ k[x], let U ′v : k[x]→ k[x] be the restriction of Ux to k[x]. Prove

U ′( f g)(x) =U ′f (x)U′g(x) (1) MULU

for all f ,g ∈ k[t] and conclude

[U ′v,U′w] = 0, U ′Uvw =U ′2v U ′w, U ′vn =U ′nv (2) COMU

for all v,w ∈ k[x] and all n ∈ N.(b) Assume there are integers n > d > 0 and scalars αd , . . . ,αn−1 ∈ k such that αd ∈ k× and

αdxd +αd+1xd+1 + · · ·+αn−1xn−1 + xn = 0.


Show that there is a unique element v ∈ k2d [x] satisfying Uxd v = x2d . Conclude Uv = 1 onk2d [x] and that c := v2 is an idempotent in k2d [x] satisfying Uc = 1 on k2d [x]. (Hint. ApplyExc. 31 (a) and Exc. 149 (a).)

(c) Let ϕ : J→ J′ be a surjective homomorphism of power-associative para-quadratic algebrasover k and suppose Ker(ϕ) ⊆ J is a nil ideal (Exc. 148). Conclude from (b) that everyidempotent c′ ∈ J′ can be lifted to J, i.e., there exists an idempotent c∈ J satisfying ϕ(c)= c′.

pr.OUTCEN 152. The outer centroid. Let J be a para-quadratic algebra over k and define the outer centroid ofJ, denoted by Centout(J), as the centralizer of the multiplication algebra of J. We say J is outercentral if the natural map α 7→ α1J from k to the outer centroid of J is an isomorphism. Prove:

(a) If J is outer simple, then its outer centroid is a(n associative) division ring.(b) If k is a field and J is finite-dimensional and outer simple over k, then EndCent0(J)(J) =

Mult(J).(c) Assume k is a field and J is finite-dimensional over k. Then J is outer central and outer

simple if and only if it is non-zero and Mult(J) = Endk(J).(d) For a finite-dimensional para-quadratic algebra over a field, the following conditions are

equivalent.

(i) J is outer central and outer simple.(ii) Every base field extension of J is outer simple.(iii) The scalar extension of J to the algebraic closure of the base field is outer simple.

pr.COS 153. Orthogonal systems of idempotents. Let J be a para-quadratic algebra over k. A finite family(c1, . . . ,cr) of elements in J is called an orthogonal system of idempotents if each ci, 1 ≤ i ≤ r, isan idempotent, and the following relations hold, for all i, j, l = 1, . . . ,r.

Uci c j = cicic j= ci c j = 0 (i 6= j), cic jcl= 0 (i, j, l mutually distinct). (3) PACOSY

An orthogonal system (c1, . . . ,cr) of idempotents in J is said to be complete if ∑ri=1 ci = 1J . Prove:

(a) If (c1, . . . ,cr) is an orthogonal system of idempotents in J, then ∑ri=1 ci is an idempotent, and

(c1, . . . ,cr,1J−r

∑i=1

ci)

is a complete orthogonal system of idempotents in J.(b) If J = A(+) for some unital flexible k-algebra A, then the (complete) orthogonal systems of

idempotents in J and in A are the same.

31. Jordan algebras and basic identities.

s.JABASAs experimental studies carried out by Jacobson in the nineteen-fifties suggest, themost promising way of extending the theory of linear Jordan algebras to arbitrarycommutative rings from those containing 1

2 consists in axiomatizing properties ofthe U-operator. The fruitfulness of this approach is underscored by the fact thatthe explicit formulas for the U-operator in our examples of linear Jordan algebras(Exc. 24 (b), (29.10.1, (29.10.2)), as opposed to the ones for the bilinear product((5.3.1), 29.2, 29.8), are defined over the integers and hence make sense over anycommutative ring. Unfortunately, however, the question of which specific propertiesof the U-operator should be singled out as axioms remained a mystery for a long

31 Jordan algebras and basic identities. 219

time. But then, in 1966, McCrimmon [75] introduced the concept of what he calleda quadratic Jordan algebra. He showed that this concept is equivalent to the conceptof a unital linear Jordan algebra over rings containing 1

2 and that it gives rise to afar reaching structure theory, culminating eventually in the Zel’manov-McCrimmonenumeration [84] of non-degenerate prime quadratic Jordan algebras.

Our aim in the present section will be to define quadratic Jordan algebras (hence-forth referred to simply as Jordan algebras) and to show that in the presence of 1

2they are categorically isomorphic to unital linear Jordan algebras. Using some ba-sic identities, we derive a few elementary properties of Jordan algebras and extendthe standard examples previously obtained in the linear case to the more generalquadratic setting.

Throughout we let k be an arbitrary commutative ring. In deriving the elemen-tary properties of Jordan algebras required in the present volume, we mostly followJacobson [45].

ss.COJA31.1. The concept of a Jordan algebra. By a Jordan algebra over k we meana para-quadratic k-algebra J with U-operator U and base point 1J satisfying thefollowing identities in all scalar extensions.

UUxy =UxUyUx, (1) JAUU

UxVy,x =Vx,yUx. (2) JAUVE

Equation (1) is called the fundamental formula. We write k-jord for the category ofJordan algebras over k, viewed as a full subcategory of k-paquad, the category ofpara-quadratic algebras over k. By definition, Jordan algebras remains stable underbase change.

Let J be a Jordan algebra over k. The triple (resp. circle) product associated withJ in its capacity as a para-quadratic algebra will be referred to as the Jordan tripleproduct (resp. the Jordan circle product) of J. Setting x = 1J in (2) and observing(30.1.2), (30.1.6), we conclude V1J ,x = Vx, hence 1Jxy = x1Jy = x y for allx,y ∈ J. Thus Jordan algebras are unital para-quadratic algebras.

Before we can proceed, we require a rather obvious but still useful observation.ss.UFRBACH

31.2. Unimodular free base change. Let R ∈ k-alg and assume that 1R can beextended to basis (ti)i∈I of R as a k-module. Since tensor products commute withdirect sums, the natural map M→MR, for any k-module M, is an embedding, andidentifying M ⊆MR canonically, we conclude

MR =⊕i∈I

(tiM) (1) EMART

as a direct sum of k-modules. Moreover, any linear map f : M → N between k-modules M,N may be recovered from its R-linear extension fR : MR → NR viarestriction, and it follows that the assignment f 7→ fR defines an injection fromHomk(M,N) to HomR(MR,NR). We identify Homk(M,N)⊆HomR(MR,NR) accord-


ingly and claim that the R-linear map⊕i∈I

(ti Homk(M,N)

)= Homk(M,N)R −→ HomR(MR,NR)

extending the inclusion Homk(M,N) → HomR(MR,NR) is injective. Indeed, givena family ( fi) of elements in Homk(M,N) with finite support, the map in question,thanks to the preceding identifications, sends ∑ ti fi = ∑( fi⊗ ti) to ∑ ti fiR = ∑ ti fi,and if this is zero, then so is ∑ ti fi(x) for all x ∈M, which by (1) implies fi = 0 forall i and proves the assertion. Summing up we have shown that after the appropriateidentifications, the following inclusions hold:

Homk(M,N)⊆ Homk(M,N)R =⊕i∈I

(ti Homk(M,N)

)⊆ HomR(MR,NR). (2) HOMEM

Moreover, the elements of Homk(M,N)R act on MR according to the rule

(ti f )(t jx) = tit j f (x) ( f ∈ Homk(M,N), i, j ∈ I, x ∈M) (3) TEIEF

since (ti f )(t jx) = (ti fR)(x ⊗ t j) = ti fR(x ⊗ t j) = ti( f (x)⊗ t j) = f (x)⊗ (tit j) =tit j f (x).

ss.JABAS31.3. Basic identities. Let J be a Jordan algebra over k. The following identitieshold strictly in J, i.e., for all x,y,z,w in every scalar extension.


U1J = 1J , (1) UONE

UUxy =UxUyUx, (2) UUXY

UUxy,Uxz =UxUy,zUx, (3) UUXYZU

UxVy,x =Vx,yUx =Ux,Uxy, (4) UXVY

Vx =V1J ,x =U1J ,x, (5) VXUX

UUxy,xyz =UxUyUx,z +Ux,zUyUx, (6) UUXYZ

UUxy,Uzy +Uxyz =UxUyUz +UzUyUx +Ux,zUyUx,z (7) UUXYUZ

UUx,wy,Ux,zy +UUxy,Uz,wy =Ux,wUyUx,z +UxUyUz,w +Uz,wUyUx +Ux,zUyUx,w, (8) UUXW

Ux,zVy,x +UxVy,z =Vz,yUx +Vx,yUx,z =Uz,Uxy +Ux,Ux,zy, (9) UXZVY

2Ux =V 2x −Vx2 , (10) TWOUX

UxVx =VxUx =Ux,x2 , (11) UXVX

VxVy +Vy,x =VyVx +Vx,y, (12) VXVY

Ux,yx = xxy= x2 y, (13) UXYX

Ux,y =VxVy−Vx,y, (14) UXYV

Uxy +UUxy,y =UxUy +UyUx +VxUyVx, (15) UXCY

UxVy +Ux,yVx =VxUx,y +VyUx, (16) UXVYU

(Uxy) y = x (Uyx), (17) UXYC

VxVy,x +UxVy =VyUx +Vx,yVx, (18) VXVYX

VUxy =VxVy,x−VyUx =Vx,yVx−UxVy, (19) VUXYVX

VUxy,y =Vx,Uyx, (20) VUXYS

VUxy,z +VUxz,y =Vx,yxz, (21) VUXYZ

UUxy,z =Ux,zVy,x−Vz,yUx =Vx,yUx,z−UxVy,z, (22) UUXYZV

Vx,yxz+UxUy,z =Vx,yVx,z +VUxz,y =Vx,zVx,y +VUxy,z,, (23) VXYXZ

Vxyz,y =Vx,Uyz +Vz,Uyx, (24) VXYZY

Vx,yVz,y =Vx,Uyz +Ux,zUy, (25) VXYVZY

Vx,yUz +UzVy,x =Uz,xyz, (26) VXYUZ

Vx,yVx,z =VUxy,z +UxUy,z, (27) VXYVXZ

[Vx,y,Vu,v] =Vxyu,v−Vu,yxv, (28) VXYVUV

VUxy,zUx =UxVy,Uxz, (29) VUXYZU

VUxy,zVx,y =UxUyVx,z +Vx,UyUxz, (30) VUXYZV

Uxyz+UUxUyz,z =UxUyUz +UzUyUx +Vx,yUzVy,x, (31) UXYZ

UUxUyz,xyz =UxUyUzVy,x +Vx,yUzUyUx. (32) UUXUY

Proof. Since Jordan algebras are stable under base change, it suffices to verify theseidentities for all x,y,z,u,v.w ∈ J. Here (1), (5) hold since J is unital para-quadratic,


while (2) and the first equation in (4) are valid by the definition of a Jordan algebra.Applying this part of (4) to z, viewing the result as a linear map acting on y, usingthe fact that the Jordan triple product is symmetric in the outer variables and writingagain y for z, the second equation of (4) drops out. On the other hand, (3) followsimmediately by linearizing (2) with respect to y. Passing to the polynomial ring k[t],which is free as a k-module with basis (ti)i∈N, we may apply 31.2 to obtain naturalidentifications

J ⊆ Jk[t] =⊕i∈N

(tiJ), Endk(J)⊆ Endk(J)k[t] =⊕i∈N

(ti Endk(J)

)⊆ EndR(JR)

as direct sums of k-modules. Replacing x by x+ tz, x,z ∈ J, in (2) and comparingcoefficients of t, t2, respectively, we conclude that (6), (7) hold. Similarly, replacingx by x+ tw, x,w ∈ J, in (6) (resp. by x+ tz, x,z ∈ J, in (4)) yields (8) (resp. (9)).Specializing y and z to 1J in (7), we obtain Vx2 + 4Ux = 2Ux +V 2

x , hence (10). Onthe other hand, since Jordan algebras are unital para-quadratic, setting y = 1J in (4)yields (11). Next we put x = 1J in (9) and observe (5). Writing x for z, we end upwith (12), which, when applied to z, amounts to

x (y z)+yxz= y (x z)+xyz.

Putting z = x and applying (10) we conclude

x (x y)+Ux,yx = 2x2 y+2Uxy = 2x2 y+ x (x y)− x2 y.

Thus (13) holds, which linearizes to xzy+zxy= (x z)y and yields (14) afterinterchanging x and y. Next we put z = 1J in (7) to obtain (15), while linearizing(11) yields (16). Applying this to y and using (13), we conclude

2Uxy2 +x(x y)y= xxyy+ y (Uxy) = x (x y2)+ y (Uxy)

=V 2x y2 + y (Uxy).

Now (10) yields with

y (Uxy) = x(x y)y+(2Ux−V 2x )y

2 = x(x y)y−Vx2y

= x(x y)y− x2 y2

an equation whose right-hand side is symmetric in x and y. Hence so is the left, andwe have proved (17). Linearizing gives

y (Uxz)+ z (Uxy) = x (Uy,zx) = x (Vy,xz),

and viewing this as a linear map in z, we arrive at

VUxy =VxVy,x−VyUx. (33) VUXYV


Setting z = 1J in (9), we obtain (18), which combines with (33) to imply (19). Using(14), (19) and (15), we can now compute

VUxy,y−Vx,Uyx =VUxyVy−UUxy,y−VxVUyx +Ux,Uyx

= (VxVy,x−VyUx)Vy− (UxUy +UyUx +VxUyVx−Uxy)

−Vx(Vy,xVy−UyVx)+(UyUx +UxUy +VyUxVy−Uyx)

= 0.

This proves (20), which linearizes to (21). Applying this to w ∈ J, we obtain

(Uxy)zw+(Uxz)yw= xyxzw,

which in turn may be viewed as a linear map in z and, writing z for w, yields the firstequation of (22), while the second now follows from (9). We now apply (9) to u andobtain

xyxuz+Uxyzu= zy(Uxu)+xyxuz= zu(Uxy)+xuxyz.

Viewing this as a linear map in z and replacing u by z, we arrive at (23). Next weapply (20) to z and linearize the resulting equation, (Uxy)yz = x(Uyx)z, withrespect to x:

xywyz= w(Uyx)z+x(Uyw)z. (34) XYWYZ

Viewing (34) as a linear map in z and replacing w by z gives (24), while viewing itas a linear map in w and interchanging x with z gives (25). Applying first (22) andthen (9), we obtain, after interchanging x with z,

Vx,yUz +UzVy,x =Ux,zVy,z−UUzy,x +UzVy,x =Uz,xyz,

hence (26). Combining (23) with (21) yields

Vx,yVx,z +VUxz,y =Vx,yxz+UxUy,z =VUxy,z +VUxz,y +UxUy,z,

hence (27). Linearizing (26) at z in the direction u and applying the result to v, weobtain

xyuvz+uyxvz= uvxyz+xyuvz,

which, when viewed as a linear map in z, amounts to (28). Letting the left-hand sideof (29) act on w and observing (3), we conclude

VUxy,zUxw =UUxy,Uxwz =UxUy,wUxz =UxVy,Uxzw,

and (29) is proved. Applying the left-hand side of (29) to w and observing (6) implies

VUxy,zVx,yw =UUxy,xywz =UxUyUx,wz+Ux,wUyUxz = (UxUyVx,z +Vx,UyUxz)w,


hence (30). Consulting (22), (27) and (22) again, we now compute

Vx,yUzVy,x =Ux,zVy,zVy,x−UUzy,xVy,x =Ux,zVUyz,x +Ux,zUyUx,z−Ux,UzyVy,x

=Ux,zUyUx,z +UUxUyz,z +Vz,UyzUx−UUxy.Uzy−VUzy,yUx,

which by (20) implies

Vx,yUzVy,x =Ux,zUyUx,z +UUxUyz,z−UUxy,Uzy,

hence combines with (7) to yield

Uxyz+UUxUyz,z =Uxyz+Vx,yUzVy,x +UUxy,Uzy−Ux,zUyUx,z

=UxUyUz +UzUyUx +Vx,yUzVy,x,

and this is (31). Finally, in order to derive (32), we linearize (6) with respect to yand obtain

UUxu,xyz+UUxy,xuz =UxUy,uUx,z +Ux,zUy,uUx.

Here we replace u by Uyz. Then a computation involving (4), (20), (22) and (6) (withUzy in place of z) yields

UUxUyz,xyz =UxUy,UyzUx,z +Ux,zUy,UyzUx−UUxy,x(Uyz)z

=UxUyVz,yUx,z +Ux,zVy,zUyUx−UUxy,xy(Uzy)

=UxUy(UUzy,x +UzVy,x)+(UUzy,x +Vx,yUz)UyUx−UUxy,xy(Uzy)

=UxUyUzVy,x +Vx,yUzUyUx +UxUyUUzy,x +Ux,UzyUyUz−UUxy,xy(Uzy)

=UxUyUzVy,x +Vx,yUzUyUx,

which completes the proof of (32).ss.COLIJO

31.4. The connection with unital linear Jordan algebras. Assume that 2 is in-vertible in k. Let J be a unital linear Jordan algebra over k. Then its identity elementand the U-operator (29.9.1) convert J into a para-quadratic algebra Jquad, whichby the advanced identities (29.11.1), (29.11.6), (29.11.8) combined with the usualscalar extension argument (cf. Prop. 29.4) is a Jordan algebra.

Conversely, let J be a Jordan algebra over k Then the bilinear multiplication

xy :=12

Ux,y1J =12x1Jy= 1

2Vxy

gives J the structure of a non-associative k-algebra Jlin with left multiplicationLx =

12Vx for x ∈ J. By definition and (31.3.1), (31.3.5), Jlin is commutative with

identity element 1J , and its squaring agrees with the one of J. By (31.3.11), thelinear operators Vx, Ux commute, hence by (31.3.10) so do Vx and Vx2 . This shows[Lx,Lx2 ] = 0, and from (29.3.1) we conclude that Jlin is a unital linear Jordan alge-bra. One checks easily that the two constructions J 7→ Jquad and J 7→ Jlin are inverse


to each other, and that a linear map between unital linear Jordan algebras is a homo-morphism of unital linear Jordan algebras if and only if it is one of Jordan algebras.Summing up, we have thus proved the following result.

t.EQLIQUA31.5. Theorem. Assume that 2 is invertible in k. Then the constructions presentedin 31.4 yield inverse isomorphisms between the categories of unital linear Jordanalgebras over k and Jordan algebras over k.

c.COLIQUA31.6. Convention. Assume that 2 is invertible in k. If there is no danger of confu-sion, we will always use Thm. 31.5 to identify Jordan algebras over k in the senseof 31.1 with unital linear Jordan algebras over k in the sense of 29.1; accordingly,the terms “unital linear Jordan algebra over k” and “Jordan algebra over k” willbe used interchangeably. Note that under the preceding identification, the advancedidentities 29.11 valid in linear Jordan algebras translate equivalently to appropriateidentities in the long list of formulas derived in 31.3.

The reader may wonder why, in defining Jordan algebras over arbitrary baserings, we have insisted on an identity element. The reason is that, without this re-striction, the algebraic structure which ensues becomes computationally much moredifficult to deal with. We refer to McCrimmon [80] for details.

e.ASS31.7. Examples: associative algebras. Let A be a unital associative k-algebra. Weclaim that the para-quadratic algebra A(+) of 30.4 with base point 1A(+) = 1A andU-operator

Uxy = xyx (x,y ∈ A) (1) UOPAS

is in fact a Jordan algebra. Indeed, since associativity is preserved by scalar exten-sions, it suffices to verify (31.1.1) and (31.1.2) over the base ring, so let x,y,z ∈ A.Then

UUxyz = (xyx)z(xyx) = x(y(xzx)y

)x =UxUyUxz,

UxVy,xz = x(yxz+ zxy)x = xy(xzx)+(xzx)yx =Vx,yUxz,

as desired. Recall from (30.4.2) that the Jordan triple and circle products of A(+) arerespectively given by

xyz= xyz+ yzx, x y = xy+ yx (2) TRICAS

for all x,y ∈ A. Recall further that, if 2 ∈ k is invertible, the identifications of 31.6show A(+) = A+ as unital linear Jordan algebras over k.

e.ASIN31.8. Examples: associative algebras with involution. Let (B,τ) be an associa-tive algebra with involution over k. By definition, B is unital and

H(B,τ) = x ∈ B | τ(x) = x


obviously is a subalgebra of B(+) and hence is a Jordan algebra (Exc. 154).In particular, if (B,τ) = (Aop⊕A,εA) as in 11.4, with εA the exchange involution,

then the diagonal embedding A→ A⊕A determines an isomorphism

A(+) ∼−→ H(Aop⊕A,εA), x 7−→ x⊕ x, (1) SYMEX

of Jordan algebras.ss.SPEQUA

31.9. Special and exceptional Jordan algebras. A Jordan algebra over k is saidto be special if it is isomorphic to a subalgebra of A(+), for some unital associativek-algebra A. Otherwise, it is said to be exceptional. If 2 ∈ k is invertible, one checkseasily that these notions are equivalent to the ones defined in 29.2 (a).

e.ALT31.10. Examples: alternative algebras. Let A be a unital alternative k-algebra.Then the left Moufang identity (14.3.1) implies that the left multiplication of A,

L : A(+) −→ Endk(A)(+), x 7−→ Lx,

is a homomorphism of para-quadratic algebras and obviously injective. Hence A(+)

is a special Jordan algebra. Recall that we have encountered the U-operator of A(+)

already in 14.5, where it was referred to as the U-operator of A.Now let (B,τ, p) be an alternative algebra with isotopy involution of type ε =±

over k. Then Lemma 17.1 shows that

Sym(B,τ, p) = x ∈ B | τ(x) = x

is a subalgebra of B(+), hence a special Jordan algebra.e.POQUAM

31.11. Examples: pointed quadratic modules. Let (M,q,e) be a pointed quadraticmodule over k and write x 7→ x for its conjugation. Then we claim that that the basepoint e and the quadratic map x 7→Ux from M to Endk(M) given by

Uxy := q(x, y)x−q(x)y (x,y ∈M) (1) UOPQUA

give M the structure of a Jordan algebra over k, denoted by J := J(M,q,e) andcalled the Jordan algebra of (M,q,e). Indeed, writing t for the trace of (M,q,e), weobtain Uey = t(y)e− y = t(y)e− (t(y)e−y) = y for all y ∈M, hence Ue = 1M . ThusJ is a para-quadratic k-algebra. In order to show that this para-quadratic algebra is,in fact, a Jordan algebra, we first note that the construction of J out of (M,q,e) iscompatible with base change, so it suffices to verify (31.1.1) and (31.1.2) over k.First of all, (1) implies

Vx,yz = xyz= q(x, y)z+q(z, y)x−q(x,z)y (2) TRIQUA

and a straightforward verification shows


q(Uxy) = q(x)2q(y), (3) QUUXY

q(Uxy,z) = q(x, y)q(x,z)−q(x)q(y, z), (4) QUUXYZ

Uxx = q(x)x, (5) UXBAR

Vx,yx = 2(q(x, y)x−q(x)y

). (6) VXYX

Combining these identities with the fact that the conjugation leaves q invariant, wecan now compute

UxUyUxz =UxUy(q(x, z)x−q(x)z

)=Ux

(q(x, z)q(y, x)y−q(x, z)q(y)x−q(x)q(y,z)y+q(x)q(y)z

)= q(x, y)q(x, z)q(x, y)x−q(x)q(x, y)q(x, z)y−q(x)q(y)q(x, z)x

−q(x)q(y,z)q(x, y)x+q(x)q(y,z)q(x)y

+q(x)q(y)q(x, z)x−q(x)q(y)q(x)z

= q(x, y)(q(x, y)q(x, z)−q(x)q(y,z)

)x

−q(x)(q(x, y)q(x, z)−q(x)q(y,z)

)y−q(x)2q(y)z

= q(Uxy, z)(q(x, y)x−q(x)y

)−q(Uxy)z

= q(Uxy, z)Uxy−q(Uxy)z =UUxyz.

Similarly,

UxVy,xz =Ux(q(y, x)z+q(z, x)y−q(y,z)x

)= q(x, y)q(x, z)x−q(x, y)q(x)z

+q(x, z)q(x, y)x−q(x, z)q(x)y−q(x)q(y,z)x

=(2q(x, y)q(x, z)−q(x)q(y,z)

)x−q(x)

(q(x, y)z+q(x, z)y

)= 2q(x, y)q(x, z)x−2q(x)q(x, z)y

−q(x)q(x, y)z−q(x)q(z, y)x+q(x)q(x, z)y

=Vx,y(q(x, z)x−q(x)z

)=Vx,yUxz.

Thus (31.1.1), (31.1.2) hold in J and the proof is complete.ss.JOPOID

31.12. The Jordan algebra of a pointed quadratic module: identities. Letting(M,q,e) be a pointed quadratic module over k, we write t for its trace, x 7→ x for itsconjugation and J := J(M,q,e) for the corresponding Jordan algebra. The followingidentities, valid for all x,y,z ∈ J and all n ∈N, are either easily verified or have beenchecked before, and are collected here for convenience.


t(x) = q(1J ,x), (1) JOTN

q(1J) = 1, t(1J) = 2, (2) JOA1

x = t(x)1J− x, 1J = 1J , x = x, (3) JOCONJ

Uxy = q(x, y)x−q(x)y, (4) JOQUADU

x2 = t(x)x−q(x)1J , (5) JOQUAD

x3 = t(x)x2−q(x)x, (6) JOQUAT

Vx,yz = xyz= q(x, y)z+q(y,z)x−q(z,x)y, (7) JOQUADV

x y = t(x)y+ t(y)x−q(x,y)1J , (8) JOQUADL

Uxx = q(x)x, Uxx2 = q(x)21J , x+ x = t(x)1J , (9) JOCONJC

q(x) = q(x), t(x) = t(x), (10) JOCONJO

q(Uxy) = q(x)2q(y), q(xn) = q(x)n, (11) JOQUU

t(x2) = t(x)2−2q(x), (12) JOTQUAD

t(x y) = 2(t(x)t(y)−q(x,y)

), (13) JOBILT

q(x, y) = t(x)t(y)−q(x,y), (14) JONOCO

Uxy =Uxy. (15) JOCONJM

Note the analogy of these formulas with the ones of 19.5.The Jordan algebras of pointed quadratic modules are often referred to collec-

tively as Jordan algebras of Clifford type.ss.QUAZEJO

31.13. Quadratic Z-structures as Jordan algebras. Let D be one of the subal-gebras R,C,H,O of the Graves-Cayley octonions. We have seen in Thm. 5.10 thatJ := Her3(D) is a unital linear Jordan algebra over R and hence may be viewedcanonically as a real Jordan algebra by means of the identification 31.6. Now sup-pose Λ ⊆ J is a quadratic Z-structure of J. By its very definition (cf. 6.5), Λ is aZ-subalgebra of J as a (quadratic) Jordan algebra. Hence Λ is a quadratic Jordanalgebra over Z in its own right but, in general, not a linear one. For example, if Mis a Z-structure of D in the sense of 3.6 (d), then Λ := Her3(M) by 6.3 is not closedunder the bilinear Jordan product of J.

ss.TOPOAS31.14. Towards power-associativity. We wish to show in analogy to Cor. 29.6 thatJordan algebras are power-associative. Combining the fundamental formula (31.3.2)and its linearization (31.3.3) with the recursive definition of powers in (30.8.1), wesee by induction that any Jordan algebra J over k satisfies the identities

Uxn =Unx , UxUxm,xnUx =Uxm+2,xn+2 (1) UXN

for all x ∈ J and all m,n ∈ N.p.JOAPOA

31.15. Proposition. Jordan algebras are power-associative: for all Jordan alge-bras J over k, all x ∈ J and all m,n, p ∈ N we have


Uxmxn = x2m+n, (1) UXMN

xmxnxp= 2xm+n+p. (2) TRIXMN

In particular, k[x]⊆ J is a Jordan subalgebra.

Proof. Equation (1) follows immediately from (31.14.1) by induction on m. Equa-tion (2) will now be proved by induction on l := m+ n+ p. We first note that (2)follows from (1) for m = p and is symmetric in m, p. Hence we may always assumeif necessary that m < p. Moreover, (2) is obvious if two of the exponents m,n, pare zero. Summing up, not only the induction beginning l = 0 is trivial, but alsothe cases l = 1,2 are. Let us now assume l ≥ 3 and that (2) holds for all expoentsm′,n′,n′ ∈ N having m′+n′+ p′ < l. We consider the following cases.Case 1. m = 0. Then (2) amounts to xn xp = 2xn+p, hence is symmetric in n, p andobvious for n = 0 or n = p. Thus we may assume 1 ≤ n < p. But the assertion canalso be written in the form xn1Jxp= 2xn+p, so we have reduced Case 1 toCase 2. m≥ 1.Case 2.1. m = 1. Then p ≥ 2, and (30.8.1), (31.3.4) combine with the inductionhypothesis to imply

xxnxp=Vx,xnUxxp−2 =UxVxn,xxp−2 =Uxxnxxp−2= 2Uxxn+p−1 = 2x1+n+p,

as claimed.Case 2.2. m ≥ 2. Then p ≥ m ≥ 2 and the induction hypothesis combined with(31.14.1) yields

xmxnxp=UUxxm−2,Uxxp−2xn =UxUxm−2,xp−2Uxxn =Uxxm−2xn+2xp−2

= 2Uxxm+n+p−2 = 2xm+n+p,

which completes the induction.

Exercises.

pr.CHRAJO 154. Show for a para-quadratic algebra J over k that the following conditions are equivalent.

(i) J is a Jordan algebra.(ii) The identities

UUxy =UxUyUx, (3) RUUXY

UxVy,x =Vx,yUx, (4) RUXVY

UUxy,Ux,zy =UxUyUx,z +Ux,zUyUx, (5) RUUXYZ

UUxy,Uzy +UUx,zy =UxUyUz +UzUyUx +Ux,zUyUx,z, (6) RUUXYUZ

UUx,wy,Ux,zy +UUxy,Uz,wy =Ux,wUyUx,z +UxUyUz,w +Uz,wUyUx +Ux,zUyUx,w, (7) RUUXW

Ux,zVy,x +UxVy,z =Vz,yUx +Vx,yUx,z (8) RUXZVY

hold in J.(iii) The identities (3), (4) hold in Jk[t].


Conclude that subalgebras and homomorphic images of Jordan algebras (in the category k-paquad)are Jordan algebras.

pr.OUTIJO 155. Let J be a Jordan algebra over k. Show that

(a) a k-submodule I ⊆ J is an outer ideal (resp. an ideal) if and only if UJI ⊆ I (resp. UIJ+UJI ⊆I),

(b) an element c ∈ J is an idempotent if and only if c2 = c,(c) the extreme radical of J is an ideal. Conclude that the centroid of a simple Jordan algebra is

a field.

pr.CENCENT 156. Assume 12 ∈ k and let J be a unital linear Jordan algebra over k. Show that the centroid of

Jquad is a unital commutative associative subalgebra of Endk(J) and that the left multiplication ofJ induces an isomorphism Cent(J)∼= Cent(Jquad).

pr.CHAJOCLI 157. Let J be a Jordan algebra over k and q : J → k a quadratic form with q(1J) = 1, so that(M,q,e) is a pointed quadratic module over k, where M stands for the k-module underlying J ande = 1J . Write t for the trace of (M,q,e) and prove that J = J(M,q,e) if and only if the equations

x2− t(x)x+q(x)e = 0 = x3− t(x)x2 +q(x)x (9) QUACUB

hold strictly in J.

pr.NIPOQUA 158. Let (M,q,e) be a pointed quadratic module over k and J = J(M,q,e) the corresponding Jordanalgebra. Write t for the trace of (M,q,e) and show that an element x ∈ J is nilpotent if and only ift(x) and q(x) are nilpotent elements of k. Conclude that

Nil(J) = x ∈M | q(x),q(x,y) ∈ Nil(k) for all y ∈M.

pr.EVPOQUA 159. (a) Suppose 2 = 0 in k and consider the situation of Exc. 149 (b) for n = 2. Decide whetherthere exists a pointed quadratic module (M,q,e) over k such that J(M,q,e)∼= J2 = k[t](+)/I2.(b) Show that, over an appropriate commutative ring, there exists a pointed quadratic module whosecorresponding Jordan algebra contains an element x satisfying x2 = 0 6= x3.

pr.IDPOQUA 160. Let (M,q,e) be a pointed quadratic module over k and J = J(M,q,e). Write t for the traceof (M,q,e) and show that, if k 6= 0 is connected, then c ∈ J is an idempotent 6= 0,1J in J if andonly if t(c) = 1 and q(c) = 0. Conclude that, for any commutative ring k and any element c ∈ J,the following conditions are equivalent.

(i) c is an idempotent satisfying cR 6= 0,1JR for all R ∈ k-alg, R 6= 0.(ii) c is an idempotent satisfying cp 6= 0,1Jp for all prime ideals p⊆ k.(iii) q(c) = 0, t(c) = 1.(iv) c is an idempotent and the elements c,1J− c are unimodular.

If these conditions are fulfilled, we call c an elementary idempotent of J.

pr.DECPOQUA 161. Let (M,q,e) be a pointed quadratic module over k and J := J(M,q,e). Write t for the trace of(M,q,e) and prove for c ∈ J that the following conditions are equivalent.

(i) c is an idempotent in J.(ii) There exists a complete orthogonal system (ε(0),ε(1),ε(2)) of idempotents in k, giving rise

to decompositionsk = k(0)⊕ k(1)⊕ k(2), J = J(0)⊕ J(1)⊕ J(2)

as direct sums of ideals, where k(i) = kε(i) and J(i) = ε(i)J = Jk(i) as the Jordan algebracorresponding to the pointed quadratic form (M,q,e)k(i) for i = 0,1,2, such that

c = 0⊕ c(1)⊕1J(2) ,

where c(1) is an elementary idempotent of J(1).

32 Power identities. 231

In this case, the idempotents ε(i), i = 0,1,2, in (ii) are unique and given by

ε(0) :=

(1−q(c)

)(1− t(c)

), ε

(1) :=(1−q(c)

)t(c), ε

(2) := q(c). (10) VEPI

pr.COBRPL 162. Let C be a flexible conic algebra over k. Show that the para-quadratic algebra C(+) agreeswith the Jordan algebra of the pointed quadratic module (C,nC,1C) if and only if C is alternative.

pr.POJOCAT 163. Let k-poquainj be the category of pointed quadratic modules over k whose underlying k-modules are projective, with injective homomorphisms of pointed quadratic modules as mor-phisms, and similarly, let k-jordinj the category of Jordan algebras over k, with injective homo-morphisms of para-quadratic algebras as morphisms. Prove that the assignment(

ϕ : (M,q,e)−→ (M′,q′,e′))7−→

(ϕ : J(M,q,e)−→ J(M′,q′,e′)

)defines a faithful and full embedding from k-poquainj to k-jordinj.

pr.POQUAIN 164. Let (V,q,e) be a non-degenerate pointed quadratic module of dimension 3 over a field F ofcharacteristic not 2. Show that there exist a quaternion algebra B over F and an involution τ of Bsuch that J(V,q,e) ∼= H(B,τ). Show further that (B,τ) is uniquely determined by this condition.(Hint. Use Exc. 110.)

32. Power identities.

s.POIDBy showing in Cor. 29.6 that linear Jordan algebras over a commutative ring contain-ing 1

2 are power-associative, we derived what may be called a local property: everysubalgebra on a single generator is associative. The proof of this result has been re-duced to yet another property of linear Jordan algebras that, though no longer local,could at least be called semi-local: given any element x in a linear Jordan algebra J,the left multiplication operators of arbitrary elements in k1[x]⊆ J acting on all of Jcommute by pairs.

The local property of a linear Jordan algebra to be power-associative has beenextended to the setting of Jordan algebras in Prop. 31.15. It is the purpose of thepresent section to accomplish the same objective for the semi-local analogue ofthis property alluded to above. More specifically, fixing a Jordan algebra J over anarbitrary commutative ring k throughout this section, the main result we wish toestablish reads as follows.

t.UPOL

32.1. Theorem. Let x ∈ J. Then

U( f g)(x) =U f (x)Ug(x)

for all f ,g ∈ k[t].r.UPOL

32.2. Remark. The proof of this result given in Jacobson [45, Cor. 3.3.3] rests on ageneral principle [45, 3.3.1] that may be regarded as a weak version of Macdonald’stheorem [45, Thm. 3.4.15]. The proof we are going to provide below will workinstead with explicit and elementary manipulations of some of the basic identitiesin 31.3 valid in arbitrary Jordan algebras.


ss.FIRE32.3. A first reduction. In order to derive Thm. 32.1, we write the polynomialsf ,g ∈ k[t] in the form

f = ∑i∈N

αiti, g = ∑m∈N

βmtm,

where the families (αi)i∈N,(βm)m∈N ∈ kN both have finite support. With N2 =N×N,this implies

f g = ∑(i,m)∈N2

αiβmti+m,

hence

( f g)(x) = ∑(i,m)∈N2

αiβmxi+m,

and endowing N2 with the lexicographic ordering, we apply (31.14.1) to obtain theexpansion

U( f g)(x) = ∑(i,m)∈N2

α2i β

2mU i+m

x + ∑(i,m),( j,n)∈N2,(i,m)<( j,n)

αiα jβmβnUxi+m,x j+n

= ∑i,m∈N

α2i β

2mU i+m

x + ∑i, j,m,n∈N,i< j

αiα jβmβnUxi+m,x j+n

+ ∑i,m,n∈N,m<n

α2i βmβnUxi+m,xi+n .

Thus we have

U( f g)(x) = ∑i,m∈N

α2i β

2mU i+m

x + ∑i,m,n∈N,m<n

α2i βmβnUxi+m,xi+n (1) UPRO

+ ∑i, j,m∈N,i< j

αiα jβ2mUxi+m,x j+m

+ ∑i, j,m,n∈N,i< j,m<n

αiα jβmβn(Uxi+m,x j+n +Uxi+n,x j+m).

On the other hand, from

f (x) = ∑i∈N

αixi, g(x) = ∑m∈N

βmxm

we deduce


U f (x)Ug(x) =(∑i∈N

α2i U i

x + ∑i, j∈N,i< j

αiα jUxi,x j)(

∑m∈N

β2mUm

x + ∑m,n∈N,m<n

βmβnUxm,xn)

= ∑i,m∈N

α2i β

2mU i+m

x + ∑i,m,n∈N,m<n

α2i βmβnU i

xUxm,xn

+ ∑i, j,m∈N,i< j

αiα jβ2mUxi,x jUm

x + ∑i, j,m,n,i< j,m<n

αiα jβmβnUxi,x jUxm,xn .

Comparing this with (1), we see that Thm. 32.1 will be a consequence of the identi-ties (32.4.5), (32.4.6) below.

p.LIPO32.4. Proposition. For all x ∈ J and all i, j,m,n ∈N, the following identities hold.

Vxn+2 =V 2x Vxn −UxVxn −VxUx,xn , (1) VEN

Ux,xn+2 =Ux(VxVxn −Ux,xn), (2) ULEN

Vxn+1 =VxVxn −Ux,xn , (3) VEXEN

Vxm,xn =Vxm+n , (4) VXMXN

UxiUxm,xn =Uxi+m,xi+n =Uxm,xnUxi , (5) PROU

Uxi,x jUxm,xn =Uxi+m,x j+n +Uxi+n,x j+m . (6) PROUL

The proof of this result will be preceded by the following lemma.l.UXCOM

32.5. Lemma. For x ∈ J, the linear operators Uxi , Uxm,xn (i,m,n ∈ N) commute bypairs, and we have

Vxn+2 =VxVxnVx−VxUx,xn −VxnUx, (1) VEXVEN

Ux,xn+2 =Ux(VxnVx−Ux,xn) (2) UXVEN

for all n ∈ N.

Proof. We first apply (31.3.19) to obtain Vxn+2 = VUxxn = VxVxn,x −VxnUx and(31.3.14) yields (1). Similarly, since Ux,xn+2 =Ux,Uxxn =UxVxn,x by (31.3.4), anotherapplication of (31.3.14) gives (2). It remains to prove

[Uxi ,Uxm,xn ] = 0, (3) UXIJ

[Uxi,x j ,Uxm,xn ] = 0 (4) UXIM

for all i, j,m,n ∈ N.We begin with (3), note by symmetry and (31.14.1) that we may assume i = 1,

m < n and argue by induction on l := m+n. For l = 0, there is nothing to prove. Forl = 1, we have m = 0, n = 1, and the assertion comes down to [Ux,Vx] = 0, whichholds by (31.3.11). Now suppose l ≥ 2 and assume (3) holds for all natural numbersm′,n′ in place of m,n having m′+n′ < l. Then we consider the following cases.10. m= 0, n= l. Then the assertion comes down to [Ux,Vxl ] = 0, which follows from(1) for n = l−2 and the induction hypothesis since, in particular, [Ux,Ux,xl−2 ] = 0.


20. 0 < m < n. Then l ≥ 3. Here m = 1 implies n = l− 1, and the assertion comesdown to [Ux,Ux,xl−1 ] = 0, which follows from (2) for n = l− 3 since [Ux,Vxl−3 ] =0 = [Ux,Ux,xl−3 ] by the induction hypothesis. On the other hand, if n > m ≥ 2, then(31.3.3) gives Uxm,xn = UUxxm−2,Uxxn−2 = UxUxm−2,xn−2Ux, and the assertion followsfrom the induction hypothesis. This completes the proof (3).

Turning to (4), we note that the assertion is symmetric in (i, j) and (m,n) and that,by (3), we may assume i< j, m< n. Now we argue by induction on l := i+ j+m+n.The cases l = 0,1 are obvious by what we have just noted, while for l = 2 we mayassume i = m = 0, j = n = 1, in which case the assertion reduces to the triviality[Vx,Vx] = 0. Now let l ≥ 3 and assume the assertion holds for all natural numbersi′, j′,m′,n′ in place of i, j,m,n having i′+ j′+m′+ n′ < l. Then we consider thefollowing cases.10. i = m = 0. Then we have to show [Vx j ,Vxn ] = 0. For j = 1, hence n = l− j ≥ 2,the assertion follows from (1) for n−2 in place of n since the induction hypothesisimplies [Vx,Ux,xn−2 ] = 0. On the other hand, if j≥ 2, the assertion follows again from(1), this time for n = j−2, and the induction hypothesis since [Ux,x j−2 ,Vxn ] = 0.

20. i = m = 1. This implies j ≥ 2, and we have to show [Ux,x j ,Ux,xn ] = 0, whichfollows from (2) for n = j−2 since the induction hypothesis implies [Vx j−2 ,Ux,xn ] =0 = [Ux,x j−2 ,Ux,xn ]. Now, by symmetry, the only remaining case is

30. j > i≥ 2. Then equation (31.3.3) yields Uxi,x j =UUxxi−2,Uxx j−2 =UxUxi−2,x j−2Ux,which commutes with Uxm,xn by (3) and the induction hypothesis.

pr.LIPO32.6. Proof of Proposition 32.4. Combining the first part of Lemma 32.5 with(32.5.1), we obtain (32.4.1), while (32.4.2) agrees with (32.5.2).

Next we prove (32.4.5) for i = 1, m = 0, i.e.,

UxVxn =Ux,xn+1 , (1) UXVEX

and (32.4.3) simultaneously by induction on n. Both equations are obvious for n= 0.Now suppose n > 0 and assume (32.4.3) and (1) both hold for all natural numbers< n. Then (32.4.2) for n−1 in place of n and the induction hypothesis for (32.4.3)imply

Ux,xn+1 =Ux(VxVxn−1 −Ux,xn−1) =UxVxn ,

hence (1). Combining both parts of the induction hypothesis with (32.4.1) for n−1in place of n, we now obtain

VxVxn −Ux,xn =V 2x Vxn−1 −VxUx,xn−1 −UxVxn−1 =Vxn+1 ,

hence (32.4.3). This completes the proof of (32.4.3) and (1).We are now in a position to prove (32.4.5), where (31.14.1) allows us to assume

i = 1, and first deduce from Lemma 32.5 that it suffices to establish the first oneof the two equations. By symmetry and (31.14,1) we may assume m < n and thenargue by induction on m. The case m = 0 has been settled in (1). For m > 0, weapply (31.3.3) and the induction hypothesis to conclude


Uxm+1,xn+1 =UUxxm−1,Uxxn−1 =UxUxm−1,xn−1Ux =Uxm,xnUx =UxUxm,xn ,

as claimed.Turning to (32.4.4), we first deduce from Lemma 32.5 that the linear operators

Vxm , Vxn commute. Hence (31.3.12) implies Vxm,xn =Vxn,xm , so our assertion is sym-metric in m and n. We now argue by induction on m. For m = 0 there is nothing toprove. For m = 1, we apply (31.3.14) and (32.4.3) to obtain Vx,xn =VxVxn −Ux,xn =Vxn+1 , hence the assertion. Now suppose m ≥ 2 and assume the assertion holds forall m′ ∈ N in place of m satisfying m′ < m. Combining (31.3.21) with Prop. 31.15,the case m = 1 and the induction hypothesis, we deduce

Vxm,xn =VUxxm−2,xn =Vx,xm−2xxn−VUxxn,xm−2 = 2Vx,xm+n−1 −Vxn+2,xm−2

= 2Vxm+n −Vxm−2,xn+2 = 2Vxm+n −Vxm+n =Vxm+n ,

and the proof of (32.4.4) is complete.Finally, turning to (32.4.6), we may assume i < j and m < n by symmetry,

Lemma 32.5 and (32.4.5).Then we argue by induction on l := i + m. For l = 0,i.e., i = m = 0, we have to prove Vx jVxn =Vx j+n +Ux j ,xn . But Vx jVxn =Ux j ,xn +Vx j ,xn

by (31.3.14), and the assertion follows from (32.4.4). Next suppose l = i+m > 0and assume (32.4.6) for all natural numbers i′, j′,m′,n′ in place of i, j,m,n satisfyingi′+m′ < l. Since Uxi,x j and Uxm,xn commute by Lemma 32.5, we may assume i > 0.Then (32.4.5) and the induction hypothesis yield

Uxi,x jUxm,xn =UxUxi−1,x j−1Uxm,xn =UxUxi+m−1,x j+n−1 +UxUxi+n−1,x j+m−1

=Uxi+m,x j+n +Uxi+n,x j+m ,

which completes the induction and the proof of Prop. 32.4.

With the proof of Prop. 32.4, we have also established Thm. 32.1. The remainderof this section will be devoted to a useful application. We begin with an auxiliaryresult, where we use the notation already employed in Exc. 149.

p.LIKEX32.7. Proposition. For x ∈ J and R ∈ k-alg such that R = k[x] as k-modules, thefollowing conditions are equivalent.

(i) R(+) = k[x] as Jordan algebras.(ii) The powers of x in R and in J coincide.(iii) f (x) = 0 implies (t f )(x) = 0, for all f ∈ k[t].(iv) Ix := f ∈ k[t] | f (x) = 0 ⊆ k[t] is an ideal.(v) Ix = I0

x := f ∈ k[t] | f (x) = (t f )(x) = 0.

If these conditions are fulfilled, R is unique. In fact, the evaluation map εx : k[t]→ Jof Exc. 149 (a) induces canonically an isomorphism

εx : k[t]/Ix∼−→ R, f + Ix 7−→ f (x), (1) LINEV

of unital commutative associative k-algebras.


Proof. (i)⇒ (ii). By 30.8, the powers of x in R and R(+) coincide.(ii)⇒ (iii). Denote the multiplication of R by juxtaposition. Assume f = ∑αiti ∈

k[t] with αi ∈ k satisfies f (x) = 0. From (ii) we deduce (t f )(x) = ∑αixi+1 =x∑αixi = x f (x) = 0. Thus (iii) holds.

(iii) ⇒ (v). We trivially have I0x ⊆ Ix, while the reverse inclusion follows from

(iii).(v)⇒ (iv). By Exc. 149 (a), Ix = I0

x is an ideal in k[t]. Thus (iv) holds.(iv) ⇒ (i). The evaluation εx : k[t]→ k[x] (Exc. 149) induces a k-linear bijec-

tion εx : k[t]/Ix → k[x]. Let R ∈ k-alg be the unique k-algebra having R = k[x] ask-modules and making εx : k[t]/Ix

∼→ R an isomorphism in k-alg. But then εx is anisomorphism of Jordan algebras from (k[t]/Ix)

(+) = k[t](+)/Ix not only to R(+) butalso to k[x]⊆ J. Hence (i) holds.

Uniqueness of R follows from the fact that it is spanned by the powers of x in Jas a k-module. The rest is clear

Remark. The preceding arguments show that the proposition holds, more generally,for para-quadratic algebras that are power-associative at x.

ss.LOLI32.8. Local linearity. Our Jordan algebra J is said to be linear at x ∈ J if thereexists a unital commutative associative k-algebra R, necessarily unique, such thatR = k[x] as k-modules and the equivalent conditions (i)−(iv) of Prop. 32.7 hold. Byabuse of language, we simply write R = k[x] for this k-algebra and have

( f g)(x) = f (x)g(x) ( f ,g ∈ k[t]). (1) LOCLINX

Finally, we say J is locally linear if it is linear at x, for every x ∈ J.e.LOCIOJ

32.9. Examples. (a) If 12 ∈ k, then every Jordan algebra over k is locally linear.

This follows from Exc. 149 (a) combined with Prop. 32.7.(b) Every special Jordan algebra is locally linear. Indeed, if A is a unital associativek-algebra and J ⊆ A(+) is a subalgebra, then for any x ∈ J the meanings of k[x] in Jand in A are the same.(c) There are pointed quadratic modules (M,q,e) over appropriate base rings suchthat the associated Jordan algebra, J(M,q,e), is not locally linear (Exc. 159 (b)).

ss.ABZEDI32.10. Absolute zero divisors. An element x ∈ J is called an absolute zero divisor(in J) if Ux = 0. By abuse of language we say that J has no absolute zero divisorsif Ux = 0 implies x = 0, for all x ∈ J, in other words, if 0 is the only absolute zerodivisor of J.

32.11. Theorem (McCrimmon [78, Prop. 1]). J is locally linear provided it has not.AZLOCabsolute zero divisors.

Proof. Let x ∈ J and f ∈ Ix (cf. Prop. 32.7 (iii)). Then Thm. 32.1 implies U( f g)(x) =U f (x)Ug(x) = 0 for all g ∈ k[t], forcing ( f g)(x) = 0 by hypothesis, hence f g ∈ Ix.

33 Inverses, isotopes and the structure group. 237

Thus Ix ⊆ k[x] is an ideal, whence J is linear at x, by Prop. 32.7 (iii). Thus J islocally linear.

Exercises.

pr.ABZECLI 165. Absolute zero divisors in Jordan algebras of Clifford type. Let (M,q,e) be a pointed quadraticmodule over k and J := J(M,q,e) the corresponding Jordan algebra of Clifford type.(a) Show that if k is reduced, then x ∈ J is an absolute zero divisor if and only if x ∈ Rad(q), i.e.,q(x) = q(x,y) = 0 for all y ∈ J.(b) Deduce from (a) that the absolute zero divisors of J are contained in the nil radical of J.Conclude that J is locally linear if Nil(J) = 0 but not in general.

33. Inverses, isotopes and the structure group.

s.INISThe present section is devoted to three fundamental concepts that have dominatedthe theory of Jordan algebras since ancient times. To begin with, the notions of in-vertibility and inverses arise naturally out of the analogy connecting the U-operatorof Jordan algebras with the left (or right) multiplication operator of associative alge-bras. Isotopes, on the other hand, have been discussed earlier for alternative algebrasbut unfold their full potential only in the setting of Jordan algebras. And, finally, thestructure group derives its importance not only from the connection with isotopesbut, more significantly, from the one with exceptional algebraic groups that will bediscussed more fully in later portions of the book.

Throughout this section, we let k be a commutative ring and J,J′,J′′ be Jordanalgebras over k.

ss.COIN33.1. The concept of invertibility. The idea of defining invertibility of an elementin a linear Jordan algebra by properties (e.g., as in the alternative or associative case,by the bijectiveness) of its left multiplication operator is not a particularly usefulone, see Exc. 171 below for details. Instead, it turns out to be much more prof-itable to do so by properties of the U-operator, an approach that has the additionaladvantage of making sense also for arbitrary Jordan algebras.

Accordingly, an element x ∈ J is said to be invertible (in J) if there exists anelement y ∈ J, called an inverse of x (in J), such that

Uxy = x, Uxy2 = 1J . (1) ININ

We will see in a moment that an inverse of x in J, if it exists, is unique. Moreprecisely, we can derive the following characterization of invertibility.

p.CHAIN33.2. Proposition. For x ∈ J, the following conditions are equivalent.

(i) x is invertible.(ii) Ux is bijective.(iii) Ux is surjective.


(iv) 1J ∈ Im(Ux).

In this case, x has a unique inverse, written as x−1 and given by

x−1 =U−1x x. (1) FOIN

Proof. (i) ⇒ (ii). Let y ∈ J be an inverse of x. Then (33.1.1) and the fundamentalformula (31.3.2) imply UxUy2Ux = 1J . Hence Ux, having a left and a right inverse inEndk(J), is bijective.

(ii)⇒ (iii)⇒ (iv). Clear.(iv)⇒ (ii). Let z ∈ J satisfy Uxz = 1J . The UxUzUx = 1J , and as before it follows

that Ux is bijective.(ii)⇒ (i). Put y :=U−1

x x ∈ J. Then Uxy = x, UxUyUx =UUxy =Ux, and since Uxis bijective, we conclude Uy =U−1

x . Hence Uxy2 =UxUy1J =UxU−1x 1J = 1J . Thus

(33.1.1) holds, and x is invertible with inverse y. Combining (33.1.1) with condition(ii), we see that the remaining assertions of the proposition also hold,

p.PROIN33.3. Proposition. Let x,y ∈ J.(a) If x is invertible, then so is x−1 and

(x−1)−1 = x, Ux−1 =U−1x , Vx−1 =U−1

x Vx =VxU−1x . (1) FININ

(b) x and y are both invertible if and only if Uxy is invertible. In this case,

(Uxy)−1 =Ux−1y−1. (2) UIN

Proof. (a) From the fundamental formula and (33.1.1) we deduce UxUx−1Ux =UUxx−1 = Ux, which implies Ux−1 = U−1

x since Ux is bijective by Prop. 33.2. Thusx−1 is invertible, and (33.2.1) yields (x−1)−1 = U−1

x−1x−1 = Uxx−1 = x. Finally,(31.3.4) implies UxVx−1,x = Ux,Uxx−1 = 2Ux, hence Vx−1,x = 2 · 1J . Now (31.3.19)yields Vx =VUxx−1 =VxVx−1,x−Vx−1Ux = 2Vx−Vx−1Ux, hence Vx =Vx−1Ux and, sim-ilarly, Vx =UxVx−1 . This completes the proof of (a).

(b) By Prop. 33.2, the element Uxy ∈ J is invertible iff the linear map UUxy =UxUyUx is bijective iff so are Ux and Uy iff x and y are both invertible. In this case,(33.2.1) gives (Uxy)−1 =U−1

UxyUxy =U−1x U−1

y U−1x Uxy =U−1

x y−1, as claimed.

ss.SEIN33.4. The set of invertible elements. We denote by J× the set of invertible ele-ments in J. By Prop. 33.3 (b), it contains the identity element and is closed underthe para-quadratic operation (x,y) 7→Uxy. Note for a subalgebra J′ ⊆ J that, if anelement x ∈ J′ is invertible in J′, then it is so in J and the two inverses are the same.

We say that J is a Jordan division algebra if J 6= 0and all its non-zero elementsare invertible: J× = J \0. We will see in Exc. 171 combined with Exc. 168 belowthat, in the presence of 1

2 , Jordan division algebras are not the same as linear Jordanalgebras that are division algebras in the sense of 9.7.

The following result is an immediate consequence of the definitions.


p.HOIN33.5. Proposition. If ϕ : J → J′ is a homomorphism, then ϕ(J×) ⊆ J′× andϕ(x−1) = ϕ(x)−1 for all x ∈ J×.

e.ALIN33.6. Examples: alternative algebras. Let A be a unital alternative algebra over k.We claim that an element x ∈ A is invertible in the Jordan algebra A(+) if and onlyif it is so in A, in which case the two inverses coincide. This follows immediatelyfrom Propositions 14.6 and 33.2. In particular, A(+) is a Jordan division algebra ifand only if A is an alternative division algebra.

e.ASININ33.7. Examples: associative algebras with involution. Let (B,τ) be an associa-tive k-algebra with involution and J := H(B,τ) the Jordan algebra of τ-symmetricelements in B. Then τ(x−1) = τ(x)−1 for all x ∈ Band×, by Lemma 17.1 for p = 1.Hence x ∈ J is invertible in J if and only if it is so in B, in which case its inversesin J and B coincide. In particular, if B is an associative division algebra, then J is aJordan division algebra.

But the converse of this implication does not hold: Let A be an associative di-vision algebra and ε the exchange involution on B := Aop⊕A. Then H(B,ε) is aJordan division algebra by (31.8.1) and 33.6 but B is not an associative divisionalgebra.

ss.JOISOT33.8. Isotopes. Let p ∈ J be invertible. On the k-module J we define a new para-quadratic algebra over k, depending on p and written as J(p), by the U-operatorU (p) : J → Endk(J), x 7→ U (p)

x := UxUp and the base point 1J(p) := 1(p) := p−1,

which by (33.3.1) does indeed satisfy the relation U (p)1(p) = 1J(p) . For x,y,z ∈ J,

the triple and circle product associated with J(p) are given by xyz(p) = U (p)x,z y =

Ux,zUpy = x(Upy)z and x (p) y = x1(p)y(p) = x(Up p−1)y = xpy, respec-tively. Summing up, writing V (p) for the V -operator of J(p), we obtain the formulas

1J(p) = 1(p) = p−1, (1) UNIP

U (p)x =UxUp, (2) UOPP

U (p)x,y =Ux,yUp, (3) BUOPP

xyz(p) = x(Upy)z, (4) TRIP

x(p) y = xpy, (5) CIRP

V (p)x,y =Vx,Upy, (6) VIP

V (p)x =Vx,p (7) LVIP

for all x,y,z ∈ J. The para-quadratic algebra J(p) is called the p-isotope (or simplyan isotope) of J. Note that passing to isotopes is

(i) unital: J(1J) = J.


(ii) functorial: if ϕ : J → J′ is a homomorphism, then ϕ : J(p) → J′(ϕ(p)) is ahomomorphism of para-quadratic algebras.

(iii) compatible with base change: (J(p))R = (JR)(pR) for all R ∈ k-alg.

For n ∈ N, the n-th power of x ∈ J performed in the p-isotope J(p) will be denotedby x(n,p). For example x(2,p) =U (p)

x 1(p) =UxUp p−1, hence

x(2,p) =Ux p. (8) SQUAP

The first fundamental fact to be derived in the present context is that isotopes ofJordan algebras are Jordan algebras.

t.ISTJO33.9. Theorem. J(p) is a Jordan algebra over k, for all p ∈ J×.

Proof. Since passing to isotopes is compatible with base change (33.8 (iii)), it suf-fices to prove (31.1.1), (31.1.2) for J(p). Let x,y∈ J. Then (31.1.1) for J and (33.8.2)imply

U (p)

U(p)x y

=UUxUpyUp =UxUpUyUpUxUp =U (p)x U (p)

y U (p)x ,

hence (31.1.1) for J(p). In order to accomplish the same for the identity (31.1.2), weapply (31.1.2) and (31.3.29), (33.8.2), (33.8.6) for J to conclude

UpU (p)x V (p)

y,x =UpUxUpVy,Upx =UUpxVy,Upx =VUpx,yUUpx =VUpx,yUpUxUp

=UpVx,UpyUxUp =UpV (p)x,y U (p)

x ,

and canceling Up yields (31.1.2) for J(p).

t.ISTJIT33.10. Theorem. Let p ∈ J be invertible.(a) An element x ∈ J is invertible in J(p) if and only if it is so in J. In this case, itsinverse in J(p) is given by x(−1,p) =U−1

p x−1.

(b) Let q ∈ J(p)× = J×. Then (J(p))(q) = J(Upq).

Proof. (a) The first part follows immediately from (33.8.2) combined with Prop. 33.2.Moreover, applying (33.2.1) to J(p) and J, we conclude

x(−1,p) =U (p)−1x x =U−1

p U−1x x =U−1

p x−1,

as claimed.(b) From (33.8.1) and (a) we deduce (1(p))(q) = q(−1,p) =Up−1q−1 = (Upq)−1 =

1(Upq). Moreover, for x ∈ J we apply (33.8.2) repeatedly and obtain

(U (p))(q)x =U (p)

x U (p)q =UxUpUqUp =UxUUpq =U (Upq)

x .

Summing up, we have proved (b).


c.ISTJYM

33.11. Corollary. Setting p−2 := (p−1)2, we have (J(p))(p−2) = J for all p ∈ J×.

Proof. (J(p))(p−2) = J(Up(p−1)2) = J by (33.1.1).

e.ALIST33.12. Examples: alternative algebras. Let A be a unital alternative algebra overk and p ∈ A(+)× = A× (33.6). Writing Rp for the right multiplication operator of pin A, we claim that

Rp : A(+)(p) ∼−→ A(+) (1) PLUSP

is an isomorphism of Jordan algebras. Indeed, Rp1A(+)(p) = Rp p−1 = 1A(+) , so Rppreserves identity elements. Moreover, letting x,y ∈ A and applying (14.5.4), weobtain

RpU (p)x y = RpUxUpy = RpUxLpRpy =UxpRpy =URpxRpy,

hence the assertion.In particular, we can now conclude that isotopes of special Jordan algebras are

special. Indeed, if J is special, then there exists a unital associative algebra A overk and an injective homomorphism ϕ : J → A(+). Therefore we deduce for p ∈ J×

that ϕ is also an injective homomorphism from J(p) to A(+)(ϕ(p)) ∼= A(+), Thus J(p)

is special.ss.ISTJALT

33.13. The connection with isotopes of alternative algebras. Let A be a unitalalternative algebra over k and p,q ∈ A×. Consulting (16.5.1) and Prop. 16.6, weconclude

A(p,q)(+) = A(+)(pq). (1) ISTJALT

In particular, as is already implicit in Lemma 16.10, passing to unital isotopes ofalternative algebras does not change the Jordan structure:

Ap(+) = A(+) (p ∈ A×). (2) JUNIST

e.ISTALTIN33.14. Examples: associative algebras with involution. Let (B,τ) be an associa-tive k-algebra with involution and p ∈ H(B,τ)× = H(B,τ)∩B× (33.7).Then it isreadily checked that

τp : B−→ B, x 7−→ τ

p(x) := p−1τ(x)p (1) PETWIN

is an involution of B satisfying

H(B,τ p) = H(B,τ)p; (2) PETSYM

this also follows from Prop. 17.7. In view of 33.12, we therefore conclude that

Rp : H(B,τ)(p) ∼−→ H(B,τ p) (3) HABEP


is an isomorphism of Jordan algebras that fits into the commutative diagram

B(+)(p)Rp

∼= // B(+)

H(B,τ)(p)?

OO

Rp

∼= // H(B,τ p).?

OO

Remark. The second fundamental fact to be observed in the present context is thatisotopes of J, though they are always Jordan algebras, will in general not be isomor-phic to J. For examples along these lines, see Exc. 169 (d),(e) below.

ss.HOMOT33.15. Homotopies. Homotopies of Jordan algebras are homomorphisms into ap-propriate isotopes. More precisely, a homotopy from J to J′ is a map η : J→ J′ suchthat η : J→ J′(p′) is a homomorphism, for some p′ ∈ J′×. In this case, Prop. 33.5and Thm. 33.10 (a) imply η(J×) ⊆ J′(p′)× = J′×, and p′ is uniquely determinedsince η(1J) = 1J′(p′) = p′−1, hence

p′ = η(1J)−1. (1) HOMOTONE

We will see in Prop. 33.16 below that compositions of homotopies are homotopies.Hence Jordan k-algebras under homotopies form a category, denoted by k-jordhmt.The isomorphisms in this category are precisely the bijective homotopies and arecalled isotopies. The isotopies from J to itself are called autotopies.

p.HOMOTIST33.16. Proposition. (a) η : J→ J′ is a homomorphism if and only if η is a homo-topy preserving units: η(1J) = 1J′ .(b) Let p ∈ J×, p′ ∈ J′×. Then η : J → J′ is a homotopy if and only if so isη : J(p)→ J′(p′).(c) If η : J→ J′ and η ′ : J′→ J′′ are homotopies, then so is η ′ η : J→ J′′.

Proof. (a) A homomorphism is a homotopy preserving units. Conversely, letη : J→ J′ be a homotopy preserving units. By (33.15.1), therefore, η : J→ J′(p′)isa homomorphism, with p′ = η(1J)

−1 = 1J′ .(b) Assume first that η : J → J′ is a homotopy. Then some q′ ∈ J′× makes

η : J→ J′(q′) a homomorphism. Hence a repeated application of Thm. 33.10 (b) im-

plies that so is η : J(p)→ J′(p′)(q′1), with q′1 =Up′−1Uq′η(p). Thus η : J(p)→ J′(p′)

is a homotopy. Conversely, if η : J(p) → J′(p′) is a homotopy, then what we havejust shown implies that so is η : J = J(p)(p−2)→ J′(p′)(p′−2) = J′.

(c) Some p′ ∈ J′× makes η : J→ J′(p′) a homomorphism. But η ′ : J′(p′)→ J′′ isa homotopy by (a), so some p′′ ∈ J′′× makes η ′ : J′(p′)→ J′′(p′′) a homomorphism.Hence so is η ′ η : J→ J′′(p′′), implying (c).


p.CHAUTOT33.17. Proposition. For all linear maps η : J → J′, the following conditions areequivalent.

(i) η is an isotopy from J to J′.(ii) η is bijective and there exists a bijective linear map η] : J′→ J such that

Uη(x) = ηUxη] (1) USH

for all x ∈ J.

In this case, η] is uniquely determined and satisfies

η] = η

−1Uη(1J). (2) ETASH

Proof. The final assertion follows immediately from (1) for x = 1J .(i)⇒ (ii). Some p′ ∈ J′× makes η : J→ J′(p′) an isomorphism. For x,y ∈ J we

therefore conclude η(Uxy) = U (p′)η(x)η(y) = Uη(x)Up′η(y), hence ηUx = Uη(x)Up′η ,

and we obtain (1) by setting η] := η−1Up′−1 .(ii) ⇒ (i). Put p′ := η(1J)

−1 ∈ J′×. Then η(1J) = p′−1 and (1) yields Up′ =

U−1η(1J)

= (ηη])−1 = η]−1η−1. Applying (1) once more, we therefore deduce

U (p′)η(x)η(y) =Uη(x)Up′η(y) = ηUxη

]η]−1

η−1

η(y) = η(Uxy)

for all x,y∈ J. Thus η : J→ J′(p′) is an isomorphism, making η : J→ J′ an isotopy.

ss.STRUG

33.18. The structure group. The group of autotopies of J is denoted by Str(J) andcalled the structure group of J. By Prop. 33.17 for J′= J, it consists of all η ∈GL(J)such that there exists an η] ∈ GL(J) satisfying

Uη(x) = ηUxη] (1) STREL

for all x ∈ J. Combining this with the fundamental formula (31.1.1) and Prop. 33.2,we see that Ux ∈ Str(J) for all x ∈ J×; in fact, we then have U ]

x =Ux. The subgroupof Str(J) generated by the linear operators Ux, x ∈ J×, is called the inner structuregroup of J, denoted by Instr(J). Combining (1) with (33.17.2), we obtain ηUxη−1 =Uη(x)Uη(1J)−1 for all η ∈ Str(J) and all x ∈ J×. Thus Instr(J)⊆ Str(J) is actually anormal subgroup. It is also readily checked that the assignment η 7→ η] determinesan involution of Str(J), i.e., an anti-automorphism of period 2.

t.STRINV33.19. Theorem. (a) Both the structure group and the inner structure group remainunchanged when passing to isotopes:

Str(J(p)) = Str(J), Instr(J(p)) = Instr(J) (p ∈ J×). (1) STRIST


(b) The structure group of J acts canonically on J×, and we have

η(x)−1 = η]−1(x−1) (η ∈ Str(J), x ∈ J×). (2) INVSTR

(c) The stabilizer of 1J in Str(J) is Aut(J), the automorphism group of J.(d) For p,q ∈ J× and η ∈ Endk(J), the following conditions are equivalent.

(i) η : J(p)→ J(q) is an isomorphism.(ii) η ∈ Str(J) and η(p−1) = q−1.(iii) η ∈ Str(J) and η](q) = p.

Proof. (a) follows immediately from Prop. 33.16 (b) and (33.8.2).(b) The first part follows immediately from (33.18.1) combined with Prop. 33.2.

Moreover, given x ∈ J×, an application of (33.18.1) and (33.2.1) yields η(x)−1 =U−1

η(x)η(x) = η]−1U−1x η−1η(x), hence (2).

(c) follows immediately from Prop. 33.16 (a).(d) (i) ⇔ (ii). By Prop. 33.16, condition (i) holds if and only if η ∈ Str(J) and

η(p−1) = η(1(p)) = 1(q) = q−1.(ii) ⇔ (iii). For η ∈ Str(J), the condition η(p−1) = q−1 by (2) is equivalent toq = η(p−1)−1 = η]−1(p), hence to η](q) = p.

c.ORSTR33.20. Corollary. For p,q ∈ J×, the isotopes J(p) and J(q) are isomorphic if andonnly if p and q belong to the same orbit of J× under the action of the structuregroup of J.

c.CLIST33.21. Corollary. Let J be an algebraic Jordan algebra over an algebraicallyclosed field F of characteristic not 2. Then the inner structure group of J acts tran-sitively on J×, and any two isotopes of J are isomorphic.

Proof. Let p∈ J×. By Exc. 33, there exists a q∈ J× such that p = q2 =Uq1J . Hencep belongs to the orbit of 1J under the inner structure group of J. This proves the firstassertion, while the second one now follows immediately from Cor. 33.20.

Remark. Cor. 33.21 does not hold in characteristic 2, see Exc. 169 (e) below.ss.METHIST

33.22. On the methodological importance of isotopes. By deriving the resultsof the present section, we have brought to the fore the close analogy connecting theU-operator of Jordan algebras with the left (or right) multiplication operator of asso-ciative algebras. There are important differences, however. For example, the trivialobservation that the group of left multiplications induced by invertible elements ofa unital associative algebra A is transitive on A× has no analogue in the Jordan set-ting. In fact, combining Cor. 33.20 with Exc. 169 (d) below, it follows that thereare Jordan algebras over fields where not even the full structure group, let alone theinner one, acts transitively on their invertible elements.

Isotopes may be regarded as a substitute for this deficiency. For example, one isoften confronted with the task of proving an identity for Jordan algebras involvinginvertible elements x,y,z, . . . . This task can sometimes be simplified by passing to


an appropriate isotope, which would then allow one to assume that, e.g., x is theidentity element. Here is a typical result where the procedure just described turnsout to be successful.

33.23. Theorem (Jacobson [45, Prop. 1.7.10]). Let x,y be elements of J and assumet.TRINVthat y is invertible. If two of the elements x,x+Uxy,x+y−1 are invertible, then so isthe third and

(x+Uxy)−1 +(x+ y−1)−1 = x−1. (1) TRINV

Proof. We first treat the case y = 1J by showing that if two of the elements x,x+x2,1J + x are invertible, then so is the third and

(x+ x2)−1 +(1J + x)−1 = x−1. (2) TRINU

We begin by applying Thm. 32.1 to deduce

Ux+x2 =UxU1J+x. (3) TRUNU

Hence our first intermediate assertion holds. Moreover, combining (31.3.11) with(31.15.2), we obtain Ux,x2x−1 = VxUxx−1 = Vxx = 2x2, which together with (3) im-plies

Ux+x2((x+ x2)−1 +(1J + x)−1)= x+ x2 +Ux(1J + x) = x+2x2 + x3

=Uxx−1 +Ux,x2x−1 +Ux2x−1 =Ux+x2x−1,

hence (2).Next let y ∈ J× be arbitrary. Passing to the y-isotope J(y), which satisfies J(y)× =

J× by Thm. 33.10 (a), the relations

x+Uxy = x+ x(2,y), x+ y−1 = 1(y)+ x

and the special case treated before prove the first part of the theorem. Moreover, byThm. 33.10 (a) again and (2) for J(y),

(x+Uxy)−1 +(x+ y−1)−1 =Uy((x+ x(2,y))(−1,y)+(1(y)+ x)(−1,y))

=Uyx(−1,y) = x−1.

33.24. Corollary. (Hua’s identity). Let x,y ∈ J be invertible such that x− y−1 isc.UINVinvertible as well. Then so is x−1− (x− y−1)−1 and

Uxy = x−(x−1− (x− y−1)−1)−1

.

Proof. Thm. 33.23 for −x in place of x shows that −x+Uxy is invertible and


(−x+Uxy)−1 =−(x−1− (x− y−1)−1).

The assertion follows.

Exercises.

pr.EVINV 166. Evaluation at invertible elements. Let x be an invertible element of J.

(a) Define

x−n := (x−1)n (1) NEPO

for all positive integers n (see Cor. 33.11 for n = 2) and prove

Uxn =Unx , (2) NEUXN

Uxm xn = x2m+n, (3) NEUXMN

(xm)n = xmn, (4) NEIT

xmxnxp= 2xm+n+p, (5) NETRI

Vxm,xn =Vxm+n , (6) NEVXMN

UxiUxm,xn =Uxi+m,xi+n =Uxm,xnUxi , (7) NEUUXMN

Uxi,x jUxm,xn =Uxi+m,x j+n +Uxi+n,x j+m (8) NEUUXIJMN

for all i, j,m,n ∈ Z.(b) Write k[t, t−1] for the ring of Laurent polynomials in the variable t over k and

ε×x : k[t, t−1](+)→ J

for the unique linear map sending tn to xn for all n ∈ Z. Show that ε×x is a homomorphism ofJordan algebras and that k[x,x−1] := Im(ε×x ) is the subalgebra of J generated by x and x−1.

(c) Write f (x) := ε×x ( f ) for f ∈ k[t, t−1] and show that

U( f g)(x) =U f (x)Ug(x) (9) NEUFG

for all f ,g ∈ k[t, t−1]. Conclude that if J has no absolute zero divisors, then k[x,x−1] carriesthe structure of a unique algebra R ∈ k-alg such that k[x,x−1] = R(+) as Jordan algebras.

pr.INVLIN 167. Invertibility in linear Jordan algebras. Let J be a linear Jordan algebra over a commutativering containing 1

2 . Elements x,y ∈ J are said to operator commute if [Lx,Ly] = 0.

(a) Prove for x,y ∈ J that x2 and y operator commute if and only if so do x and xy.(b) Prove for x ∈ J that the following conditions are equivalent.

(i) x is invertible.(ii) There exists an element y ∈ J such that xy = 1J and x,y operator commute.(iii) There exists an element y ∈ J such that xy = 1J and x2y = x.

Show further that if these conditions are fulfilled, then y as in (ii) (resp. (iii)) is unique andequal to x−1.

Remark. The characterization of invertibility in (b) (ii) is due to Koecher (unpublished).

pr.INVPOI 168. Invertibility in pointed quadratic modules. Let (M,q,e) be a pointed quadratic module withconjugation x 7→ x over k. Prove that an element x ∈ M is invertible in the Jordan algebra J :=J(M,q,e) if and only if q(x) ∈ k is invertible in k. In this case,


x−1 = q(x)−1x, q(x−1) = q(x)−1. (10) INVPOI

If x is invertible in J, we also say x is invertible in (M,q,e) with inverse x−1. Thus the Jordanalgebra of a pointed quadratic module (M,q,e) over a field is a Jordan division algebra if and onlyif the quadratic form q is anisotropic.

pr.ISTPOI 169. Isotopes of pointed quadratic modules. Let (M,q,e) be a pointed quadratic module over k,with trace t and conjugation x 7→ x.

(a) Let f ∈M be invertible in (M,q,e) (Exc. 168). Show that

(M,q,e)( f ) := (M,q( f ),e( f )), q( f ) := q( f )q, e( f ) := f−1 (11) MQEF

is a pointed quadratic module over k with trace

t( f ) : M −→ k, x 7−→ t( f )(x) := q( f ,x), (12) TEF

and conjugation

x 7−→ x( f ) := q( f )−1q( f ,x) f − x. (13) CONEF

Show further

J((M,q,e)( f ))= (J(M,q,e)

)( f ). (14) JEF

We call (M,q,e)( f ) the f -isotope (or simply an isotope) of (M,q,e).(b) Let f ∈ M be invertible in (M,q,e). Prove that the invertible elements of (M,q,e) and

(M,q,e)( f ) are the same, and that((M,q,e)( f ))(g) = (M,q,e)(U f g)

for all invertible elements g of (M,q,e), where U stands for the U-operator of the Jordanalgebra J(M,q,e).

(c) Prove that if M is projective as a k-module, then the structure group of J(M,q,e) agreeswith the group of similarity transformations of the quadratic module (M,q), consisting bydefinition of all bijective linear maps η : M→M such that qη = αq for some α ∈ k×.

(d) Let k := R and M := R3 with the canonical basis (e1,e2,e3). Put e := e1 and q := 〈S〉quadwith

S :=

1 0 00 −1 00 0 −1

.

Find an isotope of J := J(M,q,e) which is not isomorphic to J.(e) Let k := F be a field of characteristic 2 and assume (M,q,e) is traceless in the sense that

t = 0. Furthermore, let f ∈ M be anisotropic relative to q and assume f /∈ Rad(Dq). Thenput J := J(M,q,e) and prove that the isotope J( f ) is not isomorphic to J. Finally, give anexample where the preceding hypotheses are fulfilled even when F is algebraically closed.

pr.FOUROP 170. A useful formula for the U-operator. (cf. Braun-Koecher [15, IV, Satz 3.8]) Let x,y be invert-ible elements of J. Prove

UxUx−1+y−1Uy =Ux+y.

(Hint. Pass to the isotope J(y) and apply Exc. 166 (9).)

pr.LININV 171. Linear invertibility. Let J be a linear Jordan algebra over a commutative ring k containing12 . An element x ∈ J is said to be linearly invertible if the left multiplication operator Lx : J→ Jis bijective. In this case we call x−1 := L−1

x 1J the linear inverrse of x in J. Prove that if x is


linearly invertible, then it is invertible, and its linear inverse and its ordinary inverse are the same.Finally, prove that invertible elements of J need not be linearly invertible by showing for anypointed quadratic module (M,q,e) over a field of characteristic not 2 that J := J(M,q,e) is a lineardivision algebra in the sense of 9.7 if and only if q is anisotropic and dimF (J)≤ 2.

For more on the connection between linear and ordinary Jordan division algebras, see Petersson[97].

pr.STRHOT 172. Strong homotopies. A linear map η : J → J′ is called a strong homotopy if some p ∈ J×

makes η : J(p)→ J′ a homomorphism.

(a) Prove that strong homotopies are homotopies. Conversely, if η : J→ J′ is a homotopy andη(J×) = J′×, show that η is a strong homotopy.

(b) Give an example of a homotopy which is not a strong homotopy.

pr.VALJO 173. (Petersson [94, 96]) Discrete valuations of Jordan division rings.1 Let J be a Jordan divi-sion ring, i.e., a Jordan division algebra over Z, and write F for the centroid of J. Recall fromExc. 155 (c) that F is a field and that J may canonically be regarded as a Jordan algebra over F .

By a discrete valuation of J we mean a map λ : J→ Z∞ := Z∪∞ satisfying the followingconditions, for all x,y ∈ J.

λ (x) = ∞ ⇐⇒ x = 0, (15) VALDEF

λ (Uxy) = 2λ (x)+λ (y), (16) UMULT

λ (x+ y)≥ minλ (x),λ (y). (17) NARTRI

For the rest of this exercise, fix a discrete valuation λ of J and observe by Thm. 32.11 that J islocally linear. Then prove:

(a) For all x ∈ J× and all n ∈ Z, we have λ (xn) = nλ (x).(b) For all x ∈ J and all v,w ∈ F [x], we have λ (vw) = λ (v)+λ (w). Conclude that

λ0 : F −→ Z∞, a 7−→ λ0(a) := λ (a1J),

is a discrete valuation of of F , with valuation ring, valuation ideal, and residue field respec-tively given by

o0 := a ∈ F | λ0(a)≥ 0,p0 := a ∈ F | λ0(a)> 0,F := o0/p0.

(c) The sets

O := x ∈ J | λ (x)≥ 0 ⊆ J,

P := x ∈ J | λ (x)> 0 ⊆O,

J :=O/P

are respectively an o0-subalgebra of J, an ideal in O, a Jordan division algebra over F . Butnote that, even if J is a linear Jordan algebra over F (i.e., F has characteristic not 2), O neednot be one over o0, and J need not be one over F .

(d) λ (J×) is a subgroup of Z, called the value group of λ .(e) For p ∈ J×, the map

λ(p) : J(p) −→ Z∞, x 7−→ λ

(p)(x) := λ (p)+λ (x),

1 In this exercise, the reader is assumed to be familiar with the rudiments of valuation theory.

34 The Peirce decomposition 249

is a discrete valuation of J(p) having the same value group as λ . We call λ (p) the p-isotopeof λ . Given another element q ∈ J×, show further (λ (p))(q) = λ (Upq).

(f) λ satisfies the Jordan triple product inequality

λ (xyz)≥ λ (x)+λ (y)+λ (z) (18) LATRI

for all x,y,z ∈ J. (Hint. Reduce to the case y = 1J . Then derive and use the formula

(x y)2 =Uxy2 +Uyx2 + x (Uyx), (19) CIRSQU

valid for all x,y in arbitrary Jordan algebras.)

pr.ALJODI 174. Let J be an algebraic Jordan division algebra over a field F . Prove:

(a) For x ∈ J, the subalgebra F [x]⊆ J is a finite algebraic field extension of F .(b) If F is algebraically closed, then J ∼= F(+).(c) For F = R, there exists a pointed quadratic module (M,q,e) over R such that J ∼= J(M,q,e)

and q is positive definite.

34. The Peirce decomposition

s.PEIDECThe Peirce decomposition in its various guises belongs to the most powerful tech-niques in the structure theory of Jordan algebras. Initiated by Jordan-von Neumann-Wigner in their study of euclidean Jordan algebras, it dominated the scene way intothe nineteen-sixties, losing a certain amount of its appeal only when research beganto focus on Jordan algebras without finiteness conditions where idempotents (theprincipal ingredient of the Peirce decomposition) are in short supply.

The present section is devoted to those properties of the Peirce decompositionthat are relevant for our subsequent applications to cubic Jordan algebras. Our treat-ment of the subject relies quite heavily on the approach of Loos [67, § 5] (after beingspecialized from Jordan pairs to algebras), which in turn owes much to the one ofSpringer [120, § 10].

Throughout we let k be a commutative ring and J a Jordan algebra over k. Re-call from Exc. 155 (b) that an idempotent in J is an element c satisfying c2 = c.The additional hypothesis c3 = c that has to be imposed in the setting of arbitrarypara-quadratic algebras (cf. 30.9), implying cn = c for all positive integers n, holdsautomatically in the Jordan case.

l.QUAID34.1. Lemma. Let c be an idempotent in J and put d := 1J−c. Writing D := k⊕kfor the split quadratic etale k-algebra,

Θc : D−→ Endk(J), γ⊕δ 7−→Θc(γ⊕δ ) :=Uγc+δd (γ,δ ∈ k)

is a quadratic map that is unital and permits composition:

Θc(1D) = 1J , Θc((γ1⊕δ1)(γ2⊕δ2)

)=Θc(γ1⊕δ1)Θc(γ2⊕δ2) (1) THUNCO

for all γi,δi ∈ k, i = 1,2. In particular,


Θc(γ⊕δ ) ∈ GL(J) (γ,δ ∈ k×). (2) THINV

Moreover, Θc is compatible with base change: ΘcR = (Θc)R for all R ∈ k-alg.

Proof. The first equation of (1) is obvious. In order to prove the second, we leti= 1,2 and note Θc(γi⊕δi)=Uδi1J+(γi−δi)c =U fi(c), where fi := δi+(γi−δi)t∈ k[t].Hence Thm. 32.1 implies Θc(γ1⊕δ1)Θc(γ2⊕δ2) =U( f1 f2)(c), where

f1 f2 = δ1δ2 +((γ1−δ1)δ2 +δ1(γ2−δ2)

)t+(γ1−δ1)(γ2−δ2)t2.

Evaluating at c, we conclude

( f1 f2)(c) = δ1δ21J +((γ1−δ1)δ2 +δ1(γ2−δ2)+(γ1−δ1)(γ2−δ2)

)c

= δ1δ21J +((γ1−δ1)γ2 +δ1(γ2−δ2)

)c

= δ1δ21J +(γ1γ2−δ1δ2)c

= (γ1γ2)c+(δ1δ2)d,

and the second equation of (1) is proved. The remaining assertions are now obvious.

34.2. Theorem (Singular Peirce decomposition). Let c be an idempotent in J andt.PEDESIput d := 1J− c. Then the following statements hold.(a) The linear endomorphisms of J defined by

E2 := E2(c) :=Uc, E1 := E1(c) =Uc,d =Vc−2Uc, E0 := E0(c) :=Ud (1) PEPRSI

satisfy the relation

UtcR+dR = E0R + tE1R + t2E2R (2) UCDEXP

for all R ∈ k-alg and all t ∈ R.(b) The Ei, i = 0,1,2, are orthogonal projections of J, called its Peirce projectionsrelative to c, and their sum is the identity. Hence they give rise to a decomposition

J = J2⊕ J1⊕ J0 (3) PEDESI

as a direct sum of submodules Ji := Ji(c) := Im(Ei) for i = 0,1,2, called the Peircecomponents of J relative to c.(c) We have

J2 = Im(Uc), J1⊕ J0 = Ker(Uc), (4) JTWO

J0 = Ker(Uc)∩Ker(Vc), (5) JZERO

Ji ⊆ x ∈ J | c x = ix (i = 0,1,2), (6) JI

J1 = x ∈ J | c x = x. (7) JONE


(d) Setting Ji := 0 for i ∈ Z \ 0,1,2, the following composition rules hold, forall i, j, l = 0,1,2.

UJiJ j ⊆ J2i− j, (8) UJIJJ

JiJ jJl ⊆ Ji− j+l , (9) JIJJJL

J2J0J= 0= J0J2J. (10) JTWZE

Proof. (a) Expanding Uc,d with respect to d gives the second equation for E1 in (1),while (2) is equally obvious.

(b) R := k[s, t], the polynomial ring in two independent variables s, t over k, isfree as a k-module with basis (sit j)i, j∈N. Hence the identifications of 31.2 imply

J ⊆ JR =⊕i, j≥0

(sit jJ

), Endk(J)⊆ Endk(J)R =

⊕i, j≥0

(sit j Endk(J)

)⊆ EndR(JR)

as direct sums of k-modules. With this in mind, Lemma 34.1 combined with (2)yields

2

∑i, j=0

sit jEiE j = (2

∑i=0

siEi)(2

∑j=0

t jE j) =ΘcR(s⊕1R)ΘcR(t⊕1R)

=ΘcR

((s⊕1R)(t⊕1R)

)=ΘcR

((st)⊕1R

)=

2

∑i=0

sitiEi.

Comparing coefficients of sit j, we obtain EiE j = δi jEi for i, j = 0,1,2, and the firstassertion of (b) follows. The remaining ones are obvious.

(c) Since Uc = E2 by (1), equation (4) is obvious. Applying (1) once more, weconclude J0 =Ker(E2)∩Ker(E1) =Ker(Uc)∩Ker(Vc−2Uc), and (5) follows. From(1) we also deduce Vc = E1 + 2Uc = 2E2 + E1 = ∑

2j=0 jE j, and applying this to

xi ∈ Ji, we obtain cxi =Vcxi = ∑ jE jxi = ixi, giving (6) and the inclusion from leftto right in (7). Conversely, suppose x∈ J satisfies cx = x and write x = x2+x1+x0,xi ∈ Ji. Then (6) implies x = 2x2 + x1 and comparing Ji-components for i = 0,1,2,we conclude x2 = x0 = 0, hence x = x1 ∈ J1. This completes the proof of (7).

(d) Let R = k[t, t−1] be the k-algebra of Laurent polynomials in the variable tover k, which is free as a k-module with basis (ti)i∈Z. Hence 31.2 yields the identi-fications

J ⊆ JR =⊕i∈Z

(tiJ), Endk(J)⊆ Endk(J)R =⊕i∈Z

(ti Endk(J)

)⊆ EndR(JR)

as direct sums of k-modules. Since t ∈ R×, we deduce from (2) and Lemma 34.1that the map

Utc+d =2

∑j=0

t jE j : JR→ JR


is bijective with inverse Ut−1c+d . Moreover, for x = ∑2l=0 xl , xl ∈ Jl , l = 0,1,2, we

compute

Utc+dx =2

∑j,l=0

t jE jxl =2

∑l=0

tlxl ,

which shows

Ji = x ∈ J |Utc+dx = tix= x ∈ J |Ut−1c+dx = t−ix. (11) JIUT

Now let i, j, l ∈ 0,1,2 and xi ∈ Ji, y j ∈ J j, zl ∈ Jl . Then the fundamental formulaand (11) imply

Utc+dUxiy j =Utc+dUxiUtc+dUt−1c+dy j =UUtc+dxiUt−1c+dy j

=Utixit− jy j = t2i− jUxiy j,

hence Uxiy j ∈ J2i− j. This proves (8). Similarly,

Utc+dxiy jzl=Utc+dUxi,zlUtc+dUt−1c+dy j

=UUtc+dxi,Utc+dzlUt−1c+dy j =Utixi,tlzlt− jy j

= ti− j+lxiy jzl,

which proves xiy jzl ∈ Ji− j+l , hence (9). It remains to prove the first equation of(10) since the second one then follows after the substitution c 7→ d. For xi ∈ Ji,i = 0,2, we must show Vx2,x0 = 0. Applying (31.3.21), (31.3.13) and (1), (4), (6), wefirst obtain Vc,x0 = VUcc,x0 = Vc,ccx0−VUcx0,c = Vc,cx0 = 0 from (1), (6) and thenVx2,x0 =VUcx2,x0 =Vc,x2cx0−VUcx0,x2 = 0 since x2cx0 ∈ J0 by (9). This completesthe proof.

c.PEIBAS34.3. Corollary. Let c be an idempotent in J and put d := 1J− c.(a) For all R ∈ k-alg, there are natural identifications

(JR)i(cR) = Ji(c)R (i = 0,1,2).

(b) d is an idempotent in J satisfying Ji(d) = J2−i(c) for i = 0,1,2.(c) The k-module J2(c) (resp. J0(c)) is a Jordan algebra over k with unit elementc (resp. d) whose U-operator is derived from the U-operator of J by restriction.Moreover, J2(c)⊕ J0(c) is a direct sum of ideals and a subalgebra of J.(d) The Peirce components Ji := Ji(c) of J (i = 0,1,2) satisfy the bilinear composi-tion rules

Ji Ji ⊆ Ji, Ji J1 ⊆ J1, J2 J0 = 0, J1 J1 ⊆ J2⊕ J0 (1) PEICIR

for i = 0,2.

Proof. (a) This follows immediately from (34.2.3) and the fact that scalar extensionsof k-modules commute with direct sums.


(b) This follows immediately from Thm. 34.2 (a), (b).(c) We put c2 := c, c0 := d and let i = 0,2. From (34.2.8) we deduce UJiJi ⊆ Ji, so

restricting the U-operator of J to Ji gives a quadratic map U : Ji→ Endk(Ji) whichby Thm. 34.2 sends ci to the identity on Ji. In this way, therefore, Ji becomes apara-quadratic algebra over k, which is in fact a Jordan algebra since the definingidentities (31.1.1), (31.1.2) hold not only on Ji but by (a) on all scalar extensionsas well. By (34.2.8) we have UJ2J0 +UJ0J2 = 0, which together with (34.2.10)implies that J2⊕J0 is a direct sum of ideals. As such it is a Jordan algebra in its ownright with identity element c1 + c0 = 1J , hence a subalgebra of J.

(d) For i = 0,2, j = 0,1,2, we apply (34.2.9), (34.2.10) and obtain Ji J j ⊆JiJiJ j+ JiJ2−iJ j ⊆ J j hence first two relations of (1), while the third onenow follows by symmetry: J2 J0 ⊆ J0 ∩ J2 = 0. On the other hand, J1 J1 ⊆J1J2J1+ J1J0J1 ⊆ J0 + J2 by (34.2.9), giving also the fourth relation, and theproof of (1) is complete.

Remark. J2(c), though a Jordan algebra, in general is not a subalgebra of J; in fact,it is one if and only if c = 1J .

ss.PEDELI34.4. The Peirce decomposition in linear Jordan algebras. Suppose k contains12 and let c be an idempotent in J. We claim

Ji(c) = x ∈ J | c x = ix (i = 0,1,2). (1) PEEIG

By (34.2.6), (34.2.7), the left-hand side is contained in the right and we have equalityfor i = 1. Now assume i = 0,2 and that x ∈ J satisfies c x = ix.Writing x = x2 +x1 +x0, x j ∈ J j(c) for j = 0,1,2, we conclude ix2 + ix1 + ix0 = cx = 2x2 +x1. Fori = 0, this implies 2x2 = x1 = 0, hence x = x0 ∈ J0(c) since k contains 1

2 . By thesame token, i = 2 implies x1 = 2x0 = 0, hence x = x2 ∈ J2(c). This completes theproof.

Viewing J as a unital linear Jordan algebra over k via Thm. 31.5, with bilinearmultiplication xy = 1

2 x y = 12Vxy, the left multiplication operator of c in J is Lc =

12Vc, and (1) may be rewritten as

Ji := Ji(c) = x ∈ J | cx =i2

x (i = 0,1,2), (2) PELI

so the Peirce components of c are basically the “eigenspaces” of Lc with respect tothe “eigenvalues” 0, 1

2 ,1. For this reason, the i-th Peirce component of c, i = 0,1,2,in the classical literature is usually denoted by J i

2(c). But we will never adhere to this

convention. Instead we confine ourselves to rewriting the composition rules (34.3.1)in the language of linear Jordan algebras: for i = 0,2 we have

J2i ⊆ Ji, JiJ1 ⊆ J1, J2J0 = 0, J2

1 ⊆ J2⊕ J1.

p.PEPOQUA34.5. Proposition. Let (M,q,e) be a pointed quadratic module over k. An elementc ∈ M is an elementary idempotent in J := J(M,q,e) (cf. Exc. 160) if and only if


(c,d) with d := e− c is a hyperbolic pair in the quadratic module (M,q). In thiscase, d is an elementary idempotent of J as well and

J2(c) = kc, J1(c) = (kc⊕ kd)⊥, J0(c) = kd, (1) PECOPO

where “⊥” stands for orthogonal complementation relative to the bilinearization ofq, are the Peirce components of J relative to c.

Proof. Write t for the trace and x 7→ x for the conjugation of (M,q,e). If (c,d) is ahyperbolic pair of (M,q), then q(c) = 0 and t(c) = q(c,c+d) = 2q(c)+q(c,d) = 1.Thus c ∈ J is an elementary idempotent. Conversely, let this be so. Then q(d) =q(e−c)= 1−t(c)+q(c)= 0 and q(c,d)= q(c,e−c)= t(c)−2q(c)= 1. Thus (c,d)is a hyperbolic pair of (M,q). It remains to determine the Peirce components of Jrelative to c. To begin with, let x∈ J2(c). Then (31.12.4) implies x=Ucx= q(c, x)c−q(c)x= q(c, x)c∈ kc, and the first equation of (1) holds. But then J0(c) = J2(d) = kdsince d is an elementary idempotent. It remains to prove J1(c) = (kc⊕kd)⊥. Lettingx ∈ J and applying (31.12.8), we obtain cx = t(c)x+ t(x)c−q(c,x)e = x+ t(x)c−q(c,x)e, and (34.2.7) yields

J1(c) = x ∈M | t(x)c = q(c,x)e. (2) ONEPO

Suppose first x ∈ J1(c). Then (2) implies q(c,x) = t(c)q(c,x) = q(q(c,x)e,c) =q(t(x)c,c) = 2t(x)q(c) = 0, hence t(x)c = q(c,x)e = 0, forcing t(x) = 0 since c isunimodular, and then q(d,x) = t(x)−q(c,x) = 0. Thus x ∈ (kc⊕ kd)⊥. Conversely,let x ∈ (kc⊕ kd)⊥. Then q(c,x) = q(d,x) = 0 implies t(x) = q(c,x)+ q(d,x) = 0,and (2) implies x ∈ J1(c).

The Peirce decomposition of a Jordan algebra relative to a single idempotent is oftennot fine enough for the intended applications. We therefore wish to replace it by aPeirce decomposition relative to a complete orthogonal system of as many idempo-tents as possible. In order to accomplish this, we require a number of preparationsthat we are now going to address.

Orthogonality of a family of idempotents has been defined in 30.9 and Exc. 153for arbitrary para-quadratic algebras. In the case of Jordan algebras, there is a simplecharacterization in terms of the Peirce decomposition.

p.ORIDJO34.6. Proposition. For idempotents c1,c2 ∈ J, the following conditions are equiv-alent.

(i) c1 ⊥ c2.(ii) c2 ∈ J0(c1).(iii) c1 ∈ J0(c2).

Proof. Orthogonality being a symmetric relation on idempotents, it suffices to es-tablish the equivalence of (i) and (ii). Combining (30.9.1) with (31.3.13), we seethat (i) holds if and only if Uc1c2 = Uc2c1 = Vc1c2 = 0. In this case, (34.2.5) im-plies c2 ∈ J0(c1). Conversely, suppose c2 ∈ J0(c1). Then Uc1c2 =Vc1c2 = 0, but alsoUc2c1 ∈UJ0(c1)J2(c1) = 0 by Cor. 34.3 (b). Thus (i) holds. .


c.ORCOJO34.7. Corollary. A finite family (c1, . . . ,cr) (r ∈ Z, r > 0) of idempotents in J is anorthogonal system of idempotents if and only if ci ⊥ c j for 1≤ i, j ≤ r, i 6= j.

Proof. Assume ci ⊥ c j for 1≤ i, j ≤ r, i 6= j and let i, j, l ∈ 1, . . . ,r. The first setof relations in Exc. 153, (3), holds by the definition of orthogonality (cf. (30.9.1)).For the second set, let i, j, l be mutually distinct. From Prop. 34.6 and (34.2.10)we deduce cic jcl ∈ J2(ci)J0(ci)J) = 0. Thus all of Exc. 153, (3), holds, so(c1, . . . ,cr) is an orthogonal system of idempotents. The converse is obvious., againby Exc. 153, (3).

p.COORPE34.8. Proposition. Let Ω = (c1, . . . ,cr) be a complete orthogonal system of idem-potents in J and define linear maps

Eii := Eii(Ω) :=Uci , Ei j := Ei j(Ω) :=Uci,c j (1≤ i, j ≤ r, i 6= j). (1) COORPE

Then Ei j = E ji for 1≤ i, j≤ r, and the Ei j, 1≤ i≤ j≤ r, are orthogonal projectionsof J, called its Peirce projections relative to Ω , such that ∑1≤i≤ j≤r Ei j is the identity.

Proof. The first assertion is obvious, and we clearly have ∑i≤ j Ei j =U∑ci =U1J =1J , so we need only show that the Ei j, 1≤ i≤ j ≤ r, are orthogonal projections. Wedo so in two steps.10. Let 1≤ i, j≤ r, i 6= j. Exc. 153 (a) shows that ci+c j ∈ J is an idempotent, forcingJ′ := J2(ci+c j) by Cor. 34.3 (b) to be a Jordan algebra with identity element ci+c j.Applying Thm. 34.2 to J′, ci in place of J, c, respectively, we see that Eii,Ei j,E j jare orthogonal projections on J′ vanishing identically on J1(ci + c j)⊕ J0(ci + c j).Hence they are orthogonal projections on all of J that map J to J′.20. The proof will be complete once we have shown

Ei jElm = 0 (1≤ i, j, l,m≤ r, i, j 6= l,m). (2) SPEPRO

First suppose i = j in (2). By 10, we may assume l 6= m and i /∈ l,m. Then cl ⊥ci ⊥ cm, which implies cl + cm ⊥ ci by Prop. 34.6 and then

EiiElmJ+ElmEiiJ ⊆UJ0(cl+cm)J2(cl + cm)+J2(cl + cm)J0(cl + cm)J= 0.

This shows (2) not only for for i = j but also for l = m. We are left with the casei 6= j, l 6= m. By symmetry, we may assume i /∈ l,m, which implies

Ei jElmJ ⊆ ciJ2(cl + cm)c j ⊆ J0(cl + cm)J2(cl + cm)J= 0

and completes the proof of (2).

c.QUACOM34.9. Corollary. Let Ω = (c1, . . . ,cr) be a complete orthogonal system of idempo-tents in J and put D := k⊕·· ·⊕ k ∈ k-alg (r summands) as a direct sum of ideals.Then

ΘΩ : D−→ Endk(J), x 7−→U∑ri=1 γici ,


for x = γ1⊕ ·· · ⊕ γr ∈ D, γi ∈ k, 1 ≤ i ≤ r, is a quadratic map that is unital andpermits composition:

ΘΩ (1D) = 1J , ΘΩ (xy) =ΘΩ (x)ΘΩ (y) (1) QUACOM

for all x,y ∈ D. In particular, ΘΩ (x) ∈ GL(J) for all x ∈ D×. Moreover, ΘΩ iscompatible with base change: (ΘΩ )R =ΘΩR , ΩR := (c1R, . . . ,crR), for all R∈ k-alg.

Proof. All assertions are obvious except, possibly, the property of ΘΩ permit-ting composition. Bus this follows immediately from Prop. 34.8 since ΘΩ (x) =∑i≤ j γiγ jEi j(Ω).

ss.PEITRI34.10. Peirce triples. By a Peirce triple, we mean a(n ordered) triple of unorderedpairs of positive integers. Thus a Peirce triple has the form

(i j, lm,np) :=(i, j,l,m,n, p

)for integers i, j, l,m,n, p > 0. A Peirce triple (i j, lm,np) will always be identifiedwith the Peirce triple (np, lm, i j). It is said to be connected if (up to the identificationjust defined) it can be written in the form (i j, jm,mp). For example, the Peirce triple(43,23,12) is connected but (54,23,12) is not.

34.11. Theorem. (Multiple Peirce decomposition). Let Ω = (c1, . . . ,cr) be a com-t.PEDECOMplete orthogonal system of idempotents in J.(a) Setting Ji j := Ji j(Ω) := Im(Ei j(Ω)) in the notation of Prop. 34.8, we haveJi j = J ji for 1≤ i, j ≤ r and

J =⊕

1≤i≤ j≤r

Ji j (1) PEDECOM

as a direct sum of submodules, called the Peirce components of J relative to Ω .Furthermore, the following relations hold, for all i, j = 1, . . . ,r:

J2(ci) = Jii, J1(ci) = ∑l 6=i

Jil , J0(ci) = ∑l,m 6=i

Jlm, (2) JCI

Ji j = J1(ci)∩ J1(c j) (i 6= j). (3) JIJ

(b) More generally, if I ⊆ 1, . . . ,r, then cI := ∑i∈I ci is an idempotent in J with thePeirce components

J2(cI) = ∑i, j∈I

Ji j, J1(cI) = ∑i∈I, j/∈I

Ji j, J0(cI) = ∑i, j/∈I

Ji j. (4) JCMULT

(c) The composition rule

Ji jJ jlJlm ⊆ Jim (5) IJJLLM


holds for all i, j, l,m = 1, . . . ,r. Moreover, if the Peirce triple (i j, jl, i j) is connected,there exists a unique m = 1, . . . ,r such that i j = lm, and we have

UJi j J jl ⊆ Jim. (6) UIJJL

In particular,

UJi j Ji j ⊆ Ji j. (7) UIJIJ

Finally, if (i j, lm,np) (resp. (i j, lm, i j)) is not connected, then

Ji jJlmJnp= 0 (resp. UJi j Jlm = 0). (8) UIJLM

Proof. (a), (b). The first assertion of (a) and eqn. (1) follow immediately fromProp. 34.8, eqn. (3) is a consequence of (2), and (2) is a special case of (4). Inorder to complete the proof of (a) and (b) , it therefore suffices to establish (4). Tothis end, we put dI := 1J− cI = ∑i/∈I ci and apply Thm. 34.2 to obtain

J2(cI) = Im(UcI ) = Im( ∑i, j∈I,i≤ j

Ei j) = ∑i, j∈I

Ji j,

J1(cI) = Im(UcI ,dI ) = Im( ∑i∈I, j/∈I

Ei j) = ∑i∈I, j/∈I

Ji j,

J0(cI) = Im(UdI ) = ∑i, j/∈I

Ji j.

(c) Let R := k[t±11 , . . . , t±1

r ] the ring of Laurent polynomials over k in the variablest1, . . . , tr, which is free as a k-module with basis (ti1

1 · · · tirr )i1,...,ir∈Z. Thus the identi-

fications of 31.2 imply

J ⊆ JR =⊕

i1,...,ir∈Z(ti1

1 · · · tirr J),

Endk(J)⊆ Endk(J)R =⊕

i1,...,ir∈Z

(ti11 · · · t

irr Endk(J)

)⊆ EndR(JR)

as direct sums of k-modules. We now claim

w :=r

∑λ=1

tλ cλ ∈ J×R and w−1 =r

∑λ=1

t−1λ

cλ . (9) WEINV

Indeed, since tλ ∈ R× for 1≤ λ ≤ r, Cor. 34.9 shows not only Uw ∈GL(JR) but alsoUw′ = U−1

w , where w′ = ∑ t−1λ

cλ . Hence w ∈ J×R by Prop. 33.2, and since cλ ∈ Jλλ

by (2), we conclude

Uww′ = ∑i≤ j

tit jEi j ∑λ

t−1λ

cλ =r

∑i=1

t2i t−1

i ci =r

∑i=1

tici = w,


which in turn implies w′ =U−1w w = w−1 by (33.2.1). Next we claim

Ji j = x ∈ J |Uwx = tit jx= x ∈ J |Uw−1x = t−1i t−1

j x. (10) PEWEDE

Indeed, given x ∈ J, the expansion Uwx = ∑λ≤µ tλ tµ Eλ µ x, where Eλ µ x ∈ Jλ µ , com-bined with (1) yields the first equation of (10), while the second follows from thefirst. We now put

T := ti11 · · · t

irr | i1, . . . , ir ∈ Z, T0 := tit j | 1≤ i, j ≤ r

and claim

(∗) Every t ∈ T such that Uwx = tx for some non-zero element x ∈ J belongs to T0.

In order to see this, we write x = ∑i≤ j xi j, xi j ∈ Ji j, and deduce

∑i≤ j

txi j = tx =Uwx = ∑i≤ j

tit jxi j

from (10), and comparing Peirce components, the assertion follows. Now fix inte-gers i, j, l,m,n, p = 1, . . . ,r and x ∈ Ji j, y ∈ Jlm, z ∈ Jnp. Then we claim

Uwxyz= tit jt−1l t−1

m tntpxyz, (11) UWETRI

UwUxy = t2i t2

j t−1l t−1

m Uxy. (12) UWEUOP

In order to prove (11), we apply (9), (10), (31.3.3) and obtain

Uwxyz=UwUx,zUwUw−1y =UUwx,UwzUw−1y =Utit jx,tntpzt−1l t−1

m y,

hence (11). Similarly,

UwUxy =UwUxUwUw−1y =UUwxUw−1y =Utit jxt−1l t−1

m y,

and this yields (12). Now (5) follows by specializing l = j, n = m = l, p = m in(11) and applying (10). Next suppose the Peirce triple (i j, jl, i j) is connected. ByExc. 178 below, there is a unique m = 1, . . . ,r such that i j = lm, and specializingl = j, m= l in (12) gives t2

i t2j t−1l t−1

m = titit jt−1l = titm, hence (6), which for l = i spe-

cializes to (7). It remains to establish (8). To this end, assume first that (i j, lm,np)is not connected. If i, j∩ l,m = /0 = l,m∩ n, p, then tit jt−1

l t−1m tntp /∈ T0,

and (∗) implies xyz = 0. Otherwise, since the k-module Ji jJlmJnp remains un-affected by the identification of Peirce triples described in 34.10, Exc. 178 be-low allows us to assume l = n = j, m = l, p = n, with l 6= i, j,n, which impliestit jt−1

l t−1m tntp = tit jt−1

l tn /∈ T0, hence again xyz = 0. This proves the first part of(8). As to the second, assume that (i j, lm, i j) is not connected. If i, j∩l,m= /0,the t2

i t2j t−1l t−1

m /∈ T0, and we conclude Uxy = 0 from (∗). Otherwise, we may assumel = j, put m := l and have i 6= l 6= j, hence t2

i t2j t−1l t−1

m = t2i t jt−1

l /∈ T0, This againimplies Uxy = 0 and completes the proof of (8).

c.PECOBA


34.12. Corollary. Let Ω = (c1, . . . ,cr) be a complete orthogonal system of idem-potents in J.(a) For R ∈ k-alg, ΩR := (c1R, . . . ,crR) is a complete orthogonal system of idempo-tents in JR and there are canonical identifications

(JR)i j(ΩR) = Ji j(Ω)R (1≤ i, j ≤ r). (1) PECOBA

(b) For 1≤ i≤ r, the k-module Jii(Ω) is a Jordan algebra over k with unit elementci whose U-operator is derived from the U-operator of J by means of restriction.Moreover

⊕ri=1 Jii(Ω) is a direct sum of ideals and a subalgebra of J.

(c) The Peirce components of J relative to Ω satisfy the bilinear composition rules

Ji j(Ω) J jl(Ω)⊆ Jil(Ω), Ji j(Ω) Ji j(Ω)⊆ Jii(Ω)+ J j j(Ω)

for 1≤ i≤ j ≤ l ≤ r.

Proof. This is established in exactly the same manner as Cor. 34.3 has been derivedfrom Thm. 34.2.

Exercises.

pr.LOLIC 175. Show that J is linear at c, for every idempotent c ∈ J.

pr.ISTTWOZER 176. Let c∈ J be an idempotent and suppose v∈ J1(c) is invertible in J. Prove that Uv via restrictioninduces an isotopy from J2(c) onto J0(c). Moreover, this isotopy is an isomorphism if and only ifv2 = 1J .

pr.JTWOSP 177. Let c be an idempotent in J and put Ji := Ji(c) for i = 0,1,2. Then prove that

ϕ : J2 −→ Endk(J1), x 7−→Vx|J1,

is a homomorphism of Jordan algebras. Conclude that, if the Jordan algebra J2 is simple andJ1 6= 0, then J2 is special.

Remark. By a theorem of McCrimmon [81], if J is simple, then so is J2. Moreover, if c 6= 0,1J ,then J1 6= 0 by Cor. 34.3 (b). Thus for a simple Jordan algebra J and an idempotent c 6= 0,1J inJ, the Jordan algebra J2(c) is special.

pr.PETRCO 178. Show for a Peirce triple T := (i j, lm,np) that the following conditions are equivalent.

(i) T is not connected.(ii) Always up to the identification of 34.10, either i, j∩l,m= /0 or

T = (i j, jm, jp), m 6= i, j, p.

Conclude that a Peirce triple (i j, jl, i j) is connected if and only if l = i or l = j, in which case thereis a unique positive integer m such that i j = lm.

pr.MUPEAL 179. The multiple Peirce decomposition for alternative algebras (cf. Schafer [114]). Let A be aunital alternative algebra over k and Ω := (c1, . . . ,cr) a complete orthogonal system of idempotentsin A.


(a) Put

Ei j := Ei j(Ω) := Lci Rc j (1≤ i, j ≤ r) (2) ELCEAR

and show that (Ei j)1≤i, j≤r is a family of orthogonal projections of A whose sum is the iden-tity. Conclude

A =⊕

1≤i, j≤r

Ai j (3) PECOAL

as a direct sum of k-modules, where Ai j := Ai j(Ω) = Im(Ei j) for 1 ≤ i, j ≤ r. The Ai j arecalled the Peirce components of A relative to Ω .

(b) Show

Ai j = x ∈ A | clx = δilx, xcl = δ jlx (1≤ l ≤ r) (4) DEPECO

for 1≤ i, j ≤ r.(c) Prove that the Peirce components of A relative to Ω satisfy the following composition rules,

for all i, j, l,m = 1, . . . ,r.

Ai jA jl ⊆ Ail , (5) AIJAJM

A2i j ⊆ A ji, (6) AIJAIJ

Ai jAlm = 0(

j 6= l, (i, j) 6= (l,m))

(7) AIJALM

Finally, prove x2 = 0 for all x ∈ Ai j , 1≤ i, j ≤ r, i 6= j.(d) Note by Exc. 153 (b) that Ω is a complete orthogonal system of idempotents in A(+), with

Peirce components A(+)i j := A(+)

i j (Ω) (1≤ i≤ j ≤ r), and prove

A(+)ii = Aii (1≤ i≤ r), A(+)

i j = Ai j⊕A ji (1≤ i < j ≤ r).

pr.CONCOM 180. Connectedness in complete orthogonal systems of idempotents. (Jacobson [45, Prop. 5.3.3])Let Ω = (c1, . . . ,cr) be a complete orthogonal system of idempotents in J, with Peirce componentsJi j = Ji j(Ω), 1≤ i, j ≤ r. Fix indices i, j = 1, . . . ,r with i 6= j and prove:

(a) For vi j ∈ Ji j , the following conditions are equivalent.

(i) vi j ∈ J2(ci + c j)×.

(ii) Uvi j J×ii ⊆ J×j j , Uvi j J

×j j ⊆ J×ii .

(iii) Uvi j ci ∈ J×j j , Uvi j c j ∈ J×ii .

In this case, ci and c j are said to be connected by vi j . We say ci and c j are connected ifvi j ∈ Ji j exists connecting ci and c j . And finally, we say Ω is connected if ci and c j areconnected, for all i, j = 1, . . . ,r distinct.

(b) For vi j ∈ Ji j , the following conditions are equivalent.

(i) v2i j = ci + c j .

(ii) Uvi j ci = c j , Uvi j c j = ci

In this case, ci and c j are said to be strongly connected by vi j . We say ci and c j are stronglyconnected if vi j ∈ Ji j exists strongly connecting ci and c j . And finally, we say Ω is stronglyconnected if ci and c j are strongly connected, for all i, j = 1, . . . ,r distinct.

(c) Let 1 ≤ l ≤ r and assume i 6= l 6= j. If ci and c j are (strongly) connected by vi j ∈ Ji j , andc j and cl are (strongly) connected by v jl ∈ J jl , then ci and cl are (strongly) connected byvi j v jl ∈ Jil .


pr.ISTCOM 181. Isotopes and complete orthogonal systems of idempotents. (Jacobson [45, Prop. 5.3.4]) LetΩ = (c1, . . . ,cr) be a complete orthogonal system of idempotents in J, with Peirce componentsJi j := Ji j(Ω), 1≤ i, j ≤ r. Following Cor. 34.12 (b), consider the subalgebra of J defined by

DiagΩ (J) :=r⊕

i=1

Jii (8) DIAGOM

as a direct sum of ideals. For 1≤ i≤ r, let pi ∈ J×ii and put

p :=r

∑i=1

pi ∈ DiagΩ (J)× ⊆ J×. (9) DIAINV

(a) Put c(p)i := p−1

i ∈ Jii (the inverse of pi in Jii) for 1≤ i≤ r and show that

Ω(p) := (c(p)

1 , . . . ,c(p)r )

is a complete orthogonal system of idempotents in the isotope J(p) such that

J(p)i j := (J(p))i j(Ω

(p)) = Ji j (10) PECOP

for 1≤ i, j ≤ r. In particular, DiagΩ (p) (J(p)) = DiagΩ (J)(p).

(b) Let also qi ∈ J×ii for 1≤ i≤ r and q := ∑ri=1 qi ∈DiagΩ (J)×. Then show (Ω (p))(q) = Ω (Upq).

(c) Assume c1 and ci are connected for 2 ≤ i ≤ r (cf. Exc. 180). Show that there exist pi ∈ J×ii ,1≤ i≤ r, such that c(p)

1 and c(p)i with p := ∑

rj=1 p j are strongly connected in their capacity

as members of the complete orthogonal system Ω (p) of idempotents in J(p).

pr.LIFCOM 182. Lifting of complete orthogonal systems of idempotents. Let ϕ : J→ J′ be a surjective homo-morphism of Jordan algebras and suppose Ker(ϕ) ⊆ J is a nil ideal. Show that every completeorthogonal system (c′1, . . . ,c

′r) of idempotents in J′ can be lifted to J: there exists a complete or-

thogonal system (c1, . . . ,cr) of idempotents in J such that ϕ(ci) = c′i for 1≤ i≤ r.

pr.CLIFSIM 183. Simple Jordan algebras of Clifford type. (cf. Jacobson-McCrimmon [48, Theorem 11]) Let(M,q,e) be a non-zero pointed quadratic module over a field F and suppose q is non-degenerate.Show that the Jordan algebra J(M,q,e) is either simple or isomorphic to (F⊕F)(+).

Chapter 6Cubic Jordan algebras

c.CUNOS

Just as octonions fit naturally into the more general picture of composition algebras,the most efficient approach to Albert algebras consists in viewing them as specialcases of cubic Jordan algebras. These, in turn, are best understood by means of cubicnorm structures which, therefore, will have to be studied first. In the subsequentsections, we derive their most basic properties before we will be able to investigatecubic Jordan algebras in general, and Albert algebras in particular, over arbitrarycommutative rings.

35. From cubic norm structures to cubic Jordan algebras

s.CUNOJAThe natural habitat of cubic Jordan algebras is provided by the concept of a cubicnorm structure, which will be introduced and investigated in the first half of thepresent section by closely following the treatment of McCrimmon [77]. This inves-tigation will then be used in the second half to initiate the study of cubic Jordanalgebras. We conclude the section by showing that the categories of cubic normstructures and of cubic Jordan algebras are isomorphic.

Throughout k stands for an arbitrary commutative ring. Relaxing on the conven-tions of the previous section, we do not distinguish anymore between the notationfor a polynomial law f : M→ N and that for its induced set maps f = fR : MR →NR,R ∈ k-alg. We begin with a preliminary concept of a more technical nature thatwill be included here only for convenience.

ss.CUAR35.1. Cubic arrays. By a cubic array over k we mean a k-module X together with

• a distinguished element 1X ∈ X (the base point),• a quadratic map ] = ]X : X → X , x 7→ x] (the adjoint, dependence on X most of

the time being understood),• a cubic form NX : X → k (the norm)

such that the following conditions are fulfilled.

(i) 1X ∈ X is unimodular.(ii) The base point identities hold:

1]X = 1X , NX (1X ) = 1. (1) BAID

Given another cubic array X ′ over k, a homomorphism from X to X ′ is defined asa linear map ϕ : X → X ′ preserving base points, adjoints and norms: ϕ(1X ) = 1X ′ ,ϕ(x]X ) = ϕ(x)]X ′ for all x ∈ X , and NX ′ ϕ = NX as polynomial laws over k. In

263

264 6 Cubic Jordan algebras

this way, we obtain the category of cubic arrays over k, denoted by k-cuar. ForR ∈ k-alg, the R-module XR together with the base point 1XR := (1X )R ∈ XR, theadjoint ]= ]XR =(]X )R : XR→XR defined as the R-quadratic extension of the adjointof X , and the norm NXR := NX ⊗R : XR→ R as a cubic form over R is a cubic arrayover R, called the scalar extension or base change of X from k to R.

ss.TRACU35.2. Traces of cubic arrays. Let X be a cubic array over k. We write

x× y = (D])(x,y) = (x+ y)]− x]− y] (1) BILAD

for the bilinearization of the adjoint and, with regard to the norm, combine the no-tational simplifications of Exc. 58 with (13.14.2) and Cor. 13.20 to obtain not only

NX (x,y) = (DNX )(x,y) = (D2NX )(y,x) = (∂yNX )(x) (2) DICU

but also

NX (x,y,z) = (∂y∂zNX )(x) = (Π (1,1,1)NX )(x,y,z) (3) TOLINO

for the total linearization of the cubic form NX , which is therefore trilinear andtotally symmetric in x,y,z (cf. 13.8 (c)). Recall further from Exc. 58 that

NX (x+ y) = NX (x)+NX (x,y)+NX (y,x)+NX (y), (4) EXPNO

NX (x,y,z) = NX (x+ y,z)−NX (x,z)−NX (y,z). (5) LINOFI

The linear map

TX : X −→ k, y 7−→ TX (y) := NX (1X ,y), (6) LITR

is called the linear trace of X , while the quadratic form

SX : X −→ k, y 7−→ SX (y) := NX (y,1X ) (7) QUATR

is called the quadratic trace of X . And finally, we define the bilinear trace of X asthe symmetric bilinear form TX : X×X → k given by

TX (y,z) = (∂yNX )(1X )(∂zNX )(1X )− (∂y∂zNX )(1X ) (8) BILTR

= NX (1X ,y)NX (1X ,z)−NX (y,z,1X ),

which up to a sign agrees with the logarithmic Hessian of NX at 1X :

TX (y,z) =−(∂y∂zlogNX )(1X ) =−∂y

(∂zNX

NX

)(1X ). (9) LOHES

In view of (5)−(7), we may rewrite (8) as

TX (y,z) = TX (y)TX (z)−SX (y,z). (10) TEES

35 From cubic norm structures to cubic Jordan algebras 265

Combining all this with Euler’s differential equation (13.14.3) and (35.1.1), we ob-tain

TX (1X ) = SX (1X ) = 3, TX (y,1X ) = TX (y) (11) TESONE

since (5) and (7) yield SX (y,1X ) = NX (y,1X ,1X ) = N(1X ,1X ,y) = 2N(1X ,y), hence

SX (y,1X ) = 2TX (y). (12) ESLI

If ϕ : X → X ′ is a homomorphism of cubic arrays, then the chain rule (13.15.4) and(5) yield

NX ′(ϕ(x),ϕ(y)

)= NX (x,y), NX ′

(ϕ(x),ϕ(y),ϕ(z)

)= NX (x,y,z). (13) ENEXFI

Combining this with (6)−(8) and the property of ϕ to preserve base points, weconclude that ϕ preserves (bi-)linear and quadratic traces:

TX ′(ϕ(y)

)= TX (y), TX ′

(ϕ(y),ϕ(z)

)= TX (y,z), SX ′

(ϕ(y)

)= SX (y). (14) TEPHI

ss.REGEL35.3. Regular elements and non-singularity. Let X be a cubic array over k.(a) An element x ∈ X is said to be regular if NX (x) ∈ k is invertible.(b) X is said to be non-singular if it is finitely generated projective as a k-moduleand its bilinear trace is non-singular as a symmetric bilinear form, i.e., if it inducesa linear isomorphism from X onto its dual module X∗ in the usual way.Both notions defined in (a), (b), respectively, are invariant under base change: ifx ∈ X is regular, then so is xR ∈ XR, and if X is non-singular, then so is XR, for allR ∈ k-alg.

ss.CUNO35.4. The concept of a cubic norm structure. By a cubic norm structure overk we mean a cubic k-array X satisfying the following identities strictly, i.e., in allscalar extensions:

1× x = TX (x)1X − x (unit identity), (1) CUNI

NX (x,y) = TX (x],y) (gradient identity), (2) CGRADI

x]] = NX (x)x (adjoint identity). (3) CADJI

A homomorphism of cubic norm structures is defined as a homomorphism of themas cubic arrays. Thus cubic norm structures over k form a full subcategory, denotedby k-cuno, of k-cuar. By definition, cubic norm structures are stable under basechange, so if X is a cubic norm structure over k, then the cubic array XR over R is,in fact, a cubic norm structure, for all R ∈ k-alg.

ss.CUNOSU35.5. Cubic norm substructures. Let X be a cubic array over k. By a cubic sub-array of X we mean a cubic array Y such that Y ⊆ X is a k-submodule and the inclu-sion i : Y → X is a homomorphism of cubic arrays. This is equivalent to requiring


(i) 1X ∈ Y , (ii) Y ] ⊆ Y , i.e., Y is stabilized by the adjoint of X , and (iii) NX i = NYas polynomial laws over k, i.e., (NX )R iR = (NY )R as set maps YR → R, for allR ∈ k-alg. Thus any submodule Y of X , with corresponding inclusion i : Y → X ,that contains 1X and is stabilized by the adjoint of X may canonically be regarded asa cubic subarray of X , with base point 1Y = 1X , adjoint ] : Y → Y given by restrict-ing the adjoint ] : X → X of X to Y , and norm NY := NX |Y := NX i. In this case,the (bi-)linear and quadratic trace of Y by (35.2.14) is just the corresponding objectof X restricted to Y (resp. to Y ×Y ).

Now suppose X is a cubic norm structure over k and let Y ⊆X be a cubic subarray.Then it follows either from Cor. 13.9 or from Exc. 186 that Y is, in fact, a cubic normstructure, called a cubic norm substructure of X . If E ⊆ X is an arbitrary subset,the smallest cubic norm substructure of X containing E is called the cubic normsubstructure generated by E.

Our next aim will be to derive a number of basic identities for cubic norm structures.In order to do so, we require two preparations, the first one connecting cubic normstructures with para-quadratic algebras, the second one spelling out a sufficient con-dition for all regular elements of a cubic array to be unimodular.

ss.COPAR35.6. Connecting with para-quadratic algebras. Let X be a cubic array over k.We define a U-operator, i.e., a quadratic map U : X → Endk(X), x 7→Ux, by

Uxy := TX (x,y)x− x]× y (x,y ∈ X), (1) CUNOP

which in analogy to (30.1.3), (30.1.5) linearizes to the associated triple product

xyz :=Vx,yz :=Ux,zy = TX (x,y)z+TX (y,z)x− (z× x)× y (2) CUNTRI

for x,y,z∈X . Moreover, if X is a cubic norm structure, then the unit identity (35.4.1)shows U1X = 1X . Hence the k-module X together with the U-operator (1) and thebase point 1X ∈ X forms a para-quadratic algebra over k, which we denote by J(X)and will show in due course to be Jordan algebra. Passing from X to J(X) is clearlycompatible with base change: J(XR) = J(X)R for all R ∈ k-alg.

l.REUNIM35.7. Lemma. Let X be a cubic array over k such that the identity

xx]y= 2NX (x)y

holds strictly. Then every regular element of X is unimodular.

Proof. Since 1X ∈X is unimodular by definition, there exists a linear form λ : X→ ksuch that λ (1X ) = 1. Now suppose x ∈ X is regular. Thanks to Euler’s differentialequation combined with the hypothesis of the lemma, the linear form

λx : X −→ k, y 7−→ NX (x)−1(NX (x,y)−λ (yx]1X))

satisfies λx(x) = 1. Hence x is unimodular.


ss.BACUNO35.8. Basic identities for cubic norm structures. Let X be a cubic norm structureover k. Using the abbreviations 1 := 1X , N := NX , T := TX , S := SX , we claim thatthe following identities hold strictly in X (in compiling the list below, we do nothesitate to include identities that have been introduced or derived earlier).


1] = 1, N(1) = 1, (1) BAPO

T (1) = S(1) = 3, T (y,1) = T (y), S(y,1) = 2T (y), (2) TSONE

1× x = T (x)1− x, (3) UNI

N(x,y) = T (x],y), (4) GRADI

x]] = N(x)x, (5) ADJI

N(x+ y) = N(x)+T (x],y)+T (x,y])+N(y), (6) EXPF

T (x× y,z) = T (x,y× z), (7) DGRAD

x]× (x× y) = N(x)y+T (x],y)x, (8) DADJ

(x× y)× (x× z)+ x]× (y× z) = (9) DDADJ

= T (x],y)z+T (x],z)y+T (x× y,z)x,

x]× y]+(x× y)] = T (x],y)y+T (x,y])x, (10) SADJ

T (x],x) = 3N(x), (11) EUL

S(x) = T (x]), (12) ESTESH

S(x,y) = T (x× y) = T (x)T (y)−T (x,y), (13) BQUAT

x]× x =(T (x)S(x)−N(x)

)1−S(x)x−T (x)x], (14) CADJ

Uxy = T (x,y)x− x]× y, (15) UNOP

Ux(x× y) = T (x],y)x−N(x)y, (16) UBADJ

xx]y= 2N(x)y, (17) TRISH

N(x]) = N(x)2, (18) NADJ

x× (x]× y) = N(x)y+T (x,y)x], (19) ADJD

(Uxy)] =Ux]y] = T (x],y])x]−N(x)x× y], (20) UADJ

N(Uxy) = N(x)2N(y), (21) JCOMP

x] = x2−T (x)x+S(x)1, (22) ADJJA

x× y = x y−T (x)y−T (y)x+(T (x)T (y)−T (x,y)

)1, (23) LADJJA

T (x y) = 2T (x,y), (24) TEXCY

(x× y)× (z×w)+ (y× z)× (x×w)+(z× x)× (y×w)

= T (x× y,z)w+T (y× z,w)x+T (z×w,x)y (25) LDDADJ

+T (w× x,y)z,

x×(y× (x× z)

)= T (x,y)x× z+T (x],z)y+T (y,z)x]− (x]× y)× z (26) UTI

= (Uxy)× z+T (x],z)y+T (y,z)x],

x×(y× (z×w)

)+ z×

(y× (w× x)

)+w×

(y× (x× z)

)= T (x,y)z×w+T (z,y)w× x+T (w,y)x× z (27) LUTI

+T (x× z,w)y,

x× (x× y) =(T (x)S(x,y)+S(x)T (y)−T (x],y)

)1− (28) LCADJ

S(x,y)x−S(x)y−T (x)x× y−T (y)x]− x]× y,

N(x× y) = T (x],y)T (x,y])−N(x)N(y), (29) LNADJ

xyz=Vx,yz =Ux,zy = T (x,y)z+T (y,z)x− (z× x)× y, (30) EXTRI

T (Uxy,z) = T (y,Uxz), (31) TRAU

T(xyz,w

)= T

(z,yxw

), (32) TRATRI

T (x y,z) = T (x,y z). (33) TRACI


Proof. It clearly suffices to verify these identities for all x,y,z,w ∈ X . Moreover,thanks to Prop. 13.22, we may always assume if necessary that some of the vari-ables involved are regular. Now, while the base point identities (1) are valid evenin arbitrary cubic arrays and (3)−(5) belong to the very definition of a cubic normstructure, (2) has been observed already in (35.2.11), (35.2.12). Next, combining(35.2.4) with the gradient identity (4), we obtain (6). Differentiating the gradientidentity gives T (x× y,z) = N(x,y,z), which is totally symmetric in x,y,z. Hence(7) follows. To establish (8), we differentiate the adjoint identity (5)by making useof the chain and the product rule. Linearizing still further and using (7) yields(9), while (10) follows from (5) combined with the second order chain and prod-uct rules (13.15.6), (13.15.7). To derive (11), we combine the gradient identity (4)with Euler’s differential equation (13.3). The gradient identity (4) for y = 1 and(2) imply (12). To establish (13), we linearize (12) and apply (35.2.10). Similarly,specializing y = 1 in (8) and observing (3), we obtain (14), while (15) is just arepetition of (35.6.1). To derive (16), we combine (7) with (1), (8) and concludeUx(x× y) = T (x,x× y)x− x] × (x× y) = T (x× x,y)x−N(x)y− T (x],y)x, hence(16) since x× x = 2x]. Next, (2), (8) and (11) immediately imply (17). Combiningthis with Lemma 35.7, we can therefore conclude that all regular elements of X areunimodular. To establish (18), we assume that x is regular, apply the adjoint identity(5) and obtain

N(x])x] = x]]] = N(x)2x]. (34) SHAIT

Taking adjoints again, we deduce N(x])2x = N(x)4x, hence N(x])2 = N(x)4 ∈ k×

since x is unimodular. This shows that x] is regular, and (18) follows from (34).Turning to (19), we again assume that x is regular, replace x by x] in (8), observe (18)and obtain N(x)x× (x]× y) = N(x])y+N(x)T (x,y)x] = N(x)(N(x)y+ T (x,y)x]),hence (19). Now (20) follows by expanding the left-hand side and observing (19),(10) as well as the adjoint identity (5). In (21), we may assume that Uxy is regularsince U11 = 1. Then (20) and the adjoint identity yield N(Uxy)Uxy = (Uxy)]] =Ux]]y

]] = N(x)2N(y)Uxy, so the assertion follows from the unimodularity of Uxy. Toderive (22), we note that x2 =Ux1 = T (x,1)x− x]×1, whence the unit identity (3)and (12) lead to the desired conclusion. (23) follows immediately from linearizing(22) and observing (13). Combining (23) with (13), we deduce

T (x)T (y)−T (x,y) = T (x× y) = T (x y)−2T (x)T (y)+3(T (x)T (y)−T (x,y)

),

hence (24). Next, linearizing (9), we obtain (w× y)× (x× z)+ (x× y)× (w× z)+(x×w)× (y× z) = T (x×w,y)z+T (x×w,z)y+T (x×y,z)w+T (w×y,z)x, and (7)yields (25). Similarly, (19) and (4) imply z×(x]×y)+x×((x×z)×y) = T (x],z)y+T (z,y)x] + T (x,y)x× z, hence the first part of (26), while the remaining one nowfollows from (15). Linearizing (26) gives x× (y× (w× z)) +w× (y× (x× z)) =T (w,y)x×z+T (x,y)w×z+T (x×w,z)y+T (y,z)x×w−((x×w)×y)×z, and afterinterchanging w with z, we obtain (27). Similarly, (14) implies (x×y)×x+x]×y =(T (y)S(x)+T (x)S(x,y)−T (x],y))1−S(x,y)x−S(x)y−T (x)x× y−T (y)x], hence(28). Relation (29) is less obvious. With independent indeterminates s, t we apply


(18) and compare coefficients of s3t3 in

N(s2x]+ stx× y+ t2y]) = N((sx+ ty)]

)= N(sx+ ty)2

=(s3N(x)+ s2tT (x],y)+ st2T (x,y])+ t3N(y)

)2.

Expanding the very left-hand side by means of Exc. 58, (6), we conclude

N(x× y)+T(x]× (x× y),y]

)= 2N(x)N(y)+2T (x],y)T (x,y]),

where the second summand on the left by (8) and the Euler differential equation (11)agrees with N(x)T (y,y])+T (x],y)T (x,y]) = 3N(x)N(y)+T (x],y)T (x,y]). Hence(29) holds. (30) has already been noted in (35.6.2). To derive (31), we expand theleft-hand side by (15) and, using (7), obtain the expression T (Uxy,z) = T (T (x,y)x−x]×y,z) = T (x,y)T (x,z)−T (x],y×z), which is symmetric in y,z; hence (31) holds.Similarly, (30) yields T (xyz,w) = T (x,y)T (z,w)+ T (y,z)T (x,w)− T ((z× x)×y,w), which by (7) remains unchanged under the substitution x↔ y, z↔ w. Thisyields (32), while (33) follows from (23) and the fact that the right-hand side of T (xy,z) = T (x×y,z)+T (x)T (y,z)+T (y)T (x,z)−T ((T (x)T (y)−T (x,y))1,z) = T (x×y,z)+T (x)T (y,z)+T (y)T (z,x)+T (z)T (x,y)−T (x)T (y)T (z) is totally symmetricin x,y,z.

After these preparations, we are ready for the first main result of this section.

35.9. Theorem (McCrimmon [77]). Let X be a cubic norm structure over k. Thent.CUNOJOthe para-quadratic algebra J(X) of 35.6 with base point 1X and U-operator givenby

Uxy = TX (x,y)x− x]× y (1) CUNUOP

is a Jordan algebra over k such that the identities

x3−TX (x)x2 +SX (x)x−NX (x)1X = 0 = x4−TX (x)x3 +SX (x)x2−NX (x)x, (2) UNIV

Uxx] = NX (x)x, Uxx]2 = NX (x)21X , (3) UNADJ

hold strictly in J(X). We call J(X) the Jordan algebra associated with or corre-sponding to X .

Proof. For J = J(X) to be a Jordan algebra we must show that the identities

U1X x = x, (4) UU

UUxyz =UxUyUxz, (5) FF

Uxyxz= xyUxz (6) IF

hold for all x,y,z ∈ X . Here (4) follows immediately from the unit identity (35.8.3).To establish (5), we abbreviate 1 := 1X , N :=NX , T := TX , S := SX and first combine(35.8.19) with (35.8.7) to conclude


T (x× y,x]× z) = N(x)T (y,z)+T (x],y)T (x,z) = T(x,(x]× z)× y

). (7) TADJD

Next we expand the left-hand side of (5), using the definition of the U-operator (1)and (35.8.20). A short computation gives

UUxyz =(T (x,y)T (x,z)−T (x]× y,z)

)Uxy

−T (x],y])x]× z+N(x)(x× y])× z.

Similarly, expanding the right-hand side of (5) and applying (35.8.7), (35.8.16),(35.8.19), we obtain

UxUyUxz =(T (x,y)T (x,z)−T (x]× y,z)

)Uxy+N(x)T (y],z)x

+N(x)T (x,z)y]− x]×(y]× (x]× z)

).

To establish (5), we therefore have to prove

x]×(y]× (x]× z)

)= N(x)T (y],z)x+N(x)T (x,z)y]

+T (x],y])x]× z−N(x)(x× y])× z.

But this follows immediately from (35.8.26) combined with the adjoint identity(35.8.5). Finally, we must prove (6), which is less troublesome. One simply expandsthe left-hand side, using (1), (35.6.2), and applies (35.8.16) to obtain

Uxyxz= T (x,y)Uxz+(T (x,y)T (x,z)−T (x]× y,z)

)x

−T (x,z)x]× y+N(x)y× z.

Similarly, one expands the right-hand side, using (35.8.19), and arrives at the sameexpression. Thus J is a Jordan algebra, and it remains to verify (2), (3). ObservingT (x,x) = T (x)2−2S(x) by (35.8.13), we further obtain, using (35.8.14),

x3 =Uxx = T (x,x)x− x]× x

=(T (x)2−2S(x)

)x−(T (x)S(x)−N(x)

)1+S(x)x+T (x)x],

which by (35.8.22) yields the first equation of (2). Now we are in a position totackle (3). From (1), (35.8.11) and (35.8.5) we deduce Uxx] = T (x,x])x− x]× x] =3N(x)x− 2x]] = 3N(x)x− 2N(x)x, giving the first equation of (3). As to the sec-ond, we apply the first, (35.8.22), (35.8.12), (35.8.5) and obtain Uxx]2 = Ux(x]]+T (x])x]−S(x])1) = N(x)x3 +S(x)N(x)x−T (x)N(x)x2, and the first equation of (2)completes the proof of (3)

Applying (3)) and again (35.8.22), we now conclude

x4 =Uxx2 =Uxx]+T (x)Uxx−S(x)Ux1

= N(x)x+T (x)x3−S(x)x2,

hence the second equation of(2). In both cases, strictness is clear.


35.10. Corollary (McCrimmon [77]). Let X be a cubic norm structure over k andc.REGINVJ = J(X) the Jordan algebra associated with X. An element x ∈ X is invertible in Jif and only if it is regular in X, i.e., NX (x) is invertible in k. In this case,

x−1 = NX (x)−1x], (x−1)] = NX (x)−1x, NX (x−1) = NX (x)−1. (1) CUINV

Proof. Put 1 := 1X , N := NX . If x ∈ J×, then (35.8.21) implies 1 = N(1) =N(Uxx−2) = N(x)2N(x−2), hence N(x) ∈ k×. Conversely, assume N(x) ∈ k× andput y := N(x)−1x]. From (35.9.3) we then deduce Uxy = x, Uxy2 = 1, hence x ∈ J×

and the first equation of (1). Taking adjoints, we immediately obtain the second.In order to prove the third, we again apply (35.8.21) to obtain N(x) = N(Uxx−1) =N(x)2N(x−1) and therefore N(x−1) = N(x)−1.

ss.ISCUNO35.11. Isotopes of cubic norm structures. Isotopes of Jordan algebras have acounterpart on the level of cubic norm structures which we now proceed to dis-cuss. Let X be a cubic norm structure over k. Given a regular element p in X andwriting p−1 for its inverse in J = J(X) (cf. Cor. 35.10), we claim that the k-module Xtogether with the base point 1X(p) := 1(p)

X ∈ X (p), the adjoint x 7→ x(],p) (a quadratic

map X → X), and the norm NX(p) := N(p)X (a cubic form X (p)→ k) defined respec-

tively by

1X(p) := 1(p)X := p−1, (1) ISOUN

x(],p) := NX (p)Up−1x] = NX (p)U−1p x], (2) ISOAD

NX(p)(x) := N(p)X (x) = NX (p)NX (x) (3) ISONO

in all scalar extensions is a cubic norm structure over k, denoted by X (p) and calledthe p-isotope of X. Moreover, the (bi-)linear trace TX(p) =: T (p)

X and the quadratic

trace SX(p) =: S(p)X of X (p) are given by

TX(p)(y,z) = T (p)X (y,z) = TX (Upy,z), (4) ISOBILT

TX(p)(y) = T (p)X (y) = TX (p,y), (5) ISOLINT

SX(p)(y) = S(p)X (y) = TX (p],y]) (6) ISOQUAT

for all for all y,z ∈ X. Indeed, dropping the subscripts for convenience, (35.10.1)combined with (35.8.17) and Lemma 35.7 shows that X (p) is a cubic array over kwhose bilinear trace by (35.2.8), (35.8.7), (1) and the gradient identity (35.8.4) hasthe form

T (p)(y,z) = N(p)(p−1,y)N(p)(p−1,z)−N(p)(p−1,y,z)

= N(p)2T (p−1],y)T (p−1],z)−N(p)T (p−1× y,z)

= T (p,y)T (p,z)−T (p]× y,z).


Hence (4) holds, as do (5), (6). The defining conditions, (35.8.3), (35.8.4), (35.8.5),of a cubic norm structure are now straightforward to check, using the relation y×(p)

z = N(p)U−1p (y× z) for the bilinearization of (], p). As an example, verifying the

adjoint identity, we apply (35.10.1), (35.8.20) and obtain

x(],p)(],p) = N(p)U−1p(N(p)Up−1x]

)]= N(p)3U−1

p Up−1]x]]

= N(p)3U−1p UN(p)−1 px]] = N(p)U−1

p Upx]]

= N(p)N(x)x = N(p)(x)x.

35.12. Proposition (McCrimmon [77]). Let X be a cubic norm structure over k.p.ITIS

(a) If p ∈ X is a regular element, then J(X (p)) = J(X)(p) is the p-isotope of theJordan algebra associated with X.

(b) If p,q ∈ X are regular, then so is Upq and (X (p))(q) = X (Upq).

Proof. (a) Both algebras live on the same k-module and have the same unit element,so it remains to show that they have the same U-operator as well. To do so, wewrite U ′ for the U-operator of J(X (p)), abbreviate T := TX and obtain, combining(35.8.20), (35.11.4)with the fundamental formula,

U ′xy = T (p)(x,y)x− x(],p)×(p) y = T (Upx,y)x−N(p)2U−1p((Up−1x])× y

)= T (Upx,y)x−U−1

p((Up]x

])× y)= T (Upx,y)x−U−1

p((Upx)]× y

)=U−1

p(T (Upx,y)Upx− (Upx)]× y

)=U−1

p UUpxy =U−1p UpUxUpy =UxUpy =U (p)

x y,

which is exactly what we had to prove.(b) The first part follows from (35.8.21) and Cor. 35.10, the rest from (a) and a

straightforward computation.

In view of the preceding results, particularly Thm. 35.9, it is natural to ask for anintrinsic characterization of those Jordan algebras over k that may be written in theform J(X), for some cubic norm structure X . The key to such a characterization,taking (35.8.21), (35.9.2) as a guide, is provided by the most important concept ofthe present chapter.

ss.COCUJO35.13. The concept of a cubic Jordan algebra. In analogy to the concept of aconic algebra as presented in 19.1, we define a cubic Jordan algebra over k as aJordan k-algebra J together with a cubic form NJ : J→ k (the norm) such that thefollowing conditions hold.

(i) 1J ∈ J is unimodular.(ii) The norm of J satisfies NJ(1J) = 1 and permits Jordan composition: the equa-

tion

NJ(Uxy) = NJ(x)2NJ(y) (1) JJCOMP


holds strictly in J.(iii) For all R ∈ k-alg and all x ∈ JR, the monic cubic polynomial

mJ,x(t) := NJ(t1JR − x) ∈ R[t]

satisfies the equations

mJ,x(x) = (tmJ,x)(x) = 0.

By definition, cubic Jordan algebras are invariant under base change. If J′ is anothercubic Jordan algebra over k, a homomorphism ϕ : J → J′ is defined as a homo-morphism of Jordan algebras preserving norms in the sense that NJ′ ϕ = NJ aspolynomial laws over k. In this way we obtain the category of cubic Jordan algebrasover k, denoted by k-cujo. Note that k-cujo is a subcategory of k-jord , the categoryof Jordan algebras over k, but not a full one, as we shall see in 35.22 below.

In order to make condition (iii) above more explicit, we imitate the procedure of35.2 to define the linear trace of J as the linear map

TJ : J −→ k, x 7−→ TJ(x) := NJ(1,x), (2) JLITR

as well as the quadratic trace of J as the quadratic form

SJ : J −→ k, x 7−→ SJ(x) := NJ(x,1). (3) JQUATR

Then condition (iii) above is equivalent to the strict validity of the equations

x3−TJ(x)x2 +SJ(x)x−NJ(x)1J = 0, (4) JUNVT

x4−TJ(x)x3 +SJ(x)x2−NJ(x)x = 0 (5) JUNVF

in J. Note that (4) implies (5) if 12 ∈ k since J is then a linear Jordan algebra. Fur-

thermore, (2), (3) combined with Euler’s differential equation imply

TJ(1J) = SJ(1J) = 3. (6) JTESONE

Inspired by (35.8.22), we define the adjoint of J as the quadratic map x 7→ x] fromJ to J given by

x] := x2−TJ(x)x+SJ(x)1, (7) JADJ

dependence on J being understood. The adjoint linearizes to

x× y := (x+ y)]− x]− y] = x y−TJ(x)y−TJ(y)x+SJ(x,y)1. (8) JBIADJ

Combining (7) with (6), we conclude 1]J = 1J , so the k-module J together with thebase point 1J , the adjoint ] and the norm NJ is a cubic array over k, denoted by X(J).In particular, we have the bilinear trace of X(J), which we call the bilinear trace of


J, denoted by TJ : J× J→ k. Summing up, we conclude that not only the norm andadjoint but also the (bi-)linear and quadratic trace of J and X(J) are the same.

Clearly, every homomorphism of cubic Jordan algebras preserves not only normsbut also (bi-)linear and quadratic traces. If there is no danger of confusion, we willalways extend the notational conventions spelled out for cubic arrays in 35.1 to cubicJordan algebras by writing their identity elements as 1, their norms as N, and their(bi-)linear and quadratic traces as T,S, respectively.

Before we can proceed, we need a lemma.l.ESTESH

35.14. Lemma. Let X be a cubic array over k such that the unit identity and theadjoint identity hold strictly in X:

1× x = TX (x)1X − x, x]] = NX (x)x. (1) UNIADJ

Then

SX (x) = TX (x]) (2) STCL

for all x ∈ X.

Proof. Applying the second order chain and product rules to the second equation of(1), we obtain

x]× y]+(x× y)] = NX (x,y)y+NX (y,x)x.

Putting y = 1X and combining with the first equation of (1), we obtain

SX (x)1X +TX (x)x = TX (x])1X − x]+(TX (x)1X − x

)]= TX (x])1X − x]+TX (x)21X −TX (x)

(TX (x)1X − x

)+ x]

= TX (x])1X +TX (x)x,

and since 1X is unimodular, the assertion follows. t.CHACUJ

35.15. Theorem. Let J be a cubic Jordan algebra over k. Then the cubic arrayX := X(J) of 35.13 is a cubic norm structure over k such that J = J(X).

Proof. We always drop the subscript “X” for convenience and then proceed in sev-eral steps, where we try to read some of the arguments in our previous results back-wards.10. An element x ∈ J is invertible if and only if x is regular in X. In this case,x−1 = N(x)−1x] and N(x−1) = N(x)−1. Indeed, assume first x ∈ J× Since N per-mits Jordan composition, we obtain 1 = N(1) = N(Uxx−2) = N(x)2N(x−2), henceN(x)∈ k×, and N(x) =N(Uxx−1) =N(x)2N(x−1) implies N(x−1) =N(x)−1. Beforeproving the converse of our assertion, assume for the time being that x is arbitrary.Combining (35.13.5) with (35.13.7) we obtain

Uxx] = N(x)x, (1) UXESH


while Thm. 32.1 and (35.13.4) yield UxUx] = N(x)21J , hence

Uxx]2 = N(x)21. (2) UXESC

Now suppose N(x)∈ k×. Then (1) and (2) show that y :=N(x)−1x] satisfies Uxy= x,Uxy2 = 1, hence x ∈ J× and y = x−1.20. The following identities hold strictly in J.

1× x = T (x)1− x, (3) JUNI

x]] = N(x)x, (4) JADJI

N(x]) = N(x)2, (5) JNADJ

S(x) = T (x]), (6) JESTESH

xx]y= 2N(x)y = x]xy, (7) JTRISH

Ux(x× y) = N(x,y)x−N(x)y, (8) JUBADJ

(Uxy)] =Ux]y], (9) JUADJ

N(Uxy,Uxz) = N(x)2N(y,z), (10) ENUX

N(x)Uxy = N(x],y)x−N(x)x]× y. (11) ENXU

We put y = 1 in (35.13.8) and obtain 1× x = 2x−T (x)1−3x+S(x,1)1, which im-plies (3) since S(x,1) = T (x)T (1)−T (x,1) = 2T (x) by (35.2.10) (35.13.6). Turningto (4), (5), Prop. 13.22 and 10 allow us to assume that x is invertible. Then 10 im-plies x]] = (N(x)x−1)] = N(x)2(x−1)] = N(x)2N(x−1)(x−1)−1 = N(x)x, hence (5).But then (1) implies N(x)4 = N(N(x)x) = N(Uxx]) = N(x)2N(x]), and (5) drops outas well. (6) follows from (3), (4) and Lemma 35.14. For the first part of (7) wecombine (35.13.7) with (32.4.4) and (35.13.4) and obtain

xx]y=Vx,x]y =(Vx,x2 −T (x)Vx,x +S(x)Vx

)y

=(Vx3 −T (x)Vx2 +S(x)Vx

)y =

(x3−T (x)x2 +S(x)x

) y

= N(x)1 y = 2N(x)y,

as desired. The second equation follows analogously. In order to derive (8), wedifferentiate (1) in the direction y, which yields N(x,y)x+N(x)y =Ux,yx]+Ux(x×y) = xx]y+Ux(x×y), and (8) follows from (7). In (9), we apply Prop. 13.22 to thepolynomial law g : J×J→ Endk(J) given by g(x,y) :=Uxy. Hence we may assumethat x and y are both invertible. Then so is Uxy by Prop. 33.3, and 10 combined with(33.3.2) implies

(Uxy)] = N(Uxy)(Uxy)−1 = N(x)2N(y)Ux−1y−1 =UN(x)x−1N(y)y−1 =Ux]y].

Equation (10) follows by fixing x and differentiating (35.13.1) at y in the di-rection z. Finally, in order to derive (11), we replace x by x] in (1) and obtainN(x)Ux]x = N(x)2x], hence Ux]x = N(x)x] first for x invertible and then in full


generality. Differentiating in the direction y, we conclude N(x,y)x]+N(x)x× y =Ux],x×yx+Ux]y = x]x(x× y)+Ux]y = 2N(x)x× y+Ux]y by (7). Thus

Ux]y = N(x,y)x]−N(x)x× y.

Here we replace x by x] to deduce N(x)2Uxy = N(x)N(x],y)x−N(x)2x× y, whichyields (11) first for x ∈ J× and the in full generality.30. For p ∈ J×, the k-module J together with the base point 1(p) := p−1, the adjointx 7→ x(],p) := N(p)Up−1x] and the norm N(p) := N(p)N : J→ k is a cubic array X (p)

whose linear and quadratic traces are given by

T (p)(x) = N(p)N(p−1,x), S(p)(x) = N(x, p]). (12) TEPE

Moreover, X (p) strictly satisfies the unit and adjoint identities:

1(p)×(p) x = T (p)(x)1(p)− x, x(],p)(],p) = N(p)(x)x, (13) UNEP

where ×(p) stands for the bilinearized adjoint of X (p). The straightforward verifica-tion that X (p) is a cubic array satisfying (12) is left to the reader. It therefore remainsto check (13). First of all, 10 and (8), (12) imply

1(p)×(p) x = N(p)Up−1(p−1× x) = N(p)N(p−1,x)p−1−N(p)N(p−1)x

= N(p)−1N(p],x)p−1− x = T (p)(x)1(p)− x,

while 10, (9), (4), (35.13.1) yield

x(],p)(],p) =(N(p)Up−1x]

)(],p)= N(p)Up−1

((N(p)Up−1x]

)])= N(p)Up−1

((N(p)−1Up]x

])])

= N(p)−1Up−1((Upx)]]

)= N(p)−1U−1

p(N(Upx)Upx

)= N(p)N(x)x = N(p)(x)x,

as claimed.40. We can now show that X is a cubic norm structure. In view of (3), (4), we onlyhave to verify the gradient identity. To this end, let p ∈ J× Then 40 and (35.14.2)imply S(p)(x) = T (p)(x(],p)). But S(p)(x) = N(x, p]) by (12), while (12) and (10)imply

T (p)(x(],p)) = N(p)N(

p−1,N(p)Up−1x]))= N(p)2N(Up−1 p,Up−1x]) = N(p,x]).

Thus N(x, p]) = N(p,x]), and in view of Prop. 13.22, we have shown that the iden-tity

N(x,y]) = N(y,x])

holds strictly in J. Differentiating in the direction y, we conclude


N(x,y× z) = N(y,z,x]).

Putting z = 1 and applying (35.13.3), (3), (35.1.10), (6), we therefore obtain

T (y)S(x)−N(x,y) = T (y)N(x,1)−N(x,y) = N(x,T (y)1− y) = N(x,y×1)

= N(y,1,x]) = N(x],y,1) = S(x],y) = T (x])T (y)−T (x],y)

= S(x)T (y)−T (x],y,)

and the adjoint identity is proved. It remains to show that J = J(X) is the Jordan al-gebra associated with X . In (11), the gradient identity implies N(x],y) = T (x]],y) =N(x)T (x,y), and the formula Uxy= T (x,y)x−x]×y drops out for x invertible, hencein full generality. But this means J = J(X).

c.CUJUNIMO35.16. Corollary. Every invertible element of a cubic Jordan algebra is unimodu-lar.

Proof. Let J be a cubic Jordan algebra over k Since Thm. 35.15 implies J = J(X)for some cubic norm structure X , the assertion follows from Cor. 35.10 combinedwith Lemma 35.7 and (35.8.17).

ss.TISCAT35.17. Towards an isomorphism of categories. Let J be a cubic Jordan algebraover k. Then Thm. 35.15 shows J = J(X(J)). Conversely, let X be a cubic normstructure over k. By Thm. 35.9 combined with (35.8.21), J(X), always consideredtogether with the norm of X : NJ(X) = NX , is a cubic Jordan algebra, and (35.8.22)implies X =X(J(X)). Now let ϕ : X→X ′ be a homomorphism of cubic norm struc-tures. Since ϕ preserves not only base points, adjoints and norms but also linear andquadratic traces, it is a homomorphism ϕ : J(X)→ J(X ′) of cubic Jordan algebras.Conversely let ϕ : J → J′ be a homomorphism of cubic Jordan algebras. Since ϕ

preserves units, norms, linear and quadratic traces, it follows from (35.13.7) thatit preserves adjoints as well. Thus ϕ : X(J)→ X(J′) is a homomorphism of cubicnorm structures. Summing up, we have shown the following result.

c.ISNOJO35.18. Corollary. The formalism set up in 35.17 yields an isomorphism of cate-gories between cubic norm structures and cubic Jordan algebras over k.

ss.CONOJO35.19. Convention. From now on we will use Cor. 35.18 to identify cubic normstructures and cubic Jordan algebras over k canonically. Thus the two terms willhenceforth be employed interchangeably.

ss.COCUAL35.20. Cubic alternative algebras. By a cubic alternative algebra over k we meana unital alternative algebra A over k together with a cubic form NA : A→ k (thenorm) such that the following conditions are fulfilled.

(i) 1A ∈ A is unimodular.(ii) NA(1A) = 1.


(iii) For all R ∈ k-alg and all x ∈ AR, the monic polynomial

mA,x(t) = NA(t1AR− x) ∈ R[t]

annihilates x:mA,x(x) = 0.

As in the case of cubic Jordan algebras, we then define the linear and the quadratictrace of A by

TA : A−→ k, x 7−→ TA(x) := NA(1A,x), (1) CUALLT

SA : A−→ k, x 7−→ SA(x) := NA(x,1A), (2) CUALQT

respectively. Condition (iii) may then be rephrased by saying that the equation

x3−TA(x)x2 +SA(x)x−NA(x)1A = 0 (3) CAHALT

holds strictly in A. If A′ is another cubic alternative k-algebra, a homomorphismϕ : A→ A′ of cubic alternative k-algebras is defined as a homomorphism of unitalk-algebras that preserves norms: NA′ ϕ = NA as polynomial laws over k. In thisway we obtain the category of cubic alternative algebras over k, denoted by k-cual.If A in the preceding discussion is even associative, then we speak of a cubic asso-ciative algebra over k. Cubic associative k-algebras may and always will be viewedcanonically as a full subcategory, denoted by k-cuas, of cubic alternative k-algebras.Similar conventions apply to commutative associative algebras.

A cubic alternative algebra A as above is said to be multiplicative if the equation

NA(xy) = NA(x)NA(y) (4) COALMU

holds strictly in A. By definition arbitrary cubic alternative algebras (resp. multi-plicative ones) are stable under base change. Moreover, combining Exc. 86 withExc. 192 below, it follows that cubic alternative algebras need not be multiplicative,and their norms need not be uniquely determined by the algebra structure alone.

p.CUALJO35.21. Proposition. Let A be a multiplicative cubic alternative algebra over k.Then A(+) together with the norm NA(+) := NA is a cubic Jordan algebra over kwhose linear and quadratic trace agree with those of A: TA(+) = TA, SA(+) = SA.Moreover, the bilinear trace of A(+) is given by TA(+)(x,y) = TA(xy) for all x,y ∈ A.

Proof. Equation (35.13.4) agrees with (35.20.3), while (35.13.5) follows after mul-tiplying (35.20.3) by x in A. Moreover, from (35.20.4) we deduce that N := NA =NA(+) permits Jordan composition: N(Uxy) = N(xyx) = N(x)2N(y) holds strictlyin J := A(+). Thus J together with N is a cubic Jordan algebra whose linear andquadratic trace by definition are the same as those of A. With T := TA (the lineartrace of A, hence of J) and T ′ := TJ (the bilinear trace of J), it remains to showT ′(x,y) = T (xy) for all x,y ∈ A. By Prop. 13.22, we may assume p := y ∈ A×. From(35.11.5) combined with Prop. 35.12 and Thm. 35.15 we therefore conclude that


J(p) is a cubic Jordan algebra with linear trace T (p) given by T (p)(x) = T ′(x, p) forall x ∈ A. On the other hand, since A is multiplicative, Rp : J(p) → J by 33.12 isan isomorphism of cubic Jordan algebras. As such it preserves linear traces, whichimplies T (xp) = T (p)(x) = T ′(x, p), as claimed.

e.SPLIQUJO35.22. Cubic structures on the split quadratic etale algebra. Let E := k⊕ k bethe split quadratic etale k-algebra as in 24.18. Define a cubic form N : C→ k by

N(r1⊕ r2) := r1r22 (1) NOKAKA

for R ∈ k-alg and r1,r2 ∈ R. If also s1,s2 ∈ R, then

N(r1⊕ r2,s1⊕ s2) = s1r22 +2r1r2s2,

hence

T (r1⊕ r2) = 2r1 + r2, S(r1⊕ r2) = 2r1r2 + r21. (2) TRKAKA

It follows immediately from (1) that N(1C) = 1 and N is multiplicative. Moreover,a straightforward verification shows

x3−T (x)x2 +S(x)x−N(x)1C = 0

for x = r1⊕ r2 ∈CR = R⊕R. Thus C together with the norm N is a cubic alternativealgebra over k, which we denote by

(k⊕ k)cub.

Note that

x] = r1r2⊕ r21 (3) KAKASH

for x = r1⊕ r2 ∈CR = R⊕R, which follows immediately from (2) and (35.8.22).On the other hand, we may define a cubic form N′ : C→ k by

N′(r1⊕ r2) = r1r22

for R ∈ k-alg and r1,r2 ∈ R. Arguing as before, it follows that C together with thenorm N′ is a cubic alternative algebra over k as well, which we denote by

(k⊕ k)opcub.

Thus the algebra structure alone of a cubic alternative algebra will in general notdetermine its norm uniquely, even if the underlying module is free of finite rank.Returning to our example, the conjugation of C is an automorphism of C but neitherone of (k⊕k)cub nor one of (k⊕k)op

cub; in fact, it is an isomorphism from the formerto the latter:

ιk⊕k : (k⊕ k)cub∼−→ (k⊕ k)op

cub.


In particular, k-cual is not a full subcategory of k-alt, and neither is k-cujo one ofk-jord, as may be seen from the present example by passing from C to C(+).

e.MATTHR35.23. Example: 3× 3-matrices. Mat3(k) is a cubic associative k-algebra withnorm (resp. linear trace) given by the ordinary determinant (resp. trace) of matricessince det(13,x) = tr(x) for all x ∈Mat3(k). On the other hand, the bilinear trace ofMat3(k)(+), viewed as a cubic Jordan algebra, may be read off from Prop. 35.21:T (x,y) = tr(xy). In particular, the cubic Jordan matrix algebra Mat3(k)(+) is non-singular. On the other hand, the adjoint of Mat3(k)(+) is the usual adjoint of matri-ces.

e.CUBET

35.24. Examples: cubic etale algebras. Let E be a unital commutative associativek-algebra that is finitely generated projective as a k-module and finite etale (in thesense of 22.18) as an algebra over k. Then N(x) := det(Lx) for x in every scalarextension of E, where L stands for the left multiplication of E, defines a cubic formN = NE : E → k which converts E into a multiplicative cubic alternative k-algebra;we speak of a cubic etale algebra in this context. The linear trace of E is then givenby T (x) = tr(Lx), while the bilinear one by Prop. 35.21 has the form T (xy) = tr(Lxy).It follows from 22.18, therefore, that E(+), viewed as a cubic Jordan algebra over k,is non-singular.

A particularly simple case is that of the split cubic etale k-algebra defined byE = k⊕ k⊕ k as a direct sum of ideals, with identity element 1E = 1⊕ 1⊕ 1 andnorm N = NE : E→ k given by

N(x) = ξ1ξ2ξ3 (1) NXXI

for x = ξ1⊕ ξ2⊕ ξ3 ∈ ER, R ∈ k-alg. The linear and quadratic trace of E have theform

T (x) = ξ1 +ξ2 +ξ3, (2) LITXXI

S(x) = ξ2ξ3 +ξ3ξ1 +ξ1ξ2, (3) QUATXXI

while the adjoint and the bilinear trace of E(+) are given by

x] = ξ2ξ3⊕ξ3ξ1⊕ξ1ξ2, (4) ADJXXI

T (x,y) = ξ1η1 +ξ2η2 +ξ3η3 (5) BILTXXI

for y = η1⊕η2⊕η3 ∈ ER. It is sometimes convenient to identify E canonically withthe algebra of 3×3-diagonal matrices over k:

E = Diag3(k) = diag(γ1,γ2,γ3) | γ1,γ2,γ3 ∈ k, (6) DIATH

viewed as a cubic commutative associative subalgebra of Mat3(k). For

Γ = diag(γ1,γ2,γ3) ∈ Diag3(k)


we have

N(Γ ) = det(Γ ) = γ1γ2γ3, (7) NGAM

T (Γ ) = tr(Γ ) = γ1 + γ2 + γ3, (8) TGAM

S(Γ ) = γ2γ3 + γ3γ1 + γ1γ2, (9) SGAM

Γ] = diag(γ2γ3,γ3γ1,γ1γ2). (10) SHGAM

Exercises.

pr.NSPCF 184. Non-singular pointed cubic forms (Springer [118], McCrimmon [77]). By a pointed cubicform over k we mean a k-module X together with a distinguished element 1 = 1X ∈ X (the basepoint) and a cubic form N = NX : X → k (the norm) such that N(1) = 1. We then define theassociated (bilinear) trace of X as the symmetric bilinear form T = TX : X×X → k given by

T (y,z) := N(1,y)N(1,z)−N(1,y,z) (y,z ∈ X).

A homomorphism of pointed cubic forms is a linear map preserving norms and base points. Apointed cubic form X with base point 1, norm N and trace T is said to be non-singular if X isfinitely generated projective as a k-module and T is non-singular as a symmetric bilinear form, i.e.,it induces an isomorphism from X onto its dual X∗ in the usual way.

Now let X be a non-singular pointed cubic form over k as above.

(a) Show that there is a unique map x 7→ x] from X to X satisfying the gradient identity

T (x],y) = N(x,y) (x,y ∈ X).

Conclude that X together with 1, ] and N is a cubic array satisfying the unit identity. Inparticular, X is a cubic norm structure if and only if the adjoint identity holds.

(b) Let ϕ : X → X ′ be a surjective homomorphism of pointed cubic forms. Show that ϕ is anisomorphism of cubic arrays.

pr.HOCUJO 185. Let ϕ : J→ J′ be a k-linear map of cubic Jordan algebras over k. Prove that ϕ is

(a) a homomorphism of cubic Jordan algebras if it preserves unit elements and adjoints,(b) an isomorphism of cubic Jordan algebras if J is non-singular and ϕ is surjective preserving

unit elements and norms.

pr.CHACUN 186. Show that a cubic array X over k is a cubic norm structure if and only if the identities

1× x = T (x)1− x, (11) RANI

N(x,y) = T (x],y), (12) RAGRA

x]] = N(x)x, (13) RADJI

x]× y]+(x× y)] = T (x],y)y+T (x,y])x, (14) QUAPO

x]× (x× y) = T (x],y)x+N(x)y. (15) CUPO

hold for all x,y,z ∈ X .

pr.RACUNO 187. Rational cubic norm structures (Tits-Weiss [123], Muhlherr-Weiss [88]). The following ex-ercise characterizes cubic norm structures in terms of conditions that avoid scalar extensions. By arational cubic norm structure over k we mean a k-module X together with

• a distinguished element 1 ∈ X (the base point),• a map X → X , x 7→ x] (the adjoint),


• a bilinear map X×X → X , (x,y) 7→ x× y (the bilinearized adjoint),• a symmetric bilinear form T : X×X → k (the bilinear trace),• a map N : X → k (the norm)

such that the following conditions are fulfilled.

(i) 1 ∈ X is unimodular.(ii) The following equations hold for all α ∈ k and all x,y,z ∈ X .

(αx)] = α2x], (16) HOAD

N(αx) = α3N(x), (17) HONO

(x+ y)] = x]+ x× y+ y], (18) ABESH

N(x+ y) = N(x)+T (x],y)+T (x,y])+N(y), (19) NOAB

T (x,x]) = 3N(x), (20) TEASH

x]] = N(x)x, (21) ASH

x]× y]+(x× y)] = T (x],y)y+T (x,y])x, (22) ASHBE

x]× (x× y) = N(x)y+T (x],y)x, (23) ASHT

1] = 1, (24) ONESH

1× y = T (1,y)1− y. (25) ONESHB

A homomorphism of rational cubic norm structures over k is defined as a linear map preservingbase points, (bilinearized) adjoints, traces and norms in the obvious sense. In this way we obtainthe category of rational cubic norm structures over k, denoted by k-racuno. Show for a rationalcubic norm structure X over k, with 1, ],×,T,N as above, that x 7→ x] is a quadratic map and thatthere exists a unique cubic form N : X → k which makes the k-module X together with 1, ], N acubic norm structure over k, written as X , having bilinear trace T and satisfying Nk = N. Concludethat the assignment X 7→ X on objects and the identity on morphisms yield an isomorphism ofcategories from k-racuno to k-cuno.

pr.GENTWO 188. (Bruhne [17]) Let X be a cubic norm structure over k. Show that the cubic norm substructureof X generated by arbitrary elements x,y ∈ X is spanned as a k-module by

1,x,x],y,y],x× y,x]× y,x× y],x]× y]. (26) TWOGEN

pr.NORID 189. Norm ideals. Let X be a cubic array over k. By a norm ideal of X we mean a k-submoduleI ⊆ X such that the following conditions hold.

(i) 1 and I are linearly separated, i.e., there exists a linear form λ : X → k such that λ (1) = 1and λ (I) = 0.

(ii) N(x) = 0 for all x ∈ I.(iii) I]+X× I ⊆ I.

Prove:

(a) If ϕ : X → X ′ is a homomorphism of cubic arrays, then Ker(ϕ)⊆ X is a norm ideal.(b) Conversely, if X is a cubic norm structure and I ⊆ X is a norm ideal, there is a unique way

of viewing the quotient module X1 := X/I as a cubic norm structure over k such that thecanonical projection π : X → X1 is a homomorphism of cubic norm structures with kernel I.(Hint. Show T (x,y) = 0 for all x ∈ I, y ∈ X and apply Exc. 187.)

pr.TRACUJO 190. Traceless cubic Jordan algebras. Let J be a cubic Jordan algebra over k which is traceless inthe sense that its bilinear trace vanishes identically. Show 3 = 0 in k and that J, viewed as a linearJordan algebra, is commutative alternative.


pr.CUBNIL 191. (Petersson-Racine [101], Loos [71]) Let J be a cubic Jordan algebra over k. Prove that anelement x ∈ J is nilpotent if and only if T (x),S(x),N(x) ∈ k are nilpotent. Conclude that the nilradical of J (Exc. 148) can be described as

Nil(J) = x ∈ J | ∀y ∈ J : T (x,y),T (x],y),N(x) ∈ Nil(k), (27) NILARB

Nil(J) = x ∈ J | ∀y ∈ J : T (x,y),N(x) ∈ Nil(k) (if12∈ k), (28) NILTWO

Nil(J) = x ∈ J | ∀y ∈ J : T (x,y),T (x],y) ∈ Nil(k) (if13∈ k), (29) NILTHR

Nil(J) = x ∈ J | ∀y ∈ J : T (x,y) ∈ Nil(k) (if16∈ k). (30) NILSIX

pr.CUADUN 192. In this exercise, cubic Jordan (resp. alternative) algebras will be constructed out of Jordanalgebras of Clifford type (resp. out of conic alternative algebras).

(a) Let (M,q,e) be a pointed quadratic module with trace t and conjugation u 7→ u over k, andlet J := J(M,q,e) be the corresponding Jordan algebra of Clifford type. Show that

J := k(+)⊕M

as a direct sum of ideals in the category of para-quadratic k-algebras is a cubic Jordan algebraover k, with norm N : J→ k given by

N(r⊕u) := rq(u) (R ∈ k-alg, r ∈ R, u ∈ J) (31) NOHAJ

Show further that the (bi-)linear and quadratic trace of J as well as its adjoint have the form

T (α⊕u,β ⊕ v) = αβ +q(u, v), (32) BITHAJ

T (α⊕u) = α + t(u), (33) LITHAJ

S(α⊕u) = αt(u)+q(u), (34) QITHAJ

(α⊕u)] = q(u)⊕α u (35) ADHAJ

for all α,β ∈ k, u,v ∈ J.(b) Let C be a conic alternative algebra over k. Show that

C := k⊕C

as a direct sum of ideals is a cubic alternative k-algebra, with norm N : C→ k given by (31),and trace and quadratic trace given by (33), (34), respectively. Show further that C arisesfrom C by adjoining a unit element and that C is multiplicative if and only if C is. Finally,explain the connection of this exercise with Example 35.22.

pr.PSUNOS 193. Cubic norm pseudo-structures (Petersson-Racine [101]). By a cubic norm pseudo-structureover k we mean a k-module X together with an element 1 ∈ X , a quadratic map x 7→ x] from Xto X and a cubic form N : X → k such that all conditions of 35.1 and 35.4 hold, with the possibleexception of the base point 1 being unimodular. As in 35.2, we can then speak of the (bi-)linearand the quadratic trace of X . Similarly, by a rational cubic norm pseudo-structure over k, we meana k-module X together with data 1, ],×,T,N as in Exc. 187 such that equations (16)−(25) of thatexercise hold for all α ∈ k, x,y,z ∈ X but the base point 1 ∈ X may fail to be unimodular.

(a) Show for a rational cubic norm pseudo-structure X over k that there exists a unique cu-bic form N : X → k making X together with base point and adjoint a cubic norm pseudo-structure.

(b) Let V,W be vector spaces over a field F and suppose V is finite-dimensional, with basis(e1, . . . ,en). Given a quadratic map v 7→ v] from V to V , with bilinearization (v,v′) 7→ v× v′,

36 Building up cubic norm structures 285

a symmetric bilinear map σ : V ×V →W and arbitrary elements w1, . . . ,wn ∈W , show thatthe following conditions are equivalent.

(i) There exists a homogeneous polynomial law µ : V →W of degree 3 such that µ(ei) =wi for 1≤ i≤ n and µ(v,v′) = σ(v],v′) holds strictly in V .

(ii) σ(e]i ,ei) = 3wi for 1≤ i≤ n and the trilinear map

V ×V ×V −→ (v1,v2,v3) 7−→ σ(v1× v2,v3),

is totally symmetric.

In this case, µ is unique.(c) Use (a), (b) to give an example of a cubic norm pseudo-structure X such that the relation

N(x]) = N(x)2 does not always hold and the bilinear trace of X is different from zero. (Hint.The proof of [101, Thm. 2.10] contains a gap that should be filled by means of (a).)

36. Building up cubic norm structures

s.BUCUNOIn this rather technical section, a number of basic tools will be provided that turnout to be of crucial importance in various constructions of cubic Jordan algebras.These tools are best described in the language of cubic norm structures. Accord-ingly, throughout this section, we fix an arbitrary commutative ring k and a cubicnorm structure X over k. Generally speaking, we will be concerned with the ques-tion of which additional ingredients are needed in order to understand X in terms ofa given cubic norm substructure of X . These ingredients are based on a number ofuseful identities that are our main focus in the present section. In order to describetheir range of validity in a precise manner, a peculiar conceptual framework will beneeded that we now proceed to consider.

ss.ORCOCU36.1. The orthogonal complement of a cubic norm substructure. Let X0 be acubic norm substructure of X and write

X⊥0 := u ∈ X | ∀x0 ∈ X0 : T (x0,u) = 0 (1) DOSH

for the orthogonal complement of X0 relative to the bilinear trace of X . From (35.8.7)we obtain a well defined bilinear action

X0×X⊥0 −→ X⊥0 , (x0,u) 7−→ x0 .u :=−x0×u (2) EXEXPE

that will be of fundamental importance later on. It seems worth noting, however,that the ingredients of this action are in general not compatible with base change:for R ∈ k-alg, the scalar extension X0R need not be a cubic norm substructure of X ,and even if it is, (X0R)

⊥, its orthogonal complement in XR relative to the bilineartrace, need not be the same as (X⊥0 )R. To remedy this deficiency, the followingconcept will be introduced.

ss.PUCUNO


36.2. Pure cubic norm substructures. Let X0 be a cubic norm substructure of Xand R∈ k-alg. If the inclusion i : X0 → X extends to an R-linear injection iR : X0R→XR, then X0 is said to be R-pure. In this case, X0R can and always will be canonicallyregarded as a cubic norm substructure of XR over R. Furthermore, even though the R-module (X0R)

⊥ may still be distinct from (X⊥0 )R, the extended element uR = u⊗1R ∈XR, for any u ∈ X⊥0 , belongs to (X0R)

⊥. We say that X0 is pure if it is R-pure, for allR ∈ k-alg.

If R is a flat k-algebra (so the functor −⊗R from k-modules to R-modules isexact [11, I,§2,Definition 1]), then every cubic norm substructure of X is R-pure. Onthe other hand, if X0 ⊆ X is a cubic norm substructure which is a direct summand ofX as a k-module, then X0 is pure.

ss.XXPER36.3. The action of X0 on X⊥0 . Let X0 be a cubic norm substructure of X and writeN0 for the norm of X0.. We claim that the following relations hold, for all R ∈ k-algmaking X0 R-pure and all x0,y0 ∈ X0R, all u ∈ X⊥0R.

1 .u = u, (1) UNISH

Ux0u = x]0 .u, (2) UACT

x0 .(y0 .(x0 .u)

)= (Ux0y0) .u, (3) UOSH

x0 .(x]0 .u) = N0(x0)u = x]0 .(x0 .u), (4) EXDOT

N(x0 .u) = N0(x0)N(u). (5) ENDOT

We may clearly assume R = k. Then (1) follows immediately from (36.1.2) and theunit identity (35.8.3), while the definition of the U-operator (35.8.15) combined with(36.1.2) and the relation T (x0,u) = 0 gives (2). To derive (3), we apply (35.8.26)with z = u, observe T (x]0,u) = T (y0,u) = 0 and obtain x0 .(y0 .(x0 .u)) = −x0 ×(y0× (x0×u)) =−(Ux0y0)×u = (Ux0y0) .u, hence (3). Since T (x0,u) = T (x]0,u) =0, we obtain (4) immediately from (36.1.2) and (35.8.7), (35.8.19). To prove (5), weobserve that (35.8.29) reduces to N(x0 .u) = −N(x0× u) = N(x0)N(u), and this is(5).

Relations (1), (3) above are equivalent to saying that the linear map σ : J0 :=J(X0)→ Endk(X⊥0 )(+) given by σ(x0)u := x0 .u for x0 ∈ J0, u ∈ X⊥0 is a homomor-phism of Jordan algebras. Thus if k = F is a field, J0 is simple as a Jordan algebraover F and X⊥0 6= 0, then J0 is special.

ss.SORT36.4. Strong orthogonality. With X0 as in 36.3, an element u ∈ X is said to bestrongly orthogonal (or strongly perpendicular) to X0 if u and u] both belong to X⊥0 .Following Petersson-Racine [105, 4.4], we call

X⊥⊥0 := u ∈ X | u ∈ X⊥0 , u] ∈ X⊥0 .

the strong orthogonal complement of X0 in X . Note that X⊥⊥0 ⊆ X will not be a k-submodule in general and may, in fact, be empty. On the other hand, if u ∈ X isstrongly orthogonal to X0, so is u] by the adjoint identity. We now claim that the


following relations hold, for all R ∈ k-alg making X0 R-pure, all x0,y0,z0 ∈ X0R andall u ∈ (X0R)

⊥⊥.

(x0 .u)] = x]0 .u], (1) SHDOT

(x0 .u)× (y0 .u) = (x0× y0) .u], (2) DOSHDO

(x0 .u)×u] =−N(u)x0 = (x0 .u])×u, (3) DOTSH

T (x0 .u,y0 .u) = 0, (4) EXVEVE

T (x0 .u,y0 .u]) = N(u)T0(x0,y0), (5) TDOTSH(x0 .(y0 .u)

)×u = T0(x0,y0)u]− y0 .(x0 .u]), (6) PUSHA(

x0 .(y0 .u))×u]+u×

(x0 .(y0 .u])

)= −N(u)x0 y0, (7) CIRC(

x0 .(y0 .u))× (x0 .u]) = −N(u)Ux0y0, (8) UXY(

x0 .(y0 .u))× (z0 .u])+

(z0 .(y0 .u)

)× (x0 .u]) = −N(u)x0y0z0, (9) LUXY(

x0 .(y0 .u))×u] = u×

(y0 .(x0 .u])

)= (x0 .u)× (y0 .u]), (10) XVYV

(x0 .u)×((x0 .u)× (y0 .u])

)= N(u)(Ux0y0) .u. (11) XVXVY

Proof. As before, we may clearly assume R= k. The relations T (x]0,u)= T (x0,u])=0 combined with (35.8.10) imply (1), which immediately yields (2) after lineariz-ing with respect to x0, while the first relation of (3) follows from (x0 .u)× u] =−(x0 × u)× u] and (35.8.8); the same argument replacing (35.8.8) by (35.8.19)also gives the second. To establish (4), (5), we apply (36.1.1), (35.8.7) and (2)to obtain T (x0 .u,y0 .u) = −T (x0 .u,y0 × u) = −T ((x0 .u)× u,y0) = −T ((x0 ×1) .u],y0) = 0 and T (x0 .u,y0 .u]) = T (x0 × u,y0 × u]) = T ((x0 × u)× u],y) =N(u)T0(x0,y0). Turning to (6), we put x := u, y := x0, z := y0 in (35.8.26) and obtainu×(x0 .(y0 .u)) = u×(x0×(u×y0)) = T (u,x0)u×y0+T (u],y0)x0+T (x0,y0)u]−(u]× x0)× y0 = −T (x0,u)y0 .u+T (y0,u])x0 +T (x0,y0)u]− y0 .(x0 .u]); since u isstrongly perpendicular to X0, (6) follows. Similarly, setting x := u], y := x0, z := y0,w := u in (35.8.27) yields

u]× (x0× (y0×u))+ y0× (x0× (u×u]))+u× (x0× (u]× y0))

= T (u],x0)y0×u+T (y0,x0)u×u]+T (u,x0)u]× y0

+T (u]× y0,u)x0.

But u×u]=−N(u)1 by (3), which implies T (u]×y0,u)=T (y0,u×u])=−N(u)T (y0),and we conclude(

x0 .(y0 .u))×u]−N(u)(1× x0)× y0 +u×

(x0 .(y0 .u])

)=−N(u)T (x0,y0)1−N(u)T (y0)x0,

where (1× x0)× y0 = T (x0)1× y0− x0× y0 = T (x0)T (y0)1− T (x0)y0− x0× y0.Thus

288 6 Cubic Jordan algebras(x0 .(y0 .u)

)×u]+u×

(x0 .(y0 .u])

)=−N(u)

(x0× y0 +T (x0)y0 +T (y0)x0−

(T (x0)T (y0)−T (x0,y0)

)1)

=−N(u)x0 y0

by (35.8.23). This proves (7). In (8) we may assume by Lemma 13.23 that x0 is in-vertible. Then (36.1.4) implies u= x0 .v, v= x−1

0 .u∈ X⊥⊥0 by (1), and using (36.3.3),(1), (36.3.4), (3), (36.3.5), we obtain(

x0 .(y0 .u))× (x0 .u]) =(

x0 .(y0 .(x0 .v)

))×(x0 .(x0 .v)])

= ((Ux0y0) .v)×(x0 .(x

]0 .v

]))

= N0(x0)((Ux0y0) .v)× v]

= −N0(x0)N(v)Ux0 y0 =−N(x0 .v)Ux0y0 =−N(u)Ux0y0,

hence (8), which implies (9) after linearization. Comparing (9) for z0 = 1 with (7)we obtain (x0 .(y0 .u))×u]+(y0 .u)×(x0 .u]) =−N(u)x0 y0 = (x0 .(y0 .u))×u]+u× (x0 .(y0 .u])), giving the second relation of (10) with x0 and y0 interchanged. Onthe other hand, linearizing (36.3.3) we obtain (x0 .(y0 .u))×u]+(y0 .(x0 .u))×u] =((x0 y0) .u)×u] =−N(u)x0 y0 by (3), and comparing with (7), we also obtain thefirst relation of (10) with x0 and y0 interchanged. Finally, to establish (11), we apply(35.8.28) to x0 .u,y0 .u] instead of x,y, respectively. Since x0 .u,y0 .u] ∈ X⊥⊥0 by (1),we have T (x0 .u) = T (y0 .u]) = S(x0 .u) = 0, T ((x0 .u)],y0 .u]) = T (x]0 .u

],y0 .u]) =0 by (4), S(x0 .u,y0 .u]) =−T (x0 .u,y0 .u]) =−N(u)T0(x0,y0) by (5), and (x0 .u)]×(y0 .u]) = (x]0 .u

])× (y0 .u]) = (x]0× y0) .u]] = N(u)(x]0× y0) .u by (1), (2). Hence(35.8.28) yields (x0 .u)× ((x0 .u)× (y0 .u])) = N(u)(T0(x0,y0)x0 − x]0 × y0) .u =N(u)(Ux0y0) .u, as desired.

ss.COMCUB36.5. Complemented cubic norm substructures. By a complemented cubic normsubstructure of X we mean a pair (X0,V ) such that X0 ⊆ X is a cubic norm substruc-ture, V ⊆ X is a k-submodule and the relations

X = X0⊕V, V ⊆ X⊥0 , X0 .V ⊆V (1) OSP

hold. This concept is clearly compatible with base change, so if (X0,V ) is a comple-mented cubic norm substructure of X , then (X0,V )R := (X0R,VR) is one of XR, forall R ∈ k-alg. In particular, X0 is pure.

r.EXTCOM36.6. Remark. If X0 ⊆ X is a non-singular cubic norm substructure, then we de-duce from Lemma 12.10 that (X0,X⊥0 ) is the unique complemented cubic norm sub-structure of X extended from X0. Conversely, suppose X itself is non-singular and(X0,V ) is a complemented cubic norm substructure of X . Then X0 is non-singularand V = X⊥0 .


For the remainder of this section, we fix a complemented cubic norm substruc-ture (X0,V ) of X and write N0 for the norm, T0 (resp. S0) for the (bi-)linear (resp.quadratic) trace of X0.

ss.SPLAD36.7. Splitting the adjoint. We use (36.5.1) to follow Springer’s approach totwisted compositions (Springer [119, p. 95], Springer-Veldkamp [121, 6.5], Knus-Merkurjev-Rost-Tignol [59, §38A, p. 527]) (where k is a field of characteristic not2) or Petersson-Racine [102, (3.1)] (with a sign change and X0 coming from a cubicetale k-algebra) to define quadratic maps Q : V → X0, H : V →V by

u] =−Q(u)+H(u) (u ∈V, Q(u) ∈ X0, H(u) ∈V ). (1) QUH

We claim that the identities

(x0 +u)] =(x]0−Q(u)

)+(− x0 .u+H(u)

), (2) SPLAD

N(x0 +u) = N0(x0)−T0(x0,Q(u)

)+N(u), (3) SPLINO

T (y0 + v,z0 +w) = T0(y0,z0)+T0(Q(v,w)

), (4) SPLIBILT

T (y0 + v0) = T0(y0), (5) SPLITR

S(x0 +u) = S0(x0)−T0(Q(u)

)(6) SPQATR

hold strictly for x0,y0,z0 ∈ X0, u,v,w ∈V . Indeed, N(x0 +u) = N0(x0)+T (x]0,u)+T (x0,u])+N(u) = N0(x0)− T0(x0,Q(u))+N(u) by (36.5.1), (1) and (x0 + u)] =x]0+x0×u+u] = x]0−x0 .u−Q(u)+H(u) by (36.1.2) and (1). Finally, T (y0+v,z0+w) = T0(y0,z0)+ T (v,w), and (35.8.13) yields T (v,w) = T (v)T (w)− T (v×w) =−T (v×w) = T (Q(v,w)−H(v,w)) = T0(Q(v,w)), giving (4), which immediatelyimplies first (5) and then (6).

ss.IDQH36.8. Identities for Q and H. The following identities hold strictly for all x0,y0,z0 ∈X0, u,v,w ∈V .


Q(x0 .u) =Ux0Q(u), (1) QXV

Q(x0 .u,y0 .u) =Ux0,y0Q(u), (2) QXYV

Q(x0 .u,x0 .v) =Ux0Q(u,v), (3) QXVW

Q(x0 .u,y0 .v)+Q(x0 .v,y0 .u) =Ux0,y0Q(u,v), (4) QXYVW

H(x0 .u) = x]0 .H(u), (5) HXV

H(x0 .u,y0 .u) = (x0× y0) .H(u), (6) HXYV

H(x0 .u,x0 .v) = x]0 .H(u,v), (7) HXVW

H(x0 .u,y0 .v)+H(x0 .v,y0 .u) = (x0× y0) .H(u,v), (8) HXYVW

Q(H(u)

)= Q(u)], (9) QHV

H(H(u)

)= N(u)u−Q(u) .H(u), (10) HHV

Q(x0 .u,H(u)

)= N(u)x0 = Q

(u,x0 .H(u)

), (11) QXVH

Q(H(u),v

)+Q

(u,H(u,v)

)= T0

(Q(H(u),v

))1, (12) QHVWT

H(x0 .u,H(u)

)= T0

(x0,Q(u)

)u−Q(u) .(x0 .u), (13) HXVT

H(u,x0 .H(u)

)=(x0×Q(u)

).u, (14) HVX

T0(x0,Q(u,v)

)= T0

(Q(x0 .u,v)

), (15) TXQV

Q(H(u,v)

)+Q

(H(u),H(v)

)= Q(u,v)]+Q(u)×Q(v), (16) QHVW

H(H(u,v)

)+H

(H(u),H(v)

)= T0

(Q(u,H(v)

))u+T0

(Q(H(u),v

))v− (17) HHVW

Q(u) .H(v)−Q(u,v) .H(u,v)−Q(v) .H(u),

N(H(u)

)= N(u)2−2N0

(Q(u)

). (18) NHV

Proof. (35.8.10) yields (x0 .u)] = (x0× u)] = T (x]0,u)u+ T (x0,u])x0− x]0× u] =−T0(x0,Q(u))x0 + x]0 ×Q(u)− x]0 ×H(u) = −Ux0Q(u) + x]0 .H(u), and compar-ing X0- and V -components by means of (36.7.1), we obtain (1),(5). (Repeatedly)linearizing (1) (resp. (5)) implies (2)−(4) (resp. (6)−(8)). Similarly, by the ad-joint identity and (36.7.2), N(u)u = u]] = (−Q(u)+H(u))] = Q(u)]−Q(H(u))+(Q(u) .H(u) + H(H(u))), which leads to (9),(10). Applying (35.8.8) we obtainN(u)x0−T0(x0,Q(u))u=N(u)x0+T (u],x0)u= u]×(u×x0)=−(x0 .u)×(−Q(u)+H(u)) = −Q(u) .(x0 .u)+Q(x0 .u,H(u))−H(x0 .u,H(u)), hence (13) and the firstrelation of (11). Similarly, (35.8.19) implies N(u)x0 = N(u)x0 + T (u,x0)u] = u×(u] × x0) = u× (−Q(u)× x0 + H(u)× x0) = (x0 ×Q(u)) .u− u× (x0 .H(u)) =(x0×Q(u)) .u+Q(u,x0 .H(u))−H(u,x0 .H(u)) implies (14) and the second rela-tion of (11). Linearizing (11) for x0 = 1 with respect to u and applying (36.7.4) givesQ(v,H(u)) + Q(u,H(u,v)) = T (u],v)1 = T (H(u),v)1 = T0(Q(H(u),v))1, hence(12). To derive (15), we apply (1) and (35.8.7) to obtain T0(x0,Q(u,v))=−T (x0,u×v) = −T (x0×u,v) = T (x0 .u,v) = T0(Q(x0 .u,v)) by (36.7.4). Next (16), (17) willfollow from (9), (10), respectively, by using the second order chain and productrules of the differential calculus for polynomial laws; we omit the details. And fi-


nally, turning to (18), we combine (35.8.18) with (36.7.1),(36.7.3),(9) and obtain

N(u)2 = N(u]) = N(−Q(u)+H(u)

)= −N0

(Q(u)

)+T0

(Q(u),Q

(H(u)

))+N

(H(u)

)= −N0

(Q(u)

)+T0

(Q(u),Q(u)]

)+N

(H(u)

)= −N0

(Q(u)

)+3N0

(Q(u)

)+N

(H(u)

)by Euler’s differential equation (35.8.11), and (18) follows.

ss.BUP36.9. The build-up. The identities derived in the preceding subsection provide theopportunity of building up new cubic norm structures out of old ones. Let X0 be acubic norm structure over k, with base point 1, adjoint x0 7→ x]0, norm N0, (bi-)lineartrace T0 and quadratic trace S0. Suppose we are given

(i) a k-module V ,(ii) a bilinear action X0×V →V , (x0,u) 7→ x0 .u,(iii) quadratic maps Q : V → X0, H : V → V such that Q(u,H(u)) ∈ R1R for all

u ∈VR, R ∈ k-alg.

Since 1 ∈ X0 is a unimodular vector, condition (iii) yields a unique cubic formN : V → k such that

N(u)1R = Q(u,H(u)

)(u ∈VR,R ∈ k-alg). (1) NHAT

Put X := X0⊕V as a k-module, identify X0,V canonically as submodules of X anddefine a cubic form N : X → k as well as a quadratic map ] : X → X by requiringthat

(x0⊕u)] :=(x]0−Q(u)

)⊕(− x0 .u+H(u)

), (2) SHAEX

N(x0⊕u) := N0(x0)−T0(x0,Q(u)

)+ N(u) (3) NOEX

hold strictly for all x0 ∈ X0, u ∈ V . Inspecting (2), (3), we see that the k-module Xtogether with the base point 1, adjoint ] and norm N is a cubic array over k whoseadjoint bilinearizes to

(x0⊕u)× (y0⊕ v) =(x0× y0−Q(u,v)

)⊕(− x0 .v− y0 .u+H(u,v)

)(4) BISHAEX

for all x0,y0 ∈ X0, u,v ∈V . We also claim that the (bi-)linear and the quadratic traceof X are given by

T (y0⊕ v,z0⊕w) = T0(y0,z0)+T0(Q(v,w)

), (5) BITREX

T (y0⊕ v) = T0(y0) (6) LITREX

S(x0⊕u) = S0(x0)−T0(Q(u)

)(7) QUATREX

for x0,y0.z0 ∈ X0, u,v,w ∈V . Indeed, differentiating (3) implies


N(x0⊕u,y0⊕ v) = N0(x0,y0)−T0(x0,Q(u,v)

)−T0

(Q(u),y0

)+ N(u,v), (8) DINOTR

where putting x0 = 1, u = 0 (resp. y0 = 1, v = 0) yields (6) (resp. (7)). Now (5)follows by linearizing (5) and applying (35.2.10).

36.10. Proposition (Petersson-Racine [102, Lemma 3.3]) . For X as defined in 36.9p.EXCUNOto be a cubic norm structure over k it is necessary and sufficient that the followingidentities hold in all scalar extensions.

1 .u = u, (1) CUNISH

x]0 .(x0 .u) = N0(x0)u, (2) CEXDOT

Q(x0 .u) =Ux0Q(u), (3) CQXV

H(x0 .u) = x]0 .H(u), (4) CHXV

Q(H(u)

)= Q(u)], (5) CQHV

H(H(u)

)= N(u)u−Q(u) .H(u), (6) CHHV

Q(x0 .u,H(u)

)= N(u)x0, (7) CQXVH

Q(H(u),v

)+Q

(u,H(u,v)

)= T0

(Q(H(u),v

))1, (8) CQHVWT

T0(x0,Q(u,v)

)= T0

(Q(x0 .u,v)

), (9) CTXQV

H(x0 .u,H(u)

)= T0

(x0,Q(u)

)u−Q(u) .(x0 .u). (10) CHXVT

Proof. Assume first that X is a cubic norm structure. Then the set-up described in36.9 shows that (X0,V ) is complemented cubic norm substructure of X . Moreover,the identities (1)−(10), being a subset of the ones assembled in 36.3 and 36.8, holdstrictly for x0 ∈ X0, u,v ∈ V . Conversely, let this be so. We must show the unit,gradient and adjoint identity. The unit identity is the least troublesome since theunit identity for X0 combined with (36.9.2), (36.9.6) and (1) imply 1× (x0⊕ u) =1× x0 − 1 .u = T0(x0)1− x0 − u = T (x0 ⊕ u)1− (x0 ⊕ u). In order to derive thegradient identity, we differentiate (36.9.1) and combine the result with (8) to obtainN(u,v)1 = Q(v,H(u))+Q(u,H(u,v)) = T0(Q(H(u),v))1, hence

N(u,v) = T0

(Q(H(u),v

)). (11) DIHAN

Now (36.9.8), the gradient identity for X0 and (11), (9), (36.9.5), (36.9.2) imply


N(x0⊕u,y0⊕ v) = T0(x]0,y0)−T0

(x0,Q(u,v)

)−T0

(Q(u),y0

)+T0

(Q(H(u),v

))= T0

(x]0−Q(u),y0

)−T0

(Q(x0 .u,v)

)+T0

(Q(H(u),v

))= T0

(x]0−Q(u),y0

)+T0

(Q(− x0 .u+H(u),v

))= T

((x]0−Q(u)

)⊕(− x0 .u+H(u)

),y0⊕ v

)= T

((x0⊕u)],y0⊕ v

),

as claimed. Finally, we apply the adjoint identity for X0 and (36.9.2), (5), (3), (7),(2), (4), (10), (6), (35.8.15), (36.9.3) to derive

(x0⊕u)]] =((

x]0−Q(u))⊕(− x0 .u+H(u)

))]=((

x]0−Q(u))]−Q

(− x0 .u+H(u)

))⊕(−(x]0−Q(u)

).(− x0 .u+H(u)

)+H

(− x0 .u+H(u)

))=(

x]]0 − x]0×Q(u)+Q(u)]−Q(x0 .u)+Q(x0 .u,H(u)

)−Q

(H(u)

))⊕(

x]0 .(x0 .u)− x]0 .H(u)−Q(u) .(x0 .u)+Q(u) .H(u)+H(x0 .u)

−H(x0 .u,H(u)

)+H

(H(u)

))=(N0(x0)x0− x]0×Q(u)−Ux0Q(u)+ N(u)x0

)⊕(

N0(x0)u−T0(x0,Q(u)

)u+ N(u)u

)=(N0(x0)x0−T0

(x0,Q(u)

)x0 + N(u)x0

)⊕(

N0(x0)u−T0(x0,Q(u)

)u+ N(u)u

)= N(x0⊕u)(x0⊕u),

hence the adjoint identity for X .

ss.PASIST36.11. Passing to isotopes. Let X0 be a cubic norm substructure of X and p aregular element of X0. Then X (p)

0 , the p-isotope of X0, is a cubic norm substructureof X (p), and (35.11.4) shows

X (p)⊥0 = X⊥0 (1) PPERP

as k-submodules of X . Moreover, the natural action of X (p)0 on X (p)⊥

0 as defined in(36.1.2) is given by the formula

x0 ·(p) u = p .(x0 .u) (2) EXPEPE


for x0 ∈ X (p)0 and u ∈ X (p)⊥

0 . Indeed, (35.11.2) and (36.3.2) imply

x0 ·(p) u = − x0×(p) u =−N(p)Up−1(x0×u) = N(p)Up−1(x0 .u)

= N(p−1)−1 p−1] .(x0 .u) = (p−1)−1 .(x0 .u) = p .(x0 .u),

as claimed. Similarly, one checks that u ∈ X (p) is strongly orthogonal to X (p)0 if and

only if it is strongly orthogonal to X0, in which case

u(],p) = p .u]. (3) SHAPER

ss.COCUIS36.12. Complemented cubic norm substructures under isotopy. Let (X0,V ) bea complemented cubic norm substructure of X and p a regular element of X0. From(36.5.1), (36.11.1) and (36.11.2) we conclude that (X0,V )(p) := (X (p)

0 ,V ) is a com-plemented cubic norm substructure of X (p). Writing Q(p),H(p) for the analogues ofQ,H as defined in 36.7, with (X0,V ) replaced by (X0,V )(p), we claim

Q(p)(u) = N(p)Up−1Q(u), H(p)(u) = p .H(u) (u ∈V ). (1) IQUH

Indeed, applying (36.7.1), (35.11.2), (36.1.2) we get

−Q(p)(u)+H(p)(u) = u(],p) = N(p)Up−1u] =−N(p)Up−1Q(u)+N(p)Up−1H(u)

= −N(p)Up−1Q(u)+N(p−1)−1(p−1)] .H(u)

= −N(p)Up−1Q(u)+(p−1)−1 .H(u)

= −N(p)Up−1Q(u)+ p .H(u),

and comparing the components in X0,V , respectively, the assertion follows.

Exercises.

pr.EICO 194. Cubic solutions to the eiconal equation (Tkachev [124]). Let k be a commutative ring contain-ing 1

6 . By an eiconal triple over k we mean a triple (V,Q,N) consisting of a quadratic space (V,Q)over k and a cubic form N : V → k such that the quadratic map H : V →V uniquely determined bythe strict validity of

Q(H(u),v

)=

13

N(u,v) (2) HADJ

in V ×V strictly satisfies the eiconal equation

Q(H(u)

)= Q(u)2 (3) EICEQ

in V . Prove:

(i) If (V,Q,N) is an eiconal triple over k, then the k-module X := k⊕V together with the basepoint 1 ∈ X , the adjoint X → X , x 7→ x] and the norm N : X → k respectiveoy defined by

37 Elementary idempotents and cubic Jordan matrix algebras 295

1 := 1⊕0, (4) EICBA

(r⊕u)] =(r2−Q(u)

)⊕(− ru+H(u)

), (5) EICAD

N(r⊕u) := r3−3rQ(u)+N(u) (6) EICNO

for R ∈ k-alg, r ∈ R, u ∈VR, is a non-singular cubic norm structure over k.(ii) Conversely, let X be a non-singular cubic norm structure over k. Then X0 := k = k1⊆ X is

a non-singular cubic norm substructure, and if we put V := X⊥0 to define Q : V → k by thecondition Q(u)+u] ∈V for all u ∈V , then (V,Q,N|V ) is an eiconal triple over k.

(iii) The constructions presented in (i), (ii) are inverse to each other.(iv) If (Rn,Q,N) is an eiconal triple over the field of real numbers, then the eiconal equation (3)

takes on the co-ordinate form

n

∑i, j=1

qi j ∂N∂xi

(u)∂N∂x j

(u) = 9( n

∑i, j=1

qi juiu j)2 (7) EICCO

for

u =

u1...

un

∈ Rn,

where Q = (qi j) ∈ Symn(R)∩GLn(R) corresponds to the quadratic form Q and Q−1 =(qi j).

37. Elementary idempotents and cubic Jordan matrix algebras

s.ELIDMAElementary idempotents have been a useful tool in our study of composition alge-bras. As will be seen in the present section, the natural extension of this conceptto cubic Jordan algebras turns out to be even more momentous. After introduc-ing the concept itself, we describe the Peirce decomposition relative to elementaryidempotents, and to complete orthogonal systems thereof, called elementary frames,purely in terms of cubic norm structures. We then proceed to define cubic Jordanmatrix algebras, containing the diagonal elementary frame as a particularly usefulconstituent. This is underscored by the Jacobson co-ordinatization theorem, to beproved at the very end of this section, which basically says that every cubic Jor-dan algebra containing an elementary frame which is connected (cf. Exc. 180) isisomorphic to a cubic Jordan matrix algebra.

Throughout this section, we let k be a commutative ring and, unless other ar-rangements have been made, let J be a cubic Jordan algebra over k, with adjointx 7→ x], norm N = NJ , (bi-)linear trace T = TJ and quadratic trace S = SJ .

ss.COELID37.1. The concept of an elementary idempotent. An element e ∈ J is called anelementary idempotents if e] = 0 and T (e) = 1. In this case, S(e) = T (e]) = 0,and (35.8.22) implies e2 = e, so e is indeed an idempotent and a unimodular oneat that. In particular, the adjoint identity yields N(e)e = e]] = 0, hence N(e) = 0.We also have T (e,e) = T (e)2−2S(e) by (35.8.13), hence T (e,e) = 1. Note by thebase point identities (35.8.1) that 1 = 1J is an elementary idempotent if and only if


k = 0 (hence J = 0). The property of being an elementary idempotent is clearlypreserved by homomorphisms and scalar extensions of cubic Jordan algebras.

37.2. Proposition (cf. Racine [108], Petersson-Racine [104]). Let e be an elemen-p.PELEtary idempotent of J.(a) The complementary idempotent f = 1− e satisfies f ] = e and T ( f ) = 2.(b) The Peirce components of J relative to e are orthogonal with respect to T andcan be described as

J2(e) = ke, (1) PETWO

J1(e) = x ∈ J | T (x) = 0, e× x = 0, (2) PEONE

J0(e) = x ∈ J | e× x = T (x) f − x. (3) PEZERO

(c) x] = S(x)e for all x ∈ J0(e).

Proof. (a) f ] = (1− e)] = 1]−1× e+ e] = 1−T (e)1+ e = e and T ( f ) = T (1)−T (e) = 3−1 = 2.

(b) Let xi ∈ Ji := Ji(e) for i = 0,1,2. We must show T (xi,x j) = 0 for i, j = 0,1,2distinct. First of all, Thm. 34.2 and (35.8.31) yield T (x2,x1 + x0) = T (Uex2,x1 +x0) = T (x2,Ue(x1+x0)) = 0, hence T (x2,x1) = T (x2,x0) = 0. Similarly, T (x0,x1) =T (U f x0,x1) = T (x0,U f x1)) = 0 since Ji( f ) = J2−i by Cor. 34.3 (b). This proves thefirst part of (b).

As to the explicit description of the Peirce components, we have J2(e) = Im(Ue)by (34.2.4) and Uex = T (e,x)e− e]× x = T (e,x)e ∈ ke for x ∈ J, hence J2(e)⊆ ke.Here the relation Uee = e3 = e gives equality. To prove (2), (3), we apply (35.8.23)to obtain e×x = ex−T (e)x−T (x)e+(T (e)T (x)−T (e,x))1= ex−x+T (x) f −T (e,x)1, hence

e x = x+ e× x−T (x) f +T (e,x)1 (x ∈ J). (4) EEEX

But x ∈ J1 if and only if e x = x by (34.2.7), and from (4) we conclude

x ∈ J1 ⇐⇒ e× x = T (x) f −T (e,x)1. (5) CHARONE

Now suppose x ∈ J1. Then T (e,x) = T ( f ,x) = 0 by the first part, hence T (x) =T (e,x)+T ( f ,x) = 0, and (5) yields e× x = 0 as well. Conversely, suppose e× x =0 and T (x) = 0. Taking traces of the first equation, we conclude 0 = T (e× x) =T (e)T (x)− T (e,x) = −T (e,x), and (5) shows x ∈ J1. This proves (2). Turning to(3), let x ∈ J0. Then e x = 0 by (34.2.6) and T (e,x) = 0 by the first part, forcinge× x = T (x) f − x by (4). Conversely, if this relation holds, then T (x) = 2T (x)−T (x) = T (x)T ( f )−T (x) = T (e× x) = T (e)T (x)−T (e,x) = T (x)−T (e,x), whichimplies T (e,x) = 0 and then T (x) = T (1,x) = T (e,x)+T ( f ,x) = T ( f ,x). Thus (a)gives U f x = T ( f ,x) f − f ]× x = T (x) f − e× x = x, forcing x ∈ J0, and the proof of(b) is complete.


(c) For x ∈ J0 = J2( f ) we have x =U f x, and (35.8.20) combined with (a) impliesx] = (U f x)] = U f ]x

] = Uex]. Hence x] ∈ J2 = ke, and we find a scalar α ∈ k withx] = αe. Taking traces, (c) follows.

37.3. Corollary (Faulkner [28, Lemma 1.5]). With the notation and assumptionsc.FAULEof Prop. 37.2, write M0 for J0(e) as a k-module and S0 for the restriction of S toM0. Then (M0,S0, f ) is a pointed quadratic module over k and J0(e) = J(M0,S0, f )as Jordan algebras. Moreover, trace and conjugation of (M0,S0, f ) are respectivelygiven by the restriction of T to J0(e) and by the assignment x 7→ T (x) f − x.

Proof. The relation S0( f ) = T ( f ]) = T (e) = 1 shows that (M0,S0, f ) is a pointedquadratic module over k with trace x 7→ S0( f ,x) = T ( f ×x) = T ( f )T (x)−T ( f ,x) =2T (x)− T (x) = T (x) since T (e,x) = 0 by Prop. 37.2 (b). Hence the trace of(M0,S0, f ) is as indicated and, consequently, so is its conjugation, denoted by x 7→ x.Moreover, for x,y ∈ M0, we compute S0(x,y) = T (x× y) = T (x)T (y)− T (x,y) =T (x, f )T (y)− T (x,y) = T (x,T (y) f − y) = T (x, y), which combines with (37.2.3)and Prop. 37.2 (c) to imply that the U-operator of J0 agrees with Uxy = T (x,y)x−x]× y = S0(x, y)x−S0(x)e× y = S0(x, y)x−S0(x)(T (y) f − y) = S0(x, y)x−S0(x)y,hence with the U-operator of J(M0,S0, f ).

In our subsequent computations, we conveniently subscribe to what we callss.TERCYC

37.4. The ternary cyclicity convention. Unless explicitly stated otherwise, in-dices i, j, l (or m,n, p) are always tacitly assumed to vary over all cyclic permu-tations (i jl) (or (mnp)) of (123). For example, given arbitrary elements xi jl of anarbitrary k-module, this convention allows us to write ∑xi jl for x123 + x231 + x312.

p.COMPOR37.5. Proposition. Elementary idempotents e1,e2 ∈ J are orthogonal if and only ife1×e2 = 1−e1−e2 =: e3. In this case, (e1,e2,e3) is a complete orthogonal systemof elementary idempotents in J, and with the corresponding Peirce decompositionJ = ∑(Jii + J jl), the following statements hold.(a) ei× e j = el .(b) Jii = kei.(c) The Peirce components of J relative to (e1,e2,e3) are orthogonal with respect tothe bilinear trace.(d) The linear trace of J vanishes on J23 + J31 + J12.(e) x y = x× y for all x ∈ Ji j, y ∈ J jl .(f) x2 =−S(x)(e j + el) for all x ∈ J jl .

Proof. The very first assertion follows from (37.2.3) and the following chain ofequivalent conditions.

e1,e2 are orthogonal ⇐⇒ e2 ∈ J0(e1)

⇐⇒ e1× e2 = T (e2)(1− e1)− e2

⇐⇒ e1× e2 = e3.


In this case, T (e3) = T (1)−T (e1)−T (e2) = 3−1−1 = 1, and applying (35.8.10)we obtain e]3 = (e1× e2)

] = T (e]1,e2)e2 + T (e1,e]2)e1− e]1× e]2 = 0. Hence e3 is

an elementary idempotent, making (e1,e2,e3) a complete orthogonal system of thedesired kind.

(a) is now clear by symmetry.(b) This is just (34.11.2) and (37.2.1).(c), (d) By (34.11.4) we have Ji j ⊆ J1(ei)∩J0(el). Therefore T (ei,Ji j)=T (el ,Ji j)=

T (Ji j,J jl) = 0, giving (c) and T (Ji j) = T (∑el ,Ji j) = 0, hence (d).(e) By (c), (d), T (x) = T (y) = T (x,y) = 0, and (35.8.23) yields the assertion.(f) We have x ∈ J0(ei) by (34.11.2), and Cor. 37.3 combined with (d) implies

x2 = T (x)x−S(x)(1− ei) =−S(x)(e j + el).

ss.COELF37.6. The concept of an elementary frame. Adapting the terminology of Loos[67, 10.12] to the present set-up, we define an elementary frame of J to be a completeorthogonal system of elementary idempotents of length 3 in J. By Prop. 37.5, anelementary frame of J has length 3 unless k = 0, in which case J = 0 and thelength can be arbitrary.

If (e1,e2,e3) is an elementary frame of J, then we write

J = ∑(Jii + J jl) (1) ELPE

for the corresponding Peirce decomposition. Combining Prop. 37.2 (b) with (34.11.2),(34.11.3), we conclude

Jii = kei, J jl = x ∈ J | T (x) = 0, e j× x = el× x = 0. (2) PECOEL

Elementary idempotents are intimately tied up with a particularly important class ofcubic Jordan algebras, called cubic Jordan matrix algebras. Since the formal defini-tion of this concept looks rather artificial at first, we begin by describing its intuitivebackground.

In Exc. 180, we have introduced the concept of (strong) connectedness for or-thogonal systems of idempotents. In the present set-up, this concept allows a naturalcharacterization by means of data belonging exclusively to the underlying cubicnorm structure.

p.CONEL

37.7. Proposition. Let (e1,e2,e3) be an elementary frame of J with the Peirce de-composition J = ∑(kei + J jl). For elements u jl ∈ J jl , the following conditions areequivalent.

(i) e j and el are connected (resp. strongly connected) by u jl .(ii) S(u jl) ∈ k× (resp. S(u jl) =−1).

Proof. By Cor. 37.3, we have J2(e j +el) = J0(ei) = J(M0,S0, f ), where M0 = J0(ei)as k-modules, S0 = S|M0 and f = 1− ei = e j + el . Now e j and el are connectedby u jl if and only if u jl ∈ J2(e j + el)

×, which by Exc. 168 happens if and only if


S(u jl) = S0(u jl) ∈ k×. On the other hand, u2jl =−S(u jl)(e j +el by Prop. 37.5 (f), so

e j and el are strongly connected by u jl if and only if S(u jl) =−1.

ss.TWIMAT37.8. Twisted matrix involutions. Let C be a conic algebra over k whose con-jugation is an involution, and which is faithful as a k-module, allowing us toidentify k ⊆ C canonically as a unital subalgebra. If m is a positive integer, thenMatm(k) ⊆ Matm(C) by Exc. 35 is a nuclear subalgebra. Twisting the conjugatetranspose involution of Matm(C) in the sense of 11.8 by means of a diagonal matrix

Γ = diag(γ1, . . . ,γm) ∈ GLm(k) (1) TWIGAM

in the sense of 11.9, we therefore conclude that the map

Matm(C)−→Matm(C), x 7−→ Γ−1xt

Γ (2) GATWIS

is an involution, called the Γ -twisted conjugate transpose involution of Matm(C).The elements of Matm(C) remaining fixed under this involution are called Γ -twistedhermitian matrices, and simply hermitian matrices if Γ = 1m is the m×m unitmatrix. The Γ -twisted hermitian matrices of Matm(C) having diagonal entries in kform a k-submodule of Matm(C) which we denote by

Herm(C,Γ ); (3) HERCEG

in particular, we put

Herm(C) := Herm(C,1m). (4) HERC

If 12 ∈ k, then Herm(C,Γ ) is the totality of all Γ -twisted hermitian matrices in

Matm(C), the condition on the diagonal entries being automatic since H(C, ιC) =k1C = k by (19.6.3).

Now assume m = 3. Writing ei j, 1 ≤ i, j ≤ 3 for the ordinary matrix unitsof Mat3(k) ⊆ Mat3(C), we obtain a natural set of generators for the k-moduleHer3(C,Γ ) by considering the quantities

u[ jl] := γlue jl + γ juel j (u ∈C, j, l = 1,2,3 distinct). (5) UJAYL

Indeed, a straightforward verification shows that x∈Mat3(C) belongs to Her3(C,Γ )if and only if it can be written in the form (necessarily unique)

x = ∑(ξieii +ui[ jl]) (ξi ∈ k, ui ∈C, i = 1,2,3). (6) ELHERTH

Thus we have a natural identification

Her3(C,Γ ) = ∑(keii⊕C[ jl]) = (k⊕C)⊕ (k⊕C)⊕ (k⊕C) (7) HERTHMO

as k-modules.


With a few extra hypotheses on C, we wish to define a cubic norm structure onthe k-module Her3(C,Γ ) in a natural way that commutes with base change. Actu-ally, we will be able to do so without any conditions on C as a k-module, and withoutassuming that the diagonal matrix Γ ∈Mat3(k) be invertible. This will be accom-plished by formalizing the preceding set-up in a slightly different manner. We beginby introducing a convenient terminology.

ss.COPA37.9. Co-ordinate pairs. By a pre-co-ordinate pair over k we mean a pair (C,Γ )consisting of a multiplicative conic alternative k-algebra C and a diagonal matrixΓ ∈ Mat3(k). In this case, it will always be tacitly assumed that Γ has the formΓ = diag(γ1,γ2,γ3), with γi ∈ k, 1 ≤ i ≤ 3. If Γ is invertible, we speak of a co-ordinate pair over k. For Γ = 13, we identify C = (C,13) and refer to this as aco-ordinate algebra over k. If C is an octonion algebra, the terms octonionic (pre-)co-ordinate pair, octonionic co-ordinate algebra, respectively, will be used, dittofor C being a quaternion or quadratic etale k-algebra.

ss.THERCU37.10. Towards a hermitian cubic norm structure. Let (C,Γ ) be a pre-co-ordinate pair over k. By Prop. 20.2, C is norm-associative (i.e., the identities(19.11.1)−(19.11.5) hold in C) and its conjugation is an involution. Guided by theformulas of 37.8, but abandoning their interpretation by means of hermitian matri-ces, we consider the k-module of all formal expressions

∑(ξieii +ui[ jl])

for ξi ∈ k, ui ∈C and i = 1,2,3. Thanks to the formal character of these expressions,and in analogy of (37.8.7), this k-module, denoted by X(C), may be written as

X(C) = ∑(keii +C[ jl]) = (k⊕C)⊕ (k⊕C)⊕ (k⊕C), (1) FORHER

and hence its dependence on C is compatible with base change: X(C)R = X(CR)for all R ∈ k-alg. Note, however, that the diagonal matrix Γ ∈Mat3(k) has not yetentered the scene, which will happen only after we have given X(C) the structure ofa cubic array over k. In order to do so, we consider elements

x = ∑(ξieii +ui[ jl]), y = ∑(ηieii + vi[ jl]) (2) ELFOHE

of X(C)R, with xi,ηi ∈ R, ui,vi ∈CR for i = 1,2,3, to define base point, adjoint andnorm on X(C) by the formulas

1 = 1X := ∑eii, (3) MABA

x] = ∑

((ξ jξl− γ jγlnC(ui)

)eii +

(−ξiui + γiu jul

)[ jl]), (4) MADJ

N(x) := NX (x) = ξ1ξ2ξ3−∑γ jγlξinC(ui)+ γ1γ2γ3tC(u1u2u3), (5) MANO

the very last expression on the right of (5) being unambiguous by (19.12.1). Onechecks easily that these formulas make X(C) a cubic array over k, which we denote


by Her3(C,Γ ) and which is clearly compatible with base change: Her3(C,Γ )R =Her3(CR,ΓR) for all R ∈ k-alg. Moreover, the adjoint bilinearizes to

x× y = ∑

((ξ jηl +η jξl− γ jγlnC(ui,vi)

)eii (6) MABADJ

+(−ξivi−ηiui + γiu jvl + v jul

)[ jl]),

and we claim that the (bi-)linear and quadratic trace of X are given by

T (x,y) = ∑(ξiηi + γ jγlnC(ui,vi)

), (7) MABIT

T (x) = ∑ξi, (8) MALIT

S(x) = ∑(ξ jξl− γ jγlnC(ui

)). (9) MAQIT

Indeed, differentiating (5) at x in the direction y, we obtain, using (19.12.1),

N(x,y) = ∑

(ξ jξlηi− γ jγl

(ηinC(ui)+ξinC(ui,vi)

)+ γ1γ2γ3tC(uiu jvi)

), (10) DIMANO

and setting x = 1 gives (8), while setting y = 1 yields (9). On the other hand, lin-earizing (9) we deduce

S(x,y) = ∑(ξ jηl +η jξl− γ jγlnC(ui,vi)

). (11) MABQIT

Finally, combining (11) with (8) and (35.2.10), we end up with (7).

37.11. Theorem (Freudenthal [32, 33], McCrimmon [77]) . Let (C,Γ ) be a pre-t.CUJOMAco-ordinate pair over k. Then the cubic array Her3(C,Γ ) of 37.10 is a cubic normstructure ocer k.

Proof. The unit (resp. gradient) identity follows from (37.10.3), (37.10.6), (37.10.8)(resp. (37.10.4), (37.10.7), (37.10.11)) by a straightforward verification. It remainsto prove the adjoint identity over k. To this end we put

x] = ∑(ξ ]i eii +u]i [ jl]), (1) EXXISH

where ξ]i ∈ k and u]i ∈C can be read off from (37.10.4). We must show ξ

]]i = N(x)ξi

and u]]i = N(x)ui. We begin with the former by repeatedly applying (37.10.4):

ξ]]i = ξ

]j ξ

]l − γ jγlnC(u

]i )

=(ξlξi− γlγinC(u j)

)(ξiξ j− γiγ jnC(ul)

)− γ jγlnC(−ξiui + γiu jul).

Expanding the terms on the right-hand side of the last equation and combining(19.5.7), (20.1.1), (19.5.5) with the fact that the expression tC(u1u2u3) is invariantunder cyclic permutations of its arguments, we conclude


ξ]]i = ξ1ξ2ξ3ξi− γiγ jξlξinC(ul)− γlγiξiξ jnC(u j)+ γ1γ2γ3γinC(u j)nC(ul)

− γ jγlξ2i nC(ui)+ γ1γ2γ3ξinC(ui,u jul)− γ1γ2γ3γinC(u jul)

=(ξ1ξ2ξ3− γ jγlξinC(ui)− γlγiξ jnC(u j)− γiγ jξlnC(ul)+ γ1γ2γ3tC(u1u2u3)

)ξi

= N(x)ξi,

as desired. Similarly, we expand

u]]i = −ξ]i u]i + γiu

]ju

]l =−

(ξ jξl− γ jγlnC(ui)

)(−ξiui + γiu jul

)+ γi(−ξl ul + γluiu j

)(−ξ ju j + γ jului

)= ξ1ξ2ξ3ui− γiξ jξlu jul− γ jγlξinC(ui)ui + γ1γ2γ3nC(ui)u jul

+ γiξ jξl ul u j− γiγ jξl ul(ului)− γlγiξ j(uiu j)u j + γ1γ2γ3(uiu j)(ului).

Here we combine the fact that the conjugation of C is an involution with Kirmse’sidentities (20.3.1) to conclude

u]]i =(ξ1ξ2ξ3−∑γnγpξmnC(um)

)ui + γ1γ2γ3

(nC(ui)u jul +(uiu j)(ului)

),

where the middle Moufang identity (14.3.3) and (20.3.2) yield

nC(ui)u jul +(uiu j)(ului) = nC(ui)u jul +ui(u jul)ui = nC(ui,u jul)ui

= tC(u1u2u3)ui.

Thus u]]i = N(x)ui, and the proof of the adjoint identity is complete.

ss.COCUJM37.12. The concept of a cubic Jordan matrix algebras. The cubic Jordan alge-bra corresponding to the cubic norm structure Her3(C,Γ ) of Thm. 37.11 will alsobe denoted by J := Her3(C,Γ ) and is called a cubic Jordan matrix algebra. Thejustification of this terminology derives from the fact that, if 1C ∈C is unimodularand Γ ∈ GL3(k), then the elements of J by 37.8 may be identified canonically withthe Γ -twisted 3× 3 hermitian matrices having entries in C and scalars down thediagonal, i.e., with the matrices

x = ∑(ξieii +ui[ jl]) =

ξ1 γ2u3 γ3u2γ1u3 ξ2 γ3u1γ1u2 γ2u1 ξ3

(1) ELCUJM

for ξi ∈ k, ui ∈C, 1≤ i≤ 3, and this identification is compatible with base change.Note, however, that the Jordan structure of J is not as closely linked to ordinarymatrix multiplication as one would naively expect. On the positive side, it followsfrom Exc. 197 below that the squaring of J and that of Mat3(C) coincide on J. Henceso do the circle product of J and the symmetric matrix product (x,y) 7→ xy+ yx ofMat3(C). Thus, if 1

2 ∈ k, then its linear Jordan structure make J a unital subalgebraof Mat3(C)+; in particular, the euclidean Albert algebra Her3(O) of 5.5 is a (veryimportant) example of a cubic Jordan matrix algebra over the reals. On the other


hand, returning to the case of an arbitrary base ring, if C is properly alternative, thenthe k-algebra Mat3(C) is not flexible, and the U-operator Uxy of J will in general notbe the same as (xy)x or x(yx) (Exc. 197 (b)). On the other hand, the explicit formula(7) of that exercise immediately implies the following useful observation.

p.ASCOOR37.13. Proposition. Let (C,Γ ) be a co-ordinate pair over k such that 1C ∈ C isunimodular. If C is associative, then the Jordan algebra Her3(C,Γ ) is a subalgebraof Mat3(C)(+).

ss.CUJOCO37.14. Cubic Jordan matrix algebras and composition algebras. The most im-portant cubic Jordan matrix algebras have the form J = Her3(C,Γ ), where (C,Γ )is a co-ordinate pair and C is a composition algebra. Since C is finitely generatedprojective as a k-module, so is J, by (37.10.1) and we deduce from (37.10.7) that Jis non-singular if and only if C is. Here is a special case.

p.HERKEI37.15. Proposition. The map ϕ : Mat3(k)(+)→ Her3(k⊕ k) defined by

ϕ(x) := ∑(ξiieii +(ξ jl⊕ξl j)[ jl]

)for x = (ξi j)1≤i, j≤3 ∈Mat3(k) is an isomorphism of cubic Jordan algebras.

Proof. ϕ is a linear bijection sending 13 to 1J with J := Her3(k⊕ k). By Ex. 35.23and Exc. 185, therefore, it suffices to show that ϕ preserves norms. Actually, sinceϕ commutes with base change, we need only show NJ(ϕ(x)) = det(x) for all x =(ξi j) ∈Mat3(k). To this end we put C := k⊕ k and note nC(α ⊕β ) = αβ , tC(α ⊕β ) = α +β for α,β ∈ k. Hence (37.10.5) implies

NJ(ϕ(x)

)= NJ

(∑(ξiieii +(ξ jl +ξl j)[ jl]

))= ξ11ξ22ξ33−∑ξiiξ jlξl j +ξ23ξ31ξ12 +ξ32ξ13ξ21

= ξ11ξ22ξ33 +ξ12ξ23ξ31 +ξ13ξ21ξ32

−ξ31ξ22ξ13−ξ32ξ23ξ11−ξ33ξ21ξ12

= det(x),

as claimed. ss.EICJM

37.16. Elementary identities in cubic Jordan matrix algebras. Let (C,Γ ) be apre-co-ordinate pair over k. Then one checks that Her3(C,Γ )satisfies the identities


e]ii = 0, T (eii) = 1, eii× e j j = ell , (1) EXELFR

e j×ui[ jl] = el×ui[ jl] = 0, ei×ui[ jl] =−ui[ jl], (2) DITI

ui[ jl]] = − γ jγlnC(ui)eii, (3) OFFTEQ

ui[ jl] v j[li] = ui[ jl]× v j[li] = γluiv j[i j], (4) OFFTI

Uui[ jl]vi[ jl] = γ jγl(uiviui)[ jl], (5) UJLJL

Uui[ jl]e j j = γ jγlnC(ui)ell , (6) UJLJJ

Uui[ jl]ell = γ jγlnC(ui)e j j, (7) UJLLL

ui[ jl]u j[li]ul [i j]= γ1γ2γ3tC(u1u2u3)e j j (8) TRJLIJ

for all ui,vi ∈C, i = 1,2,3.

We are now in a position to investigate more closely the role played by elementaryframes in cubic Jordan matrix algebras.

p.PEDEDI37.17. Proposition. Let (C,Γ ) be a pre-co-ordinate pair over k. Then (e11,e22,e33)is an elementary frame in J := Her3(C,Γ ), called its diagonal frame, whose Peircecomponents are given by

Jii = keii, J jl =C[ jl]. (1) PEDEDI

Moreover, for ui ∈C, the orthognal idempotents e j j and ell ae connected by ui[ jl] ∈J jl if and only if ui ∈C× and γ j,γl ∈ k×.

Proof. That the diagonal matrix units form an elementary frame of J follows im-mediately from (37.16.1) combined with Prop. 37.5. It remains to establish (1), thefirst relation being obvious by (37.6.2). As to the second, let

x = ∑(ξmemm +um[np]) ∈ J (ξm ∈ k, um ∈C, m = 1,2,3).

Then (37.6.2), (37.16.1), (37.16.2) imply

x ∈ J jl ⇐⇒ T (x) = 0, e j j× x = ell× x = 0

⇐⇒ ∑ξm = 0,

ξiell +ξleii−u j[li] = 0 = ξ jeii +ξie j j−ul [i j]

⇐⇒ ξ1 = ξ2 = ξ3 = 0, u j = ul = 0⇐⇒ x = ui[ jl]

⇐⇒ x ∈C[ jl].

The final statement follows immediately from Prop. 37.5 and (37.10.9), which im-plies S(ui[ jl]) =−γ jγlnC(ui).

c.DIACON37.18. Corollary. The diagonal frame of Her3(C,Γ ) is connected if and only ifΓ ∈ GL3(k), i.e., (C,Γ ) is a co-ordinate pair over k.


ss.CUCOSY37.19. Co-ordinate systems. By a co-ordinate system of a cubic Jordan algebraJ over k we mean a quintuple S = (e1,e2,e3,u23,u31) ∈ J5 such that (e1,e2,e3) isan elementary frame of J, inducing the corresponding Peirce decomposition J =

∑(kei + J jl), and for i = 1,2, the orthogonal idempotents e j,el are connected byu jl ∈ J jl . We refer to (e1,e2,e3) as the elementary frame belonging to S. A pair(J,S) consisting of a cubic Jordan algebra J over k and a co-ordinate system S ofJ will be called a co-ordinated cubic Jordan algebra.

e.COHER37.20. Example. Let (C,Γ ) be a co-ordinate pair over k. Then Prop. 37.17 showsthat

D(C,Γ ) := (e11,e22,e33,1C[23],1C[31]) (1) DIACOS

is a co-ordinate system of the cubic Jordan algebra Her3(C,Γ ), called its diagonalco-ordinate system. We write

Her3(C,Γ ) :=(

Her3(C,Γ ),D(C,Γ ))

(2) COHER

for the corresponding co-ordinated cubic Jordan algebra.

In the remainder of this section, we will show that, conversely, every co-ordinatedcubic Jordan algebra is isomorphic to a cubic Jordan matrix algebra, under an iso-morphism matching the given co-ordinate system of the former with the diagonalone of the latter. This will be the content of the Jacobson co-ordinatization theo-rem 37.25 below.

In order to accomplish this result, we begin with a series of preparations.l.EXPQUAT

37.21. Lemma. We have

S(x× y,z) = T (x)T (y)T (z)−T (x,y)T (z)−T (x× y,z)

for all x,y,z ∈ J.

Proof. Applying (35.8.13) twice, we obtain

S(x×y,z) = T (x×y)T (z)−T (x×y,z) = T (x)T (y)T (z)−T (x,y)T (z)−T (x×y,z),

as claimed. p.DECNOAD

37.22. Proposition. Let (e1,e2,e3) be an elementary frame of J and J = ∑(kei +J jl) the corresponding Peirce decomposition of J. Then J0 := ∑kei is a non-singularcubic subalgebra of J canonically isomorphic to E(+), where E stands for the splitcubic etale k-algebra. Moreover,

Ji j× J jl ⊆ Jli, (1) OFFDARP

and given


x = ∑(ξiei +u jl), y = ∑(ηiei + v jl) ∈ J

with ξi,ηi ∈ k, u jl ,v jl ∈ J jl , i = 1,2,3, the following relations hold.

N(x) = ξ1ξ2ξ3 +∑ξiS(u jl)+T (u23×u31,u12), (2) EXPNOR

x] = ∑

((ξ jξl +S(u jl)

)ei +

(−ξiu jl +uli×ui j

)), (3) EXPAD

x× y = ∑

((ξ jηl +η jξl−T (u jl ,v jl)

)ei (4) EXPBAT

+(−ξiv jl−ηiu jl +uli× vi j + vli×ui j

)[ jl]),

T (x,y) = ∑ξiηi +∑T (u jl ,v jl), (5) EXPBT

T (x) = ∑ξi, (6) EXPLT

S(x) = ∑(ξ jξl +S(u jl)

), (7) EXPQT

S(x,y) = ∑(ξ jηl +η jξl−T (u jl ,v jl)

). (8) EXPQTB

Proof. Relation (1) follows from Ji j×J jl = Ji j J jl (by Prop. 37.5 (e)) and the Peircerules. For the remaining assertions, we proceed in three steps.10. Noting that J0 =∑kei is a direct sum of ideals by Cor. 34.12 (b), our first aim willbe to show that J0 and E(+) are canonically isomorphic. Since e]i = 0 by definitionand N(ei) = 0 by 37.1, combining equation (6) of Exc. 58 with Prop. 37.5 (a) yieldsN(∑ξiei) = ξ1ξ2ξ3T (e1× e2,e3) = ξ1ξ2ξ3T (e3,e3), so 37.1 again yields

N(∑ξiei

)= ξ1ξ2ξ3 (9) NODIAG

in all scalar extensions. Thus the map ϕ : E(+)→ J0 defined by

ϕ(ξ1⊕ξ2⊕ξ3) := ∑ξiei

for ξ1,ξ2,ξ3 ∈ k is an isomorphism not only of Jordan algebras but, in fact, of cubicones. Hence (35.24.4), (35.24.5) yield(

∑ξiei)]

= ∑ξ jξlei, (10) SHDIAG

T(∑ξiei,∑ηiei

)= ∑ξiηi. (11) TDIAG

20. We now apply the formalism of 36.7, derive the relation

V := X⊥0 = J23 + J31 + J12 (12) PERPDIAG

from Prop. 37.5 (c) and recall from Remark 36.6 that (X0,V ) is a complementedcubic norm substructure of J, so it makes sense to compute the quadratic maps Q,Hof 36.7 in the special case at hand. Following (12), let

u = u23 +u31 +u12 ∈V, u jl ∈ J jl . (13) ELPERP


Since J jl ⊆ J0(ei) by (34.11.2), Prop. 37.2 (c) implies

u] =(u23 +u31 +u12

)]= u]23 +u]31 +u]12 +u23×u31 +u31×u12 +u12×u23

= S(u23)e1 +S(u31)e2 +S(u12)e3 +u23×u31 +u31×u12 +u12×u23.

Combining this with (1) and (36.7.1), we deduce

Q(u) = −∑S(u jl)ei, (14) QPERP

H(u) = ∑uli×ui j. (15) HPERP

We also wish to know how N acts on V = X⊥0 . Using (36.8.11) (for x0 = 1), (15)and linearizing (14), we obtain

N(u)1 = Q(u,H(u)

)=−∑S(u jl ,uli×ui j)ei,

where Lemma 37.21 and Prop. 37.5 (d) imply

S(uli×ui j,u jl) =−T (uli×ui j,u jl) =−T (u23×u31,u12)

since the expression T (x× y,z) is totally symmetric in its arguments. Thus

N(u) = T (u23×u31,u12). (16) NOPERP

30. It is now easy to complete the proof of the proposition. Combining (36.7.3),(36.7.2)with (9),(11),(16), we obtain

N(x) = N(∑ξiei +u

)= N

(∑ξiei

)−T

(∑ξiei,Q(u)

)+N(u)

= ξ1ξ2ξ3 +∑ξiS(u jl)+T (u23×u31,u12),

giving (4), and

x] =(∑ξiei +u

)]=((

∑ξiei)]−Q(u)

)+(∑ξiei×u+H(u)

).

But u jl ∈ J0(ei), while uli,ui j ∈ J1(ei). Hence ei× u = −u jl by (37.2.2),(37.2.3),Prop. 37.5 (d), and (10),(14),(15) imply

x] = ∑(ξ jξl +S(u jl)

)ei +∑

(−ξiu jl +uli×ui j

),

giving (3). Applying Prop. 37.5 (c), we obtain (5), which immediately implies (6)and combines with (3) to yield (7). Linearizing (7), we obtain

S(x,y) = ∑(ξ jηl +η jξl +S(u jl ,v jl)

).

Since T (u jl) = 0 by (6), we deduce S(u jl ,v jl) = −T (u jl ,v jl) from (35.8.13), and(8) follows, as does (4) by linearizing (3).


l.ESPE37.23. Lemma. Under the assumptions of Prop. 37.22, the relations

S(u jl ,v jl) = −T (u jl ,v jl), (1) ESTE

u jl× (u jl×uli) = −S(u jl)uli, (2) BSHAQT

(uli×ui j)×ui j = −S(ui j)uli, (3) BASHART

u jl× (v jl×uli)+ v jl× (u jl×uli) = T (u jl ,v jl)uli, (4) BSHABQT

(uli×ui j)× vi j +(uli× vi j)×ui j = T (ui j,vi j)uli, (5) BSHABRT

S(u jl×uli) = −S(u jl)S(uli) (6) QTCOMP

hold for all u jl ,v jl ∈ J jl , uli ∈ Jli.

Proof. (1) is an immediate consequence of (37.22.8). In order to establish (2),we first note Jli ⊆ J1(e j + el), J jl ⊆ J2(e j + el). Hence Prop. 37.5 combined with(37.22.1) and Exc. 177 yields

u jl× (u jl×uli) = u jl (u jl uli) = u2jl uli =−S(u jl)(e j + el)uli =−S(u jl)uli,

hence (2). An analogous computation yields (3). Linearizing (2) (resp. (3)) and com-bining with (1) gives (4) (resp. (5)). Finally, in order to establish (6), we apply(35.8.10) to conclude

S(u jl×uli) = T((u jl×uli)

])= T

(T (u]jl ,uli)uli +T (u jl ,u

]li)u jl−u]jl×u]li

),

which by Prop. 37.22 reduces to

S(u jl×uli) =−S(u jl)S(uli)T (ei× e j) =−S(u jl)S(uli)T (el) =−S(u jl)S(uli),

as claimed. p.CONALT

37.24. Proposition. Let (J,S) be a co-ordinated cubic Jordan algebra over k, withS= (e1,e2,e3,u23,u31) the corresponding co-ordinate system, and let J = ∑(kei +J jl) be the Peirce decomposition of J relative to the elementary frame belonging toS. Then S(u jl) ∈ k× for i = 1,2, and with

ω := ωJ,S := S(u23)−1S(u31)

−1 ∈ k×, (1) CONOM

the k-module

C :=CJ,S := J12 (2) ONETWOCO

becomes a multiplicative conic alternative k-algebra with multiplication, norm, unitelement, trace, conjugation respectively given by


uv := ω(u×u23)× (u31× v), (3) CONPRO

nC(u) = −ωS(u), (4) CONNO

1C = u12 := u23×u31, (5) CONUN

tC(u) = ωT (u12,u), (6) CONTR

u = ωT (u12,u)u12−u (7) CONBAR

for all u,v ∈C. Moreover,

(C,Γ ) = (CJ,S,ΓJ,S) =: Cop(J,S) (8) TRICON

with

Γ := ΓJ,S := diag(γ1,γ2,γ3), γ1 =−S(u31), γ2 =−S(u23), γ3 = 1. (9) CONMA

is a co-ordinate pair over k, called the co-ordinate pair associated with (J,S).

Proof. We have S(u jl) ∈ k× by Prop. 37.7; in particular, ω ∈ k× exists, and (3) by(37.22.1) defines a non-associative algebra structure on C = J12. It remains to showthat C is a multiplicative conic alternative algebra with norm, unit element, trace,conjugation as indicated. Let u,v ∈C. The element u12 ∈C by Lemma 37.23, (1),(3) satisfies

u12v = ω((u23×u31)×u23

)× (u31× v)

= −S(u23)ωu31× (u31× v) = S(u23)S(u31)ωv = v,

uu12 = ω(u×u23)×(u31× (u23×u31)

)= −S(u31)ω(u×u23)×u23 = S(u23)S(u31)ωu = u.

Thus C is unital with unit element 1C = u12, and (5) holds. Defining the quadraticform nC : C→ k by (4), we derive the relation nC(1C)=−ωS(u23×u31)=ωS(u23)S(u31),and (1) yields

nC(1C) = 1. (10) NOUN

In order to distinguish the squaring in J from the one in C, we write u·2 for the latter.Applying Lemma 37.23, (35.8.7), (1), we then obtain

u·2 = ω(u×u23)× (u31×u)

= ωT (u23,u31×u)u−ω(u× (u31×u)

)×u23

= ωT (u12,u)u+ωS(u)u31×u23 = ωT (u12,u)u−nC(u)1C,

where (37.23.1) and (4) yield nC(1C,u) = −ωS(u12,u) = ωT (u12,u). Thus C is aconic k-algebra with norm, unit, trace given by (4), (5), (6), respectively. Relation(7) is now clear. Next we show alternativity, again by appealing to the relations ofLemma 37.23 and (6):


u(uv) = ω2(u×u23)×

(u31×

((u×u23)× (u31× v)

))= ω

2T (u31,u×u23)(u×u23)× (u31× v)−

ω2(u×u23)×

((u×u23)×

(u31× (u31× v)

))= tC(u)uv−ω

2S(u×u23)S(u31)v = tC(u)uv+ω2S(u23)S(u31)S(u)v

= tC(u)uv−nC(u)v =(tC(u)u−nC(u)1C

)v = u·2v,

(vu)u = ω2((

(v×u23)× (u31×u))×u23

)× (u31×u)

= ω2T (u23,u31×u)(v×u23)× (u31×u)−

ω2((

(v×u23)× (u31×u))× (u31×u)

)×u23

= tC(u)vu−ω2S(u31×u)S(u23)v = tC(u)vu+ω

2S(u31)S(u23)S(u)v

= tC(u)vu−nC(u)v = v(tC(u)u−nC(u)1C

)= vu·2.

Finally, by (37.23.6), (3), (4), the norm of C permits composition:

nC(uv) = −ω3S((u×u23)× (u31× v)

)= ω

3S(u×u23)S(u31× v) = ω2S(u)S(v) = nC(u)nC(v).

Summing up we have thus shown that C is a multiplicative conic alternative algebraover k.

37.25. Theorem (The Jacobson co-ordinatization theorem) . Let (J,S) be a co-t.JACOordinated cubic Jordan algebra over k. With the notation of Prop. 37.24, the mapφJ,S : Her3(C,Γ )→ J defined by

φJ,S(x) := ∑(ξiei + v jl), (1) CONPHI

for

x = ∑(ξieii + vi[ jl]) ∈ Her3(C,Γ ) (ξi ∈ k, vi ∈C, i = 1,2,3), (2) CONEX

where

v23 :=−S(u31)−1u31× v1, v31 :=−S(u23)

−1u23× v2, v12 := v3, (3) CONPE

is an isomorphism of cubic Jordan algebras matching the diagonal co-ordinate sys-tem of Her3(C,Γ ) with the co-ordinate system S of J.

Proof. Note by (37.24.2) that C = J12 as k-modules. Since the conjugation of anyconic algebra leaves its norm invariant, (37.24.4) shows

S(v) = S(v) (v ∈C = J12). (4) CONSH

Putting Φ := ΦJ,S and defining Ψ : J→ Her3(C,Γ ) by


Ψ(∑(ξiei + v jl)

):= ∑(ξieii + vi[ jl])

for ξi ∈ k, v jl ∈ J jl , i = 1,2,3, where

v1 := u31× v23, v2 := u23× v31, v3 := v12,

(37.23.2), (37.23.3) imply Φ Ψ = 1J and Ψ Φ := 1Her3(C,Γ ). Thus Φ is a bi-jective linear map. It remains to show that φ is a homomorphism of cubic Jordanalgebras. Since φ obviously preserves unit elements, the assertion will follow fromExc. 185 (a) once we have shown that φ preserves adjoints. In order to do so, we de-note by x]

′the adjoint of x ∈ Her3(C,Γ ) as given by (2) and deduce from (37.10.4)

that

x]′= ∑(ηieii +wi[ jl]), (5) EXSHAPR

where

ηi = ξ jξl− γ jγlnC(vi), (6) EXSHABE

wi = −ξivi + γiv jvl . (7) EXSHAWE

Hence (2), (3) imply

φ(x]′) = ∑(ηiei +w jl), (8) CONPHIPR

where

w23 :=−S(u31)−1u31× w1, w31 :=−S(u23)

−1u23× w2, w12 := w3. (9) CONPEWE

Before we can proceed, we require the identity

S(v jl) =−γ jγlnC(vi), (10) ESVE

which follows by combining (3), (4), (37.23.6), (37.24.1), (37.24.4) and (37.24.9)with the computations

S(v23) = S(u31)−2S(u31× v1) =−S(u31)

−1S(v1) =−S(u23)ωS(v1) =−γ2γ3nC(v1),

S(v31) = S(u23)−2S(u23× v2) =−S(u23)

−1S(v2) =−S(u31)ωS(v2) =−γ3γ1nC(v2),

S(v12) = S(v3) =−ω−1nC(v3) =−γ1γ2nC(v3).

Applying (10) and (6), we now conclude

ηi = ξ jξl +S(v jl). (11) BETQT

On the other hand, combining (9), (7), (3), (37.24.9) gives


w23 = −S(u31)−1u31× w1 = S(u31)

−1ξ1u31× v1−S(u31)

−1γ1u31× (v2v3)

= −ξ1v23 +u31× (v2v3),

where Lemma 37.23 and Prop. 37.24 imply

u31× (v2v3) = ωu31×((v2×u23)× (u31× v3)

)= ωT (u31,v2×u23)u31× v3−ω(v2×u23)×

(u31× (u31× v3)

)= tC(v2)u31× v3 +S(u23)

−1(v2×u23)× v3

= −S(u23)−1(tC(v2)(u23×u31)×u23− v2×u23

)× v3

= −S(u23)−1((

tC(v2)1C− v2)×u23

)× v3

= −S(u23)−1(u23× v2)× v3 = v31× v12.

Summing up,

w23 =−ξ1v23 + v31× v12.

Similarly,

w31 = −S(u23)−1u23× w2 = ξ2S(u23)

−1u23× v2−S(u23)−1

γ2u23× (v3v1)

= −ξ2v31 +u23× (v3v1),

where

u23× (v3v1) = ωu23×((v3×u23)× (u31× v1)

)= ωT (u23,u31× v1)v3×u23−ω(u31× v1)×

((v3×u23)×u23

)= tC(v1)v3×u23 +S(u31)

−1(u31× v1)× v3

= −S(u31)−1(tC(v1)(u23×u31)×u31−u31× v1

)× v3

= −S(u31)−1((

tC(v1)1C− v1)×u31

)× v3

= −S(u31)−1(v1×u31)× v3 = v23× v12.

Thus

w31 =−ξ2v31 + v12× v23.

Since

w12 = w3 =−ξ3v3 + γ3v1v2 =−ξ3v12 + v2v1

= −ξ3v12 +ω(v2×u23)× (u31× v1) =−ξ3v12 + v23× v31,

our computations can be unified to

w jl =−ξiv jl + vli× vi j. (12) CONWJL


Inserting (11) and (12) into (8), we may apply (37.22.3) and (1) to obtain

φ(x]′) = ∑

(ξ jξl +S(v jl)

)ei +∑

(−ξiv jl + vli× vi j

)=(∑ξiei +∑v jl

)]= φ(x)].

Hence φ preserves adjoints and thus is an isomorphism of cubic Jordan algebras.It remains to show that φ matches the respective co-ordinate systems. To this end,

we have to prove φ(1C[ jl]) = u jl for i = 1,2, which follows from (1), (3), (37.23.2),(37.24.5) and

φ(1C[23]) = −S(u31)−1u31×1C =−S(u31)

−1u31× (u23×u31) = u23,

φ(1C([31]) = −S(u23)−1u23×1C =−S(u23)

−1u23× (u23×u31) = u31,

completing the proof of the entire theorem.

r.JACOEX37.26. Remark. Versions of the Jacobson co-ordinatization theorem which inmany ways are much more general than the one presented here may be found in theliterature., see, e.g., Jacobson [43, 44, 45] and McCrimmon [75] for details. Givenany integer m≥ 3, the most important difference is that instead of co-ordinate pairsone has to consider quadruples (D,τ,D0,∆) consisting of

• a unital alternative k-algebra D,• an involution τ : D→ D,• a unital subalgebra D0 of H(D,τ) contained in the nucleus of D and being D-

ample in the sense that uD0τ(u)⊆ D0 for all u ∈ D,• and an invertible diagonal matrix ∆ ∈Matm(D0).

With considerable effort, it can then be shown that the k-module

Herm(D,τ,D0,∆)

of all ∆ -twisted m×m hermitian matrices with entries in D and diagonal ones in D0,carries the structure of a Jordan algebra over k provided m = 3 or D is associative.Conversely, given a Jordan algebra J over k whose extreme radical is zero and aconnected orthogonal system Ω = (e1, . . . ,em) of idempotents in J, the Jacobsonco-ordinatization theorem says that there exist a quadruple (D,τ,D0,∆) as aboveand an isomorphism from J onto Herm(D,τ,D0,∆) matching the orthogonal systemΩ of the former with the diagonal one of the latter.

Here the condition on the extreme radical, being automatic for the algebrasHerm(D,τ,D0,∆), cannot be avoided. One may therefore wonder why it is absentfrom our version of the Jacobson co-ordinatization theorem. The answer rests onour formal definition of cubic Jordan matrix algebras in 37.10, 37.12 which, for aco-ordinate pair (C,Γ ), allows a concrete base change invariant interpretation ofHer3(C,Γ ) in terms of Γ -twisted hermitian matrices only if 1C ∈C is unimodular.Indeed, dropping this hypothesis, the extreme radical of Her3(C,Γ ) may very wellbe different from zero, while otherwise it is not (Exc. 204).


Exercises.

pr.IDCUJO 195. Idempotents in cubic Jordan algebras. Let J be a cubic Jordan algebra over k. An idempotente ∈ J is said to be co-elementary if the complementary idempotent 1− e is elementary. Now lete be any idempotent in J. Prove that there exists a complete orthogonal system (ε(i))0≤i≤3 ofidempotents in k, giving rise to decompositions

k = k(0)⊕ k(1)⊕ k(2)⊕ k(3), J = J(0)⊕ J(1)⊕ J(2)⊕ J(3)

as direct sums of ideals, where k(i) = ε(i)k, J(i) = ε(i)J = Jk(i) as cubic Jordan algebras over k(i) for0≤ i≤ 3 such that

e = 0⊕ e(1)⊕ e(2)⊕1J(3) ,

e(1) is an elementary idempotent of J(1) and e(2) is a co-elementary idempotent of J(2). Showfurther that the ε(i) are unique and given by

ε(0) = N(1− e) = 1−T (e)+S(e)−N(e), (1) DECIZERO

ε(1) = T (e)−2S(e)+3N(e), (2) DECIONE

ε(2) = S(e)−3N(e), (3) DECITWO

ε(3) = N(e). (4) DECITHREE

pr.FERLE 196. Ferrar’s lemma [28, Lemma 1.10]. Let u1,u2,u3 be elements of a cubic Jordan algebra J overk such u]i = 0 for i = 1,2,3. Prove that q := ∑ui is invertible in J if and only if T (u1× u2,u3) isinvertible in k, and that, in this case, (u1,u2,u3) is an elementary frame in the isotope J(p), p := q−1.Conclude that three elementary idempotents in J adding up to 1 form an elementary frame of J.

pr.UOPMA 197. The U-operator and matrix multiplication. Let (C,Γ ) be a co-ordinate pair over k and assumethat 1C ∈C is unimodular. Let

x = ∑(ξieii +ui[ jl]), y = ∑(ηieii + vi[ jl]) ∈ J := Her3(C,Γ )

with ξi,ηi ∈ k, ui,vi ∈ C for i = 1,2,3 and write 1,x,x2,x3, . . . for the powers of x in J, whiledenoting matrix multiplication In Mat3(C) by juxtaposition. Then prove

x2 = xx, (5) SQUAMA

x3 = x(xx)+ γ1γ2γ3[u1,u2,u3]13 = (xx)x− γ1γ2γ3[u1,u2,u3]13, (6) CUBMA

Uxy = x(yx)− [x,x,y]+ γ1γ2γ3([u1,u2,v3]+ [u2,u3,v1]+ [u3,u1,v2]

)13 (7) UOPMA

= (xy)x+[y,x,x]− γ1γ2γ3([u1,u2,v3]+ [u2,u3,v1]+ [u3,u1,v2]

)13,

[x,x,x] = 2γ1γ2γ3[u1,u2,u3]13, (8) ASXXX

x]x =(N(x)1C + γ1γ2γ3[u1,u2,u3]

)13, (9) SHAMA

xx] =(N(x)1C− γ1γ2γ3[u1,u2,u3]

)13. (10) MASHA

Finally, writing x] asx] = ∑(ξ ]

i +u]i [ jl])

with ξ]i ∈ k, u]i ∈C for i = 1,2,3, conclude that

[u]1,u]2,u

]3] =−N(x)[u1,u2,u3]. (11) USHAI

pr.DIAFIX 198. Let (C,Γ ) and (C,Γ ′) with

Γ = diag(γ1,γ2,γ3), Γ′ = diag(γ ′1,γ

′2,γ′3) ∈ Diag3(k)

×


be two co-ordinate pairs over k.(a) View Diag3(k) via 35.24 as the split cubic etale k-algebra and consider the following conditions,for any ∆ = diag(δ1,δ2,δ3) ∈ Diag3(k)

×.

(i) γ ′jγ′l = δ 2

i γ jγl for i = 1,2,3 and γ ′1γ ′2γ ′3 = δ1δ2δ3γ1γ2γ3, i.e., Γ ′] = ∆ 2Γ ] and N(Γ ′) =

N(∆)N(Γ ).(ii) γ ′i = δ jδlδ

−1i γi for i = 1,2,3, i.e., Γ ′ = ∆ ]∆−1Γ .

(iii) The mapϕC,∆ : Her3(C,Γ )

∼−→ Her3(C,Γ ′)

defined by

ϕC,∆

(∑(ξieii +ui[ jl])

)= ∑

(ξieii +(δ−1

i ui)[ jl])

(12) DIAFEL

for ξi ∈ k, ui ∈C, i = 1,2,3 is an isomorphism of cubic Jordan algebras.

Show that the implications(i) ⇐⇒ (ii) =⇒ (iii)

hold, and that all three conditions are equivalent if 1C ∈C is unimodular.(b) Conclude from (a) that the isomorphism class of Her3(C,Γ ) does not change if

(i) Γ is multiplied by an invertible scalar,(ii) each diagonal entry of Γ is multiplied by an invertible square,(iii) Γ is replaced by an appriate diagonal matrix in Mat3(k) of determinant 1.

pr.DIISJO 199. Diagonal isotopes of cubic Jordan matrix algebras. Let (C,Γ ) be a co-ordinate pair over k.Prove that

p := ∑γieii ∈ Her3(C,Γ )

is invertible and that the map

ϕ : Her3(C,Γ )(p) ∼−→ Her3(C)

defined by

ϕ(∑(ξieii +ui[ jl])

):= ∑

((γiξi)eii +(γ jγlui)[ jl]

)(13) DIIS

for ξi ∈ k, ui ∈C, i = 1,2,3 is an isomorphism of cubic Jordan algebras.

pr.ISCOPA 200. Isotopes of pre-co-ordinate pairs. Let (C,Γ ) be a pre-co-ordinate pair over k and p,q ∈C×.Put

Γ(p,q) := diag(γ(p,q)

1 ,γ(p,q)2 ,γ

(p,q)3 ) := diag

(nC(pq)−1

γ1,nC(q)γ2,nC(p)γ3),

so that(C,Γ )(p,q) := (C(p,q),Γ (p,q))

is a pre-co-ordinate pair over k, and show that the map

ϕ : Her3(C(p,q),Γ (p,q))∼−→ Her3(C,Γ )

defined by

ϕ(∑(ξieii +ui[ jl])

):= ∑(ξieii +u′i[ jl]) (14) ISOISC

for ξi ∈ k, ui ∈C, i = 1,2,3, where

u′1 := (pq)u1(pq), u′2 := u2 p, u′3 := qu3, (15) OFFDI

is an isomorphism of cubic Jordan algebras.


pr.OUTCENCU 201. Let (C,Γ ) be a co-ordinate pair over k, Show that the cubic Jordan matrix algebra J =Her3(C,Γ ) over k is outer central.

pr.IDCUJM 202. Ideals of cubic Jordan matrix algebras. Let (C,Γ ) be a co-ordinate pair over k and suppose1C ∈C is unimodular, so that we obtain a natural identification k ⊆C as a unital subalgebra whichis stable under base change. Prove:(a) The outer ideals of the Jordan algebra J := Her3(C,Γ ) are precisely of the form

H3(I0, I,Γ ) := ∑(I0eii + I[ jl]) = ∑(ξieii +ui[ jl]) | ξi ∈ I0, ui ∈ I, 1≤ i≤ 3, (16) HIIG

where I is an ideal in (C, ιC) (viewed as an algebra with involution) and I0 is an ideal in k, containedin I∩ k and weakly I-ample in the sense that it contains the trace of arbitrary elements in I:

tC(I)⊆ I0 ⊆ I∩ k. (17) WIAMP

(b) The ideals of J are precisely of the form (16), where I as in (a) is an ideal in (C, ιC) and I0 isan ideal in k, contained in I∩k and I-ample in the sense that it contains norm and trace of arbitraryelements in I:

knC(I)+ tC(I)⊆ I0 ⊆ I∩ k. (18) IAMP

(c) For I0, I as in (a), we have 2(I∩ k)⊆ I0, hence I0 = I∩ k if 12 ∈ k.

(d) If C is a non-singular composition algebra, then the outer ideals of J are ideals and have theform aJ with a varying over the ideals of k.(e) If k = F is a field and C is a pre-composition algebra over F , then the outer ideals of J =Her3(C,Γ ) are precisely of the form 0, Rad(T ) (the radical of the bilinear trace), and J. Inparticular, J is a simple Jordan algebra. Moreover, J is outer simple if and only if C is a non-singular composition algebra over F .

pr.ABZECU 203. Absolute zero divisors in cubic Jordan algebras. Let J be a cubic Jordan algebra over k.(a) Show that if x ∈ J is an absolute zero divisor, then so is x].(b) Assume that k is reduced and consider the following conditions on x ∈ J.

(i) x is an absolute zero divisor.(ii) x] = 0 and T (x,y) = 0 for all y ∈ J.(iii) N(x) = 0 and T (x,y) = 0 for all y ∈ J.

Then prove that the implications

(i) ⇐⇒ (ii) =⇒ (iii)

hold.(c) Prove that every absolute zero divisor of J is contained in the nil radical of J.(d) (Weiss) Show that, contrary to what has been claimed in Petersson-Racine [103, p. 214], con-ditions (i), (ii), (iii) of part (b) are not equivalent, even if k = F is a field and J is a simple Jordanalgebra.(e) Show that the nil radical of J is zero if and only if k is reduced and J has no absolute zerodivisors.

pr.EXRACO 204. The extreme radical of co-ordinated cubic Jordan algebras. Let S= (e1,e2,e3,u23,u31) be aco-ordinate system of the cubic Jordan algebra J over k and write J = ∑(kei + J jl) for the Peircedecomposition of J relative to the elementary frame (e1,e2,e3). Prove without recourse to theJacobson co-ordinatization theorem that the extreme radical of J may be described as


Rex(J) =∑ξiei | ξi ∈ k, ξiJi j = 0 for i = 1,2,3

(19) CHAREX

⊆∑ξiei | ξ 2

i = 2ξi = 0 for i = 1,2,3.

Conclude for a co-ordinate pair (C,Γ ) over k that the extreme radical of Her3(C,Γ ) is zero if1C ∈C is unimodular but not in general.

pr.JACOCAT 205. A categorical set-up for the Jacobson co-ordinatization theorem. (a) Let (C,Γ ) and (C′,Γ ′)be co-ordinate pairs over k. We define a homomorphism from (C,Γ ) to (C′,Γ ′) as a pair (η ,∆)consisting of

(i) a homomorphism η : C→C′ of conic k-algebras,(ii) a matrix ∆ ∈ Diag3(k)

× such that Γ ′ = ∆ ]∆−1Γ .

In this way we obtain the category of co-ordinate pairs over k, denoted by k-copa.(b) Let (J,S) and (J′,S′) be co-ordinated cubic Jordan algebras over k and write

S= (e1,e2,e3,u23,u31) ∈ J5, S′ = (e′1,e′2,e′3,u′23,u

′31) ∈ J′5.

We define a homomorphism from (J,S) to (J′,S′) as a triple (ϕ;δ1,δ2) consisting of

(i) a homomorphism ϕ : J→ J′ of cubic Jordan algebras satisfying ϕ(ei) = e′i for i = 1,2,3,(ii) scalars δ1,δ2 ∈ k× such that ϕ(u jl) = δ

−1i u′jl for i = 1,2.

In this way we obtain the category of co-ordinated cubic Jordan algebras over k, denoted byk-cocujo.(c) Let (η ,∆) : (C,Γ )→ (C′,Γ ′) with ∆ = diag(δ1,δ2,δ3) ∈ Diag3(k)

× be a homomorphism ofco-ordinate pairs over k. Define

Her3(η ,∆) : Her3(C,Γ )−→ Her3(C′,Γ ′)

by

Her3(η ,∆)(∑(ξieii +ui[ jl])

):= ∑

(ξieii +

(δ−1i η(ui)

)[ jl])

(20) HERET

for ξi ∈ k, ui ∈C, i = 1,2,3 and show that

Her3(η ,∆) := (Her3(η ,∆);δ1,δ2) : Her3(C,Γ )−→Her3(C′,Γ ′)

is a homomorphism of co-ordinated cubic Jordan algebras over k, giving rise to a functor

Her3 : k-copa−→ k-cocujo.

(d) Let (ϕ;δ1,δ2) : (J,S)→ (J′,S′) be a homomorphism of co-ordinated cubic Jordan algebrasover k. Write J = ∑(kei + J jl) for the Peirce decomposition of J relative to the elementary framebelonging to S, ditto for J′, and ϕ12 for the linear map J12 → J′12 induced by ϕ via restriction.Then prove that

Cop(ϕ;δ1,δ2) : Cop(J,S)−→ Cop(J′,S′),

where

Cop(ϕ;δ1,δ2) := (δ1δ2ϕ12,∆), ∆ = diag(δ1,δ2,δ3), δ3 := δ1δ2, (21) COTFI

is a homomorphism of co-ordinate pairs over k, giving rise to a functor

Cop: k-cocujo−→ k-copa.

(e) Let (C,Γ ) be a co-ordinate pair over k and put


(C′,Γ ′) := Cop(Her3(C,Γ )

). (22) COTHER

Show C′ =C[12] as k-modules and that

ΨC,Γ := (ψC,Γ ,ΛC,Γ ) : (C,Γ )∼−→ (C′,Γ ′),

where

ψC,Γ : C −→C′, u 7−→ γ3u[12], (23) PSICO

ΛC,Γ := diag(λ1,λ2,λ3), λ1 = λ2 = 1, λ3 = γ3, (24) LAMCO

is an isomorphism of co-ordinate pairs over k.(f) Let (J,S) be a co-ordinated cubic Jordan algebra over k and put

(J′,S′) := Her3(

Cop(J,S)). (25) HERCOT

Show with the terminology of Thm. 37.25 that

ΦJ,S := (φJ,S;1,1) : (J′,S′) ∼−→ (J,S) (26) PHIJAY

is an isomorphism of co-ordinated cubic Jordan algebras.(g) Conclude that the functors

k-copaHer3 // k-cocujoCop

oo

give an equivalence of categories.

Solutions

Chapter 1

Section 1.

1. Putting [x1,x2,x3] =: a⊕u for some a ∈ C, u ∈ C3 and using (1.4.1), we obtainsol.ASSGC

a⊕u = [a1⊕u1,a2⊕u2,a3⊕u3]

=((a1⊕u1)(a2⊕u2)

)(a3⊕u3)− (a1⊕u1)

((a2⊕u2)(a3⊕u3)

)=((a1a2− ut

1u2)⊕ (u2a1 +u1a2 + u1× u2))(a3⊕u3)

− (a1⊕u1)((a2a3− ut

2u3)⊕ (u3a2 +u2a3 + u2× u3)),

and an application of (1.1.1) yields

a = a1a2a3− ut1u2a3−a1ut

2u3− a2ut1u3− (u1×u2)

t u3−a1a2a3

+a1ut2u3 + ut

1u3a2 + ut1u2a3 +ut

1(u2×u3)

= det(u1,u2,u3)−det(u1,u2,u3),

x = u3a1a2−u3ut2u1 +u2a1a3 +u1a2a3 +(u1× u2)a3 +(u2× u3)a1

+(u1× u3)a2 +(u1×u2)× u3−u3a2a1−u2a3a1− (u2× u3)a1

−u1a2a3 +u1ut2u3− (u1× u3)a2− (u1× u2)a3− u1× (u2×u3)

= −u3ut2u1 +(u1× u2)(a3− a3)+(u2× u3)(a1− a1)+(u3× u1)(a2− a2)

+(u1×u2)× u3 +u1ut2u3 +(u2×u3)× u1

= ∑(ui× u j)(ak− ak)+∑(ui×u j)× uk.

Thus the associator formula holds, While the C-component of the right-hand side is trivially alter-nating in x1,x2,x3, so is its C3-component since it is the sum over all cyclic permutations of (123)of a trilinear expression, namely, (u1×u2)× u3 +(u1× u2)(a3− a3), that vanishes for x1 = x2.

2. Let x = a⊕ u,y = b⊕ v with a,b ∈ C, u,v ∈ C3 and assume x 6= 0 = xy. We must show y = 0.sol.DIVOFirst of all, (1.4.1) implies

ab = ut v, va+ub+ u× v = 0. (27) ZEDI

Multiplying the second equation with ut from the left and applying the first combined with (1.1.1),(1.5.4), we obtain

0 = ut(u× v)+ ut ub+ ut va = (‖u‖2 +‖a‖2)b = ‖x‖2b,

hence b = 0. Thus (27) reduces to

ut v = 0, u× v =−va. (28) ZEDIB

Here the second equation gives 0 = vt(u× v) =−‖v‖2a, so if a 6= 0 then v = 0. Hence we are leftwith the case a= 0, which implies u 6= 0. On the other hand, (28) yields (u× v)×u= vut u− uut v=‖u‖2v, hence again v = 0.

319


3. (a) For a,b ∈ C, u,v ∈ C3, x = a⊕u,y = b⊕ v ∈O, we apply (1.4.1), (1.5.7) and concludesol.INVAS

xy = (ab− ut v)⊕ (va+ub+ u× v) = (ab− ut v)⊕ (−va−ub− u× v)

=(ba− (−v)

t(−u)

)⊕((−u) ¯b+(−v)a+(−v)× (−u)

)=(b⊕ (−v)

)(a⊕ (−u)

)= yx,

which proves (a).(b) Let x = a⊕u, y = b⊕v, with a,b ∈C, u,v ∈C3. From (1.4.1) and (1.4.5) we conclude that

tO(xy) = ab+ ab− ut v−ut v = ab+ ab− ut v− vt u is symmetric in x,y, giveing the first equation.Moreover, theC-component of the associator [x1,x2,x3] in Exc. 1 is purely imaginary. Hence thesecond equationfollows from (1.5.5). Combined with (1.5.13) and (a) it implies

nO(xy,z) = tO((xy)z

)= tO

(x(yz)

)= nO(x,yz) = nO(x,zy),

hence the first of the remaining equations. The second one follows analogously.(c) The first set of equations follows immediately from (1.5.10) and the definition of the conju-

gation. As to the remaining ones, we apply (1.5.12), (1.5.10), (1.5.6) to deduce

xyx = x(x y)− x2y = tO(x)xy+ tO(y)x2−nO(x,y)x− tO(x)xy+nO(x)y

= tO(x)tO(y)x−nO(x)tO(y)1O−nO(x,y)x+nO(x)y

= nO(x, tO(y)1O− y

)x−nO(x)

(tO(y)1O− y

)= nO(x, y)x−nO(x)y,

which completes the proof of (c).

4. We begin by proving (1). If tr stands for the trace form of Mat3(C), then for any A ∈Mat3(C),sol.MIMOUwe find a scalar λ (A) ∈ C such that

tr((

(u× v)t w+(v×w)t u+(w×u)t v)A)= λ (A)det(u,v,w) (u,v,w ∈ C3) (29) CROMAT

since the left-hand side is an alternating trilinear function of its arguments u,v,w∈C3. Specializingu := e1, v := e2, w := e3 and observing eiet

i = eii in terms of the ordinary matrix units for 1≤ i≤ 3,we deduce λ (A) = tr(A), hence

tr((

(u× v)t w+(v×w)t u+(w×u)t v)A)= tr

((det(u,v,w)13

)A).

Since the symmetric bilinear form (X ,Y ) 7→ tr(XY ) on Mat3(C) is easily seen to be non-degenerate,(1) holds.

It remains to prove the Moufang identities. Since the conjugation of O is an algebra involution(Exc. 3 (a)), the third Moufang identity immediately follows from the first.

We begin by proving the first Moufang identity, which by Exc. 3 (c) comes down to showing

x(y(xz)

)= nO(x, y)xz−nO(x)yz. (30) FIEX

Since this relation is bilinear in (y,z), it suffices to put x = a⊕ u, a ∈ C, u ∈ C3, and to considerthe following cases.Case 1. y = b⊕0, z = c⊕0, b,c ∈ C. Then (1.4.1) yields

x(y(xz)

)= x(

y((a⊕u)(c⊕0)

))= x((b⊕0)(ac⊕uc)

)= (a⊕u)(abc⊕ubc)

= (a2bc− ut ubc)⊕u(ab+ ab)c.

On the other hand, applying (1.5.1), (1.5.2) and (1.5.7) we obtain


nO(x, y)xz−nO(x)yz = nO(a⊕u, b⊕ v)(a⊕u)(c⊕0)−nO(a⊕u)()b⊕0)(c⊕0)

= (ab+ab)(ac⊕uc)− (aa+ ut u)(bc⊕0)

= (aabc+a2bc− aabc− ut ubc)⊕u(abc+ abc)

= (a2bc− ut ubc)⊕u(ab+ ab)c.

Thus (257) holds.Case 2. y = b⊕0, z = 0⊕w, b ∈ C, w ∈ C3. Basically arguing as before, we obtain

x(y(xz)

)= x(

y((a⊕u)(0⊕w)

))= x((

b⊕0)((−ut w)⊕ (wa+ u× w)

))=(

a⊕u)(

(−but w)⊕(wab+(u× w)b

))=(−abut w− ut wab− ut(u× w)b

)⊕(wa2b+(u× w)ab−ubut w+(u× w)ab+ u× (u×w)b

).

Here (1.1.1) and the Grassmann identity (1.1.3) imply ut(u× w) = 0 as well as u× (u×w) =(w×u)× u = uut w−wut u, hence

x(y(xz)

)= − ut w(ab+ ab)⊕

(wa2b+(u× w)(ab+ ab)−w(ut u)b

)On the other hand,

nO(x, y)xz−nO(x)yz = nO(a⊕u, b⊕0)(a⊕u)(0⊕w)−nO(a⊕u)(b⊕0)(0⊕w)

=(ab+ab

)((−ut w)⊕ (wa+ u× w)

)− (aa+ ut u)(0⊕wb)

= − (ut w)(ab+ ab)⊕(w(aab+ a2b)+(u× w)(ab+ ab)−waab−wut ub

)= − (ut w)(ab+ ab)⊕

(wa2b+(u× w)(ab+ ab)−wut ub

)Comparing with the preceding expression for x(y(xz)) settles Case 2.Case 3. y = 0⊕ v, z = c⊕0, v ∈ C3, c ∈ C. Then

x(y(xz)

)= x(

y((a⊕u)(c⊕0)

))= x((0⊕ v)(ac⊕uc)

)=(a⊕u

)(− vt uc⊕

(vac+(v× u)c

))=(−avt uc− ut vac− ut(v× u)c

)⊕(vaac+(v× u)ac−uvt uc+(u× v)ac+ u× (v×u)c

).

Here ut(v× u) = 0 and u× (v×u) = (u× v)× u = vut u−uut v, which implies

x(y(xz)

)=(−ac(ut v+ vt u)

)⊕(v(aa+ ut u)c−u(ut v+ vt u)c

).

On the other hand,

nO(x, y)xz−nO(x)yz = −nO(a⊕u,0⊕ v)(a⊕u)(c⊕0)+nO(a⊕u)(0⊕ v)(c⊕0)

= − (ut v+ vt u)(ac⊕uc)+(aa+ ut u)(0⊕ vc)

=(−ac(ut v+ vt u)

)⊕(−u(ut v+ vt u)c+ v(aa+ ut u)c

),

and the discussion of Case 3 is complete.Case 4. y = 0⊕ v, z = 0⊕w, v,w ∈ C3. This is the most delicate case of them all.


x(y(xz)

)= x(

y((a⊕u)(0⊕w)

))= x((

0⊕ v)((−ut w)⊕ (wa+ u× w)

))= (a⊕u)

((− vt wa− vt(u× w)

)⊕(− vut w+(v× w)a+ v× (u×w)

))Since v× (u×w) = (w×u)× v = uvt w−wvt u, we conclude

x(y(xz)

)= (a⊕u)

((− vt wa− vt(u× w)

)⊕(uvt w−wvt u− vut w+(v× w)a

))=(− aavt w−avt(u× w)− ut uvt w+ ut wvt u+ ut vut w− ut(v× w)a

)⊕(uvt wa−wvt ua− vut wa+ aa(v× w)−uvt wa−uvt(u× w)

+(u× u)vt w− (u× w)vt u− (u× v)ut w+ u× (v×w)a).

Here u× (v×w) = (w× v)× u = vut w−wut v, so that we obtain

x(y(xz)

)=(ut vut w+ ut wvt u− ut uvt w− aavt w

)⊕(−wvt ua−wut va (31) LHLM

+ aa(v× w)−uvt(u× w)− (u× v)ut w− (u× w)vt u).

On the other hand,

nO(x, y)xz−nO(x)yz = −nO(a⊕u,0⊕ v)(a⊕u)(0⊕w)+nO(a⊕u)(0⊕ v)(0⊕w)

= − (ut v+ vt u)((−ut w)⊕ (wa+ u× w)

)+(aa+ ut u)

((−vt w)⊕ (v× w)

)=(ut vut w+ vt uut w− aavt w− ut uvt w

)⊕(−wut va−wvt ua

− (u× w)ut v− (u× w)vt u+ aa(v× w)+ ut u(v× w))

Comparing this with (31) we see that the C-components are the same. Since ut w = wt u and ut v =vt u, equality of the C3-components is equivalent to

−uvt(u× w)− (u× v)wt u− (u× w)ut v =−(u× w)ut v− (u× w)vt u+(v× w)ut u,

which in turn is equivalent to(det(u, v, w)13

)u =−udet(v, u, w) =−uvt(u× w) =

((u× v)wt +(v× w)ut +(w× u)vt)u.

But this relation holds by (1), and the discussion of Case 4 is complete.We now turn to the second Moufang identity. Arguing as before, it will be enough to consider

the following cases.Case 1. y = b⊕0, z = c⊕0, b,c ∈ C. Then (1.4.1) yields

(xy)(zx) =((a⊕u)(b⊕0)

)((c⊕0)(a⊕u)

)= (ab⊕ub)(ca⊕uc)

= (a2bc− bcut u)⊕ (uabc+uabc),

x(yz)x =((a⊕u)⊕ (bc⊕0)

)x = (abc⊕ubc)(a⊕u)

= (a2bc−bcut u)⊕ (uabc+uuabc),

hence the assertion.Case 2. y = b⊕0, z = 0⊕w, b ∈ C, w ∈ C3. Then (1.4.1) and (1.1.1) yield


(xy)(zx) =((a⊕u)(b⊕0)

)((0⊕w)(a⊕u)

)= (ab⊕ub)

((−wt u)⊕ (wa+ w× u)

)=(−abwt u−abut w− but(w× u)

)⊕((waab+(w× u)ab−ubwt u+(u× w)ab+ u× (w×u)b

)= (−abwt u−abut w)⊕

(waab−ubwt u+ u× (w×u)b

),

x(yz)x =((a⊕u)

((b⊕0)(0⊕w)

))x =

((a⊕u)(0⊕wb)

)x

=((−but w)⊕

(wab+(u× w)b

))(a⊕u)

=(−abut w−abwt u− b(u×w)t u

)⊕(−ubwt u+waab+(u× w)ab+(w× u)ab+(u×w)× ub

)= (−abut w−abwt u)⊕

(−ubwt u+waab+ u× (w×u)b

),

and a comparison gives the assertion.Since ιO is an algebra involution by Exc. 3 (a), the case y = 0⊕v, z = c⊕0 with v ∈C3, c ∈C

immediately reduces to Case 2. Thus we are left withCase 3. y = 0⊕ v, z = 0⊕w, v,w ∈ C3. Then we put

(xy)(zx) = a0⊕u0, x(yz)x = a1⊕u1 (ai ∈ C, ui ∈ C3, i = 1,2) (32) DEMO

and must show a0 = a1, u0 = u1. Since

xy = (a⊕u)(0⊕ v) = (−ut v)⊕ (va+ u× v), (33) LEFA

zx = (0⊕w)(a⊕u) = (−wt u)⊕ (wa+ w× u), (34) REFA

we conclude

a0 = ut vwt u−a2vt w−avt(w× u)−a(u× v)t w− (u× v)t(w× u),

where the last summand may be written as

(u× v)t(w× u) = ut((u× v)× w)= ut(vwt u−uwt v) = ut vwt u− ut uwt v.

Thus

a0 = ut uwt v−a2vt w−a(

det(u,v,w)+det(u,v,w)). (35) ANU

Using (33), (34) to compute the C3-component of (xy)(zx), we obtain

u0 = −wavt u− (w× u)vt u− vawt u− (u× v)wt u+(v× w)aa

+ v× (w×u)a+(u× v)× wa+(u× v)× (w×u).

Here the last three terms may be written as

v× (w×u)a = (u×w)× va = wvt ua−uvt wa,

(u× v)× wa = vwt ua−uwt va,

(u× v)× (w×u) = v(w×u)t u−u(w×u)t v =−udet(u,v,w).

Thus

u0 = −uvt wa−uwt va+(v× w)aa−((u× v)wt +(w× u)vt)u−udet(u,v,w). (36) UNU

We now apply Exc. 3 (c) to tackle the expression


x(yz)x = nO(x,yz)x−nO(x)yz. (37) REMO

Since yz = (−vt w)⊕ (v× w), we deduce from (1.5.7) that

yz = (−wt v)⊕ (−v× w). (38) COYZ

When combined with (1.5.2), this implies

nO(x,yz) = nO(yz,x) =−vt wa− wt va− (v×w)t u− (v× w)t u

= − vt wa− wt va−det(u,v,w)−det(u,v,w), (39) BNYZ

hence, in view of (37), (38), (35)

a1 = nO(x,yz)a+nO(x)wt v

= − vt wa2− wt vaa−det(u,v,w)a−det(u,v,w)a+ wt vaa+ wt vut u = a0,

giving our first claim. It remains to prove the second, i.e., u0 = u1. Again by (37), (38), we obtain

u1 = nO(x,yz)u+nO(x)(v× w)

= −uvt wa−uwt va−udet(u,v,w)−udet(u,v,w)+(v× w)aa+(v× w)ut u,

and comparing with (36) yields

u1−u0 =((u× v)wt +(v× w)ut +(w× u)vt −det(u, v, w)13

)u,

which is zero by (1), and the assertion follows.

Section 2.

5. Put N := (s, t)∈Z×Z | 1≤ s, t ≤ 7, s 6= t and note that M (resp. N) has 21 (resp. 42) elements.sol.MUTADefining maps ϕ± : M→ N by

ϕ+(r, i) := (r+ i,r+3i), ϕ−(r, i) := (r+3i,r+ i) ((r, i) ∈M, integers mod 7), (40) CTPM

it suffices to show that (i) the assignments (40) for ϕ± do indeed take values in N, (ii) ϕ± are bothinjective, and (iii) their images in N are disjoint. We prove these assertions one at a time.

(i) Let (r, i) ∈ M and assume r+ i ≡ r+ 3i mod 7. Then 2i ≡ 0 mod 7, hence i ≡ 0 mod 7, acontradiction.

(ii) Let (r, i),(s, j)∈M. If ϕ+(r, i)=ϕ+(s, j), then r+ i≡ s+ j mod 7 and r+3i≡ s+3 j mod 7.Taking differences, we conclude 2i ≡ 2 j mod 7, hence i ≡ j mod 7 and then i = j. This impliesr ≡ s mod 7, hence r = s, and we have show that ϕ+ is injective. But ϕ− agrees with the switch(s, t) 7→ (t,s) of N, which is bijective, composed with ϕ+. Thus ϕ− is injective as well, whichcompletes the proof of (ii).

(iii) Let (r, i),(s, j) ∈M and suppose ϕ+(r, i) = ϕ−(s, j). Then r+ i ≡ s+ 3 j mod 7, r+ 3i ≡s+ j mod 7, and taking differences yields 2i≡−2 j mod 7, hence i+ j≡ 0 mod 7, a contradiction.This proves (iii).

6. (i)⇒ (iii). Combining (1.5.10) with (2.1.1), we deduce tO(ur) = 0, nO(cr) = 1 for 1 ≤ r ≤ 7,sol.CHACSso ur ∈O0 has euclidean norm 1, while (4) follows immediately from (2.1.2). Thus (ii) holds.

(iii) ⇒ (iv). For 1 ≤ r ≤ 7, i = 1,2,4, indices mod7, we combine (4) with (1.5.13), (1.5.9)and obtain nO(ur+i,ur+3i) = tO(ur+iur+3i) = −tO(ur+iur+3i) = tO(ur) = 0. By Exc. 5, therefore,(ur)1≤r≤7 forms an orthonormal basis of O0, giving the final assertion of the problem. The secondset of equations in (5) is just (4) for r ≥ 4, i = 4, and implies nO(u1u2,u3) = nO(u4,u3) = 0. Thus(iii) holds.


(iv)⇒ (i). Systematically counting indices mod 7, we establish the desired implication by prov-ing the following intermediate assertions.10. nO(ur) = 1 for 1 ≤ r ≤ 7. By (iii) this is clear for r ≤ 3, and for r ≥ 4 follows from (5) byinduction since the norm of O by Theorem 1.7 permits composition.20. nO(ur,us) = 0 for 1≤ r,s≤ 4 distinct. By (iii), we may assume s = 4, r ≤ 2. Then 1.8 impliesnO(ur,u4) = nO(ur,u1u2) = nO(ur)tO(u3−r) = 0, as claimed.30. Let 1 ≤ r ≤ 7 and i = 1,2,4 such that ur = ur+iur+3i. If ur+i has trace zero, then ur hastrace zero if and only if ur+i,ur+3i are orthogonal. Moreover, if this is so and ur+i as well asur+3i both have trace zero, then ur = −ur+3iur+i. Since ur+i = −ur+i, we deduce from (1.5.13)that tO(ur) = −nO(ur+i,ur+3i), giving the first assertion. Since, under the final conditions stated,ur,ur+i,ur+3i are all skew relative to the conjugation of O, and this is an algebra involution, weconclude ur =−ur =−ur+3iur+i =−ur+3iur+i, as claimed.40. ur ∈O0 for 1≤ r≤ 7. By (iii), this is automatic for r≤ 3, while it follows immediately from 20,30 (with i = 4) for r = 4,5,6. It remains to discuss the case r = 7. We have u7 = u4u5, and 30 com-bined with 1.8 yields nO(u4,u5) = nO(u1u2,u2u3) =−nO(u2u1,u2u3) =−nO(u2)nO(u1,u3) = 0.Hence the first part of 30 yields tO(u7) = 0.50. (ur)1≤r≤7 is an orthonormal basis of O0 and u2

r =−u0, urus =−usur for 1≤ r,s≤ 7 distinct.The first part follows immediately from 10, 30, 40 combined with Exc. 5, while the second part isnow a consequence of 10 and (1.5.10), (1.5.13).60. Let 1≤ r ≤ 7 and i = 1,2,4 such that ur+iur+3i = ur . Then

us+ jus+3 j = us (s = r+ i, j ∈ 1,2,4, j ≡ 2i mod 7), (41) RTWI

us+ jus+3 j = us (s = r+3i, j ∈ 1,2,4, j ≡ 4i mod 7). (42) RFOI

Since ur+3iur+i =−ur by 50 and O is an alternative algebra, ur+3iur =−u2r+3iur+i = ur+i, and (41)

follows. (42) is proved similarly.In view of 60, the proof of the implication (iv)⇒ (i) will be complete once we have shown

70. ur+4ur+5 = ur for 1≤ r ≤ 7. By (5), this is clear for 4≤ r ≤ 7, and 60 yields

us+1us+3 = us (1≤ s≤ 4), (43) ESONE

us+2us+6 = us (2≤ s≤ 5). (44) ESTWO

Now let 1≤ r ≤ 2. Then (5) combined with 50, the middle Moufang identity (cf. Exc. 4) and (43)yields

ur+4ur+5 = (ur+1ur+2)(ur+2ur+3) = (ur+2ur+1)(ur+3ur+2) = ur+2(ur+1ur+3)ur+2

= ur+2urur+2 =−ur+2(ur+2ur) =−u2r+2ur = ur,

so we are left with the case r = 3. Applying (43) for s = 1 and (44) for s = 3, we obtain

u7u8 = (u4u5)u1 = (u4u5)(u2u4) = u4(u5u2)u4 = u4(u3+2u3+6)u4

= u4u3u4 =−u4(u4u3) =−u24u3 = u3,

as claimed.Summing up, we have shown not only that conditions (i), (iii), (iv) are equivalent but also the

final statement of the problem. It remains to establish(i) ⇔ (ii). The first equation of (3) is the same as (2.1.1). As to the second, the alternative

laws combined with the fact that O is a division algebra yield the following chain of equivalentconditions.


(urur+1)ur+3 =−1O ⇐⇒ (urur+1)u2r+3 =−ur+3⇐⇒ urur+1 = ur+3 (45) FELGRO

⇐⇒ u(r+3)+4u(r+3)+3·4 = ur+3.

Similarly, the third equation of (3) may be rephrased as

ur(ur+1ur+3) =−1O ⇐⇒ u2r (ur+1ur+3) =−ur ⇐⇒ ur+1ur+3 = ur. (46) SELGRO

Hence (i) implies (ii). Conversely, suppose (ii) holds. Then the first equation of (3) combinedwith (1.5.10) shows that the elements ur , 1 ≤ r ≤ 7, belong to O0 and have length 1, while(45), (46) yield urur+1 = ur+3, ur+1ur+3 = ur . Combining with (1.5.9), (1.5.13) we deducenO(ur,ur+1) = tO(ur ur+1) = −tO(urur+1) = tO(ur+3) = 0 and, similarly, nO(ur+1,ur+3) = 0. Inparticular, u1,u2,u3 form an orthonormal system in O0 such that nO(u1u2,u3) = nO(u3,u4) = 0,and we have ur = ur−3ur−2 for 4≤ r ≤ 7. Since, therefore, condition (iv) holds, so does (ii).

7. O+ is clearly a commutative real algebra with identity element 1O+ = 1O. Moreover, the squar-sol.OPLUings in O and O+ coincide. Hence any ϕ ∈ Aut(O+) fixes 1O and preserves squares in O. From(1.5.10) we therefore deduce t(x)x = n(x)1O for all x∈O, where t := tO ϕ− tO, n := nO ϕ−nO.Thus O is the union of the subspaces Ker(t) and R1O. This implies first t = 0 and then n = 0, so ϕ

is an orthogonal transformation of O fixing 1O, hence stabilizing O0 = (R1O)⊥. The assignmentϕ 7→ ϕ|O0 clearly gives an injective group homomorphism from Aut(O0) to O(O0). Conversely,the unique linear extension fixing 1O of ψ ∈ O(O0) belongs to the orthogonal group of O andleaves the trace invariant, hence is an automorphism of O0. Moreover, the maps in question areinverse isomorphisms not only of abstract groups but, in fact, of topological ones since both arerestrictions of linear maps and hence automatically continuous. Finally, Aut(O) is trivially a sub-group of Aut(O+) which is closed since it may be realized via

Aut(O) =⋂

x,y∈Oϕ ∈ GL(O) | ϕ(xy) = ϕ(x)ϕ(y)

as the intersection of a family of closed subsets of GL(O).

Section 3.

8. Put n := dimR(V ).sol.CHALAT(i)⇒ (iii). Obvious.(iii)⇒ (ii). Since L0 spans V as a real vector space, so does L. Since L is an additive subgroup

of L1, it is finitely generated and torsion-free, hence free of finite rank with rk(L)≤ rk(L1) = n.(ii) ⇒ (i). Let ε = (e1, . . . ,er), r ≤ n, be a Z-basis of L. Since L spans V , so does ε , and we

conclude that ε is an R-basis of V , forcing L⊆V to be a lattice.

9. M +√

2M ⊆ A is a free abelian subgroup of rank 2dimR(A) which is obviously closed undersol.LASTRUmultiplication but not a lattice, hence not a Z-structure of A.

10. Since M is clearly a discrete additive subgroup of D, it follows that M′ := M∩R is a discretesol.REONEadditive subgroup of R containing 1. Thus M′ = Zv for some 0 6= v ∈R. Hence M′ is a Z-structureof R. As such it contains Z. On the other hand, v2 ∈M′ implies v2 = rv for some integer r, hencev = r ∈ Z, and we also conclude M′ ⊆ Z, as desired.

Section 4.

11 . By Thm. 4.2, (1H, i, j,h) is an R-basis of H that is associated withHur(H). Hence Hur(H)sol.HURCONconsists of all linear combinations


m1H+ni+ pj+qh = (m+q2)1H+(n+

q2)i+(p+

q2)j+

q2

k

with m,n, p,q ∈ Z. Since the coefficients of the linear combination on the right either all belong toZ or to 1

2 +Z depending on whether q is even or odd, the assertion follows.

12. Write x ∈ L in the form x = α1H+β i+γj+δh with α,β ,γ,δ ∈R. We must show α,β ,γ,δ ∈sol.MAXHURZ. Replacing x by x− [x], [x] := [α]1H + [β ]i+ [γ]j+ [δ ]h ∈ Hur(H) ⊆ L, if necessary, we mayassume 0≤ α,β ,γ,δ < 1 and then must show α = β = γ = δ = 0. By hypothesis, the quantities

nH(x) = α2 +β

2 + γ2 +(α +β + γ +δ )δ , (47) NHX

tH(x) = 2α +δ , (48) THX

nH(i,x) = 2β +δ , (49) NHIX

nH(j,x) = 2γ +δ , (50) NHJX

nH(h,x) = α +β + γ +2δ (51) NHHX

are all integers. Multiplying (51) by 2 and subtracting the sum of (48), (49), (50) from the result,we conclude δ ∈ Z, hence δ = 0, and then 2α,2β ,2γ,α2 + β 2 + γ2 ∈ Z by (47)−(50). Thusα,β ,γ ∈ 0, 1

2, forcing α = β = γ = 0 as well.

13. We begin by deriving a general formula for the determinant of a block matrix. Let A∈GLp(R),sol.DETFOB ∈Matp,q(R), C ∈Matq,p(R) and D ∈Matq(R). Then we claim

det((

A BC D

)) = det(A)det(D−CA−1B). (52) DEBLO

To see this, we make use of the well known fact that (52) holds for B = 0 or C = 0. Hence

det((

A BC D

)) = det(A)det(

(A BC D

)(A−1 0

0 1q

)) = det(A)det(

(1p B

CA−1 D

))

= det(A)det((

1p 0−CA−1 1q

)(1p B

CA−1 D

)) = det(A)det(

(1p B0 −CA−1B+D

))

= det(A)det(D−CA−1B),

as claimed. Turning to the matrix S of our problem, we now conclude

det(S) = det(r ·1p)det(s ·1q−T2(r−1 ·1p)T1

)= rp det(s ·1q− r−1T2T1)

= rp−q det(rs−T2T1) = rp−q charT2T1 (rs),

as claimed. Next suppose T1 = T , T2 = T t , with T ∈Matp,q(R) being comprised of column vectorsv1, . . . ,vq ∈ R that are mutually orthogonal but may have different euclidean lengths. Then

T2T1 = T t T =

vt1...

vtq

(v1, . . . ,vq) = (vtiv j)1≤i, j≤q = diag(‖v1‖2, . . . ,‖vq‖2).

Thus the preceding determinant formula reduces to

det(S) = rp−qq

∏i=1

(rs−‖vi‖2).

In particular, if ‖vi‖ =√

a for some a > 0 and all i = 1, . . . ,q, the second assertion stated in ourproblem drops out. Finally, assuming ‖vi‖2 < rs for all i = 1, . . . ,q, we claim that the matrix S ispositive definite. Since it is symmetric, it will be enough to show that all its principal minors are


positive. Let 1≤ i≤ p+q. For i≤ p, the the i-th principal minor of S is 1, while for p < i≤ p+q,it is the determinant of the matrix

S′ :=(

r ·1p T ′

(T ′)t s ·1i−p

),

where T ′ = (v1, . . . ,vi−p) ∈Matp,i−p(R). But by what we have just seen, S′ has determinant

det(S′) = rp−qp−i

∏j=1

(rs−‖v j‖2)> 0,

and the assertion follows. Since the final statement of the problem is a special case of this assertion,the solution is complete.

14. (a) Since the norm of D permits composition by Thm. 1.7, the set of units in M is closed undersol.UNITmultiplication and contains 1D.

If (i) holds, then (1.5.11) implies uu = nD(u)1D = 1D = uu. Thus (i) implies (ii) and (iii).Conversely, suppose (ii) holds. Then 1 = nD(uv) = nD(u)nD(v), and since both factors on theright are integers by Prop. 3.9, we conclude u ∈ M×. This shows that (i) and (ii) are equivalent.The equivalence of (i) and (iii) is proved similarly. Moreover, for any v satisfying (iii) we applyExc. 3 (c) and obtain u = 1Du = (vu)u = nD(u)v = v (by (i)), and the proof is complete.

(b) (1, i) is an orthonormal basis of C associated withGa(C). Hence Ga(C)× = ±1,±i.Similarly, (1H, i, j,k) is an orthonormal basis of H associated withGa(H). Hence Ga(H)× =±1H,±i,±j,±k. And, finally, any Cartan-Schouten basis (ur)0≤r≤7 of O is an orthonormal basisassociated withGaε (O). Hence Gaε (O)× = ±ur | 0≤ r ≤ 7. Now we claim

Hur(H)× = ±1H,±i,±j,±k,12(±1H± i± j±k). (53) HUNI

To prove this, we note that the first eight elements of the right-hand side belong to Ga(H)× ⊆Hur(H)×. The remaining sixteen may all be obtained from adding a unit of Ga(H) to h, hencebelong to Hur(H). Since they also have norm 1, the right-hand side of (53) belongs to the left.Conversely, suppose u = α01H+α1i+α2j+α3k with αi ∈ R, 0 ≤ i ≤ 3, is a unit in Hur(H). ByExc. 11, there are two cases.Case 1. αi ∈ Z, 0≤ i≤ 3. Then u ∈ Ga(H)× ⊆ Hur(H)×.Case 2. αi =

12 +βi, βi ∈ Z, 0≤ i≤ 3. Then

1 = nH(u) =3

∑i=0

(14+βi +β

2i ) = 1+

3

∑i=0

(β 2i +βi),

and we conclude ∑3i=0(β

2i +βi) = 0, where the summands are all non-negative. Thus β 2

i = −βi,i.e., βi =−1,0 for 0≤ i≤ 3.. But this means αi =± 1

2 for 0≤ i≤ 3, and u belongs to the right-handside of (53).

15. That the εi, 1 ≤ i ≤ 4, form an orthonormal basis of H relative to 〈x,y〉 follows immediatelysol.HURDEfrom the fact that the vectors 1H, i, j,k of H are orthogonal of norm 1. Moreover, the latter vectorscan be linearly expressed by the former according to the formulas

ε1− ε2 = j, ε2− ε3 =−h, ε3− ε4 = 1H, ε3 + ε4 = i.

In particular, the second equation of the problem holds. Given ξi ∈ R, 1≤ i≤ 4, we also obtain


4

∑i=1

ξiεi =12(ξ1j−ξ1k−ξ2j−ξ2k+ξ31H+ξ3i−ξ41H+ξ4i

)=

12((ξ3−ξ4)1H+(ξ3 +ξ4)i+(ξ1−ξ2)j− (ξ1 +ξ2)k

).

Using (4.1.1), we therefore deduce

4

∑i=1

ξiεi =12((ξ3−ξ4)1H+(ξ3 +ξ4)i+(ξ1−ξ2)j− (ξ1 +ξ2)(2h−1H− i− j)

)=

12((ξ1 +ξ2 +ξ3−ξ4)1H+(ξ1 +ξ2 +ξ3 +ξ4)i+2ξ1j

)+(ξ1 +ξ2)h

=12(ξ1 +ξ2 +ξ3−ξ4)1H+

12(ξ1 +ξ2 +ξ3 +ξ4)i+ξ1j+(ξ1 +ξ2)h,

which by Thm. 4.2 yields the following chain of equivalent conditions.

4

∑i=1

ξiεi ∈ Hur(H) ⇐⇒ ξ1,ξ1 +ξ2 ∈ Z, ξ1 +ξ2 +ξ3±ξ4 ∈ 2Z

⇐⇒ ξ1,ξ2 ∈ Z,4

∑i=1

ξi ∈ 2Z, 2ξ4 ∈ 2Z

⇐⇒ ξ1,ξ2,ξ3,ξ4 ∈ Z,4

∑i=1

ξi ∈ 2Z.

Thus the second equation of the problem holds. Finally, we note nH(εi) =12 for 1 ≤ i ≤ 4, so for

ξi ∈ Z, ∑ξi ∈ 2Z, we obtain

4

∑i=1

ξiεi ∈ Hur(H)× ⇐⇒ 12

4

∑i=1

ξ2i = 1 ⇐⇒

4

∑i=1

ξ2i = 2,

which amounts to ξi =±1 for precisely two indices i = 1,2,3,4 and ξi = 0 for the remaining ones.But this means the units of Hur(H) have the form stated at the very end of the problem.

16. (a) This follows immediately from the fact that Cartan-Schouten bases are orthonormal relativesol.ROOCto nO (Exc. 6). For example, nO(ε1)=

14 (nO(u0)+nO(u2)=

12 , which implies 〈ε1,ε1〉= 2nO(ε1)=

1.(b) We write L for the additive subgroup of O generated by the quantities assembled in (1). We

claim

Cox(O)⊆ L⊆ 12

8

∑i=1

Zεi, (54) LEIGHT

where the second inclusion is obvious. In order to prove the first, we note that

u0 = − ε1 + ε2, u1 =−ε3 + ε4, u2 = ε1 + ε2, u3 =−ε3− ε4, (55) UZER

u4 = − ε5 + ε6, u5 = ε5 + ε6, , u6 = ε7 + ε8, u7 =−ε7 + ε8

all belong to L. Moreover, (55) and (4.4.3)–(4.4.7) imply


p =12(−ε1 + ε2− ε3 + ε4 + ε1 + ε2− ε3− ε4) = ε2− ε3 ∈ L, (56) PEP

u1p =12(ε1− ε2− ε3 + ε4− ε5 + ε6− ε7 + ε8) ∈ L, (57) ONEPEP

u2p =12(ε1− ε2 + ε1 + ε2 + ε5− ε6 + ε5 + ε6) = ε1 + ε5 ∈ L, (58) TWOPEP

u4p =12(ε3− ε4 + ε1 + ε2− ε5 + ε6− ε7− ε8) ∈ L. (59) FOURPEP

Thus the basis E ′ exhibited in (4.5.2), which is associated withCox(O) by Thm. 4.5, is entirelycontained in L, and the first inclusion of (54) holds. From this and Exc. 8 we deduce that L is alattice in O. Since Cox(O) is unimodular by Thm. 4.5, the proof of (b) will be complete once wehave shown that the lattice L is integral quadratic (Cor. 3.14). Since nO(εi) =

12 for 1 ≤ i ≤ 8 by

(a), equation (1) yields

nO(±εi ± ε j) = nO(1

2

8

∑i=1

siεi)= 1 (60) UNEI

for 1≤ i < j≤ 8 and all families s = (si) ∈ ±18. Writing M := 1,2, . . . ,8 and M±s := i ∈M |si =±1, it remains to show

nO(±εi ± ε j,±εk ± εl) ∈ Z, nO(1

2

8

∑i=1

siεi,12

8

∑i=1

tiεi)∈ Z (61) MIXEI

for all 1 ≤ i < j ≤ 8, 1 ≤ k < l ≤ 8 and all families s = (si), t = (ti) ∈ ±18 such that |M−s |and |M−t | are both even. The first relation of (61) is obvious. In order to prove the second, we lets = (si) ∈ ±18 be arbitrary and note |M+

s |+ |M−s |= 8, so |M±s | are either both even or both odd.Moreover, −s := (−si) ∈ ±18 and M±−s = M∓s . Now, letting also t = (ti) ∈ ±18 be arbitraryand setting st := (siti) ∈ ±18, we obtain

nO(1

2

8

∑i=1

siεi,12

8

∑i=1

tiεi)=

14

8

∑i=1

siti =14(|M+

st |− |M−st |)= 2− 1

2|M−st |,

and must show that if |M±s | and |M±t | are even, then so is |M±st |. Suppose not. Then

|M+st |= |M+

s ∪M+t |+ |M−s ∪M−t |

is odd. Replacing s by −s, t by −t if necessary, we may assume that |M+s ∪M+

t | is odd and |M−s ∪M−t | is even. But then

|M+s |= |M+

s ∪M+t |+ |M+

s ∪M−t |shows that |M+

s ∪M−t | is odd, whence

|M−t |= |M+s ∪M−t |+ |M−s ∪M−t |

is odd as well, a contradiction.(c) Write L′ for the right-hand side of (2). Inspecting (55)–(59), we conclude Cox(O) ⊆ L′ ⊆

12 ∑Zεi, so L′ is a lattice in O by Exc. 8. As before, we will be through once we have shown that L′

is integral quadratic. Let x = ∑ξiεi ∈ L′ with ξ1, . . . ,ξ8 ∈R. Then (2) implies 2ξ1 ∈Z, ξi = ξ1+ni,ni ∈ Z for 1≤ i≤ 8, and

8

∑i=1

ξi =8

∑i=1

(ξ1 +ni) = 4(2ξ1)+8

∑i=1

ni

belongs to 2Z. Thus ∑ni ∈ 2Z, and N := i ∈ M | ni ≡ 1 mod 2 contains an even number ofelements. Now nO(x) = 1

2 ∑ξ 2i and


8

∑i=1

ξ2i =

8

∑i=1

(ξ1 +ni)2 =

8

∑i=1

(ξ 21 +2ξ1ni +n2

i ) = 2(2ξ1)2 +2ξ1

8

∑i=1

ni +8

∑i=1

n2i

≡8

∑i=1

n2i ≡ ∑

i∈Nn2

i ≡ |N| ≡ 0 mod 2.

Hence nO(x) ∈ Z, as claimed.(d) By (b), (c) and (60), the quantities in (1) are all units in Cox(O). Conversely, suppose

u = ∑8i=1 ξiεi, ξi ∈ R, 1≤ i≤ 8, is a unit in Cox(O). Then (a) implies

8

∑i=1

ξ2i = 2nO(u) = 2. (62) UNO

If ξ1 is an integer, then all ξi, 1≤ i≤ 8, are, and (62) shows that u belongs to the first type of (1).On the other hand, if ξ1 belongs to 1

2 +Z, then all ξi, 1≤ i≤ 8, do, and (62) shows ξ 2i = 1

4 , henceξi = ± 1

2 for all i = 1, . . . ,8. Since ∑ξi ∈ 2Z by (2) the number if indices i = 1, . . . ,8 having ξinegative must be even, so u is of the second type in (1).

The units of Cox(O) belonging to the first type of (1) are in bijective correspondence with thequadruples of subsets of M consisting of two elements, hence are 4 ·

(82

)= 4 · 8·7

2 = 4 · 28 = 112in number. On the other hand, the units of Cox(O) belonging to the second type of (1) are inbijective correspondence with the subsets of M consisting of an even number of elements, henceare 28

2 = 27 = 128 in number. All in all, therefore, Cox(O) has exactly 112+128 = 240 units.

17. The inclusions Ga(O)⊆Kir(O)⊆ 12 Ga(O) follow immediately from the definition and showsol.KILA

by Exc. 8 that L := Kir(O) is a lattice in O.Since (ur)0≤r≤7 is an orthonormal basis of O, we have nO(Ga(O),vi)⊆ Z for 1≤ i≤ 4, and a

straightforward verification shows

nO(vi) = 1, nO(v1,v2) = 0, nO(v j,vk) = 1 (63) NOVI

for 1 ≤ i, j,k ≤ 4, j 6= k, j,k 6= 1,2. Combining this with the definition of L, we concludenO(L)⊆ Z, so L is a unital integral quadratic lattice in O.

Next we consider the family

E1 := (1O,u1,u2,u3,v1,v2,v3,v4)

of elements in L. From u4 = 2v1−1O−u1−u2, u5 = 2v4−1O−u2−u3, u7 =−2v3+1O+u1+u3,u6 = −2v2 +u3 +u5−u7 we deduce that E1 spans O as a real vector space, hence is a basis, andthat the additive map from Z8 to L determined by E1 is surjective, hence bijective. Thus E1 is anR-basis of O associated withL.

We now proceed to determine the discriminant of L by computing the determinant of DnO(E1).A straightforward verification using (63) shows

DnO(E1) =

(2 ·14 B

Bt D

),

where

B =

1 0 1 11 0 1 01 0 0 10 1 1 1

, D =

2 0 1 10 2 1 11 1 2 11 1 1 2

.

Now we use Exc. 13. (52)), and obtain


det(DnO(E1)

)= det(2 ·14)det

(D−Bt(

12·14)B

)= det(2 ·D−Bt B)

= det(

1 0 0 00 3 1 10 1 1 00 1 0 1

) = 3−1−1 = 1.

Thus L is a unimodular integral quadratic lattice in O.It remains to show that L is not a Z-structure of O, equivalently, that it is not closed under

multiplication. To this end, we consider the product v1v3. A short computation gives

v1v3 =12(u1 +u2 +u3 +u5). (64) PROTH

Assuming v1v3 ∈ Kir(O), we would find integers αr,βi, 0≤ r ≤ 3, 1≤ i≤ 4, such that

v1v3 =3

∑r=0

αrur +4

∑i=1

βivi. (65) LICOV

Comparing coefficients of u4, we obtain β1 = 0; comparing coefficients of u1, we obtain 12 =

α1 +12 β3, hence α1 =

1−β32 , and β3 is odd. Comparing coefficients of u2, we obtain 1

2 = α2 +12 β4,

hence α2 =1−β4

2 , and β4 is odd. Comparing coefficients of u3, we obtain 12 =α3+

12 (β2+β3+β4),

hence α3 =− 12 (β3 +β4)+

1−β22 , and β2 is odd since both β3,β4 are odd. And, finally, comparing

coefficients of u5, we obtain 12 = 1

2 (β2 + β4), hence β2 + β4 = 1, which is a contradiction sinceβ2,β4 are both odd. Thus v1v3 does not belong to Kir(O), as desired.

18. The idea of the solution is to imitate our construction of the Coxeter octonions with a differentsol.ALCOmodel of the Hamiltonian quaternions inside O and a different vector q ∈O in place of p such thatthe resulting Z-structure agrees with R. In doing so, repeated use will be made of the multiplcationrules for Cartan-Shouten bases as depicted in Fig. 1 of 2.5.

On a more formal level, these multiplication rules as described in (2.1.1) and (2.1.2) show thatthe linear bijection ϕ : O→O determined by

ϕ(1O) = 1O, ϕ(ur) = ur+2 (1≤ r ≤ 7, indices mod 7)

is an automorphism of O. With H⊆O as given in (4.4.1) we therefore conclude that

B := ϕ(H) = R1O+Ru3 +Ru4 +Ru6 ⊆O

is a model of the Hamiltonian quaternions matching 1H = 1O, i = u3, j = u4, k = u6 and

Ga(B) := ϕ(

Ga(H))= Z1O+Zu3 +Zu4 +Zu6.

Observing (4.4.3), we also put

q := ϕ(p) =12(1O+u3 +u4 +u5) (66) KUFIP

and deduce from Thm. 4.5 that

R′ := ϕ(

Cox(O))= Ga(B)+Ga(Bq

is a Z-structure of O isomorphic to Cox(O). It therefore remains to show R′ = R.We first note that Ga(O) is contained in R (by definition) but also in R′ (By Thm. 4.5). More-

over, invoking (4.4.4)−(4.4.7), we deduce


u3q = ϕ(u1p) =12(−1O+u2 +u3 +u6),

u4q = ϕ(u2p) =12(−1O+u4−u6 +u7),

u6q = ϕ(u4p) =12(−u1−u3 +u4 +u6).

Now an inspection shows

v1 + v4 ≡12(1O+u3 +u4 +u5)≡ q mod Ga(O),

v2 + v4 ≡12(1O+u2 +u3 +u6)≡ u3q mod Ga(O),

v1 + v2 + v3 + v4 ≡12(1O+u4 +u6 +u7)≡ u4q mod Ga(O),

v1 + v2 + v4 ≡12(u1 +u3 +u4 +u6)≡ u6q mod Ga(O).

Sinceϕ(E ′) = (1O,u3,u4,u6,q,u3q,u4q,u6q)

by (4.5.2) is a basis of O associated with R′, the preceding relations imply R′ ⊆ R. But they alsoyield

u4q−u6q ≡ v3 mod Ga(O),

u6q−u3q ≡ v1 mod Ga(O),

u6q−q ≡ v2 mod Ga(O),

q+u3q−u6q ≡ v4 mod Ga(O)

and hence imply R⊆ R′. This shows R = R′ and the proof is complete.

19. Let (ur)0≤r≤7 be a Cartan-Schouten basis of O. In the language of Exc. 15 we havesol.HUCO

q :=12(1O+u3 +u4 +u6) =

12(−ε1 + ε2− ε3− ε4− ε5 + ε6 + ε7 + ε8),

whence part (d) of that exercise shows that q is a unit in Cox(O). Since a straightforward verifi-cation yields u3u4 = u6, u4u6 = u3, u6u3 = u4, we conclude from Thm. 4.2 that the assignments1H 7→ 1O, i 7→ u3, j 7→ u4, h 7→ q determine an additive embedding Hur(H) → Cox(O) that pre-serves multiplication, hence is an embedding of Z-algebras.

Section 5.

20. Computing the square of the matrix (5.4.2) in a straighforward manner yields (5.9.3). Lineariz-sol.FORMUing and dividing by 2 immediately yields (5.9.4). Taking the trace of (5.9.4) and applying (5.8.3),we obtain

T (x• y) = ∑(αiβi +

12

nO(u j,v j)+12

nO(ul ,vl))

= ∑αiβi +12(∑nO(ui,vi)+∑nO(ui,vi)

),

and (5.9.5) follows. Writing

z = ∑(γieii +wi[ jl]

)(γi ∈ R, wi ∈O, 1≤ i≤ 3),


we first note by (1.5.13) and Exc. 3 (b) that, for all u,v,w ∈O, the expression

nO(uv,w) = tO(uvw) is invariant under cyclic permutations of the variables. (67) TCYC

Combining (5.9.4) with (5.9.5), we therefore obtain

T((x• y)• z

)= ∑

((αiβi +

12

nO(u j,v j)+12

nO(ul ,vl))γi

+12

nO((α j +αl)vi +(β j +βl)ui +u jvl + v jul ,wi

))= ∑

(αiβiγi +

12(γ j + γl)nO(ui,vi)+

12(α j +αl)nO(vi,wi)

+12(β j +βl)nO(wi,ui)+

12(tO(uiv jwl)+ tO(wiv jul)

)).

This expression obviously being symmetric in x,z, we end up with (5.9.5).We now turn to (5.9.7). Combining (5.8.3) with (5.9.5) we deduce

T (x)T (y)−T (x• y) = (∑αi)(∑βi)−∑(αiβi +nO(ui,vi)

)= ∑

(αiβ j +βiα j−nO(ui,vi)

),

which by (5.8.7) agrees with S(x,y). Similarly, by (5.9.3), (5.8.3), 5.8.6) and (5.8.4) we obtain

x2−T (x)x+S(x)13 = ∑

((α

2i +nO(u j)+nO(ul)

)eii +

((α j +αl)ui +u jul

)[ jl])

−(∑αi

)∑(αieii +ui[ jl]

)+∑

(α jαl −nO(ui)

)(∑eii

)= ∑

((α

2i +nO(u j)+nO(ul)−α

2i −αiα j−αlαi

+αiα j +α jαl +αlαi−nO(ui)−nO(u j)−nO(ul))eii

+((α j +αl)ui +u jul −αiui−α jui−αlui

)[ jl])

= ∑

((α jαl −nO(ui)

)eii +

(−αiui +u jul

)[ jl])= x],

which completes the proof of (5.9.8).Differentiating (5.8.2) at x in the direction y and observing (67), we obtain

DN(x)(y) = β1α2α3 +α1β2α3 +α1α2β3−∑(βinO(ui)+αinO(ui,vi)

)+ tO(v1u2u3)+ tO(u1v2u3)+ tO(u1u2v3)

= ∑(α jαlβi−nO(ui)βi−αinO(ui,vi)+ tO(u julvi)

),

while combining (5.9.5) with (5.8.4) and (67) yields

T (x] • y) = ∑

((α jαl −nO(ui)

)βi +nO

(−αiui +u jul ,vi

))= ∑

(α jαlβi−nO(ui)βi−αinO(ui,vi + tO(u julvi)

),

which proves the first equation of (5.9.9), while the second one derives from (5.9.8).In order to prove (5.9.10), we apply (5.9.4) and (5.8.4) to compute


x• x] = ∑

((αi[α jαl −nO(ui)]+

12

nO(u j,−α ju j +ului)+12

nO(ul ,−αlul +uiu j))eii

+12((α j +αl)[−αiui +u jul ]+ [αlαi−nO(u j)+αiα j−nO(ul)]ui

+u j[−αlul +uiu j]+ [−α ju j +ului]ul)[ jl]).

Since the conjugation of O is an involution and (uiu j)u j = nO(u j)ui, ul(ului) = nO(ul)ui byExc.???, we conclude

x• x] = ∑(α1α2α3−

3

∑r=1

αrnO(ur)+ tO(u1u2u3))eii = N(x)13,

and plugging in (5.9.8) gives (5.9.10).Finally, Differentiating x3 = (x • x) • x at x in the direction y, we find (y • x) • x+(x • y) • x+

(x• x)• y = 2x• (x• y)+ x2 • y. Performing the same procedure on the right-hand side of (5.9.10),we therefore get (5.9.11).

21. We have Her1(O)∼= R+, while Her2(O) sits in the euclidean Albert algebra viasol.HEMAO

Her2(O)∼=α1 u3 0

u3 α2 00 0 0

| α1,α2 ∈ R, u3 ∈O

as a non-unital subalgebra. By Thm. 5.10, therefore, Hern(O) is a Jordan algebra for 1 ≤ n ≤ 3.It remains to show that it is not Jordan for n ≥ 4. We begin by differentiating the Jordan identityu(u2v) = u2(uv) at u in the direction w and obtain

w(u2v)+2u((uw)v

)= 2(uw)(uv)+u2(wv).

Differentiating at u again, this time in the direction x, changing notation and rearranging terms,we end up with the fully linearized Jordan identity, which therefore holds in arbitrary real Jordanalgebras.

Now consider the commutative real algebra J := Hern(O) for some integer n ≥ 4. For 1 ≤i, j, l,m≤ n mutually distinct and u,v ∈O, we will make use of the obvious relations

eii •u[lm] =12(δil +δim)u[lm] = u[lm]• eii,

u[i j] = u[ ji],

u[i j]• v[ jl] =12(uv)[il],

u[i j]• v[lm] = 0.

Assuming now J is a Jordan algebra, we combine the preceding relations with the fully linearizedJordan identity and conclude

14((uv)w

)[14] =

(u[12]• v[23]

)•((e33 + e44)•w[34]

)= −

(v[23]• (e33 + e44)

)•(u[12]•w[34]

)−((e33 + e44)•u[12]

)•(v[23]•w[34]

)+u[12]•

((v[23]• (e33 + e44)

)•w[34]

)+ v[23]•

(((e33 + e44)•u[12]

)•w[34]

)+(e33 + e44)•

((u[12]• v[23])•w[34]

)=

18(u(vw)

)[14]+

18((uv)w

)14].


Thus we obtain the contradiction that O is associative. Hence J cannot be a Jordan algebra.

22. (a) By hypothesis, deg(µx) = dimR(R[x]) = 3. Hence µ := det(t1R[x]−L0x ) is monic of degreesol.INVRE

3 as well. Moreover, µ(x) = µ(L0x)13 = 0, which proves the first assertion. The remaining ones

now follow from (5.13.1).(b) We begin with preparations for a Zariski density argument. From (5.9.10) we deduce x4 =

T (x)x3−S(x)x2 +N(x)x, hence

x4 =(T (x)2−S(x)

)x2−

(T (x)S(x)−N(x)

)x+T (x)N(x)13. (68) EXFO

Given αi,βi ∈ R for 0≤ i≤ 2, we therefore conclude

( 2

∑i=0

αixi)( 2

∑i=0

βixi)= 2

∑i=0

γixi

with

γ1 = α0β0 +α2β2T (x)N(x),

γ1 = α0β1 +α1β0−α2β2(T (x)S(x)−N(x)

), (69) PROX

γ2 = α0β2 +α1β1 +α2β0 +α2β2(T (x)2−S(x)

).

Now assume 13∧x∧x2 6= 0 in∧3(J) and put u := ∑

2i=0 αixi. Then (69) implies that R[u] =R[x] if

and only if

∆(u) := det(

1 α0 α20 +α2

2 T (x)N(x)0 α1 2α0α1−α2

2(T (x)S(x)−N(x)

)0 α2 2α0α2 +α2

1 +α22(T (x)2−S(x)

)) 6= 0.

On the other hand, by (68), we have R[x2] = R[x] if and only if T (x)S(x) 6= N(x).Then consider the set of 7-tuples

(α0,α1,α2,β0,β1,β2,x) ∈ R6× J

such that, setting u = ∑2i=0 αixi, v = ∑

2i=0 βixi, we have 13∧ x∧ x2 6= 0 in

∧3(J), T (x)S(x) 6= N(x),and the quantities ∆(u), ∆(v) and ∆(u• v) are all different from zero. By the preceding formulas,this set is Zariski-open R6× J. But it is also Zariski-dense since, for example, it contains

(ξ0,ξ1,ξ2,ξ0,ξ1,ξ2,

2

∑i=0

ξieii)

for real numbers ξi such that the six quantities ±ξi, 0≤ i≤ 2, are mutually distinct.Hence it suffices to prove the first claim under the additional hypotheses 13 ∧ x∧ x2 6= 0 in∧3(J) and R[x] =R[u] =R[v] =R[u•v]. But then, since R[x] is commutative associative, (a) gives

N(u• v) = det(L0u•v) = det(L0

uL0v) = det(L0

u)det(L0v) = N(u)N(v).

It remains to show that the equation N(x•y) = N(x)N(y) does not hold for all x,y ∈ J. To this end,put

x := ∑αieii, y := ∑vi[ jl] (αi ∈ R, viıO, 1≤ i≤ 3).

Then (5.8.2) and (5.9.4) imply


N(x) = α1α2α3,

N(y) = tO(v1v2v3),

x• y =12 ∑(α j +αl)vi[ jl]

henceN(x• y) =

18(α2 +α3)(α3 +α1)(α1 +α2)tO(v1v2v3),

which in general will be distinct from N(x)N(y).(c) If x is invertible in J, then, by definition, it is so in R[x], and x−1 ∈R[x] satisfies x•x−1 = 13.

Now (b) implies N(x)N(x−1)= 1 and, in particular, N(x) 6= 0. Conversely, let this be so. Combining(5.9.8) with (5.9.10), we obtain x] ∈ R[x] and x • x] = N(x)13. But this implies x • y = 13 withy := N(x)−1x] ∈ R[x], so x is invertible in J, and the inverse has the desired form.

(d) By Zariski density, it suffices to prove both identitiesfor x invertible. Taking norms on bothsides of (c), we obtain N(x)−1 = N(x−1) = N(x)−3N(x]), hence N(x]) = N(x)2. In particular, x] isinvertible with inverse x]−1 = N(x])−1x]] = N(x)−2x]]. Hence

x = (x−1)−1 =(N(x)−1x]

)−1= N(x)x]−1 = N(x)−1x]],

and also the adjoint identityfollows.

23. First let ϕ be an automorphism of J. Then ϕ fixes 13 and in order to prove N(ϕ(x)) = N(x) forsol.AUREx ∈ J, we may assume, by Zariski densitiy, that 13∧x∧x2 6= 0 in

∧3(J). Since ϕ preserves powers,(5.9.10) implies

T (x)ϕ(x)2−S(x)ϕ(x)+N(x)13 = ϕ(x3) = ϕ(x)3

= T(ϕ(x)

)ϕ(x)2−S

(ϕ(x)

)ϕ(x)+N

(ϕ(x)

)13,

and since 13,ϕ(x),ϕ(x)2 are linearly independent, we conclude N(ϕ(x)) = N(x), as claimed.Conversely, suppose ϕ preserves 13 and N. Then ϕ also preserves the directional derivative of

N at x in the direction y, i.e., in view (5.9.9) we have

T(ϕ(x)] •ϕ(y

)= DN

(ϕ(x)

)(ϕ(y)

)= DN(x)(y) = T (x] • y).

Setting x = 13 implies T ϕ = T , while setting y = 13 implies S ϕ = S. Returning with this to(5.9.10), we conclude that ϕ preserves cubes: ϕ(x3) = ϕ(x)3. Linearizing this gives 2ϕ(x • (x •y))+ϕ(x2 • y) = 2ϕ(x)• (ϕ(x)•ϕ(y))+ϕ(x)2 •ϕ(y), and setting y = 13 shows that ϕ preservessquares: ϕ(x2) = ϕ(x)2. Linearizing this yet again, we see that ϕ preserves the product of J, andthe proof of the first part is complete.

As to the second part, the linear map ϕ is clearly bijective and preserves the unit. Assume thatϕ is an automorphism of J. Then (5.9.3) yields

e22 + e33 = ϕ(1O[23]2) = ϕ(1O[23])2 = u−1[23]2 = nO(u)−1(e22 + e33),

and we conclude nO(u) = 1. Conversely, let this be so. By the first part of the problem, it sufficesto show that ϕ preserves the norm. To this end, we compute

N(ϕ(x)

)= N

(∑αieii +(u−1u1)[23]+ (u2u−1)[31]+ (uu3u)[12]

)= α1α2α3−α1nO(u−1u1)−α2nO(u2u−1)−α3nO(uu3u)+ tO

((u−1u1)(u2u−1)(uu3u)

).

But u has norm 1, nO permits composition, and the Moufang identities (Exc. 4) combined withExc. 3 (b) give


tO((u−1u1)(u2u−1)(uu3u)

)= tO

((u−1(u1u2)u−1)(uu3u

))= tO

(u−1((u1u2)[u−1(uu3u)]

))= tO

(u−1((u1u2)u3

)u)= tO(u1u2u3).

Hence N(ϕ(x)) = N(x), and the problem is solved.

Section 6.24. (a) This follows by a straightforward computation.sol.UORE(b) Linearizing (5.9.8) and applying (5.9.7), we obtain

x× y = 2x• y−T (x)y−T (y)x+T (x)T (y)13−T (x• y)13,

and combining this with (5.9.8), (5.9.9), (5.9.11) yields

x]× y = 2x] • y−T (x])y−T (y)x]+T (x])T (y)13−T (x] • y)13)

= 2x2 • y−2T (x)x• y+2S(x)y−S(x)y−T (y)x]+S(x)T (y)13−DN(x)(y)13

= 2x2 • y−2T (x)x• y+S(x)y−T (y)x2 +T (x)T (y)x−S(x)T (y)13 +S(x)T (y)13

− x2 • y+2T (x)x• y+T (y)x2−S(x)y−S(x,y)x−2x• (x• y)

= −Uxy+T (x• y)x,

as claimed.

25 . Let A,B be commutative real algebras and ϕ : A→ B a linear map. If ϕ is a homomorphism,sol.OPLSPthen it clearly preserves squares. Conversely, let this be so. Then ϕ(x2) = ϕ(x)2 for all x ∈ A, andlinearizing gives

ϕ(xy) =12

(ϕ((x+ y)2− x2− y2))= 1

2

((ϕ(x)+ϕ(y)

)2−ϕ(x)2−ϕ(y)2)= ϕ(x)ϕ(y).

Thus ϕ is a homomorphism.Now define the left multiplication of O as the linear map L : O+→ EndR(O)+ sending x ∈O+

to Lx : O→ O, y 7→ xy. Since Lx2 = L2x for all x ∈ O by 1.6, L preserves squares and hence is a

homomorphism of commutative algebras. It is also injective since Lx = 0 implies x = Lx1O = 0.Thus L maps O+ isomorphically onto a subalgebra of EndR(O), and we conclude from 5.7 (a) thatO+ is a special Jordan algebra. As in Exc. 24 (a), the verification of the formula for the U-operatoris straightforward: for x,y ∈O we obtain, using alternativity 1.6,

Uxy = 2x• (x• y)− x2 • y =12(x(xy+ yx)+(xy+ yx)x− x2y− yx2)

=12(x(xy)− x2y+ x(yx)+(xy)x+(yx)x− yx2)= xyx,

as claimed. Finally, the Jordan algebra O+ is not euclidean since, given a Cartan-Shouten basis(ui)0≤i≤7 of O, we have 12

O+u2i = 0 for i = 1, . . . ,7.

26. For 0≤ i≤ n we have e0 •ei =12 (1Aei+ei1A) = ei. Now let 1≤ i, j≤ n, i 6= j. By orthnormal-sol.GAULI

ity, ei has trace 0 and norm 1, which by (1.5.10) implies ei • ei = e2i =−1A =−e0, while (1.5.12)

implies

ei • e j =12

ei e j =12(tD(ei)e j + tD(e j)ei−nD(ei,e j)

)= 0.

Thus Λ is closed under the multiplication of D+ and hence is a linear Z-structure of that algebra.But note that, in general, Λ will not be closed under the multiplication of D. Finally, by Ex. 3.16


and Exc. 6, the Gaussian integers of D are the Z-linear span of some orthonormal basis of D,yielding the final assertion of the problem.

27. Exc. 22 (b) implies N(c)=N(c2)=N(c)2, hence N(c)∈0,1. If N(c)= 1, then c is invertiblesol.IDREJO(Exc. 22 (c)), and the relations 13− c ∈ R[c], c(13− c) = 0 yield c = 13, a contradiction. ThusN(c) = 0. Using this and (5.9.10), we deduce c = c3 = (T (c)−S(c))c, hence

S(c) = T (c)−1. (70) IDCU

Now (5.9.8) yields c] = (1− T (c))c + S(c)13 = S(c)(13 − c), hence S(c) = T (c]) = S(c)(3−T (c)) =−S(c)2 +2S(c). Thus S(c)2 = S(c), forcing S(c) ∈ 0,1, and (70) implies the assertion.

28. Settingsol.EXPNOx = ∑

(αieii +ui[ jl]

), y = ∑

(βieii + vi[ jl]

)with αi,βi ∈ R, ui,vi ∈ D, 1 ≤ i ≤ 3 and applying (5.8.2), (5.8.3), (5.9.5) and the fact that theexpression tO(uvw) = nO(uv,w) remains invariant under cyclic permutations of the variables, wecompute

N(x+ y) = (α1 +β1)(α2 +β2)(α3 +β3)−∑(αi +βi)nO(ui + vi)

+ tO((u1 + v1)(u2 + v2)(u3 + v3)

)= α1α2α3 +∑α jαlβi +∑αiβ jβl +β1β2β3

−∑(αinO(ui)+αinO(ui,vi)+αinO(vi)+βinO(ui)+βinO(ui,vi)+βinO(vi)

)+ tO(u1u2u3)+∑ tO(u julvi)+∑ tO(uiv jvl)+ tO(v1v2v3)

=(α1α2α3−∑αinO(ui)+ tO(u1u2u3)

)+∑

((α jαl −nO(ui)

)βi +nO(−αiui +u jul ,vi)

)+∑

(αi(β jβl −nO(vi)

)+nO(ui,−βivi + v jvl)

)+(β1β2β3−∑βinO(vi)+ tO(v1v2v3)

)= N(x)+T (x] • y)+T (x• y])+N(y),

as desired.

Chapter 2

Section 8.

29. For x ∈ A and ε =±, we putsol.HOSI

Mε (x) := y ∈ A | f (xy) = ε f (x) f (y),

which is a submodule of A. By hypothesis we have A = M+(x)∪M−(x), so some ε = ± hasMε (x) = A. Now put

Nε := x ∈ A | ∀y ∈ A : f (xy) = ε f (x) f (y)= x ∈ A |Mε (x) = A,

which is again a submodule of A, and from what we have just proved, we obtain A = N+ ∪N−.This gives Nε = A for some ε = ±, and if ε = + (resp. ε = −), f (resp. − f ) is a homomorphismfrom A to B.


30. For any subset X ⊆ A and any homomorphism ϕ : A→ B of k-algebras, one makes the follow-sol.NILRADing observations by straightforward induction:

(a) ϕ(Monm(X)) = Monm(ϕ(X)) for all integers m > 0.(b) Monm(Y )⊆Monmn(X) for all Y ⊆Monn(X) and all integers m,n > 0.

If x ∈ A is nilpotent, then (a) implies that ϕ(x) ∈ B is nilpotent. Hence if A is nil, so are I andA′ := A/I. Conversely, suppose I and A′ are nil. Writing π : A→ A′ for the canonical epimorphism,any x∈A makes π(x)∈A′ nilpotent, so by (a), Monn(x) meets I for some integer n> 0. But sinceI is nil, we conclude that some y∈Monn(x) is nilpotent, leading to a positive integer m such that0 ∈Monmn(x) by (b). Thus x is nilpotent, forcing A to be a nil algebra. Now write n := Nil(A)for the sum of all nil ideals in A. Given x ∈ n, there are finitely many nil ideals n1, . . . ,nr ⊆ A suchthat x ∈ n1 + · · ·+nr . Thus we only have to show that the sum of finitely many nil ideals in A isnil. Arguing by induction, we are actually reduced to the case r = 2. But then the isomorphism(n1 +n2)/n1 ∼= n2/(n1∩n2) combines with what we have proved earlier to yield the assertion.

31. (a) Since xm is a linear combination of powers xi, d ≤ i < m, the k-module kd [x] is finitelysol.IDgenerated. Moreover, right multiplication by x in the power associative algebra A gives a linearmap ϕ : kd [x]→ kd [x] whose image is kd+1[x]. But kd+1[x] = kd [x] since αm−d ∈ k×. Therefore,ϕ is surjective, hence bijective, and so is ϕd . This proves existence and uniqueness of c ∈ kd [x]satisfying cxd = xd , and the relation c2xd = c(cxd) = cxd shows that c is an idempotent.(b) Let x ∈ A be a pre-image of c′ under ϕ . Then x2−x is nilpotent, so some integer n > 0 has 0 =(x2−x)n = x2n−·· ·+(−1)nxn. Now (a) yields an idempotent c∈ kn[x] satisfying cxn = xn. Writingu := ϕ(u) for u ∈ A, we conclude cc′ = cc′n = cxn = xn = c′. On the other hand, c = ∑i≥n αixi forsome scalars αi ∈ k, i≥ n, so with α := ∑i≥n αi we obtain c = αc′ = αc′2 = cc′ = c′.

Section 9.

32. If I j are ideals in A j for 1 ≤ j ≤ n, then I := I1 ⊕ ·· · ⊕ In is an ideal in A since AI + IA =sol.IDDIRSUM∑ j(A jI j + I jA j)⊆∑ j I j = I. Conversely, let I be an ideal in A and e j the identity element of A j for1≤ j ≤ n. Then I j = e jI = Ie j is clearly an ideal in A j such that I = ∑ j I j .

The assertion fails without the assumption that A be unital: suppose the A j have trivial multipli-cation. Then so has A, and any decomposition of A into the direct sum of n submodules not relatedto the A j in any way is a decompos ition onto a direct sum of ideals.

To complete the solution, let us assume that the A j , 1 ≤ j ≤ n, are simple k-algebras. By thefirst part of the exercise, the ideals of A have the form

⊕j∈N A j , where N varies over the 2n subsets

of 1, . . . ,n. Hence the A j , 1≤ j≤ n, are precisely the minimal ideals of A. This description beingindependent of the decomposition chosen, the assertion follows.

33 .(a) The polynomials µ1, . . . ,µr have greatest common divisor 1. Hence f1, . . . , fr ∈ F [t] withsol.ALGELthe desired properties exist. We therefore conclude ∑

ri=1 ci = 1A. Moreover, for 1≤ i, j ≤ r, i 6= j,

the polynomial µiµ j is a multiple of µx, hence kills x. This proves cic j = 0 and then ci = ci1A =

∑rl=1 cicl = c2

i . Thus (c1, . . . ,cr) is a complete orthogonal system of idempotents in R := F [x]. Nowput

v := x−r

∑i=1

αici =r

∑i=1

(x−αi1A)ci ∈ R. (71) MINIL

Then the first relation of (2) trivially holds and

vn =r

∑i=1

(x−αi1A)nci =

( r

∑i=1

(t−αi)nµi fi

)(x).

For 1≤ i≤ r,(t−αi)

nµi fi = (t−αi)

ni µi(t−αi)n−ni fi = µx(t−αi)

n−ni fi


is a multiple of µx and hence kills x. This shows vn = 0, and the proof of (2) is complete.(b) For I ⊆ 1, . . . ,r, the quantity cI is clearly an idempotent in R. Conversely, let c be an

idempotent in R. The c = f (x) for some f ∈ F [t]. Since R is commutative associative, we have

Nil(R) = u ∈ R | u is nilpotent. (72) NILCAS

Hence (2) implies, for all m ∈ C,

xm =m

∑i=1

αmi ci + vm, vm ∈ Nil(R)

and then

g(x) =r

∑i=1

g(αi)ci + vg, vg ∈ Nil(R),

for all g ∈ F [t]. In particular, setting βi := f (αi) for 1≤ i≤ r,

c = f (x) =r

∑i=1

βici + v f , c2 = f 2(x) =r

∑i=1

β2i ci + v f 2 .

But c = c2, so ∑(β 2i −βi)ci = v f 2 − v f is nilpotent, which obviously implies β 2

i = βi, hence βi ∈0,1 for 1≤ i≤ r. Put

I := i ∈ Z | 1≤ i≤ r, βi = 1.Then, setting w := v f ∈ Nil(R),

cI +w = c = c2 = cI +2cIw+w2,

and we conclude

(1A−2cI)w = w2, (1A−2cI)2 = 1A. (73) CECI

In particular, 1A− 2ci is invertible in R. Let m ∈ Z satisfy m ≥ 1 and wm = 0 6= wm−1. Assumingm ≥ 2, we conclude from (73) that (1− 2cI)wm−1 = wm = 0, hence wm−1 = 0, a contradiction.Thus m = 1, forcing w = 0 and c = cI .

(c) (i)⇒ (ii). Let 1≤ i≤ r and put v := ((t−αi)µi)(x) ∈ R. Then

vni =((t−αi)

ni µnii

)(x) = (µxµ

ni−1i )(x) = 0,

and (i) implies v = 0.Thus µx = (t− αi)ni µi divides (t− αi)µi, forcing ni = 1 since µi is not

divisible by t−αi.(ii)⇒ (iii). This follows immediately from (2).(iii)⇒ (i). Let v ∈ Nil(R). Then

v = f (x) =r

∑i=1

f (αi)ci

for some f ∈ F [t] is nilpotent, which can happen only if f (α1) = · · · = f (αr) = 0 and henceamounts to v = 0.

(d) Write µx = ∑dj=0 γ jt j with d = ∑

ri=1 ni and γ0, . . . ,γd ∈ F . Then the condition αi 6= 0 for all

i = 1, . . . ,r is equivalent to γ0 = µx(0) 6= 0, hence implies xy = 1A with y =−γ−10 ∑

dj=1 γ jx j−1] ∈ R,

so x is invertible in R. Conversely, let this be so and assume γ0 = 0. The x∑dj=1 γ jx j−1 = µx(x) = 0,

and since x is invertible in R, the polynomial ∑dj=1 γ jt j−1 ∈ F [t] of degree at most d−1 kills x, in

contradiction to µx being the minimum polynomial of x. Ths γ0 6= 0.


(e) Let βi ∈ F× satisfy β 2i = αi for 1≤ i≤ r. Setting z := ∑

ri=1 βici, we obtain z−2x = 1A +w,

for some w ∈ Nil(R). Thus we may assume x = 1A +w and wm = 0 6= wm−1 for some positiveinteger m. The case m = 1 being obvious, we may actually assume m≥ 2. Let

y =m−1

∑j=0

γ jw j ∈ R (γ0 . . . ,γm−1 ∈ F).

The minimum polynomial of w is tm, and we conclude that 1A,w, . . . ,wm−1 are linearly independentover F . Hence y2 = x if and only if

1A +w =m−1

∑j=0

( j

∑l=0

γlγ j−l)w j,

which in turn is equivalent to

γ20 = 2γ0γ1 = 1,

j

∑l=0

γlγ j−l = 0 (2≤ j < m).

This system of quadratic equations can be solved recursively by

γ0 = 1, γ1 =12, γ j =−

12

j−1

∑l=1

γlγ j−l (2≤ j < m).

This proves the assertion.

Remark. The equation y2 = x = 1A +w, w ∈ R nilpotent, can be solved more concisely by

y =∞

∑l=0

( 12l

)wl ,

combined with the observation that( 1

2l

)∈Q has exact denominator a power of 2 (depending on l).

34. We write D (resp. E) for the set of decompositions of A into the direct sum of n complementarysol.CENIDSPLIDEideals (resp. of complete orthogonal systems of n central idempotents) in A. Let (e j)1≤ j≤n ∈E. For1 ≤ j, l ≤ n, centrality implies (Ae j)(Ael) = A2(e jel), which is zero for j 6= l and Ae j for j = l.Moreover, x = ∑xe j ∈∑Ae j for every x ∈ A, and given x j ∈ A for 1≤ j ≤ n satisfying ∑x je j = 0,we multiply with el , 1≤ l ≤ n, and deduce xlel = 0. All this boils down to A = (Ae1)⊕·· ·⊕ (Aen)as a direct sum of ideals. Thus the assignment (e j) 7→ (Ae j) gives a map from E to D.

Conversely, let (I j)1≤ j≤n ∈ D. Then each I j , 1 ≤ j ≤ n, is a unital k-algebra in its own right,and setting e j := 1A j , we deduce ∑e j = 1A, e jel = δ jle j for 1≤ j, l ≤ n, so (e j)1≤ j≤n is a completeorthogonal system of idempotents in A, which are clearly central since Cent(A) = Cent(I1)⊕·· ·⊕Cent(In) as a direct sum of ideals. Thus the assignment (I j)1≤ j≤n 7→ (e j)1≤ j≤n determines a mapfrom D to E, and it is readily checked that the two maps constructed are inverse to each other.

35. For 1≤ i, j, l,m, p,q≤ n, a straightforward verification showssol.NUCMAT

[aei j,belm,cepq] = δ jlδmp[a,b,c]eiq, (74) ASMA

[aei j,belm] = δ jl(ab)eim−δim(ba)el j. (75) COMMA

The inclusion Matn(Nuc(A))⊆Nuc(Matn(A)) follows immediately from (74). Conversely, let x =(ai j) = ∑ai jei j ∈ Nuc(Matn(A)). Then (74) for m = p implies 0 = [x,belm,cmp] = ∑i[ail ,b,c]eiq,hence [ail ,A,A] = 0. Similarly, 0 = [belm,cemq,x] = ∑ j[b,c,aq j]el j , forcing [A,A,aq j] = 0.And, finally, 0 = [belm,x,cepq] = [b,amp,c]elq implies [b,amp,c] = 0. Summing up, we concludeai j ∈ Nuc(A) for 1≤ i, j ≤ n, solving the first part of the problem.


Concerning the second one, let x = (ai j) = ∑ai jei j ∈ Cent(Matn(A)). By the first part, ai j ∈Nuc(A) for 1 ≤ i, j ≤ n. Moreover, (75) for l = m yields 0 = [x,ell ] = ∑i aileil −∑ j al jel j , henceai j = 0 for 1≤ i, j≤ n, i 6= j. Now let l 6=m in (75). Then [x,belm] = (allb)elm−(bamm)elm, forcingall = amm ∈ Cent(A), and we conclude x ∈ Cent(A)1n. Conversely, it is clear that every element ofCent(A)1n belongs to Cent(Matn(A)).

36. Let t : A→ k be an associative linear form and suppose the corresponding bilinear form σ :=σtsol.ASLIFO(cf. 8.9) is non-singular (cf. 12.9). By hypothesis, there exists a unique element e ∈ A such thatt(x) = σ(e,x) = t(ex) for all x ∈ A. Now the relations t((ex)y) = t(e(xy)) = t(xy) and t((xe)y) =t(y(xe)) = t((yx)e) = t(e(yx)) = t(yx) = t(xy) for all x,y ∈ A combined with non-degeneracy of σ

imply ex = x = xe for all x ∈ A. Hence A is unital with 1A = e.

37 Let D be a finite-dimensional non-associative division algebra over F and suppose D hassol.FIDIAdimension n > 1. Picking linearly independent vectors x,y ∈ D, the polynomial det(Lx+ty) ∈ F [t]has degree n, hence a root α ∈ F . But then left multiplication by the non-zero element x+αy ∈ Dhas a non-trivial kernel, contradicting the property of D being a division algebra. Thus n = 1, andwe conclude D = Fe for some non-zero element e ∈ D. Therefore e2 = αe for some α ∈ F×, andit is readily checked that F → D, ξ 7→ α−1ξ e is an isomorphism of F-algebras.

Section 10.

38. Let m,n ∈ Z such that u := mx+ny ∈ Nuc(A). Then 0 = [u,x,x] = m(x2x− xx2)+n((yx)x−sol.CEBAyx2) = 2m(x− x)− 2ny = −2ny, hence n = 0, and 0 = [u,x,y] = m(x2y− x(xy)) = 2my, hencem = 0. Thus Cent(A) = Nuc(A) = Z1A; in particular, A is central. On the other hand, passingto the base change AR, R = Z/2Z and setting u := uR for u ∈ A, we conclude that 1A, x, y is anR-basis of AR with multiplication table x2 = xy = yx = y2 = 0. It follows from this at once thatAR is commutative associative, so while Cent(AR) = Nuc(AR) = AR is a free R-module of rank 3,(Cent(A))R = (Nuc(A))R = R1A is a free R-module of rank 1.

39. We proceed in several steps.sol.FIB10. Let can : k→ κ(p), α 7→ α(p), be the natural map. Then the diagram

k can//

ϕ

κ(p) = k⊗κ(p)

ϕ⊗1κ(p)

k′

ψ

// k′⊗κ(p)

commutes since

(ϕ⊗1κ(p)) can(α) = (ϕ⊗1κ(p))(1⊗α(p)

)= 1k′ ⊗α(p) = 1k′ ⊗ (α ·1κ(p))

= (α ·1k′ )⊗1κ(p) = ϕ(α)⊗1κ(p) = ψ ϕ(α)

for all α ∈ k. Applying the contra-variant functor Spec to this diagram, we conclude that also

Spec(k′⊗κ(p)

)Spec(ψ)

//

Spec(ϕ⊗1κ(p))

Spec(k′)

Spec(ϕ)

Spec

(κ(p)

)Spec(can)

// Spec(k)

commutes. But Spec(κ(p)) consists of a single point sent by Spec(can) to can−1(0)=Ker(can)=p. Hence the image of Spec(ψ) belongs to the fiber Spec(ϕ)−1(p). Thus Spec(ψ) induces canoni-


cally a continuous mapΨ : Spec

(k′⊗κ(p)

)−→ Spec(ϕ)−1(p).

20. Let p′ ∈ Spec(ϕ)−1(p). Then ϕ−1(p′) = p, and consulting (10.4.6), we find a unique homo-morphism

σp′ : k′⊗κ(p)−→ κ(p′)of κ(p)-algebras making a commutative diagram

kϕ //

canp

k′ψ //

canp′

k′⊗κ(p)

σp′

kpϕ ′ //

canp(p)

k′p′

canp′ (p′)

κ(p)

ϕ

// κ(p′).

(76) SIGMAPE

Thus Ker(σp′ ) ∈ Spec(k′⊗κ(p)), and we obtain a map

Φ : Spec(ϕ)−1(p)−→ Spec(k′⊗κ(p)

)given by

Φ(p′) := Ker(σp′ ) (p′ ∈ Spec(ϕ)−1(p)).

The definitions and (76) imply

Ψ Φ(p′) = ψ−1(

σ−1p′ (0)

)= (σp′ ψ)−1(0) = α ′ ∈ k′ | α ′(p′) = 0= p′

for all p′ ∈ Spec(ϕ)−1(p), hence Ψ Φ = 1Spec(ϕ)−1(p).

30. Let q ∈ Spec(k′⊗κ(p)). Then

Φ Ψ(q) = Φ(p′) = Ker(σp′ ), p′ := ψ−1(q) ∈ Spec(ϕ)−1(p).

By (10.4.6) again, we obtain a diagram

kϕ //

canp

k′ψ //

canp′

k′⊗κ(p)

σp′

jj

canq

kp

ϕ ′ //

canp(p)

k′p′ψ ′ //

canp′ (p′)

(k′⊗κ(p)

)q

canq(q)

κ(p)

ϕ // κ(p′)ψ // κ(q)

where the four squares commute. But so does the diagram


k′ψ //

canp′ (p′)canp′

k′⊗κ(p)

canq(q)canq

σp′zz

κ(p′)ψ

// κ(q),

which by (76) is clear for the upper triangle but holds true also for the lower one because ψ σp′

and canq(q)canq are both κ(p)-linear satisfying ψ σp′ ψ = canq(q)canq ψ , hence ψ σp′ =canq(q) canq. We therefore conclude

Φ Ψ(q) = Ker(ψ σp′ ) = Ker(

canq(q) canq)= q.

Summing up we have shown that Ψ is bijective with inverse Φ .40. By 10 and 30, it remains to show that Φ is continuous, equivalently, that Ψ is open. In order tosee this, we first deduce from (10.3.1) and 10.2 the natural identifications

k′⊗κ(p) = (k′⊗ kp)⊗kp (kp/pp) = k′p⊗ (kp/pp) = k′p/ppk′p

such that, with the natural maps i : k′→ k′p, π : k′p→ k′p/ppk′p, the triangle

k′

i

ψ

&&k′p π

// k′p/ppk′p = k′⊗κ(p)

commutes. The proof will be complete once we have shown

Ψ

(D(π( f/s)

))= D( f )∩Spec(ϕ)−1(p) (77) PSIDEPI

for all f ∈ k′, s ∈ k \ p. Let q ∈ Spec(k′⊗κ(p)) = Spec(k′p/ppk′p). Then the chain of equivalentconditions

q ∈ D(π( f/s)

)⇐⇒ π( f/s) /∈ q ⇐⇒ f/s /∈ π

−1(q) (78) QUDEPI

⇐⇒ f /∈ i−1(π−1(q)

)= ψ

−1(q) ⇐⇒ ψ−1(q) ∈ D( f )

shows that the left-hand side of (77) is contained in the right. Conversely, let p′ ∈ D( f ) ∩Spec(ϕ)−1(p). Then q :=Ψ−1(p′) ∈ Spec(k′⊗κ(p)) and ψ−1(q) = p′ ∈ D( f ). Consulting (78),we conclude q ∈D(π( f/s)), hence p′ =Ψ(q) ∈Ψ(D(π( f/s))), and the proof of (77) is complete.

40. (a) We may assume χ = 0. Let u1, . . . ,um be a finite set of generators for the k-module M. Forsol.LOCHOMp ∈U and 1≤ i≤ m, we have ψ(ui)/1 = ψ(ui)p = ψp(uip) = 0 in Np. Hence there are f ∈ k \p,n ∈ N such that f nψ(ui) = 0 in N for all i = 1, . . . ,m. But this means p ∈ D( f ) and ψ(ui)/1 = 0in N f for all i = 1, . . . ,m, which in turn, by 10.5, amounts to ψq = 0 for all q ∈ D( f−). Hence wehave shown p ∈ D( f )⊆U , and U is Zariski-open in X .

(b) Again let u1, . . . ,um be a finite set of generators for A as a k-module. Given i = 1, . . . ,m, wedefine k-linear maps ψi,χi : A→ B by ψi(v) := ϕ(uiv), χi(v) := ϕ(ui)ϕ(v), v ∈ A. By (a),

Ui := p ∈ X | ψip = χip

is Zariski-open in X . Hence so is V =⋂m

i=1 Ui.

41. In the solution to this problem, free use will be made of the formalism for the Zariski topologysol.IDEPARas explained, for example, in [11, II, § 4.3].


(a) For p ∈ X , the chain of equivalent conditions

p ∈ D(ε)⇐⇒ ε /∈ p⇐⇒ εp /∈ pp⇐⇒ εp ∈ k×p ⇐⇒ εp = 1p⇐⇒ (1− ε)p = 0

gives the first displayed equality, while the second one follows from the first and the fact that localrings are connected, i.e., contain only the trivial idempotents 0 and 1. The third equality is nowobvious. To establish the final assertion of (a), we note for p ∈ X that (eip)i∈I is an orthogonalsystem of idempotents in kp, whence (∑ei)p 6= 0 if and only if eip 6= 0 for some i = 1, . . . ,r. Thisproves

⋃D(εi) = D(∑εi), and the union on the left is disjoint since D(εi)∩D(ε j) = D(εiε j) =

D(0) = /0 for i, j ∈ I, i 6= j.(b) Given a complete system (εi)i∈I of orthogonal idempotents in k, the subsets Ui := D(εi) ⊆ X ,i ∈ I, are open in X and by (a) satisfy

⋃Ui = D(∑ei) = D(1) = X , where the union on the very left

is disjoint and the Ui are empty for almost all i ∈ I. Thus the assignment (εi) 7→ (Ui) gives a mapof the right kind, which is injective since the Ui by (a) determine the εi locally, hence globally. Itremains to show that the map in question is surjective, so let (Ui)i∈I be a decomposition of X intodisjoint open subsets almost all of which are empty.

First, view X as a geometric space with structure sheaf O . Since the Ui form a disjoint opencover of X , we may apply the sheaf axioms to find, for each i ∈ I, a unique global section εi ∈H0(X ,O) = k which restricts to the identity element of H0(Ui,O) and to zero on H0(U j,O) for allj ∈ I, j 6= i. It is then readily checked that the εi form a complete orthogonal system of idempotentsin k, forcing the D(εi) to be an open disjoint cover of X . On the other hand, given p ∈ Ui, weobtain εip = 1p, which implies Ui ⊆ D(εi) by (a). Summing up, we have equality and surjectivityis proved.

Adopting a more direct approach based on the proof of [11, II, §4, Prop. 15], let i ∈ I. ThenUi and Yi :=

⋃j 6=i U j are closed subsets of X , hence may be written in the form Ui = V (ai), Yi =

V (bi) for some ideals ai,bi ⊆ k. Since V (ai + bi) = V (ai)∩V (bi) = Ui ∩Yi = /0, we concludeai+bi = k, hence αi+βi = 1 for some αi ∈ ai, βi ∈ bi. This implies αi =αi(αi+βi) =α2

i +αiβi ≡α2

i mod aibi, so αi ∈ ai becomes an idempotent modulo aibi. On the other hand, X = Ui ∪Yi =V (ai)∪V (bi)=V (aibi), so aibi⊆ k is a nil ideal. Now Exc. 31 (b) yields an idempotent ε ′i ∈ ai suchthat ε ′i ≡ αi mod aibi. Thus εi := 1− ε ′i ∈ k is an idempotent as well satisfying εi ≡ βi mod aibi,hence ε ′ ∈ bi. We conclude Ui =V (ai)⊆V (ε ′i ) = D(εi), Yi =V (bi)⊆V (εi) = D(ε ′i ) and D(εi)∩D(ε ′i ) = /0 by (a). Therefore Ui = D(εi), and it remains to show that (εi)i∈I is a complete orthogonalsystem of idempotents. For i, j ∈ I, i 6= j, we deduce D(εiε j) = D(εi)∩D(ε j) = Ui ∩U j = /0, sothe idempotent εiε j is also nilpotent, hence zero. Thus the εi are orthogonal. For the same reason,εi = 0 for almost all i ∈ I Now (a) gives

X =⋃

Ui =⋃

D(εi) =⋃

V (1− εi) =V(∏(1− εi)

)=V (1−∑εi),

forcing the idempotent 1−∑εi to be nilpotent, hence zero.

42. Setting X := Spec(k), the subsetssol.RANDEC

Ui := p ∈ Spec(k) | rkp(M) = i ⊆ X (i ∈ N) (79) RANKI

form an open disjoint cover of X and since X is quasi-compact [11, II,§4, Proposition 12], Ui isempty for almost all i ∈ N. This proves (a) while, in order to prove (b), we apply Exercise 41 toobtain a complete orthogonal system (εi)i∈N of idempotents in k such that Ui = D(εi) for all i ∈N.Given j ∈N, let π j : k→ k j be the canonical projection induced by (1) and let p j ∈ Spec(k j). Then

p := Spec(π j)(p j) = π−1j (p j) = p j⊕

⊕i6= j

ki ∈ X ,

and (10.4.7) showsM jp j = Mp⊗kp k jp j .


Since Mp is free of rank j over kp, it follows that M jp j is free of rank j over k jp j . This provesexistence. In order to establish uniqueness, let (ηi)i∈N be any complete orthogonal system of idem-potents in k, giving rise, in analogy to (1), (2), to the decompositions

k =⊕i∈N

li, li = kηi (i ∈ N), (80) KLEI

M =⊕i∈N

Ni, Ni = N⊗ li (i ∈ N), (81) MNEI

and making each Ni a projective li-module of rank i. Again by Exercise 41, the Vi := D(ηi) areempty for almost all i∈ I and form an open disjoint cover of X , so it suffices to show Vi ⊆Ui for alli ∈ N. Fixing j ∈ N and q ∈ Vj , we apply §9, Exercise 32, and find a decomposition q =

⊕i∈N qi,

where qi = q∩ li is an ideal in li. But η j /∈ q while ηiη j = 0 ∈ q for all i 6= j, forcing ηi ∈ q∩ li = qiand then qi = li. Summing up we have shown

q= q j⊕⊕i6= j

li,

so q j is a prime ideal in l j and Spec(ρ j)(q j) = q, where ρ j : k→ l j is the canonical projectioninduced by (80). Since

N jq j = Mq⊗kq l jq j

has rank j over l jq j , the free kq-module Mq must have rank j over kq. This proves q ∈U j by (79),and the solution is complete.

43. If ϕ : A→ A′ is a unital algebra homomorphism, then so is ϕ(p) : A(p)→ A′(p), for eachsol.LOCSIALGp ∈ Spec(k). But since A(p) is simple by hypothesis, ϕ(p) is injective and a dimension argumentshows that it is, in fact, an isomorphism. Hence, by Nakayama’s lemma [65, X, § 4], ϕp : Ap→ A′pis an isomorphism, whence ϕ is one [11, II, Thm. 1].

Section 11.

44. (a) The right-hand side is clearly contained in the left, so it remains to establish the reverse in-sol.GRANTOclusion. To this end, put (A′,G′) = (A,G)⊗k′ and write a′ ∈Cent(A′,G′) in the form a′ = ∑ai⊗α ′iwith ai ∈ A and α ′i ∈ k′ linearly independent over k. For x,y ∈ A, we have ∑[ai,x,y]⊗ α ′i =[a′,x⊗1k′ ,y⊗1k′ = 0 and, similarly, ∑[x,ai,y]⊗α ′i =∑[x,y,ai]⊗α ′i =∑[x,ai]⊗α ′i = 0. Moreover,for g ∈G, ∑g(ai)⊗α ′i = (g⊗1k′ )(a′) = a′ = ∑ai⊗α ′i . Since the α ′i are linearly independent overk, we therefore obtain [ai,x,y] = [x,ai,y] = [x,y,ai] = [ai,x] = 0, g(ai) = ai, hence ai ∈ Cent(A,G)for all i, which implies a′ ∈ Cent(A′,G′), as claimed.(b) By Prop. 10.15, the right multiplication of A induces an isomorphism R : Cent(A)→EndMult(A)(A).Hence it suffices to show that the elements of Cent(A,G) correspond to the Mult(A,G)-linearmaps A→ A under this isomorphism, which follows from the relations Rg(a)g = gR(a) for alla ∈ Cent(A), g ∈ G.(c) If (A,G) is a simple k-algebra with a group of antomorphisms, then M(A,G) acts irreduciblyon A, so Schur’s lemma and (b) yield the assertion.(d) Since A, by simplicity of (A,G), is an irreducible faithful M(A,G)-module, the first part followsfrom the Jacobson-Chevalley density theorem [65, Chap. XVII, § 3, Cor. 3 of Thm. 1] and imme-diately implies necessity in the second part. Conversely, suppose A 6= 0 and M(A,G) = Endk(A).Then A is an irreducible M(A,G)-module, forcing (A,G) to be simple and Cent(A,G) to be a field.Also, by (b), Endk(A) = M(A,G) = EndCent(A,G)(A) and a dimension count yields sufficiency aswell.


45 . (i) ⇒ (ii). Let k′ be any extension field of k and put (A′,G′) = (A,G)⊗ k′ as a k′-algebrasol.GRANTOCSwith a group of antomorphisms. A moment’s reflection shows M(A′,G′) = M(A,G)⊗ k′. Now theassertion follows from the second part of Exc. 44 (d).

(ii)⇒ (iii). Obvious.(iv)⇒ (i). This follows immediately from Exc. 44 (a), (c).

46 . We follow standard arguments, reproduced in [15, I Satz 5.5], for example. First note thatsol.TPGRANTOk is a field, by Exc. 44 (c) and central simplicity of (A,G). Now let J ⊆ A⊗A′ be a non-zeroideal of (A,G)⊗ (A′,G′) and consider the least positive integer n such that there exist elementsx1, . . . ,xn ∈ A,x′1, . . . ,x

′n ∈ A′ satisfying

0 6= x :=n

∑i=1

xi⊗ x′i ∈ J. (82) MIRE

Then the elements x′1, . . . ,x′n are linearly independent over k. Furthermore, a straightforward veri-

fication shows that the totality of elements u1 ∈ A such that

n

∑i=1

ui⊗ x′i ∈ J

for some u2, . . . ,un ∈ A forms an ideal I of (A,G) which is non-zero since x1 ∈ I and x1 6= 0 bythe minimality of n. The simplicity of (A,G) therefore implies I = A, so in (82) we may assumex1 = 1. But then, for all u,v ∈ A,

n

∑i=2

[xi,u,v]⊗ x′i = [x,u⊗1,v⊗1] ∈ J.

Again the minimality of n yields [x,u⊗1,v⊗1] = 0 and then [xi,u,v] = 0 for 2≤ i≤ n. Similarly,[u,xi,v] = [u,v,xi] = [xi,u] = 0, forcing xi ∈ Cent(A) for 2 ≤ i ≤ n. Applying g⊗ 1 for any g ∈ Gto (82) and again observing x1 = 1, we conclude

n

∑i=2

(xi−g(xi)

)⊗ x′i = x− (g⊗1)(x) ∈ J ,

and the minimality of n implies ∑ni=2(xi−g(xi))⊗x′i = 0, hence xi ∈ Cent(A,G) for 2≤ i≤ n. But

(A,G) was assumed to be central simple, so Cent(A,G) = k1A, and we conclude x = 1A⊗ x′ ∈ Jfor some x′ ∈ A′. Now put

I′ = u′ ∈ A′ | 1A⊗u′ ∈ J.Again it is straightforward to check that I′ is an ideal in (A′,G′) which is non-zero since it containsx′ 6= 0. The simplicity of (A′,G′) therefore implies I′ = A′, hence 1A⊗A′ = 1A⊗1A′ ∈ J and J = A,as claimed.

Section 12.

47. The implications (iii)⇒ (i)⇒ (ii) being obvious, it suffices to prove (ii)⇒ (iii), which we dosol.SPLHYPLAby following a suggestion of O. Loos. Letting (e+,e−) be a hyperbolic pair of hL and writing e± =v∗±⊕ v±, v± ∈ L, v∗± ∈ L∗, we apply (12.17.1) and obtain 〈v∗++ v∗−,v++ v−〉 = hL(e++ e−) = 1.Therefore v++ v− ∈ L is a unimodular vector, hence a basis of L over k, and (iii) follows.

48. The implications (ii)⇒ (i) being obvious, let us assume that (u1,u2) is a hyperbolic pair in h.sol.HYPPAWriting u j = α1 je1 +α2 je2 (αi j ∈ k, i, j = 1,2), we conclude

α1 jα2 j = 0, α11α22 +α21α12 = 1 ( j = 1,2). (83) COHY


Thus (ε+,ε−), ε+ := α11α22, ε− := α21α12, is a complete orthogonal system of idempotents in k,giving rise to a decomposition of k as in (ii) such that (2) holds. Moreover, setting γ+ :=α11ε+ ∈ k×+with inverse γ

−1+ = α22ε+, γ− := α21ε− ∈ k×− with inverse γ

−1− = α12ε−, we apply (83) and obtain

u1+ = ε+u1 = α11α22(α11e1 +α21e2) = γ+e1+,

u2+ = ε+u2 = α11α22(α12e1 +α22e2) = γ−1+ e2+,

u1− = ε−u1 = α21α12(α11e1 +α21e2) = γ−e2−,

u2− = ε−u2 = α21α12(α12e1 +α22e2) = γ−1− e1−,

hence (3), (4).

49. We argue by induction on n. The case n = 1 has been settled in Cor. 12.5. Now suppose n > 1sol.MULQUAand assume that the assertion holds for n−1. For any un ∈Mn, the map

Fun : M1×·· ·×Mn−1 −→M, (u1, . . . ,un−1) 7−→ F(u1, . . . ,un−1,un) (84) EFUN

is obviously k-(n− 1)-quadratic. Hence by the induction hypothesis, it has a unique R-(n− 1)-quadratic extension

(Fun )R : M1R×·· ·×Mn−1,R −→MR.

Now let xi ∈MiR be arbitrary for 1≤ i < n and put x := (x1, . . . ,xn−1). We claim that the map

Fx : Mn −→MR, un 7−→ Fx(un) := (Fun )R(x). (85) EFEX

is k-quadratic. In order to see this, we put

V := Quad(M1, . . . ,Mn−1;M)

and conclude from (84) that the map Q : Mn→V , un 7→ Fun is k-quadratic. On the other hand, (85)implies Fx(un) = Q(un)R(x), so Fx is Q composed with two k-linear maps: the natural map

Quad(M1, . . . ,Mn−1;M)−→ Quad(M1R, . . . ,Mn−1,R;MR), G 7−→ GR,

given and k-linear by the induction hypothesis, and the evaluation at x. This proves our claim.By Cor. 12.5 we therefore obtain an R-quadratic map

(Fx)R : MnR −→ (MR)R := M⊗R⊗R,

which composes with the R-linear map

µ : M⊗R⊗R−→M⊗R = MR, u⊗ r⊗ s 7−→ u⊗ rs (86) MUR

to yield a map FR : M1R×·· ·×Mn−1,R×MnR→MR via

FR(x1, . . . ,xn−1,xn) := µ((Fx1,...,xn−1 )R(xn)

)(87) EFR

for xi ∈MiR, 1≤ i≤ n. By construction, FR is quadratic in xn. Hence, writing xn = ∑rlulnR, rl ∈ R,uln ∈Mn, we obtain, after a straightforward computation,

(FR)xn = ∑r2l (Fuln )R + ∑

l<mrlrm

(Fuln+umn )R− (Fuln )R− (Fumn )R

),

which shows that the left-hand side of (87) is R-quadratic in xi, 1 ≤ i < n. Thus FR is an R-n-quadratic map. Next we show that (5) commutes. Indeed, given ui ∈Mi, 1≤ i≤ n, we apply (87),Cor. 12.5, (86), (85), the induction hypothesis and (84) to obtain


FR(u1R, . . . ,un−1,R,unR) = µ((Fu1R,...,un−1,R )R(unR)

)= µ

((Fu1R,...,un−1,R (un)

)R

)= Fu1R,...,un−1,R (un) = (FunR )R(u1R, . . . ,un−1,R)

= Fun (u1, . . . ,un−1)R = F(u1, . . . ,un−1,un)R,

and commutativity of (5) is proved.It remains to establish uniqueness. Let G,H : M1R× ·· ·×MnR → MR be R-n-quadratic maps

both rendering the diagram

M1×·· ·×MnF //

can

M

can

M1R×·· ·×MnR

G,H// MR.

(88) QUAUN

commutative. For un ∈Mn, using obvious notation, both

GunR ,HunR : M1R×·· ·×Mn−1,R −→MR

are R-(n−1)-quadratic maps rendering the diagram

M1×·· ·×MnF //

can

M

can

M1R×·· ·×MnR

GunR ,HunR

// MR.

commutative. From the uniqueness part of the induction hypothesis, we therefore conclude GunR =HunR . Now let xi ∈MiR for 1≤ i≤ n and write xn = ∑rlunlR, rl ∈ R, uln ∈Mn. Then what we havejust shown implies

G(x1, . . . ,xn−1,xn) = ∑r2l G(x1, . . . ,xn−1,unlR)+ ∑

l<m

(G(x1, . . . ,xn−1,ulnR +umnR)

−G(x1, . . . ,xn−1,ulnR)−G(x1, . . .xn−1,umnR))

= ∑r2l GunlR (x1, . . . ,xn−1)+ ∑

l<m

(G(uln+umn)R (x1, . . . ,xn−1)

−GulnR (x1, . . . ,xn−1)−GumnR (x1, . . . ,xn−1)

= H(x1, . . . ,xn−1,xn)

This proves uniqueness and completes the induction. Finally, a straightforward verification usinguniqueness of FR shows that the assignment F 7→ FR is k-linear.

50. We first assume that M is free and let (ei)i∈I be an ordered basis of M, so I is a totally orderedsol.QUABILset with respect to some ordering. Let β : M×M→ k be the unique bilinear form on M given bythe values

β (ei,e j) =

q(ei) for i = j ∈ I,q(ei,e j) for i, j ∈ I, i < j,0 for i, j ∈ I, i > j

on the basis vectors. Then, for any x = ∑ξiei ∈M, ξi ∈ k, we have

β (x,x) = β (∑ξiei,∑ξiei) = ∑i, j∈I

ξiξ jβ (ei,e j) = ∑ξ2i q(ei)+∑

i< jξiξ jq(ei,e j) = q(x),


as claimed. Now let M be an arbitrary projective module. Then there exists a k-module M′ makingN := M⊕M′ free. Let q′ : M′ → k be the zero quadratic form on M′ and consider the quadraticform Q := q⊕ q′ : N → k. By the special case just treated we find a bilinear form B : N×N → ksuch that Q(y) = B(y,y) for all y ∈ N. But this implies q(x) = β (x,x), where β : M×M→ k is therestriction of B to M×M.

51. (i)⇒ (iii). Obvious.sol.SEQUAFI(iii) ⇒ (ii). If x ∈ V satisfies q(x) = q(x,y) = 0 for all y ∈ V , then by linearity xK ∈ VK sat-

isfies (qK)(xK) = (qK)(xK ,y′) = 0 for all y′ ∈ VK , and (iii) gives xK = 0, hence x = 0. Thus qis non-degenerate, and it remains to show dimF (Rad(Dq)) ≤ 1. Since the radical of a symmetricor skew-symmetric bilinear form over a field is compatible with base change, it suffices to showdimK(Rad(DqK)) ≤ 1. But qK is non-degenerate by hypothesis, so its restriction to Rad(DqK)is anisotropic. On the other hand, K being algebraically closed, all quadratic forms of (possiblyinfinite) dimension > 1 over K are isotropic. The assertion follows.

(ii)⇒ (i). By hypothesis, there exists an element u ∈V satisfying Rad(Dq) = Fu, and if u 6= 0,then q(u) 6= 0 since q is non-degenerate. Now let L/k be any field extension. Then Rad(DqL)) =LuL. If x′ ∈VL satisfies (qL)(x′)= (qL)(x′,y′)= 0 for all y′ ∈VL then, in particular, x′ ∈Rad(DqL)=LuL, so some a ∈ L has x′ = a(uL). This implies 0 = qL(x′) = a2q(u), hence a = 0 or q(u) = 0,which in turn yields a= 0 or u= 0, hence x′= 0. Therefore qL is non-degenerate, and q is separable.

52. (a) Let x ∈V . The assertion is obvious if x = 0 or x is anisotropic. Hence we may assume that xsol.LIQUAis isotropic. Then there is a hyperbolic plane H ⊆V containing x, and the assumption n≥ 3 impliesthat H⊥ 6= 0 is a quadratic space in its own right. Let y ∈H⊥ be anisotropic. Then so are−y andx+ y, and x = (x+ y)− y is a decomposition of the desired kind.

(b) We may assume q(x1,x2) = 0. Let H ⊆ V be a hyperbolic subspace of dimension 4. Byhypothesis, q is non-singular and isotropic, hence universal, on H⊥ 6= 0. There are two cases.Case 1. x1 and x2 are linearly dependent. Then x2 =αx1 for some α ∈ F×, and we are in character-istic 2. The Witt Extension Theorem [27, Thm. 8.3] allows us to assume x1 ∈ H. Pick y1 ∈ H suchthat q(x1,y1) 6= 0, and y3 ∈H⊥ such that y := y1+y3 ∈V is anisotropic. Then q(x1,y 6= 0 6= q(y,x2).Case 2. x1 and x2 are linearly independent. Write H = H1 ⊥ H2 with hyperbolic planes H1,H2.Since q is universal on each of the subspaces H1,H2, the Witt Extension Theorem again justifiesthe assumption xi ∈Hi for i = 1,2. Pick yi ∈Hi such that q(xi,yi) 6= 0 for i = 1,2 and y3 ∈H⊥ suchthat y := y1 +y2 +y3 is anisotropic. Then q(x1,y) = q(x1,y1) 6= 0 6= q(y2,x2) = q(y,x2), as desired.

We now show that in (a) the assumption n ≥ 3 cannot be avoided. Indeed, let (V,q) be thehyperbolic plane over F2, the field with two elements. V contains precisely two isotropic vectorsrelative to q, denoted by e1,e2, which form a hyperbolic pair, and precisely one anisotropic one,denoted by a = e1 +e2. Hence any finite sum of anisotropic vectors in (V,q) is either zero or equalto a, so neither e1 nor e2 can be written in this form.

Finally, we show that in (b) the assumption of the Witt index being at least 3 cannot be avoided.Indeed, with (V,q) as in the previous paragraph, put (W,Q) = (V,q) ⊥ (V,q). Then W containsprecisely four anisotropic vectors relative to Q, namely a⊕ei, e j⊕a for i, j = 1,2. Moreover, Q(a⊕ei,a⊕e j) = 1−δi j and Q(a⊕ei,e j⊕a) = 0. Hence x1 := a⊕e1 and x2 := e1⊕a are anisotropic aswell as orthogonal, and there is a unique anisotropic vector y ∈W satisfying Q(x1,y) 6= 0, namely,y = a⊕ e2. But Q(y,x2) = 0, so the conclusion of (b) does not hold.

Section 13

53. Setting T = (t1, . . . , tn) as usual, x = (x1, . . . ,xn), y = (xn+1, . . . ,xn+p), xi ∈MR for 1≤ i≤ n+ psol.ZERLINand applying Thm. 13.7, we obtain


∑ν∈Nn

Tν (Π (ν ,0) f )R(x,y) = ∑(ν ,ν ′)∈Nn+p

tν11 · · · t

νnn 0νn+1 · · ·0νn+p (Π (νν ′) f )R(x,y)

= fR[T](n

∑i=1

tixi +p

∑i=1

0xn+i) = fR(n

∑i=1

tixi)

= ∑ν∈Nn

Tν (Π ν f )R(x),

and comparing coefficients, the assertion follows.

54. For d ∈ N define fd := Π (d) f , which by Thm. 13.7 and 13.8 (a) is a homogeneous polynomialsol.HOCOlaw M→ N of degree d over k. Moreover, the family ( fd)d≥0 is pointedly finite, and (13.7.1) forn = 1, r1 = 1R implies f = ∑d≥0 fd . This proves the existence. Conversely, let ( fd)d≥0 be anyfamily of polynomial laws over k with the desired properties. For R ∈ k-alg, r ∈ R, x ∈ MR wededuce

fR(rx) = ∑d≥0

fdR(rx) = ∑d≥0

rd fdR(x),

and the uniqueness property of Thm. 13.7 yields fd = Π (d) f for all d ∈ N. This shows uniquenessand solves the first part of the problem.

It remains to exhibit an example of a polynomial law f having fd 6= 0 for all d ∈N. To this end,consider a free k-module M of countably infinite rank and let (ei)i∈N be a basis of M over k. Ford ∈ N and R ∈ k-alg, define a set map fdR : MR→ R by

fdR(∑i∈N

rieiR) := rdd ,

where (ri)i∈N is an arbitrary sequence of finite support in R. The family ( fdR)R∈k-alg is clearly ahomogeneous polynomial law fd : M→ k of degree d over k, and the family ( fd)d≥0 is pointedlyfinite. Hence, by 13.6, f := ∑d≥0 fd : M→ k exists as a polynomial law over k and has the desiredproperty.

55 . The family of set maps wR : MR → NR for w ∈ N clearly varies functorially with R ∈ k-algsol.COPOLAand hence is a polynomial law w : M→ N such that wR(rx) = wR = r0wR(x) for all r ∈ R, x ∈MR.Thus w is homogeneous of degree 0. Conversely, let f : M→ N be any homogeneous polynomiallaw of degree 0 over k. Then w := fk(0) ∈ N, and for all R ∈ k-alg, x ∈MR, we apply (13.2.2) toobtain wR = fk(0)R = fR(0) = fR(0x) = 00 fR(x) = fR(x), hence wR = fR and then w = f .

56. (a) The first part is obvious. To establish the second, we begin by treating the case n = 1, so letsol.MULIf : M→ N be a homogeneous polynomial law of degree 1. Then 13.8 (c) implies Π ν f = 0 for allν ∈ N2 \(1,0),(0,1), and (13.7.1) yields

fk(αx+βy) = α(Π (1,0) f )k(x,y)+β (Π (0,1) f )k(x,y)

for all α,β ∈ k and all x,y ∈ M. Specializing α = 1, β = 0 and α = 0, β = 1, we conclude(Π (1,0) f )k(x,y) = fk(x), (Π (0,1) f )k(x,y) = fk(y). Hence fk : M→ N is a k-linear map. Since f ⊗R : MR → NR, for R ∈ k-alg, is a homogeneous polynomial law of degree 1 over R, it follows,therefore, that fR : MR→NR is an R-linear map. Moreover, for r ∈R, x∈M, we deduce fR(x⊗r) =fR(rxR) = r fR(xR) = r fk(x)R = fk(x)⊗ r = ( fk ⊗ 1R)(x⊗ r), hence fR = ( fk)R. Thus f is thepolynomial law derived from the scalar extensions of the linear map fk. This settles the case n = 1.

Now let n be arbitrary and µ : M1×·· ·×Mn→ N be a multi-homogeneous polynomial law ofmulti-degree 1 = (1, . . . ,1). For 1≤ i≤ n and fixed v j ∈M j (1≤ j≤ n, j 6= i), it is straightforwardto verify, using (13.4.1), that fi : Mi→ N defined by

fiR(x) := µR(v1R, . . . ,vi−1,R,x,vi+1,R, . . . ,vnR)


for R ∈ k-alg, x ∈ MR, is a homogeneous polynomial law of degree 1. Therefore, by the specialcase treated above, fik : Mi→ N is a k-linear map, forcing µk : M1×·· ·×MN → N to be k-multi-linear. Applying this to the extended polynomial law µ ⊗R : M1R× ·· ·×MnR → NR over R, weconclude that µR : M1R×·· ·×MnR→ NR is an R-multi-linear map. Hence, given ri j ∈ R, vi j ∈M jfor 1≤ j ≤ n and finitely many indices i, we obtain

µR(∑i

ri1vi1R, . . . ,∑i

rinvinR) = ∑i1,...,in

ri11 · · ·rinnµk(vi11, . . . ,vinn)R

= (µk⊗R)(∑i

ri1vi1R, . . . ,∑i

rinvinR).

This proves µR = µk⊗R and completes the solution of (a).(b) Let Q : M→ N be a quadratic map. We show that Q : M→ N is a polynomial law over k

by letting ϕ : R→ S be a morphism in k-alg and x,x′ ∈M, r,r′ ∈ R. The Cor. 12.5 implies

(1N ⊗ϕ)QR(x⊗ r) = (1N ⊗ϕ)QR(rxR) = (1N ⊗ϕ)(r2Q(x)R

)= (1N ⊗ϕ)

(Q(x)⊗ r2)

= Q(x)⊗ϕ(r)2 = ϕ(r)2Q(x)S = ϕ(r)2QS(xS)

= QS(ϕ(r)xS

)= QS

(x⊗ϕ(r)

)= QS (1M⊗ϕ)(x⊗ r)

and, similarly, (1N ⊗ϕ) QR(x⊗ r,x′⊗ r′) = QS (1M ⊗ϕ)(x⊗ r,x′⊗ r′). This proves that Q isindeed a polynomial law over k and obviously homogeneous of degree 2. Conversely, let f : M→Nbe a homogeneous polynomial law of degree 2 over k and consider the set map Q := fk : M→ N.By 13.8 (c) and (13.7.1), we have

Q(αx+βy) = α2(Π (2,0) f )k(x,y)+αβ (Π (1,1) f )k(x,y)+β

2(Π (0,2) f )k(x,y)

for all α,β ∈ k and all x,y ∈M. After specializing α = 1, β = 0 and α = 0, β = 1 and observing(13.14.1), this may be rewritten as

Q(αx+βy) = α2Q(x)+αβ (D f )k(x,y)+β

2Q(y),

where (D f )k = (Π (1,1) f )k is k-bilinear by 13.8 (a) and (a). This shows that Q is a quadratic mapwith DQ = (D f )k. For R ∈ k-alg, the R-quadratic map QR : MR → NR is uniquely determined by(12.5.1), which proves QR = fR, hence Q = f .

57. (a) Applying Thm. 13.7, we deducesol.HIDIDE

∑ν1,...,νp≥0

tν11 · · · t

νpp

(∑

νp+1,...,νn≥0tνp+1

p+1 · · · tνnn (Π (ν1,...,νn) f )R(x1, . . . ,xp,x, . . . ,x)

)= ∑

ν∈NnTν (Π ν f )(x1, . . . ,xp,x, . . . ,x) = fR[T]

( p

∑j=1

t jx j +(n

∑j=p+1

t j)x)

= ∑ν1,...,νp,νp+1≥0

tν11 · · · t

νpp (

n

∑j=p+1

t j)νp+1 (Π (ν1,...,νp,νp+1) f )R(x1, . . . ,xp,x)

= ∑ν1,...,νp≥0

tν11 · · · t

νpp

(∑i≥0

(n

∑j=p+1

t j)i(Π (ν1,...,νp,i) f )R(x1, . . . ,xp,x)

).

Comparing coefficients and using the multi-monomial formula (Abramowitz-Segun [1, p. 883]),this implies, for all ν1, . . . ,νp ∈ N,


∑νp+1,...,νn≥0

tνp+1p+1 · · · t

νnn (Π (ν1,...,νn) f )R(x1, . . . ,xp,x, . . . ,x)

= ∑i≥0

(n

∑j=p+1

t j)i(Π (ν1,...,νp,i) f )R(x1, . . . ,xp,x)

= ∑i≥0

(∑

νp+1,...,νn≥0,νp+1+···+νn=i

i!νp+1! · · ·νn!

tνp+1p+1 · · · t

νnn (Π (ν1,...,νp,i) f )R(x1, . . . ,xp,x)

)= ∑

νp+1,...,νn≥0tνp+1

p+1 · · · tνnn(νp+1 + · · ·+νn)!

νp+1! · · ·νn!(Π (ν1,...,νp,νp+1+···+νn) f )R(x1, . . . ,xp,x),

and comparing coefficients again completes the proof of (a).(b) Let ϕ : R→ S be a morphism in k-alg and x ∈MR. Since Dp f is a polynomial law over k,

an application of (13.4.2) yields

(∂[p]y f )S (1M⊗ϕ)(x) = (Dp f )S

((1M⊗ϕ)(x),yS

)= (Dp f )S

((1M⊗ϕ)(x),(1M⊗ϕ)(yR)

)= (Dp f )S (1M×M⊗ϕ)(x,yR) = (1N ⊗ϕ) (Dp f )R(x,yR)

= (1N ⊗ϕ) (∂ [p]y )R(x).

Thus ∂[p]y f is a polynomial law over k and the map ∂

[p]y : Polk(M,N)→ Polk(M,N) is clearly

k-linear.Next combine Prop. 13.19 for y1 = · · ·= yp = y with (a) to conclude((∂y)

p f)

R(x) = (∂y · · ·∂y f )R(x) = ∑i≥0

(Π (i,1,...,1) f )R(x,yR, . . . ,yR) = p! ∑i≥0

(Π (i,p) f )R(x,yR)

= p!(Dp f )R(x,yR) = p!(∂ [p]y f )R(x),

which implies (∂y)p = p!∂ [p]

y as linear maps Polk(M,N)→ Polk(M,N). This completes the solutionof (b).

(c) Let R ∈ k-alg, t, t1, t2 be independent variables and T = (t1, t2). For x,y1,y2 ∈MR we com-bine the Taylor expansion formula (13.13.3) with Thm. 13.7 and obtain

∑n≥0

tn(Dn f )R[T](x, t1y1 + t2y2) = fR[t,T](x+ t(t1y1 + t2y2)

)= fR[t,T]

(x+(tt1)y1 +(tt2)y2

)= ∑

ν0,ν1,ν2≥01ν0 (tt1)

ν1 (tt2)ν2 (Π (ν0,ν1,ν2) f )R(x,y1,y2)

= ∑ν1,ν2≥0

tν1+ν2 tν11 tν2

2 ∑i≥0

(Π (i,ν1,ν2) f )R(x,y1,y2)

= ∑n≥0

tnn

∑j=0

t j1tn− j

2 ∑i≥0

(Π (i, j,n− j) f )R(x,y1,y2)

Comparing coefficients of tn, we therefore end up with the formula

(Dn f )R[T](x, t1y1 + t2y2) =n

∑j=0

t j1tn− j

2 ∑i≥0

(Π (i, j,n− j) f )R(x,y1,y2) (89) DIFENX

This formula will now be applied in the special case n = 2. From Exc. 53 combined with 13.8 (a)we deduce, for all i ∈ N,

(Π (i,2,0) f )R(x,y1,y2) = (Π (i,2) f )R(x,y1),

(Π (i,0,2) f )R(x,y1,y2) = (Π (i,2,0) f )R(x,y2,y1) = (Π (i,2) f )R(x,y2).


Hence (89) for n = 2 implies

(D2 f )R[T](x, t1y1 + t2y2) = t21 ∑

i≥0(Π (i,2) f )R(x,y1)+ t1t2 ∑

i≥0(Π (i,1,1) f )R(x,y1,y2)

+ t22 ∑

i≥0(Π (i,2) f )R(x,y2)

= t21(D

2 f )R(x,y1)+ t1t2 ∑i≥0

(Π (i,1,1) f )R(x,y1,y2)+ t22(D

2 f )R(x,y2).

In particular, applying Prop. 13.19,

(∂[2]y+z f )R(x) = (D2 f )R(x,yR + zR)

= (D2 f )R(x,yR)+∑i≥0

(Π (i,1,1) f )R(x,yR,zR)+(D2 f )R(x,zR)

=((∂

[2]y f )R +(∂y∂z f )R +(∂

[2]z f )R

)(x),

which completes the proof of (c).

58 . By (13.14.1) we have D2 f = Π (1,2) f , and (13.8.1) implies (D2 f )(x,y) = Π (1,2) f )(x,y) =sol.EXOCUB(Π (2,1) f )(y,x) = (D f )(y,x), hence

f (y,x) = (D2 f )(x,y). (90) EFYEX

Combining (13.14.2) for d = 3 with (13.13.3) and (2), (90), we obtain (3). In order to derive (4),we first let x,y ∈M. Then (90), Exc. 57 (c) and Prop. 13.19 imply

f (x+ y,z) = (D2 f )(z,x+ y) = (∂[2]x+y f )(z)

= (∂[2]x f )(z)+(∂x∂y f )(z)+(∂

[2]y f )(z)

= (D2 f )(z,x)+(Π (1,1,1) f )(x,y,z)+(D2 f )(z,y)

= f (x,z)+ f (x,y,z)+ f (y,z),

as claimed. The general case x,y ∈MR now follows by looking at f ⊗R rather than f . Combining(3) for t = 1 and (4), we now obtain

f (x+ y+ z) − f (x+ y)− f (y+ z)− f (z+ x)+ f (x)+ f (y)+ f (z)

= f (x+ y)+ f (x+ y,z)+ f (z,x+ y)+ f (z)− f (x+ y)− f (y)− f (y,z)

− f (z,y)− f (z)− f (z)− f (z,x)− f (x,z)− f (x)+ f (x)+ f (y)+ f (z)

= f (x,z)+ f (x,y,z)+ f (y,z)+ f (z,x)+ f (z,y)+ f (z)− f (y)− f (y,z)− f (z,y)

− f (z)− f (z)− f (z,x)− f (x,z)− f (x)+ f (x)+ f (y)+ f (z)

= f (x,y,z),

and this is (5). Before dealing with (6), we prove

f (n

∑i=1

rixi,x) =n

∑i=1

r2i f (xi,x)+ ∑

1≤i< j≤nrir j f (xi,x j,x) (91) EFRIXI

for x ∈MR by induction on n. For n = 1 there is nothing to prove. For n > 1, the induction hypoth-esis and (4) imply


f (n

∑i=1

rixi,x) = f (n−1

∑i=1

rixi,x)+ f (n−1

∑i=1

rixi,rnxn,x)+ f (rnxn,x)

=n−1

∑i=1

r2i f (xi,x)+ ∑

1≤i< j<nrir j f (xi,x j,x)+

n−1

∑i=1

rirn f (xi,xn,x)+ r2n f (xn,x)

=n

∑i=1

r2i f (xi,x)+ ∑

1≤i< j≤nrir j f (xi,x j,x),

as claimed. Now we can prove (6), again by induction on n. The case n = 1 again being obvious,let us assume n > 1 and that (6) holds for n−1. Then this and (91) yield

f (n

∑i=1

rixi) = f (n−1

∑i=1

rixi)+ f (n−1

∑i=1

rixi,rnxn)+ f (rnxn,n−1

∑i=1

rixi)+ f (rnxn)

=n−1

∑i=1

r3i f (xi)+ ∑

1≤i, j<n,i6= jr2

i r j f (xi,x j)+ ∑1≤i< j<l<n

rir jrl f (xi,x j,xl)

+n−1

∑i=1

r2i rn f (xi,xn)+ ∑

1≤i< j<nrir jrn f (xi,x j,xn)+

n−1

∑j=1

r2nr j f (xn,x j)+ r3

n f (xn)

=n

∑i=1

r3i f (xi)+ ∑

1≤i, j≤n,i6= jr2

i r j f (xi,x j)+ ∑1≤i< j<l≤n

rir jrl f (xi,x j,xl).

This completes the induction and the proof of (a).(b) (i) ⇒ (ii). From (i) we deduce gF (ei) = 0 for 1 ≤ i ≤ n, while (5) implies gF (x,y,z) = 0

for all x,y,z ∈ Fn. Moreover, from Euler’s differential equation (13.14.3 we conclude gF (x,x) =(Dg)F (x,x) = 3gF (x) = 0. It remains to show that F consists of two elements and gF (x0,y0) 6= 0for some x0,y0 ∈ Fn. Replacing f by g and specializing t 7→ α ∈ F× in (3), we conclude gF (x,y)+αgF (y,x) = 0. Assuming |F |> 2, this implies gF (x,y) = 0 for all x,y ∈ Fn. But If gF (x,y) = 0 forall x,y ∈ Fn, then (7) and (6) for g in place of f and eiR in place of xi for 1 ≤ i ≤ n would lead tothe contradiction g = 0 as a polynomial law over F . Thus (ii) holds.

(ii) ⇒ (iii). By (7) and morcumap(91) for g in place of f , the map Fn ×Fn → F , (x,y) 7→gF (x,x) is an alternating F-bilinear form, forcing

S :=(gF (ei,e j)

)1≤i, j≤n ∈Matn(F)

to be an alternating matrix and S 6= 0 by (ii). Now let

x =

r1...

rn

=n

∑i=1

rieiR ∈ Rn = (Fn)R (r1, . . . ,rn ∈ R).

The (6) and (7) imply

gR(x) = gR(n

∑i=1

rieiR) =n

∑i=1

r3i gF (ei)+ ∑

1≤i, j≤n,i6= jr2

i r jgF (ei,e j)+ ∑1≤i< j<l≤n

rir jrlgF (ei,e j,el)

= ∑1≤i, j≤n

rigF (ei,e j)r2j = xt Sx2,

as claimed.(iii)⇒ (i). Since F ∼= F2 consists of two elements, the elements of Fn as an F-algebra are all

idempotents. Hence gF (x) = xt Sx = 0 for all x∈ Fn, whence the set map gF : Fn→ F is identicallyzero. On the other hand, since S 6= 0, there are elements x0,y0 ∈Fn such that xt

0Sy0 6= 0, and passing


from F ∼=F2 to the separable quadratic extension K =F(θ)∼=F4 with θ ∈K satisfying θ 2 = θ +1,we deduce

gK(x0 +θy0) = (x0 +θy0)t S(x2

0 +θ2y2

0)

= (x0 +θy0)t S(x0 +θ

2y0) = (x0 +θy0)t S(x0 +θy0)+(x0 +θy0)

t Sy0

= xt0Sy0 6= 0.

Thus gK 6= 0, so g : Fn→ F is a non-zero cubic form.

59 . (a) If f : M→ N and f ′ : M′ → N′ are polynomial laws over k, a morphism from f to f ′ insol.CUMAk-polaw is defined as a pair of linear maps ϕ : M→M′, ψ : N→ N prime making a commutativediagram

Mf //

ϕ

N

ψ

M′

f ′// N′

(92) MORPOL

of polynomial laws over k.(c) Since g is quadratic in the first variable, the expression g(x,y,z) is surely trilinear. Moreover,for x,y,z ∈M, we repeatedly use condition (iii) to compute

f (x+ y+ z)− f (x+ y)− f (y+ z)− f (z+ x)+ f (x)+ f (y)+ f (z)

= g(x+ y,z)+g(z,x+ y)+ f (z)− f (y)−g(y,z)−g(z.y)− f (z)

− f (z)−g(z,x)−g(x,z)− f (x)+ f (x)+ f (y)+ f (z)

= g(x,y,z),

and since the very left-hand side is totally symmetric in x,y,z, so is the right. This proves thefirst part of (a). As to the second, (10) follows immediately by linearizing g(y,x) = 0 for x ∈Rad( f ,g) with respect to y ∈ M and using the total symmetry of (8). Hence (iii) implies thatRad( f ,g) is a submodule of M. Finally, for a linear map π as indicated, the maps f1 : M1 → Nand g1 : M1×M1 → N given by f1(π(x)) := f (x) and g1(π(x),π(y)) := g(x,y), respectively, forx,y ∈M in view of (9) are well defined, and it is straightforward to check that ( f1,g1) : M1→ N1is a cubic map with the desired properties. This completes the proof of (c).

(d) If ( f ,g) : M→ N and ( f ′,g′) : M′→ N′ are cubic maps over k, a morphism from ( f ,g) to( f ′,g′) in k-cumap is defined as a pair of linear maps ϕ : M→M′, ψ : N→ N′ making commuta-tive diagrams

Mf //

ϕ

N

ψ

M×Mg //

ϕ×ϕ

N

ψ

M′

f ′// N′, M′×M′

g′// N′

(93) MORCUMAP

of set maps.Now let f : M→ N be a homogeneous polynomial law of degree 3 over k. We combine equa-

tions (3), (4) of Exc. 58 with Euler’s differential equation (13.14.3) to conclude that ( fk,(D f )k)is a cubic map from M to N. If (ϕ,ψ) : f → f ′ is a morphism in k3-holaw, then the chain rules(13.15.4), (13.15.5) applied to the commutative diagram (92) show that (ϕ,ψ) is also a morphismfrom ( fk,(D f )k) to ( f ′k,(D f ′)k) in k-cumap. Thus we have obtained a functor from k3-holaw to


k-cumap. We now proceed to define a functor in the opposite direction that is inverse to the pre-vious one, and first claim that the assignment f 7→ ( fk,(D f )k) gives a linear bijection from thehomogeneous polynomial laws M→ N of degree 3 over k onto the cubic maps M→ N over k. Themap in question is clearly k-linear. But it is also injective since fk = (D f )k = 0 for a homogeneouspolynomial law f : M→ N of degree 3 over k by equation (6) of Exc. 58 implies f = 0. It thusremains to show that it is surjective as well, so let ( f ,g) : M → N be any cubic map. Our taskwill be define set maps fR : MR → NR that vary functorially with R ∈ k-alg, hence give rise to apolynomial law f : M→ N over k, are homogeneous of degree 3, and have fk = f , (D f )k = g.

We begin by extending g to a unique R-quadratic-linear map gR : MR×MR→ NR such that

gR(xR,yR) = g(x,y)R (x,y ∈M). (94) RQUALI

In order to do so, let y ∈M. By Prop. 12.4, the k-quadratic map g(−,y) : M→ N extends uniquelyto an R-quadratic map g(−,y)R : MR → NR such that g(−,y)R(xR) = g(x,y)R for all x ∈M. Thuswe obtain a map

g′ : MR×M −→ NR, (x,y) 7−→ g′(x,y) := g(−,y)R(x),

which is clearly k-linear in the second variable and R-quadratic in the first. For x ∈MR, therefore,g′(x,−) : M→NR is k-linear and hence induces canonically an R-linear map g′(x,−)R : MR→NR.Summing up, we can now define gR : MR×MR → NR by gR(x,y) := g′(x,−)R(y) for x,y ∈ MR,which satisfies (94) since for x,y ∈M we have

gR(xR,yR) = g′(xR,−)R(yR) = g′(xR,−)(y) = g′(xR,y) = g(−,y)R(xR) = g(x,y)R.

For finitely many ri,sl ∈ R, xi,yl ∈M, we now conclude

gR(∑i

rixiR,∑l

slylR) = ∑i

∑l

r2i slg(xi,yl)R +∑

i< j∑

lrir jslg(xi,x j,yl)R (95) TIGIJ

from the fact that g is R-quadratic-linear and (94) holds. In particular, g is uniquely determined bythese two conditions.

We now turn to the construction of f . First we note

f (∑i

αixi) = ∑i

α3i f (xi)+∑

i6= jα

2i α jg(xi,x j)+ ∑

i< j<lαiα jαlg(xi,x j,xl) (96) EXCUMA

for finitely many αi ∈ k, xi ∈ M. The proof proceeds by induction on the number of non-zerosummands along exactly the same lines as the one of formula (6) in Exc. 58. Let us now begin byassuming that M is free as a k-module, with basis (ei)i∈I , indexed by a totally ordered set I. In thespirit of (96), we then define a set map fR : MR→ NR by

fR(∑i

rieiR) := ∑i

r3i f (ei)R +∑

i6= jr2

i r jg(ei,e j)R + ∑i< j<l

rir jrlg(ei,e j,el) (97) TIFCUB0

for all families (ri)i∈I with finite support in R. By definition, the set maps fR vary functorially withR ∈ k-alg, hence define a polynomial law f : M→ N over k which is obviously homogeneous ofdegree 3. For x = ∑i rieiR,y = ∑i sieiR ∈MR, where the families (ri),(si) in R have finite support,we obviously have

(D f )R(x,y) = 3∑i

r2i si f (ei)R +∑

i6= j(r2

i s j +2rir jsi)g(eI ,e j)R (98) DETIF

+ ∑i< j<l

(rir jsl + r jrlsi + rlris j)g(ei,e j,el)R. (99)


On the other hand,combining condition (iv) with (95) and the total symmetry of g(ei,e j,el) ini, j, l ∈ I, we obtain

gR(x,y) = ∑i

∑l

r2i slg(ei,el)R +∑

i< j∑

lrir jslg(ei,e j,el)

= (∑l=i

+∑l 6=i

)r2i slg(ei,el)R +( ∑

l<i< j+ ∑

l=i< j+ ∑

i<l< j+ ∑

i< j=l+ ∑

i< j<l)rir jslg(ei,e j,el)R

= 3∑i

r2i si f (ei)R +∑

i6= j(r2

i s j +2rir jsi)g(ei,e j)R

+ ∑i< j<l

(rir jsl + r jrlsi + rlris j)g(ei,e j,el)R.

Thus ( fR,(D f )R) = ( fR, gR), and this is a cubic map MR→ NR over R. In particular, for R = k, weconclude ( fk,(D fk)) = ( f ,g), which completes the proof in case M is free as a k-module.

Now assume that M is arbitrary. Then there exists a short exact sequence

0 // L ι // F π // M // 0

of k-modules, where F is free. Since the functor −⊗R is right exact, we obtain an induced exactsequence

LRιR // FR

πR // MR // 0.

Now observe that ( f1,g1) := ( f π,g (π×π)) is a cubic map from F to N. By the special casetreated before, we therefore obtain a homogeneous polynomial Law f1 : F → N of degree 3 andan R-quadratic-linear map g1R : FR×FR→ NR satisfying ( f1R,(D f1)R) = ( f1R, g1R) and f1k = f1,g1k = g1. We now claim

Ker(πR)⊆ Rad( f1R, g1R). (100) KEPI

The last displayed exact sequence shows that any element of Ker(πR) can be written as ιR(z), forsome z = ∑i riuiR ∈ LR, ri ∈ R, ui ∈ L. Applying (96) to the R-cubic map ( f1R, g1R), we conclude

f1R(ιR(z)

)= f1R

(∑

iriι(ui)R

)= ∑

ir3

i f1(ι(ui)

)R +∑

i6= jr2

i r jg1(ι(ui), ι(u j)

)R + ∑

i< j<lrir jrlg1(ι(ui), ι(u j), ι(ul)

)R,

where f1(ι(ui)) = ( f π ι)(ui) = 0, g1(ι(ui), ι(u j)) = g((π ι)(ui),(π ι)(u j)) = 0 and, simi-larly, g1(ι(ui), ι(u j), ι(ul)) = 0. An analogous computation, involving (95), shows g1R(ιR(z),y) =g1R(y, ιR(z)) = 0 for all y ∈ FR. Hence ιR(z) ∈ Rad( f1R, g1R), and the proof of (100) is com-plete. By (a), therefore, we find a cubic map ( fR, gR) : MR → NR such that fR πR = f1R andgR (πR×πR) = g1R. Since the linear maps πR, for all R ∈ k-alg, are surjective, and the f1R varyfunctorially with R, so do the fR. In other words, f : M→N is a homogeneous polynomial law overk of degree 3. By construction, we have fk π = f1k = f1 = f π , hence fk = f , while (13.15.5)implies (D f )k (π×π) = (D( f π))k = (D f1)k = g1k = g1 = g (π×π), hence (D f )k = g. Thusthe map in question is also surjective, which completes the proof of our intermediate assertion.

To every object ( f ,g) of k-cumap we can thus associate a unique object f ∗g := f in k3-holawsuch that ( f ∗ g)k = f and D( f ∗ g)k = g, and we will be through once we have shown that everymorphism (ϕ,ψ) : ( f ,g)→ ( f ′,g′) in k-cumap is also a morphism (ϕ,ψ) : f ∗ g→ f ′ ∗ g′ ink3-holaw. By (92) and (93), we have

360 6 Cubic Jordan algebras(ψ ( f ∗g)

)k = ψ ( f ∗g)k = ψ f = f ′ ϕ = ( f ′ ∗g′)k ϕ =

(( f ′ ∗g′)ϕ

)k,(

D(ψ ( f ∗g)

))k(x,y) = ψ

(D(( f ∗g)

)k(x,y)

)= ψ

(g(x,y)

)= g′

(ϕ(x),ϕ(y)

)=(D( f ′ ∗g′)

)k

(ϕ(x),ϕ(y)

)=(

D(( f ′ ∗g′)ϕ

))k(x,y).

Hence the uniqueness part of our intermediate assertion shows that (ϕ,ψ) : f ∗g→ f ′ ∗g′ is indeeda morphism in k3-holaw.

60. Fixing R ∈ k-alg, there is clearly a unique k-linear mapsol.NONSC

ΨR : N⊗Pol(M,k)−→Map(MR,NR)

given byΨR(v⊗ f )(x) = v⊗ fR(x)

for v ∈ N, f ∈ Pol(M,k), x ∈MR. For the first part of the problem, it will be enough to show thatΨRvaries functorially with R, so let ϕ : R→ S be a morphism in k-alg. Since f is a scalar polynomiallaw, hence the diagram

MRfR//

1M⊗ϕ

R

ϕ

MS

fS// S

commutes, so doesMR

ΨR(v⊗ f )//

1M⊗ϕ

NR

1N⊗ϕ

MS

ΨS(v⊗ f )// NS.

Hence we have shown that a k-linear map Ψ : N ⊗ Pol(M,k)→ Pol(M,N) satisfying (11) doesindeed exist and is obviously unique. It remains to prove that Ψ is an isomorphism if N is finitelygenerated projective.

To this end, N still being arbitrary, we write Ψ N :=Ψ to indicate dependence on N and consideranother k-module N′. Then the linear projections π : N⊕N′→ N, π ′ : N⊕N′→ N′, canonicallyregarded as homogeneous polynomial laws of degree 1 via Exercise 56, induce k-linear maps

π∗ : Pol(M,N⊕N′)−→ Pol(M,N), π′∗ : Pol(M,N⊕N′)−→ Pol(M,N′)

given by f 7→ π f , f 7→ π ′ f , respectively, which in turn yield an isomorphism

π∗⊕π′∗ : Pol(M,N⊕N′) ∼−→ Pol(M,N)⊕Pol(M,N′).

After the natural identification M⊗ (N⊕N′) = M⊗N⊕M⊗N′, it is straightforward to check thatthe diagram

(N⊕N′)⊗Pol(M,k)Ψ N⊕N′

//

can ∼=

Pol(M,N⊕N′)

∼= π∗⊕π ′∗

(N⊗Pol(M,k))⊕ (N′⊗Pol(M,k))Ψ N⊕Ψ N′

// Pol(M,N)⊕Pol(M;N′)

(101) PSINPR


commutes. Now suppose N is finitely generated projective and choose a k-module N′ makingN⊕N′ free of finite rank. By a repeated application of (101), we are reduced to the case N = k.But then Ψ =Ψ k = 1Pol(M,k), and the problem is solved.

61. We require the following easy lemma.sol.SEPOLALemma. Assume N is a K-module, N′ is a K′-module and ϕ : M′→M, ψ : N′→ N are σ -semi-linear maps. Then there exists a unique σ -semi-linear map

ϕ⊗σ ψ : M′⊗K′ N′ −→M⊗K N

such that(ϕ⊗σ ψ)(x′⊗K′ y

′) = ϕ(x′)⊗K ψ(y′)

for all x′ ∈M′, y′ ∈ N′.

Proof. Uniqueness is obvious. To prove existence, we note that ϕ (resp. ψ) may be viewed as aK′-linear map ϕ ′ : M′ → Mσ (resp. ψ ′ : N′ → Nσ ) satisfying ϕ ′(x′) = ϕ(x′)σ for x′ ∈ M′ (resp.ψ ′(y′) = ψ(y′)σ for y′ ∈ N′). Hence

ϕ′⊗K′ ψ

′ : M′⊗K′ N′ −→Mσ ⊗K′ N

σ

exists as a K′-linear map. On the other hand, one checks easily, using (13.24.3), that the map

Mσ ×Nσ −→ (M⊗K N)σ , (x,y) 7−→(σM(x)⊗K σN(y)

)σ

is K′-bilinear, hence canonically induces a K′-linear map

φ : Mσ ⊗K′ Nσ −→ (M⊗K N)σ .

Thus(ϕ⊗σ ψ)′ := φ (ϕ ′⊗K′ ψ

′) : M′⊗K′ N′ −→ (M⊗K N)σ

is K′-linear and therefore determines a σ -semi-linear map

ϕ⊗σ ψ := σM⊗K N (ϕ⊗σ ψ)′ : M′⊗K′ N′ −→M⊗K N

with the desired properties. Using this lemma, it suffices to put ϕR := ϕ⊗σ σR to obtain a σ -semi-linear map M′Rσ →MR

satisfying (12), uniqueness again being obvious. For r,r1 ∈ R and x′ ∈M′, we apply (12) and obtain

ϕR(rσ

1 (x′⊗K′ r

σ ))= ϕR

(x′⊗K′ (r1r)σ

)= ϕ(x′)⊗K (r1r)

= r1(ϕ(x′)⊗K r

)= σR(rσ

1 )ϕR(x′⊗K′ rσ ),

showing that ϕR is indeed σR-semi-linear.We show that ϕ := (ϕR)R∈k-alg is a σ -semi-polynomial law by letting λ : R→ S be any mor-

phism in K-alg to compute, for x′ ∈M′, r ∈ R,

(1M⊗K λ )ϕR(x′⊗K′ rσ ) = (1M⊗K λ )(ϕ(x′)⊗K r) = ϕ(x′)⊗K λ (r)

= ϕS(x′⊗K′ λ (r)

σ)= ϕS

(x′⊗K′ λ

σ (rσ ))

= ϕS (1M′ ⊗K′ λσ )(x′⊗K′ r

σ ).

Hence the set maps ϕR : M′Rσ → MR vary functorially with R ∈ K-alg, whence ϕ : M′ → M is aσ -semi-polynomial law over k.

(b) Let λ : R→ S be a morphism in K-alg. The fact that λ σ : Rσ → Sσ is a morphism in K′-alggives rise to the commutative diagram


M′′Rστ

( fg)R

%%gRσ

//

1M′′⊗K′′λστ

M′RσfR//

1M′⊗K′λσ

MR

1M⊗K λ

M′′Sστ

( fg)S

99gSσ

// M′Sσ

fS// MS,

and the first assertion follows. As to the second, let χ : K′′′ → K′′ be another morphism in k-alg,M′′′ a K′′′-module and h : M′′′→M′′ a χ-semi-polynomial law over k. Then(

( f g)h)

R = ( f g)R hRστ = fR gRσ h(Rσ )τ = fR (gh)Rσ =(

f (gh))

R.

Hence ( f g)h = f (gh) as στχ-semi-polynomial laws M′′′→M over k.(c) For R ∈ k-alg, x′′ ∈M′′, r ∈ R we compute

(ϕ ψ)R(x′′⊗K′′ rστ ) = (ϕ ψ)(x′′)⊗K r = ϕ

(ψ(x′′)

)⊗K r

= ϕR(ψ(x′′)⊗K′ r

σ)= ϕR

(ψRσ

(x′′⊗K′′ (r

σ )τ))

= (ϕR ψRσ )(x′′⊗K′′ rστ ),

and this means (ϕ ψ)∼ = ϕ ψ .(d) We have σσ−1 = 1K , σ−1σ = 1K′ and (13.24.3) implies

1M = (σσ−1)M = σM (σ−1)Mσ ,

1Mσ = (σ−1σ)Mσ = (σ−1)Mσ σ

(Mσ )σ−1 = (σ−1)Mσ σM .

By (c), therefore, (σM)∼ ((σ−1)Mσ )∼ is the polynomial law over K given by the identity of M,while ((σ−1)Mσ )∼ (σM)∼ is the polynomial law over K′ given by the identity of Mσ . Hence ifg is a polynomial law satisfying the requirements of (d), then g = ((σ−1)Mσ )∼ f . On the otherhand, defining g in this way and observing that(

(σ−1)Mσ

)∼ : M = (Mσ )σ−1 −→Mσ

is a σ−1-semi-polynomial law over k, it follow from (b that g : M′→Mσ is an ordinary polynomiallaw over K′ such that f = (σM)∼ g.

Chapter 3

Section 14.

63. Since the associator of A is alternating, the first part follows immediately from (8.5.1). Thesol.COMALTsecond part is now obvious.

64. The implications (i)⇒(ii)⇒(iii)⇒(iv) and (i)⇒(v) are obvious, while (iv)⇒(iii) follows fromsol.UNALTthe fact that Lx,Rx,Ux = LxRx commute by pairs. We are thus left with the following implications.

(iv)⇒(i). Given y ∈ A, we find an element z ∈ A satisfying xzx = y. Moreover, since (iii) holdsby what we have just seen, there are e, f ∈ A such that ex = x = x f . Now the Moufang identities


yield ey = e(xzx) = ((ex)z)x = xzx = y, y f = (xzx) f = x(z(x f )) = xzx = y, so e (resp. f ) is a left(resp. right) unit for A. Hence A is unital with 1A = e = f .

(v)⇒(i). Since Lx is surjective, some e ∈ A has xe = x and any z ∈ A can be written as z = xwfor some w ∈ A. Hence x(ez) = x(e(xw)) = (xex)w = x2w = x(xw) = xz, forcing e to be a left unitfor A since Lx is also injective. Passing to Aop, we find a right unit for A as well, and (i) holds.

65. The implication (ii)⇒(i) is obvious, while (i)⇔(iii) becomes (i)⇔(ii) in Aop. Thus we onlysol.ONESINVALThave to prove (i)⇒(ii). Using the Moufang identities in operator form (14.3.4), (14.3.5), we obtainE2 = LxLyLxLy = LxyxLy = LxLy = E and, similarly, F2 = F . Furthermore, by (14.5.4), EF =LxLyRyRx = LxUyRx =Uxy =U1A = 1A, hence 1A = E2F = EEF = E, and this is (ii).

66 . Since xy ∈ A is invertible, so is the linear operator Uxy, by Prop. 14.6. But Uxy = RyUxLysol.INVPRODALTby (14.5.4), so Ly is injective. On the other hand, the left Moufang identity and (14.6.2) implyLy(x(y(xy)−1)) = y[x(y(xy)−1)] = [y(xy)](xy)−1 = y = Ly1A, and since Ly is injective, we concludex(y(xy)−1) = 1A, so x is right invertible with y(xy)−1 as a right inverse. Reading this in Aop showsthat y is left invertible with (xy)−1x as a left inverse.

67. For i, j = 1,2 and x ∈ A, the expression cixc j is unambiguous since it takes place in the unitalsol.PEIRCEALTsubalgebra of A generated by c and x, which is associative by Cor. 15.5. We now claim that thefollowing relations hold.

Lci Lc j = δi jLci , Rci Rc j = δi jRci , Lci Rc j = Rc j Lci . (102) PEMUL

For i = j, this follows immediately from (14.1.4)−(14.1.6), while for i 6= j it suffices to observeLc j = 1A−Lci , Rc j = 1A−Rci to arrive at the desired conclusion. Now let i, j, l,m = 1,2 and putEi j := Lci Rc j . Then (102) yields Ei jElm = Lci Rc j Lcl Rcm = Lci Lcl Rc j Rcm = δilδ jmLci Rc j = δilδ jmEi j ,so the Ei j are indeed orthogonal projections that add up to ∑i, j=1,2 Ei j = ∑i, j=1,2 Lci Rc j =Uc1+c2 =U1A = 1A. This proves (1), where the Ai j = Ai j(c) are defined by the first among the relations in (2).The remaining ones are now obvious. Next we establish (3)−(5). In order to do so, we let xi j ∈ Ai j ,ylm ∈ Alm etc. and first prove

cix jl = δi jx jl , xi jcl = δ jlxi j. (103) IDPE

Indeed, by (2) the first relation is clear for i = j, while for i 6= j we obtain cix jl = x jl − c jx jl =x jl−x jl = 0, as claimed. The second relation is proved analogously. Combining now the linearizedright alternative law (14.2.2) with (103), we obtain (xi jy jl)cl = xi j(y jlcl + cly jl)− (xi jcl)y jl =xi jy jl +δ jlxi jy jl−δ jlxi jy jl = xi jy jl and, similarly, ci(xi jy jl) = xi jy jl . This proves (3). Furthermore,for i 6= j, the left Moufang identity and (103) yield xiiy jl = (cixiici)y jl = ci(xii(ciy jl)) = 0 and, sim-ilarly, yl jxii = 0. Thus (4) holds. Finally, since c j(xi jyi j) = (c jxi j +xi jc j)yi j−xi j(c jyi j) = xi jyi j =

(xi jyi j)ci, we also have (5). In particular, x2i j = c jx2

i j = (c jxi j)xi j = 0, while the remaining assertionA2

i j = 0 for A associative is obvious.

Section 16.

68. By Prop. 14.6, x ∈ A is invertible in A(p,q) iff U (p,q)x = UxUpq (by (16.5.1)) is bijective iff Uxsol.INVISOTALT

is bijective iff x is invertible in A, in which case its inverse in A(p,q) is x(−1,p,q) = (U (p,q)x )−1x =

U−1pq U−1

x x =U−1pq x, as claimed.

69 . By definition, f ,g,h are linear bijections from A to B satisfying f (xy) = g(x)h(y) for allsol.ALBISOTx,y ∈ A. Setting u := g−1(1B), v := h−1(1B), we conclude f (uy) = h(y), f (xv) = g(x), equiva-lently, f Lu = h, f Rv = g. Thus Lu,Rv : A→ A are linear bijections, forcing A to be unital(Exc. 64) and u,v ∈ A to be invertible (Prop. 14.6), with inverses p := v−1, q := u−1 This im-plies f ((xp)(qy)) = g(xp)h(qy) = f ((xv−1)v) f (u(u−1y)) = f (x) f (y). Thus f : A(p,q) → B is anisomorphism; in particular, B is alternative.


70 . See [99, 2.2-2.8]. In particular 1Str(A) = (1A.1A,1A) is the unit element of Str(A) andsol.STRGRALT(p,q,g)−1 = (g−1(q−1 p−2),g−1(q−2 p−1),g−1) is the inverse of (p,q,g) ∈ Str(A).

71. The equation (Lux)u · u−2(Luy) = uxu · u−2(uy) = u(x(u(u−1y))) = u(xy) = Lu(xy) for allsol.EXTLRMULTx,y ∈ A shows that Lu : A→ A(u,u−2) is an isomorphism. Hence Lu ∈ Str(A). Reading this in Aop

also yields Ru ∈ Str(A). Invoking (??.1) and Cor. 15.5, we now compute

LuLvLu = (u,u−2,Lu)(v,v−2,Lv)Lu =(uu−2uvu,u−2uv−2uu−2,LuLv

)Lu

= (vu,u−1v−2u−1,LuLv)(u,u−2Lu) = (w,z,LuLvLu) = (w,z,Luvu),

where

w = vuu−1v−2u−1uvuvu = uvu,

z = u−1v−2u−1uvu−2vuu−1v−2u−1 = u−1v−1u−2v−1u−1 = (uvu)−2.

Thus LuLvLu = (uvu,(uvu)−2,Luvu) = Luvu. Reading this in Aop, we also obtain RuRvRu = Ruvu.Since L1A = 1Str(A) = R1A , the relations (Lu)

−1 = Lu−1 , (Ru)−1 = Ru−1 are now clear. Next we

compute

LuRv = (u,u−2,Lu)(v−2,v,Rv) = (uu−2uv−2u,u−2uvuu−2,LuRv) = (v−2u,u−1vu−1,LuRv),

as claimed. A similar computation or passing to Aop yields the analogous equation for RvLu. Therest is clear.

72. See [99, 3.2, 3.3]sol.UNITSTRGRALT

73. (a) The extended right multiplication by pq as defined in Exc. 71 yields an isomorphismsol.UNITARBISOTALT

Rpq : A ∼−→ A((pq)−2,pq),

which by functoriality 16.4 and Prop. 16.3 may also be regarded as an isomorphism

Rpq : A(p,q) ∼−→ (A((pq)−2,pq))(p2q,qpq) = A(p′,q′)

with

p′ = (pq)−2(pq)p2q(pq)−2 = q−1 p−1 pq−1 p−1 = q−2 p−1,

q′ = (pq)(qpq)(pq)−2(pq) = pq2.

Thus Rpq : A(p,q) → Apq2is an isomorphism, which may also be verified directly, using (14.7.1)

and the Moufang identities:

(Rpqx)(pq2)−1 · (pq2)Rpqy) =(x(pq)

)(q−2 p−1) ·

((pq)q

)(y(pq)

)=(x(pq)

)(q−2 p−1) · (pq)(qy)(pq)

=[([(

x(pq))(q−2 p−1)

](pq)

)(qy)

](pq)

=[(

x((pq)(q−2 p−1)(pq)

))(qy)

](pq)

=((xp)(qy)

)(pq) = Rpq

((xp)(qy)

).

(b) The argument uses (16.9.2) to conclude A(pq)r = (Apq)r = ((Ap)q)r = (Ap)qr = Ap(qr), whichby (16.9.3) yields (pq)r = up(qr) for some u ∈ Nuc(A)×. But the argument is faulty since withB := Ap we have ((Ap)q)r = (Bq)r = Bs, where s is the product of q and r not in A but in B = Ap:


s = (qp−1)(pr), forcing Bs = (Ap)(qp−1)(pr) = At , t = p[(qp−1)(pr)] = [p(qp−1)p]r = (pq)r, incomplete agreement with the previous computation. Moreover, the conclusion does not hold, whichfollows from looking at the real octonions . . . .

74. The solution to Exc. 73 shows that right multiplication by pq yields an isomorphism fromsol.NUCISOTALTA(p,q) to an appropriate unital isotope of A. It therefore suffices to show Nuc(Ap) = Nuc(A) andCent(Ap) = Cent(A) for all p ∈ A×; actually, since A = (Ap)p−1

, it will be enough to verify theinclusions Nuc(A) ⊆ Nuc(Ap), Cent(A) ⊆ Cent(Ap). Writing [−,−,−]p (resp. [−,−]p) for theassociator (resp. the commutator) of Ap, this will follow once we have shown

[x,y,z]p = [x,yp−1, pz]− [x, p−1,(py)z], [x,y]p = [x,y]+ [x, p−1, py]− [y, p−1, px] (104) ASSP

for all x,y ∈ A. In order to derive the first relation of (104), we compute

[x,y,z]p =((

(xp−1)(py))

p−1)(pz)− (xp−1)

(p((yp−1)(pz)

))=(x(yp−1)(pz)− (xp−1)

((py)z

)= [x,yp−1, pz]+ x

((yp−1)(pz)

)− (xp−1)

((py)z

),

where the second summand on the right agrees with

x[

p−1(

p((yp−1)(pz)

))]= x(

p−1((py)z))

=−[x, p−1,(py)z]+ (xp−1)((py)z

).

Inserting this into the previous equation yields the first relation of (104). For the second, we obtain[x,y]p = (xp−1)(py)− (yp−1)(px) = [x, p−1, py] + xy− [y, p−1, px]− yx = [x,y] + [x, p−1, py]−[y, p−1, px], as claimed.

Section 17.75. In order to get the terminology straight, let us denote by paltinvε

k the category of ε-pointedsol.CAPIalternative k-algebras with involution as defined in Exc. 75. By contrast, let us denote by altistε

kthe category of alternative k-algebras with isotopy involution as defined in 17.4.

We begin by defining a functor Φ : paltinvεk → altistε

k . Let ((B,τ),q) be an object of paltinvεk .

Then Cor. 17.8 shows thatΦ

(((B,τ),q

)):= (Bq,τq,q)

is an object of altistεk . If ϕ : ((B,τ),q)→ ((B′,τ ′),q′) is a morphism in paltinvε

k , then it is straight-forward to check that

ϕ : (Bq,τq,q)−→ (B′q′,τ ′q

′,q′)

is a morphism in altistεk . Hence we obtain a functor Φ of the desired kind.

Next we define a functor Ψ : altistεk → paltinvε

k in the opposite direction. Let (B,τ,q) bean object of altistε

k , i.e., an alternative k-algebra with isotopy involution. Then Prop. 17.7 andLemma 17.1 show with p := q−1 that

Ψ((B,τ,q)

):=((Bp,τ p),q

)is an object of paltinvε

k . If ψ : (B,τ,q)→ (B′,τ ′,q′) is a morphism in altistεk and p := q−1, p′ :=

q′−1, then ψ(p) = p′, and it is straightforward to check that

ψ :((Bp,τ p), p

)−→

((B′p

′,τ ′p

′),q′)

is a morphism in paltinvεk . Hence we obtain a functor Ψ of the desired kind.

Now we simply apply (17.7.5) to conclude that Ψ Φ is the identity on paltinvεk and Φ Ψ is

the identity on altistεk . Summing up


Φ : paltinvεk∼−→ altistε

k

is an isomorphism of categories with inverse Ψ .

76. The map εA has clearly period two and fixes p. For the first part of the problem, it thereforesol.SWISINVsuffices to show (17.2.1) for εA in place of τ . To this end, let x,x′,y,y′ ∈ A. Then(

εA(y⊕ y′)p−1)(pεA(x⊕ x′))=((y′⊕ y)(q⊕q)

)((q−1⊕q−1)(x′⊕ x)

)=((qy′)⊕ (yq−1)(qq)

)((x′q−1)⊕ (q−1q−1)(qx)

)=((qy′)⊕ (yq)

)((x′q−1⊕ (q−1x)

)= (x′q−1)(qy′)⊕

((yq)q−1)(q(q−1x)

)= (x′q−1)(qy′)⊕ yx

= εA((yx)⊕ (x′q−1)(qy′)

)= εA

((x⊕ x′)(y⊕ y′)

),

as desired. Turning to the second part of the problem, we first note

Sym(Aop⊕Aq,εA, p) = (x,x) | x ∈ A.

On the other hand, given x,x′ ∈ A, we have(εA(x⊕ x′)p

)(x⊕ x′) =

((x′⊕ x)(q−1⊕q−1)

)(x⊕ x′) =

((q−1x′)⊕ (xq−1)(qq−1)

)(x⊕ x′)

=((q−1x′)⊕ (xq−1)

)(x⊕ x′) = x(q−1x′)⊕

((xq−1)q−1)(qx′)

= x(q−1x′)⊕ (xq−2)(qx′).

Combining, and using (16.9.2), we obtain the following chain of equivalent conditions:

∀x,x′ ∈ A :(εA(x⊕ x′)p

)(x⊕ x′) ∈ Sym(Aop⊕Aq,εA, p) ⇐⇒∀x,x′ ∈ A : (xq−2)(qx′) = x(q−1x′)

⇐⇒∀x,x′ ∈ A : (xq−2)(q2x′) = xx′

⇐⇒ A = Aq2 ⇐⇒ q2 ∈ Nuc(A).

This solves the problem.

Chapter 4

Section 19.

77. (a) We begin by deriving the following identities:sol.NILRADCON

nC(x2) = nC(x)2, (105) NOSQ

nC(x,x2) = tC(x)nC(x), (106) NORASQ

tC(x2) = tC(x)2−2nC(x). (107) TRSQ

Indeed, the identities of 19.5 yield nC(x2) = nC(tC(x)x−nC(x)1C) = tC(x)2nC(x)− tC(x)2nC(x)+nC(x)2, hence (105), while nC(x,x2) = nC(x, tC(x)x− nC(x)1C = 2tC(x)nC(x)− tC(x)nC(x) yields(106) and tC(x2) = tC(tC(x)x− nC(x)1C) = tC(x)2− 2nC(x) is (107). Combining these identitieswith those of (19.5) and writing y,z∈ k[x] as y=α01C+α1x, y= β01C+β1x with α0,α1,β0,β1 ∈ k,we now compute


nC(yz)−nC(y)nC(z) = nC((α01C +α1x)(β01C +β1x)

)−nC(α01C +α1x)nC(β01C +β1x)

= nC(α0β01C +(α0β1 +α1β0)x+α1β1x2)

−(α

20 +α0α1tC(x)+α

21 nC(x)

)(β

20 +β0β1tC(x)+β

21 nC(x)

)= α

20 β

20 +α0β0(α0β1 +α1β0)tC(x)+α0β0α1β1tC(x2)

+(α0β1 +α1β0)2nC(x)+(α0β1 +α1β0)α1β1nC(x,x2)+α

21 β

21 nC(x2)

−(α

20 +α0α1tC(x)+α

21 nC(x)

)(β

20 +β0β1tC(x)+β

21 nC(x)

)= α

20 β

20 +α0β0(α0β1 +α1β0)tC(x)+α0α1β0β1tC(x)2−2α0α1β0β1nC(x)

+(α0β1 +α1β0)2nC(x)+(α0β1 +α1β0)α1β1tC(x)nC(x)+α

21 β

21 nC(x)2

−(α

20 +α0α1tC(x)+α

21 nC(x)

)(β

20 +β0β1tC(x)+β

21 nC(x)

)= α

20 β

20 +α

20 β0β1tC(x)+α0α1β

20 tC(x)+α0α1β0β1tC(x)2 +α

20 β

21 nC(x)

+α21 β

20 nC(x)+α0α1β

21 tC(x)nC(x)+α

21 β0β1tC(x)nC(x+α

21 β

21 nC(x)2

−(α

20 +α0α1tC(x)+α

21 nC(x)

)(β

20 +β0β1tC(x)+β

21 nC(x)

)= 0.

This completes the proof of (a).(b) If x is invertible in k[x] with inverse x−1 ∈ k[x], then (a) yields

nC(x)nC(x−1) = nC(xx−1) = nC(1C) = 1,

hence nC(x)∈ k×. Conversely, if nC(x)∈ k×, then y := nC(x)−1x by (19.5.6) satisfies xy = 1C = yx.Thus x is invertible in k[x] with inverse y.

(c) If tC(x),nC(x) ∈ Nil(k), the elements tC(x)x,nC(x)1C ∈ k[x] are nilpotent and commute.But then x2 = tC(x)x−nC(x)1C is nilpotent. Hence so is x. Conversely, assume x is nilpotent. Thenxn = 0 for some positive integer n, and we show by induction on n that tC(x) and nC(x) are nilpotent.If n = 1, there is nothing to prove. If n > 1, then we put m := 1 for n = 2 and m := [ n

2 ] + 1 > n2

for n > 2. This implies 2m ≥ n, hence (x2)m = 0, and m < n. Thanks to the induction hypothesis,therefore, the scalars tC(x2) = tC(x)2−2nC(x) (by (107)) and nC(x2) = nC(x)2 (by (105)) are bothnilpotent. Hence nC(x) is nilpotent and so is tC(x)2 = tC(x2)+2nC(x). But then tC(x) itself must benilpotent, which completes the induction.

78. n := nC′ ϕ : C→ k is a quadratic form such that Dn = (DnC′ ) (ϕ×ϕ). Since ϕ preservessol.CONALGHOMNORunits, we conclude n(1C) = 1 and t := (Dn)(1C,−) = tC′ ϕ . Now let x ∈C. Then

ϕ(t(x)x−n(x)1C

)= tC′

(ϕ(x)

)ϕ(x)−nC′

(ϕ(x)

)1C′ = ϕ(x)2 = ϕ(x2),

and since ϕ is injective, x2 = t(x)x− n(x)1C . Thus not only nC but also n is a norm for the conicalgebra C. But by the hypotheses on C, its norm is unique (Prop. 19.15), which implies nC = n =nC′ ϕ , as desired.

If we drop the assumption that ϕ be injective, the asserted conclusion is false: let C := k⊕k, C′ := k and ϕ : k⊕ k→ k be the projection onto the first summand. Then 19.2 (a),(b) yieldsnk(ϕ(α⊕β )) = nk(α) = α2 for all α,β ∈ k, while nk⊕k(α⊕β ) = αβ . Thus ϕ does not preservenorms, hence, though being a unital algebra homomorphism, is not one of conic algebras.

79. (a) One verifies immediately that C is unital with identity element e. Defining nC : C→ k bysol.CONMO

nC(αe+ x) := α2 +αT (x)+B(x,x)

for α ∈ k, x ∈Mλ , we obviously obtain a quadratic form with bilinearization

nC(αe+ x,βe+ y) = 2αβ +αT (y)+βT (x)+B(x,y)+B(y,x)


for α,β ∈ k, x,y ∈Mλ . We have nC(e) = 1, while tC := (DnC)(−,e) satisfies

tC(αe+ x) = 2α +T (x)

for α ∈ k, x ∈Mλ . Hence

tC(αe+ x)(αe+ x)−nC(αe+ x)e =(2α +T (x)

)(αe+ x)−

(α

2 +αT (x)+B(x,x))e

=(α

2−B(x,x))e+((

α +T (x))x+αx

)= (αe+ x)2,

which shows that C is a conic k-algebra with norm nC and trace tC .(b) We first show that (T,B,K) is a conic co-ordinate system. The defining conditions of such

a system are obviously fulfilled, with the following two exceptions. (i) the bilinear map K takesvalues in Mλ , and (ii) K is alternating. In order to establish (i), we let x,y ∈ Mλ = Ker(λ ) andobtain

λ (x× y) = tC(x)λ (y)−λ (xy)+λ (xy)λ (e) =−λ (xy)+λ (xy) = 0,

hence x× y ∈Mλ . In order to establish (ii), we compute

x× x = tC(x)x− x2 +λ (x2)e = nC(x)e+λ(tC(x)x−nC(x)e

)e = nC(x)e−nC(x)e = 0,

so K is indeed alternating. We now denote by “·” the product in the conic algebra C′ :=Con(T,B,K)and obtain, for α,β ∈ k, x,y ∈Mλ ,

(αe+ x) · (βe+ y) =(αβ +λ (xy)

)e+((

α + tC(x))y+βx− tC(x)y+ xy−λ (xy)e

)= αβe+αy+βx+ xy = (αe+ x)(βe+ y),

nC′ (αe+ x) = α2 +αtC(x)−λ (x2) = α

2 +αtC(x)− tC(x)λ (x)+nC(x)λ (e)

= α2 +αtC(x)+nC(x) = nC(αe+ x).

Summing up, we have proved C = Con(TC,BC,KC). Finally, let (T,B,K) be a conic co-ordinatesystem and C := Con(T,B,K). Consulting the multiplication formula for C (resp. the formula forthe trace of C), we conclude

TC(x) = tC(x) = T (x),

BC(x,y) = −λ (xy) = B(x,y)

KC(x,y) = tC(x)y− xy+λ (xy)e = T (x)y− xy+λ (xy)e

=(B(x,y+λ (xy)

)e+(T (x)y−T (x)y+K(x,y)

)= K(x,y).

Thus (TC,BC,KC) = (T,B,K) and we have proved that the two constructions in (a), (b) are inverseto each other.

80. By assumption we have 1C ∧ x∧ x2 = 0 for all x ∈C. Replacing x by x+αy, x,y ∈C, α ∈ k,sol.CONFIELDwe obtain (since k contains more than two elements)

1C ∧ y∧ x2 +1C ∧ x∧ (x y) = 0 (x,y ∈C). (108) LIQA

Now put

U := 0∪u ∈C \ k1C | u2 ∈ k1C (109) TRAZE

We claim that U is a vector subspace of C. Clearly being closed under scalar multiplication, wemust show u+v ∈U for all u,v ∈U such that u,v are linearly independent. Then 1C,u are linearlyindependent, and assuming v = α1C +βu for some α,β ∈ k, we obtain α21C + 2αβu+β 2u2 =


v2 ∈ k1C , hence α = 0 or β = 0, a contradiction. Thus 1C,u,v are linearly independent. On theother hand, setting x = u, y = v (resp. x = v, y = u) in (108), and observing (109), we obtainu v = α1C + βu = γ1C + δv for some α,β ,γ,δ ∈ k, hence u v ∈ k1C . But this amounts to(u+ v)2 ∈ k1C , so we have proved that U is closed under addition and thus, altogether, a vectorsubspace of C.

Now let x ∈C \ k1C . Then x2 = α1C +βx for some α,β ∈ k, which implies (x− β

2 1C)2 ∈ k1C ,

hence x− β

2 1C ∈U , and we have shown C = k1C ⊕U as a direct sum of subspaces. Now definea quadratic form nC : C→ k by nC(α1C +u) := α2 +nC(u) for α ∈ k, u ∈U , where nC(u) ∈ k isdetermined by u2 =−nC(u)1C . This implies

nC(α1C +u,β1C + v) = 2αβ +nC(u,v), u v =−nC(u,v)1C (α,β ∈ k, u,v ∈U),

hence tC(α1C +u) := nC(1C,α1C +u) = 2α . Summing up, we obtain

(α1C +u)2 = α21C +2αu+u2 = 2α(α1C +u)−

(α

2 +nC(u))1C

= tC(α1C +u)(α1C +u)−nC(α1C +u)1C.

Thus C is a conic algebra over k with norm nC .

81 . Since conic algebras are stable under base change, they clearly satisfy the Dickson condi-sol.DICONtions. Conversely, suppose C satisfies the Dickson condition. By hypothesis, the vector 1C ∈C isunimodular, so we find a submodule M ⊆C such that

C = k1C⊕M (110) UNIM

as a direct sum of k-modules. Since the assignment x 7→ x2 defines a quadratic map M→ C, theprojections to the direct summands k1C ∼= k and M of the decomposition (110) give rise to quadraticmaps n : M→ k and s : M→M such that

x2 = s(x)−n(x)1C (x ∈M). (111) SQUADEC

In particular, n is a quadratic form over k that will eventually become the norm (restricted to M)of the prospective conic k-algebra C. On the other hand, by the Dickson condition, we also haves(x) ∈ kx for x ∈M, and we would like to think of s(x) as t(x)x with t(x) ∈ k becoming the traceof x in the prospective conic k-algebra C. But since the annihilator of x may not be zero, we don’teven know at this stage how to make t(x) a quantity that is well defined, let alone a linear form. Forthis reason, we bring the hypotheses on the module structure of C. and the strict Dickson conditioninto play.

Let us first reduce to the case that M is a free k-module. Otherwise, M is finitely generatedprojective by hypothesis, so there exists a finite family of elements f ∈ k that generate the unitideal in k and make M f a free k f -module of finite rank, for each such f . Assuming the free casehas been settled, it follows that all C f are conic, and Prop. 19.15 ensures that the norms nC f glueto give a quadratic form n : C→ k. It is then clear that C is a conic algebra with norm n since thisis so after changing scalars from k to each k f .

For the rest of the proof, we may therefore assume that M is a free k-module, possibly of infiniterank, with basis (ei)i∈I . We let T = (ti)i∈I be a family of independent variables and write k[T] forthe corresponding polynomial ring. For a non-empty finite subset E ⊆ I, we consider the submodule

ME := ∑i∈E

kei ⊆M, (112) EME

which is a direct summand, and the element

xE := ∑i∈E

ei⊗ ti ∈MEk[T] ⊆Mk[T]. (113) EXE


We claim the annihilator of xE in k[T] is zero; indeed, for f (T) ∈ k[T] the relation f (T)xE = 0implies ∑ei ⊗ (ti f (T)) = 0, hence ti f (T) = 0 for all i ∈ E and then f (T) = 0 since ti is not azero divisor in k[T]. Invoking the strict Dickson condition, we therefore find a unique polynomialg(T) ∈ k[T] such that

s(xE) = g(T)xE .

Specializing this relation with an additional variable u to uT, we obtain g(uT)uxE = s(uxE) =u2s(xE) = (ug(T))uxE , so the polynomial g(T) ∈ k[T] is homogeneous of degree 1. Thus thereexists a unique linear form tE : ME → k satisfying the relation s(x) = tE(x)x for all x∈ME becauseany such tE will be converted into the linear homogeneous polynomial g(T) after extending scalarsfrom k to k[T]. Here the uniqueness condition implies that the linear forms tE as E varies over thenon-empty finite subsets of I glue to give a linear form t : M → k such that s(x) = t(x)x for allx ∈M. Combining with (111), we obtain the relation

x2− t(x)x+n(x)1C = 0 (114) SQUAM

for all x ∈M. We now extend t,n as given on M to all of C by

t(α1C + x) = 2α + t(x), n(α1C + x) = α2 +αt(x)+n(x) (α ∈ k, x ∈M)

and then conclude from a straightforward computation that (114) holds for all x ∈C. Since we alsohave t = n(1C,−), it follows that C is a conic algebra.

82. For the first part, assume k 6= 0 is connected and let c ∈C. If nC(c) = 0 and tC(c) = 1, then csol.IDEMPis an idempotent by (19.5.1) which cannot be zero since tC(c) = 1, and cannot be 1 since nC(c) = 0.Conversely, let c ∈C be an idempotent 6= 0,1C . From Exc. 77 (a) we deduce that nC(c) ∈ k is anidempotent. If nC(c) = 1, then c is invertible in k[c] (Exc. 77 (b)), whence the relation c(1C−c) = 0implies c = 1C , a contradiction. Hence nC(c) = 0, and (19.5.1) reduces to c = c2 = tC(c)c. Takingtraces, we conclude that tC(c) ∈C is an idempotent, which cannot be zero since this would implyc = 0. Thus tC(c) = 1. We now proceed to establish the equivalence of (i)–(iv).

(i)⇒ (ii). This is clear.(ii)⇒ (iii). For any prime ideal p ⊆ k, the ring kp is local, hence connected, and the first part

of the problem yields nC(c)p = nCp (cp) = 0, tC(c)p = tCp (cp) = 1p, and since this holds true forall p ∈ Spec(k), condition (iii) follows.

(iii)⇒ (iv). c is clearly an idempotent. Moreover, the second condition of (iii) shows that c isunimodular, and since tC(1C− c) = 2−1 = 1, so is 1C− c.

(iv)⇒ (i). Unimodularity is stable under base change, so for R ∈ k-alg, R 6= 0, the elementscR and 1CR − cR are both different from zero. Hence (i) holds.

83. We need an auxiliary result.sol.LILIDLemma. Every finitely generated nil ideal I ⊆ k is nilpotent, i.e., there exists a positive integer nsuch that In = 0.

Proof. Assume α1, . . . ,αm ∈ I generate I as an ideal and let p ∈ Z, p > 0, satisfy αpi = 0 for all

i = 1, . . . ,m. For any positive integer n, the ideal In ⊆ k is generated by the expressions αi1 · · ·αin ,where 1≤ i1 ≤ ·· · ≤ in ≤ m. If for all i = 1, . . . ,m,

qi := | j ∈ Z|1≤ j ≤ n, i j = i| ≤ p−1,

then n = ∑mi=1 qi ≤m(p−1). Thus for n > m(p−1), some i = 1, . . . ,m has qi ≥ p. But this implies

first αi1 · · ·αin = 0 and then In = 0.

In the situation of our exercise, an element x ∈ IC actually belongs to I′C for some ideal I′ ⊆ Cwhich is finitely generated and belongs to I. Hence xn ∈ I′nC for all positive integers n, where bythe lemma I′n will eventually be zero. Thus x is nilpotent, and we have shown that IC ⊆C is a nilideal. By §8, Exercise 31, therefore, we find an idempotent c ∈C satisfying c = c′. We claim thatc is elementary, i.e., by §19, Exercise 82, that nC(c) = 0 and tC(c) = 1. Passing to the base change


from k to k, we conclude

nC(c) = nC(c′) = 0, tC(c) = tC(c

′) = 1k,

hence nC(c) ∈ I, tC(c) ∈ 1+ I ⊆ k× since I ⊆ k is a nil ideal. Thus, thanks to §19, Exercise 77 (a),nC(c) ∈ k is a nilpotent idempotent, forcing nC(c) = 0, and applying the trace to (19.5.1) for x = c,we see that tC(c)∈ k is an invertible idempotent, forcing tC(c) = 1. Summing up, c is an elementaryidempotent of C.

84. We begin with a short digression: let ε ∈ k be an idempotent. Then R := kε becomes a membersol.DECIDof k-alg with identity elements 1R = ε and k-algebra structure given by the homomorphism α 7→αε

from k to R. For a k-module M, scalar multiplication k×M → M restricts to a bi-additive mapR×εM→ εM making εM an R-module. We claim that there is a natural identification MR = εM asR-modules such that xR = x⊗ε = εx for all x∈M. In order to see this, we note that the map x 7→ εxfrom M to εM is k-linear, hence gives rise to an R-linear map MR→ εM satisfying xR = x⊗ε 7→ εx.This map is clearly surjective, and it suffices to show that it is injective as well. Accordingly, letx∈M and assume εx= 0. Then, setting ε ′ := 1−ε , we have xR = x⊗ε = (ε ′x)⊗ε = x⊗(εε ′) = 0,as desired.

Returning now to the exercise itself, we prove(ii)⇐ (i). Obvious.(i) ⇒ (ii). If c ∈ C is an idempotent, then (19.5.1) implies tC(c) = tC(c2) = tC(c)2− 2nC(c),

hence

tC(c)(1− tC(c)

)=−2nC(c). (115) IDNOTR

The quantities ε(i), i = 0,1,2, as defined in (10) obviously satisfy ∑ε(i) = 1, while the fact thatnC(c) ∈ C is an idempotent by Exc. 77 (a) combined with (115) implies ε(i)ε( j) = 0 for i, j =0,1,2 distinct. Thus they form a complete orthogonal system of idempotents in k. Our preliminaryremark shows C(i) =Ck(i) and c = ∑c(i), where c(i) = ck(i) = ε(i)c for i = 0,1,2. We now computec(0) = ε(0)c=(1−nC(c))(1−tC(c))c, where (1−tC(c))c= c2−tC(c)c=−nC(c)1C , which impliesc(0) = 0. Next we turn to

c(1) = ε(1)c =

(1−nC(c)

)tC(c)c =

(1−nC(c)

)(c+nC(c)1C

)=(1−nC(c)

)c,

which together with (115) implies

tC(c(1)) = ε(1)tC(c) =

(1−nC(c)

)tC(c)2 =

(1−nC(c)

)(tC(c)+2nC(c)

)=(1−nC(c)

)tC(c) = ε

(1) = 1k(1) ,

nC(c(1)) = ε(1)nC(c) =

(1−nC(c)

)tC(c)nC(c) = 0.

Thus c(1) ∈C(1) is an elementary idempotent. Finally,

c(2) = ε(2)c = nC(c)c,

which implies nC(c(2)) = ε(2)nC(c) = ε(2) = 1C(2) . Thus c(2) ∈ C(2) is an invertible idempotentin the sense of Exc. 77 (b), forcing c(2) = 1C(2) , as claimed. The statement just proved impliestC(c(2)) = 2 in k(2). This can also be proved directly by noting(

2− tC(c))ε(2) = 2nC(c)− tC(c)nC(c) = nC(c,c2)− tC(c)nC(c)

which is zero by (106) of the solution to Exc. 77.It remains to prove uniqueness of the ε(i), so let η(i), i = 0,1,2, be any complete orthogonal

system of idempotents in C satisfying mutatis mutandis the conditions of (ii), and define the ε(i),i = 0,1,2, as in (10). Then nC(c(i)) = nC(η

(i)c) = η(i)nC(c) = nC(c)k(i) = nC(i) (c(i)) and, similarly,


tC(c(i)) = tC(i) (c(i)) for i = 0,1,2. Since c(1) ∈C(1) is an elementary idempotent and c(2) = 1C(2) byhypothesis, this implies

nC(c) = nC(c(1))+nC(c(2)) = nC(1) (c(1)))+nC(2) (c(2)) = η(2),

tC(c) = tC(c(1))+ tC(c(2)) = tC(1) (c(1))+ tC(2) (c(2)) = η(1)+2η

(2),

hence η(2) = nC(c)= ε(2) and η(1) = tC(c)−2nC(c). But the idempotents η(1),η(2) are orthogonal,which implies tC(c)nC(c) = 2nC(c), and we conclude η(1) = tC(c)(1−nC(c)) = ε(1) by (10). Sincethe orthogonal systems formed by the η’s and by the ε’s are both complete, we also have η(0) =ε(0), as claimed.

85. If C is a conic k-algebra with trivial conjugation, then 2x = tC(x)1C for all x ∈C by (19.5.4).sol.TRICOSince the trace form of C is surjective, we find an element u ∈C satisfying tC(u) = 1. This implies2u = 1C , hence x = 2ux = tC(ux)1C . But C is a faithful k-module. Hence C ∼= k as conic algebras.The converse is obvious.

Section 19.

86. For the subsequent computations it is important to notesol.NONUNNOR

(α.β0)ε = (αβ0)ε (α ∈ k, β0 ∈ k0) (116) DRODOT

since, for α =α0+α ′0ε , α0,α′0 ∈ k0, the right-hand side becomes (α0β0+(α ′0β0)ε)ε = (α0β0)ε =

(α.β0)ε , as claimed.Turning to the exercise itself, we begin by showing associativity. Writing (ei)1≤i≤3 for the

canonical basis of k30 over k0, we deduce from (1) that [k3

0,k30] = k0e3 and [e3,k3

0] = 0, hence[[k3

0,k30],k

30] = 0= [k3

0, [k30,k

30]]. Now let α,β ,γ ∈ k and u,v,w ∈ k3

0 . Then((α⊕u)(β ⊕ v)

)(γ⊕w) =

((αβ )⊕ (α.v+β .u+[u,v])

)(γ⊕w)

= (αβγ)⊕((αβ ).w+(γα).v+(βγ).u+ γ.[u,v]

+α.[v,w]+β .[u,w]+ [[u,v],w]),

(α⊕u)((β ⊕ v)(γ⊕w)

)= (α⊕u)

((βγ)⊕ (β .w+ γ.v+[v,w])

)= (αβγ)⊕

((αβ ).w+(γα).v+α.[v,w]

+ (βγ).u+β .[u,w]+ γ.[u,v]+ [u, [v,w]]),

and the preceding observation shows [[u,v],w] = [u, [v,w]] = 0. Hence C is associative.The algebra C is obviously unital with identity element 1C = 1⊕0, and the quadratic form nz

obviously satisfies nz(1C) = 1. Moreover, (116) yields

nz(α⊕u) = α2 +(αzt u)ε (α ∈ k, u ∈ k3

0),

and we have

nz(α⊕u,β ⊕ v) = 2αβ +(αzt v+β zt u)ε (α,β ∈ k, u,v,∈ k30),

hence

tz(α⊕u) := nz(1C,α⊕u) = 2α +(zt u)ε.

Comparing(α⊕u)2 = α

2 +2(α.u)

with


tz(α⊕u)(α⊕u)−nz(α⊕u)1C =(2α +(zt u)ε

)(α⊕u)−

(α

2 +(αzt u)ε)(1⊕0)

=(2α

2 +(αzt u)ε−α2− (αzt u)ε

)⊕((

2α +(zt u)ε).u)

= α2⊕2(α.u),

we see that C is indeed a conic k-algebra with norm nz and trace tz. We now turn to the equivalenceof (i)–(iv). By Props. 20.2 and 19.13, we have (i)⇒ (ii)⇒ (iii). Thus it remains to show (iii)⇒(iv)⇒ (i).

(iii)⇒ (iv). Since the conjugation of C is an involution and Ann(C) = 0, we conclude fromProp. 19.9 that tz(xy) = nz(x, y) for all x,y ∈C. Writing x = α ⊕ u, y = β ⊕ v, α,β ∈ k, u,v ∈ k3

0 ,we apply (116) and obtain

tz(xy) = tz((αβ )⊕ (α.v+β .u+[u,v])

)= 2αβ +

(zt(α.v+β .u+[u,v])

)ε

= 2αβ +(α.(zt v)+β .(zt u)+ zt [u,v]

)ε = 2αβ +(αzt v+β zt u+ zt [u,v])ε,

y = tz(β ⊕ v)1C− (β ⊕ v) =(2β +(zt v)ε

)(1⊕0)− (β ⊕ v) =

(β +(zt v)ε

)⊕ (−v),

nz(x, y) = nz

(α⊕u,

(β +(zt v)ε

)⊕ (−v)

)= 2αβ +2αzt vε +

(−αzt v+β zt u+(zt v)εu

)ε

= 2αβ +(αzt v+β zt u)ε.

Comparing, we conclude zt [u,v] = 0 for all u,v ∈ k30 , and specializing u = e1, v = e2, (ei)1≤i≤3

being the k0-basis of unit vectors in k30 , yields δ3 = 0, hence (iv).

(iv)⇒ (i). For α,β ∈ k, u,v ∈ k30 , we compute

nz((α⊕u)(β ⊕ v)

)= nz

((αβ )⊕ (α.v+β .u+[u,v])

)= (αβ )2 +

((α2

β )zt v+(αβ2)zt u+(αβ )zt [u,v]

)ε,

nz(α⊕u)nz(β ⊕ v) =(α

2 +(αzt u)ε)(

β2 +(β zt v)ε

)= α

2β

2 +(α2β zt v+αβ

2zt u)ε.

By definition of the algebra A, the e1- and e2-components of [u,v] are both zero, as is the e3-component of z, by (iv). Hence zt [u,v] = 0, and we conclude that nz permits composition, i.e., theconic algebra C with norm nz is multiplicative.

87. By Prop. 20.2 (a), C is norm-associative. We putsol.NIRA

N := x ∈C | nC(x),nC(x,y) ∈ Nil(k) for all y ∈C

It is straightforward to check that N ⊆ C is a k-submodule. We claim that it is, in fact, an ideal.In order to see this, let x ∈ N and y,z ∈ C. By multiplicativity, nC(xy) = nC(x)nC(y) is nilpotent,while norm-associativity yields the same for nC(xz,y) = nC(x,yz) and nC(zx,y) = nC(x, zy). Thusxz and zx both belong to N, forcing N to be an ideal in C. Since nC(x) and tC(x) = nC(x,1C) arenilpotent, so is x by Exc. 77 (c). Hence N ⊆C is a nil ideal and as such contained in the nil radical.Conversely, let x∈Nil(C). Then x is nilpotent, as is xy∈Nil(C) for all y∈C. Consulting Exc. 77 (c)again, we see that nC(x) and nC(x,y) = tC(xy) belong to Nil(k), which implies x∈N and completesthe proof.

88. For the first part we must show that the submodule M of C spanned by the elements 1C,x,y,xysol.ARTCONALGis, in fact, a subalgebra, i.e., it is closed under multiplication. In order to see this, we apply thealternative laws, the identities of 19.5, and (20.3.2) to obtain


z2 = tC(z)z−nC(z)1C ∈M (z ∈ x,y,xy),

x(xy) = x2y = tC(x)xy−nC(x)y ∈M,

(xy)x = nC(x, y)x−nC(x)y = nC(x, y)x−nC(x)tC(y)1C +nC(x)y,

yx = x y− xy = tC(x)y+ tC(y)x−nC(x,y)1C− xy ∈M.

By symmetry, therefore, M is closed under multiplication, giving the first part of the problem. As tothe second, it suffices to verify the associative law on the generators of M, which is straightforward.

89. Abbreviating 1 := 1C , n := nC , t := tC and expanding the norm of an associator, we obtainsol.NORMA

n([x1,x2,x3]

)= n((x1x2)x3

)−n((x1x2)x3,x1(x2x3)

)+n(x1(x2x3)

),

where multiplicativity of the norm yields

n([x1,x2,x3]

)= 2n(x1)n(x2)n(x3)−n

((x1x2)x3,x1(x2x3)

). (117) NASL

Turning to the second summand on the right of (117), we obtain, by (20.1.4),

n((x1x2)x3,x1(x2x3)

)= n(x1x2,x1)n(x3,x2x3)−n

((x1x2)(x2x3),x1x3

).

Since C is norm associative by Prop. 20.2, we may apply apply (19.11.2) to the first summand onthe right, which yields

n((x1x2)x3,x1(x2x3)

)= t(x2)

2n(x3)n(x1)−n((x1x2)(x2x3),x1x3

). (118) NAST

Manipulating the expression (x1x2)(x2x3) by means of (19.5.5) and the middle Moufang identity(14.3.3), we obtain

(x1x2)(x2x3) = (x1x2) (x2x3)− (x2x3)(x1x2)

= t(x1x2)x2x3 + t(x2x3)x1x2−n(x1x2,x2x3)1− x2(x3x1)x2,

where associativity of the trace (19.12.1) and (19.11.5),(20.3.2) yield

(x1x2)(x2x3) = t(x1x2)x2x3 + t(x2x3)x1x2−n(x1x2,x2x3)1− (119) FOURP

t(x1x2x3)x2 +n(x2)x3x1.

Here we use (19.11.3),(19.5.4) to compute

n(x1x2,x2x3) = n(x1,x2x3x2) = t(x2)n(x1,x2x3)−n(x1,x2x3x2),

and (19.11.5),(20.3.2) give

n(x1x2,x2x3) = t(x2)n(x1,x2x3)− t(x2x3)n(x1,x2)+ t(x3x1)n(x2)

= t(x1)t(x2)t(x2x3)− t(x2)t(x1x2x3)− t(x1)t(x2)t(x2x3)+

t(x1x2)t(x2x3)+ t(x3x1)n(x2),

hence

n(x1x2,x2x3) = t(x1x2)t(x2x3)− t(x2)t(x1x2x3)+ t(x3x1)n(x2).

Inserting this into (119), and (119) into the second term on the right of (118), we conclude


n((x1x2)(x2x3),x1x3

)= t(x1x2)n(x2x3,x1x3)+ t(x2x3)n(x1x2,x1x3)−

t(x1x2)t(x2x3)t(x3x1)+ t(x2)t(x3x1)t(x1x2x3)−

t(x3x1)2n(x2)− t(x1x2x3)n(x2,x1x3)+

n((x3x1)

∗,x1x3)n(x2)

= t(x1x2)t(x1x∗2)n(x3)+ t(x2x3)t(x2x∗3)n(x1)−t(x1x2)t(x2x3)t(x3x1)+ t(x2)t(x3x1)t(x1x2x3)−

t(x3x1)2n(x2)− t(x2)t(x3x1)t(x1x2x3)+

t(x1x2x3)t(x2x1x3)+ t(x3x21x3)n(x2),

where we may use (19.12.1),(19.5.1),(19.11.5) to expand

t(x3x21x3)n(x2)− t(x3x1)

2n(x2) = t(x23x2

1)n(x2)− t(x3x1)2n(x2)

= t([t(x3)x3−n(x3)1][t(x1)x1−n(x1)1]

)n(x2)−

t(x3x1)2n(x2)

= t(x3)t(x1)t(x3x1)n(x2)− t(x1)2n(x2)n(x3)−

t(x3)2n(x1)n(x2)+2n(x1)n(x2)n(x3)−

t(x3x1)2n(x2)

= t(x3x1)t(x3x1)n(x2)− t(x1)2n(x2)n(x3)−

t(x3)2n(x1)n(x2)+2n(x1)n(x2)n(x3).

Inserting the resulting expression

n((x1x2)(x2x3),x1x3

)= ∑ t(xix j)t(xix∗j)n(xl)− t(x1x2)t(x2x3)t(x3x1)+

t(x1x2x3)t(x2x1x3)− t(x1)2n(x2)n(x3)−

t(x3)2n(x1)n(x2)+2n(x1)n(x2)n(x3)

into (118) and (118) into (117), the assertion follows.

90. From Prop. 16.3 combined with 16.7 we deduce that C(p,q) is a unital alternative k-algebra withsol.ISTQALTidentity element 1(p,q)

C = (pq)−1. Defining nC(p,q) := nC(pq)nC , Prop. 20.4 yields nC(p,q) (1(p,q)C ) = 1,

and we have nC(p,q) (x,y) = nC(pq)nC(x,y) for all x,y ∈C, which implies

tC(p,q) (x) := nC(p,q) (1(p,q)C ,x) = nC(pq)nC((pq)−1,x

)= nC(pq,x).

Writing x(m,p,q) for the m-th power of x ∈ C in C(p,q), we apply (16.2.1), the middle Moufangidentity (14.3.3) and (20.3.2) to conclude

x(2,p,q) = (xp)(qx) = x(pq)x = nC(x, pq)x−nC(x)pq = tC(p,q) (x)x−nC(p,q) (x)(pq)−1

= tC(p,q) (x)x−nC(p,q) (x)1(p,q)C .

Thus C(p,q) is a conic alternative k-algebra with norm and trace as indicated. Moreover,

ιC(p,q) (x) = tC(p,q) (x)1(p,q)C − x = nC(pq,x)(pq)−1− x = nC(pq)−1(nC(pq,x)pq−nC(pq)x

)= nC(pq)−1 pqx pq,

giving the desired formula for the conjugation of C(p,q). Finally, if C is multiplicative, then


nC(p,q) (x .p,q y) = nC(pq)nC((xp)(qy)

)= nC(pq)nC(x)nC(p)nC(q)nC(y)

= nC(pq)nC(x)nC(pq)nC(y) = nC(p,q) (x)nC(p,q) (y)

for all x,y ∈C, and C(p,q) is multiplicative as well.

91. By Prop. 20.2, the conjugation of C is an involution, and since q has trace zero, we havesol.CONISq = −q. Hence Lemma 17.3 shows that (C,τ,q) is an alternative algebra with isotopy involutionof negative type, i.e., an object of the category altistε

k with ε = −. By the solution to Exc. 75, thepointed alternative k-algebra with involution corresponding to (C,τ,q) is

Ψ((C,τ,q)

)=((Cp,τ p),q

),

where Prop. 20.4 yields p := q−1 = nC(q)−1q = −nC(q)−1q, hence Cp = Cq. Similarly, (17.7.2)implies τ p = τq and since τ(q) =−q, we deduce from (17.7.2) again that

τq(x) =−q−1

τ(qx) =−q−1(q−1qxq) = q−2xq2 = x

for all x ∈ C since q±2 ∈ k1C . Thus τ p = τq = ιC , and we conclude that the pointed alternativealgebra with involution corresponding to (C,τ,q) via Exc. 75 is ((Cq, ιC),q). But actually it is also((C, ιC),q) since we have an isomorphism

ϕ : ((C, ιC),q)−→ ((Cq, ι),q), x 7−→ q−1xq.

Indeed, ϕ obviously fixes q, and since q3 = q2q =−nC(q)q, we have Cq =Cq3, forcing ϕ : C→Cq

by Exc. 72 (b) to be an isomorphism such that

ϕ(x) = qxq−1 = qxq−1 = q−1xq = ϕ(x)

for all x∈C. Thus ϕ : (C, ιC)→ (Cq, ιC) is an isomorphism of alternative algebras with involution,as claimed.

Section 21.

92 . We put C := Cay(B,µ) and let u1,u2,v1,v2 ∈ B. Using the fact that B is norm associative bysol.CDMULProp. 20.2 (a) and applying (21.3.1), we obtain

nC((u1 + v1 j)(u2 + v2 j)

)= nC

((u1u2 +µ v2v1)+(v1u2 + v2u1)

)= nB(u1u2 +µ v2v1)−µnC(v1u2 + v2u1)

= nB(u1)nB(u2)+µnB(u1u2, v2v1)+µ2nB(v2)nB(v1)

−µnB(v1)nB(u2)−µnB(v1u2,v2u1)−µnB(v2)nB(u1)

=(nB(u1)−µnB(v1)

)(nB(u2)−µnB(v2)

)+µnB

(v2(u1u2),v1

)−µnB

(v1,(v2u1)u2)

)= nB(u1 + v1 j)nB(u2 + v2 j)−µnB

([v2,u1,u2],v1

),

from which the assertion can be read off immediately.

93. For u1,u2,v1,v2 ∈ B, we combine the hypothesis that a belongs to the nucleus of B with thesol.CDSCAPARfact that the conjugation of B is an involution (Prop. 20.2) (a) to compute


ϕ(u1 + v1 j)ϕ(u2 + v2 j) =(u1 +(av1) j

)(u2 +(av2) j

)= (u1u2 +µav2av1)+

((av2)u1 +(av1)u2

)j

=(u1u2 +µ v2(aa)v1

)+(a(v2u1 + v1u2)

)j

=(u1u2 +nB(a)µ v2v1

)+(a(v2u1 + v1u2)

)j

= ϕ

((u1u2 +nB(a)µ v2v1

)+(v2u1 + v1u2) j

)= ϕ

((u1 + v1 j)(u2 + v2 j)

).

Thus ϕ is an obviously unital algebra homomorphism. Moreover, setting C := Cay(B,nB(a)µ),C′ := Cay(B,µ), we have, for all u,v ∈ B,

nC′ ϕ(u+ v j) = nC′(u+(av) j

)= nB(u)−µnB(av) = nB(u)−nB(a)µnB(v)

= nC(u+ v j)

since B is multiplicative. Thus ϕ is a homomorphism of conic algebras. The final statement isobvious.

94. (iii) in means the same in C as in Cop, while (i) in C is equivalent to (ii) in Cop. Hence it sufficessol.ZERDIVto show (i)⇔ (iii).

(i) ⇒ (iii). Assume f := nC(x) is not a zero divisor in k. We must show that x is not a rightzero divisor in C. By 21.7, f is not a zero divisor in C, whence it follows from Bourbaki [11,§ 2, Prop. 4] that the natural map z 7→ z f = z/1 from C to C f is injective. On the other hand,nC f (x f ) = nC(x) f = f f ∈ k×f , forcing x f to be invertible in C f (Prop. 20.4). Now suppose y ∈ Csatisfies xy = 0. Then x f y f = (xy) f = 0 in C f , hence y f = 0, which by what we have just seenimplies y = 0. Thus x is not a right zero divisor of C.

(iii) ⇒ (i). Assume x is not a right zero divisor of C. We must show that nC(x) is not a zerodivisor in k, so let α ∈ k satisfy αnC(x) = 0. Then 0 = αxx = x(α x) implies α x = 0, hence αx =α x = 0. But this means x(α1C) = 0, hence α1C = 0, and since 1C ∈C is unimodular, we concludeα = 0, as desired.

95. We put C := Cay(Bop,µ), C′ := Cay′(B,µ) and define ϕ : C→C′op bysol.VARCD

ϕ(u+ v j) := u+ j′v (u,v ∈ B).

Then ϕ is a k-linear bijection, and indicating the products in Bop as well as in C′op by “·”, but theones in B,C and C′ by juxtaposition, we obtain, for all u1,u2,v1,v2 ∈ B,

ϕ(u1 + v1 j) ·ϕ(u2 + v2 j) = (u2 + j′v2)(u1 + j′v1) = (u2u1 +µv1v2)+ j′(u2v1 +u1v2)

= (u1 ·u2 +µ v2 · v1)+ j′(v1 · u2 + v2 ·u1)

= ϕ((u1 ·u2 +µ v2 · v1)+(v1 · u2 + v2 ·u1) j

)= ϕ

((u1 + v1 j)(u2 + v2 j)

).

Hence ϕ is an algebra isomorphism. Transporting the norm of C to C′ by means of ϕ , therefore,gives a quadratic form nC′ on C′ making it a conic algebra isomorphic to C under ϕ . Since thenorms of B and Bop coincide, we have nC = nB ⊥ (−µ)nB, which obviously implies

nC′ (u+ j′v) = nB(u)−µnB(v),

tC′ (u+ j′v) = tB(u),

u+ j′v = u− j′v

for all u,v ∈ B.Finally, let us assume that the conjugation of B is an involution. Then we claim that


ψ : Cay(B,µ)op ∼−→ Cay(Bop,µ), u+ v j 7−→ u+ v j,

is an isomorphism of conic algebras. Since ψ is obviously bijective and preserves norms, it will beenough to show that it is an algebra homomorphism. Indeed, extending the previous conventionsto Cay(B,µ)op, we obtain, for all u1,u2,v1,v2 ∈ B,

ψ(u1 + v1 j)ψ(u2 + v2 j) = (u1 + v1 j)(u2 + v2 j) = (u1 ·u2 +µv2 · v1)+(v2 ·u1 + v1 · u2) j

= (u2u1 +µ v1v2)+(u1v2 + u2v1) j

= (u2u1 +µ v1v2)+ v2u1 + v1u2 j

= ψ((u2u1 +µ v1v2)+(v2u1 + v1u2) j

)= ψ

((u2 + v2 j)(u1 + v1 j)

)= ψ

((u1 + v1 j) · (u2 + v2 j)

),

as claimed. Combining we obtain an isomorphism ϕ ψ : Cay(B,µ)op→ Cay′(B,µ)op, hence anisomorphism

ϕ ψ : Cay(B,µ) ∼−→ Cay′(B,µ), u+ v j 7−→ u+ j′v,

of conic algebras.

Section 22.

96. (i)⇒ (ii). Since (C,n,1C) by (22.1.1) is a pointed quadratic module, the implication followssol.COMPUNfrom Lemma 12.14.

(ii)⇒ (iii). Obvious.(iii)⇒ (i). By (22.1.2), (22.1.3), the quantity ε := n(1C)∈ k is an idempotent with εn(x) = n(x),

εn(x,y) = n(x,y) for all x,y∈C. Therefore the idempotent ε ′ := 1−ε ∈ k satisfies n(ε ′1C) = ε ′ε =0, n(ε ′1C,y) = ε ′εn(1C,y) = 0 for all y∈C, which implies ε ′1C = 0 by non-degeneracy of n, henceε ′ = 0 since 1C is faithful. Thus n(1C) = 1.

97 . Let K be a unital F-algebra of dimension 2. Then K is quadratic, hence commutative asso-sol.CONILciative (19.4), and we have K = F [u] for any u ∈ K \F1K . Writing µu ∈ F [t] for the minimumpolynomial of u over F , which is monic of degree 2, we have the following possibilities.10. µu is irreducible and separable. The K/F is a separable quadratic field extension, and we are inCase (i).20. µu is reducible and separable. Then µu = (t−α)(t−β ) for some α,β ∈ F , α 6= β . Here theChinese Remainder Theorem [65, II, Cor. 2.2] implies

K ∼= F [t]/(µu)∼= F [t]/(t−α)⊕F [t]/(t−β )∼= F⊕F

as a direct sum of ideals, and we are in Case (ii).30. µu is irreducible and inseparable. Then char(F) = 2 and K/F is an inseparable quadratic fieldextension. Hence we are in Case (iii).40. µu is reducible and inseparable. Then µu = (t−α)2 for some α ∈ F , which implies K = F [v],where v := u−α1K satisfies v2 = 0 6= v. Hence the map F [ε]→K sending ε to v is an isomorphismof F-algebras, and we are in Case (iv).For the second part of the problem, assume that F is perfect of characteristic 2 and let C be a conicF-algebra without nilpotent elements 6= 0. It will be enough to prove Ker(tC) = F1C , where theinclusion “⊇” is obvious. Conversely, let u ∈ Ker(tC) and assume u /∈ F1C . Then K := F [u] ⊆ Cis a two-dimensional subalgebra, so by what we have just proved, it satisfies one of the conditions(i)−(iv) above. If (i) or (ii) hold, then the trace of K kills 1K = 1C but is not identically zero.Thus Ker(tK) = F1C , forcing tC(u) = tK(u) 6= 0, a contradiction. Case (iii) cannot hold since F isperfect, but neither can Case (iv) since C does not contain nilpotent elements different from zero.This completes the proof.


98. Let I 6=C be a two-sided ideal of (C, ιC). For x ∈C we obtain nC(x)1C = xx ∈ I, which impliessol.COSIMnC(x) = 0. Moreover, for y∈C we obtain nC(x,y)1C = xy+yx ∈ I, which implies nC(x,y) = 0. ButnC is non-degenerate. Thus x = 0, and we have shown I = 0, so (C, ιC) is simple as an algebrawith involution. Here Prop. 11.5 shows that either C is simple, or (C, ιC) ∼= (Aop⊕A,ε) for somesimple unital F-algebra A, where ε stands for the exchange involution. Identifying C = Aop⊕Aby means of this isomorphism, we put c := 1Aop , c′ := 1A and conclude from Exc. 82 that c,c′

are elementary idempotents in C such that 1C = c+ c′. For x ∈ Aop, we now obtain 2x = c x =tC(c)x+ tC(x)c−nC(c,x)1C = x+ tC(x)c−nC(c,x)1C , hence

x =(tC(x)−nC(c,x)

)c−nC(c,x)c′.

This shows nC(c,x) = 0 and x = tC(x)c ∈ Fc, and we have proved Aop ∼= F , hence A∼= F as well.Thus C ∼= F⊕F is split quadratic etale.

99. We first show that the norm of a conic divsion algebra over a field F is anisotropic. Indeed,sol.CDDIVALGlet C be a conic division algebra over F and 0 6= x ∈ C. Then F [x] ⊆ C is a unital subalgebra ofdimension at most 2, and left multiplication by x gives a linear injection from F [x] to itself, whichtherefore is a bijection. Hence x is invertible in F [x], which implies nC(x) 6= 0 by Exc. 77 (b) andproves our claim. For the rest of the proof, we assume that F is a field of arbitrary characteristic,µ ∈ F is a non-zero scalar, B is an octonion algebra over F and C = Cay(B,µ).

(a) Isotropic non-singular quadratic forms over F are known to be universal. The assumptionµ /∈ nB(B×) therefore implies that nB is anisotropic, forcing B to be an octonion division algebraby Thm. 22.8 combined with Prop. 20.4. Now (21.3.1) implies

x1x2 = (u1u2 +µ v2v1)+(v2u1 + v1u2) j,

and we conclude

x1x2 = 0 ⇐⇒ u1u2 =−µ v2v1, v2u1 =−v1u2. (120) ZECRI

Suppose first x1x2 = 0. If u1 = 0, then v1 6= 0 and (120) implies v2 = u2 = 0, hence x2 = 0, acontradiction. Thus u1 6= 0. If u2 = 0, then (120) again implies v2 = 0. This contradiction showsu2 6= 0. Since B is a division algebra, the first equation of (120) now implies v1 6= 0 6= v2. We havethus shown ui 6= 0 6= vi for i = 1,2. Next, since the norm of B is multiplicative, (120) yields firstnB(u1)nB(u2) = µ2nB(v1)nB(v2) and then

nB(u1)2nB(u2) = µ

2nB(v1)nB(v2)nB(u1) = µ2nB(v1)

2nB(u2).

Thus nB(u1) = ±µnB(v1). But nB(u1) = µnB(v1) is impossible since this and multiplicativity ofnB would yield the contradiction µ ∈ nB(B×). Hence (i) holds. Turning to (ii), we use (120), (i)and Kirmse’s identities (20.3.1) to compute

(u1u2)v1 = −µ(v2v1)v1 =−µnB(v1)v2 = nB(u1)v2 = u1(u1v2) =−u1v1u2 =−u1(u2v1),

and this is (ii). Finally, (iii) is equivalent to the second equation of (120). Conversely, supposeui 6= 0 6= vi for i = 1,2 and (i)−(iii) hold. Then so does the second equation of (120), and (i), (ii)imply

(u1u2)v1 =−u1(u2v1) =−u1v1u2 = u1(u1v2) = nB(u1)v2 =−µnB(v1)v2 =−µ(v2v1)v1.

But v1, being different from zero, is invertible in B. Hence the first equation of (120) holds as well.This completes the proof of (a).

(b) We know already that if C is a division algebra, then nC is anisotropic. Conversely, supposenC = nB⊕ (−µ)nB (cf. Remark 21.4) is anisotropic. Assuming µ = nB(u) for some u ∈ B× wouldlead to the contradiction nC(u+ j) = 0. Thus µ /∈ nB(B×). If C were not a division algebra, somexi = ui + vi j, ui,vi ∈ B, i = 1,2, would satisfy the conditions of (a). In particular, since we are in


characteristic 2, condition (i) would imply nC(x1) = nB(u1)+µnB(v1) = 0, in contradiction to nCbeing anisotropic. Hence C is a division algebra.

(c) Suppose first x,y ∈ A, z ∈ A⊥ for some quaternion subalgebra A⊆ B and x y = 0, equiva-lently, xy = −yx. Then we may identify B = Cay(A,ν) = A⊕Ai for some non-zero scalar ν ∈ F(Thm. 22.15 (b)) and have A⊥ =Ai, hence z= ui for some u∈A. Now we use (21.3.1) and compute

(xy)z = (xy)(ui) = (uxy)i,

x(yz) = x(y(ui)

)= x((uy)i

)= (uyx)i =−(uxy)i.

Hence (xy)z=−x(yz), as desired. Conversely, let this be so. Since char(F) 6= 2 and B is alternative,we have x /∈ F1B, y /∈ F [x]. Let A be the (unital) subalgebra of B generated by x,y. Since B is adivision algebra, so is A, forcing its norm to be anisotropic, hence non-singular. Since, therefore,A is a composition algebra of dimension at least 3 and at most 4 (Exc. 88), it must, in fact, be aquaternion algebra. As before, we may assume B = Cay(A,ν) = A⊕Ai for some ν ∈ F×. Thenz = u+ vi, u,v ∈ A, and we conclude

(xy)z = xyu+(vxy)i,

−x(yz) = − x(yu+(vy)i

)=−xyu− (vyx)i.

Comparing coefficients, this implies xyu = v(x y) = 0, hence u = 0. But then v 6= 0, forcingx y = 0 and z = vi ∈ A⊥.

(d) Suppose first that C is a division algebra. Then its norm is anisotropic, and we have seen in(b) that this implies µ /∈ nB(B×). Assume now −µ = nB(z) for some element z ∈ B of trace zero.Pick any non-zero element y ∈ F [z]⊥ and write A′ for the subalgebra of B generated by y,z. Theargument produced in the proof of (c) shows that A′ is a quaternion subalgebra of B. Pick any non-zero element x ∈ A′⊥ ⊆ F [y]⊥. Then the subalgebra A of B generated by x,y is again a quaternionalgebra, and we have not only nB(z,1B) = nB(z,x) = nB(z,y) = 0 but also (by norm associativity)nB(z,xy) = nB(zy,x)∈ nB(A′,x) = 0. Thus z∈A⊥, while (19.5.5) yields xy= 0. Setting u1 := x,u2 := y, v1 := u1z−1 =−nB(z)−1u1z ∈ A⊥, v2 :=−(v1u2)u−1

1 , we have v1 =−v1 ∈ A⊥, whence itis clear that conditions (i)−(iii) of (a) hold. Hence (a) produces zero divisors in C, a contradiction.Conversely, suppose µ /∈ nB(B×), −µ /∈ nB(B0 ∩B×). If C were not a division algebra, we wouldfind elements ui,vi ∈ B, i= 1,2 satisfying the conditions of (a). Then (c) would lead to a quaternionsubalgebra A⊆ B such that u1,u2 ∈ A, v1 ∈ A⊥ and u1 u2 = 0. This would imply v1 =−v1 ∈ A⊥

and −µ = nB(z), z := u1v−11 ∈ A⊥ and, in particular, tB(z) = 0. This contradiction completes the

proof.(e) The algebra in question is the same as Cay(O,−1). The norm of O is positive definite, and

so is its restriction to the trace zero elements. Hence 1 =−(−1) cannot avoid being the norm of atrace zero element of O.

100. Let C be a finite-dimensional alternative division algebra over R. Then Exc. 64 shows that C issol.FROBTHEunital. Moreover, for 0 6= x ∈C, the unital subalgebra R[x]⊆C is finite-dimensional, commutativeassociative and without zero divisors. Thus R[x]/R is a finite algebraic field extension, whichimplies R[x] = R1C or R[x] ∼= C as R-algebras. In particular, the elements 1C,x,x2 are linearlydependent over R, and we deduce from Exc. 80 that C is a conic algebra. Since the non-zeroelements of C are invertible, Prop. 20.4 shows that the norm of C is anisotropic. On the otherhand, it represents 1 and hence (by the intermediate value theorem) must be positive definite.Now Thm. 22.8 shows that C is a composition division algebra over R. By Cor. 22.16, therefore,C ∼= Cay(R,α1, . . . ,αn) for some n = 0,1,2,3 and α1, . . . ,αn ∈ R×. Here the assumption αi > 0for some i = 0, . . . ,n would contradict Remark 21.4 in conjunction with the property of nC to bepositive definite. Thus αi < 0 for 0 ≤ i ≤ 3, and invoking Exc. 93, we are actually reduced to thecase αi =−1, 0≤ i≤ n, which means C ∼= R,C,H,O for n = 0,1,2,3, respectively.

101. For the first part of the problem, put C := Cay(B,µ) and write ϕ : C→Cp for the map definedsol.ISOTby


ϕ(u+ v j) := (p−1up)+ v j (u,v ∈ B),

which is obviously a linear bijection preserving units but also norms since B was assumed to bemultiplicative. It therefore suffices to show that ϕ is an algebra homomorphism. In order to do so,write “·” for the product in Cp and let u1,u2,v1,v2 ∈ B. Combining (21.3.1) and associativity of Bwith the fact that its conjugation is an involution (Prop. 20.2 (a)), we obtain

ϕ(u1 + v1 j) ·ϕ(u2 + v j) =((p−1u1 p)+ v1 j

)·((p−1u2 p)+ v2 j

)=((

(p−1u1 p)+ v1 j)

p−1)(

p((p−1u2 p)+ v2 j

))=(

p−1u1 +(v1 p−1) j)(

u2 p+(v2 p) j)

= (p−1u1u2 p+µ pv2v1 p−1)+(v2 pp−1u1 + v1 p−1 pu2) j.

But p = nB(p)p−1, p−1 = nB(p)−1. Hence

ϕ(u1 + v1 j) ·ϕ(u2 + v j) =(

p−1(u1u2 +µ v2v1)p)+(v2u1 + v1u2) j

= ϕ((u1u2 +µ v2v1)+(v2u1 + v1u2) j

)= ϕ

((u1 + v1 j)(u2 + v2 j)

),

as desired.Turning to the second part of the problem, let C be an octonion algebra over k and suppose p,q∈

C× have the property that pq2 belongs to some quaternion subalgebra B⊆C. By Exc. 73 (a) and itssolution, it will be enough to show for p∈B× and B as above that C and Cp are isomorphic. In orderto do so, we note C = B⊕B⊥ as a direct sum of submodules (Lemma 12.10) and define ϕ : C→Cp

by ϕ(u+v) := (p−1up)+v for u∈B, v∈B⊥. We claim that ϕ is an isomorphism of conic algebras.This assertion is local on k, so we may assume that k is a local ring. By Thm. 22.15 (a), we mayidentify C = Cay(B,µ) = B⊕B j, where the relation B⊥ = B j can be read off from (21.3.4) and thenon-singularity of nB. Thus our new map ϕ agrees with the one defined in the first part and hencemust be an isomorphism of conic algebras.

102. (a) We begin by showing that for a quadratic k-algebra R and u ∈ R,sol.AZUQUAD

nR(u− u) = 4nR(u)− tR(u)2, (121) UMIBA

which follows from

nR(u− u) = nR(u)−nR(u, u)+nR(u) = 2nR(u)− tR(u)2 +nR(u,u) = 4nR(u)− tR(u)2.

Now suppose that u− u is invertible in R. Then Prop. 22.5 and (121) show that D := k[u]⊆ R is aquadratic etale subalgebra. By Lemma 12.10, therefore, we obtain the decomposition R = D⊕D⊥

as k-modules and conclude, by comparing ranks, that R = D is quadratic etale.Next suppose k is a semi-local ring and D is a quadratic etale k-algebra. Then Thm. 22.15 (b)

yields an element u ∈ D of trace 1 such that D′ := k[u] ⊆ D is a quadratic etale subalgebra. Asbefore, we have D = D′ ⊕D′⊥ as k-modules, and comparing ranks yields D = D′ = k[u]. NowProp. 22.5 combined with (121) shows nD(u− u) ∈ k×, hence u− u ∈ D×.

Finally, returning to an arbitrary base ring k and a quadratic etale k-algebra D, it remains toshow H(D, ιD) = k1D. In order to do so, we may assume D = k[u] for some u ∈ D such thatu− u ∈ D×. In particular, by Prop. 22.5, D is free (of rank 2) as a k-module with basis 1D,u. Weclearly have k1D ⊆ H(D, ιD). Conversely, write x ∈ H(D, ιD) as x = α1D +βu for some α,β ∈ k.Then x = x = α1D +β u implies β (u− u) = 0, hence β = 0 (since invertible elements, being theimage of 1D under a linear bijection, are unimodular), and we end up with x = α1D ∈ k1D.

(b) Let B be a quaternion algebra over k. We must show that an element x∈B satisfying [x,y] = 0for all y ∈ B is a scalar multiple of 1C . This assertion being local on k, we may assume that k is alocal ring. By Cor. 22.16, we may further assume B = Cay(D,µ) = B⊕B j for some quadratic etale


k-algebra D and some µ ∈ k. From Thm. 22.10 we deduce µ ∈ k×, while Part (a) of our problemyields an element u∈D making u− u invertible. Now write x = u1+v1 j, u1,v1 ∈D. By hypothesiswe have [x,u2 + v2 j] = 0 for all u2,v2 ∈ D, which by (21.12.1) implies

0 = [u1,u2]+µ(v2v1− v1v2) = µ(v2v1− v1v2).

Since µ is invertible, we conclude v2v1 = v1v2 for all v2 ∈ D. Setting v2 = 1B gives v1 = v1, whilesetting v2 = u now gives (u− u)v1 = 0, hence v1 = 0 since u− u is invertible. Next applying(21.12.2), we obtain 0 = v2(u1− u1)− v1(u2− u2) = v2(u1− u1) for all v2 ∈ B, which for v2 = 1Bamounts to u1 = u1, hence by (a) to u1 ∈ k1B. Thus we have x ∈ k1C , as desired.

Finally, let C be an octonion algebra over k. We have

k1C ⊆ x ∈C | ∀y ∈C : xy = yx∩Nuc(C),

so it remains to show that both sets taking part in the intersection on the right-hand side belongto k1C . Since these are local questions, we may assume that k is a local ring and then obtain, byCor. 22.16, C = Cay(B,µ) = B⊕B j up to isomorphism, for some quaternion subalgebra B ⊆ Cand some scalar µ ∈ k× (Thm. 22.10). Now let x = u1 + v1 j ∈C with u1,v1 ∈ B and first assume[x,u2 + v2 j] = 0 for all u2,v2 ∈ B. From (21.12.1), (21.12.2), we conclude

[u1,u2]+µ(v2v1− v1v2) = 0 = v2(u1− u1)− v1(u2−u2) (122) OCOM

for all u2,v2 ∈ B. By what we have seen before, the first of these equations for v2 = 0 showsu1 ∈ Cent(B) = 1B. Now the second one implies v1(u2− u2) = 0 for all u2 ∈ B. By Thm. 22.15 (b),we find a quadratic etale subalgebra D⊆ B, which in turn, by part (a) of this exercise, contains anelement u2 making u2− u2 invertible. Hence v1 = 0 and thus x ∈ k1C . Next assume x ∈ Nuc(C).Applying (21.12.3), for u2 = v3 = 0, we obtain (v2v1)u3 = (u3v2)v1 for all v2,u3 ∈ B. Settingv2 = 1B, this implies v1u3 = u3v1 for all u3 ∈ B, hence v1 ∈ Cent(B) = k1B. Thus v1 = β1B forsome β ∈ k, and our original equation reduces to (β v2)u3 = u3(β v2) for all v2,u3 ∈ B This meansβv2 ∈ Cent(B) = k1B for all v2 ∈ B. But 1B, being unimodular, may be extended to basis of B as ak-module, and picking for v2 one of the basis vectors distinct from 1B, we conclude β = 0, hencev1 = 0. Now we consider (21.12.4) for u2 = u3 = v2 = 0 and v3 = 1B. Then [u1,u2] = 0 for allu2 ∈ B, which yields u1 ∈ Cent(B) = k1B, hence x ∈ k1C , and completes the proof.

103 . If k allows a decomposition of the desired kind, then ϕ is clearly an automorphism of D.sol.ETAUTConversely, let this be so. We first treat the case that k is a local ring. Then D is free of rank2 as a k-module and Theorem 22.15 (b) implies D = k[u] for some element u ∈ D of trace 1.By Proposition 22.5, therefore, (1D,u) is a basis of D, and there are α,β ∈ k such that ϕ(u) =α1D +βu. But ϕ , being an automorphism of D, preserves norms, traces and unit. Hence

2α +β = tD(ϕ(u)

)= tD(u) = 1,

α2 +αβ +β

2nD(u) = nD(ϕ(u)

)= nD(u),

and we conclude β = 1−2α ,

α2 +α−2α

2 +(1−4α +4α2)nD(u) = nD(u).

Thus (α −α2)(1− 4nD(u)) = 0. On the other hand, since D is quadratic etale, Proposition 22.5shows 1− 4nD(u) ∈ k×. Hence α ∈ k is an idempotent. But as a local ring, k is connected. Wetherefore conclude α = 0, β = 1 or α = 1, β =−1, which implies ϕ = 1D in the first case, ϕ = ιDin the second.

Now let k be arbitrary and put X := Spec(k). Since D is finitely generated as a k-module,

X+ := p ∈ X | ϕp = 1Dp, X− := p ∈ X | ϕp = ιDp


by § 10, Exercise 40, are Zariski-open subsets of X . Moreover, they are disjoint by Exercise 102 (a)and cover X by the special case just treated. Hence § 10, Exercise 41, yields a complete orthogonalsystem (ε+,ε−) of idempotents in k such that X± = D(ε±). Put k± := ε±k. Then k = k+ ⊕ k−as a direct sum of ideals, and with the canonical projection π+ : k→ k+, we consult (10.6.6) toconclude p := Spec(π+)(p+) = p+⊕ k− ∈ D(ε+) for any prime p+ ∈ k+. Now 10.5, 10.6 showthat ϕ+p+ = ϕp⊗kp k+p+ = 1D+p+

. Thus ϕ+ = 1D+ and, similarly, ϕ− = ιD− .

104. We begin by establishing the following claim.sol.IDEALSCOALClaim. Let I ⊆ C be an ideal satisfying I ∩ k = 0. Then I = 0 provided I is stable underconjugation or C has rank r 6= 2.Proof. Localizing if necessary, we may assume that C has a rank, denoted by r. The case r = 1being obvious, we may also assume that C is non-singular.

Suppose first I is stable under conjugation and let x ∈ I. Then x ∈ I = I, forcing tC(x) = x+x ∈I∩ k = 0. But I ⊆C is an ideal, so xz ∈ I for all z ∈C, and we conclude nC(x,z) = tC(x, z) = 0,hence x = 0 by non-singularity.

Next consider the case r > 2. Since I∩ I ⊆C is an ideal stable under conjugation and satisfyingI ∩ I ∩ k ⊆ I ∩ k = 0, the previous case yields I ∩ I = 0. Now let x ∈ I be non-zero. Thenx = tC(x)− x does not belong to I, forcing tC(x) 6= 0. Thus the linear map tC : I → k is injective.Given z,w ∈C, the relations

tC((xz)w

)= tC

(x(zw)

), tC(xz) = tC(zx)

therefore imply (xz)w = x(zw), xz = zx, so x belongs to the centre of C, which is k by Exc. 102 (b).Hence I ⊆ I∩ k = 0, which proves our claim.Turning to the problem in its proper sense, we put

I(k) := a | a⊆ k is an ideal,I(C, ιC) := I | I ⊆ (C, ιC) is an ideal,

I(C) = I | I ⊆C is an ideal.

In particular, I(C, ιC) consists of the ideals in C that are stable under conjugation. Besides theinclusion j : I(C, ιC) → I(C), we now define maps

Φ : I(k)−→ I(C, ιC), a 7−→Φ(a) := aC,

Ψ : I(C)−→ I(k), I 7−→Ψ(I) := I∩ k,

which may all be depicted in the following diagram:

I(k)Φ // I(C, ιC)q Q

j

I(C).

Ψ

ZZ

The theorem will follow once we have shown

Ψ j Φ = 1I(k), Φ Ψ j = 1I(C,ιC), (123) PHPS

j Φ Ψ = 1I(C) (if C has rank r 6= 2). (124) JPHPS


To do so, let a be an ideal in k. Since 1C ∈C is unimodular, we find a k-submodule M ⊆C suchthat

C = k⊕M (125) CM

This gives aC = a⊕aM, hence

(Ψ j Φ)(a) = (aC)∩ k = (a⊕aM)∩ k = a

and we have established the first relation of (123). We derive the second one along with (124) byletting I be an ideal of (C, ιC) (resp. of C if C has rank r > 2). Then (Φ Ψ j)(I) = (I∩ k)C ⊆ I(resp. ( j Φ Ψ)(I) = (I∩ k)C ⊆ I), so it remains to show

I ⊆ (I∩ k)C. (126) IKC

Case 1. I∩ k = 0.Then we have to prove I = 0, which follows from our claim above.Case 2. a := I∩ k is arbitrary.Then we pass to the base change k′ := k/a, so we look at the composition algebra C′ :=C⊗ k′ =C/aC over k′, denote by I′ = I/aC the image of I under the canonical map

C −→C′, x 7−→ x′,

which belongs to I(C′), and claim that a′ := I′∩ k′ satisfies

a′ = 0. (127) APR

To see this, we first note that (125) implies

C′ = k′⊕M′, M′ := M⊗ k′ = M/aM. (128) CMP

Given α1 ∈ a′⊆ I′, we can therefore find elements ξ ∈ k, y∈M satisfying x := ξ +y∈ I and x′=α1.In view of (128), this yields ξ ′ = α1, y′ = 0, hence y ∈ aM ⊆ aC ⊆ I and ξ = x− y ∈ I ∩ k = a.But then α1 = ξ ′ = 0 and the proof of (127) is complete. Now we are in a position to apply case 1,which yields I′ = 0, hence I ⊆ aC, as claimed in (126).

105 . We begin by proving (2) and first note that not only c1 = c but also c2 = 1C − c is ansol.PEIRCEELIDelementary idempotent (cf. Exc. 82, particularly (iv)). Let i = 1,2 and x ∈Cii. Then cix = x = xci,and (20.3.2) yields x = cixci = nC(ci, x)ci− nC(ci)x = nC(ci, x)ci ∈ Cii, and (2) holds. Inview ofthis, arbitrary elements x,y ∈C can be written as in (3). Expanding the norm of x in the obviousmanner, we obtain

nC(x) = nC(α1c1 + x12 + x21 +α2c2)

= α21 nC(c1)+α1nC(c1,x12)+α1nC(c1,x21)+α1α2nC(c1,c2)+nC(x12)

+nC(x12,x21)+α2nC(x12,c2)+nC(x21)+α2nC(x21,c2)+α22 nC(c2).

Applying (20.1.3), (20.1.2) and observing nC(c1) = nC(c2) = 0, we obtain, for i, j= 1,2,

nC(ci) = 0,

nC(ci,xi j) = nC(cici,cixi j) = nC(ci)nC(ci,xi j) = 0,

nC(ci,x ji) = nC(cici)nC(x jici) = nC(ci,x ji)nC(ci) = 0,

nC(c1,c2) = nC(c,1C− c) = tC(c)−2nC(c) = 1,

nC(xi j) = nC(cixi j) = nC(ci)nC(xi j) = 0.


Inserting these relations into the preceding equation, we end up with (4), while linearizing (4) gives(5), which for y = 1C yields (6). Finally, (7) follows immediately from the definition.

Now let C be a composition algebra. We must show for i, j = 1,2 that DnC determines aduality between Ci j and C∗ji. Let xi j ∈Ci j and suppose nC(xi j,C ji) = 0. Then (5) yields

nC(xi j,C) = nC(xi j,kci +Ci j +C ji + kc j) = nC(xi j,C ji) = 0

hence xi j = 0 since nC is non-singular. Next let λ ji : C ji → k be a linear form and λ : C → kthe linear extension of λ ji that kills kc1⊕Ci j ⊕ kc2. By non-singularity of nC , there exists x ∈ Csuch that λ = nC(x,−). Writing x in the form (3) and choosing y ji ∈C ji, we apply (5) and obtainλ ji(y ji) = λ (y ji) = nC(x,y ji) = nC(xi j,y ji). Summing up, DnC does indeed determine a dualitybetween Ci j and C ji. In order to prove the final statement, we must show tC(x2z) = tC(xy2) = 0 forall x,y,z ∈Ci j . This follows from the final statement in § 15, Exercise 67.

106. (a) Since scalar multiplication by D acts on the second factor of D⊗D, the map ϕ is a unitalsol.MISPLEThomomorphism of D-algebras. Moreover, for u ∈ D we have

ϕ(u⊗1 : D−1D⊗ u) = (u⊕ u)− (u⊕ u) = (u− u)⊕0,

ϕ(1D⊗u−u⊗1D) = (u⊕u)− (u⊕ u) = 0⊕ (u− u).

In order to prove that ϕ is an isomorphism, we may assume that k is a local ring. By Exc. 102 (a),the element u may be so chosen that u− u∈D×. The preceding equations therefore show that 1D⊕0 and 0⊕ 1D both belong to Im(ϕ). But D is generated by these elements as a D-algebra. Henceϕ is surjective. On the other hand, combining Exc. 102 (a) with Prop. 22.5, we conclude that D isa free k-module with basis (1D,u). Hence D⊗D is a free D-module with basis (1D⊗1D,u⊗1D).Hence an arbitrary element of Ker(ϕ) can be written in the form d0(1D⊗ 1D)+ d1(u⊗ 1D) withd0,d1 ∈ D, and we conclude

0 = ϕ(d0(1D⊗1D)+d1(u⊗1D)

)= ϕ(1D⊗d0 +u⊗d1)

= (d0⊕d0)+(ud1⊕ ud1) = (d0 +ud1)⊕ (d0 + ud1).

Thus d0 + ud1 = d0 + ud1 = 0. Subtracting gives (u− u)d1 = 0, hence d1 = 0 and then d0 = 0 aswell. This shows that ϕ is injective, and the proof of (a) is complete.

(b) By definition, we have σ ′ ϕ = ϕ σ ′. Hence, for x,d ∈ D,

σ′((xd)⊕ (xd)

)= σ

′(ϕ(x⊗d)

)= ϕ

(σ(x⊗d)

)= ϕ(x⊗ d) = (xd)⊕ (xd) = xd⊕ xd,

which proves the assertion since, by (a), any element of D⊕D can be written as a finite sum ofexpressions of the form (xd)⊕ (xd) with x,d ∈ D.

107 . (a) Let x,y ∈ R. Since R by Prop. 20.2 (b) is a multiplicative conic algebra, we may applysol.QUACOMQUA(19.11.5), (19.5.10) and (20.1.3) to obtain

tR(e(xy)

)= tR

((ex)(ey)

)= tR(ex)tR(ey)−nR(ex,ey) = tR(ex)tR(ey)−nR(e)nR(x,y) = tR(ex)tR(ey).

Hence tR Le : R→ k is an algebra homomorphism, which by Exc. 82 is unital if and only if e isan elementary idempotent.

(b) For x,y ∈ R, we combine (a) with the multiplicativity of nR and obtain

q(x)q(y) =(tR(ex2)+ εnR(x)

)(tR(ey2)+ εnR(y)

)= tR(ex2)tR(ey2)+ εtR(ex2)nR(y)+ εnR(x)tR(ey2)+ ε

2nR(xnR(y)

= tR(e(xy)2)+ tR(εex2)nR(y)+nR(x)tR(εey2)+ εnR(xy)

= tR(e(xy)2)+ εnR(xy) = q(xy).


Hence q permits composition.(c) The condition is clearly sufficient. Conversely, suppose q permits composition. We put

e1 := 1⊕0, e2 := 0⊕1 and

ε1 := q(e1), ε2 := q(e1,e2), ε3 := q(e2).

Then (8) holds. Moreover, since q permits composition, (ε1,ε3) is an orthogonal system of idempo-tents such that ε1ε2 = q(e1)q(e1,e2) = q(e2

1,eee2) = 0 and, similarly, ε3ε2 = 0. On the other hand,ε := q(1D) = ∑εi is an idempotent in k satisfying q(x) = εq(x), q(x,y) = εq(x,y) for all x,y ∈ D.In particular, we have εεi = εi for i = 1,2,3, whence ε2 = ε − ε1− ε3 must be an idempotent aswell.

(d) Let q : D→ k be a quadratic form permitting composition. Since, therefore, its scalar ex-tension qD : D⊗D→D permits composition, so does q′ := qD ϕ−1 : D⊕D→D, where ϕ is theisomorphism of Exc. 106 (a). Hence (c) yields an orthogonal system (d1,d2,d3) of idempotents inD such that

q(a⊕b) = d1a2 +d2ab+d3b2

for all a,b ∈ D. With σ as in Exc. 106 (b), we now claim

qD σ = ιD qD. (129) QUDE

In order to prove this, we let x,x′,d,d′ ∈ D and compute

(qD σ)(x⊗d) = qD(x⊗ d) = q(x)d2,

(ιD qD)(x⊗d) = qD(x⊗d) = q(x)d2 = q(x)d2,(qD (σ ×σ)

)(x⊗d,x′⊗d′) = qD(x⊗ d,x′⊗ d′) = q(x,x′)dd′,

(ιD qD)(x⊗d,x′⊗d′) = q(x,x′)dd′ = q(x,x′)dd′.

Hence (129) holds. Using this, and keeping the notation of Exc. 106 (b), we conclude

q′ σ′ = qD ϕ

−1 ϕ σ ϕ−1 = qD σ ϕ

−1 = ιD qD ϕ−1 = ιD q′.

deviating slightly from the notation used in (c), we now put e1 := 1D⊕0, e2 := 0⊕1D and obtain

d1 = q′(e1) = (ιD q′)(e1) =

hence d3 = d1. Similarly,

d2 = (ιD q′)(e1,e2) = q′(σ′(e1),σ

′(e2))= q′(e2,e1) = d2.

Thus d2 ∈H(D, ιD) = k1D (by Exc. 102 (a)). Since 1D is unimodular, we find a unique idempotentε ∈ k such that d2 = ε1D. Then e := d1 is an idempotent in D such that εe = d2d1 = 0, while therelation nD(e)1D = ee = d1d3 = 0 implies nD(e) = 0. Finally, for x ∈ D, we compute

q(x)1D = qD(x⊗1D) = (q′ ϕ)(x⊗1D) = q′(x⊕ x) = ex2 + εxx+ ex2

=(tD(ex2)+ εnD(x)

)1D,

and the assertion follows.

107 . We begin by elaborating on the exact sequence alluded to in the problem and define asol.QUACOMCOMlinear map Φ : Endk(M) → Quad(M) by f 7→ Q f , where Q f : M → k is the quadratic formx 7→ Q( f (x),x). We then claim that the sequence

0 // Alt(M,Q) // Endk(M)Φ // Quadk(M) // 0


is exact, equivalently, that the linear map Φ is surjective. Let q : M→ k be any quadratic form.By §12, Exercise 50, there exists a bilinear form b : M×M → k (possibly not symmetric) suchthat q(x) = b(x,x) for all x ∈ M, and since Q is non-singular, we find an element f ∈ Endk(M)satisfying b(x,y) = Q( f (x),y) for all x,y ∈ M. This implies q = Φ( f ), and surjectivity of Φ isproved.

q = εnC for some idempotent ε ∈ k clearly implies that q permits composition. Conversely,suppose q : C→ k is a quadratic form permitting composition. The case r = 1 being trivial, we mayassume r > 2, so C is either a quaternion or an octonion algebra; in any event, it is non-singular.Since q permits composition, we obtain

q(x) = εq(x). (x ∈C, ε := q(1C) = ε2 ∈ k) (130) PCUNIT

We now proceed in several steps.10. Our first aim will be to show that if the bilinearization of q vanishes, so does q itself. Indeed, ifq(x,y) = 0 for all x,y ∈C, then q : C→ k is a (possibly non-unital) homomorphism of Z-algebras,forcing

I := Ker(q) = x ∈C | q(x) = 0 ⊆C

to be an ideal. From Exc. 104 we therefore deduce I = aC for some ideal a⊆ k. It suffices to showa= k. Otherwise, the base change C′ =C/I =C⊗k′ would continue to be a composition algebra ofrank r > 2 over the non-zero ring k′ = k/a, and q would induce an injection C′→ k of Z-algebras,which would make C′ a commutative associative k′-algebra, a contradiction.20. Since the norm of C is non-singular, the preceding exact sequence yields a k-linear map f : C→C satisfying

q(x) = nC(

f (x),x)

(x ∈C), (131) REPENDQ

which implies

q(x,y) = nC(g(x),y), g := f + f ∗, (132) REPENDQL

for all x,y ∈ C, where f ∗ stands for the adjoint of f relative to DnC . Obviously, g is uniquelydetermined by q. Combining (132) with (19.11.4) and the property of q to permit composition, wenow obtain nC(xg(xy),z) = nC(g(xy),xz) = q(xy,xz) = q(x)q(y,z) = nC(q(x)g(y),z), hence

xg(xy) = q(x)g(y) (133) FGG

since nC is non-singular. Linearizing (133) and consulting (132) again, we conclude

xg(zy)+ zg(xy) = nC(g(x),z)g(y), (134) FGGL

where we set z = 1C to conclude

g(xy) = λ (x)g(y)+ xg(y), λ := tC g− tC. (135) FGGUNT

For y = 1C , this implies

g(x) = λ (x)u+ xu, u := g(1C). (136) FGGUUNT

Now observe that Cop is a composition algebra over k having the same norm as C and satisfyingall of our previous hypotheses. Also, q : Cop→ k permits composition and passing from C to Cop

neither affects g nor λ and u. Hence (136) yields

g(x) = λ (x)u+ux. (137) FGGUONT

Comparing (136) and (137), we obtain ux = xu for all x ∈C, so Exc. 102 (b) implies


u = ε′1C (ε ′ ∈ k), (138) UUNT

and (136) converts itself into

g(x) = ε′λ (x)1C + ε

′x. (139) GAID

Combining (139) with (136), a short computation gives

ε′[λ (x)y+λ (y)x+

(λ (x)λ (y)−λ (xy)

)1C] = 0. (140) LAHO

We wish to prove ε ′λ = 0 in C∗ = Homk(C,k). Localizing if necessary, we may assume that1C extends to a basis e0 = 1C,e1, . . . ,er−1 of C as a k-module. Setting x = ei, y = e j in (140)(1≤ i, j ≤ r−1, i 6= j), we conclude ε ′λ (ei) = 0, while ε ′λ (1C) = 0 follows from (136), (138) byspecializing x to 1C . Summing up, all this yields ε ′λ = 0 and (139) reduces to

g = ε′1C. (141) GID

30. Setting y = 1C in (133) and observing (141), we conclude

ε′nC(x) = ε

′q(x), (142) EPQNOR

which implies

εε′ = ε

′ (143) EPPRIUNT

for x = 1C . On the other hand, linearizing (142) and consulting (141) again yields nC(ε′x,y) =

ε ′nC(x,y)= ε ′q(x,y)= ε ′nC(ε′x,y)= nC(ε

′2x,y), and ε ′= ε ′2 is an idempotent in k. By (143), so isε1 = ε−ε ′, allowing us to analyze the base change of q from k to k1 = kε1. Since ε ′1k1 = ε ′ε1 = 0,we conclude g⊗ k1 = 0 from (141), so q1 = q⊗ k1 has zero bilinearization, hence by 10 must bezero itself. Using (130), (143), this implies

0 = q1(1Ck1) = εε1 = ε

2− εε′ = ε− ε

′,

hence ε = ε ′. Now we may use (130), (142) to conclude εnC(x) = εq(x) = q(x) for all x ∈C, andthe proof is complete.

109. We first treat the case that k = F is a field. Then R is a two-dimensional F-algebra. If R issol.COVQUATALGquadratic etale, then Thm. 22.15 (a) yields a scalar µ ∈ k× such that the inclusion R →C extendsto an embedding from Cay(B,µ) = B⊕B j to C. Its image, therefore, is a quaternion subalgebraof C generated by R and the image of j in C. Hence we may assume that R is not quadratic etale.Then Exc. 97 implies that R/F is either an inseparable quadratic field extension of characteristic 2or the F-algebra of dual numbers. In any event, setting u0 := 1C , there is an element u1 ∈C suchthat R = Fu0⊕Fu1 and nC(u0,u1) = tC(u1) = 2α = 0, where α := nC(u1). For any β ∈ F , wecan therefore find an element u2 ∈C such that nC(u0,u2) = tC(u2) = 1 and nC(u1,u2) = β . Settingγ := nC(u2) and u3 := u1u2, we conclude

nC(u0,u3) = nC(1C,u1u2) = nC(u1,u2) =−nC(u1,u2) =−β ,

nC(u1,u1) = 2α = 0,

nC(u1,u3) = nC(u1,u1u2) = nC(u1)tC(u2) = α,

nC(u2,u2) = 2nC(u2) = 2γ,

nC(u2,u3) = nC(u2,u1u2) = nC(u2)tC(u1) = 0,

nC(u3,u3) = 2nC(u1u2) = 2nC(u1)nC(u2) = 2αnC(u2) = 0.

This implies


det((

nC(ui,u j))

0≤i, j≤3

)= det(

2 0 1 −β

0 0 β α

1 β 2γ 0−β α 0 0

) = det(

0 −2β 1−4γ −β

0 0 β α

1 β 2γ 00 α +β 2 2βγ 0

)

= det(

−2β 1−4γ −β

0 β α

α +β 2 2βγ 0

)

= α(α +β2)(1−4γ)+(α +β

2)β 2 +4αβ2γ

= (α +β2)(α +β

2) = (α +β2)2

since 2α = 0. Setting β = 1 (resp. β = 0) for α = 0 (resp. α 6= 0), the above determinant will bedifferent from zero, forcing (ui)0≤i≤3 to be linearly independent and ∑Fui ⊆C by Exc. 88 to be aquaternion subalgebra containing R.

Next assume that k = F1⊕·· ·⊕Fm is a direct sum of finitely many fields Fj , 1≤ j ≤ m. ThenC =C1⊕·· ·⊕Cm with octonion algebras C j =C⊗Fj over Fj for 1≤ j≤m, and R = R1⊕·· ·⊕Rmwith quadratic algebras R j = R⊗Fj over Fj for 1 ≤ j ≤ m. Then, clearly, R j = Fj[a j] ⊆ C j (forsome a j ∈C j , 1≤ j≤m) is a quadratic subalgebra, so the preceding case yields an element b j ∈C jmaking B j := Fj1C j +Fja j +Fjb j +Fja jb j ⊆C j a quaternion subalgebra containing R j . Now put

a := a1⊕·· ·⊕am, b := b1⊕·· ·⊕bm, B := B1⊕·· ·⊕Bm

The R = k[a] and B = k1C + ka+ kb+ kab⊆C is a quaternion subalgebra generated by R and b.Finally, we are able to treat the general case. The property of R to be a direct summand of C

as a k-module is preserved under scalar extensions. Hence, setting k := k/Jac(k), C := C⊗ k =C/Jac(k)C, it follows from the previous case that R := R⊗ k = R/Jac(k)R ⊆ C is a quadraticsubalgebra generated by some element a′ ∈ R and that there is an element b′ ∈ C making B′ :=k1C + ka′+ kb′+ ka′b′ ⊆ C a quaternion subalgebra containing R. Lift a′ (resp. b′) to elements a(resp. b) in C. Then B := k1C + ka+ kb+ kab, the unital subalgebra of C generated by a,b, is aquaternion algebra containing R since the determinant of DnB evaluated at the bunch of generatorsindicated above becomes a unit after reduction mod Jac(k), hence must have been one all along.

110. (a) Let σ be a reflection of C and write B± := x ∈C | σ(x) =±x. Then B+ = Fix(σ)⊆Csol.RINVis a unital subalgebra and since 1

2 ∈ k, we have C = B+⊕B− as a direct sum of submodules. Infact, this sum is even an orthogonal one since σ preserves nC (Exc. 78), so for x± ∈ B± we obtainnC(x+,x−) = nC(σ(x+),σ(x−)) =−nC(x+,x−), forcing nC(x+,x−) = 0. It follows that Fix(σ) =B+ ⊆C is a composition subalgebra. In order to prove that its rank is r

2 , we may assume that k is alocal ring. Then Thm. 22.15 (a) yields an element µ ∈ k× such that Cay(B+,µ) = B+⊕B+ j maybe viewed as a subalgebra of C. In particular, j ∈ C×, and since σ is an automorphism of C, weconclude jB± ⊆ B∓. This shows that B+ and B− have the same rank. But the sum of the two ranksis r. Thus rk(B±) = r

2 .Conversely, suppose B⊆C is a composition subalgebra of rank r

2 , automatically non-singularsince 1

2 ∈ k. Hence C = B⊕B⊥ as direct sum of submodules, and we have

BB⊆C, BB⊥ ⊆ B⊥, B⊥B⊆ B⊥, B⊥B⊥ ⊆ B. (144) BEBEP

Here the first inclusion is obvious, while the second and third one follow from the fact that C isnorm associative, particularly from (19.11.3), (19.11.4). Finally, in order to establish the fourth in-clusion, we may assume that k is a local ring, in which case Thm. 22.15 (a) yields an identificationC = Cay(B,µ) = B⊕B j for some µ ∈ k× and then B⊥ = B j by (21.3.4). But now the assertionfollows from (21.3.1). By (144), the map σB := 1B⊕ (−1B⊥ ) is a reflection of C, and it is straight-forward to check that the assignments σ 7→ Fix(σ), B 7→ σB define inverse bijections between theset of reflections of C and the set of composition subalgebras of C having rank r

2 .


Let ρ ∈ Aut(C). If σ is a reflection of C, then so is σρ := ρ−1 σ ρ and Fix(σρ ) =ρ−1(Fix(σ)). On the other hand, if B ⊆C is a composition subalgebra of rank r

2 , then σρ−1(B) =

(σB)ρ . This shows that two reflections of C are conjugate under Aut(C) if and only if their fixed

algebras are, and the proof of (a) is complete.(b) Let τ be an involution of C. Since C and Cop have the same unit, norm and trace, Exc. 78

shows that they are preserved by τ . For x ∈C, this implies

τ(x) = τ(tC(x)1C− x

)= tC

(τ(x)

)1C− τ(x) = τ(x).

Thus τ commutes with ιC .Now let τ be an involution of C distinct from ιC . Since τ , as we have just seen, commutes with

ιC , the map στ := τ ιC = ιC τ is a reflection of C. Conversely, let σ be a reflection of C. Beingan automorphism of C, σ preserves conjugation, forcing τσ := σ ιC = ιC σ to be an involu-tion other than ιC . It is clear that the assignments τ 7→ στ and σ 7→ τσ define inverse bijectionsbetween the set of involutions of C distinct from ιC and the set of reflections of C. Moreover, theidentities τσρ = ρ−1 τσ ρ and σρ−1τρ = (στ )

ρ are immediately verified. Invoking (a), theaforementioned bijections, therefore, induce canonically inverse bijections between the isomor-phism classes of involutions of C other than ιC and the conjugacy classes under Aut(C) of thecomposition subalgebras of C having rank r

2 .Finally, let τ 6= ιC be an involution of C and B ⊆C the corresponding composition subalgebra

of rank r2 . Then τ = σB ιC , and for x ∈ B, y ∈ B⊥, we obtain τ(x+ y) = σB(x− y) = x+ y, hence

H(C,τ) = H(B, ιB)⊕B⊥ = k1B⊕B⊥ ∼= k⊕B⊥,

and since B⊥ is finitely generated projective of rank r2 , the final,assertion follows.

Section 24.

111. (a) We begin with the followingsol.INQUAClaim. If k = F is a field and M 6= 0, then the map M→ F, x 7→ h(x,x), is not identically zero.In order to see this, we argue indirectly and assume h(x,x) = 0 for all x ∈ M. Linearizing thisrelation yields

tD(h(x,y)

)= h(x,y)+h(x,y) = h(x,y)+h(y,x) = 0

for all x,y ∈ M, where we may replace x by xa, a ∈ D, to obtain nD(a,h(x,y)) = tD(ah(x,y)) =tD(h(xa,y)) = 0 for all a ∈ D, x,y ∈M. But since the norm of D is non-singular, this leads to thecontradiction h = 0 and proves our claim.

Returning to our local ring k, with maximal ideal m and residue field k := k/m, we argue byinduction on n and have nothing to prove for n = 0. If n > 0, we pass to the base change from kto k and put (M, h) := (M,h)k, which is a hermitian space of rank n over the field k. Hence ourpreceding claim yields an element x ∈ M = M/mM such that h(x, x) 6= 0. Lifting x to an elemente1 ∈M under the natural map M→ M, we deduce α1 := h(e1,e1) ∈ k×. Now we put

N := (e1D)⊥ = y ∈| h(e1,y) = 0= x ∈M | h(x,e1) = 0

and see by a straightforward verification that M = (e1D)⊕N = (e1D)⊥ N relative to h. It followsthat (N,h|N×N) is a hermitian space of rank n− 1 over D, and applying the induction hypothesiscompletes the proof.

(b) We may clearly assume that k is a local ring. Let the ei ∈M, αi ∈ k, 1≤ i≤ n, be as in (a).Then

M = (e1D)⊕·· ·⊕ (enD)

as a direct sum of free D-submodules of rank 1. Since D is a free k-module of rank 2, this showsthat M is a free k-module of rank 2n. On the other hand, the relations h(eid,e jd′) = δi jαidd′ for


d,d′ ∈ D, 1 ≤ i, j ≤ n show that the preceding decomposition is an orthogonal one relative to h,hence relative to q as well, and q(eid) = h(eid,eid) = αidd = αinD(d) for d ∈ D, 1≤ i≤ n. Thus

q∼= α1nD ⊥ ·· · ⊥ αnnD ∼= 〈α1, . . . ,αn〉⊗nD

is the finite orthogonal sum of of non-singular quadratic forms over k, hence must be non-singularitself.

112. We put C :=Ter(D;M,h,∆) and have C =D⊕M as k-modules. After the natural identificationsol.HERISOTD⊆C, we obtain p±1 = p±1⊕0, and writing “·” for the product in Cp, we conclude, for a,b ∈ D,x,y ∈M,

(a⊕ x) · (b⊕ y) =((a⊕ x)(p−1⊕0)

)((p⊕0)(b⊕ y)

)=((ap−1)⊕ (xp−1)

)((pb)⊕ (yp)

)=(ap−1 pb−h(xp−1,yp)

)⊕(ypap−1 + xp−1 pb+(xp−1)×h,∆ (yp)

)=(ab− p−1h(x,y)p

)⊕(ya+ xb+(x×h,∆ y)p−1 p

)Hence

(a⊕ x) · (b⊕ y) =(ab−h(x,y)

)⊕(ya+ xb+(x⊗h,∆ y)p−1 p

)(a,b ∈ D, x,y ∈M). (145) PROP

We will now be able to compute the constituents of Cp = Ter(D;Mp,hp,∆ p) by appealing toThm. 24.11 as follows.

As a k-module, Mp is the orthogonal complement of D = Dp in Cp relative to the bilinearizednorm of Cp, which by Exc. 90 agrees with the bilinearized norm of C. Thus Mp = M as k-modules.Consulting (145), the right action of Dp = D on Mp = M may be read off from

(x,a) 7−→ (0⊕ x) · (a⊕0) = 0⊕ (xa),

which amounts to Mp = M as right D-modules.For x,y ∈M, we deduce from (145) that hp(x,y) is the negative of the D-component of

(0⊕ x) · (0⊕ y) =(−h(x,y)

)⊕((x×h,∆ y)p−1 p

), (146) PROPER

hence agrees with h(x,y). Thus hp = h.Finally, let a ∈ D×. Then a∆ :

∧3 M ∼→ D is a volume element of M, and 24.8 shows

h(x×h,a∆ y,z) = a∆(x∧ y∧ z) = ah(x×h,∆ y,z) = h((x×h,∆ y)a,z

)for all x,y,z ∈M, which is equivalent to

x×h,a∆ y = (x×h,∆ y)a (x,y ∈M, a ∈ D×). (147) ADEL

Now by Thm. 24.11, x×h,∆ p y is the M-component of (0⊕ x) · (0⊕ y), hence by (146) agreeswith (x×h,∆ y)p−1 p. Since, fixing h, the hermitian vector product uniquely determines the volumeelement it corresponds to, we conclude from this and (147) that ∆ p = pp−1∆ .

Summing up, we have proved

Cp = Ter(D;M,h, pp−1∆). (148) HERP

Finally, we wish to understand what all this means for the algebra C = Zor(k) of Zorn vectormatrices over k. By 24.17, we have

C = Ter(D;D3,h,∆), D = k⊕ k, D3 = k3⊕ k3, h =〉sesq13,

∆(e1∧ e2∧ e3) = 1, ei = ei⊕ ei (1≤ i≤ 3),


where (ei)1≤i≤3 is the canonical basis of unit vectors in k3. The quantity p−1 p is essentially anarbitrary element of D having norm 1, hence can be written in the form γ⊕ γ−1 for some γ ∈ k×

Writing x,y ∈M as x = u1⊕u2, y = v1⊕ v2, we may apply (24.17.3) and obtain

(x×h,∆ p y) = (x×h,∆ y)p−1 p =((u2× v2)⊕ (u1× v1)

)(γ⊕ γ

−1)

=(γ(u2× v2)

)⊕(γ−1(u1× v1)

).

Repeating the compuations of 24.17, we therefore conclude that Zor(k)p is the k-module(k k3

k3 k

)under the multiplication(

α1 u2u1 α2

)(β1 v2v1 β2

)=

(α1β1−ut

2v1 α1v2 +β2u2 + γ−1(u1× v1)β1u1 +α2v1 + γ(u2× v2) α2β2−ut

1v2

)for αi,βi ∈ k, ui,vi ∈ k3, i = 1,2. Since p belongs to the split quaternion subalgebra(

k ke1ke1 k

)⊆ Zor(k),

it follows from Exc. 101 that Zor(k) and Zor(k)p are isomorphic.

114. We proceed in several steps.sol.PRESPLI10. We write (ei)1≤i≤3 for the canonical basis of unit vectors in k3 and put

E :=(

1 00 0

), X1 :=

(0 −e1e1 0

), X2 :=

(0 −e2e2 0

).

Then (24.17.4) implies

X3 := X1X2 =

(0 e3e3 0

), X2

1 =

(1 0,0 1

)= 1C = X2

2 ,

X1X2X1 = X3X1 =

(0 e3e3 0

)(0 −e1e1 0

)=

(0 e2−e2 0

)=−X2.

Moreover, since E is an elementary idempotent, we have

E = 1C−E =

(0 00 1

),

and using (20.3.2), we see XiEXi = nC(Xi, E)Xi−nC(Xi)E which is E for i = 1,2 and−E for i = 3.We have thus established all the relations of (4). Now observe that an easy computation yields

EXi =

(0 −ei0 0

), EX3 =

(0 e30 0

)(i = 1,2),

which shows that the quantities

E, E,X1,X2,X3,EX1,EX2,EX3

form a basis of C as a k-module. In particular, C is generated by E,X1,X2 as a k-algebra.20. Now let A be any unital k-algebra and suppose that e,x1,x2 ∈ A satisfy the relations


e2 = e, x21 = x2

2 = 1A, x1x2x1 =−x2, (149) SMAPR

x1ex1 = x2ex2 = e := 1A− e, (x1x2)e(x1x2) =−e.

Then we put

ε12 := 1, ε21 :=−1, x3 := x1x1. (150) SIGN

Since (149), (150) yield x2x1x2 = x2x1x2x21 = −x2x2x1 = −x and x2x1 = x2

1x2x1 = −x1x2 = −x3,we obtain

ε ji =−εi j, e2 = e, x2i = 1A, xix jxi =−x j, (151) UNIF

xiexi = e, (xix j)e(xix j) =−e, xix j = εi jx3 (i, j= 1,2).

Next we put

B := ke+ ke+3

∑i=1

kxi +3

∑i=1

kyi ⊆ A (yi := exi, 1≤ i≤ 3) (152) BEX

and claim that the spanning elements of B as a k-module displayed in (152) satisfy the followingmultiplication rules.

e2 = e, e2 = e, x2i = 1C, x2

3 =−1C, y2i = y2

3 = 0 (i = 1,2), (153) SQR

xiyi = e, yixi = e, x3y3 =−e, y3x3 =−e (i = 1,2), (154) XYE

ee = 0, exi = yi, eyi = yi (1≤ i≤ 3), (155) ELE

ee = 0, xie = xi− yi, yie = 0 (1≤ i≤ 3), (156) ERI

exi = xi− yi, eyi = 0 (1≤ i≤ 3), (157) BELE

xie = yi, yie = yi (1≤ i≤ 3), (158) BERI

xix j = εi jx3, xix3 = εi jx j, xiy j = εi j(x3− y3), xiy3 = εi j(x j− y j) (i, j= 1,2),(159) XLE

x3xi = − εi jx j, y jxi =−εi j(x3− y3), y3xi =−εi j(x j− y j) (i, j= 1,2),(160) XRI

x3yi = − εi j(x j− y j), yix3 = εi j(x j− y j) (i, j= 1,2),(161) XTHR

yiy j = εi j(x3− y3), yiy3 = εi j(x j− y j), y3yi =−εi j(x j− y j) (i, j= 1,2).(162) YLE

Suppose these relations have been proved. Then they hold, mutatis mutandis, for E,X1,X2 ∈C aswell, and it follows that the linear map C→ A acting on the basis vectors of C exhibited in 10

according to

E 7→ e, E 7→ e, Xi 7→ xi, EXi 7→ yi (1≤ i≤ 3)

is the unique unital homomorphism satisfying the conditions of the problem. The kernel of thishomomorphism is an ideal in C, which, by Exc. 104, has the form aC for some ideal a⊆ k. HenceB∼=C/aC ∼=C⊗ (k/a)∼= Zor(k/a) as k-algebras.30. It remains to establish (153)−(162), which we now proceed to do by making use, among otherthings, of Artin’s theorem (Cor. 15.5) allowing us to drop parentheses in products with at most twodistinct factors.

Proof of (153). The first four relations are clear in view of (151). Moreover, by (149), (150),x2

3 = x1x2x1x2 =−x22 =−1C , y2

i = exiexi = ee = 0 for i = 1,2 and y23 = ex3ex3 =−ee = 0, which

completes the proof.


Proof of (154). For i = 1,2, the definitions and (151) yield xiyi = xiexi = e, yixi = ex2i = e,

x3y3 = x3ey3 =−e, y3x3 = ex23 =−e, as desired.

Proof of (155). The first relation is obvious, the next three hold by definition (cf. (152)), whilethe remaining ones follow from eyi = eexi = exi = yi for i = 1,2,3.

Proof of (156). The first equation is again obvious. As to the next three, we apply (151), (153) toobtain xie= xiex2

i = exi = xi−exi = xi−yi for i= 1,2 and x3e=−x3ex23 = ex3 = x3−ex3 = x3−y3,

as claimed.Proof of (157) We have exi = xi− exi for 1 ≤ i ≤ 3, hence the first three equations of (157),

while (156) yields yie = yi− yie = yi for 1≤ i≤ 3, which completes the proof.Proof of (158). From (156) we deduce xie = xi−xie = yi, yie = yi−yie = yi for 1≤ i≤ 3, and

the assertion follows.Proof of (159). Let i, j = 1,2. The first equation is already in (151). By the same token,

xix3 = εi jxixix j = εi jx j yields the second equation. Applying (151), (157) and the left Moufangidentity, we obtain xiy j = xi(ex j) = εi jxi(e(xix3)) = εi j(xiexi)x3 = εi j ex3 = εi j(x3−y3) and xiy3 =xi(ex3) = εi jxi(e(xix j)) = εi j(xiexi)x j = εi j ex j = εi j(x j− y j), which completes the proof of (159).

Proof of (160). Let i, j = 1,2. Then (159) yields x3xi = εi jxix jxi =−εi jx j , while this, (156),(151), (159), (158) imply y jxi = x jxi − (x je)xi = ε jix3 + εi j((x3xi)e)xi = εi j(x3(xiexi)− x3) =εi j(x3e− x3) = −εi j(x3 − y3). Finally, applying (156), (160), (159), (158), we obtain y3xi =x3xi− (x3e)xi =−εi jx j− ε ji((x jxi)e)xi = εi j(x j(xiexi)− x j) = εi j(x j e− x j) =−εi j(x j− y j).

Proof of (161). Let i, j= 1,2. Then (153), (159), (156), (160) yield

x3yi = − x3(exi)x23 =−[x3(exi)x3]x3 =−[(x3e)(xix3)]x3 =−εi j[(x3e)x j]x3

= − εi jx3x jx3 + εi j[(ex3)x j]x3 = εi jε jixix3 + εi je(x3x jx3) =−εi jx j− εi jε jie(xix3)

= − εi jx j + εi jex j =−εi j(x j− y j),

yix3 = [xi(xie)xi]x3 = xi[(xie)(xix3)] = εi jxi[(xie)x j] = εi jxi[(xi− yi)x j]

= εi j(x2

i x j− xi(yix j))= εi j

(x j + ε jixi(x3− y3)

)= εi jx j− xix3 + xiy3

= εi jx j− εi jx j + εi j(x j− y j) = εi j(x j− y j),

hence (161).Proof of (162). Again, let i, j = 1,2. Then (156), (160), (155) yield yiy j = yix j −

(exi)(x je) = εi j(x3 − y3)− e(xix j)e = εi j(x3 − y3 − ex3e) = εi j(x3 − y3), giving the first equa-tion of (162). Similarly, yiy3 = yix3− (exi)(x3e) = εi j(x j− y j)− e(xix3)e = εi j(x j− y j− ex je) =εi j(x j − y j). And finally, y3yi = y3(xi − xie) = y3xi − (ex3)(xie) = −εi j(x j − y j)− e(x3xi)e =−εi j(x j− y j)+ εi jex je =−εi j(x j− y j), which completes the entire proof.

113. (i)⇒ (ii). Let x,y,z ∈M1: Then (i) impliessol.TERFU

h2(χ(x)×h2,∆2 χ(y),χ(z)

)= ∆2

(χ(x)∧χ(y)∧χ(z)

)=(∆2 (

∧3χ))(x∧ y∧ z)

= ∆1(x∧ y∧ z) = h1(x×h1,∆1 y,z) = h1(χ(x×h1,∆1 )y),χ(z)

).

But h2 is non-singular and χ is bijective. Hence (ii) holds.(ii)⇒ (iii). We put ϕ := 1D⊕χ and use (ii) to compute, for all a,b ∈ D, x,y ∈M1,

ϕ(a⊕ x)ϕ(b⊕ y) =(a⊕χ(x)

)(b⊕χ(y)

)=(

ab−h2(χ(x),χ(y)

))⊕(χ(y)a+χ(x)b+χ(x)×h2,∆2 χ(y)

)=(

ab−h2(χ(x),χ(y)

))⊕(χ(ya+ xb+ x×h1,∆1 y)

)= ϕ

((a⊕ x)(b⊕ y)

).

Thus ϕ is an algebra homomorphism. Being obviously bijective, it satisfies (iii).


(iii)⇒ (i). With ϕ := 1D⊕χ as in the previous paragraph, we first let x ∈M1, b ∈D, apply (iii)and obtain

0⊕χ(xb) = ϕ(0⊕ (xb)

)= ϕ

((0⊕ x)(b⊕0)

)= ϕ(0⊕ x)ϕ(b⊕0)

=(0⊕χ(x)

)(b⊕0) = 0⊕

(χ(x)b

).

Hence χ is D-linear. On the other hand, let x,y ∈M1. Then(−h2

(χ(x),χ(y)

))⊕(χ(x)×h2,∆2 χ(y)

)=(0⊕χ(x)

)(0⊕χ(y)

)= ϕ(0⊕ x)ϕ(0⊕ y)

= ϕ((0⊕ x)(0⊕ y)

)= ϕ

((−h1(x,y)

)⊕ (x×h1,∆1 y)

)=(−h1(x,y)

)⊕χ(x×h1,∆1 y).

Thus h2(χ(x),χ(y)) = h1(x,y) and χ(x×h1,∆1 y) = χ(x)×h2,∆2 χ(y). The first equation shows thatχ : (M1,h1)→ (M2,h2) is an isometry, while the second one amounts to(

∆2 (∧3

χ))(x∧ y∧ z) = ∆2

(χ(x)∧χ(y)∧χ(z)

)= h2

(χ(x)×h2,∆2 χ(y),χ(z)

)= h2

(χ(x×h1,∆1 y),χ(z)

)= h1(x×h1,∆1 y,z) = ∆1(x∧ y∧ z),

hence to ∆2 (∧3

χ) = ∆1 and (i) holds.

Section 25.

115. For the first part of the problem, we begin by assuming that c is absolutely primitive. Thensol.DECAPIDwe have to show that a decomposition (1) with the properties stated thereafter exists. In order to doso, we consider the following cases.Case 1. C has rank r, for some r = 1,2,4,8.Case 1.1. r = 1. Then C = k, and ε := c ∈ k is an absolutely primitive idempotent. By Prop. 25.4,The ideal kε ⊆ k is free of rank one with basis ε as a k-module. Hence the equation (1− ε)ε = 0implies c = ε = 1. But then the assertion follows by setting k(1) := 0, k(2) := k, C(1) := 0,C(2) := k, c1 := 0, c2 := 1 = c.Case 1.2. r > 1. Then Lemma ?? shows that the idempotent c ∈ C is elementary, so we havenC(c) = 0, tC(c) = 1. Employing the language of Exc. 84, we therefore conclude ε(0) = ε(2) = 0,ε(1) = 1. This gives k(0) = k(2) = 0, k(1) = k, C(0) =C(2) = 0, C(1) =C, and the decompositionC =C(1)⊕C(2), c = c(1)⊕1k(2) yields the assertion.Case 2. C is arbitrary. Since the rank function of C is locally constant (see 10.4), there exists adecomposition of X := Spec(k) into the disjoint union of finitely many non-empty open subsetsUi ⊆ X , 1≤ i≤ n, such that the rank of C is constant on each Ui for 1≤ i≤ n. Here Exc. 41 yieldsa complete orthogonal system (εi)1≤i≤n of n on-zero idempotents in k such that Ui = D(εi) =Spec(kε ) for 1≤ i≤ n. Observing the beginning of the solution to Exc. 84, we also obtain a naturalidentification Ci := Cεi = C⊗ kεi = εiC as kεi-modules. Applying Case 1 to this set-up, we nowfind, for 1 ≤ i ≤ n, a decomposition kεi = (kεi)

(1) ⊕ (kεi)(2) as a direct sum of ideals, as well

as a composition algebra C(1)i over (kεi)

(1) and an elementary idempotent c(1)i ∈ C(1)i such that

Ci =C(1)i ⊕ (kεi)

(2) and ci := εic = c(1)i ⊕1(2)kεi. Now it suffices to put

k( j) :=n

∑i=1

(kεi)( j), C(1) :=

n

∑i=1

C(1)i , c(1) :=

n

∑i=1

c(1)i ( j = 1,2)

and to note that, since the c(1)i ∈C(1)i are elementary idempotents for 1≤ i≤ n, so is c(1) ∈C(1).


Conversely suppose we have a decomposition (1) with the properties stated thereafter, parti-cularly (2). Then c(1) ∈ C(1), being an elementary idempotent, is an absolutely primitive one byProp. 25.6, while the idempotent 1k(2) ∈ k(2) is absolutely primitive for trivial reasons. In order toshow that c ∈C is absolutely primitive, it will therefore be enough to establish the followingClaim. Let k = k1⊕k2 be a direct sum of ideals, Ai a ki-algebra and ci ∈ Ai an absolutely primitiveidempotent for i = 1,2. Viewing A := A1⊕A2 as a k-algebra in a natural way, c := c1⊕ c2 is anabsolutely primitive idempotent in A.The straightforward verification of this claim is left to the reader.

It remains to show that the rank of C(1) is nowhere equal to 1. Otherwise we would haveC(1)p = k(1)p for some prime ideal p⊆ k, whence Case 1.1 above would imply that

c(1)p = 1k(1)p

is be an elementary idempotent of k(1)p , a contradiction.

116. (a) (i)⇒ (ii). We view c canonically as an idempotent linear map c : k⊕L0→ k⊕L0. Withsol.ELIDREDQUATLc := Im(c), Lc := Ker(c) = Lc, we then have

k⊕L0 = Lc⊕ Lc = Lc⊕Lc (163) IDEC

as a directi sum of submodules. Hence Lc and Lc are both finitely generated projective k-modules.For p ∈ Spec(k), we have cp 6= 0 6= cp since c is elementary, so (Lc)p = Lcp and (Lc)p = Lcpare both free kp-modules of positive rank. On the other hand, (163) yields rkp(Lc)+ rkp(Lc) = 2,which altogether implies rkp(Lc) = rkp(Lc) = 1. Hence Lc is a line bundle, as claimed.

(ii)⇒ (iii). Regardless of whether the idempotent c is elementary or not, (163) holds, forcingLc and Lc to be finitely generated projective k-modules. NOw suppose as in (ii) that Lc is a linebundle. Then [12, III, § 7, Cor. of Prop. 10] implies

L0 ∼= k⊗L0 ∼=∧2

(k⊕L0)∼=∧2

(Lc⊕Lc)∼= Lc⊗Lc.

Thus Lc ∼= L0⊗L∗c is a line bundle as well, and we have established (3). At this stage, we requirethe following well knownFact. Let M,N be finitely generated projective k-modules. Then there is a natural identificationHomk(M,N) = N⊗M∗ such that (y⊗ x∗)(x′) = (〈x∗,x′〉y for all x′ ∈M, y ∈ N, x∗ ∈M∗.Using this and (3), we now obtain a chain of natural isomorphism as follows:

B∼= Endk(k⊕L0)∼= Endk(lc⊕Lc)∼= Endk

(LcLc

)∼=(

Homk(Lc,Lc) Homk(Lc,Lc)Homk(Lc,Lc) Homk(Lc,Lc)

)∼=(

Lc⊗L∗c Lc⊗L∗cLc⊗L∗c Lc⊗L∗c

)∼=(

k L∗0⊗L⊗2c

L0⊗L∗⊗2c

)∼= B′ := Endk(k⊕L)

with L := L0⊗L∗⊗2c an appropriate line bundle over k. Writing Φ : B→ B′ for the composite of

these isomorphisms, and following step-by-step their effect on the idempotent c, we conclude

Φ(c) =(

1 00 0

), (164) PHICE

which completes the proof of (iii). Now suppose L is any line bundle over k such that there existsan isomorphism Φ : B→ B′ := Endk(k⊕L) satisfying (164). Then L∼= B′21(Φ(c))∼= B21(c), L∗ ∼=


B′12(Φ(c))∼= B12(c), which not only shows that L is unique up to isomorphism but also completesthe proof of (4).

(iii)⇒ (i). Since Φ(c) is obviously elementary, so is c.(b) Let g ∈ GL(k⊕L0) = B× such that d = gcg−1. Then

Ld = d(

kL0

)= gcg−1

(k

L0

)= gc

(k

L0

)= g(Lc).

Hence g determines via restriction an isomorphism Lc∼→ Ld . Conversely, suppose Lc and Ld are

isomorphic. Then, by (3), so are Lc and Ld . Let

ρ : Lc∼−→ Ld , ρ : Lc

∼−→ Ld

be isomorphisms. Then so is

g := ρ⊕ ρ : k⊕L0 = Lc⊕Lc∼−→ Ld ⊕Ld = k⊕L0,

which amounts to g ∈ GL(k⊕ L0) = B×. Moreover, for x ∈ Ld , y ∈ Ld we have ρ−1(x) ∈ Lc,ρ−1(y) ∈ Lc, hence

gcg−1(x+ y) = gc(ρ−1(x)+ ρ

−1(y))= (ρ⊕ ρ)

(ρ−1(x)

)= x.

But this means gcg−1 = d, and c,c are conjugate under inner automorphisms of B. Now supposeϕ(c) = d for some automorphism ϕ of B. Then B12(c)∼= B12(d), which by (4) implies L⊗2

c∼= L⊗2

d .Conversely, let this be so and put L := L0⊗L∗⊗2

c . Then (iii) yields isomorphisms Φ : B ∼→ B′ :=Endk(k⊕L), Ψ : B ∼→ B′ such that

Φ(c) =(

1 00 0

)=Ψ(d).

Hence ϕ :=Ψ−1 Φ is an automorphism of B sending c to d.(c) By (3), the line bundle Lc is a direct summand of k⊕ L0. Conversely, let L be a direct

summand of k⊕L0, so k⊕L0 = L⊕L′ for some submodule L′ ⊆ k⊕L0. Then the projection fromk⊕L0 onto L alongside L′ yields an idempotent c ∈ B such that L = Lc, and by (a), c is elementary.

117 . (a) Let xi/1 be a basis of L fi over k fi , for i = 1, . . . ,n. Given x ∈ L, there exist m ∈ N andsol.LITWOα1, . . . ,αn ∈ k such that x/1 = (αi/ f m

i )(xi/1) = (αixi)/ f mi in L fi for all i = 1, . . . ,n. Hence for

some integer p≥m and all i = 1, . . . ,n, f pi x = βixi, βi := f p−m

i αi. Since the f p1 , . . . , f p

n continue togenerate k as an ideal, we find g1, . . . ,gn ∈ k such that ∑ f p

i gi = 1. But this implies x = ∑βigixi, soL = ∑kxi is generated by x1, . . . ,xn.

(b) (i)⇒ (ii). Since L is generated by two elements, we obtain a short exact sequence

s

0 // L′ // k⊕ k // L // 0

of k-modules, which splits since L is projective. Thus L⊕L′ ∼= k⊕ k is free of rank 2, and takingdeterminants (= second exterior powers), we obtain L⊗L′ ∼= k, i.e., L′ ∼= L∗.

(ii)⇒ (iii). This is Exc. 116 for L0 := k.(iii)⇒ (iv). Write

c =(

α β

γ δ

)with α,β ,γ,δ ∈ k. Since c has trace 1 and determinant 0, we have

α +δ = 1, αδ = βγ. (165) EQEL


From Exc. 116 we deduce

L∼= Lc = Im(c) = k(

α

γ

)+ k(

β

δ

).

Observing

β

(α

γ

)= α

(β

δ

), δ

(α

γ

)= γ

(β

δ

), (166) COLELIDMAT

we now put f1 = α , f2 = δ . Then k f1 + k f2 = k by (165), and (166) shows that L f1 is the free k f1 -

module of rank 1 with basis(

α/1β/1

), while L f2 is the free k f2 -module of rank 1 with basis

(β/1δ/1

).

Thus (iv) holds.(iv)⇒ (i). This is just a special cse of (a).Now suppose L satisfies one (hence all) of the preceding four conditions. Then, by (ii), so does

L∗, allowing us to assume n > 0. But then (iv) holds for L⊗n. In partiuclar, by (ii), an elemen-tary idempotent c(n) ∈ Mat2(k) satisfying L⊗n ∼= Lc(n) exists for any n ∈ Z and is unique up toconjugation by inner automorphisms (Exc. 116 (b)). Since L⊗0 ∼= k is free of rank 1, we may putc(0) =

(1 00 0

). Moreover, combining (ii) with (3) in Exc. 116, we may also put c(−1) := c = 12− c,

and it will be enough to treat the case n > 0. Writing c as in the proof of the implication (iii)⇒(iv),we claim that

αnαn +δ

nδn = 1, (167) TREN

where

αn :=n−1

∑i=0

(2n−1

i

)α

n−1−iδ

i, δn :=n−1

∑i=0

(2n−1n+ i

)α

n−1−iδ

i,

and

c〈n〉 :=(

αnαn β nδnγnαn δ nδn

)∈Mat2(k) (168) CEEN

is an elementary idempotent satisfying Lc〈n〉∼= L⊗n

c .We begin by proving (167), which follows from the computation

αnαn +δ

nδn =

n−1

∑i=0

(2n−1

i

)α

2n−1−iδ

i +n−1

∑i=0

(2n−1n+ i

)α

n−1−iδ

n+i

=n−1

∑i=0

(2n−1

i

)α

2n−1−iδ

i +2n−1

∑j=n

(2n−1

j

)α

2n−1− jδ

j

=2n−1

∑i=0

(2n−1

i

)α

2n−1−iδ

i = (α +δ )2n−1 = 1.

Combining (167) with (166), we conclude that c(n) as defined in (168) has trace 1 and determinant0, hence is an elementary idempotent. Moreover,

βn(

αn

γn

)= α

n(

β n

δ n

), δ

n(

αn

γn

)= γ

n(

β n

δ n

). (169) COLEN

We now put

f (n) := αnαn, g(n) := δ

nδn (170) EFEN


and have f (n)+g(n) = 1 by (167). Setting L := Lc, Lc(n) , we must show L⊗n ∼= L(n). To this end, weexamine the situation of k f (n) and kg(n) .

Over k f (n) , the quantities α,αn are both invertible. By (166), therefore, L f (n) is the free k f (n) -module with basis

(αγ

). Consequently, (L⊗n) f (n) = (L f (n) )

⊗n is the free k f (n) -module with basis(αγ

)⊗n. By the same token, (L(n)) f (n) is the free k f (n) -module with basis(

αnαnγnαn

)=(

αn

γn

)αn, which

is the same as the free k f (n) -module with basis(

αn

γn

). Summing up, we find a unique isomorphism

Φ(n) : (L⊗n) f (n)

∼−→ (L(n)) f (n) such that Φ(n)((

α

γ

)⊗n)=

(αn

γn

).

Similarly, by (170), the quantities δ ,δ (n) are invertible over kg(n) , and ther e is a unique isomor-phism

Ψ(n) : (L⊗n)g(n)

∼−→ (L(n))g(n) such that Ψ(n)((

β

δ

)⊗n)=

(β n

δ n

).

Over k f (n)g(n) , all the quantities α,αn,δ ,δn,β ,γ become invertible, and we have(

β

δ

)= γ−1δ

(αγ

),

hence(

β

γ

)⊗n= γ−nδ n

(αγ

)⊗n, but also(

β n

δ n

)= γ−nδ n

(αn

γn

). Hence the isomorphisms Φ (n), Ψ (n)

agree over k f (n)g(n) and thus glue to an isomorphism L⊗n ∼→ L(n). This completes the proof.

Remark. The final proof of this exercise would have become much more natural if we had usedthe identification of Pic(k) with H1

Zar(k,GL1).

118. (i)⇒ (ii). Let L be any line bundle over k that is a direct summand of k⊕L0. By Exc. 116 (c),sol.ISOREDQUATthere exists an elementary idempotent c ∈ B such that L ∼= Lc. Using (i) to select an isomorphismΦ : B ∼→ B′, we obtain an elementary idempotent c′ := Φ(c) ∈ B′ and, again by Exc. 116, we findin L′ := Lc′ a line bundle over k that is a direct summand of k⊕L′0. Now (4) of Exc. 116

L∗0⊗L⊗2 ∼= L∗0⊗L⊗2c∼= B12(c)∼= B′12(c

′)∼= L′∗0 ⊗L′⊗2,

and (5) holds.(ii) ⇒ (iii). This is clear since line bundles L over k satisfying the requirements of (ii) exist,

e.g., L := k⊕0 ⊆ k⊕L0.(iii)⇒ (i). Let L, L′ be line bundles over k that are direct summands of k⊕L0, k⊕L′0, respec-

tively, and satisfy (5). By Exc. 116 (c), there exist elementary idempotents c ∈ B, c′ ∈ B′ such thatL∼= Lc, L′ ∼= Lc′ . Now (5) yields

M := L∗0⊗L⊗2c∼= L′∗0 ⊗L⊗2

c′ ,

and Exc. 116 (a) produces isomorphisms Φ : B ∼→C := Endk(k⊕M), Φ ′ : B′ ∼→C sending c,c′ to(1 00 0

), respectively. In particular, Φ ′−1 Φ is an isomorphism from B to B′, and (i) holds.

119. We apply Exc. 118 to B := Endk(k⊕L0), L0 := L, and B′ := Mat2(k) = Endk(k⊕L′0), L′0 := k.sol.FQUATSPLThen M := k⊕0 ⊆ k⊕L0 is a free submodule of rank 1 and a direct summand at the same time.If B is split, then B ∼= B′ and condition (ii) of Exc. 118 yields a line bundle M′ ⊆ k⊕L′0 = k⊕ kwhich is a direct summand of k⊕ k and satisfies L0⊗M′⊗2 ∼= L′0⊗M⊗2. This means L0 ∼= M′∗⊗2.Moreover, k⊕ k ∼= M′⊕M′′ for some line bundle M′′, and taking determinants (= seond exteriorpowers), we obtain M′ ⊗M′′ ∼= k. Hence M′∗ ∼= M′′, being a homomorphic image of k⊕ k, isgenerated by two elements.

Conversely, suppose L0 = L ∼= M′⊗2 for some line bundle M′ on two generators over k. ThenExc. 117 (b) implies M′⊕M′∗ ∼= k⊕k. Since L0⊗M′∗⊗2 ∼= L⊗M′∗⊗2 ∼= k∼= L′0⊗M⊗2, we deducefrom Exc. 118 that B and B′ = Mat2(k) are isomorphic, i.e., B is split.

Finally, suppose we are given a line bundle L on two generators over k that is not a square inPic(k). By what we have just shown, the quaternion algebra


B := Endk(k⊕L) =(

k L∗

L k

)is reduced but not split. It will be free as a k-module once we have shown that L⊕ L∗ is a freek-module. But this follows immediately from Exc. 117 (b).

120. By Exc. 119, the quaternion algebrasol.FAIWITCAN

B := Endk(k⊕L′) =(

k L′∗

L′ k

)is split since L′ is the square (in Pic(k)) of some line bundle on two generators over k. Computingthe norm of B in two ways by means of (25.9.1), we deduce h ⊥ h ∼= nB ∼= h ⊥ (−hL′ )∼= h ⊥ hL′

and hence obtain the first assertion. As to the second, it suffices to consult Exc. 47, which showsthat the hyperbolic plane hL′ is not split.

121. Localizing if necessary, we may assume that M is free (of rank 3). Let (ei)1≤i≤3 be a basis ofsol.GRASSSPLM and (e∗i )1≤i≤3 the corresponding dual basis of M∗. Setting α := θ(e1∧e2∧e3), β := θ ∗−1(e∗1∧e∗2∧ e∗3), we apply (25.11.1) to conclude αβ = 1. In particular, α and β are both invertible.

Letting (i jl) vary over the cyclic permutations of (123) and s = 1,2,3, we now apply (25.11.2)to obtain 〈ei×θ e j,es〉= θ(ei∧e j∧es)=αδls = 〈αe∗l ,es〉 and 〈e∗s ,e∗i ×θ e∗j〉= θ ∗−1(e∗i ∧e∗j ∧e∗s )=βδls = 〈βe∗l ,es〉, which amounts to

ei×θ e j = αe∗l , e∗i ×θ e∗j = βel . (171) TWIBA

Now observe that our asserted equations are alternating in u,v (resp. u∗,v∗). For the first equation,we may therefore assume u = ei, v = e j , w∗ = e∗s . Then (171) yields

(ei×θ e j)×θ e∗s = αe∗l ×θ e∗s =

αβe j for s = i,−αβei for s = j,0 for s = l

=

e j for s = i,−ei for s = j,0 for s = l

= 〈e∗s ,ei〉e j−〈e∗s ,e j〉ei,

as claimed. Similarly, for the second equation, we may assume u∗ = e∗i , v∗ = e∗j , w = es and obtain

(e∗i ×θ e∗j)×θ es = βel ×θ es =

αβe∗j for s = i,−αβe∗i for s = j,0 for s = l

=

e∗j for s = i,−e∗i for s = j,0 for s = l

= 〈e∗i ,es〉e∗j −〈e∗j ,es〉e∗i .

This completes the proof.

122. Let D := k⊕ k be the split quadratic etale k-algebra, (M,h) a ternary hermitian space oversol.TEREMD, ∆ :

∧3(M)∼→ D a volume element of M and C := Ter(D,M,h,∆) the octonion algebra over

k arising from these data by means of the ternary hermitian construction. We wish to relate C toan octonion algebra of appropriately defined twisted Zorn vector matrices. In order to do so, we


perform the following steps.10. We put c1 := 1⊕0 ∈ D, c2 := c1 = 0⊕1 ∈ D and have

D = kc1⊕ kc2

as a direct sum of ideals. Moreover, kci = Dci for i = 1,2 are D-algebras in a natural way, andkci ∼= k as k-algebras.20. From 10 and what we have seen in the solution to Exc. 84. we deduce

M = M1⊕M2, Mi := M⊗D (kci) = Mci (i = 1,2) (172) SPLIM

as a direct sum of k-submodules. In obvious notation, we now claim∧3

D(M) =

∧3

k(M1)⊕

∧3

k(M2) (173) SPLIDE

as right D-modules, where D acts diagonally on the right-hand side. Indeed, since taking exteriorpowers is compatible with base change [12, III, § 7, Prop. 8], we apply (172) and obtain∧3

k(Mi) =

∧kci

(M⊗D (kci)

)=∧3

D(M)⊗D (kci) =

∧3

D(M)ci (i = 1,2), (174) SPLIWE

hence the assertion. Note by (172) that Mi for i = 1,2 is a finitely generated projective k-module ofrank 3.30. We now examine how the hermitian form h behaves with respect to the decomposition (172).For i, j = 1,2, xi ∈Mi, y j ∈M j , the equation

h(xi,y j) = h(xici,y jc j) = cih(xi,y j)c j = h(xi,y j)(1− ci)c j = (1−δi j)h(xi,y j)c j

amounts to

h(Mi,Mi) = 0, h(M3−i,Mi)⊆ kci (i = 1,2). (175) HASPLI

In particular, there exists a k-bilinear form β : M1 ×M2 → k such that h(x1,y2) = β (x1,y2)c2,hence

h(y2,x1) = h(x1,y2) = β (x1,y2)c2 = β (x1,y2)c2 = β (x1,y2)c1,

for all x1 ∈M1, y2 ∈M2. Summing up, and consulting (175)again, we conclude

h(x1⊕ y1,x2⊕ y2) = β (x2,y1)c1 +β (x1,y2)c2 = β (x2,y1)⊕β (x1,y2) (176) HBSPLI

for xi ∈M1, yi ∈M2, i = 1,2.40. Suppose y2 ∈ M2 satisfies β (M1,y2) = 0. Then (176) yields h(x1⊕ y1,0⊕ y2) = 0 for allx1 ∈M1, y1 ∈M2, hence y2 = 0. On the other hand, let λ : M1→ k be any k-linear form. Then

λ ⊕0: M = M1⊕M2 −→ k⊕ k = D

is a D-linear form, and non-singularity of h yields unique elements x1 ∈ M1, y1 ∈ M2 such that(λ ⊕ 0)(x2 ⊕ y2) = h(x1 ⊕ y1,x2 ⊕ y2) for all x2 ∈ M1, y2 ∈ M2. By (176) this means λ (x2) =β (x2,y1) for all x2 ∈M1, and we have shown that the natural map

M2∼−→M∗1 , y 7−→ β (−,y),

induced by β is an isomorphism. Since the ternary hermitian construction is functorial (Exc. 113),we may identify M2 = M∗1 , and then have β (x,y∗) = 〈y∗,x〉 for all x ∈M1, y∗ ∈M∗1 . Hence (172)and(176) imply


M = M1⊕M∗1 , h(x1⊕ y∗1,x2⊕ y∗2) = 〈y∗1,x2〉⊕〈y∗2,x1〉 (xi ∈M1,y∗i ∈M∗1 , i = 1,2). (177) HADU

Incidentally, it is straightforward to check that, conversely, the right-hand side of (177) determinesa non-singular hermitian form on the right D-module M1⊕M∗1 .50. We now turn to the volume element ∆ of M. By (173), we obtain induced volume elementsθ :

∧3(M1)∼→ k, θ ′ :

∧(M∗1 )

∼→ k such that

∆((x1⊕ y∗1)∧ (x2⊕ y∗2)∧ (x3⊕ y∗3)

)= θ(x1∧ x2∧ x3)⊕θ

′(y∗1∧ y∗2∧ y∗3) (178) SPLIDEL

for all xi ∈M1, y∗i ∈M∗1 , i = 1,2,3. Note that∧3(M∗1 ) = (

∧3(M1))∗ under the identification 24.6.

Since the ∆ -determinant of h is 1, consulting 24.7 and (24.4.1) yields

θ(x1∧ x2∧ x3)θ′(v∗1∧ v∗2∧ v∗3)⊕θ(u1∧u2∧u3)θ

′(y∗1∧ y∗2∧ y∗3)

=(θ′(v∗1∧ v∗2∧ v∗3)⊕θ(u1∧u2∧u3)

)(θ(x1∧ x2∧ x3)⊕θ

′(y∗1∧ y∗2∧ y∗3))

=(θ(u1∧u2∧u3)⊕θ ′(v∗1∧ v∗2∧ v∗3)

)(θ(x1∧ x2∧ x3)⊕θ

′(y∗1∧ y∗2∧ y∗3))

= ∆((u1⊕ v∗1)∧ (u2⊕ v∗2)∧ (u3⊕ v∗3)

)∆((x1⊕ y∗1)∧ (x2⊕ y∗2)∧ (x3⊕ y∗3)

)=(∧3

h)((u1⊕ v∗1)∧ (u2⊕ v∗2)∧ (u3⊕ v∗3),(x1⊕ y∗1)∧ (x2⊕ y∗2)∧ (x3⊕ y∗3)

)= det

((h(ui⊕ v∗i ,x j⊕ y∗j)

)1≤i, j≤3

)= det

((〈v∗i ,x j〉⊕〈y∗j ,ui〉

)1≤i, j≤3

)= det

((〈v∗i ,x j〉)1≤i, j≤3

)⊕det

((〈y∗j ,ui〉)1≤i, j≤3

).

Comparing the first components of both sides, we conclude θ(x1 ∧ x2 ∧ x3)θ′(v∗1 ∧ v∗2 ∧ v∗3) =

det((〈v∗i ,x j〉)1≤i, j≤3), which in view of (25.11.1) shows θ ′ = θ ∗−1.Thus, taking into account (178)and the identification (173), we finally end up with

∆ = θ ⊕θ∗−1. (179) DELTH

60. The preceding formula will enable us to decompose the vector product with respect to h,∆relative to the splitting (177). In order to do so, we let xi ∈M1, y∗i ∈M∗1 for i = 1,2,3 and defineu ∈M1, v∗ ∈M∗1 by

(x1⊕ y∗1)×h,∆ (x2⊕ y∗2) = u⊕ v∗.

Then . . . yield

θ(x1∧ x2∧ x3)⊕θ∗−1(y∗1∧ y∗2∧ y∗3) = ∆

((x1⊕ y∗1)∧ (x2⊕ y∗2)∧ (x3⊕ y∗3)

)= h((x1⊕ y∗1)×h,∆ (x2⊕ y∗2),x3⊕ y∗3

)= h(u⊕ v∗,x3⊕ y∗3) = 〈v∗,x3〉⊕〈y∗3,u〉.

By . . . , therefore,

〈v∗,x3〉= θ(x1∧ x2∧ x3) = 〈x1×θ x2,x3〉, 〈y∗3,u〉= θ∗−1(y∗1∧ y∗2∧ y∗3) = 〈y∗3,y∗1×θ y∗2〉,

and since x3,y∗3 are arbitrary, we deduce the final decomposition formula

(x1⊕ y∗1)×h,∆ (x2⊕ y∗2) = (y∗1×θ y∗2)⊕ (x1×θ x2) (xi ∈M1,y∗i ∈M∗1 , i = 1,2). (180) SPLIVE

70. Finally, let αi,βi ∈ k, xi ∈M1, y∗i ∈M∗1 for i = 1,2. Applying . . . , we obtain

37 Elementary idempotents and cubic Jordan matrix algebras 403((α1⊕β1)⊕ (x1⊕ y∗1)

)((α2⊕β2)⊕ (x2⊕ y∗2)

)=[(α1⊕β1)(α2⊕β2)−h(x1⊕ y∗1,x2⊕ y∗2)

]+[(x1⊕ y∗1)(α2⊕β2)+(x2⊕ y∗2)α1⊕β1 +(x1⊕ y∗1)×h,∆ (x2⊕ y∗2)

]=[(α1α2−〈y∗1,x2〉)⊕ (β1β2−〈y∗2,x1〉)

]+[(x1α2 + x2β1 + y∗1×θ y∗2)⊕ (y∗1β2 + y∗2α1 + x1×θ x2)

].

Comparing this with (25.11.3), we see that the map

Φ : Ter(k⊕ k,M,h,∆)∼−→ Zor(M1,θ)

defined by

Φ((α⊕β )⊕ (x⊕ y∗)

)=

(α y∗

x β

)for α,β ∈ k, x ∈M1, y∗ ∈M∗1 is an isomorphism of octonion algebras. This done, it is now clearthat and how Thm. 25.12 follows immediately from Thm. 24.13.

123 . Let C be a reduced octonion algebra over the Dedekind domain k. By Thm. 25.12, theresol.OCTDEDDOMexist a finitely generated projective k-module M of rank 3 and a volume element θ of M suchthat C ∼= Zor(M,θ) in the sense of 25.11. By [65, III, Exc. 13 (c)], we can find an ideal a ⊆ ksatisfying M ∼= k2⊕a. But

∧3(M)∼= k by means of θ . By [12, III, § 7, Cor. of Prop. 10], therefore,∧2(k2)⊗a∼= k⊗a∼= a is free of rank 1, forcing M to be free of rank 3. This yields identificationsM = M∗ = k3 that match 〈y∗,x〉 with xt y∗ for x,y∗ ∈ k3 and then lead to an isomorphism C ∼=Zor(M,θ)∼= Zor(k). Thus C is split.

124. For an ideal a⊆ k, I := aC ⊆C is even a two-sided ideal. Conversely, passing from C to Copsol.DEDOC

if necessary, it will be enough to show that any right ideal I ⊆C has the form as indicated. To thisend, we consider the following cases.Case 1. k = F is a field. Then we must show I = 0 or I = C. If I contains invertible elements,then, clearly, I =C. Otherwise, since C is conic alternative, Prop. 20.4 shows that I ⊆C is a totallyisotropic subspace relative to nC .Case 1.1. tC vanishes identically on I. Then (19.5.4) shows I = I, so the conjugation of C being aninvolution implies that I ⊆C is, in fact, a two-sided ideal. But then I = 0 by Exc. 104.Case 1.2. tC 6= 0 on I. Then some c ∈ I satisfies tC(c) = 1, nC(c) = 0. Thus c ∈C is an elementaryidempotent, and since we are working over a field, Thm 25.12 yields an identification C = Zor(k)such that

c =(

1 00 0

).

Write Ci j , i, j = 1,2, for the Peirce components of C relative to c. Then C11 = kc⊆ I, C12 = cC12 ∈ I,which implies

C21 =

(0 0k3 0

)=

(0 0

k3× k3 0

)=

(0 k3

0 0

)(0 k3

0 0

)=C12C12 ⊆ I.

But now c ∈C22 =C21C12 ∈ I, hence 1C = c+ c ∈ I, and we conclude I =C, as desired.Case 2. k = o is a discrete valuation ring. We write p ⊆ o for the corresponding valuation idealand κ = o/p for the corresponding residue field. We may assume I 6= 0 and must show I = C.Since

⋂i≥0 p

i = 0 and C is a free o-module of finite rank, we conclude⋂

i≥0 piC = 0 as well.

Thus there exists a natural number r which is maximal with respect to the property I ⊆ prC. Inparticular, I′ := p−rI ⊆ o is a right ideal, and the assumption I′ ⊆ pC would imply the contradictionI = prI′ ⊆ pr+1C. Hence I′ * pC. Consequently, (I′ + pC)/pC is a non-zero right ideal in theoctonion algebra C(p) over the field κ , which by Case 1 must be all of C(p). Thus I′+pC =C, andNakayama’s lemma [65, X, Lemma 4.2] implies I′ =C, hence I = prI′ = prC.Case 3. The general case. By Exc. 104 it will be enough to show that I ⊆C is, in fact a two-sided


ideal, equivalently, that Ip ⊆Cp is a two-sided ideal for any prime ideal p ⊆ k. In order to do so,we may assume Ip 6= 0, in which case the assertion follows immediately from Case 2.

Section 26

125. The cases s = 1 and s = r are obvious. Hence we may assume 1 < s < r, which by Cor. 22.17sol.SKONOCOimplies that C,B,B′ are all non-singular. Arguing by induction on r− s, we let ϕ : B ∼→ B′ be anisomorphism. Then C = B⊥ B⊥ = B′ ⊥ B′⊥, hence

nB ⊥ nC|B⊥ ∼= nC ∼= nB′ ⊥ nC|B′⊥

On the other hand, ϕ is an isometry from nB onto nB′ . Thus, by Witt cancellation (Cor. 26.6), nC|B⊥and nC|B′⊥ are isometric. Since s < r, we have Supp(B⊥) = Spec(k), and applying Lemma 22.14,we find an element l ∈ B⊥ such that nC(l) ∈ k×, which in turn leads to an element l′ ∈ B′⊥ satis-fying nC(l′) = nC(l) =:−µ . Write B1 (resp. B′1) for the subalgebra of C generated by B (resp. B′)and l (resp. l′). Then Cor. 21.8 yields unique isomorphisms h : Cay(B,µ) = B⊕B j ∼→ B1 (resp.h′ : Cay(B′,µ) = B′⊕B′ j′ ∼→ B′1) extending the identity of B (resp. B′) and sending j (resp. j′) tol (resp. l′). The isomorphisms thus obtained fit into the diagram

Cay(B,µ)h∼=//

Cay(B,ϕ) ∼=

B1

ϕ1∼=

Cay(B′,µ)h′

∼= // B′1,

which can be completed uniquely to a commutative square by the dotted isomorphism ϕ1 : B1∼→

B′1 as indicated. Since B1 and B′1 both have rank 2s, the induction hypothesis applies to ϕ1 andcompletes the proof. To conclude the solution to the problem, let B⊆O be a non-zero subalgebraand let 0 6= x ∈ O. Then nO(x) 6= 0 and B contains the element x2 = tO(x)x− nO(x)1O, hence1O. Thus B is a unital subalgebra of O on which nO continues permit composition and to beanisotropic, hence non-singular as well since the characteristic is not 2. Summing up, therefore,B⊆O is a composition division subalgebra. Form 26.11 we now conclude that B is isomorphic toone of the algebras R,C,H,O, and the Skolem-Noether theorem completes the proof.

126. Since k is an integral domain, Exc. 94 yields a non-zero element u∈C such that nC(u)= 0. Onsol.COMPRIthe other hand, k being a PID, C is free of finite rank as a k-module. Letting (ei)1≤i≤r be a k-basisof C, we write u = ∑

ri=1 αiei for some α1, . . . ,αr ∈ k not all zero. In fact, dividing by their greatest

common divisor, we may assume that they are mutually prime, so we can find β1, . . . ,βr ∈ k suchthat ∑

ri=1 αiβi = 1. Since C has rank > 1, it is non-singular, allowing us to consider the DnC-dual

basis ( fi)1≤i≤r of C relative to (ei). Setting v := ∑ri=1 βi fi ∈ C, we therefore obtain nC(u,v) = 1.

Thus c := uv satisfies tC(c) = 1, nC(c) = nC(u)nC(v) = 0 and hence is an elementary idempotent.By Prop. 25.6, our composition algebra C is therefore reduced. By Cor. 25.8, it is split if the rankis 2. But by Prop. 25.10 and Thm. 25.12, the same conclusion holds also in ranks 4 and 8 sincefinitely generated projective k-modules are free.

127. By Prop. 26.2 (b), f preserves not only norms and units, but also traces and conjugations.sol.UNINOR(a) Applying (20.3.2), we obtain

f (Uxy) = nC(x, y) f (x)−nC(x) f (y) = nC(

f (x), f (y))

f (x)−nC(

f (x))

f (y) =U f (x) f (y),

hence (1). In order to prove (2), we first note

x2 =Ux1C,(Ux+y−Ux−Uy

)z = x(yz)+ z(yx),


so by (1), f preserves squares and the expression x(yz)+ z(yx). Abbreviating the left-hand side of(2) by A and using the fact that the associator of C is alternating, we can now compute

A = f (xy)2− f (xy)(

f (y) f (x))−(

f (x) f (y))

f (xy)+ f (x) f (y)2 f (x)

= f((xy)2)− f (xy)

(f (y) f (x)

)− f (x)

(f (y) f (xy)

)− [ f (x), f (y), f (xy)]+ f (xy2x)

= [ f (x), f (xy), f (y)]+ f((xy)2− (xy)(yx)− x

(y(xy)

)+ xy2x

)= [ f (x), f (xy), f (y)]

since, thanks to Artin’s Theorem (Cor. 15.5), the subalgebra of C generated by two elements isassociative. This completes the proof of (2). In order to complete the proof of (a), it will thereforesuffice to show that the right-hand side of (2) is symmetric in x,y. To see this, we note that, sincesquares are preserved by f , so is the circle product. Hence f (x y) = f (x) f (y), which implies

[ f (y), f (yx), f (x)] = [ f (y), f (x y), f (x)]− [ f (y), f (xy), f (x)]

= [ f (y), f (x) f (y), f (y)]+ [ f (x), f (xy), f (y)] = [ f (x), f (xy), f (y)],

again by Artin’s Theorem, and the assertion follows.(b) Since f preserves norms and traces, we have tC′ (c′) = tC′ ( f (c)) = tC(c) = 1, nC′ (c′) =

nC′ ( f (c)) = nC(c) = 0, so c′ ∈ C′ is an elementary idempotent. Now, as we have seen in (a), fpreserves the circle product: f (xy) = f (x) f (y) for all x,y∈C. In particular, f (cy) = c′ f (y).Hence, writing Ci j , i, j = 1,2, for the Peirce components of C relative to c, it will be enough toshow

C12 +C21 = x ∈C | c x = x. (181) CIROFF

For x ∈C, we let x = x11 + x12 + x21 + x22 be its Peirce decomposition relative to c. Then

c x = cx+ xc = x11 + x12 + x11 + x21 = 2x11 + x12 + x21,

and comparing Peirce components we see that c x = x if and only if x11 = x22 = 0. Hence (181)is proved.

128 . (a) We begin by assuming that the case of a local ring has been settled and then let k besol.UNORQUAarbitrary. We put X := Spec(k) and deduce from § 10, Exercise 40 (b), that

X+ := p ∈ X | fp : Bp→ B′p is an isomorphism,X− := p ∈ X | fp : Bp→ B′p is an anti-isomorphism

are Zariski-open subsets of X . Since a quaternion algebra is not commutative, they are also disjoint,and our assumption implies that they cover X . Now § 10, Exercise 41, yields a complete orthogonalsystem (ε+,ε−) of idempotents in k satisfying X± = D(ε±). Now it suffices to put k± = kε± and toinvoke (10.6.6), which implies for any p± ∈ Spec(k±) that p := p±⊕ k∓ ∈ D(ε±) makes f±p± =( fp)k±p± (by (10.4.7)) an isomorphism in case of the plus-sign and an anti-isomorphism in case ofa minus-sign. Hence f+ is an isomorphism and f− is an anti-isomorphism.

We are thus left with the case that k is a local ring and must show that f is either an isomorphismor an anti-isomorphism. Theorem 22.15 (b) yields a quadratic etale subalgebra D = k[u] ⊆ B, forsome u ∈ B having trace 1. Since f preserves units, norms, and traces by Proposition 26.2, we con-clude that D′ := f (D) = k[u′], u′ := f (u), is a quadratic etale subalgebra of B′ (Proposition 22.5),and f |D : D→ D′ is an isomorphism (Proposition 26.4). Now apply Theorem 22.15 (a) to reach Bfrom D by means of the Cayley-Dickson construction: there exists a unit µ ∈ k× such that the in-clusion D → B extends to an identification B = Cay(D,µ) = D⊕D j, j ∈D⊥, nB( j) =−µ . Settingj′ := f ( j) ∈ D′ ⊆ B′, we obtain nB′ ( j′) = −µ , and from Proposition 21.6 we conclude that f |Dextends to a homomorphism g : B→ B′ of conic algebras satisfying g( j) = j′. By Corollary 21.8,therefore, g is an isomorphism from B onto the subalgebra of B′ generated by D′ and j′. Countingranks we conclude that g is in fact an isomorphism from B onto B′. Hence f1 := g−1 f : B→ B is


a unital norm equivalence inducing the identity on D and satisfying f1( j) = j. Since f1 stabilizesD⊥ = D j, we find a k-linear bijection ϕ : D→ D such that f1(v j) = ϕ(v) j for all v ∈ D. Thenϕ(1D) = 1D, and since nB permits composition, ϕ leaves nD invariant and is thus a unital normequivalence of D, hence an automorphism (Proposition 26.4 (b)). By § 22, Exercise 103, therefore,we are left with the following cases.Case 1. ϕ = 1D. Then f1 = 1B, and f = g : B→ B′ is an isomorphism.Case 2. ϕ = ιD. By § 21, Exercise 93, the map ψ : B→ B defined by ψ(v+w j) := v−w j for allv,w ∈ D is an automorphism, and one checks that f1 = ψ Int( j) ιB, where Int( j) stands for theinner automorphism x 7→ jx j−1 of B affected by j. Hence f = g ψ Int( j) ιB : B→ B′ is ananti-isomorphism.

(b) The implications (i)⇒ (ii)⇒ (iii) are obvious, so let us prove (i) under the assumption (iii).By Corollary 26.3, there exists a unital norm equivalence f : B→ B′. With the notation of (3), weapply (a) to conclude that g := f+⊕ ( f− ιB− ) : B→ B′ is an isomorphism.

129. By Prop. 26.4 (a), every automorphism of B is a unital norm equivalence. Hence it suffices tosol.ELIDNOshow that, conversely, if f : B→ B is a unital norm equivalence sending e11 to c, then an automor-phism of B exists having the same property. From Exc. 127 (b) we deduce that c is an elementaryidempotent in B. Hence Exc. 116 (a) yields a line bundle L over k and an isomorphism

Φ : B ∼−→ B′ := Endk(k⊕L) =(

k L∗

L k

)such that Φ(c) =

(1 00 0

).

Viewing B12(c)⊕ B21(c) (resp. L∗ ⊕ L) canonically as quadratic submodules of (B,nB) (resp.(B′,nB′ ),

Φ : B12(c)⊕B21(c)∼−→ L∗⊕L

is an isometry and by (25.9.1), (12.17.1), there is a natural identification of L∗⊕L with the hyper-bolic plane hL. On the other hand, f , being a unital norm equivalence, by Exc. 127 (b) induces anisometry from the split hyperbolic plane k⊕ k ∼= B12(e11)⊕B21(e11) onto B12(c)⊕B21(c). Hencethe hyperbolic plane hL is split, which by Exc. 47 forces the line bundle L to be free of rank 1.Thus we may identify L = k and in this way view Φ as an automorphism of B sending e11 to c.

130. By Exc. 72, ϕ := LpRp−1 is an isomorphism from C to Cq with q := p−3.sol.OCTNOREQU(a) By Prop. 26.2 (a), ϕ is a norm similarity fixing 1C and hence a unital norm equivalence.(b) ϕ is an automorphism of C if and only if C = Cq, which by (16.9.3) is equivalent to q ∈

Nuc(C) = k1C (Exc. 102 (b)), hence to p3 = q−1 ∈ k1C .(c) Assume ϕ is an anti-automorphism of C. Then Cop =Cq, so we have yx=(xq−1)(qy), equiv-

alently, y(xq) = x(qy), for all x,y ∈C. Setting x = 1C gives yq = qy for all y ∈C. By Exc. 102 (b),this implies q ∈ k1C and then xy = yx for all x,y ∈C, a contradiction.

131 . Let p,q ∈ C×. By Exc. 90, C(p,q) is a multiplicative conic alternative k-algebra with normsol.ISTISnC(p,q) = nC(pq)nC . By multiplicativity, Lpq is an isometry from nC(p,q) to nC . Hence C(p,q) is acomposition algebra which is non-singular if C is. Moreover, if k is a semi-local ring, the normequivalence theorem 26.7 shows that C and C(p,q) are isomorphic.

132. Let xi = ui + vi j, ui,vi ∈O, i = 1,2 be non-zero elements of S, Then Exc. 99 (a) for µ =−1sol.ZEPAshows that x1x2 = 0 if and only if ui 6= 0 6= vi for i = 1,2 and

nO(u1) = nO(v1), (u1u2)v1 =−(u1u2)v1, v2 = (v1u2)u−11 . (182) ZEROC

Now let (wi)1≤i≤7 be a Cartan-Shouten basis of O in the sense of 2.1. Combining (21.3.1) withFig. 1 of 2.5, we compute

(w1 +w3 j)(w2−w6 j) = (w1w2− (−w6)w3)+(w3w2−w6w1)

= (w1w2−w6w3)+(−w3w2−w6w1),


where w6w3 = w4 = w1w2 and w6w1 = w5 = −w3w2. Thus the quantities a := w1 +w3 j, b :=w2−w6 j have S-norm 2 and satisfy ab = 0, so we conclude (a,b) ∈ Zer(S), while (182) (or theproperty of (wi) being a Cartan-Shouten basis) yields

w6 =−(w3w2)w−11 =−(w3w2)w1. (183) DUSIX

We must show that the action of G on Zer(S) is simply transitive. Let (x1,x2) ∈ Zer(S) and writexi = ui + vi j with ui,vi ∈ O for i = 1,2 as before. Then (182) shows nO(ui) = nO(vi) for i = 1,2,whence the relations nS(xi) = 2 imply nO(ui) = nO(vi) = 1. Consulting Exc. 99 (b), we also getnO(u1u2,v1) = 0. By Exc. 6, therefore, the quantities u1,u2,v1 can be extended to a Cartan-Shoutenbasis of O. Hence there exists an automorphism σ of O sending u1,u2,v1 respectively to w1,w2,w3.Consulting (182), (183), we conclude that σ also sends v2 to w6. But this means σ((x1,x2)) =(a,b), and we have proved that the action is transitive. It remains to show that σ(a) = a, σ(b) = bimplies σ = 1O. But this is clear since we then have σ(wi) = wi for i = 1,2,3, and w1,w2,w3 byExc. 6 generate the octonion algebra O.

133. We begin by proving the following general statement.sol.ZOFIClaim. Let (V,Q) be a hyperbolic quadratic space of dimension 2n over Fq. Then the number ofanisotropic vectors in (V,Q) is

qn−1(q−1)(qn−1).

In order to see this, we identify (V,Q) = (Fnq⊕Fn

q,x⊕ y 7→ xt y) and note for x ∈ Fnq that

|y ∈ Fnq | xt y = 0|=

qn for x = 0,qn−1 for x 6= 0.

Thus |v ∈V | Q(v) = 0|= qn +(qn−1)qn−1, and we conclude that (V,Q) contains precisely

q2n−qn− (qn−1)qn−1 = qn(qn−1)− (qn−1)qn−1 = (qn−qn−1)(qn−1) = qn−1(q−1)(qn−1)

anisotropic vectors. The claim is proved.We now turn to the assertions of our problem. Since the anisotropic vectors of C relative to nCare just its invertible elements (Prop. 20.4), and the norm of the split octonions is hyperbolic(Cor. 25.14), the preceding claim reduces to the first equation for n = 3.

To prove the second equation, we note that n : C×→F×q , x 7→ n(x) := nC(x), is a surjective mul-tiplicative map. Writing N := x ∈C | nC(x) = 1 for its “kernel”, we let x,y ∈C×, use Prop. 14.6and obtain

y ∈ n−1(n(x))⇔ n(y) = n(x)⇔ n(x−1y) = 1⇔ x−1y ∈ N⇔ y ∈ xN,

so the fibers of n all have the same cardinality. This shows |N|= |C×|/|F×q |= q3(q4−1).

134. (a) is trivial: if x ∈M is not minimal, the x = y+ z for some y,z ∈M such that Q(y)< Q(x)sol.MIVEand Q(z)< Q(x). Proceeding with y,z in the same manner, we eventually arrive at a decompositionof x as a finite sum of minimal elements.

(b) Again this is trivial, demanding not even the sketch of a proof.(c) We begin with a simple lemma.

Lemma. (a) If N ⊆M is a submodule, then Min(M,Q)∩N ⊆Min(N,Q|N).(b) If M = N ⊥ N′ is a decomposition of M into the orthogonal sum (relative to Q) of two submod-ules N,N′ ⊆M, then Min(M,Q) = Min(N,Q|N)∪Min(N′,Q|N′ ).In order to prove (a), let x ∈Min(M,Q)∩N and suppose we have a decomposition x = y+ z withy,z ∈ N, Q(y)< Q(x), Q(z)< Q(x). Then this decomposition takes place also in M, contradictingminimality of x in (M,Q). Thus x ∈ Min(N,Q|N). But note for x ∈ Min(N,Q|N), that it is con-


ceivable to have a decomposition x = y+ z with y,z ∈M \N and Q(y)< Q(x), Q(z)< Q(x), so ingeneral we won’t have equality in (a).

In order to prove (b), let x ∈Min(M,Q) and write x = y+ y′ with y ∈ N, y′ ∈ N′. Then Q(x) =Q(y)+Q(y′), and since x is minimal in M, we conclude Q(y) = 0 or Q(y′) = 0, hence y = 0 ory′ = 0. Thus Min(M,Q)⊆ N∪N′, which implies

Min(M,Q) = Min(M,Q)∩ (N∪N′) =(

Min(M,Q)∩N)∪(

Min(M,Q)∩N′).

By symmetry and (a), it therefore suffices to show Min(N,Q|N)⊆Min(M,Q). Let x∈Min(N,Q|N)and suppose x = y+ z for some y,z ∈M, Q(y) < Q(x), Q(z) < Q(x). Write y = u+ u′, z = v+ v′

with u,v ∈ N, u′,v′ ∈ N′. Since x belongs to N, this implies x = u+ v and Q(u) ≤ Q(y) < Q(x),Q(v)≤Q(z)<Q(x), in contradiction to x being a minimal vector of (N,Q|N). Thus x∈Min(M,Q).

We can now prove (c). Let x,y ∈Min(M,Q) be inequivalent. For u ∈ [x], v ∈ [y], the assumptionQ(u,v) 6= 0 would imply that u,v∈Min(M,Q) are equivalent. Since M[x] is spanned as a Z-moduleby the minimal elements of (M,Q) belonging to [x], ditto for M[y], this contradiction shows that M[x]and M[y] are orthogonal. But every x∈Min(M,Q) belongs to M[x], and M is spanned by Min(M,Q)as a Z-module. Thus M is the orthogonal sum of the distinct M[x]’s, x ∈Min(M,Q), and it remainsto show that each M[x] is indecomposable. Thanks to the obvious extension of (b) in the precedinglemma to more than two orthogonal summands, it will actually be enough to show that everypositive definite integral quadratic module (M,Q) all of whose minimal elements are equivalentis indecomposable. Indeed, suppose we have an orthogonal decomposition M = N ⊥ N′ relativeto Q, with submodules N,N′ ⊆ M, and let x,y ∈Min(M,Q). By part (b) of the lemma, we mayassume x ∈Min(N,Q|N), and by definition, there is a finite sequence x = x0,x1, . . . ,xk−1,xk = y ofminimal vectors in (M,Q) such that Q(xi−1,xi) 6= 0 for 1 ≤ i ≤ k. We claim xi ∈Min(N,Q|N) forall i = 0, . . . ,k and argue by induction. For i = 0, there is nothing to prove. If i > 0 and the assertionholds for i−1, then xi /∈N′ since xi−1 ∈N by the induction hypothesis and Q(xi−1,xi) 6= 0, whenceaprt (b) of the lemma implies xi ∈Min(N,Q|N), and the induction is complete. In particular, y ∈Min(N,Q|N), and we have shown Min(M,Q) = Min(N,Q|N), which obviously implies M = N,N′ = 0.

135. We put C† =C/2C =C⊗ZF2 as a conic algebra over F2 and write x 7→ x† for the natural epi-sol.ODDImorphism from C to C†. This epimorphism canonically induces a map from C× to C†×, which mayor may not be surjective. Note that, since the discriminant of C is odd, C† is a composition algebraof dimension 8 over F2, hence an octonion algebra and thus isomorphic to Zor(F2) (Cor. 25.14).Lemma 1. For x,y ∈C× we have x† = y† if and only if y =±x.Indeed, y = ±x clearly implies x† = y†. Conversely, let this be so. Then y = x + 2z for somez ∈ C. If z = 0, we are done, so we may assume z 6= 0. Since x,y ∈ C are both units, we obtain1 = nC(x + 2z) = 1 + 2nC(x,z) + 4nC(z), hence nC(x,z) = −2nC(z), and applying the Cauchy-Schwarz inequality yields

4nC(z)2 = nC(x,z)2 ≤ nC(x,x)nC(z,z) = 4nC(z).

But nC(z) is a positive integer, forcing nC(z) = 1. Thus the above inequality is, in fact, an equality,which implies z = αx for some α ∈ Q. Taking norms, we deduce α2 = 1, hence α = ±1, wherethe assumption α = 1 would lead to the contradiction that y = 3x is not a unit in C. Hence α =−1and y =−x.We can now establish the first part of the problem. By Exc. 133 for q = 2, we have |C†×| = 120,while by Lemma 1, the fiber of an arbitrary element u ∈C†× under the natural map C×→C†× iseither empty or consists of two elements. Hence |C×| ≤ 240.

For the rest of the proof, we may assume that |C×| = 240. Writing C×† for the image of C×

under the natural map C×→C†×, and putting r := |C×†|, we note that the fiber of any point in C×†

consists of two elements. Thus 2r = 240, hence r = 120 = |C†×|, and we conclude C×† =C†×. Inother words, the natural map C×→C†× is surjective.


We can now prove that the positive definite integral quadratic lattice (C,nC) is indecomposable.We have C× = x ∈C | nC(x) = 1 ⊆Min(C,nC), and for x,y ∈C×, Exc. 52 (b) yields a sequencex† = x′0,x

′1, . . . ,x

′k = y† of elements in C†× such that k ≤ 2 and nC† (x′i−1,x

′i) 6= 0 for 1 ≤ i ≤ k.

Since the natural map C× → C†× is surjective, each x′i, 1 ≤ i < k, lifts to an invertible elementxi ∈ C, so we have got a sequence x =: x0,x1, . . . ,xk := y of elements in C×, hence of minimalvectors in (C,nC), such that nC(xi−1,xi) is an odd integer and, in particular, different from 0, for1≤ i≤ k. This shows that the minimal vectors x and y of (C,nC) are equivalent, in other words, theinvertible elements of C all belong to a single equivalence class of minimal vectors, which may alsobe expressed by saying that C× ⊆C[1C ]. Now suppose that x ∈Min(C,nC) is not equivalent to 1C .From Exc. 134 (c) we deduce nC(C×,x) = 0, which implies nC† (C†×,x†) = nC† (C×†,x†) = 0.But C′ is an octonion algebra over F2 that is spanned as a vector space by C†× (Exc. 52 (a) andProp. 20.4). Thus x† = 0, hence x ∈ 2C, and we conclude that the indecomposable submodulesC[x] ⊆C, with x ∈Min(C,nC) not equivalent to 1C , all belong to 2C; in particular, they have evendiscriminant. But the discriminant of (C,nC) is odd by hypothesis, which, thanks to Exc. 134 (c),implies that (C,nC) =C[1C ] is indeed indecomposable.

136. Let R be a quadratic etale Z-algebra. Then R is a free Z-module of rank 2, and since 1R ∈ Rsol.EZEis unimodular, it can be extended to a basis (1R,u) of R. By Lemma 22.11, the trace form of R issurjective. Since tR(1R) = 2, we conclude that tC(u) = 2n+1 (for some integer n) is odd. In fact,replacing u by u−n ·1R, we may assume tR(u) = 1. Now Prop. 22.5 shows 1−4nC(u) =±1, whichimplies nC(u) = 0 or nC(u) = 1

2 . The latter option obviously being impossible, we conclude that Rhas zero divisors (Exc. 94), hence is split (Exc. 126). This solves the first part of the problem.

As to the second, assume Hur(H)∼=Cay(R,µ) for some quadratic Z-algebra R and some µ ∈Z.Combining the definition of the discriminant (3.12) with (4.2.2) and Remark 21.4, we concludewith d := disc(R) that 4 = d2µ2. On the other hand, since the norm of R ⊆ Hur(H) is positivedefinite, and R itself by the first part of this exercise is not etale, we have d ≥ 2 and hence d = 2,µ = −1. Extending 1R to a basis (1R,u) of R over Z, we thus obtain 2 = tC(u)2− 4nR(u) fromProp. 22.5, where the left-hand side is ≡ 2 mod 4, while the right-hand is ≡ 0 or ≡ 1 mod 4, acontradiction.

137. (a) Since C is not split, it has no zero divisors (Exc. 126). Hence neither has CQ, whencesol.OCZECQ is an octonion division algebra over the rationals, up to isomorphism actually the only one(Cor. 26.20). This shows CQ ∼= Cay(Q;−1,−1,−1). In particular, the norm of CQ is positive def-inite. Hence so is the norm of C. Summing up we have shown that (C,nC) is a positive definiteinner product space of rank 8 over Z. But by 4.6 (c), up to isometry there is only one. Hence, byCor. 26.3, there exists a unital norm equivalence f : C→ Cox(O). Transferring the unit element,norm and multiplication of C to Cox(O) by means of f , we may actually assume C = Cox(O) asadditive groups with 1C = 1O and nC = nCox(O). In particular |C×|= 240 (Exc. 16 (d)).

(b) Put D := Cox(O). Then C† := C/2C = C⊗Z F2 and D† = D/2D = D⊗Z F2 are bothoctonion algebras over F2, hence split (26.12) and, in particular, isomorphic. We therefore findan isomorphism ψ : C† ∼→ D† of F2-algebras. Note by the normalization provided in (a) that(C†,nC† ,1C† ) = (D†,nD† ,1D† as “pointed” quadratic spaces over F2. In particular, ψ is an or-thogonal transformation of the quadratic space (C†,nC† ). Applying [24, Prop. 14], we see that theorthogonal group of this space is generated by the orthogonal transvections

τa : x 7−→ x− nC† (a,x)nC† (a)

a (a ∈C†×),

each of which can be lifted to an orthogonal transformation from (C,nC) onto itself since, as wehave seen in the solution to Exc. 135, the natural map from C× to C†× is surjective. Hence wecan also lift ψ to an orthogonal transformation ϕ of (C,nC). In other words, ϕ : C→ Cox(O) is alinear bijection preserving norms such that the diagram


Cϕ //

can

Cox(O)

can

C/2C

ψ

// Cox(O)/2Cox(O),

commutes. Writing x† := can(x) for x ∈C× and using the fact that ψ is an algebra isomorphism,we therefore obtain, for all x,y ∈C×,(

ϕ(x · y))†

= ψ((x · y)†)= ψ(x† · y†) = ψ(x†)ψ(y†) =

(ϕ(x)

)†(ϕ(y)

)†=(ϕ(x)ϕ(y)

)†.

But for x ∈C×, the fiber of x† ∈C†× under the vertical arrows in the above diagram consists of theelements ±x. Thus ϕ(x · y) =±ϕ(x)ϕ(y).

(c) According to (b), there exists a map ε : C××C×→±1 such that ϕ(x ·y)= ε(x,y)ϕ(x)ϕ(y)for all x,y ∈C×. We claim that ε is constant. Indeed, let x,y,z ∈C×. Since ϕ preserves norms, andthese are the same for C and Cox(O), an application of (20.1.3) yields

ε(x,y)ε(x,z)nC(x)nC(y,z) = ε(x,y)ε(x,z)nC(ϕ(x)

)nC(ϕ(y),ϕ(z)

)= ε(x,y)ε(x,z)nC

(ϕ(x)ϕ(y),ϕ(x)ϕ(z)

)= nC

(ϕ(x · y),ϕ(x · z)

)= nC(x · y,x · z) = nC(x)nC(y,z).

Hence ε(x,y) = ε(x,z) provided nC(y,z) 6= 0. On the other hand, if nC(y,z) = 0, then Exc. 52 (b)combined with the surjectivity of the natural map C×→C†× yields a w∈C× satisfying nC(y,w) 6=0 6= nC(w,z), which in turn implies ε(x,y) = ε(x,w) = ε(x,z). Thus ε(x,y) does not depend ony ∈ C×. Interchanging the role of x and y, and replacing (20.1.3) by (20.1.2) in the process, thesame argument shows that ε(x,y) is also independent of x ∈C×, showing that the map ε is indeedconstant. Finally, replacing ϕ by −ϕ if necessary, we may assume ϕ(x · y) = ϕ(x)ϕ(y) for allx,y ∈C× = Cox(O)×. But since C is generated by C× as an additive group (Exc. 16) (b),(d)), thismeans that ϕ : C ∼→ Cox(O) is an isomorphism of octonion algebras.

138 . (a) We begin by proving sufficiency The assertion is clear if (i) or (ii) holds. As to (iii,sol.SPLIFICOassuming there exists an F-embedding K → C, we obtain an induced K-embedding KK → CK .If we can show that KK is not a division algebra, then neither is CK , forcing CK to be split byCorollary 25.14. Hence it suffices to show that KK = K ⊗K contains zero divisors. If K/F isseparable, this follows from Exercise 106 (a). On the other hand, if K/F is inseparable, thenchar(F) = 2 and ξ 2 ∈ F for all ξ ∈ K \F . This implies 0 6= u := 1K ⊗ ξ + ξ ⊗ 1K ∈ K⊗K andu2 = 1K ⊗ ξ 2 + ξ 2⊗ 1K = 2ξ 21K⊗K = 0. Thus KK contains non-zero nilpotent elements, and weare done.

In order to prove necessity, let us assume that K is a splitting field of F and (i), (ii) do not hold.If C is split, then it has dimension 4 or 8, and C1 := Cay(K,1) in the first case, C2 := Cay(K;1,1)in the second, is a split composition algebra over F of the same dimension as C. Thus C ∼= Ci forsome i = 1,2, and since Ci contains an isomorphic copy of K, so does C. We are left with the casethat C is a division algebra, so by Corollary 25.14 its norm is anisotropic. Pick any ξ ∈ K \F .Then ξ 2 = αξ −β with α := tK(ξ ), β := nK(ξ ). Since CK =C⊕ (C⊗ξ ) is split over K, we findelements u,v ∈C not both zero such that

0 = nC(u− v⊗ξ ) = nC(u)−ξ nC(u,v)+ξ2nC(v) =

(nC(u)−βnC(v)

)+(αnC(v)−nC(u,v)

)ξ ,


hence nC(u) = βnC(v) and nC(u,v) = αnC(v). This implies v ∈C× and with x := uv−1 we deduceβ = nC(x), α = nC(u,nC(v)−1v) = nC(u, v−1) = tC(x). Summing up, F [x] ⊆ C is a subalgebraisomorphic to K, and we have found an embedding K →C.

(b) Since the property of being a composition algebra is stable under base change, we mayassume that F itself is separably closed and must show that C is split. The case C ∼= F beingobvious, we may assume that C has dimension at least 2 and hence is non-singular. Assume C isa division algebra. Since F is separably closed, hence infinite, Zariski density leads to an elementx ∈ C \F1C having tC(x) 6= 0. But then F [x] ⊆ C is a separable quadratic field extension of F , acontradiction to F being separably closed.

Section 28

139 . (a) Let R ∈ k-alg be the split-null extension of M, living on the k-module k⊕M under thesol.MOFFmultiplication (α⊕u)(β ⊕ v) = αβ ⊕ (αv+βu) for α,β ∈ k, u,v ∈M. One check easily that Mis finitely generated as a k-module if and only if R is finitely generated as a unital k-algebra. Also,passing to the split-null extension is compatible with base change. Hence the assertion followsfrom Corollary 28.14.

(b) By [11, II, §5, Thm. 1], there is a finite family ( fi)i∈I of elements in k such that ∑i∈I k fi = kand M fi is a free k fi -module of rank r for each i∈ I. Put R :=

⊕i∈I k fi ∈ k-alg. Then MR =

⊕i∈I M fi

is a free R-module of rank r and since R is faithfully flat over k, by [11, II, §5, Prop. 3], the assertionis proved.

140. (a) Let R′ be a finitely presented R-algebra. Then there are presentationssol.TRAFIPR

0 // I // k[S]π

// R // 0 , (184) IKS

0 // J // R[T]ρ

// R′ // 0 (185) JERT

of R over k, R′ over R, respectively, with chains S = (s1, . . . ,sm), T = (t1, . . . , tn) of independentvariables and finitely generated ideals I ⊆ k[S], J ⊆ k[T]. Extending (184) from k to k[T], weconclude that the sequence

0 // I⊗ k[T] // k[S,T] = k[S]⊗ k[T]πk[T]// R⊗ k[T] = R[T] // 0 (186) IKIK

is exact, whence

ϕ := ρ πk[T] : k[S,T]−→ R′ (187) ROPIK

is a surjective morphism in k-alg which by (185) satisfies

Ker(ϕ) = π−1k[T](

Ker(ρ))= π

−1k[T](J). (188) PINV

Now pick g1, . . . ,gr ∈ R[T] such that J = ∑rj=1 R[T]g j and use (186) to find a lift h j of g j in k[S,T]

under πk[T]: πk[T](h j) = g j for 1≤ j ≤ r. By (186)−(188), this is easily seen to imply

Ker(ϕ) = I⊗ k[T]+r

∑j=1

k[S,T]h j.

But I⊗k[T] is a finite k[T]-module, forcing Ker(ϕ)⊆ k[S,T] to be a finitely generated ideal. HenceR′ is finitely presented as a k-algebra.

(b) By (a), we may assume R = k. Since the k-algebra E := k[t]/( f t− 1) is clearly finitelypresented, it suffices to show that k f and E are canonically isomorphic. Note first that the natural


projection π : k[t]→ E makes f invertible: π( f )∈ E× with inverse π(t). Hence the unit homomor-phism k→ E extends to a homomorphism ϕ : k f → E such that ϕ(1/ f ) = π(t). On the other hand,the k-homomorphism k[t]→ k f sending t to 1/ f kills f t−1 and hence induces a k-homomorphismψ : E→ k f . One checks that ϕ and ψ are inverse to one another, and the assertion follows.

141. (a) Let π : k[T]→ R be a surjective morphism in k-alg, giving rise to a presentationsol.IDFIPR

0 // I // k[T]ϕπ// R′ // 0

of R′ as a k-algebra. By Proposition 28.13, therefore, I ⊆ k[T] is a finitely generated ideal. ButI = π−1(Ker(ϕ)), and we conclude that Ker(ϕ) = π(I)⊆ R is a finitely generated ideal.

(b) Let

0 // I // k[T]π

// R // 0 (189) IKIP

be a finite presentation of R. Setting J := Ker(ϕ), this induces a presentation

0 // π−1(J) // k[T]ϕπ// R′ // 0 (190) PINJ

of R′. Apply (189) and let π( fi), fi ∈ k[T], 1≤ i≤ m, be finitely many generators of J as an idealin R. Then one checks that

π−1(J) = I +

m

∑i=1

k[T] fi,

so (190) makes R′ a finitely presented k-algebra.

142. By 27.18, k[Ma]∼= S(M∗) is the symmetric algebra of M∗, the dual of M. Since M∗ is a finitelysol.EMMAgenerated projective k-module as well, it suffices to show: if M is a finitely generated projectivek-module, then S(M) is a finitely presented k-algebra.

Applying §10, Exercise 42, we find a complete orthogonal system (εi)i∈N of idempotents ink such that Mi = M⊗ ki for each i ∈ N is a finitely generated projective module of rank i overki := kεi. Note that εi = 0, hence ki = 0 and Mi = 0, for almost all i ∈ N. Since passing to thesymmetric algebra of a module is compatible with base change [12, III, §6, Prop. 7], we conclude

S(M) =⊕i∈N

S(Mi)

as k-algebras. We are thus reduced to the case that M is finitely generated projective of rank r ∈ Nover k. But then Exercise 139 (b) produces a faithfully flat k-algebra k′ making Mk′ a free k′-module of rank r. Since S(M)k′ = S(Mk′ ), therefore being a polynomial ring in r variables [12, III,§6], hence finitely presented over k′, so is S(M) over k, by Corollary 28.14.

143. It is straightforward to verify that w is a subfunctor of Ma. In order to derive the second partsol.SGLTof the exercise, let us assume that M is finitely generated projective. For u∗ ∈M∗, the set maps

fu∗ (R) : MR −→ R, x 7−→ 〈u∗R,x−wR〉,

vary functorially with R ∈ k-alg and hence define an element fu∗ ∈ k[Ma]. Since M∗ is finitelygenerated projective as well, and the canonical pairing M∗×M→ k is non-singular, we concludethat x ∈ MR agrees with wR if and only if fu∗ (R)(x) = 0 for all u∗ ∈ M∗. By Exercises 141, 142,therefore, w⊆Ma is a finitely presented closed affine subscheme.

144 . Let K ∈ k-alg be a field. We have to prove that the quadratic form qK : MK → K over Ksol.FFSEPis non-degenerate. Write ϑR : k → R (resp. ϑK : k → K) for the unit morphism of R (resp. K).Then p := Ker(ϑK) ∈ Spec(k), and since R is faithfully flat over k, Proposition 28.8 yields a prime


ideal q ∈ Spec(R) such that p = Spec(ϑR)(q) = ϑ−1R (q). By 10.4, not only K but also κ(q) is an

extension field of κ(p). Let L be a free compositie of κ(q) and K over κ(p) [47, pp. 551−2].Since L ∈ R-alg is a field and qR is separable, the quadratic form (qK)L ∼= qL ∼= (qR)L over L isnon-degenerate. Hence so is qK over K.

145. (a) For i = 1,2, T ∈ S-alg and all k-modules P, we have the natural identificationssol.FFPOLA

(PS)T = PT = PTi = (PR)Ti

via (10.3.1), so giT is a set map from (MS)T to (NS)T , as claimed. It remains to show that theseset maps vary functorially with T . In order to see this, we regard the identity map of T as anisomorphism T → Ti, t 7→ ti, of k-algebras. Given a morphism ϕ : T → T ′ in S-alg, we thereforeobtain a morphism ϕi : Ti→ T ′i in R-alg by defining ϕi(ti) := ϕ(t)i for all t ∈ T . One checks that1MR ⊗R ϕi = 1MS ⊗S ϕ , and it follows that the diagram

(MS)T giT//

1MS⊗Sϕ

(NS)T

1NS⊗Sϕ

(MS)T ′ giT ′

// (NS)T ′

commutes. Hence gi : MS→ NS is a polynomial law over S.(b) If f : M→ N is a polynomial law over k such that f ⊗R = g, then for all T ∈ S-alg we deducethat giT = gTi = ( f ⊗R)Ti = fTi = fT is independent of i = 1,2. Thus g1 = g2. Conversely, assumeg1 = g2 =: h. We first prove uniqueness of f and let k′ ∈ k-alg. One checks that the outer squares in(3) commute. Hence so does the inner one since gRk′ = fRk′ by hypothesis. Moreover, the verticalarrows in (3) are injective by Proposition 28.4, proving uniqueness of the set maps fk′ , hence of fas a polynomial law over k.

In order to prove existence, let k′ ∈ k-alg. To simplify notation, we indicate the base changefrom k to k′ simply by a dash, for example M′ = Mk′ , R′ = R⊗ k′, S′ = S⊗ k′ = R′⊗k′ R′. Further-more, the notational conventions fixed in (a) on the level of R and S will now be employed for R′

and S′. In particular, the very definition of the R′-algebras S′i (i = 1,2) yield morphisms

σR′i : R′ −→ S′i, r′ 7−→ σ

R′i (r′) := ρ

R′i (r′)i

in R′-alg⊆ R-alg. Now we consider the following diagram.

0

0

M′

∃! fk′ //

canM′ ,R′

N′

canN′ ,R′

M′R′ 1

//

ρR′ ,M′i

(MR)R′ gR′//

1MR⊗RσR′i

(NR)R′ 1//

1NR⊗RσR′i

N′R′

ρR′ ,N′i

M′S′ 1

// (MR)S′i hS′=gS′i

// (NR)S′i 1// N′S′ ,

(191) DESSET

whose inner lower square commutes by the definition of polynomial laws, while the outer lowerones do thanks, e.g., to the following computation, for all x ∈M, r ∈ R, α ′,β ′ ∈ k′:


ρR′,M′1

((x⊗α

′)⊗k′ (r⊗β′))= (x⊗α

′)⊗k′((r⊗β

′)⊗k′ (1R⊗1k′ ))

= (x⊗α′)⊗k′

((r⊗1R)⊗β

′)= x⊗((r⊗1R)⊗α

′β′)

= x⊗((r⊗1R)⊗α

′β′)

1 = (x⊗1R)⊗R((r⊗1R)⊗α

′β′)

1

= (x⊗1R)⊗R((r⊗α

′β′)⊗k′ (1R⊗1k′ )

)1

= (x⊗1R)⊗R σR′1 (r⊗α

′β′)

= (1MR ⊗R σR′1 )((x⊗1R)⊗R (r⊗α

′β′))

= (1MR ⊗R σR′1 )(

x⊗(α′(r⊗β

′)))

= (1MR ⊗R σR′1 )((x⊗α

′)⊗k′ (r⊗β′)).

The universal property (28.5) of the equalizer canN′,R′ of ρR′,N′1 , ρ

R′,N′2 (Proposition 28.6), applied

to the map u := gR′ canM′,R′ : M′ → N′R′ , by (191) implies that there exists a unique set mapfk′ : M′→ N′ making (191) totally commutative. In other words, we have a commutative diagram

M′fk′ ]

//

canM′ ,R′

N′

canN′ ,R′

M′R′ = (MR)R′ gR′

// (NR)R′ = N′R′ .

(192) EXDES

Next we show that fk′ depends functorially on k′, so let ϕ : k′ → k′′ be a morphism in k-alg.Indicating the base change from k to k′′ by a double dash, we see that

ϕR := 1R⊗ϕ : R′ −→ R′′ (193) PHEER

is a morphism in R-alg, and one checks that the diagram

M′1M⊗ϕ

//

canM′ ,R′

M′′

canM′′ ,R′′

M′R′ = (MR)R′ 1MR⊗RϕR

// (MR)R′′ = M′′R′′

(194) EMPHER

commutes. Now consider the cube

M′fk′ //

canM′ ,R′

1M⊗ϕ

((

N′

canN′ ,R′

1N⊗ϕ

''M′′

fk′′//

canM′′ ,R′′

N′′

canN′′ ,R′′

M′R′ = (MR)R′ gR′//

1MR⊗RϕR ''

(NR)R′ = N′R′

1NR⊗RϕR ''M′′R′′ = (MR)R′′ gR′′

// (NR)R′′ = N′′R′′ ,


where by (192)−(194) all rectangles commute, with the possible exception of the top one. Bydiagram chasing, therefore, so does the top one after being composed with canN′′,R′′ . But R′′ isfaithfully flat over k′′, forcing canN′′,R′′ by Proposition 28.4 to be injective, and we have provedthat the top rectangle commutes as well. Summing up, f : M → N is a polynomial law over k.Hence the solution of the problem will be complete once we have shown f ⊗R = g. In order to seethis, let k′ ∈ R-alg and ρ : R→ k′ the corresponding unit morphism in k-alg. With the notationalsimplifications introduced before, we obtain an induced morphism ρ ′ : R′→ k′ given by

ρ′(r⊗α

′) = ρ(r)α ′ = rα′

for r ∈ R, α ′ ∈ k′. In particular, ρ ′ is a morphism in R-alg, and we obtain a diagram

M′fk′

//

canM′ ,R′

1

$$

N′

canN′ ,R′

1

zz

M′R′ = (MR)R′ gR′//

1MR⊗R ρ ′

(NR)R′ = N′R′

1NR⊗R ρ ′

M′ = (MR)k′ gk′

// (NR)k′ = N′,

where the outer triangles are easily seen to be commutative. Hence fk′ = gk′ , as desired.

146. We systematically adhere to the notation and terminology of Exercise 145.sol.FFCON(a) Let e be the unit element of CR and write g := e : CR→CR for the constant polynomial law

over R determined by e, as defined in § 13, Exercise 55, For T ∈ S-alg and i = 1,2, eTi ∈ (CR)Ti =(CS)T is the unit element of (CS)T , and we conclude eT1 = eT2 . Thus g1 = g2 as polynomial lawsCS →CS over S. Applying Exercise 145, we find a unique polynomial law f : C→C over k suchthat f ⊗R = g. Put e0 := fk(0) ∈C. Since the vertical arrows in (3) are injective, we conclude notonly e0R = e but also that C is unital with 1C = e0.

(b) Let n : CR→ R be a quadratic form over R making CR a conic R-algebra. For T ∈ S-alg andi = 1,2, there are two quadratic forms over T making (CS)T = (CR)Ti a conic T -algebra, namely,nTi . Hence Proposition 19.15 implies nT1 = nT2 . Viewing n as a homogeneous polynomial lawg : CR → R of degree 2 over R, we have g1 = nT1 = nT2 = g2. Thus Exercise 145 yields a uniquepolynomial law f : C→ k over k such that f ⊗R = g. Hence the set map n0 := fk : C→ k is aquadratic form over k, and from (3) we deduce as before that C is a conic k-algebra with norm n0whose base change to R is CR as a connic R-algebra with norm n.

147 . By Exercise 142, the k-functor M2na = (M2n)a of Example 27.18 is a finitely presentedsol.FFSPLIQ

k-scheme. Hence so is X := Hyp(Q), defined as a closed subfunctor of M2na by finitely many

equations (cf. (4) and Exercise 141 (b)). After an appropriate base change, it therefore remainsto show that the natural set map X(k)→ X(k) is surjective, where k := k/I and I ⊆ k is an idealsatisfying I2 = 0. Note that 10.2 and Corollary 12.5 imply M := Mk = M/IM, and the quadraticform q := qk : M→ k over k is given by q(x) = q(x) for all x ∈ M, where α 7→ α (resp. x 7→ x)stand for the natural maps k→ k (resp. M → M). We first note that a hyperbolic basis is indeeda basis of the underlying module and must show that any hyperbolic basis of Q := (M, q) can belifted to one of Q, so let (w′i)1≤i≤2n be a hyperbolic basis of Q. We argue by induction on n. Whilethe case n = 0 is trivial, the case n = 1 amounts to showing that a hyperbolic pair (u′,v′) of Q in thesense of 12.16 can be lifted to a hyperbolic pair of Q. To this end, we pick any elements a,b ∈Msuch that a = u′, b = v′. Then q(a),q(b) ∈ I, q(a,b) ∈ 1+ I ⊆ k×, and since I squares to zero, onechecks that

u := a−q(a,b)−1q(a)b, v :=−q(a,b)−2q(b)a+q(a,b)−1b


form a hyperbolic pair of Q lifting (u′,v′). Now assume n > 1. By what we have just seen, the hy-perbolic pair (w′n,w

′2n) of Q can be lifted to a hyperbolic pair (wn,w2n) of Q. Put V := kwn⊕ kw2n

and M0 :=V⊥ (relative to Dq). Then Lemma 12.10 implies M = M0 ⊥V , and Q0 := (M0,q|M0 ) is aquadratic space of rank 2(n−1) over k. Moreover, Q0 = (M0, q|M0

) contains (w′i)1≤i≤2(n−1) as a hy-perbolic basis, which, by the induction hypothesis, can be lifted to a hyperbolic basis (wi)1≤i≤2(n−1)of Q0. Hence (wi)1≤i≤2n is the desired lift of (w′i)1≤i≤2n.

(b) We proceed in several steps.10. For the time being, we dispense ourselves from the quadratic space Q of (a) and considerinstead arbitrary quadratic modules Q = (M,q), Q′ = (M′,q′) over k. By an isometry from Q to Q′

we mean a k-linear bijection η : M→M′ satisfying q′ η = q. We define the set

Isom(Q,Q′) := η | η : Q→ Q′ is an isometry,

which in general will be empty but gives rise to a k-functor

Isomk(Q,Q′) := Isom(Q,Q′) : k-alg→ set

by defining

Isom(Q,Q′)(R) := IsomR(QR,Q′R) (195) ISQU

for all R ∈ k-alg and by

Isom(Q,Q′)(ϕ) : Isom(Q,Q′)(R)−→ Isom(Q,Q′)(S), (196) ISOMET

IsomR(QR,Q′R) 3 η 7−→ ηS ∈ IsomS(QS,Q′S)

for all morphisms ϕ : R→ S in k-alg. The k-group functor O(Q) of Example 27.24 acts canonicallyin a simply transitive manner on Isom(Q,Q′) from the right via

IsomR(QR,Q′R)×O(QR)−→ IsomR(QR,Q′R), (η ,ζ ) 7−→ η ζ

for all R ∈ k-alg.20. After this digression, we return to our quadratic space Q = (M,q) of rank 2n over k as con-sidered in (a).We denote by h2n

k the split hyperbolic quadratic space of rank 2n over k defined onk2n =

(kn

kn

)by the quadratic form

〈S〉quad, S :=(

0 1n0 0

)(197) SPLIHYP

in the sense of (12.7.2). The canonical basis of k2n is a hyperbolic basis for h2nk and obviously

compatible with base change. It follows immediately from the definitions that the assignment

η 7−→(η(ei)

)1≤i≤2n (198) ETEI

defines a bijection

Φ = Φ(k) : Isom(h2nk ,Q)

∼−→ Hyp(Q). (199) ISHY

30. Putting X := Isom(h2nk ,Q), the set maps

Φ(R) : X(R)−→Hyp(Q)(R)

given by (198), (199) for all R ∈ k-alg are bijective and obviously compatible with base change,hence give rise to an isomorphism

Φ : X ∼−→Hyp(Q)


of k-functors. By (a), therefore, X is a smooth affine k-scheme. Moreover, X is faithful in the senseof 28.9 since non-singular even-dimensional quadratic forms over an algebraically closed field are(split) hyperbolic [27, Exercise 7.34]. Summing up, 28.17 (ii) implies that there exists a faithfullyflat etale k-algebra R satisfying X(R) 6= /0. But this means that QR is split hyperbolic. Furthermore,X is a torsor with structure group O(h2n

k ) in the etale topology, forcing

O(Q)R ∼= O(QR)∼= O(h2nk )∼= XR

to be smooth over R. By 28.17 (iii), therefore O(Q) is smooth over k.

Chapter 5

Section 30

148. (a) By induction on m. For m = 0,1 there is nothing to prove. Now let m > 1 and supposesol.PARNILthe assertion has been established for all natural numbers < m. By the induction hypothesis wehave xm =Uxxm−2 ∈Monm(x). Conversely, let x ∈Monm(x). Then x =Uyz, y ∈Monn(x),z∈Monp(x), n, p∈N, n> 0, 2n+ p=m. By the induction hypothesis, y= xn, z= xp, and powerassociativity yields x =Uxn xp = x2n+p = xm, as claimed.

(b) For any subset X ⊆ J and any homomorphism ϕ : J→ J′ of para-quadratic k-algebras, onemakes the following observations by straightforward induction:

(i) ϕ(Monm(X)) = Monm(ϕ(X)) for all m ∈ N.(ii) Monm(Y )⊆Monmn(X) for all Y ⊆Monn(X) and all m,n ∈ N.

If x ∈ J is nilpotent, then (i) implies that ϕ(x) ∈ J′ is nilpotent. Hence if I is a nil ideal in J, then I′

and I/I′ are nil ideals in J and J/I′, respectively. Conversely, let this be so. Writing π : J→ J/I′

for the canonical epimorphism, any x ∈ I makes π(x) ∈ I/I′ nilpotent, so by (i), Monn(x) meetsI for some n ∈N. But since I′ is nil, we conclude that some y ∈Monn(x) is nilpotent, leading toa positive integer m such that 0∈Monmn(x) by (ii). Thus x is nilpotent, forcing I to be a nil ideal.Now write n := Nil(J) for the sum of all nil ideals in J. Given x ∈ n, there are finitely many nilideals n1, . . . ,nr ⊆ J such that x ∈ n1 + · · ·+nr . Thus we only have to show that the sum of finitelymany nil ideals in J is nil. Arguing by induction, we are actually reduced to the case r = 2. Butthen the isomorphism (n1 +n2)/n1 ∼= n2/(n1 ∩n2) combines with what we have proved earlier toyield the assertion.

(c) It suffices to prove that Nil(k)J ⊆ J is a nil ideal. The ideal property being obvious, weare reduced to showing that Nil(k)J consists entirely of nilpotent elements. An arbitrary elementx∈Nil(k)J actually belongs to IJ, where I⊆Nil(k) is some finitely generated ideal in k. A straight-forward induction now shows xn ∈ InJ for all n ∈ N, and the assertion follows from the lemma insolution to §19, Exercise 83.

149. (a) For the first assertion, since ε(1) = 1J , we need only show ( f 2g)(x) = U f (x)g(x) forsol.EVALPOLall f ,g ∈ k[t]. By linearity, we may assume g = tn for some n ∈ N and, writing f = ∑αiti withcoefficients αi ∈ k, we obtain

f 2g = (∑α2i t2i +2 ∑

i< jαiα jti+ j)tn = ∑α

2i t2i+n +2 ∑

i< jαiα jti+ j+n,

hence, as J is power-associative,


( f 2g)(x) = ∑α2i x2i+n +∑

i< jαiα j2xi+ j+n = ∑α

2i Uxi xn +∑

i< jαiα jxixnx j

=(∑α

2i Uxi +∑

i< jαiα jUxi,x j

)xn =U

∑αixi xn =U f (x)g(x).

Thus εx is a homomorphism of para-quadratic algebras, and we conclude that I⊆ k[t](+) is an ideal.Our next aim will be to show that I0 is an ideal in k[t]. By linearity, it suffices to show for f ∈ I0

that tn f belongs to I0 for all n∈N. We do so by induction on n. For n = 0 there is nothing to prove.Now assume n > 0 and that the assertion holds for n− 1. Since εx is a homomorphism of para-quadratic algebras, the induction hypothesis yields not only (tn f )(x) = 0, but also (tn+1 f )(x) =(t2tn−1 f )(x) = Ux(tn−1 f )(x) = 0, and we have shown tn f ∈ I0. Next let I0′ be any ideal of k[t]contained in I. Given f ∈ I0′ ⊆ I, we also have t f ∈ I0′ ⊆ I, hence f (x) = (t f )(x) = 0, whichshow I0′ ⊆ I0, so I0 is indeed the largest ideal of k[t] contained in I. Finally, let f ∈ I. Thenf 2(x) = U f (x)1J = 0 and (t f 2)(x) = U f (x)x = 0, hence f 2 ∈ I0. Similarly, not only 2 f ∈ I butalso (2t f )(x) = (t f )(x) = x f (x) = 0, hence 2 f ∈ I0. The assertions about R and π making acommutative diagram as shown in the exercise are now obvious, as is the statement that k[x] ⊆ Jis a para-quadratic subalgebra. The assertion about Ker(π) = I/I(0) follows immediately from thefact that f 2 and 2 f belong to I(0) for all f ∈ I. This completes the solution to (a).

(b) Write f = αntn + h ∈ In with αn ∈ k and h ∈ tn+2k[t]. Since 2 = 0 in k and n ≥ 2, wehave f 2 = α2

n t2n + h2 ∈ tn+2k[t], which is an ideal in k[t]. Thus U f g = f 2g ∈ tn+2k[t] ⊆ In for allg ∈ k[t]. Conversely, let g = ∑βiti ∈ k[t] with coefficients β j ∈ k for j ∈ N. Then Ug f = g2 f ≡αnβ 2

0 tn mod tn+2k[t] and hence Ug f ∈ In. Since g1g2 f = 2 f g1g2 = 0 ∈ In, it follows that In isan ideal in k[t](+). But it is not an ideal in k[t] since tn belongs to In but ttn = tn+1 does not. Andfinally, if Jn ∼= A(+) for some unital flexible k-algebra A, then xn = 0 would imply the contradictionxn+1 = 0 since powers in A and A(+) coincide.

151. (a) In the para-quadratic k-algebra k[t](+) we have U f gh = ( f g)2h = f 2g2h =U f Ugh for allsol.PARALIFTf ,g,h ∈ k[t], hence

U f g =U f Ug ( f ,g ∈ k[t]). (200) UFG

Since J is power-associative, the evaluation at x (Exc. 149) is a homomorphism k[t](+) → J ofpara-quadratic algebras with image k[x]. Applying this homomorphism to (200) therefore gives(1), hence not only the first equation of (2) but also

U ′U ′f (x)g(x)=U ′( f 2g)(x) =U ′f 2(x)U

′g(x) =U ′2f (x)U

′g(x),

U ′xn =U ′tn(x) =U ′nt(x) =U ′nx ,

and the proof of (a) is complete.(b) The polynomial

f := αdtd + · · ·+αn−1tn−1 + tn ∈ k[t]

kills x and hence belongs to I := Ker(εx). From Exc. 149 (a) we therefore conclude that

f 2 = α2d t2d + · · ·+ t2n ∈ k[t]

belongs to I0, so by Exc. 149 (a) we have (ti f 2)(x) = 0, i.e.,

α2d x2d+i + · · ·+ x2n+i = 0 (i ∈ N). (201) EFSQ

Now, setting kr[x] := ∑s≥r kxs for r ∈N, we find in (kr[x])r∈N a descending chain of submodules ofk[x]. Invoking (201), we therefore conclude that k2d [x] = k2d+1[x] = k2(d+1)[x] is a finitely gener-ated k-module. Thus the linear map U ′x : k2d [x]→ k2d [x], being surjective, is in fact bijective [116,38.15]. This property carries over to (U ′x)

d = U ′xd : k2d [x]→ k2d [x], whence there exists a unique


element v ∈ k2d [x] such that U ′xd v = x2d . Here (a) implies U ′x2dU ′v =U ′x2d , and we conclude that Uv

is the identity on k2d [x]. With c := v2, this means Uc = (Uv)2 = 1 on k2d [x] and c2 = v4 =Uvv2 = c,

c3 =Ucc = c, so c ∈ k2d [x] is an idempotent of the desired kind.(c) Let x ∈ J be a pre-image of c′ under ϕ . Then x2− x is nilpotent, so some integer d > 0 has

0 = (x2− x)d = x2d −·· ·+(−1)dxd . Pick v ∈ k2d [x] as in (b) and put c := v2. Writing u := ϕ(u)for u ∈ J, we conclude Uc′ v =Uxv =Uxd v = x2d = c′. On the other hand, v = ∑i≥2d αixi for somescalars αi ∈ k, i≥ 2d, so with α :=∑i≥2d αi we obtain v= αc′, hence c′ =Uc′ v= αUc′c′ = αc′ = v.But this implies c = v2 = c′2 = c′, and the problem is solved.

152. We closely follow the arguments used in the proofs of 10.16−10.19.sol.OUTCEN(a) Since J is outer simple, Mult(J) acts irreducibly on J. Thus the assertion follows from

Schur’s lemma [47, p. 118].(b) Since Mult(J) acts faithfully and irreducibly on J, it is a primitive Artinian k-algebra [47,

Def. 4.1]. Moreover, by definition, its centralizer in Endk(J) is Cent0(J). Thus the assertion followsfrom the double centralizer theorem [47, Thm. 4.10].

(c) If J is outer central and outer simple, the assertion follows from (b). Conversely, supposeJ 6= 0 and Mult(J) = Endk(J). Then J is outer simple, and (b) implies that Cent0(J) belongs tothe centre of Endk(J). Hence J is outer central.

(d) (i)⇒ (ii). Let k′ be an extension field of k and put J′ = J⊗k′ as a para-quadratic k′-algebra.After identifying Endk′ (J′) = Endk(J)⊗ k′ canonically, a moment’s reflection shows Mult(J′) =Mult(J)⊗ k′. Hence the assertion follows from (c).(ii)⇒ (iii). Obvious.(iii)⇒ (i). If I ⊆ J is a non-trivial outer ideal, then so is I⊗ k ⊆ J⊗ k, a contradiction. Hence J isouter simple. Since the multiplication algebra of J behaves nicely under base field extensions, sodoes its centralizer. Therefore Cent0(J)⊗ k agrees with the outer centroid of J⊗ k and hence is afinite algebraic field extension of k. As such, it has degree 1, forcing Cent0(J) = k1J , and J is outercentral.

153. (a) Setting c := ∑ri=1 ci, we havesol.COS

c2 =r

∑i=1

c2i +∑

i< jci c j =

r

∑i=1

ci = c,

c3 =Ucc =r

∑i, j=1

Uci c j +∑i<l

r

∑j=1cic jcl=

r

∑i=1

c3i =

r

∑i=1

ci = c,

and we have shown that c is an idempotent in J. Hence so is cr+1 := 1J−c (30.9) and it remains toshow that (c1, . . . ,cr,cr+1) is an orthogonal system of idempotents, completeness being obvious.For 1≤ i, l ≤ r distinct, we compute


Uci cr+1 =Uci 1J−r

∑j=1

Uci c j = c2i −Uci ci = ci− c3

i = ci− ci = 0,

Ucr+1 ci =U1J−cci = ci− c ci +Ucci = ci−r

∑m=1

cm ci +r

∑m=1

Ucm ci + ∑m<ncmcicn

= ci−2ci + ci = 0,

cicicr+1= ci ci−r

∑m=1cicicm= 2c2

i −2Uci ci = 2c2i −2c3

i = 0,

cr+1cr+1ci= (1J− c)(1J− c)ci= 2ci− c ci− c ci +ccci

= 2ci−2ci−2ci +r

∑m,n=1

cmcnci=−2ci +2ci = 0,

ci cr+1 = ci 1J− ci c = 2ci−2ci = 0,

ciclcr+1= ci cl −r

∑m=1ciclcm= 0.

Hence (c1, . . . ,cr,cr+1) is a complete orthogonal system of idempotents in J.(b) If (c1, . . . ,cr) is a (complete) orthogonal system of idempotents in A, then it is clearly one

in J. Conversely, suppose it is a (complete) orthogonal system of idempotents in J. Comparing(3) with (30.9.1), we see that the idempotents ci,c j , 1 ≤ i < j ≤ r, are orthogonal in J, hence, aswe have seen in 30.9, in A. But this means that (c1, . . . ,cr) is a (complete) orthogonal system ofidempotents in A.

Section 31

154. (i)⇒ (iii). Obvious by the definition of a Jordan algebra.sol.CHRAJO(iii)⇒ (ii). This follows from the arguments of 31.3 since the identities (5)−(8) are respectively

the same as (resp. part of) (31.3.6)−(31.3.9).(ii) ⇒ (i). Let R ∈ k-alg. The left (resp. right) of (6) defines a k-3-quadratic map F (resp. G)

from J3 to Endk(J), and we have F =G by hypothesis. Thus FR =GR by Exc. 49, which by (30.12)implies that (6) holds in JR. For analogous reasons, (7), (8) holds in JR as well.Proof of (5). By linearity in z, we may assume z = wR, w ∈ J. Assume first x = uR, u ∈ J. Sinceboth sides of (5) are quadratic in y and agree for y = vR, v ∈ J, by hypothesis, it follows that (5)holds for all y ∈ JR. Now fix y ∈ JR and put

M := x ∈ JR |UxUyUx,z +Ux,zUyUx =UUxy,Ux,zy,

which is obiously closed under multiplication by scalars in R and, as we have just seen, containsall elements uR, u ∈ J. Hence (5) will follow once we have shown that M is closed under addition.But this is implied by (7) and a straightforward computation.Proof of (4). By linearity in y, we may assume y = vR, v ∈ J. Arguing as before, it suffices to showthat

M := x ∈ JR |UxVy,x =Vx,yUxis closed under addition. But this is immediate, by (8) and a straightforward computation.Proof of (3). Since (3) is quadratic in y, it suffices to prove it for y = vR, v ∈ J. Again we will bethrough once we have shown that

M := x ∈ JR |UUxy =UxUyUx

is additively closed. This follows from (5), (6) by a lengthy, but still straightforward, computation.


Since a Jordfan algebra satisfies (3)−(8), so do all its subalgebras and homomorphic images.By what we ahve just shown, therefore, they are all Jordan algebras.

155. (a) By definition it suffices to show that I is an outer ideal if UJI ⊆ I. Indeed, if this inclusionsol.OUTIJOholds, then, for all x,y ∈ J, we have Ux,yI ⊆ I and given z ∈ I, equations (31.3.14), (31.3.5) imply

xyz=Vx,yz =VxVyz−Ux,yz =Ux,1JUy,1J z−Ux,yz ∈ I,

which shows JJI ⊆ I and completes the proof.(b) If c2 = c, then , then U2

c = Uc2 = Uc by (31.14.1), hence c3 = Ucc = Ucc2 = UcUc1J =U2

c 1J =Uc1J = c2 = c. Thus c is an idempotent.(c) Let x,y ∈ J and z ∈ Z(J). Then (31.3.14) implies Vx,z = VxVz−Ux,z = VxUz,1J −Uz,x = 0

and, similarly, Vz,x = 0. From the fundamental formula (31.3.2) we deduce UUxz = 0 =UUzx. And,finally, applying (31.3.9), we obtain

UUxz,y =Uy,Uxz =Ux,yVz,x +UxVz,y−Ux,Ux,yz =−Ux,Vx,zy = 0,

UUzx,y =Uy,Uzx =Uz,yVx,z +UzVx,y−Uz,Uz,yx = 0.

Hence UJZ(J)+UZ(J)J ⊆ Z(J), and we have shown by (a) that Z(J)⊆ J is an ideal. Now supposethat J is simple. Then J 6= 0 by definition, hence U1J = 1J 6= 0, which implies Z(J) = 0 bysimplicity. The final statement now follow from Thm. 30.17

156. Let a ∈ Cent(J). Then LaLx = LxLa = Lax for all x ∈ J. Since, therefore, La and Lx commute,sol.CENCENTso do La and Ux = 2L2

x −Lx2 , for all x ∈ J. Moreover, this and a2 ∈ Cent(J) imply

ULax = 2L2ax−L(ax)2 = 2L2

aL2x −La2x2 = 2La2 L2

x −La2 Lx2 = L2aUx,

ULax,y = 2(LaxLy +LyLax−L(ax)y

)= 2La(LxLy +LyLx−Lxy) = LaUx,y

for all x,y ∈ J, whence (30.13.1) yields La ∈ Cent(Jquad). Conversely, let ϕ ∈ Cent(Jquad). By(30.13.2) we have ϕVx =Vxϕ for x∈ J, and from 31.4 we deduce ϕLx = Lxϕ , hence ϕ(xy) = xϕ(y)for all x,y ∈ J. Setting y = 1J , we obtain ϕ = La with a = ϕ(1J). In particular, La commutes withLx for all x ∈ J, whence a belongs to the centre of J. Summing up we have shown that the injectivealgebra homomorphism L : Cent(J)→ Endk(J) has image Cent(Jquad), and the assertion follows.

157. If J = J(M,q,e), then (9) holds strictly by (31.12.5), (31.12.6). Conversely, suppose (9) holdssol.CHAJOCLIstrictly. Then the second equation of (9) may be written as Uxx = t(x)x2− q(x)x, and linearizingwe may apply (31.3.13) to conclude

x2 y+Uxy =Ux,yx+Uxy = t(x)x y+ t(y)x2−q(x,y)x−q(x)y. (202) LICUB

Applying the trace (resp. q(−,y)) to the first equation of (9), we obtain

t(x2) = t(x)2−2q(x), q(x2,y) = t(x)q(x,y)−q(x)t(y),

while linearizing it yieldsx y = t(x)y+ t(y)x−q(x,y)e.

Hence on the one hand

x2 y+Uxy =Uxy+ t(x2)y+ t(y)x2−q(x2,y)e

=Uxy+(t(x)2−2q(x)

)y+ t(x)t(y)x−q(x)t(y)e− t(x)q(x,y)e+q(x)t(y)e

=Uxy+(t(x)2−2q(x)

)y+ t(x)t(y)x− t(x)q(x,y)e,

while on the other


t(x)x y+ t(y)x2−q(x,y)x−q(x)y = t(x)2y+ t(x)t(y)x− t(x)q(x,y)e

+ t(x)t(y)x−q(x)t(y)e−q(x,y)x−q(x)y

Comparing by means of (202) and writing x 7→ x for the conjugation of (M,q,e), we obtain

Uxy = q(x)y+ t(x)t(y)x−q(x)t(y)e−q(x,y)x = q(x, y)x−q(x)y.

Thus J and J(M,q,e) not only have the same unit element but also the same U-operator, whichimplies J = J(M,q,e).

158. Concerning the first part, assume that x ∈ J is nilpotent. Then (31.12.11) shows that q(x) issol.NIPOQUAnilpotent, while Exc. 150 yields a positive integer m satisfying xn = 0 for all integers n ≥ m. Wemust show that t(x) is nilpotent, and we will do so by induction on m. For m = 1, there is nothingto prove. For m = 2, we have x2 = 0 and conclude from (31.12.12) that t(x)2 = 2q(x) is nilpotent.Hence so is t(x). We may therefore assume m≥ 3. Since [m

2 ]≤m2 < [m

2 ]+1 and m > 2, we deduce

[m2]+1≤ m

2+1 = m+1− m

2< m+1−1 = m,

and if n≥ [m2 ]+1, then 2n≥ 2([m

2 ]+1)> 2 m2 =m. On the other hand, (30.8.5) yields (x2)n = x2n =

0, and the induction hypothesis implies that t(x2) is nilpotent. But then so is t(x)2 = t(x2)+2q(x)by (31.12.12), which shows that t(x) is nilpotent and completes the induction. Summing up wehave shown that if x is nilpotent, then so are t(x) and q(x). Conversely, assume that t(x) and q(x)are nilpotent. Then so are t(x)x and q(x)1J , which both belong to k[x]. Since J is power-associativeby Prop. 31.15, Exc. 150 now implies that x2 = t(x)x−q(x)1J is nilpotent. Hence so is x, and thefirst part of the problem is solved.

To establish the second part, put

N := x ∈M | q(x),q(x,y) ∈ Nil(k) for all y ∈M

and let x∈N, y,z∈M. Then (31.12.11) yields q(Uxz) = q(x)2q(z)∈Nil(k) and, similarly, q(Uzx)∈Nil(k). Moreover, since the conjugation of J leaves q invariant, the definition of the U-operatorimplies

q(Uxz+Uzx,y) = q(x, z)q(x,y)+q(z, x)q(z,y)−q(x)q(z,y)−q(z)q(x,y)

= q(x, z)q(x+ z,y)−q(x)q(y, z)−q(x, y)q(z) ∈ Nil(k).

Thus UNJ+UJN ⊆ N, and we have shown in view of Exc. 155 (a) that N ⊆ J is an ideal which, bythe first part of this problem, is nil. This proves N ⊆ Nil(J). Conversely, let x ∈ Nil(J). Then q(x)and t(x) are nilpotent and Nil(J) contains xy= t(x)y+t(y)x−q(x,y)1J . Hence t(x)y−q(x,y)1J ∈Nil(J), forcing

q(t(x)y−q(x,y1J

)= t(x)2q(y)− t(x)q(x,y)t(y)+q(x,y)2

to be nilpotent. Since, therefore, q(x,y)2 is nilpotent, so is q(x,y) and we have shown x ∈ N. Thisshows Nil(J)⊆ N and completes the proof.

159. We begin with a simple lemma.sol.EVPOQUALemma. Let (M,q,e) be a pointed quadratic module over any commutative ring and J :=J(M,q,e). If x ∈ J satisfies x2 = 0, then x3 =−q(x)x.

Proof. By (31.12.5), we have t(x)x = q(x)1J , which by taking traces implies t(x)2 = 2q(x). Ap-plying (31.12.4), we therefore obtain

x3 =Uxx = q(x, x)x−q(x)x =(t(x)2−2q(x)

)x− t(x)q(x)1J +q(x)x

=(− t(x)2 +q(x)

)x =−q(x)x,


as claimed. (a) We have I2 = kt2⊕kt4⊕kt5⊕·· · . Hence, writing x for the imagae of t under the canonical

map k[t](+) → J2, we conclude that J2 is free as a k-module, with basis 1J2 ,x,x3. In particular, x

and x3 are linearly independent. Since x2 = 0, and by the preceding lemma, therefore, J2 cannot beisomorphic to the Jordan algebra of a pointed quadratic form over k.

(b) Let k0 be a commutative ring and assume 2 = 0 in k0. Furthermore, let k = k0[ε], ε2 = 0,be the k0-algebra of dual numbers. As in Exc. 86, we may view k0 as a k-algebra by means of thehomomorphism k→ k0, α 7→ α , given by ε = 0. Now let n be a positive integer, write (e1, . . . ,en)for the canonical basis of kn over k, put M := k0⊕ kn as a k-module, define q : M→ k by

q(α⊕

n

∑i=1

αiei)

:= α2 + εα

21 +

n

∑i=2

α2i (α0,α1, . . . ,αn ∈ k),

and put e := 1k0⊕0. We claim that the map q is well defined. Indeed, if α,β ∈ k satisfy α = β , thenα = β + εγ , for some γ ∈ k, hence α2 = β 2 + ε2γ2 = β 2, as desired. It now follows trivially that(M,q,e) is a pointed quadratic module over k whose trace is identically zero. In particular, puttingJ := J(M,q,e) and x := e1, we have t(x) = 0 and q(x) = ε , hence q(x)1J = ε(1k0 ⊕0) = ε = 0 =

t(x)x, so (31.12.5) implies x2 = 0. Now the preceding lemma yields x3 =−q(x)x =−εe1 6= 0, andwe have obtained an example of the desired kind.

160 . Our solution is basically the same as the one to Exc. 82. First of all, assume that k 6= 0sol.IDPOQUAis connected, so 0 and 1 are the only idempotents of k. If c ∈ J has t(c) = 1 and q(c) = 0, then(31.12.5) shows c2 = c, so c ∈ J is an idempotent, which cannot be zero since t(c) = 1 6= 0 andcannot be 1 since q(c) = 0 6= 1. Conversely, let c ∈ J be an idempotent 6= 0,1J . Then q(c) by(31.12.11) is an idempotent in k, and since k is connected, we either have q(c) = 1 or q(c) = 0.In the former case, c = c2 = t(c)c− 1J , hence 1J = (t(c)− 1)c, and taking norms we conclude(t(c)− 1)2 = 1. But this implies c = α1J with α = t(c)− 1, α2 = 1. We therefore obtain 1J =α21J = c2 = c, a contradiction. Hence q(c) = 0. Then (31.12.5) implies c = c2 = t(c)c, and takingtraces gives t(c)2 = t(c). Thus t(c) ∈ k is an idempotent which cannot be zero since this wouldimply the contradiction c = 0. Thus t(c) = 1, as claimed.

The remainder of this exercise is established in exactly the same way as the corresponding partof Exc. 82.

161. The solution to this exercise is exactly the same as the one to Exc. 84.sol.DECPOQUA

162 . If C is alternative, then (20.3.2) shows C(+) = J(C,nC,1C). Conversely, let this be so. Forsol.COBRPLx,y ∈C we apply (19.5.5), (19.5.10), (19.5.4) and obtain

nC(x, y)x−nC(x)y = xyx = (x y)x− (yx)x = tC(x)yx+ tC(y)x2−nC(x,y)x− (yx)x

= tC(x)yx+(tC(x)tC(y)−nC(x,y)

)x− tC(y)nC(x)1C− (yx)x

= nC(x, y)x−nC(x)y−nC(x)y+(yx)x.

Comparing, we arrive at one of Kirmse’s identities (20.3.1), i.e., at (yx)x = nC(x)y. But this meansthat C is right alternative. Since it is also flexible by hypothesis, it is, in fact, alternative.

163. Homomorphism of pointed quadratic modules are clearly homomorphisms of the correspond-sol.POJOCATing Jordan algebras. Hence the assignment in question defines a functor from k-poquainj to k-jordinj, which by its very definition is faithful Before we can show that it is full, we need thefollowing lemma.Lemma. Let (M,q,e) and (M,q1,e) be pointed quadratic modules over k, with the same under-lying k-module M and the same base point e. If M is projective and J(M,q,e) = J(M,q1,e), thenq = q1.

Proof. Since e ∈ M is unimodular by Lemma 12.14, the proof of Prop. 19.15 can be extendedverbatim to this slightly different setting.


Now let (M,q,e) and (M′,q′,e′) be pointed quadratic modules over k such that M and M′ areboth projective and let ϕ : J(M,q,e)→ J(M′,q′,e′) be an injective homomorphism of Jordan al-gebras over k. Setting q1 := q′ ϕ , we conclude that (M,q1,e) is a pointed quadratic module overk (since ϕ preserves identity elements) and ϕ : (M,q1,e)→ (M′,q′,e′) is a homomorphism ofpointed quadratic modules. Hence ϕ : J(M,q1,e)→ J(M′,q′,e′) is a homomorphism of Jordan al-gebras. But ϕ was assumed to be injective. Hence J(M,q,e) = J(M,q1,e), which by the precedinglemma implies q1 = q. Hence ϕ : (M,q,e)→ (M′,q′,e′) is a homomorphism of pointed quadraticmodules. Thus the functor in question is full.

164. We put J := J(V,q,e). Since F has characteristic not 2, we may extend e= 1J to an orthogonalsol.POQUAINbasis of V relative to q, allowing us to identify q = 〈1,δ ,δε〉quad for some δ ,ε ∈ F×. Now put

K := Cay(F,−ε) = F⊕F j1, B := Cay(K,−δ ) = K⊕K j2.

as in 21.2. Then B is a quaternion algebra over F , while K ⊆ B is a quadratic etale subalgebra.Let τ be the involution of B corresponding to K by Exc. 110 (b). Then τ(a⊕ b j2) = a⊕ b j2for all a,b ∈ K, which implies H(B,τ) = W := F1B⊕K j2 = F1B⊕F j2⊕F j1 j2. By Exc. 162we have B(+) = J(B,nB,1B) and therefore H(B,τ) = J(W,nB|W ,1B). By standard properties ofthe Cayley-Dickson construction, the vectors 1B, j2, j1 j2 are orthogonal relative to nB and satisfynB(1B) = 1, nB( j2) = δ , nB( j1 j2) = nB( j1)nB( j2) = δε . Hence the pointed quadratic modules(V,q,e) and (W,NB|W ,1B) are isometric, forcing J and H(B,τ) as the corresponding Jordan alge-bras by Exc. 163 to be isomorphic.

It remains to prove that (B,τ) as a quaternion algebra with involution is unique up to isomor-phism. To this end, we require a lemma.Lemma. Let B, B′ be quaternion algebras over F and suppose W ⊆ (B,nB), W ′ ⊆ (B′,nB′ ) arenon-singular subspaces of co-dimension 1. If W and W ′ are isometric, then B and B′ are isomor-phic.

Proof. By Lemma 12.10 we have B =W ⊥ Fu, B′ =W ′ ⊥ Fu′ for some non-zero u ∈ B, u′ ∈ B′.Recall that the determinant of a quadratic space like (B,nB) (resp. (B′,nB′ )) is unique up to non-zero square factors in F . As a matter of fact, we deduce from (22.19.1) that this determinant is asquare. On the other hand, it agrees with det(W )nB(u) (resp. det(W ′)nB′ (u′)). On the other hand,W and W ′ being isometric, their determinants differ by a square factor in F . Hence so do nB(u) andnB′ (u′). But this means that B and B′ are norm-equivalent. By the norm equivalence theorem 26.7,therefore, they are isomorphic.

Now let (B′,τ ′) be another quaternion algebra with involution over F having W ′ := H(B′,τ ′)∼=J(V,q,e). By Exc. 163, the quadratic subspaces W ⊆ (B,nB) and W ′ ⊆ (B′,nB′ ) are not only non-singular but also isometric. Our preceding lemma therefore implies that B and B′ are isomorphic.We may thus assume B = B′. Let (1B, j′2, j′3) be an orthogonal basis of W ′ such that nB( j′2) = δ ,nB( j′3) = δε , with δ ,ε as above. Since (B,nB) has trivial determinant, we conclude W ′⊥ = F j′1,nB( j′1) = ε . Then F [ j′1]∼= Cay(F,−ε) = K. Now (19.5.5) implies for i = 2,3:

0 = τ′(tB( j′1) j′i + tB( j′i) j′1−nB( j′1, j′i)1B

)= τ

′( j′1 j′i) = τ′( j′1) j′i

= tB( j′1) j′i + tB( j′i)τ′( j′1)−nB

(τ′( j′1), j′i

)1B =−nB

(τ′( j′1), j′i

)1B.

Thus nB(τ′( j′1), j′i) = 0, and since we trivially have nB(τ

′( j′1),1B) = 0, we deduce τ ′( j′1) ∈W ′⊥ =F j′1, Hence τ ′( j′1) =± j′1. Here τ ′( j′1) = j′1 would imply τ ′ = 1B, a contradiction. Hence τ ′( j′1) =− j′1. Thus Thus τ ′ is the involution of B corresponding to F [ j′1]∼=K, which implies (B,τ ′)∼= (B,τ)by Exc. 110 (b).

Section 32

165 . (a) x ∈ Rad(q) implies Uxy = q(x, y)x− q(x)y = 0 for all y ∈ J, hence Ux = 0, and x is ansol.ABZECLIabsolute zero divisor of J. Conversely, let this be so. Then (31.12.9) yields q(x)2 =Uxx2 = 0, andapplying q we conclude q(x)4 = 0, hence q(x) = 0 since k is reduced. For any y ∈ J, we therefore


obtain 0 = Uxy = q(x,y)x and then q(x,y)2 = q(q(x,y)x,y) = 0. Thus q(x,y) = 0, and we haveproved x ∈ Rad(q).

(b) Here k is arbitrary. Then k := k/Nil(k) ∈ k-alg is reduced, and we have a canonical identi-fication Jk = J := J/Nil(k)J via 10.2, matching zk for z ∈ J with z, the image of z under the naturalmap J→ J. Since x is an absolute zero divisor in J, x is one in J, so by the special case just treatedwe have q(x) = q(x, y) = 0 for all y ∈ J, where q = qk is the k-quadratic extension of q. But thismeans that q(x) and q(x,y) are nilpotent elements of k, for all y ∈ J, which implies x ∈ Nil(J) byExc. 158.

Section 33166. (a) By the results of §§ 31,32, equations (2)−(8) hold for all i, j,m,n, p∈N. For n∈Z, n > 0,sol.EVIVNVwe combine (1) with (33.3.1) and obtain Ux−n =U(x−1)n =Un

x−1 = (U−1x )n =U−n

x . Hence (2) holdsfor all n ∈ Z.

By (2), the set of all integers m such that (3) holds for all integers n is a subgroup of Z. Hence,in order to prove (3), we may assume m = 1. Then we must show Uxx−n = x2−n for all n ∈ N andwe do so by induction on n. For n = 0,1, there is nothing to prove. For n≥ 2, assuming the validityof the assertion for n−2 yields Uxx−n =Ux(x−1)n =UxUx−1 (x−1)n−2 = x2−n, as claimed. Thus (3)holds.

Turning to (4), we first assume n ∈N and then argue by induction. For n = 0,1 there is nothingto prove. Now let n ≥ 2 and assume the assertion is valid for n− 2. Then (xm)n = Uxm (xm)n−2 =Uxm xmn−2m = x2m+mn−2m = xmn. as claimed. Next assume n < 0. The case just treated and (2), (3)yield U(xm)−n (xm)n = (xm)−n = x−mn =Ux−mn xmn =U(xm)−n xmn, and since U(xm)−n is bijective, theproof of (4) is complete.

Next we establish (7), where we may clearly assume i= 1. For l ∈N such that 2l+m,2l+n∈N,we obtain, using (2), (31.3.3), (3), (32.4.5),

UxUxm,xn =Ux−lUxUxlUxm,xnUxlUx−l =Ux−lUxUUxl xm,Uxl xnUx−l =Ux−lUxUx2l+m,x2l+nUx−l

=Ux−lUx2l+m+1,x2l+n+1Ux−l =UUx−l x2l+m+1,Ux−l x2l+n+1 =Uxm+1,xn+1

and, similarly, Uxm,xnUx =Uxm+1,xn+1 .Thus (7) holds.Now we can prove (5) by letting l ∈ N satisfy 2l +m,2l +n,2l + p ∈ N. Then (7) yields

Ux3lxmxnxp=Ux3lUxm,xp xn =UxlUxm,xnUxlUxl xn =UUxl xm,Uxl xnUxl xn = x2l+mx2l+nx2l+p

= 2x6l+m+n+p =Ux3l (2xm+n+p).

Thus (5) holds.Turning to (6), we use (2), (31.3.4), (3), (7) and obtain

Vxm,xnUxm+n =Vxm,xnUxmUxn =Uxm,Uxm xnUxn =Uxm,x2m+nUxn =Uxm+n,x2(m+n)

=Uxm+n,Uxm+n 1J =Vxm+n,1JUxm+n =Vxm+nUxm+n .

Hence (6) follows.Finally,in order to establish (8), we let l ∈ N satisfy l + i, l + j, l +m, l + n ∈ N. Then, by (2),

(7), (32.4.6), 31.3.3),

Uxi,x jUxm,xn =Ux−lUxlUxi,x jUxlUxm,xnUx−l =Ux−lUxl+i,xl+ jUxl+m,xl+nUx−l

=Ux−l

(Ux2l+i+m,x2l+ j+n +Ux2l+i+n,x2l+ j+m

)Ux−l

=Uxi+m,x j+n +Uxi+n,x j+m ,

as claimed.(b) Arguing as in the solution to Exc. 148 (a), it follows that ε×x : k[t, t−1](+) → J is a homo-

morphism of Jordan algebras forcing k[x,x−1] ⊆ J to be a subalgebra containing ε×x (t±1) = x±1.


Writing J′ for the subalgebra of J containing x and x−1, we therefore have J′ ⊆ k[x,x−1]. On theother hand, it follows immediately from the definitions that J′ contains all powers of x with integerexponents. Since, therefore, k[x,x−1]⊆ J′, we have equality: J′ = k[x,x−1].

(c) Arguing as in 32.3, replacing N by Z everywhere and using (7), (8), we deduce (9). Thefinal assertion may be proved in exactly the same manner as Thm. 32.11.

167. (a) By (29.3.2) we havesol.INVLIN

[Lx2 ,Ly] = 0 ⇐⇒ [Lx,Lxy] = 0.

(b) (i) ⇒ (ii). Let y := x−1. Then 2xy = x y = x1Jx−1 = 2 · 1J by Exc. 166, (5). Hencexy = 1J . Moreover, Exc. 166, (8) shows 4[Lx,Ly] = [Ux,1J ,U1 j ,y] = 0, so x and y operator commute.

(ii)⇒ (iii). We have xy = 1J , and since x and y operator commute, we also obtain x2y = y(xx) =x(yx) = x(xy) = x,

(iii)⇒ (i). Since x and xy = 1J operator commute, so do x2 and y, by (a). For the same reasony2 and x operator commute. Hence

Uxy2 = 2x(y2x)− x2y2 = x2(yy) = y(x2y) = yx = 1J ,

and x is invertible in J by Prop. 33.2.Next suppose y∈ J satisfies (ii).. Then x is invertible and Uxy= 2x(xy)−y(xx) = 2x−x(yx) = x,

hence y =U−1x x = x−1. Similarly, if y satisfies (iii), then Uxy = 2x(xy)−x2y = x and again y = x−1.

168 . If x is invertible in J, then 1 = q(1J) = q(Uxx−2) = q(x)2q(x−2) by (31.12.11), so q(x)sol.INVPOIis invertible in k. Conversely, let this be so and put y := q(x)−1x. Then (31.12.9) yields Uxy =q(x)−1Uxx= x. On the other hand, writing t for the trace of (M,q,e), we apply (31.12.10), (31.12.5)and obtain not only q(y) = q(x)−1 but also

Uxy2 = t(y)Uxy−q(y)Ux1J = q(x)−1t(x)x−q(x)−1x2 = 1J .

Hence (33.1.1) shows that x is invertible in J with inverse y. The rest is clear.

169. (a) q( f ) : M→ k is a quadratic form and e( f ) ∈M satisfies q( f )(e( f )) = q( f )q( f−1) = 1 bysol.ISTPOIExc. 168, (10). Hence (M,q,e)( f ) is a pointed quadratic module over k. For x ∈M, the definitionsimply

t( f )(x) = q( f )(e( f ),x) = q( f )q(q( f )−1 f ,x

)= q( f ,x),

x( f ) = t( f )(x)e( f )− x = q( f )−1q( f ,x) f − x,

hence (12) and (13). In (14), both sides live on the same k-module and have the same identityelement. Hence it suffices to show that they have the same U-operator. Let x,y ∈M and write Ux(resp. U ′x) for the U-operator of J(M,q,e) (resp. J((M,q,e)( f )). We first note

U f y =U f y = q( f ,y) f −q( f )y (203) UEFY

by (31.12.3), (31.12.4) and (31.12.10). Hence (12) and (13) imply

U ′xy = q( f )(x, y( f ))x−q( f )(x)y( f )

= q( f )q(x,q( f )−1q( f ,y) f − y

)x−q( f )q(x)

(q( f )−1q( f ,y) f − y

)=(q( f ,x)q( f ,y)−q( f )q(x,y)

)x−q(x)

(q( f ,y) f −q( f )y

)= q(x,q( f ,y) f −q( f )y

)x−q(x)

(q( f ,y) f −q( f )y

)= q(x,U f y

)x−q(x)U f y =UxU f y =U ( f )

x y,

hence (14).


170 . Setting 1(y) := 1J(y) as in (33.8.1), we apply (33.8.2) and obtain Ux−1+y−1Uy = U (y)1(y)+z

,sol.FOUROP

z := x−1. Here Thm. 33.10 (a) yields z(−1,y) = U−1y (x−1)−1 == U−1

y x, hence Uyz(−1,y) = x. Us-ing Exc. 166, (9), we therefore conclude

UxUx−1+y−1Uy =UUyz(−1,y)U(y)1(y)+z

=UyU(y)z(−1,y)U

(y)1(y)+z

=UyU(y)z(−1,y)+1(y)

=UyUz(−1,y)+y−1Uy

=UUyz(−1,y)+Uyy−1 =Ux+y.

171. We have xx−1 = LxL−1x 1J = 1J , and the Jordan identity (29.1.2) yields x(x2x−1) = x2(xx−1) =sol.LININV

x2, hence x2x−1 = x since Lx is bijective. We therefore conclude from Exc. 167 that x is invertiblein J with inverse x−1. Now (33.3.1) and (29.9.6) imply that Lx−1 = 1

2Vx−1 = 12U−1

x Vx = U−1x Lx is

bijective, forcing x−1 to be linearly invertible with linear (= ordinary) inverse (x−1)−1 = x.Now let F be a field of characteristic not 2 and (M,q,e) a pointed quadratic module over F . If

q is anisotropic and J := J(M,q,e) has dimension at most 2, then Exc. 97 shows that J is a fieldand, in particular, a linear division algebra.. Conversely, let J be a linear division algebra. By whatwe have just seen, J is a Jordan division algebra, forcing q to be anisotropic by Exc. 168. AssumeJ has dimension n≥ 3 and extend 1J to an orthogonal basis (e1 = 1J ,e2,e3, . . . ,en) of J relative toDq. Applying (31.12.8), we obtain

e2e3 =12(q(e1,e2)e3 +q(e1,e3)e2−q(e2,e3)e1

)= 0.

Hence e2, though invertible, is not linearly invertible, and J cannot be a linear division algebra.This contradiction shows n≤ 2 and completes the proof.

172 . (a) Let η : J → J′ be a strong homotopy, so some p ∈ J× makes η : J(p) → J′ a homo-sol.STRHOTmorphism. Hence η : J → J′ is a homotopy by Prop. 33.16 (b). Conversely, let η : J → J′ be ahomotopy such that η(J×) = J′×. Then there exists a p′ ∈ J′× such that η : J→ J′(p′) is a homo-morphism. By hypothesis, we find an element p ∈ J× satisfying η(p) = p′−2, Then 33.8 (ii) andCor. 33.11 show that

η : J(p) −→ J′(p′)(η(p)) = J′(p′)(p′−2) = J′

is a homomorphism, forcing η : J→ J′ to be a strong homotopy.(b) Let F be a field and µ ∈ F [t] be an irreducible polynomial of degree at least 2, making

K := F [t]/(µ) a finite algebraic extension field of F . We denote by π : F [t]→ K the canonicalprojection and put J := F [t](+), J′ := K(+). Pick an element p′ ∈ K \F1K . Then p′ ∈ J′× and thelinear map

η : J −→ J′, f 7−→ η( f ) := π( f )p′−1 (204) ETAM

fits into the commutative diagram

Jη //

π

J′(p′)

J′.

Rp′−1

∼=

EE

where, by (33.12.1), the horizontal arrow is a homomorphism. Thus η : J → J′ is a homotopy.Assume that this homotopy is strong. Then there exists a p ∈ J× such that η : J(p) → J′ is ahomomorphism. But Ex. 33.6 shows J× = F [t]× = F×, hence p ∈ F×. By (204), therefore,

1K = 1J′ = η(1J(p) ) = η(p−1) = p−1η(1J) = p−1 p′−1,


and we arrive at the contradiction p′ = p−11K ∈ F1K . Thus η is not a strong homotopy.

173. (a) We argue by induction if n≥ 0. For n = 0, we have to show λ (1J) = 0, which follows fromsol.VALJO(16) by setting x = y = 1J : λ (1J) = λ (U1J 1J) = 3λ (1J), and the assertion follows. For n = 1, thereis nothing to prove, and if the formula is valid for all natural numbers < n where n≥ 2, then (16)implies λ (xn) = λ (Uxxn−2) = 2λ (x)+ λ (xn−2) = 2λ (x)+ (n− 2)λ (x) = nλ (x). If n = −1, weobtain λ (x) = λ (Uxx−1) = 2λ (x)+λ (x−1), hence λ (x−1) = −λ (x). And finally, for any integern > 0, the definition (cf. Exc. 166 (1)) implies λ (x−n) = λ ((x−1)n) = nλ (x−1) = −nλ (x). Thiscompletes the proof.

(b) We have (vw)2 = v2w2 = Uvw2. Hence (a) implies 2λ (vw) = λ ((vw)2) = λ (Uvw2) =2λ (v)+λ (w2) = 2(λ (v)+λ (w)), and the first part of (b) follows. The rest is clear.

(c) is straightforward.(d) The statement is obvious if λ is trivial, i.e., if λ (x) = 0 for all x ∈ J×. Otherwise, (a)

yields a least positive integer m ∈ Γ := λ (J×), and we have Zm⊆ Γ . Conversely, let x ∈ J×. Thenλ (x) ∈ Z can be written as λ (x) = (2i+ j)m+ l, with i, j, l ∈ Z, j = 0,1, 0≤ l < m. This impliesjm+ l =−(2im−λ (x))∈Γ since Γ by (16) and (a) is closed under the operation (α,β ) 7→ 2α−β .If j = 0, we conclude l ∈ Γ , hence l = 0 by the minimality of m. On the other hand, if j = 1, weconclude m− l = (2m− (m+ l)) ∈ Γ , forcing again l = 0 by the minimality of m. In any event,λ (x) = (2i+ j)m ∈ Zm, which completes the proof.

(e) We have to verify (15)−(17) and do so by a straightforward computation: For x,y ∈ J weobtain

λ(p)(x) = ∞ ⇐⇒ λ (p)+λ (x) = ∞ ⇐⇒ λ (x) = ∞ ⇐⇒ x = 0,

λ(p)(U (p)

x y) = λ (p)+λ (UxUpy) = λ (p)+2λ (x)+2λ (p)+λ (y)

= 2(λ (p)+λ (x)

)+(λ (p)+λ (y)

)= 2λ

(p)(x)+λ(p)(y),

λ(p)(x+ y) = λ (p)+λ (x+ y)≥ λ (p)+min

λ (x),λ (y)

= min

λ (p)+λ (x),λ (p)+λ (y)

= min

λ(p)(x),λ (p)(y)

,

and finally,

(λ (p))(q)(x) = λ(p)(q)+λ

(p)(x) = λ (p)+λ (q)+λ (p)+λ (x) = 2λ (p)+λ (q)+λ (x)

= λ (Upq)+λ (x) = λ(Upq)(x).

(f) There is clearly no harm in assuming x,y,z ∈ J×. Suppose first that the assertion holds fory = 1J , so we have

λ (x z)≥ λ (x)+λ (z) (x,z ∈ J), (205) LACI

and let y ∈ J× be arbitrary. Then we pass to the y-isotopes J(y) and λ (y) and combine (33.8.5) with(205) to conclude

λ (y)+λ (xyz) = λ(y)(x(y) z)≥ λ

(y)(x)+λ(y)(z) = 2λ (y)+λ (x)+λ (z).

Hence (18) follows. We are thus reduced to the case y = 1J and have to prove (205). Fixing x ∈ J×

and letting z ∈ J× be arbitrary, the estimate

λ (x z) = λ (Ux,1J z) = λ (Ux+1J z−Uxz− z)

≥ min

2λ (x+1J)+λ (z),2λ (x)+λ (z),λ (z)

≥ min

2λ (x),0+λ (z) = min

λ (x),−λ (x)

+λ (x)+λ (z)

= λ (x)+λ (z)−max

λ (x),−λ (x)= λ (x)+λ (z)−|λ (x)|


shows that there exists a constant κ ≥ 0, depending on x, such that

λ (x z)≥ λ (x)+λ (z)−κ (z ∈ J×). (206) LACIKAP

We now consider (31.3.15) for z in place of y and apply the result to 1J . Then

(x z)2 + zUxz =Uxz2 +Uzx2 +2xUzx,

which in turn may be combined with (31.3.17) to yield

(x z)2 =Uxz2 +Uzx2 + xUzx.

This and (206) imply

2λ (x z) = λ((x z)2)= λ (Uxz2 +Uzx2 + xUzx)

≥ min

2(λ (x)+λ (z)

),2(λ (z)+λ (x)

),λ(xUzx

)≥ min

2(λ (x)+λ (z)

),2(λ (x)+λ (z)

)−κ

= 2(λ (x)+λ (z)

)−κ.

Thus

λ (x z)≥ λ (x)+λ (z)− κ

2(z ∈ J×),

and iterating this procedure, we end up with

λ (x z)≥ λ (x)+λ (z)− κ

2n (z ∈ J×,n ∈ N).

Hence, as n→ ∞, we arrive at (205).

174. (a) Since J by Thm. 32.11 is locally linear, F [x]⊆ J is a commutatuve associative F-algebra.sol.ALJODIEvery y 6= 0 in F [x] is invertible in J and hence Uy : J→ J is bijective (Prop. 33.2). Thus Uy viarestriction induces a linear injection F [x]→F [x], and a dimension argument shows that Uy : F [x]→F [x] is, in fact, bijective. Since, therefore, y ∈ F [x]×, it follows that F [x]/F is indeed a finitealgebraic field extension.

(b) For x ∈ J, the finite algebraic field extension F [x]/F has degree 1 since F is algebraicallyclosed. Hence J = F ·1J .

(c) For x ∈ J, the finite algebraic field extension R[x]/R has degree at most 2. Thus 1J ,x,x2

are linearly independent, and invoking Exc. 80, we conclude that J (viewed as a linear Jordanalgebra with bilinear product xy) is conic. Writing q for its norm and M for its underlying realvector space, (M,q,e), e := 1J , is a pointed quadratic module over R. Now (31.12.8) and (29.8.1)imply J = J(M,q,e). Since by Exc. 168, therefore, q is anisotropic, it is either positive or negativedefinite. The latter alternative being excluded since q(e) = 1 > 0, the problem is solved.

Section 34

175. Let f = ∑i≥0 αiti ∈ k[t] and suppose f (c) = 0. By Prop. 32.7, we must show (t f )(c) = 0. Withsol.LOLICd := 1J − c and α := ∑i≥1 αi ∈ k we have 0 = f (c) = α01J +αc = (α0 +α)c+α0d, and sincec,d are orthogonal idempotents, we conclude (α0 +α)c = α0d = 0, hence (t f )(c) = (∑i≥0 αi)c =(α0 +α)c = 0, as desired.

176. Write Ji := Ji(c) for i = 0,1,2 and put d := 1J−c. Applying (34.2.8), we see that Uv maps J2sol.ISTTWOZERto J0, hence via restriction induces a linear map ϕ : J2→ J0. Write w := v−1 as w = w2 +w1 +w0,wi ∈ Ji for i = 0,1,2. Then (33.2.1) and (34.2.8) again yield

v =Uvw =Uvw2 +Uvw1 +Uvw0, Uvwi ∈ J2−i (i = 0,1,2).


Comparing Peirce components and using bijectivity of Uv (Prop. 33.2), we conclude w2 = w0 = 0.Thus w ∈ J1, and by (34.2.8) combined with Cor. 34.3 (b) we have

w2 = p+q, p :=Uwd ∈ J×2 , q :=Uwc ∈ J×0 , (207) WESQU

which implies

Uv p = d, Uvq = c. (208) UVEP

Next,writing p−1 (resp. q−1) for the inverse of p (resp. q) in J2 (resp. J0), we claim

Uvd = p−1, Uvc = q−1. (209) UVED

Indeed, from (207), (208), Cor. 34.3 (b) and Exc. 166 we deduce UqUvc =UqUv2 q =UqU(w2)−1 q =

UqUp−1+q−1 q=UqUq−1 q=Uqq−1, and since Uq : J0→ J0 is bijective, the second equation of (209)

is proved. The first one follows analogously. In particular, ϕ , viewed as a linear map J2 → J(q)0preserves units. But it also preserves U-operators since for x,y ∈ J2 we may apply (208) to obtain

U (q)ϕ(x)ϕ(y) =UUvxUqUvy =UvUxUvUqUvy =UvUxUUvqy = ϕ(UxUcy) = ϕ(Uxy),

as claimed. Thus ϕ : J2 → J(q)0 is a homomorphism. Interchanging the role of c and d, we alsoobtain a homotopy ψ : J0 → J2 induced by Uw = U−1

v . Hence ϕ is bijective with inverse ψ , andwe have established the first part of the exercise.

As to the second, v2 = 1J implies w = v, and (207) yields p = c, q = d. Hence ϕ : J2 → J0is an isomorphism. Conversely, let this be so. By what we have just shown, 1J0 = ϕ(1J2 ) = q−1,forcing q = d. On the other hand, ϕ−1 : J0 → J2 is an isomorphism induced by Uv−1 , and from(207) we conclude p = Uv−1 d = ϕ−1(d) = c. Now (207) again implies first w2 = 1J and thenv2 = (w2)−1 = 1J as well.

177. From Cor. 34.3 (c) we deduce for x ∈ J2 that Vx stabilizes J1, so the map ϕ is well dfined andsol.JTWOSPlinear. Moreover, ϕ(1J2 ) = ϕ(c) = 1J1 by (34.2.7). Now let x,y ∈ J2, z ∈ J1. Then (31.3.14) yields

Uϕ(x)ϕ(y)z =VxVyVxz =Ux,yVxz+Vx,yVxz,

where Ux,yVxz ∈ J2J1J2= 0 by (34.2.9).Applying (31.3.19), we therefore conclude

Uϕ(x)ϕ(y)z =Vx,yVxz =VUxyz+UxVyz =VUxyz

since UxVyz ∈UJ2 J1 = 0 by (34.2.8). Thus ϕ is a homomorphism of Jordan algebras, and thefirst part of the problem is solved. Moreover, if J2 is simple and J1 6= 0, then Ker(ϕ) ⊆ J is anideal 6= J2 since ϕ(c) = 1J1 6= 0. Thus Ker(ϕ) = 0 and ϕ is injective.

178. (ii)⇒ (i). Clear.sol.PETRCO(i)⇒ (ii). We may assume

i, j∩l,m 6= /0 (210) NODIS

and then consider the following cases.Case 1. i, j∩l,m∩n, p= /0. We may assume l = j by (210), and since T is not connected, wehave m 6= n, p but also j 6= n, p since we are in Case 1. Thus T = (np, jm, i j) with n, p∩ j,m=/0, as in the first alternative of (ii).Case 2. i, j∩l,m∩n, p 6= /0. Then we may assume n = l = j, and since T = (i j, jm, jp) isnot connected, we necessarily have m 6= i, j, p. The first part of the problem is thereby solved.


Now let T = (i j, jl, i j) be as in the second part of the problem. If i 6= l 6= j, the first part of theproblem shows that T is not connected. Conversely, if l = i or l = j, then T is clearly connectedand there is a unique positive integer m having i j = i, j= l,m= lm.

179. Let R := k[t±11 , . . . , t±1

r ] be the ring of Laurent polynomials in the variable t1, . . . , tr . As in thesol.MUPEALproof of Thm. 34.11 (c), we have

A⊆ AR =⊕

i1,...,ir∈Z(ti1

1 · · · tirr A),

Endk(A)⊆ Endk(AR) =⊕

i1,...,ir∈Z

(ti11 · · · t

irr Endk(A)

)⊆ EndR(AR).

(a) First of all, we have

1A = L1A R1A = L∑ci R∑c j =r

∑i, j=1

Lci Rc j =r

∑i, j=1

Ei j.

It remains to show that the Ei j are orthogonal projections. Since every base change of A is flexible,we conclude

r

∑i, j=1

tit jLci Rc j = L∑ tici R∑ t jc j = R∑ t jc j L∑ tici =r

∑i, j=1

tit jRc j Lci ,

and comparing coefficients yields

Ei j = Lci Rc j = Rc j Lci (1≤ i, j ≤ r). (211) ELIARJ

Similarly,r

∑i=1

t2i Lci = L

∑ t2i ci

= L(∑ tici)2 = L2∑ tici

= (r

∑i=1

tiLci )2 =

r

∑i, j=1

tit jLci Lc j

and the analogous relation for the right multiplication imply

Lci Lc j = δi jLci , Rci Rc j = δi jRci (1≤ i, j ≤ r). (212) ELIELJ

Given i, j, l,m = 1, . . . ,r, we now apply (211), (212) and obtain

Ei jElm = Lci Rc j Rcm Lcl = δ jmLci Rc j Lcl = δ jmLci Lcl Rc j = δilδ jmLci Rc j = δilδ jmEi j.

Hence the Ei j are orthogonal projections of A, as claimed.(b) Let 1≤ i, j ≤ r and x ∈ Ai j . For 1≤ l ≤ r, we apply (212) to obtain

clx = Lcl Ei jx = Lcl Lci Rc j x = δilEi jx = δilx

and, similarly, xcl = δ jlx. Thus the left-hand side of (4) is contained in the right. Conversely, letx ∈ A and assume clx = δilx, xcl = δ jlx for l = 1, . . . ,r. Then this holds, in particular, for l = i andl = j, which implies Ei jx = ci(xc j) = x, hence x ∈ Ai j . This completes the proof of (b).

(c) Since tλ ∈ R× for 1≤ λ ≤ r, we have

w :=r

∑λ=1

tλ cλ ∈ J×R , w−1 =r

∑λ=1

t−1λ

cλ . (213) WETLA

For 1≤ i, j ≤ r, we now claim

Ai j = x ∈ A | Lwx = tix, Rwx = t jx= x ∈ A | Lw−1 x = t−1i x, Rw−1 x = t−1

j x, (214) AIJEL


where the second equation follows from the first since Lw−1 = L−1w , Rw−1 = R−1

w by (14.6.2). Inorder to prove the first, we decompose x as x = ∑xlm, xlm ∈ Alm and apply (4) to obtain

Lwx = ∑λ ,l,m

tλ cλ xlm = ∑l,m

tlxlm, Rwx = ∑λ ,l,m

tλ xlmcλ = ∑l,m

tmxlm. (215) ELWEX

Thus Lwx = tix and Rwx = t jx if and only if x = xi j , which completes the proof (214). In analogyto the proof of Thm. 34.11, we now put

T := ti11 · · · t

irr | i1, . . . , ir ∈ Z, T1 := t1, . . . , tr

and claim:

(∗) If s, t ∈ T admit an element x 6= 0 in A such that Lwx = sx and Rwx = tx, then s, t ∈ T1.

In order to see this, write x = ∑xlm, xlm ∈ Alm. Then (215) implies

∑l,m

sxlm = sx = Lwx = ∑l,m

tlxlm, ∑l,m

txlm = tx = Rwx = ∑l,m

tmxlm,

and comparing components yields unique indices i, j = 1, . . . ,r satisfying s = ti, t = t j and x = xi j .For 1≤ i, j, l,m≤ r, xi j ∈ Ai j , ylm ∈ Alm, we now obtain

Lw(xi jylm) = tit jt−1l xi jylm, Rw(xi jylm) = t−1

j tltmxi jylm (216) ELWEXI

since the left and right Moufang identities (14.3.1) and (14.3.2) combined with (14.6.2) and (214)imply

Lw(xi jylm) = w(

xi j(w(w−1ylm)

))= (wxi jw)(w−1ylm) = tit jt−1

l xi jylm,

Rw(xi jylm) =((

(xi jw−1)w)ylm

)w = (xi jw−1)(wylmw) = t−1

j tltmxi jylm,

as claimed. Specializing l 7→ j, m 7→ l in (216) yields Lw(xi jy jl) = tixi jy jl , Rw(xi jy jl) = tlxi jyl j ,hence (5). Similarly, (216) for l = i, m = j yields (6), while (7) follows from (∗) and (216) sincej 6= l and (i, j) 6= (l,m) imply that tit jt−1

l or t−1j tltm does not belong to T1. Finally, for 1≤ i, j ≤ r,

i 6= j, and x ∈ Ai j , we apply (6) to obtain x2 ∈ A ji. But then (211), (4) and right alternativity yieldx2 = E jix2 = Rci Lc j x

2 = ((c jx)x)ci = 0, as claimed.

(d) Write E(+)i j , 1≤ i≤ j≤ r, for the Peirce projections of Ω , viewed as a complete orthogonal

system of idempotents in A(+). Comparing (34.8.1) with (211), we let 1≤ i≤ j ≤ r and conclude

E(+)ii =Uci = Lci Rci = Eii, E(+)

i j =Uci,c j = Lci Rc j +Lc j Rci = Ei j +E ji.

The assertion follows.

180. Since the Peirce triples (i j, ii, i j) and (i j, j j, i j) are connected, (34.11.6) implies Uvi j Jii ⊆ J j jsol.CONCOMand Uvi j J j j ⊆ Jii.

(a) (i) ⇒ (ii). Applying Exc. 176 to J′ := J2(ci + c j), we conclude that Uvi j : Jii → J j j is anisotopy. The first inclusion of (ii) now follows from Prop. 33.5 combined with Thm. 33.10 (b). Thesecond one follows by symmetry.

(a) (ii)⇒ (iii). Clear since ci ∈ J×ii and c j ∈ J×j j .(a) (iii) ⇒ (i). Since the Peirce triples (i j, ll, i j) are not connected for l = 1, . . . ,r, i 6= l 6= j,

we have v2i j = ∑

rl=1 Uvi j cl =Uvi j ci +Uvi j c j ∈ J′×. Hence vi j ∈ J′×, and this is (a) (i). Basically the

same argument also yields (b)(i)⇒ (ii), while (b)(ii)⇒ (i) is obvious.(c) (31.3.7) for x = vi j , y = v jl , z = 1J yields


Uvi jv jl ci =Uvi jUv jl ci +Uv jlUvi j ci +Vvi jUv jlVvi j ci−UUvi j v jl ,v jl ci. (217) UVIJVJL

Since i, j, l are mutually distinct, the Peirce triples ( jl, ii, jl), ( jl, i j, jl) and (i j, jl, i j) are not con-nected. Moreover, Vvi j ci = ci vi j = vi j by (34.11.3) and (34.2.6). Summing up, therefore, (217)collapses to Uvi jv jl ci = Uv jlUvi j ci and, similarly, Uvi jv jl cl = Uvi jUv jl cl . Combining this with thecharacterization of (strong) connectedness in (a) (resp. (b)), the assertion follows.

181 . (a) Free use will be made of the Peirce multiplication rules assembled in the presentsol.ISTCOMsection. Let 1 ≤ i, j ≤ r. Setting dλ := c(p)

λfor λ = 1, . . . ,r, we apply (33.8.8) and deduce

d(2,p)i = Up−1

i∑

rλ=1 pλ = p−1

i = di, so di is an idempotents in J(p). Now assume i 6= j Since

(8) is a direct sum of ideals, we obtain U (p)di

d j = UdiU∑ pλp−1

j = Udi p j = 0, while (33.8.5)yields di (p) d j = di pd j = 0. Thus the idempotents di,d j are orthogonal in J(p). The equation∑

rλ=1 dλ = p−1 = 1J(p) shows, therefore, that Ω (p) is a complete orthogonal system of idempo-

tents in J(p). Write Elm (resp. E(p)lm ) for the Peirce projections of Ω in J (resp. Ω (p) in J(p)). Then

E(p)lm = ElmUp by (34.8.1), hence J(p)

i j = Ei jUpJ = Ei jJ = Ji j , and the proof of (a) is complete.

(b) For 1≤ i≤ r we have (c(p)i )(q) = q(−1,pi)

i =Up−1i

q−1i = (Upi qi)

−1, and the assertion follows.

(c) Suppose v1i ∈ J1i connects c1 and ci for 2≤ i≤ r. By Exc. 180 (a) we have pi :=Uv1i c1 ∈ J×ii .Put p1 := c1 ∈ J×11 and p := ∑

rλ=1 pλ ∈ DiagΩ (J)×. By the solution to Exc. 176, w1i := v−1

1i (the

inverse of v1i in J2(c1+ci)), belongs to J1i = J(p)1i and satisfies Uw1i pi = c1. Note that v2

1i = qi1+ pi,for some qi1 ∈ J×11 and hence w2

1i = q−1i1 + p−1

i . Hence (33.8.8) implies

w(2,p)1i =Uw1i p =Uw1i (p1 + pi) =Uw1i pi +Uw1i c1 = c1 +Uw2

1ipi = c1 +Up−1

ipi

= c1 + p−1i = c(p)

1 + c(p)i .

Thus c(p)1 and c(p)

i are strongly connected by w1i ∈ J(p)1i .

182. We argue by induction on r. For r = 1, the assertion agrees with Exc. 151 (c). Now let r > 1sol.LIFCOMand assume the statement holds for r− 1 in place of r. Put c′ := c′r and use Exc. 151 (c) again tofind an idempotent c = cr ∈ J satisfying ϕ(c) = c′. Write Ei (resp. Ji) for the Peirce projections(resp. components) of J relative to c and E ′i (resp. J′i ) for the Peirce projections (resp. components)of J′ relative to c′, where i = 0,1,2. We have ϕ Ei = E ′i ϕ and hence ϕ(Ji) = (ϕ Ei)(J) =E ′i ϕ(J) = E ′i (J

′) = J′i . In particular, ϕ0 := ϕ|J0 : J0→ J′0 is a surjective homomorphism of Jordanalgebras whose kernel, Ker(ϕ0) = J0∩Ker(ϕ)⊆Ker(ϕ) is a nil ideal in J0. Moreover, J′0 contains(c′1, . . . ,c

′r−1) as a complete orthogonal system of idempotents. Hence the induction hypothesis

yields a complete orthogonal system (c1, . . . ,cr−1) of idempotents in J0 such that ϕ(ci) = c′i for1≤ i < r. By Prop. 34.6 and Cor. 34.7, therefore, (c1, . . . ,cr−1,cr) is a complete orthogonal systemof idempotents in J with the desired properties.

183 . Write t for the trace and x 7→ x for the conjugation of (M,q,e). Let I ⊂ J := J(M,q,e) besol.CLIFSIMan ideal that remains stable under conjugation. We claim I = 0. In order to see this, let x ∈ I.Then I contains Uxx2 = q(x)21J (by (31.12.9)), and from I 6= J we deduce q(x) = 0. Linearizingthis, we obtai q(x,y) = 0 for all y ∈ I. Now let y ∈ J \ I. Since x ∈ I and I ⊂ J is an ideal, Uyx =q(x,y)y−q(y)x belongs to I. But so does the second summand on the right of this expression, andwe conclude q(x,y)y ∈ I, hence q(x,y) = 0. Summing up, we have shown x ∈ Rad(q) = 0, andour intermediate claim I = 0 hs been proved.

Now assume that J is not simple and let I be any proper ideal of J. Then so is I, and theideals I∩ I, I + I are stable under conjugation. This and our intermediate assertion imply J = I⊕ Ias a direct sum of ideals. As observed in 30.7, I and I are Jordan algebras in their own rightsuch that the conjugation induces an isomorphism from I onto I. Put c := 1I . Then c = 1I and1J = c+ c. Moreover, c and c are idempotents in J, and the assumption q(c)= 1 would imply c∈ J×


(Exc. 168), hence c = 1J , c = 0, a contradiction. Hence q(c) = 0 (since q(c) ∈ F by (31.12.11) isan idempotent), and (31.12.5) reduces to 0 6= c = c2 = t(c)c. Taking traces, we conclude 0 6=t(c) = t(c)2, hence t(c) = 1. Summing, we have shown that c and c are elementary idempotentsof J, whence the corresponding Peirce decomposition by Prop. 34.5 implies I = J2(c) = Fc, I =J0(c) = Fc. Thus the assignment α⊕β 7→ αc+β c defines an isomorphism from (F⊕F)(+) ontoJ.

Chapter 6

Section 35

184 . (a) For x ∈ X , the assignment y 7→ N(x,y) defines linear form on X . By non-singularity,sol.NSPCFtherefore, we find a unique element x] ∈ X such that T (x],y) = N(x,y) for all y ∈ X . On the otherhand, given y ∈ X , the assignment x 7→ N(x,yR) for R ∈ k-alg, x ∈ XR defines a polynomial lawN(−,y) : X → k which is homogeneous of degree 2, hence a quadratic form (Exc. 56 (b)). Butthen x 7→ x] mus be a quadratic map since T is non-singular, and the gradient identity holds in allscalar extensions. Next we prove that X together with 1, ],N is a cubic array. By Euler’s differentialequation and Exc. 58, (4), we have T (1,y) = N(1,1)N(1,y)−N(1,1,y) = 3N(1,y)− 2N(1,y) =N(1,y), which implies 1] = 1, as claimed. To finish the proof of (a), it remains to establish the unitidentity. Linearizing the gradient identity, we conclude T (x× y,z) = N(x,y,z) is totally symmetricin x,y,z ∈ X , where as usual x× y is the bilinearization of the adjoint. Hence

T (x,1× y) = T (x× y,1) = N(1,x,y) = N(1,x)N(1,y)−T (x,y) = T (x)T (y)−T (x,y)

= T(x,T (y)1− y

),

and the unit identity follows again from the non-singularity of T .(b) Let 1′, ]′, N′ be the base point, adjoint, norm, respectively, of X ′. We must show that ϕ

preserves adjoints. Since ϕ is linear, the chain rule in the form (13.15.5) implies N′(ϕ(x),ϕ(y)) =N(x,y) and then N′(ϕ(x),ϕ(y),ϕ(z) = N(x,y,z) in all scalar extensions of X . Since ϕ also pre-serves base points, it preserves bilinear traces as well: T ′(ϕ(x),ϕ(y)) = T (x,y). But now the gra-dient identity gives

T ′(ϕ(x]),ϕ(y)

)= T (x],y) = N(x,y) = N′

(ϕ(x),ϕ(y)

)= T ′

(ϕ(x)]

′,ϕ(y)

),

and the non-singularity of T ′ combines with the surjectivity of ϕ to yield ϕ(x]) = ϕ(x)]′, asclaimed.

185. (a) Since cubic Jordan algebras and cubic norm structures are identified via Cor. 35.18, ϕ willsol.HOCUJObe a homomorphism of cubic Jordan algebras once we have shown NJ′ ϕ = NJ as polynomiallaws over k. Moreover, since the entire set-up is stable under base change, applying Prop. 13.22to the polynomial law g := NJ′ ϕ : J→ k, we see that it suffices to prove NJ′ (ϕ(x)) = NJ(x) forall x ∈ J having ϕ(x) ∈ J′×. But then the adjoint identity combined with the fact that ϕ preservesadjoints implies NJ(x)ϕ(x) = ϕ(x]]) = ϕ(x)]] = NJ′ (ϕ(x))ϕ(x), hence the assertion since ϕ(x) byCor. 35.16 is unimodular.

(b) Since ϕ preserves unit elements and norms, it is a surjective homomorphism of pointedcubic forms in the sense of Exc. 184. By part (b) of that exercise, therefore, ϕ is an isomorphismof cubic arrays, hence of cubic Jordan algebras.

186 . If X is a cubic norm structure over k, then (11)−(13) hold by definition, while (14), (15)sol.CHACUNhave been recorded in (35.8.10), (35.8.8), respectively. Conversely, suppose (11)−(15) hold for allx,y,z ∈ X and let R ∈ k-alg. Since (11), (12), (14) are homogeneous of degree at most 2 in eachvariable, they continue to hold in XR, so it remains to verify (13), (15) over R. We first linearize


(12) to conclude that T (x× y,z) = N(x,y,z) is totally symmetric in x,y,z ∈ X . Hence linearizing(14) with respect to y implies

x]× (y× z)+(x× y)× (x× z) = T (x],y)z+T (x],z)y+T (x× y,z)x (218) ITCUPO

for all x,y,z ∈ X . Turning now to the proof of (15), linearity allows us to assume y = vR for somev ∈ X ; we then put

M := x ∈ XR | x]× (x× y) = T (x],y)x+N(x)y.By homogeneity, we have rx ∈ M for all x ∈ M, r ∈ R. Moreover, for u ∈ X , the validity of (15)over X yields

(uR)]× (uR× vR) = (u])R× (uR× vR) =

(u]× (u× v)

)R

=(T (u],v)u+N(u)v

)R = T

((uR)

],vR)uR +NR(uR)vR,

hence uR ∈M. But the elements uR, uwith ∈ X , span XR as an R-module. Therefore (15) will holdin all of XR once we have shown that M is closed under addition. In order to do so, let x,z ∈ M.Then the definition of M and (218) yield

(x+ z)]×((x+ z)× y

)= (x]+ x× z+ z])× (x× y+ z× y)

= x]× (x× y)+ x]× (z× y)+(x× z)× (x× y)+

(x× z)× (z× y)+ z]× (x× y)+ z]× (z× y)

= T (x],y)x+N(x)y+T (x],y)z+

T (x],z)y+T (x× y,z)x+T (z],y)x+T (z],x)y+

T (x× y,z)z+T (z],y)z+N(z)y

= T (x]+ x× z+ z],y)x+T (x]+ x× z+ z],y)z+(N(x)+T (x],z)+T (x,z])+N(z)

)y

= T((x+ z)],y

)(x+ z)+N(x+ z)y,

forcing x+ z ∈M, and the proof of (15) is complete. Now (13) will be treated in a similar mannerby putting

M′ := x ∈ XR | x]] = N(x)x ⊆ XR.

Since both sides of (13) are homogeneous of the same degree, N is stable under scalar multiplica-tion: x ∈M′ and r ∈ R implies rx ∈M′. Moreover, for x ∈ X we conclude, using the fact that N isa polynomial law,

(xR)]] = (x]])R =

(N(x)x

)R =

(N(x)1R

)xR = N(xR)xR,

hence xR ∈M′. But the elements xR, x ∈ X , span XR as an R-module. Therefore (13) will hold in allof XR once we have shown that M′ is closed under addition. In order to do so, let x,y ∈M′. Then(12), (14), (15) and the definition of M′ imply

(x+ y)]] = (x]+ x× y+ y])]

= x]]+ x]× (x× y)+ x]× y]+(x× y)]+(x× y)× y]+ y]]

= N(x)x+T (x],y)x+N(x)y+T (x],y)y+

T (x,y])x+T (x,y])y+N(y)x+N(y)y

=(N(x)+T (x],y)+T (x,y])+N(y)

)(x+ y) = N(x+ y)(x+ y),

forcing x+ y ∈M′, as desired.


187. Let X be a rational cubic norm structure over k, with base point 1, adjoint x 7→ x], bilinearizedsol.RACUNOadjoint (x,y) 7→ x× y, bilinear trace T and norm N. Combining (16) and (18) with the bilinearityof ×, we conclude that ] : X → X is indeed a quadratic map with bilinearization ×. Beside themap N : X→ k, we now consider the map g : X×X→ k defined by g(x,y) := T (x],y) for x,y ∈ X .Since the adjoint is a quadratic map with bilinearization ×, we deduce from (17), (19), (20) thatconditions (i)−(iv) of Exc. 59 (b) hold for (N,g), so (N,g) : X → k is a cubic map. By part (c) ofthat exercise, therefore, we find a unique cubic form N := N ∗g : X→ k such that N(x) = N(x) andN(x,y) = T (x],y) for all x,y ∈ X . Here (21) and (24) imply 1 = 1]] = N(1)1, hence N(1) = 1 since1 is unimodular. Thus X together with the base point 1, the adjoint ] and the norm N is a cubic arrayover k, denoted by X . Write T (resp. S) for the (bi-)linear (resp. the quadratic) trace of X . In view of(24), we have T (x) := N(1,x) = T (1,x) and S(x) = N(x,y) = T (1,x]) for x ∈ X , while Exc. 59 (c)implies T (x×y,z) = g(x,y,z) = T (x,y× z) for all x,y,z ∈ X . Putting x = 1 and combining (35.8.10with (25), we conclude T (y,z) = T (y)T (z)− S(y,z) = T (1,y)T (1,z)−T (1,y× z) = T (y,z). ThusT is the bilinear trace of X . Hence equation (12) of Exc. 186 holds over k But so do (11), (13)-−(15) of that exercise, by (25), (21)−(23), respectively. Thus all equations of Exc. 186 hold overk, forcing the cubic array X to be, in fact, a cubic norm structure.

If ϕ : X → Y is a homomorphism of rational cubic norm structures, then ϕ : X → Y is a linearmap preserving base points and adjoints. By Exc. 185 (a), therefore, it is a homomorphism ofcubic norm structures. Summing up, we have constructed a functor from k-racuno to k-cuno.Conversely, let X be a cubic norm structure over k, with base point 1, adjoint x 7→ x], bilinear traceT and norm N. We claim that 1, x 7→ x], (x,y) 7→ x× y, T and the set map Nk : X → k make X arational cubic norm structure over k, denoted by Rat(X), i.e., equations (16)−(25) hold. Indeed,(16)−(18) are obvious, while (19)−(25) have been established in this order in (35.8.6), (35.8.11),(35.8.5), (35.8.10), (35.8.8), (35.8.1), (35.8.3), respectively. If ϕ : X → Y is a homomorphism ofcubic norm structures, then ϕ : Rat(X)→ Rat(Y ) is trivially one of rational cubic norm structures.Thus we obtain a functor from k-cuno to k-racuno which by Exc. 59 (d) is easily seen to be inverseto the functor k-cuno→ k-racuno constructed before. this completes the proof.

188. We first linearize (35.8.28) and obtainsol.GENTWO

u× (v×w)+ v× (w×u)+w× (u× v) (219) LLADJ

=(T (u)S(v,w)+T (v)S(w,u)+T (w)S(u,v)−T (u× v,w)

)1−(

S(u,v)w+S(v,w)u+S(w,u)v)−(

T (u)v×w+T (v)w×u+T (w)u× v)

for all u,v,w ∈ X . We must show that the submodule M ⊆ X spanned by the elements assembledin (26) is stable under the adjoint map. First of all, we have 1] = 1 ∈M by the base point identities(35.8.1) and 1×M ⊆ M by the unit identity (35.8.3). Moreover, M is spanned by the elements1,s, t,s× t, s ∈ x,x], t ∈ y,y]. Hence it will be enough to show that, for all s,s′ ∈ x,x],t, t ′ ∈ y,y],

s], t],(s× t)] ∈M, (220) ESTSHA

s× s′,s× (s′× t) ∈M, (221) ESESPR

t× t ′, t× (s× t ′) ∈M, (222) ESESTE

(s× t)× (s′× t ′) ∈M. (223) ESTEES

The adjoint identity (35.8.5) implies s] ∈ N(x)x,x], t] ∈ N(y)y,y], hence not only the first twoinclusions of (220) but also s]× t] ∈ M, while the third one now follows from (35.8.10). In thefirst inclusion of (221) we may assume s = x, s′ = x], whence the assertion follows from (14). Thesecond relation of (221) follows from (28), (220) for s = s′ and from (19), (8) for s 6= s′. Next,(222) is (221) with x and y interchanged, so we are left with (223). We first note that 1×M ⊆Mand (221), (222) imply


s×M+ t×M ⊆M. (224) ESTETI

Then we combine (219) with (220)−(222) to conclude

(s× t)× (s′× t ′)≡−s′×(t ′× (s× t)

)− t ′×

((s× t)× s′

)mod M,

and (224) implies (223).

189. (a)The straightforward verification is left to the reader.sol.NORID(b) We first show T (x,y) = 0 for all x∈ X and all y∈ I. By condition (i), we find a linear form λ

on X separating 1 from I: λ (1) = 1, λ (I) = 0. From condition (iii) and the unit identity (35.8.3)we conclude that 1× y = T (y)1− y is contained in I, forcing 0 = λ (T (y)1− y) = T (y)λ (1)−λ (y) = T (y). Thus the linear trace vanishes on I. This combined with condition (iii) and (35.8.13)implies 0 = T (x× y) = T (x)T (y)− T (x,y) = −T (x,y), as claimed. Next consider the set mapsf : X → k, g : X ×X → k defined respectively by f (x) = N(x), g(x,y) = N(x,y) = T (x],y) for allx,y ∈ X . Then ( f ,g) : X → k is a cubic map,and conditions (ii), (iii) for a norm ideal combinedwith what we have just proved imply I ⊆ Rad( f ,g). Hence Exc. 59 (a) yields a unique cubic map( f1,g1) : X1 → k such that f1 π = f , g1 (π × π) = g. Put 11 := π(1). By condition (iii) fora norm ideal, the adjoint ] : X → X induces canonically a unique quadratic map ]1 : X1 → X1satisfying π(x)]1 = π(x]) for all x ∈ X . Moreover, for x,y ∈ X we have π(x)×1 π(y) = π(x× y),where ×1 stands for the bilinearization of ]1. Finally, the bilinear trace of X induces canonically asymmetric bilinear form T1 : X1×X1→ k such that T1(π(x),π(y)) = T (x,y) for all x,y∈ X . SettingN1 := f1 : X1 → k, it is now straightforward to verify that the identities (16)−(25) of Exc. 187hold for X1 since they do for X . As the linear form λ kills I, it induces canonically a linear formλ1 : X1 → k having λ1(11) = 1 and thus making 11 ∈ X1 unimodular. Summing up, Exc. 187converts X1 into a cubic norm structure over k such that π preserves base points and adjoints,hence by Exc. 185, is a homomorphism of cubic norm structures with kernel I.

190. We have 3 = T (1,1) = 0 in k, hence 2 =−1 ∈ k×. Thus J may be viewed as a linear Jordansol.TRACUJOalgebra whose bilinear multiplication by 31.4 and (35.13.8) is given by

xy =12

x y =−x y =−x× y

since not only the bilinear trace but also the linear (by (35.8.2)) and the quadratic one (by (35.8.12))are identically zero. We conclude x2 = x] (which follows also from (35.13.7)), while (29.9.1) and(35.6.1) imply

2x(xy)− x2y =Uxy = T (x,y)x− x]× y =−x]× y = x2y,

hence the left alternative law x(xy) = x2y. Being also commutative, J is in fact alternative.

191. For the first part of the problem, if T (x),S(x),N(x)∈ k are nilpotent, then so are T (x)x2,S(x)x,sol.CUBNILN(x)1 ∈ J. By (35.9.2) and Exc. 150, therefore, x3 is nilpotent, whence x is nilpotent. Conversely,assume that x is nilpotent. Since N permits Jordan composition, it commutes with taking powers,and we conclude N(x) ∈ Nil(k). Moreover, applying Exc. 150 again, there exists a positive integerm such that xn = 0 for all integers n≥m. We now show T (x),S(x) ∈Nil(k) by induction on m. Form = 1 there is nothing to prove. Next assume m > 1 and that the assertion holds for all positiveintegers < m. Since (x2)n = x2n = 0 for all integers n≥m−1, the induction hypothesis implies thatT (x2) and S(x2) are nilpotent. On the other hand, from (35.8.22) and the adjoint identity (35.8.5)we deduce

T (x)2 = 2S(x)+T (x2), S(x)2 = 2T (x)N(x)+S(x2).

Since N(x) is nilpotent, the induction hypothesis and the second equation imply that S(x) is nilpo-tent, which combines with the first equation to show that T (x) is nilpotent as well. This completesthe proof of the first part.

In order to establish the second part, we put


I := x ∈ J | ∀y ∈ J : T (x,y),T (x],y),N(x) ∈ Nil(k) (225) RIHANI

and have to show I = Nil(J). For x,y ∈ I and z ∈ J, we apply (35.8.6), (35.8.7) and obtain

N(x+ y) = N(x)+T (x],y)+T (x,y])+N(y) ∈ Nil(k),

T (x+ y,z) = T (x,z)+T (y,z) ∈ Nil(k),

T((x+ y)],z

)= T (x],z)+T (x,y× z)+T (y],z) ∈ Nil(k),

hence x+ y ∈ I. Thus I ⊆ J is a k-submodule. Next we claim

I]+ I× J ⊆ I. (226) ISHA

The inclusion I] ⊆ I follows immediately from (225), the adjoint identity (35.8.5) and (35.8.18). Itremains to show x× y ∈ I for all x ∈ I, y ∈ J. Given z ∈ J, we apply (35.8.29), (35.8.10), (35.8.7)

N(x× y) = T (x],y)T (x,y])−N(x)N(y) ∈ Nil(k),

T (x× y,z) = T (x,y× z) ∈ Nil(k),

T((x× y)]

)= T (x],y)T (y,z)+T (x,y])T (x,z)−T (x],y]× z) ∈ Nil(k).

Hence x×y∈ I and (226) holds. Now let x∈ I and y∈ J. Then Uxy = T (x,y)x−x]×y∈ I by (226),so I ⊆ J is an inner ideal. But since

T (Uyx,z) = T (x,Uyz) ∈ Nil(k),

T((Uyx)],z

)= T (Uy]x

],z) = T (x],Uy] z) ∈ Nil(k),

N(Uyx) = N(y)2N(x) ∈ Nil(k)

for all z ∈ J by (35.8.21), (35.8.31), (35.8.20) and (226), we also have Uyx ∈ I by (225), so I ⊆ J isan outer ideal and altogether, therefore, an ideal in J which by the first part of the problem consistsentirely of nilpotent elements. This proves I ⊆ Nil(J).

Conversely, let x ∈ Nil(J). Then T (x),S(x),N(x) ∈ Nil(k) by the first part of the problem, andwe have Nil(k)J⊆Nil(J) by Exc. 148 (c). Furthermore, for all y∈ J, Nil(J)3Uxy= T (x,y)x−x]×y, hence x]× y ∈ Nil(J), which by (35.8.12), (35.8.13) implies S(x)T (y)−T (x],y) = T (x]× y) ∈Nil(k). Thus T (x],y) ∈ Nil(k). Since z := Uyx = T (y,x)y]− y]× x belongs to the ideal Nil(J), sodoes x+ z, which by (35.8.7), (35.8.13), implies

T (x,y)2−2T (x],y]) = T (x,y)2−T (x× x,y]) = T (x,y)2−T (x,y]× x) = T (x,Uyx)

= T (x,z) = T (x)T (z)−S(x,z)

= T (x)T (z)+S(x)+S(z)−S(x+ z) ∈ Nil(k).

Thus T (x,y) ∈ Nil(k) and summing up we have shown x ∈ I. This completes the proof of (27).Finally, let x,y ∈ J. If 1

2 ∈ k, then T (x],y) = 12 T (x× x,y) = 1

2 T (x,x× y) by (35.8.7), so (27)implies (28). Similarly, if 1

3 ∈ k, then N(x) = 13 T (x,x]) by (35.8.11), so (27) implies (29). Finally,

(30) follows by combining (28) and (29).

192. (a) N as defined by (31) is clearly a cubic form such thatsol.CUADUN

N(r⊕u,s⊕ v) = sq(u)+ rq(u,v)

for R ∈ k-alg, r,s ∈ R, u,v ∈ JR. Defining T (resp. S) as a linear (resp. quadratic) form on J byT (x) := N(1J ,x) (resp. S(x) := N(x,1J)) for x ∈ J, we therefore conclude that (33), (34) hold. Forx = r⊕u ∈ JR, this and (31.12.5), (31.12.6) imply


x3−T (x)x2 +S(x)x−N(x)1J =(

r3−(r+ t(u)

)r2 +

(rt(u)+q(u)

)r− rq(u)

)⊕(

u3−(r+ t(u)

)u2 +

(rt(u+q(u)

)u− rq(u)e

)= 0⊕

(u3− t(u)u2 +q(u)u− r

(u2− t(u)u+q(u)e

))= 0,

x4−T (x)x3 +S(x)x2−N(x)x =(

r4−(r+ t(u)

)r3 +

(rt(u+q(u)

)r2− rq(u)r

)⊕(

u4−(r+ t(u)

)u3 +

(rt(u)+q(u)

)u2− rq(u)u

)= 0⊕

(Uu(u2− t(u)u+q(u)e

)− r(u3− t(u)u2 +q(u)u

))= 0.

Since q permits Jordan composition by (31.12.11), so does N by (31), and we have shown that Jis a cubic Jordan algebra with norm N over k such that (31), (33), (34) hold. Linearizing (34) andapplying (35.2.10), we also obtain (32). Finally, (35.8.22) implies, for x = α⊕u, α ∈ k, u ∈ J,

x] = x2−T (x)x+S(x)1J =(

α2−(α + t(u)

)α +

(αt(u)+q(u)

))⊕(

u2−(α + t(u)

)u+(αt(u)+q(u)

)e)

= q(u)⊕(

α(t(u)e−u

))= q(u)⊕α u.

This proves (35), and (a) is solved.(b) Applying Exc. 162, we see that C(+) agrees with the Jordan algebra, J := J(C,nC,1C), of

the pointed quadratic module (C,nC,1C). Thus

C(+) = k(+)⊕C(+) = k(+)⊕ J = J

is a cubic Jordan algebra over k with norm N : C→ k given by

N(r⊕u) = rnC(u) (R ∈ k-alg, r ∈ R u ∈CR). (227) NOHAC

Hence C is a cubic alternative k-algebra with norm N which by (227) is multiplicative if and onlyif C is. Finally, Example 35.22 fits into this picture by setting C := k.

193. (a) The arguments solving Exc. 187 (a) go through for rational cubic norm pseudo-structuressol.PSUNOSwithout change and prove (a).

(b) (i) ⇒ (ii). Let 1 ≤ i ≤ n. Then (i) and Euler’s differential equation imply σ(e]i ,ei) =

µ(ei,ei) = 3µ(ei) = 3wi. Moreover, linearizing the relation µ(v,v′) = σ(v],v′), we concludeσ(v1× v2,v3) = µ(v1,v2,v3), and this is totally symmetric in v1,v2,v3 ∈V .

(ii)⇒ (i). For R ∈ k-alg, we define a map µR : VR→WR by setting

µR(n

∑i=1

rieiR) :=n

∑i=1

r3i wiR + ∑

1≤i, j≤n,i6= jr2

i r jσ(e]i ,e j)R + ∑1≤i< j<l≤n

rir jrlσ(ei× e j,el)R (228) MUAR

for r1, . . . ,rn ∈ R. Then µ := (µR)R∈k-alg : V →W is obviously a homogeneous polynomial lawof degree 3 satisfying µ(ei) = wi for 1 ≤ i ≤ n. Now let v = ∑

ni=1 rieiR,v′ = ∑

ni=1 r′ieiR ∈ VR with

ri,r′i ∈ R for i = 1, . . . ,n. Then (228) implies


µ(v,v′) = ∑i

3r2i r′iwiR +∑

i6= j(2rir′ir j + r2

i r′j)σ(e]i ,e j)R

+ ∑i< j<l

(r′ir jrl + rir′jrl + rir jr′l)σ(ei× e j,el)R.

On the other handv] = ∑

ir2

i e]iR +∑i< j

rir jeiR× e jR,

and hence (ii) gives

σ(v],v′) = ∑i,l

r2i r′lσ(e]i ,el)R + ∑

i, j,l,i< jrir jr′lσ(ei× e j,el)R

= ∑i

r2i r′iσ(e]i ,ei)R +∑

i6= jr2

i r′jσ(e]i ,e j)R +∑i< j

rir′ir jσ(ei× e j,ei)R

+∑i< j

rir jr′jσ(ei× e j,e j)R +(

∑l<i< j

+ ∑i<l< j

+ ∑i< j<l

)rir jr′lσ(ei× e j,el)R

= ∑i

3r2i r′iwiR +∑

i6= j(r2

i r′j +2rir′ir j)σ(e]i ,e j)R

+ ∑i< j<l

(r′ir jrl + rir′jrl + rir jr′l)σ(ei× e j,el)R

= µ(v,v′),

and (i) holds. Finally, uniqueness of µ follows immediately from (6) of Exc. 58.(c) Since a pseudo cubic norm pseudo-structure X over k is automatically a cubic norm structure

if 13 ∈ k (λ := 1

3 T has λ (1) = 1), any example of the kind demanded in (c) must display patholo-gies of characteristic 3. With this in mind, we let K/F be a purely inseparable field extension ofcharacteristic 3 and exponent 1 and fix an element a ∈ K \F . Then F is infinite and J0 := K+ is acubic Jordan algebra over F , with norm N0 : K→ F (by Prop. 13.11 an honest-to-goodness poly-nomial function) given by N0(v)1 = v3 for all v ∈ J0. For R ∈ F-alg and v,v′ ∈ J0R, we concludeN0(v,v′)1 = 3v2v′ = 0, so the (bi-)linear and quadratic traces of J0 are both zero, and v] = v2 by(35.8.22) as well as v× v′ = 2vv′ = −vv′ for all v,v′ ∈ J0. Now let W be a vector space over F offinite dimension at least 2, pick linearly independent vectors w1,w2 ∈W and let λ : J0 →W bea non-zero linear map satisfying λ (1) = 0. Define a symmetric bilinear map σ : J0× J0 →W byσ(v,v′) := λ (vv′) for v,v′ ∈ J0. Then σ(v],v) = λ (v2v) = λ (v3) =N0(v)λ (1) = 0 for all v∈ J0, andσ(v1×v2,v3) =−λ (v1v2v3) is totally symmetric in v1,v2,v3 ∈ J0. Thus (b) yields a homogeneouspolynomial law µ : J0→W of degree 3 such that µ(1) = 0, µ(a) = w1, µ(a]) = µ(a2) = w2.

Now let k := F⊕W ∈ F-alg be the “split-null extension” of F by W , so W ⊆ k is an ideal thatsquares to zero and F acts canonically on W . In other words, the multiplication of k obeys the rule

(α⊕w)(α ′⊕w′) := (αα′)⊕ (αw′+α

′w)

for α,α ′ ∈ F , w,w′ ∈W . By letting W act trivially on J0, we may and always will view J0 as aJordan algebra J over k.

Next define a symmetric F-bilinear map T : J× J→ k

T (v,v′) := 0⊕σ(v,v′) = 0⊕λ (vv′)

for v,v′ ∈ J, which is in fact a k-bilinear form since

T((α⊕w)v,v′

)= T (αv,v′) = αT (v,v′) = (α⊕w)

(0⊕λ (vv′)

)= (α⊕w)T (v,v′)

for all α ∈ F , w ∈W , v,v′ ∈ J. Similarly, we obtain a map


N := N0⊕µ : J −→ k, v 7−→ N(v) := N0(v)⊕µ(v)

and claim that the identity element of J as base point, the (bilinearized) adjoint of J as (bilinearized)adjoint, the symmetric bilinear form T , and the map N satisfy equations (16)−(25) of Exc. 187. Toshow this, let α ∈ F , w ∈W and v,v′ ∈ J. Then(

(α⊕w)v)]

= (αv)] = α2v] = (α⊕2αw)v] = (α⊕w)2v] (229) SHARK

proves (16) and

N((α⊕w)v

)= N(αv) = α

3(N0(v)⊕µ(v))= (α⊕w)3(N0(v)⊕µ(v)

)= (α⊕w)3N(v)

proves (17). The equations (229) and((α⊕w)v

)× v′ = (αv)× v′ = α(v× v′) = (α⊕w)(v× v′)

show that the adjoint v 7→ v] is not only F-quadratic but, in fact, k-quadratic. Moreover, the bilin-earized adjoint is the k-bilinearization of the adjoint, and we have (18). Next we compute

N(v+ v′) = N0(v+ v′)⊕µ(v+ v′) =(N0(v)+N0(v′)

)⊕(µ(v)+µ(v,v′)+µ(v′,v)+µ(v′)

)=(N0(v)⊕µ(v)

)+(0⊕σ(v],v′)

)+(0⊕σ(v′],v)

)+(N0(v′)⊕µ(v′)

)= N(v)+T (v],v′)+T (v′],v)+N(v′),

and (19) holds. Since 3 = 0 in k, we have T (v],v) = 0⊕σ(v],v) = 0 = 3N(v), hence (20). Sincethe adjoint identity holds in J0, we obtain

v]] = N0(v)v =(N0(v)⊕µ(v)

)v = N(v)v,

hence (21). Similarly,

v]× v′]+(v× v′)] = − v2v′2 +(vv′)2 = 0 =(0⊕λ (v]v′)

)v′+

(0⊕λ (vv′])

)v

= T (v],v′)v′+T (v,v′])v

gives (22) and

v]× (v× v′) = v](vv′) = (v]v)v′ =(0⊕λ (v]v′)

)v+N0(v)v′ =

(N0(v)⊕µ(v)

)v′

= T (v],v′)v+N(v)v′

gives (23). While (24) is obvious, the unit identity for J0 implies 1× v =−v = (0⊕λ (v))1− v =T (1,v)1− v, hence (25).

By Exc. 187, therefore, we find a cubic form N : X → k making X a cubic norm pseudo-structure over k. Now consider the element a ∈ J. By construction we have

N(a]) = N0(a2)+µ(a2) = N0(a)2 +w2,

N(a)2 =(N0(a)+µ(a)

)2= N0(a)2 +2N0(a)µ(a) = N0(a)2−N0(a)w1,

hence N(a]) 6= N(a)2 since w1,w2 are linearly independent over F .

Section 36

194. (i) k is a cubic norm structure with base point 1, adjoint α 7→ α2, norm r 7→ r3, and bilinearsol.EICO(resp. quadratic) trace respectively given by (α,β ) 7→ 3αβ (resp. α 7→ 3α2). Scalar multiplicationgives a bilinear action k×V → V , while the eiconal triple structure provides us with quadraticmaps Q : V → k, H : V →V . Moreover, setting v = u in (2), Euler’s differential equation implies


Q(u,H(u)

)= N(u) (R ∈ k-alg, u ∈VR). (230) QUHAEN

By 36.9, therefore, formulas (4)−(6) define a cubic array X over k whose bilinear trace by (36.9.5)is non-singular. Thus the solution to (i) will be complete once we have shown that X is, in fact,a cubic norm structure. In order to do so, it suffices to show, by Prop. 36.10, that the identities(36.10.1)−(36.10.10) hold strictly in X . Here (36.10.5) is just the eiconal equation, while (36.10.7)has been established in (230) with N = N. Since (36.10.1)−(36.10.4), (36.10.9) are obvious in thespecial case at hand, we are left with (36.10.6), (36.10.8), (36.10.9), i.e., with

H(H(u)

)= N(u)u−Q(u)H(u), (231) HAHAEN

Q(u,H(u,v)

)= 2Q

(H(u),v

), (232) QUHAQU

H(u,H(u)

)= 2Q(u)u. (233) HAHAQU

We begin with (233) by linearizing (2) to conclude that

Q(H(u,v),w

)=

13

N(u,v,w) (234) LIHADJ

is totally symmetric in u,v,w ∈V . Combining this with the linearization of (3), i.e., with

Q(H(u),H(u,v)

)= 2Q(u)Q(u,v), (235) QUHAHA

we conclude Q(H(u,H(u)),v) = Q(H(u,v),H(u)) = 2Q(u)Q(u,v) = Q(2Q(u)u,v), and (233)drops out since Q is non-singular. Setting w = u and using symmetry of (234) again, we obtainQ(H(u,v),u) = Q(H(u,u),v) = 2Q(H(u),v), hence (232). And finally, linearizing (233), we de-duce

H(v,H(u)

)+H

(u,H(u,v)

)= 2Q(u,v)u+2Q(u)v.

Here we put v = H(u) apply (233), (230) and obtain

2H(H(u)

)+4Q(u)H(u) = H

(H(u),H(u)

)+2H

(u,Q(u)u

)= H

(H(u),H(u)

)+H

(u,H

(u,H(u)

))= 2Q

(u,H(u)

)u+2Q(u)H(u)

= 2N(u)u+2Q(u)H(u),

hence (231).(ii) The cubic norm substructure X0 ⊆ X is non-singular since 1

3 ∈ k. Hence (X0,V ) is a comple-mented cubic norm substructure of X . Defining a quadratic map H : V →V by H(u) := Q(u)+u]

for u ∈V , we are in the situation 36.7 and conclude that the identities (36.7.1)−(36.7.6), (36.8.1)-−(36.8.18) holds strictly in X . In particular, by (36.7.4) we have 3Q(u,v) = T (u,v), whence (V,Q)is a quadratic space over k. Moreover, for u,v ∈ V the gradient identity implies 3Q(H(u),v) =T (H(u),v) = T (u],v) = N(u,v), so the quadratic map H is connected with the cubic form N onV by (2). Finally, (36.8.9) shows that the eiconal equation (3) holds strictly in V . Summing up wehave proved that (V,Q,N|V ) is an eiconal triple over k.

(iii) follows by a straightforward verification.(iv) The gradient of N at u ∈ Rn is defined by

(grad(N)

)(u) =

∂N∂x1

(u)...

∂N∂xn

(u)

∈ Rn

and can be characterized by(grad(N)

)(u)t v = N(u,v) (v ∈ Rn).


This and (2) imply

H(u)t Qv = Q(H(u),v

)=

13(

grad(N))(u)t v,

and we conclude H(u)t Q = 13 (grad(N))(u)t , hence

H(u)t =13(

grad(N))(u)t Q−1, H(u) =

13

Q−1(grad(N))(u).

Now the left-hand side of (3) becomes

H(u)t QH(u) =19(

grad(N))(u)t Q−1QQ−1(grad(N)

)(u) =

n

∑i, j=1

qi j ∂N∂xi

(u)∂N∂x j

(u),

while the right-hand side of (2) is ( n

∑i, j=1

qi juiu j

)2.

Hence (3) follows.

Section 37

195 . (a) We provide the solution to Exc. 84 with the following supplement. Let f : M→ N be asol.IDCUJOpolynomial law over k, ε ∈ k an idempotent and x ∈M. Then the identifications

MR = εM, xR = x⊗ ε = εx (R = εk ∈ k-alg, x ∈M),

ditto for N, imply fR(εx) = fR(xR) = fk(x)R = ε fk(x), hence

fR(εx) = ε fk(x) (x ∈M). (236) EFAREP

(b) Before dealing with the problem itself, we characterize the various types of idempotents e ∈ Jin the following way.

e = 0 ⇐⇒ T (e) = S(e) = N(e) = 0, (237) CHAZER

e is elementary ⇐⇒ T (e) = 1, S(e) = N(e) = 0 ⇐⇒ T (e) = 1, S(e) = 0, (238) CHAELM

e is co-elementary ⇐⇒ T (e) = 2, S(e) = 1, N(e) = 0 (239) CHACOL

⇐⇒ T (e) = 2, S(e) = 1,

e = 1 ⇐⇒ T (e) = S(e) = 3, N(e) = 1 ⇐⇒ N(e) = 1. (240) CHAUN

In (237), the implication from left to right is obvious. Conversely, assume T (e) = S(e) = N(e) = 0.Then (35.9.2) implies e = e3 = 0. If e is elementary, then T (e) = 1, e] = 0, hence S(e) = T (e]) = 0,while N(e) = 0 has been noted in 37.1. Conversely, assume T (e) = 1, S(e) = 0. Then (35.8.22)yields e] = e2− e = 0, and e is elementary, proving (238). In (239), we have

e is co-elementary ⇐⇒ 1− e is elementary

⇐⇒ T (1− e) = 1, S(1− e) = N(1− e) = 0.

But T (1−e) = 3−T (e) and S(1−e) = S(1)−S(1,e)+S(e) = 3−2T (e)+S(e) by (35.8.2), whileN(1− e) = 1− T (e)+ S(e)−N(e) by (35.8.6), (35.8.1). This proves (239). Finally, if e = 1 in(240), then T (e) = S(e) = 3 and N(e) = 1 by (35.8.1), (35.8.2). Conversely, if N(e) = 1, thene ∈ J× by Cor. 35.10. Hence Ue is a bijective projection by Prop. 33.2 and Thm. 34.2, which im-plies Ue = 1J , hence e = e2 =Ue1 = 1.(c) Turning, finally, to the solution of the problem, we begin by showing uniqueness, so let(ε(i))0≤i≤3 be a complete orthogonal system of idempotents in k with the desired properties. For


the sake of clarity, we write T (i),S(i),N(i) for the trace, quadratic trace, norm, respectively, of J(i)

over k(i). Invoking (236)−(240), we obtain

T (e) = ∑ε(i)T (e) = ∑T (i)(ε(i)e) = ∑T (i)(e(i)) = ε

(1)+2ε(2)+3ε

(3)

and, similarly,

S(e) = ∑S(i)(e(i)) = ε(2)+3ε

(3), N(e) = ∑N(i)(e(i)) = ε(3).

Solving this system of linear equations in the unknowns ε(1),ε(2),ε(3), we obtain (2)−(4). Now (1)follows from the completeness of the orthogonal system (ε(i)).(d) In order to prve existence, we define the ε(i), 0 ≤ i ≤ 3 by (1)−(4) and must show that theyhave the desired properties. First of all, they clearly add up to 1. Next we have to show that theyare orthogonal idempotents, equivalently, that ε(i)ε( j) = 0 for 0 ≤ i, j ≤ 3, i 6= j. To begin with,since N preserves powers by (35.8.21), and e,1− e are idempotents, so are ε(0),ε(3). Moreover,ε(0)ε(3) = N(e)2N(1− e) = N(Ue(1− e)) = 0. Summing up, we have proved

ε(0)2 = ε

(0), ε(3)2 = ε

(3), ε(0)

ε(3) = 0. (241) VEPZON

Next we apply (35.9.2) and obtain e = e4 = (T (e)− S(e)+N(e))e, hence N(1− e)e = 0. SinceN(1− e) = ε(3) by (241) is an idempotent, applying T and S to the preceding equation, we end upwith

N(1− e)e = 0, T (e)N(1− e) = S(e)N(1− e) = 0. (242) ENONEE

Combining with (2), (3), (241), we conclude

ε(0)

ε(1) = ε

(0)ε(2) = 0. (243) VEZEON

On the other hand, replacing e by 1−e in (242) yields 0 = T (1−e)N(e) = 3N(e)−T (e)N(e) and0 = S(1− e)N(e) = (3−2T (e)+S(e))N(e) = 3N(e)−6N(e)+S(e)N(e), hence

T (e)N(e) = S(e)N(e) = 3N(e). (244) TEENEE

This implies ε(1)ε(3) = 3N(e)−6N(e)+3N(e) = 0, ε(2)ε(3) = 3N(e)−3N(e) = 0, and we have

ε(1)

ε(3) = ε

(2)ε(3) = 0. (245) VEONTH

In view of (241), (243), (245), the ε(i) will form a complete orthogonal system of idempotents ink once we have shown ε(1)ε(2) = 0. To this end, we first note T (e) = T (e,e)+ T (1− e,Uee) =T (e,e) + T (Ue(1− e),e) = T (e,e) by (35.8.31) and then apply (35.8.13) to conclude 2S(e) =S(e,e) = T (e)2−T (e,e), hence

2S(e) = T (e)2−T (e). (246) TWOESE

Now we expand the last two equations of (242) by using (244) and (246). We obtain 0 = T (e)−T (e)2 +T (e)S(e)−3N(e), hence

T (e)S(e) = 2S(e)+3N(e), (247) TEEESE

and 0 = S(e)−2S(e)−3N(e)+S(e)2−3N(e), hence

S(e)2 = S(e)+6N(e). (248) ESESQ

Using (244), (247), (248), we can now compute


ε(1)

ε(2) = T (e)S(e)−3T (e)N(e)−2S(e)2 +6S(e)N(e)+3S(e)N(e)−9N(e)

= 2S(e)+3N(e)−9N(e)−2S(e)−12N(e)+27N(e)−9N(e)

= 0.

Thus (ε(i))0≤i≤3 is a complete orthogonal system of idempotents in k. In particular, we have e =

∑e(i), e(i) = ε(i)e ∈ J(i) for 0 ≤ i ≤ 3. From (242) we deduce e(0) = N(1− e)e = 0, while (236)implies N(3)(e(3)) = N(3)(ε(3)e) = ε(3)N(e) = ε(3), hence e(3) = 1J(3) by (240). It remains to showthat e(1) ∈ J(1) is elementary and e(2) ∈ J(2) is co-elementary, which follows from

T (1)(e(1)) = T (1)(ε(1)e) = ε(1)T (e) = T (e)2−2T (e)S(e)+3T (e)N(e)

= T (e)2−4S(e)−6N(e)+9N(e) = T (e)−2S(e)+3N(e) = ε(1) = 1k(1) ,

S(1)(e(1)) = S(1)(ε(1)e) = ε(1)S(e) = T (e)S(e)−2S(e)2 +3S(e)N(e)

= 2S(e)+3N(e)−2S(e)−12N(e)+9N(e) = 0

and (238) in the first case, and from

T (2)(e(2)) = T (2)(ε(2)e) = ε(2)T (e) = T (e)S(e)−3T (e)N(e)

= 2S(e)+3N(e)−9N(e) = 2(S(e)−3N(e)

)= 2ε

(2) = 2 ·1k(2) ,

S(2)(e(2)) = S(2)(ε(2)e) = ε(2)S(e) = S(e)2−3S(e)N(e) = S(e)+6N(e)−9N(e)

= S(e)−3N(e) = ε(2) = 1k(2)

and (239) in the second.

196 . Applying equation (6) of Exc. 58 and the (bilinearized) gradient identity, we expandsol.FERLEN(q) = N(∑ui) = ∑N(ui) + ∑i6= j T (u]i ,u j) + T (u1 × u2,u3) = ∑N(ui) + T (u1 × u2,u3), where(35.8.18) implies N(ui)

2 = N(u]i ) = 0. Since, therefore, the scalars N(q) and T (u1× u2,u3) dif-fer by a nilpotent element in k, one of them is invertible if and only if so is the other. Thisproves the first assertion. Now assume q ∈ J× and T (u1 × u2,u3) ∈ k×. From (35.8.7) we de-duce T (ui×u j,ul) = T (u1×u2,u3). Hence the linear form x 7→ T (ui×u j,x) from J to k takes ulto an invertible element in k, forcing ul ∈ J to be unimodular. The adjoint identity (35.8.5) in thespecial form N(ul)ul = u]]l = 0 therefore yields N(ul) = 0. so we have

N(q) = T (u1×u2,u3); (249) NORQ

in particular,

p = q−1 = N(q)−1q] = T (u1×u2,u3)−1

∑(u j×ul) (250) QINVP

since the ui, 1 ≤ i ≤ 3, have u]i = 0. From (35.8.7) we deduce T (u j × ul ,u j) = 2T (u]j,ul) = 0and, similarly, T (u j× ul ,ul) = 0. Applying (35.11.2), (35.11.5) and (250), we not only conclude

u(],p)i = N(p)Uqu]i = 0 but also T (p)(ui) = T (p,ui) = N(q)−1T (∑um × un,ui) = N(q)−1T (u j ×ul ,ui) = 1 by (249). Thus u1,u2,u3 are elementary idempotents in J(p). That they do, in fact, forman elementary frame in that algebra will follow from Prop. 37.5 once we have shown u1×(p) u2 =1(p)− u1− u2 = p−1− u1− u2 = q− u1− u2. In order to do so, we first linearize (35.11.2) andapply (249), (250) to obtain

u1×(p) u2 = N(q)−1Uq(u1×u2)

= N(q)−1T (q,u1×u2)q−N(q)−1q]× (u1×u2) (251) ISORTH

= q−T (u1×u2,u3)−1

∑(u j×ul)× (u1×u2)


The term for i = 3 in the sum on the very right of (251) by (35.8.10) gives (u1×u2)× (u1×u2) =

2(u1×u2)] = 2(T (u]1,u2)u2 +T (u1,u

]2)u1−u]1×u]2) = 0. On the other hand, invoking (35.8.9) for

x = ui, y = u j , z = ul , we deduce

(ui×u j)× (ui×ul) = T (u]i ,u j)ul +T (u]i ,ul)u j +T (ui×u j,ul)ui−u]i × (u j×ul)

= T (u1×u2,u3)ui

Plugging all this this into (251) we end up with u1×(p) u2 = q−u2−u1, as claimed.Finally, in order to derive the very last statement of the problem, it suffices to note ∑ei = 1 and

apply the previous parts of the problem.

197. The tedious but straightforward verifications of (5) and the first equation of (6) are left to thesol.UOPMAreader. The former amounts to saying that the squaring of Mat3(C) restricts to the squaring of J.After linearizing, therefore, the symmetric matrix product xy+yx of Mat3(C) restricts to the circleproduct x y of J. Hence (31.15.2), (5) and the first equation of (6) imply

2x3 = x x2 = x(xx)+(xx)x = x3− γ1γ2γ3[u1,u2,u3]+ (xx)x,

which yields the second equation of (6). In order to derive (7), we note that (6) holds strictly, sowe are allowed to differentiate it at x in the direction y. Doing so for the left-hand side,i.e., forx3 = Uxx, we apply (31.3.13) and obtain Uxy+Ux,yx = Uxy+ x2 y = Uxy+ (xx)y+ yx2. Sincethe associator of C is alternating, the same procedure applied to the middle term of (6) yields theexpression

yx2 + x(yx)+ x(xy)+ γ1γ2γ3([u1,u2,v3]+ [u2,u3,v1]+ [u3,u1,v2]

)13,

and comparing, we end up with the first formula of (7). The second one follows by the same argu-ment, using the second formula of (6) instead of the first. Alternatively, we may invoke (31.3.10)to conclude

2Uxy = x (x y)− x2 y = x(xy+ yx)+(xy+ yx)x− (xx)y− y(xx)

= x(yx)+(xy)x− [x,x,y]+ [y,x,x],

so the second formula of (7) follows from this and the first. Subtracting the middle term of (6) fromthe right one, we obtain (8). Invoking (35.8.22), (35.9.2) and (5), (6), we deduce

x]x = (xx)x−T (x)xx+S(x)x = x3−T (x)x2 +S(x)x+ γ1γ2γ3[u1,u2,u3]13

=(N(x)1C + γ1γ2γ3[u1,u2,u3]1C

)13,

hence (9), while (10) follows from (9) and xx]+ x]x = x x] = 2N(x)13. Finally, turning to (11),we replace x by x] in (10) and apply (35.8.18) to conclude

x]x]] =(N(x)21C− γ1γ2γ3[u

]1,u

]2,u

]3])13.

On the other hand, the adjoint identity and (9) yield

x]x]] =UUXY ZV N(x)x]x =(N(x)21C + γ1γ2γ3N(x)[u1,u2,u3]

)13,

hence (11).

198. (a) (i)⇔ (ii). Condition (i) implies Γ ′∆ 2Γ ] = Γ ′Γ ′] = N(Γ ′)13 = N(∆)N(Γ )13 = ∆∆ ]Γ Γ ].sol.DIAFIXCanceling yields Γ ′∆ = ∆ ]Γ , hence (ii). Thus (i) implies (ii). Conversely, suppose (ii) holds. ThenΓ ′∆ = ∆ ]Γ , and taking norms we conclude N(Γ ′)N(∆) = N(∆)2N(Γ ), hence the second relationof (i). Moreover, Γ ′]∆ ] = (Γ ′∆)] = (∆ ]Γ )] = ∆ ]]Γ ] = N(∆)∆Γ ] = ∆ 2∆ ]Γ ], which implies Γ ′] =∆ 2Γ ], hence the first relation of (i). Thus (i) holds.


(ii) ⇒ (iii). Suppose (ii) holds. Hence so does (i). Since ϕ := ϕC,∆ preserves base points, itsuffices to show that it preserves adjoints as well (Exc. 185). Accordingly, let

x = ∑(ξieii +ui[ jl]) ∈ Her3(C,Γ ) (ξi ∈ k, ui ∈C, i = 1,2,3).

Then (37.10.4), (12) imply

ϕ(x]) = ∑(ξ jξl − γ jγlnC(ui)

)eii +∑

(−δ

−1i ξiui +δ

−1i γiu jul

)[ jl]. (252) EXSH

On the other hand, (12), (37.10.4) and (i), (ii) yield

ϕ(x)] = ∑(ξ jξl − γ

′jγ′l nC(δ

−1i ui)

)eii +∑

(−ξiδ

−1i ui + γ

′i (δ−1j u j)(δ

−1l ul)

)[ jl]

= ∑(ξ jξl −δ

−2i γ

′jγ′l nC(ui)

)eii +∑

(−δ

−1i ξiui +δ

−1j δ

−1l γ

′i u jul

)[ jl]

= ∑(ξ jξl − γ jγlnC(ui)

)eii +∑

(−δ

−1i ξiui +δ

−1i γiu jul

)[ jl],

and a comparison with (252) gives the assertion.Finally, let us assume that (iii) holds and 1C ∈ C is unimodular. We wish to establish (ii),

apply (37.16.4) to compute ϕ(1C[li]× 1C[i j]) = ϕ(1C[li])×ϕ(1C[i j]) and obtain δ−1i γi1C[ jl] =

δ−1j δ

−1l γ ′i 1C[ jl]. Hence γ ′i 1C = δ jδlδ

−1i γi1C , and (ii) follows.

(b) Let ε,ε1,ε2,ε3 ∈ k×. For (i), it suffices to put δ1 = δ2 = δ3 = ε in (a), while (ii) follows bysetting δi = ε jεl in (a) since this implies δ jδlδ

−1i = εlεiεiε jε

−1j ε

−1l = ε2

i . Finally, in (iii), we putδi = γ

−1i and have γ ′i = γ

−1j γ−1l γ2

i .

199. ϕ is a linear bijection which obviously preserves units. By Exc. 185 (a), therefore, it sufficessol.DIISJOto show that ϕ preserves adjoints. For

x = ∑(ξieii +ui[ jl]) ∈ J := Her3(C,Γ ),

we note p−1 = ∑γ−1i eii and apply (35.11.2), (37.10.4), Prop. 34.8, (37.17.1) to conclude

x(],p) = N(p)Up−1 x] = γ1γ2γ3U∑γ−1i eii

x] = γ1γ2γ3 ∑(γ−2i Ueii x

]+ γ−1j γ−1l Ue j j ,ell x

])

= γ1γ2γ3 ∑

(γ−2i(ξ jξl − γ jγlnC(ui)

)eii +

(γ−1j γ−1l (−ξiui + γiu jul)

)[ jl])

= ∑

(γ−1i(γ jγlξ jξl − γ

2j γ

2l nC(ui)

)eii + γi(−ξiui + γiu jul)[ jl]

),

hence

ϕ(x(],p)) = ∑

(((γ jξ j)(γlξl)−nC(γ jγlui)

)eii +

(− (γiξi)(γ jγlui)+(γlγiu j)(γiγ jul)

)[ jl])

=(∑((γiξi)eii +(γ jγlui)[ jl]

))]= ϕ(x)],

as claimed.

200. Putsol.ISCOPA(D,∆) := (C,Γ )(p,q), ∆ = diag(δ1,δ2,δ3), δi = γ

(p,q)i

for i = 1,2,3. Then C = D as k-modules, and we have

δ1 = nC(pq)−1γ1, δ2 = nC(q)γ2, δ3 = nC(p)γ3. (253) DEPEQU

Moreover, from Exc. 90 we recall

nD(u) = nC(pq)nC(u), u(p,q) = nC(pq)−1 pqu pq = nC(pq)−1(pq)u(pq) (u ∈C). (254) NOCOPE


Now let

x = ∑(ξieii +ui[ jl]) ∈ Her3(D,∆) (ξi ∈ k, ui ∈ D, i = 1,2,3) (255) EXDEDE

and write ]p,q for the adjoint of the cubic Jordan matrix algebra Her3(D,∆). Furthermore, denoteby ηi,ζi ∈ k, vi,wi ∈C the quantities satisfying

x]p,q = ∑(ηieii + vi[ jl]) ∈ Her3(D,∆), ϕ(x)] = ∑(ζieii +wi[ jl]) ∈ Her3(C,Γ ). (256) EXSHPE

Since

ϕ(x) = ∑(ξieii +u′i[ jl]) ∈ Her3(C,Γ ), ϕ(x]p,q ) = ∑(ηieii + v′i[ jl]) ∈ Her3(C,Γ ), (257) FIEX

where the u′i,v′i are to be determined by ui,vi, respectively, via (15), we have to show

ηi = ζi, v′i = wi (i = 1,2,3). (258) EZEVE

Reading (37.10.4) in Her3(D,∆) and invoking (253), (254), we compute

η1 = ξ2ξ3−δ2δ3nC(pq)nC(u1) = ξ2ξ3− γ2γ3nC(pq)2nC(u1)

= ξ2ξ3− γ2γ3nC(u′1) = ζ1,

η2 = ξ3ξ1−δ3δ1nC(pq)nC(u2) = ξ3ξ1− γ3γ1nC(p)nC(u2)

= ξ3ξ1− γ3γ1nC(u′2) = ζ2,

η3 = ξ1ξ2−δ1δ2nC(pq)nC(u3) = ξ1ξ2− γ1γ2nC(q)nC(u3)

= ξ1ξ2− γ1γ2nC(u′3) = ζ3.

This proves the first set of equations in (258). In the second one, we have to show v′i = wi fori = 1,2,3. The case i = 1 is comparatively harmless since

v′1 = (pq)v1(pq) = (pq)(−ξ1u1 +δ1(u2 p)(qu3)

(p,q))(pq)

= −ξ1(pq)u1(pq)+ γ1nC(pq)−1nC(pq)−1(pq)(

pq(u2 p)(qu3) pq)(pq)

= −ξ1u′1 + γ1u′2u′3 = w1,

as desired. But the remaining cases i = 2,3 are considerably more involved. At a crucial stage, theyrequire an application of (14.5.4) combined with the Moufang identities:

v′2 = v2 p = (−ξ2u2 +δ2(u3 p)(qu1)(p,q)

)p =

= −ξ2u2 p+ γ2nC(q)nC(pq)−1(pq)((u3 p)(qu1)

)(pq)p,

where

(pq)((u3 p)(qu1)

)(pq) =

((pq)(u3 p)

)((qu1)(pq)

)=(

p(qu3)p)(

q(u1 p)q)

= p[(qu3)

(p(q(u1 p)q

))]= p((qu3)

((pq)u1(pq)

))= p(u′3u′1).

Hence Kirmse’s identities (20.3.1) yield

v′2 = −ξ2u′2 + γ2nC(p)−1 p(u′3u′1)p =−ξ2u′2 + γ2nC(p)−1(u′3u′1 p)p

= −ξ2u′2 + γ2u′3u′1 = w2.

Similarly,


v′3 = qv3 = q(−ξ3u3 +δ3(u1 p)(qu2)

(p,q))= −ξ3qu3 + γ3nC(p)nC(pq)−1q(pq)

((u1 p)(qu2)

)(pq),

where

(pq)((u1 p)(qu2)

)(pq) =

((pq)(u1 p)

)((qu2)(pq)

)=(

p(qu1)p)(

q(u2q)q)

=[((

p(qu1)p)q)(u2 p)

]q =

(((pq)u1(pq)

)(u2 p)

)q

= (u′1u′2)q.

Thus

v′3 = −ξ3u′3 + γ3nC(q)−1q(u′1u′2)q =−ξ3u′3 + γ3nC(q)−1q(qu′1u′2)

= −ξ3u′3 + γ3u′1u′2 = w3.

201. Let α ∈ Centout(J). We have [α,Ux] = [α,Vx,y] = 0 for all x,y ∈ J and must show α = ξ 1J forsol.OUTCENCUsome ξ ∈ k. Since α commutes with the Peirce projections relative to the diagonal frame of J, itstabilizes the corresponding Peirce components, so there are ξi ∈ k and k-linear maps ϕi : C→Csuch that

αeii = ξieii, αui[ jl] = ϕi(ui)[ jl] (ξi ∈ k, ui ∈C, i = 1,2,3). (259) ALPHEI

Applying (37.16.6) and (259), we obtain γ jγlξ jell = U1C [ jl]ξ je j j = U1C [ jl]αe j j = αU1C [ jl]e j j =γ jγlαell = γ jγlξlell , hence

ξ1 = ξ2 = ξ3 =: ξ . (260) ALEQ

Similarly, since ui[ jl] 1C[li] = γl ui[i j] by (37.16.4), applying (259) again gives γlϕl(ui)[i j] =αγl ui[i j] = αV1C [li]ui[ jl] =V1C [li]αui[ jl] = ϕi(ui)[ jl]1C[li] = γlϕi(ui)[i j], hence

ϕi(ui) = ϕl(ui) (ui ∈C, i = 1,2,3) (261) PHIL

On the other hand, (259) and (37.16.5) imply γ jγlϕi(vi)[ jl] = U1C [ jl]ϕi(vi)[ jl] = U1C [ jl]αvi[ jl] =αU1C [ jl]vi[ jl] = γ jγlα vi[ jl] = γ jγlϕi(vi)[ jl], hence

ϕi(vi) = ϕi(vi) (vi ∈C, i = 1,2,3). (262) PHIBAR

Combining (261), (262), we have

ϕ1 = ϕ2 = ϕ3 =: ϕ. (263) PHIEQ

By (260) and (263), therefore, ϕ(u1)[23] = αu1[23] = αVu1[23]e22 =Vu1[23]αe22 = ξ u1[23]e22 =ξ u1[23]. Thus ϕ = ξ 1C , and we have shown α = ξ 1J , as claimed.

202. (a) We consider arbitrary elementssol.IDCUJM

x = ∑(ξieii +ui[ jl]), y = ∑(ηieii + vi[ jl])

in J, with ξi,ηi ∈ k, ui,vi ∈ C for i = 1,2,3. If I is an ideal in (C, ιC) and I0 ⊆ I ∩ k is a weaklyI-ample ideal in k, we must show that H;=H3(I0, I,Γ ) is an outer ideal in J. For u∈ I and v∈C wefirst note nC(u,v) = tC(uv)∈ I0 since I⊆C is an ideal and I0 is weakly I-ample. Now assume y∈H,i.e., ηi ∈ I0 and vi ∈ I for i = 1,2,3. Inspecting (37.10.7) and applying the preceding observation,we conclude T (x,y) ∈ I0, hence T (x,y)ξi ∈ I0 and T (x,y)ui ∈ I0C ⊆ IC ⊆ I. Thus T (x,y)x ∈ H.Moreover, since I is stabilized by ιC , and by (37.10.6), we have x× y ∈ H, which after replacingx by x] implies Uxy ∈ H. Thus H is an outer ideal in J. Conversely, let this be so. Then the Peirce


projections relative to the diagonal frame of J, i.e., Ueii and Ue j j ,ell for i = 1,2,3 stabilize H andProp. 37.17 yields

H = ∑((H ∩ keii)+(H ∩C[ jl])

). (264) HEII

We now put

I := u ∈C | u[23] ∈ H, I0 := ξ ∈ k | ξ e11 ∈ H, (265) IUCE

which are k-submodules of C,k, respectively. For u ∈ I, we apply (37.16.5) and obtain γ2γ3u[23] =U1C [23]u[23]∈H, hence u∈ I. Thus I = I. From (37.16.4) we conclude that u[ jl]×v[li] = u[ jl]v[li]belongs to H if one of the “factors” does. We now claim

H ∩C[ jl] = I[ jl]. (266) HACEJL

for i = 1,2,3. This follows from (265) for i = 1. Arguing by “cyclic induction mod 3”, suppose(266) holds for some i ∈ 1,2,3 and let u ∈C. If u ∈ I, then u ∈ I, and (37.16.4) yields γ ju[li] =1C[i j]× u[ jl] ∈ H, hence u[li] ∈ H. Conversely, if u[li] ∈ H, the H contains u[li]×1C[i j] = γiu[ jl],which implies u ∈ I, hence u ∈ I. Thus (266) holds for j instead of i, which completes its proof.Next we claim

H ∩ keii = I0eii (267) HACAPE

for i = 1,2,3. Thanks to (265), the case i = 1 is again obvious. Next suppose (267) holds forsome i ∈ 1,2,3; we must prove it for j. Let ξ ∈ k. For ξ ∈ I0 we obtain ξ eii ∈ H, so H by(37.16.6) contains U1C [i j]ξ eii = γ jγlξ e j j , hence ξ e j j . Conversely, if ξ e j j ∈ H, then H containsU1C [i j]ξ e j j = γiγ jξ eii, and we conclude ξ ∈ I0. This completes the proof of (266). Combining(264) with (266), (267), we obtain

H = ∑(I0eii + I[ jl]), (268) HAPEIR

and it remains to show that (i) I is an ideal in (C, ιC), (ii) I0 ⊆ I∩ k, and (iii) I0 is weakly I-ample.For u ∈ I, v ∈ J, we combine (37.16.4) with (266) to obtain γ1uv[23] = v[31]× u[12] ∈ H, henceuv ∈ I. Thus I, being stable under conjugation, is an ideal in (C, ιC), and (i) is proved. Turning to(ii), let ξ ∈ I0. Then (37.16.2), (34.11.3), (34.2.7) imply ξ [23] = (ξ 1C)[23] = (ξ e11)1C[23] ∈H,hence ξ ∈ I. This proves (ii). Finally, turning to weak I-ampleness, let u ∈ I. Then H containsu[23]e221C[23] = γ2γ3tC(u)e33 (by (37.16.6) linearized), which implies tC(u) ∈ I0 by (267) andcompletes the proof (iii).

(b) Let I be an ideal in (C, ιC) and I0 an ideal in k that is weakly I-ample and contained in I∩k.By (a) it will be enough to show that H := H(I0, I,Γ ) is an inner ideal of J if and only if I0 is I-ample. Suppose first that H ⊆ J is an inner ideal and let u ∈ I. Then H contains Uu[23]e22, which by(37.16.6) agrees with γ2γ3nC(u)e33, and we conclude nC(u) ∈ I0. Thus I0 is I-ample. Conversely,let this be so. For x ∈ H, i.e., ξi ∈ I0 and ui ∈ I, i = 1,2,3, we obtain T (x,y)ξi ∈ I0, T (x,y)ui ∈ I,hence T (x,y)x ∈ H. Moreover, x] ∈ H by (37.10.4) since I0 is I-ample. But then x]× y ∈ H afterinspecting (37.10.6), and we conclude Uxy ∈ H. Summing up, we have shown that H ⊆ J is aninner ideal.

(c) For ξ ∈ I∩ k we obtain 2ξ = tC(ξ 1C) ∈ I0. Thus 2(I∩ k)⊆ I0. The rest is clear.(d) Let H be an outer ideal in J. The (a) implies H = H(I0, I,Γ ) for some ideal I ⊆ (C, ιC) and

some weakly I-ample ideal I0 of k contained in I∩ k. By Exc. 104, there exists an ideal a⊆ k suchthat I = aC and I0 ⊆ I∩ k = a. Since C is non-singular, its trace form is surjective (Lemma 22.11),so some u ∈C has tC(u) = 1. For ξ ∈ a, we obtain ξ u ∈ I and therefore ξ = tC(ξ u) ∈ tC(I)⊆ I0 byweak I-ampleness. This proves I0 = a, hence H = aJ by (16).

(e) By Kaplansky’s theorem (Prop. 22.6), C is either a non-singular composition algebra or apurely inseparable field extension K/F of characteristic 2 and exponent at most 1. In the formercase, the assertion follows immediately from (d). In the latter case, we first note that the bilinearized


norm of C =K is zero and hence Rad(T ) =H(0,K,Γ ) is an outer ideal in J. Conversely, supposeH ⊆ J is a non-zero outer ideal and write H = H(I0, I,Γ ) with I0, I as in (a). Then H 6= 0 impliesI 6= 0, hence I = K, while I0 ⊆ I∩F = F is an ideal in F , weak I-ampleness being automatic. IfI0 = 0, then H = Rad(T ), and if I0 = F , then H = J. Thus J is simple but nor outer simple.

203 . (a) We have Ux = 0 and must show Ux] = 0, i.e., Ux]y = 0 for all y ∈ J. Since cubic Jordansol.ABZECUalgebras are invariant under base change, viewing Ux] as a polynomial law over k and applyingProp. 13.22 allow us to assume y ∈ J×. Then (35.10.1) for x = y−1 and (35.8.20) imply Ux]y =N(y)Ux]y

−1] = N(y)(Uxy−1)] = 0.(b) (i) ⇒ (ii). By (i) we have x2 = Ux1 = 0, and since k is reduced, Exc. 191 shows T (x) =

S(x) = N(x) = 0. By (35.8.22), therefore, x] = x2−T (x)x+S(x)1 = 0. Thus T (x,y)x = Uxy = 0,which implies T (x,y)2 = T (T (x,y)x,y)) = 0, hence T (x,y) = 0 by our hypothesis on k. Summingup, (ii) holds.

(ii)⇒ (i). Obvious, by the formula for the U-operator.(ii) ⇒ (iii). The adjoint identity yields N(x)x = x]] = 0, and applying the norm we obtain

N(x)4 = 0, hence N(x) = 0.(c) Let x ∈ J be an absolute zero divisor and assume first that k is reduced. Then (b) implies

T (x,y) = T (x],y) = N(x) = 0 for all y ∈ J, and Exc. 191 shows x ∈ Nil(J).Next assume that k is arbitrary. Then k := k/Nil(k)∈ k-alg is reduced, and we have a canonical

identification Jk = J := J/Nil(k)J via 10.2, matching zk for z ∈ J with z, the image of z under thenatural map J → J. Since x is an absolute zero divisor in J, x is one in J, so by the special casejust treated we have T (x, y) = T (x], y) = N(x) = 0 for all y ∈ J, where T = Tk (resp. N = Nk) isthe bilinear trace (resp, norm) of J over k. But this means that T (x,y), T (x],y), N(x) are nilpotentelements of k, for all y ∈ J, which means x ∈ Nil(J).

(d) Let F be a field of characteristic 2, K/F a purely inseparable field extension of exponentat most 1, Γ = diag(γ1,γ2,γ3) ∈ GL3(F) and J = Her3(K,Γ ). By Exc. 202, J is a simple Jordanalgebra over F . If u1,u2,u3 ∈ K are not all zero, neither is x := ∑ui[ jl] ∈ J. Moreover, since thebilinearized norm of K/F is zero, (37.10.5) and (37.10.7) show N(x) = 0 and T (x,y) = 0 for ally ∈ J, so x satisfies condition (iii) of (b). On the other hand,

x] = ∑(γ jγlnK(ui)eii + γiu jul [ jl]) 6= 0,

whence (b) shows that x is not an absolute zero divisor of J.(e) First assume Nil(J) = 0. We have seen in Exc. 148 (c) that Nil(k)J is a nil ideal in J. This

proves Nil(k)J = 0, and in particular Nil(k)1 = 0. But 1 ∈ J is unimodular, and we concludethat k is reduced. Now let x ∈ J be an absolute zero divisor. Then (b) implies N(x) = T (x,y) =T (x],y) = 0 for all y ∈ J. From Exc. 191 we therefore deduce x ∈Nil(J), hence x = 0, and J has noabsolute zero divisors.. Conversely, let k be reduced and suppose J has no absolute zero divisors.For x ∈ Nil(J) and y ∈ J, we have T (x,y) = T (x],y) = N(x) = 0. This implies x]] = N(x)x = 0, sox] is an absolute zero divisor by (b), forcing x] = 0 by hypothesis. But then, again by (b), x is anabsolute zero divisor, and we conclude x = 0, as desired.

204. We putsol.EXRACO

M := ∑ξiei | ξi ∈ k, ξiJi j = 0 for i = 1,2,3,

N := ∑ξiei | ξ 2i = 2ξi = 0 for i = 1,2,3.

By Exc. 155 (c), Rex(J) ⊆ J is an ideal. Thus, given x = ∑(ξiei + v jl) ∈ Rex(J), ξi ∈ k, v jl ∈ J jl ,its Peirce components relative to (e1,e2,e3) by Prop. 34.8 belong to Rex(J) as well, so we haveξiei,v jl ∈Rex(J) for 1≤ i≤ 3. Here (34.2.7) and (34.11.2) imply v jl = e j v jl =Uv jl,1 e j = 0, hencex = ∑ξiei. Similarly, ξiJi j = ξiei Ji j = Uξiei,1Ji j = 0, and we have shown Rex(J) ⊆ M. Nowsuppose x = ∑ξiei, ξi ∈ k, satisfies x ∈M, i. e., ξiJi j = 0 for 1≤ i≤ 3. Setting u12 := u23×u31,we have u jl ∈ J jl by Prop. 37.22 and S(u jl) ∈ k× by Lemma 37.23 for i = 1,2,3. In particular,ξ 2

i S(ui j) = S(ξiui j) = 0 and 2ξiS(ui j) = S(ξiui j,ui j) = 0 imply M ⊆ N, and we have Uξiei = 0.Moreover, the assignment x 7→ x× u jl by Lemma 37.23 gives a linear bijection from Ji j to Jli.


Hence ξiJli = ξiJi j×u jl = 0 and then ξiJ jl = ξiJli×ui j = 0. Summing up, we have shown

ξiJ = ∑m(kξi)em,

and from the Peirce rules we conclude

(ξiei)JJ= ei(ξiJ)J= ∑mei(kξi)emJ= eiei(ξiJ)= ∑

meiei(kξi)em

= kξieieiei= 2kξiei = 0.

But this means Uξiei,J = 0, and we have shown x = ∑ξiei ∈ Rex(J). This completes the proof of(19). Now let (C,Γ ) be a co-ordinate pair over k and put (J,S) := Her3(C,Γ ) as a co-ordinatedcubic Jordan algebra over k. By (19), and with the notation used before, the elements of Rex(J)have the form ∑ξieii, ξi ∈ k, ξiJi j = 0 for 1≤ i≤ 3. But Ji j =C[i j] by Prop. 37.17, and we concludeξiC = 0. In particular, ξi1C = 0, and if 1C ∈C is unimodular, we deduce ξi = 0. Thus the extremeradical of J is zero.

Finally, let k0 be any commutative ring in which 2 = 0 and let k := k0[ε] be the k0-algebra ofdual numbers. Then k0 may be viewed as an algebra C ∈ k-alg under the homomorphism k→ k0satisfying ε 7→ 0. The squaring α0 7→ α2

0 may be regarded as a k0-quadratic map nC : C→ k withzero bilinearization, which is in fact k-quadratic since

nC((β0 + γ0ε)α0

)= nC(β0α0) = β

20 nC(α0) = (β0 + γ0ε)2nC(α0)

for all α0,β0,γ0 ∈ C. Thus C together with nC is a multiplicative conic commutative associativek-algebra such that εC = 0, and the extreme radical of J is different from zero.

205. (a) Ifsol.JACOCAT(C,Γ )

(η ,∆)// (C′,Γ ′)

(η ′,∆ ′)// (C′′,Γ ′′)

are morphisms of co-ordinate pairs, so obviously is

(η ′,∆ ′) (η ,∆) := (η ′ η ,∆ ′∆) : (C,Γ )−→ (C′′,Γ ′′). (269) COMCOTR

It is straightforward to check that we obtain a category in this way, denoted by k-copa. For example,the identity morphism of (C,Γ ) is 1(C,Γ ) := (1C,13).

(b) If(J,S)

(ϕ;δ1,δ2)// (J′,S′)

(ϕ ′;δ ′1,δ′2)

// (J′′,S′′)

with S′′ = (e′′1 ,e′′2 ,e′′3 ,u′′23,u

′′31) are morphisms of co-ordinated cubic Jordan algebras, so is

(ϕ ′;δ′1,δ′2) (ϕ;δ1,δ2) := (ϕ ′ ϕ;δ

′1δ1,δ

′2δ2) : (J,S)−→ (J′′,S′′). (270) COMCOC

since (ϕ ′ ϕ)(ei) = ϕ ′(e′i) = e′′i for i = 1,2,3 by (i) and (ϕ ′ ϕ)(u jl) = ϕ ′(δ−1i u′jl) = δ

′−1i δ

−1i u′′jl

for i = 1,2 by (ii).It is straightforward to check that we obtain a category in this way, denoted by k-cocujo. For

example, the identity morphism of (J,S) is 1(J,S) := (1J ;1,1).(c) Let η : C→C′ be a homomorphism of multiplicative conic alternative k-algebras. We define

Her3(η) : Her3(C,Γ )−→ Her3(C′,Γ )

component-wise by

Her3(η)(∑(ξieii +ui[ jl])

):= ∑

(ξieii +η(ui)[ jl]

)(271) FUHER


for ξi ∈ k, ui ∈C, i = 1,2,3. Since η by definition preserves units, norms, traces and conjugations,we easily deduce from (37.10.3)−(37.10.5) that Her3(η) is a homomorphism of cubic Jordanalgebras. Given homomorphisms

Cη

// C′η ′

// C′′

of multiplicative conic alternative algebras, (271) easily implies Her3(η′η)=Her3(η

′)Her3(η).Now let (η ,∆) : (C,Γ )→ (C′,Γ ′) with ∆ = diag(δ1,δ2,δ3) ∈Diag3(k)

× be a homomorphismof co-ordinate pairs. We claim that

Her3(η ,∆) :=(

Her3(η ,∆);δ1,δ2) : Her3(C,Γ )−→Her3(C′,Γ ′) (272) COHE

is a homomorphism of co-ordinated cubic Jordan algebras. Indeed, since Γ ′ = ∆ ]∆ ′−1Γ by (a),(ii), Exc. 198 shows that Her3(η ,∆) = Her3(η) ϕC,∆ = ϕC′,∆ Her3(η) is a homomorphism ofthe underlying cubic Jordan algebras and by (20) satisfies Her3(η ,∆)(eii) = eii for i = 1,2,3 andHer3(η ,∆)(1C[ jl]) = (δ−1

i 1C′ )[ jl] for i = 1,2.It is straightforward to check using (20) that in this way we obtain a functor

Her3 : k-copa−→ k-cocujo. (273) FUBFHER

(d) Writing T ′,S′ for the trace, quadratic trace of J′, we put

S= (e1,e2,e3,u23,u31), ω := ωJ,S,

(C,Γ ) := Cop(J,S) = (CJ,S,ΓJ,S), Γ = diag(γ1,γ2,γ3),

S′ = (e′1,e′2,e′3,u′23,u

′31), ω

′ := ωJ′,S′ ,

(C′,Γ ′) := Cop(J′,S′) = (CJ′,S′ ,ΓJ′,S′ ), Γ′ = diag(γ ′1,γ

′2,γ′3).

to deduce from (37.24.1) and (b), (ii)

ω′ = S′(u′23)

−1S′(u′31)−1 = S′(δ1ϕ(u23))

−1S′(δ2ϕ(u31))−1

= δ−21 δ

−22 S(u23)

−1S(u31)−1,

which amounts to

ω′ = δ

−21 δ

−22 ω. (274) OMPROM

Now let v,w ∈C = J12. Then (274) combines with (37.24.3) to imply

ϕ12(v)ϕ12(w) = ω′(

ϕ(v)×u′23)×(u′31×ϕ(w)

)= δ

−21 δ

−22 δ1δ2ω

(ϕ(v)×ϕ(u23)

)×(ϕ(u31)×ϕ(w)

)= δ

−11 δ

−12 ωϕ

((v×u23)× (u31×w)

)= δ

−11 δ

−12 ϕ12(vw),

and we conclude

ϕ12(vw) = δ1δ2ϕ12(v)ϕ12(w) (v,w ∈C = J12). (275) MORMUL

In accordance with (21), we now define η := δ1δ2ϕ12 : C→C′. Given v,w∈C and applying (275),we deduce η(vw) = δ1δ2ϕ12(vw) = δ1δ2ϕ12(v)δ1δ2ϕ12(w) = η(v)η(w), so η is an algebra homo-morphism which by (37.24.5), (37.24.4), (274) satisfies η(1C)= δ1δ2ϕ(u23×u31)= δ1δ2ϕ(u23)×ϕ(u31) = u′23×u′31 = 1C′ , nC′ (η(v)) =−ω ′S′(δ1δ2ϕ(v)) =−δ 2

1 δ 22 ω ′S′(ϕ(v)) =−ωS(v) = nC(v).

Summing up we have thus shown that η : C→C′ is a homomorphism of conic algebras, so con-dition (a), (i) holds for (η ,∆). We now verify condition (a), (ii). An application of (37.24.9) yieldsγ ′1 =−S′(u′31) =−δ 2

2 S′(ϕ(u31)) =−δ 22 S(u31), γ ′2 =−S′(u′23) =−δ 2

1 S′(ϕ(u23)) =−δ 21 S(u23), and


we conclude

γ′1 = δ

22 γ1, γ

′2 = δ

21 γ2, γ

′3 = γ3 = 1. (276) GAPRGA

In view of (21), (276), we therefore have δ2δ3δ−11 γ1 = δ2δ1δ2δ

−11 γ1 = δ 2

2 γ1 = γ ′1, δ3δ1δ−12 γ2 =

δ1δ2δ1δ−12 γ2 = δ 2

1 γ2 = γ ′2, δ1δ2δ−13 γ3 = γ3 = γ ′3, which comes down to δ jδlδ

−1i γi = γ ′i for i =

1,2,3, hence to Γ ′ = ∆ ]∆−1Γ , and we have shown that condition (a), (ii) holds as well. ThusCop(ϕ;δ1,δ2) is indeed a morphism of co-ordinate pairs. It remains to show compatibility withcompositions, so let

(J,S)(ϕ;δ1,δ2)

// (J′,S′)(ϕ ′;δ ′1,δ

′2)

// (J′′,S′′),

be morphisms of co-ordinated cubic Jordan algebras. Setting ∆ ′ := diag(δ ′1,δ′2,δ′3), δ ′3 := δ ′1δ ′2,

gives∆′∆ = diag

(δ′1δ1,δ

′2δ2,(δ

′1δ1)(δ

′2δ2)

)Applying (21), we therefore conclude

Cop(ϕ ′1;δ′1,δ′2)Cop(ϕ;δ1,δ2) = (δ ′1δ

′2ϕ′12,∆

′) (δ1δ2ϕ12,∆)

= (δ ′1δ′2δ1δ2ϕ

′12 ϕ12,∆

′∆)

=((

δ′1δ1)(

δ′2δ2)(

ϕ′ ϕ

)12,diag

(δ′1δ1,δ

′2δ2,(δ

′1δ1)(δ

′2δ2)

))= Cop(ϕ ′ ϕ;δ

′1δ1,δ

′2δ2) = Cop

((ϕ ′;δ

′1,δ′2) (ϕ;δ1,δ2)

),

and the assertion follows. Thus we have indeed a functor

Cop: k-cocujo−→ k-copa.

(e) We employ the standard notation of 37.19 for the co-ordinated cubic Jordan algebra(J,S) := Her3(C,Γ ) and then combine (37.24.2) with Prop. 37.17 to conclude C′ = J12 =C[12] ask-modules, giving the first assertion. Applying (37.20.1), (37.20.2), (37.24.1), (37.10.9), we obtain

ω = S(1C[23])−1S(1C[31])−1 = (−γ2γ3n(1C))−1(−γ3γ1n(1C))

−1,

hence

ω = γ−11 γ

−12 γ

−23 . (277) OMDIAG

Now, for v,w ∈C, the multiplication in C′ by (37.24.3), (277), (37.16.4) may be expressed as

v[12]w[12] = ω(v[12]×1C[23])× (1C[31]×w[12]) = γ−11 γ

−12 γ

−23 γ2γ1v[31]× w[23]

= γ−23 γ3wv[12] = γ

−13 (vw)[12],

which amounts to

(vw)[12] = γ3v[12]w[12] (v,w ∈C). (278) PROPRI

Similarly, combining (37.24.5) and (37.20.1) with (37.16.4) yields 1C′ = 1C[23]× 1C[31] =γ31C[12], hence

1C′ = γ31C[12]. (279) UNPRI

Similarly, using (37.24.4), (37.10.9), we obtain

nC′ (v[12]) =−ωS(v[12]) =−γ−11 γ

−12 γ

−23 (−γ1γ2nC(v)),


which may be summarized to

nC′ (v[12]) = γ−23 nC(v) (v ∈C). (280) NOPRI

Invoking (278)−(280), the map ψ := ψC,Γ satisfies

ψ(vw) = γ3(vw)[12] = γ3v[12]γ3w[12] = ψ(v)ψ(w),

ψ(1C) = γ31C[12] = 1C′ ,

nC′(ψ(v)

)= nC′ (γ3v[12]) = γ

23 γ−23 nC(v) = nC(v)

for all v,w ∈ C. Hence ψ : C → C′ is an isomorphism of conic algebras. Finally, setting Γ ′ =diag(γ ′1,γ

′2,γ′3) and observing (37.24.9), (37.20.1), we deduce γ ′1 =−S(1C[31]) = γ3γ1n(1C), γ ′2 =

−S(1C[23]) = γ2γ3n(1C), which yields

γ′1 = γ3γ1, γ

′2 = γ2γ3, γ

′3 = 1. (281) GAPRI

Combining (281) with (24), we obtain λ2λ3λ−11 γ1 = γ3γ1 = γ ′1, λ3λ1λ

−12 γ2 = γ3γ2 = γ ′2, λ1λ2λ

−13 γ3 =

γ−13 γ3 = 1 = γ ′3, which may be unified to λ jλlλ

−1i γi = γ ′i . Thus, setting Λ := ΛC,Γ , we see

Γ ′ = Λ ]Λ−1Γ . Hence ΨC,Γ is an isomorphism of co-ordinate pairs, as claimed.(f) This is clear since Φ = ΦJ,S as defined in Thm. 37.25 matches exactly the diagonal co-

ordinate system of Her3(C,Γ ) with the co-ordinate system S of J.(g) It suffices to show

(i) that the ΦJ;S for all co-ordinated cubic Jordan algebras (J,S) over k determine an isomor-phism

Her3 Cop ∼−→ Idk-cocujo

of functors from k-cocujo to itself and, similarly,(ii) that the ΨC,Γ for all co-ordinate pairs (C,Γ ) over k determine an isomorphism

Idk-copa∼−→ CopHer3

of functors from k-copa to itself.

We begin with (i). Given a co-ordinated cubic norm structure (J,S) over k, Theorem 37.25 showsthat

ΦJ,S : (Her3 Cop)(J,S)∼−→ (J,S)

is an isomorphism in k-cocujo. Thus, by [47, Def. 1.3, p. 23], the proof will be complete once wehave shown that any homomorphism (ϕ;δ1,δ2) : (J,S)→ (J′,S′) of co-ordinated cubic Jordanalgebras gives rise to a commutative diagram

(Her3 Cop)(J,S)ΦJ,S

∼= //

(Her3Cop)(ϕ;δ1,δ2)

(J,S)

(ϕ;δ1,δ2)

(Her3 Cop)(J′,S′)

ΦJ′ ,S′

∼= // (J′,S′).

(282) d.HERCOT

In order to establish the commutativity of (282), we write S (resp. S′) for the quadratic trace of J(resp. J′); moreover, we put S= (e1,e2,e3,u23,u31), S′ = (e′1,e

′2,e′3,u′23,u

′31).

Applying (26), we first obtain (ϕ;δ1,δ2)ΦJ,S = (ϕ;δ1,δ2) (φJ,S;1,1), hence


(ϕ;δ1,δ2)ΦJ,S = (ϕ φJ,S;δ1,δ2). (283) VARPHIPHI

On the other hand, setting

η := δ1δ2ϕ12, ∆ := diag(δ1,δ2,δ1δ2) ∈ Diag3(k)×, (284) VARPHIDEL

we deduce from (272)), (284) that

ΦJ′,S′ Her3(

Cop(ϕ;δ1,δ2))= (φJ′,S′ ;1,1)Her3(η ,∆)

= (φJ′,S′ ;1,1)(

Her3(η ,∆);δ1,δ2)

=(φJ′,S′ Her3(η ,∆);δ1,δ2

),

so in view of (283), it suffices to show

ϕ φJ,S = φJ′,S′ Her3(η ,∆). (285) FIFIHER

In order to do so, we write (C,Γ ) :=Cop(J,S) (resp. (C′,Γ ′)=Cop(J′,S′)). Combining (37.25.1)with condition (a), (i) of Exc. 198, we obtain ϕ φJ,S(eii) = ϕ(ei) = e′i = φJ′,S′ (eii) = φJ′,S′ Her3(η ,∆)(eii) for i = 1,2,3, so it remains to show that both sides of (285) agree on vi[ jl], forall vi ∈C = J12, i = 1,2,3. Since η as defined in (284), being an isomorphism of conic algebras,preserves conjugations, so does every scalar multiple of it, in particular ϕ12. With this in mind, weapply (37.25.3), (284) and obtain

ϕ φJ,S(v1[23]) = −S(u31)−1

ϕ(u31× v1) =−S′(ϕ(u31)

)−1ϕ(u31)×ϕ(v1)

= −S′(δ−12 u′31)

−1δ−12 u′31×ϕ12(v1) =−δ

22 δ−12 S′(u′31)

−1u′31×ϕ12(v1)

= −δ2S′(u′31)−1u′31×ϕ12(v1) = δ2φJ′,S′

(ϕ12(v1)[23]

)= φJ′,S′

(δ−11 (δ1δ2ϕ12)(v1)[23]

)= φJ′,S′ Her3(η ,∆)(v1[23]).

Similarly,

ϕ φJ,S(v2[31]) = −S(u23)−1

ϕ(u23× v2) =−S′(ϕ(u23)

)−1ϕ(u23)×ϕ(v2)

= −S′(δ−11 u′23)

−1δ−11 u′23×ϕ12(v2) =−δ

21 δ−11 S′(u′23)

−1u′23×ϕ12(v2)

= −δ1S′(u′23)−1u′23×ϕ12(v2) = δ1φJ′,S′

(ϕ12(v2[31])

)= φJ′,S′

(δ−12 (δ1δ2ϕ12)(v2[31])

)= φJ′,S′ Her3(η ,∆)(v2[31]).

And, finally,

ϕ φJ,S(v3[12]) = ϕ(v3) = ϕ12(v3) = φJ′,S′(ϕ12(v3)[12]

)= φJ′,S′

((δ1δ2)

−1(δ1δ2ϕ12)(v3)[12])= φJ′,S′ Her3(η ,∆)(v3[12]).

Summing up, this completes the proof of (285), and we have shown that the diagram (282) com-mutes. Thus (i) holds.

It remains to establish (ii). To this end, we let (η ,∆) : (C,Γ )→ (C′,Γ ′)) be a morphism ofco-ordinate triples and must show that the diagram


(C,Γ )ΨC,Γ

∼= //

(η ,∆)

(CopHer3)(C,Γ )

(CopHer3)(η ,∆)

(C′,Γ ′)

ΨC′ ,Γ ′

∼= // (CopHer3)(C′,Γ ′).

(286) d.COTHER

commutes. We begin by making the specifications

Γ = diag(γ1,γ2,γ3), Γ′ = diag(γ ′1,γ

′2,γ′3), (287) GAPR

(C,Γ ) := Cop(Her3(C,Γ )

), Γ = diag(γ1, γ2, γ3), (288) CEHA

(C′,Γ ′) := Cop(Her3(C′,Γ ′)

), Γ

′ = diag(γ ′1, γ′2, γ′3). (289) CEHAPR

We abbreviate ΨC,Γ = (ψ,Λ), ΨC′,Γ ′ = (ψ ′,Λ ′) and deduce from (24)) that

Λ = diag(λ1,λ2,λ3), λ1 = λ2 = 1, λ3 = γ3, (290) LAM

Λ′ = diag(λ ′1,λ

′2,λ

′3), λ

′1 = λ

′2 = 1, λ

′3 = γ

′3. (291) LAMPR

Hence ΨC′,Γ ′ (η ,∆) = (ψ ′,Λ ′) (η ,∆), and (269)) implies

ΨC′,Γ ′ (η ,∆) = (ψ ′ η ,Λ ′∆), (292) DORI

where (23) yields

(ψ ′ η)(v) = γ′3η(v)[12] (v ∈C). (293) PSET

On the other hand, writing ∆ = diag(δ1,δ2,δ3) and invoking (291), we obtain

Λ′∆ = diag(δ1,δ2,γ

′3δ3). (294) LAPRDEL

Next we use (272) to compute

Cop(Her3(η ,∆)

)= Cop

((Her3(η ,∆);δ1,δ2

)),

where

Cop(Her3(η ,∆)

)= (η , ∆), η := δ1δ2 Her3(η ,∆)12, ∆ := diag(δ1,δ2,δ1δ2). (295) VERRI

From (η , ∆)ΨC,Γ = (η , ∆) (ψ,Λ) we therefore deduce

(CopHer3)(η ,∆)ΨC,Γ = (η ψ, ∆Λ). (296) RIDO

Given v ∈C, we observe Γ ′ = ∆ ]∆−1Γ (since (η ,∆) is a morphism in k-copa), which yields

(η ψ)(v) = η(ψC,Γ (v)

)= γ3η(v[12]) = δ1δ2γ3 Her3(η ,∆)12(v[12])

= δ1δ2δ−13 γ3η(v)[12] = γ

′3η(v)[12] = (ψ ′ η)(v).

Thus η ψ = ψ ′ η . Moreover, by (295), (290), (291)

∆Λ = diag(δ1,δ2,δ1δ2)diag(1,1,γ3) = diag(δ1,δ2,δ1δ2γ3)

= diag(δ1,δ2,δ3γ′3) = diag(1,1,γ ′3)diag(δ1,δ2,δ3) = Λ

′∆ .


Comparing now (292) with (296), we see that the diagram (286) commutes, which completes theproof of (ii).

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