“garcia de galdeano”

garcía de galdeano

seminariomatemático n. 5

PRE-PUBLICACIONES delseminario matematico 2006

Alberto ElduqueAlicia Labra

On the classification ofcommutative right-nilalgebrasof dimension at most four

Universidad de Zaragoza

ON THE CLASSIFICATION OF COMMUTATIVE

RIGHT-NILALGEBRAS OF DIMENSION AT MOST FOUR 1

Alberto Elduque,2 and Alicia Labra3

2 Departamento de Matematicas, Facultad de Ciencias,

Universidad de Zaragoza, 50009 Zaragoza, Spain

e-mail: elduque@unizar.es

3 Departamento de Matematicas, Facultad de Ciencias

Universidad de Chile, Casilla 653, Santiago-Chile

e-mail: alimat@uchile.cl

ABSTRACT

Gerstenhaber and Myung [5] classified all commutative, power-associative

nilalgebras of dimension 4. In this paper we extend Gerstenhaber and Myung’s

results by giving a classification of commutative right-nilalgebras of right-

nilindex four and dimension at most four, without assuming power-associativ-

ity. For quadratically closed fields there is, up to isomorphism, a unique such

algebra which is not power-associative in dimension 3 and seven in dimension

4.1A.M.S (2000) subject classification: 17 A 302Part of this research was done while the first author was visiting Universidad de Chile

on grants from FONDECYT 7030075 and 7050164. Support is also acknowledged from

the Spanish Ministerio de Educacion y Ciencia and FEDER (MTM 2004-08115-C04-02)

and from the Diputacion General de Aragon (Grupo de Investigacion de Algebra).3Supported by FONDECYT 1030919.

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1. INTRODUCTION

The problem of nilpotence in a commutative power-associative algebra is

known as Albert’s problem [1] (1948): Is every commutative finite dimensional

power-associative nilalgebra nilpotent?

Suttles [9] in 1972, gave an example of a commutative five dimensional

power-associative nilalgebra of nilindex 4 which is solvable and not nilpotent,

thus showing Albert’s conjecture to be false and forcing a new reformula-

tion of the conjecture. Therefore what is now really known by the Albert

conjecture is whether any commutative finite dimensional power-associative

nilalgebra is solvable.

Gerstenhaber and Myung [5] proved that any commutative and power-

associative nilalgebra of nilindex 4 and dimension 4 over a field of characte-

ristic 6= 2 is nilpotent.

Correa and Suazo [4] proved that every commutative power-associative

nilalgebra of nilindex n and dimension n is nilpotent in characteristic 6= 2, 3.

Thus, Gerstenhaber and Myung’s result follows.

Correa, Hentzel and Labra [3] proved that any commutative right nilalge-

bra of nilindex 4 and dimension 4 is nilpotent in characteristic 6= 2, 3. They

did not assume power-associativity.

More recently, Gutierrez-Fernandez [6] has shown that any commutative

nilalgebra of dimension ≤ 6 over a field of characteristic 6= 2, 3, 5 is solvable.

Moreover, in [5] Gerstenhaber and Myung determined the isomorphism

classes of commutative power-associative nilalgebras of nilindex four and di-

2

mension four. They found one family of algebras parameterized by F×/(F×)2

and 5 individual algebras, two of them being associative.

In this paper we deal only with commutative right-nilalgebras of right-

nilindex four and dimension at most four, over fields of characteristic 6= 2

and we do not assume power-associativity.

The commutative right-nilalgebras of right-nilindex 4 and dimension at

most 4 over a field of characteristic 6= 2 will be classified here, thus extending

the classification by Gerstenhaber and Myung. Moreover, no use will be made

of the results in [3], so this will give an independent proof of the main result

there.

2. RIGHT-NILPOTENCE

Let A be a nonassociative algebra and let x ∈ A. We define the right-

principal powers of x by x1 = x and xn+1 = xnx for all n ≥ 1. An element

x ∈ A is called right-nilpotent with right-nilindex n > 1 if xn = 0 and

xn−1 6= 0. A is called a right-nilalgebra with right-nilindex n if xn = 0 for

all x ∈ A and there exists some a in A with an−1 6= 0. A word of caution is

needed here: this is weaker than the definition in [6].

The algebra A is called nilpotent (respectively, right-nilpotent) in case

the descending chain of ideals (respectively right ideals) defined by A1 =

A, An =∑

r+s=nArAs =

∑n−1i=1 A

iAn−i for n > 1 (respectively A<1> = A

and A<n> = A<n−1>A for n > 1) ends up in zero. The smallest r such that

Ar = 0 (respectively A<r> = 0) is called the index of nilpotence (or index

of right-nilpotence) of A. Clearly, if A is nilpotent then A is right-nilpotent.

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Moreover, if A is commutative or anticommutative, then A5 = A3A2+A4A ⊆

A3A = A<4>. Thus, if A<4> = 0, then A is nilpotent with nilpotent index at

most 5.

More generally, it is known (see [11, Proposition 1]) that if A is a com-

mutative or anti-commutative algebra, then A2n ⊆ A<n>. Therefore if A is

right-nilpotent, it is nilpotent too.

Throughout the paper, all the algebras considered will be commutative and

defined over a ground field F of characteristic 6= 2.

The arguments in [5, Theorem 1] do not use power-associativity. There-

fore:

Theorem 1. Any commutative right-nilalgebra A of right-nilindex 3 and

dimension ≤ 4 satisfies A3 = 0. In particular, it is associative.

Remark. Note that any commutative right-nilalgebra of right-nilindex 3 is

a Jordan algebra. In fact, linearize (xx)x = 0 to obtain 2(yx)x+ (xx)y = 0.

(This can be done since the ground field contains at least 3 elements.)

Multiply this identity by x, and replace y by yx to obtain the identities

2((yx)x)x+ ((xx)y)x = 0 and 2((yx)x)x+ (xx)(yx) = 0 respectively. These

two last identities imply the Jordan identity ((xx)y)x = (xx)(yx). In partic-

ular, A is power-associative (see [8]).

3. NILINDEX 4

In what follows A will be a commutative nonassociative algebra that

satisfies strictly the identity x4 = ((xx)x)x = 0, and contains an element a

in A such that a3 6= 0. This means that A satisfies the identity x4 = 0 and

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all its linearizations. Observe that, once the ground field F contains more

than 3 elements, one can suppress the word ‘strictly’, as the linearizations of

a homogeneous identity of degree at most 4 in its variables are automatically

satisfied.

If x3 6= 0, then x, x2, x3 are linearly independent. Note that the set of

such elements is Zariski-open and non empty in A, which will occasionally

allows the use of density arguments. All references to density will refer to

density in this topology (for its definition and main features on not necessarily

finite dimensional spaces one may consult [7]).

Linearizing ((xx)x)x = 0 we get

2((yx)x)x+ (x2y)x+ x3y = 0, (1)

2(((yz)x)x+ ((yx)z)x+ ((yx)x)z + ((xz)y)x+ ((xz)x)y

)+ (x2y)z + (x2z)y = 0. (2)

Let us first derive some consequences of these linearizations:

Lemma 1. Let A be a commutative algebra satisfying strictly the identity

x4 = 0. Then for any x ∈ A:

(i) x3x2 = −(x2x2)x,

(ii) x3x3 = −(x3x2)x = (x2)3,

(iii) (x3x2)x2 = x3(x2x2) = (x2)3x = (((x2x2)x)x)x = 0.

(iv) (((x2x2)x)x2)x = (((x2x2)x)x)x2 = (x2x2)(x2x2) = 0.

Proof: The substitution y = x2 in Equation (1) gives (i), while y = x3

gives x3x3 = −(x3x2)x. Now, the substitutions y = z = x2 in Equation (2)

5

together with part (i) gives (x3x2)x = −(x2)3, thus obtaining (ii).

The substitution y = x2x2 in (1), using (i) and (ii), gives

3(x2)3x = −x3(x2x2), (3)

while the substitution y 7→ x, x 7→ x2 in (1), using (i) gives

(x3x2)x2 = −(x2)3x, (4)

and the substitution y = x3, z = x2 in (2), using (ii), gives

(x3x2)x2 = −x3(x2x2). (5)

Now, if the equations (3), (4) and the negative of (5) are added, one obtains

3(x2)3x = −(x2)3x, so (x2)3x = 0 = (x3x2)x2 = x3(x2x2), and (((x2x2)x)x)x =

−((x3x2)x)x = (x2)3x = 0 by (i) and (ii).

Finally, the substitution y 7→ x2, z 7→ x2x2 in (2), using (iii) and the

identity x4 = 0 gives

2((((x2x2)x)x2)x+ (((x2x2)x)x)x2

)+ (x2x2)(x2x2) = 0

Now, using (i) and (iii) we obtain (x2x2)x)x2)x = −((x3x2)x2)x = 0

and by (i) and (ii) ((x2x2)x)x)x2 = −((x3x2)x)x2 = (x2)4 = 0. Therefore

(x2x2)(x2x2) = 0 and (iv) follows.

The following notation will be used. Given a set S, 〈S〉 will denote the

subspace generated by S, while alg〈S〉 will be the subalgebra generated by

S.

Theorem 2. Let A be a commutative algebra of dimension 3 which satisfies

the identity x4 = 0 strictly and contains elements y with y3 6= 0. Then

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A<4> = 0, so A is nilpotent, and there is an element a ∈ A such that

{a, a2, a3} is a basis of A, and whose multiplication table is given in Table I.

Table I

a a2 a3

a a2 a3 0

a2 a3 γa3 0

a3 0 0 0

where γ = 0 or 1. Moreover, A is power-associative if and only if γ = 0.

Proof: Let a be an element in A such that a3 6= 0. Then a, a2, a3 are

linearly independent. Since dim A = 3, A = 〈a, a2, a3 〉 and the left

multiplication by a, La, defined by La(x) = ax for all x in A, is nilpotent:

L3a = 0. Extending scalars if necessary to get an infinite ground field, the

set of elements y with y3 6= 0 is dense, so we conclude that L3y = 0 for any

y ∈ A. Thus, this is true even if the ground field is finite.

If a2a2 = αa + βa2 + γa3, then (a2a2)a = αa2 + βa3 so, by Lemma

1.(i), −a3a2 = αa2 + βa3. Thus (a3 + βa)a2 = −αa2. Setting z = a3 + βa

we obtain that L−z(a2) = αa2. That is, α is an eigenvalue of the nilpotent

operator L−z. Hence α = 0. Now a2a2 = βa2 + γa3, which implies

that (a2 − γa)a2 = βa2. Since La2−γa is nilpotent we obtain that β = 0,

and a2a2 = γa3. Moreover, a3a2 = −(a2a2)a = −γa4 = 0 and Lemma 1.(ii)

implies that a3a3 = −(a2a3)a = 0, thus obtaining Table I. Now, if γ = 0,

then A3 = 0 and hence A is associative, while if γ 6= 0, substituting a by

γ−1a we get a new multiplication table with γ = 1. Finally A<4> = 0, so

A5 = 0 and A is nilpotent.

7

In order to deal with the dimension 4 case, we need the following easy

consequence of nilpotence, which can be found in [10, Theorem 6]:

Lemma 2. Every proper maximal subalgebra of a nilpotent algebra A con-

tains A2.

The next result provides an alternative proof to [3, Theorem 2], which is

valid also in characteristic 3, and provides more information on the algebra

involved:

Theorem 3. Let A be a commutative algebra of dimension 4 which is

not power-associative, satisfies strictly the identity x4 = 0 and contains an

element y with y3 6= 0. Then A<4> = 0, so A is nilpotent, and the dimension

of A2 is either 2 or 3. Moreover, (i) If dimA2 = 3, then there is an element

a ∈ A such that {a, a2, a3, a2a2} is a basis of A, whose multiplication table is

given by Table II:

Table II

a a2 a3 a2a2

a a2 a3 0 0

a2 a3 a2a2 0 0

a3 0 0 0 0

a2a2 0 0 0 0

(ii) If dimA2 = 2, then (A2)2 = A3, which has dimension 1.

Proof: Assume first that the ground field F is infinite. If there is an

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element a ∈ A such that alg〈a〉 = A, then the set of such elements is dense,

so this element can be taken with the extra assumption of a3 6= 0. Now, if

a2a2 ∈ 〈a, a2, a3〉, Lemma 1.(i)-(ii) shows that 〈a, a2, a3〉 is a subalgebra, a

contradiction. Hence {a, a2, a3, a2a2} is a basis of A. Now, by Lemma 1.(iii),

L3a = 0 and, by Zariski density, this is true for any x ∈ A.

Since 〈a, a2, a3〉 is invariant under La and has codimension 1, by nilpo-

tence of La, aA ⊆ 〈a, a2, a3〉, so a(a2a2) = αa+βa2+γa3 for some α, β, γ ∈ F .

Hence (a2a2 − βa − γa2)a = αa, and the nilpotence of La2a2−βa−γa2 forces

α = 0. Then, by Lemma 1.(i) −a3a2 = a(a2a2) = βa2 + γa3, and we get

(a3 + γa)a2 = −βa2, so by nilpotence of La3+γa we get β = 0, and hence

a2a3 = −γa3 and γ = 0. Therefore, a(a2a2) = 0 which, by Lemma 1.(i)-(ii),

forces a3a2 = 0 = a3a3 = (a2)3. Moreover, Lemma 1.(iii)-(iv) implies that

(a2a2)(a2a2) = 0 = a3(a2a2), and we obtain the multiplication in Table II. In

particular A<4> = 0, and hence A5 = 0 and A is nilpotent.

On the other hand, assume that there is no x ∈ A with A = alg〈x〉 and

let x ∈ A such that x3 6= 0, so that alg〈x〉 = 〈x, x2, x3〉 is a 3-dimensional

nilpotent subalgebra (by Theorem 2). Then alg〈x〉2 = 〈x2, x3〉 is, by Lemma

2, the only proper maximal subalgebra of alg〈x〉.

For any y ∈ A\alg〈x〉, alg〈x〉∩alg〈y〉 is a proper subalgebra of alg〈x〉 so,

by Lemma 2, alg〈x〉 ∩ alg〈y〉 ⊆ alg〈x〉2. Moreover, since alg〈x〉 has codimen-

sion 1, alg〈x〉 ∩ alg〈y〉 is a codimension 1 subalgebra of alg〈y〉 so, again by

Lemma 2, alg〈y〉2 ⊆ alg〈x〉 ∩ alg〈y〉 ⊆ alg〈x〉2, and hence y2 ∈ alg〈x〉2. By

commutativity we conclude that A2 = alg〈x〉2, which has dimension 2, and

L3x(A) ⊆ L2

x(A2) = L2

x(alg〈x〉2) = 0. Since this is valid for any x ∈ A with

x3 6= 0, and the set of these elements is dense, if follows that L3y = 0 for any

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y ∈ A. Also, for any z ∈ A, since A2 = alg〈x〉2 = 〈x2, x3〉, x2z = αx2 + βx3,

for some α, β ∈ F , so (z − βx)x2 = αx2, which implies that α = 0. But

Equation (1) shows then that x3z = −(x2z)x = −βx4 = 0. Therefore

A3 = A2A = 〈x3〉 and A<4> = x3A = 0. Since A2A2 = 〈x2x2〉 ⊆ A3 and A is

not power-associative (so A2A2 cannot be 0), we conclude that A2A2 = A3,

which has dimension 1.

Finally, if F is a finite field, we may extend scalars to conclude that

dimA2 is either 2 or 3 and that A<4> = A3A = 0. If dimA2 = 3, A2 is the

only proper maximal subalgebra of A, so for any x ∈ A \ A2, alg〈x〉 = A.

Moreover, if x3 were equal to 0, then x + x2 ∈ A \ A2 and (x + x2)3 =

(x2 +x2x2)(x+x2) = (x2)2 +(x2)2x+(x2)3 = (x2)2 by Lemma 1.(i)-(ii), and

(x2)2 6= 0 (otherwise alg〈x〉 = 〈x, x2〉 6= A). Hence, changing x by x + x2 if

necessary, there is an element a ∈ A with alg〈a〉 = A and a3 6= 0, and the

same argument used for infinite fields gives that {a, a2, a3, a2a2} is a basis of

A with the multiplication given by Table II.

Remark: In case (ii) of Theorem 3, let a ∈ A with a3 6= 0. Then A2 =

〈a2, a3〉 and L2a 6= 0 = L3

a. Hence dim kerLa = 2, so there is an element

b ∈ A \ alg〈a〉 such that ab = 0. Also, since A2A2 = A3 has dimension 1,

there is a scalar γ ∈ F with a2a2 = γa3. The multiplication table in alg〈a〉 is

then given by Table I, and γ 6= 0 (otherwise, A2A2 = alg〈a〉2 alg〈a〉2 would

be 0). If we change a by γ−1a, then a2a2 = a3 and the multiplication table

in the basis {a, a2, a3, b} is:

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Table III

a a2 a3 b

a a2 a3 0 0

a2 a3 a3 0 µa3

a3 0 0 0 0

b 0 µa3 0 αa2 + βa3

for suitable scalars α, β, µ ∈ F .

4. CLASSIFICATION

There is a unique algebra, up to isomorphism, in case (i) of Theorem 3.

Our aim in this section is to classify the algebras in Theorem 3.(ii) . Actually,

a broader class of commutative right-nilpotent algebras will be classified in

Theorem 4, by means of conditions which depend heavily on the ground

field. For quadratically closed fields, it will turn out that there are exactly 7

nonisomorphic commutative four dimensional algebras satisfying the identity

x4 = 0 and which are not power-associative.

Let W be a nonzero vector space and let ϕ, ψ be two symmetric bilinear

forms on W with ϕ 6= 0. Denote by N(W,ϕ, ψ) the commutative algebra

defined on W ⊕ F ⊕ F with multiplication

(w1, α1, β1)(w2, α2, β2) = (0, ϕ(w1, w2), ψ(w1, w2) + α1α2)

for any w1, w2 ∈ W , α1, α2, β1, β2 ∈ F .

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Proposition 1. The algebra A = N(W,ϕ, ψ) is commutative and not power-

associative. Moreover, dimA2 = 2, A2A2 = A3 has dimension 1 and A<4> =

A3A = 0. In particular, it satisfies strictly the identity x4 = 0. Conversely,

if A is a four dimensional commutative algebra which satisfies strictly the

identity x4 = 0 and is not power-associative, and if dimA2 = 2, then there

is a two dimensional vector space W , and symmetric bilinear forms ϕ 6= 0

and ψ defined on W , such that A is isomorphic to N(W,ϕ, ψ).

Proof: If A = N(W,ϕ, ψ), then A is commutative because both ϕ and

ψ are symmetric. Besides, A2 = {(0, α, β) : α, β ∈ F}, as ϕ 6= 0, and

A3 = {(0, 0, α) : α ∈ F} = A2A2, while A<4> = A3A = 0. Let u = (0, 1, 0)

and v = (0, 0, 1), so that A = W ⊕ Fu ⊕ Fv. Thus A2A2 = 〈u, v〉 and

A3 = A2A2 = 〈v〉. Let w ∈ W with ϕ(w,w) 6= 0, then w2 = ϕ(w,w)u,

w3 = 0 and w2w2 = ϕ(w,w)2v 6= 0. In particular, w2w2 6= 0 = w4 and

A is not power-associative. For the converse, take a basis {a, a2, a3, b} of A

with multiplication given by Table III, and let W = 〈a− a2, b− µa〉, u = a2,

and v = a3. Then A = W ⊕ Fu ⊕ Fv, Wu = 0 = Av, u2 = v and for any

w1, w2 ∈ W , w1w2 = ϕ(w1, w2)u + ψ(w1, w2)v for symmetric bilinear forms

ϕ, ψ on W . Since (a−a2)2 = a2−a3, ϕ 6= 0. The linear map A→ N(W,ϕ, ψ)

that takes w + εu+ δv to (w, ε, δ) is clearly an isomorphism.

Let ϕ, ϕ′ be two bilinear forms on W and W ′ respectively. Recall that a

similarity of multiplier α, Φ : (W,ϕ) → (W ′, ϕ′), is a linear isomorphism from

W onto W ′ such that ϕ′(Φ(w1),Φ(w2)) = αϕ(w1, w2), for any w1, w2 ∈ W .

The next result classifies the algebras N(W,ϕ, ψ), and hence, in particu-

lar, the algebras in item (ii) of Theorem 3.

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Theorem 4. The algebras N(W,ϕ, ψ) and N(W ′, ϕ′, ψ′) are isomorphic if

and only if there are scalars α, β ∈ F , α 6= 0, and a similarity Φ : (W,ϕ) →

(W ′, ϕ′) of multiplier α such that for every w1, w2 ∈ W , ψ′(Φ(w1),Φ(w2)) =

α2ψ(w1, w2) + βϕ(w1, w2).

Proof: Let Ψ : A = N(W,ϕ, ψ) → A′ = N(W ′, ϕ′, ψ′) be an algebra

isomorphism. Note that A2 = Fu ⊕ Fv, annA2(= {x ∈ A : xA2 = 0}) =

W ⊕ Fv, and A3 = Fv. Since Ψ is an algebra isomorphism, we have that

Ψ(annA2) = ann(A′)2, Ψ(A2) = (A′)2 and Ψ(A3) = (A′)3. Thus with

u′ = (0, 1, 0) and v′ = (0, 0, 1) in A′, from Ψ(annA2) = ann(A′)2, it follows

that there exists a linear form f : W → F such that Ψ(w)− f(w)v′ ∈ W ′ for

any w ∈ W .

Let us define Φ : A → A′ by Φ(w) = Ψ(w) − f(w)v′, for any w ∈ W ,

Φ(u) = Ψ(u) and Φ(v) = Ψ(v). Then Φ is an algebra isomorphism too

(since A′v′ = 0), but with Φ(W ) = W ′. If Φ(u) = αu′ + βv′, Φ(v) =

Φ(u2) = (Φ(u))2 = α2v′, Φ(w1w2) = ϕ(w1, w2)(αu′+βv′)+ψ(w1, w2)α

2v′ and

Φ(w1)Φ(w2) = ϕ′(Φ(w1),Φ(w2))u′ + ψ′(Φ(w1),Φ(w2))v

′ for any w1, w2 ∈ W .

Therefore we obtain that ϕ′(Φ(w1),Φ(w2)) = αϕ(w1, w2), and the restric-

tion of Φ to W is a similarity of multiplier α. Moreover, we have that

ψ′(Φ(w1),Φ(w2)) = α2ψ(w1, w2) + βϕ(w1, w2) for every w1, w2 ∈ W . Con-

versely, given such scalars α and β and such a similarity Φ of multiplier

α, Φ extends to an algebra isomorphism by means of Φ(u) = αu′ + βv′

and Φ(v) = α2v′. In fact Φ(u)2 = (αu′ + βv′)2 = α2v′ = Φ(u2) and

Φ(w1)Φ(w2) = ϕ′(Φ(w1),Φ(w2))u′ + ψ′(Φ(w1),Φ(w2))v

′ = ϕ(w1, w2)(αu′ +

βv′) + ψ(w1, w2)α2v′ = Φ(w1, w2) for any w1, w2 ∈ W .

Note that in A = N(W,ϕ, ψ), {x ∈ A : xA ⊆ A3} = A2 ⊕ kerϕ, while

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annA2 = W ⊕ A3 and (annA2)2 = W 2, which is the whole A2 if and only if

ψ is not proportional to ϕ, otherwise it has dimension 1.

Corollary 1. With W , ϕ and ψ as above, the rank of ϕ and the integer

rank(ψ : ϕ) = min{rank(ψ+βϕ) : β ∈ F} are invariants of the isomorphism

class of N(W,ϕ, ψ).

Corollary 2. If dim W = 2, then the discriminant of the form ϕ: disc(ϕ) =

(detϕ)(F×)2 and the rank of ϕ are invariants in the isomorphism class of

N(W,ϕ, ψ).

Corollary 3. For any α, β ∈ F with α 6= 0, N(W,ϕ, ψ) is isomorphic to

N(W,αϕ, α2ψ + βϕ).

We will finish the paper with refinements of Theorem 4 above whenever

possible.

Assume that dimW = 2 and that rankϕ = 1. Then Corollary 3 allows

us to scale ϕ and modify ψ, so that we can assume that there is a basis {a, b}

of W in which the coordinate matrices of ϕ and ψ are given by

ϕ :

1 0

0 0

, ψ :

0 γ

γ ε

.

The multiplication table in the basis {a, b, u, v} is

14

Table IV

a b u v

a u γv 0 0

b γv εv 0 0

u 0 0 v 0

v 0 0 0 0

Let us denote by A1(ε, γ) the algebra with this multiplication table.

Proposition 2. Let A = N(W,ϕ, ψ) be as above, with dimW = 2 and

rankϕ = 1. Then:

(i) rank(ψ : ϕ) = 0 if and only if A is isomorphic to A1(0, 0).

(ii) rank(ψ : ϕ) = 1 if and only if A is isomorphic to A1(ε, 0) with ε 6=

0. Moreover, A1(ε, 0) and A1(ε′, 0) are isomorphic if and only if ε(F×)2 =

ε′(F×)2.

(iii) rank(ψ : ϕ) = 2 if and only if A is isomorphic to A1(0, 1).

Proof: With the coordinate matrices of ϕ and ψ as before, it is clear

that

rank(ψ : ϕ) =

0 if ε = γ = 0,

1 if ε 6= 0,

2 if ε = 0 6= γ.

If rank(ψ : ϕ) = 0, case (i) follows easily from Corollary 3. If rank(ψ : ϕ) = 1,

we may change a to a − γε−1b, to get γ = 0. In this case, I = {x ∈ A :

xA ⊆ A3} = 〈b, u, v〉 and the map I/A3 × I/A3 → A3 = Fv given by

(x+A3, y+A3) 7→ xy is a bilinear symmetric form with discriminant ε(F×)2.

15

From here case (ii) follows. Finally, if rank(ψ : ϕ) = 2, then we may change

b to γ−1b to get γ = 1, thus proving (iii). (Note that the converse is trivial

in the three cases.)

Assume now that dimW = 2, rankϕ = 2 and ϕ represents 0 (this last

condition is always satisfied if F is quadratically closed); then by Corollary

3 we may assume that there is a basis {a, b} of W with coordinate matrices

for ϕ and ψ given by

ϕ :

0 1

1 0

, ψ :

ε 0

0 δ

.

The multiplication table in the basis {a, b, u, v} of A = N(W,ϕ, ψ) is

a b u v

a εv u 0 0

b u δv 0 0

u 0 0 v 0

v 0 0 0 0

Denote by A2(ε, δ) the algebra with this multiplication. Then:

Proposition 3. The algebras A2(ε, δ) and A2(ε′, δ′) are isomorphic if and

only if either ε(F×)2 = ε′(F×)2 and δ(F×)2 = δ′(F×)2, or ε(F×)2 = δ′(F×)2

and δ(F×)2 = ε′(F×)2.

Proof: Take a basis {a, b, u, v} of A2(ε, δ) (respectively {a′, b′, u′, v′} of

A2(ε′, δ′)) as above. Any similarity Φ : (W = 〈a, b〉, ϕ) → (W ′ = 〈a′, b′〉, ϕ′)

satisfies that Φ(a),Φ(b) ∈ Fa′ ∪ Fb′ (the set of isotropic vectors). Hence

there are two possibilities: either Φ = Φ1 with Φ1(a) = θa, Φ1(b) = νb, or

16

Φ = Φ2 with Φ2(a) = θb and Φ2(b) = νa, for nonzero scalars θ, ν. In both

cases, the multiplier of Φ is θν. By Theorem 4, A2(ε, δ) and A2(ε′, δ′) are

isomorphic if and only if there exists scalars α, β ∈ F with α 6= 0, and a sim-

ilarity Φ : (W,ϕ) → (W ′, ϕ′) of multiplier α, such that ψ′(Φ(w1),Φ(w2)) =

α2ψ(w1, w2) + βϕ(w1, w2).

But setting α = θν we compute

ψ′(Φ1(a),Φ1(a)) = θ2ψ′(a, a) = θ2ε′ = α2

ν2 ε′,

ψ′(Φ1(b),Φ1(b)) = ν2ψ′(b, b) = ν2δ′ = α2

θ2δ′,

ψ′(Φ1(a),Φ1(b)) = 0,

and,

ψ′(Φ2(a),Φ2(a)) = θ2ψ′(b, b) = α2

ν2 δ′,

ψ′(Φ2(b),Φ2(b)) = ν2ψ′(b, b) = α2

θ2ε′,

ψ′(φ2(a), φ2(b)) = 0,

and the Proposition follows easily.

As a direct consequence of the previous results, we get a complete and

simple classification of the commutative algebras satisfying the identity x4 =

0 over quadratically closed fields.

Corollary 4. Let F be a quadratically closed field. Then any four dimen-

sional commutative and not power-associative algebra satisfying the identity

x4 = 0 is isomorphic to one and only one of the following seven algebras:

the algebra with multiplication given by Table II, A1(0, 0), A1(1, 0), A1(0, 1),

A2(0, 0), A2(1, 0), or A2(1, 1).

Note that if F is quadratically closed, then it is infinite, so it is not

necessary to impose that the identity x4 = 0 be satisfied strictly.

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We are left with the case in which ϕ does not represent 0. Then we can

take a basis {a, b} of W with coordinate matrices

ϕ :

1 0

0 µ

, ψ :

0 γ

γ ε

, (6)

with −µ 6∈ F 2.

The multiplication table in the basis {a, b, u, v} of A = N(W,ϕ, ψ) is

a b u v

a u γv 0 0

b γv µu+ εv 0 0

u 0 0 v 0

v 0 0 0 0

Denote by A(µ; ε, γ) the algebra with this multiplication.

Proposition 4. Let A be the algebra N(W,ϕ, ψ), with dimW = 2 and

assume that ϕ does not represent 0. Then:

(i) rank(ψ : ϕ) = 0 if and only if A is isomorphic to A(µ; 0, 0) for some µ ∈

F , −µ 6∈ F 2. Moreover, the algebras A(µ; 0, 0) and A(µ′; 0, 0) are isomorphic

if and only if µ(F×)2 = µ′(F×)2.

(ii) rank(ψ : ϕ) = 1 if and only if A is isomorphic to A(µ; ε, 0) for some

0 6= ε, µ ∈ F , −µ 6∈ F 2. Moreover, the algebras A(µ; ε, 0) and A(µ′, ε′, 0) are

isomorphic if and only if µ(F×)2 = µ′(F×)2 and ε′(F×)2 equals either ε(F×)2

or −µε(F×)2.

18

(iii) rank(ψ : ϕ) = 2 if and only if A is isomorphic to A(µ; ε, γ) for some

0 6= µ, ε, γ ∈ F , −µ 6∈ F 2 and ε2 + 4µγ2 6∈ F 2.

Proof: If rank(ψ : ϕ) = 0, ψ is proportional to ϕ, so by Corollary 3, A

is isomorphic to N(W,ϕ, 0) and (i) follows from Corollary 2.

If rank(ψ : ϕ) = 1, we may change ψ by Corollary 3 and assume that

rankψ = 1. Let kerψ = Fa. Since ϕ does not represent 0, ϕ(a, a) 6= 0.

Then, by scaling ϕ (Corollary 3) we may assume that ϕ(a, a) = 1 and hence

there is a basis {a, b} of W with coordinate matrices as in (6), but with

γ = 0. Now, if A(µ; ε, 0) is isomorphic to A(µ′; ε′, 0), then µ(F×)2 = µ′(F×)2

by Corollary 2. Moreover, let qψ denote the quadratic form associated to

the bilinear form ψ. Taking the basis as above, we check that {α2ψ + βϕ :

α 6= 0, rank(α2ψ + βϕ) = 1} = (F×)2ψ ∪ (F×)2(ψ − εµϕ) and qψ(W ) = εF 2,

qψ− εµϕ(W ) = − ε

µF 2 = −µεF 2. Therefore, ε(F×)2 ∪ (−µε)(F×)2 is another

invariant of the isomorphism class of A(µ; ε, 0). Conversely, setting µ′ = µθ2

and ε′ = εν2 and taking the new basis {a, b, u, v} in A(µ; ε, 0) where a = θνa,

b = θ2

νb, u = θ2

ν2u and v = θ4

ν4v we get the multiplication table of A(µ′; ε′, 0).

On the other hand, if µ′ = µθ2 and ε′ = −µεν2 then we can get a new basis

{a, b, u, v} where, a = θµνb, b = θ2

νa, u = ( θ

µν)2(µu+εv), and v = ( θ2

µν2 )2v, with

the same multiplication table of A(µ′; ε′, 0). Therefore A(µ; ε, 0) is isomorphic

to A(µ′; ε′, 0).

Finally, since(

0 γγ ε

)+β

(1 00 µ

)has rank < 2 if and only if β(µβ+ε)−γ2 = 0,

and the polynomial µX2 + εX−γ2 has no roots if and only if ε2 +4µγ2 6∈ F 2,

(iii) follows.

Remark: Case (iii) in the Proposition above is a kind of “generic” case,

19

hence no simpler conditions for isomorphisms between these algebras can be

given than the ones in Theorem 4.

The authors thank the referee for suggestions and comments for improve-

ment of the paper.

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