Galois Theory

Torsten Wedhorn

July 13, 2015

These lecture notes contain the final third of my lecture on abstract algebra.

Notation: If not otherwise stressed, all rings and all algebras are commutative. Thesymbol K always denotes a field.

Contents

8 Field Extensions 3A Subfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3B Prime Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

9 Algebraic Extensions 5A Finite Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5B Algebraic and Transcendental Elements . . . . . . . . . . . . . . . . . . 7C Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8D Algebraic Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10E Extension of Field Homomorphisms . . . . . . . . . . . . . . . . . . . . 11F Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

10 The Fundamental Theorem of Galois Theory for Finite Extensions 14A Normal extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14B Fundamental Theorem of Galois Theory for Finite Extensions . . . . . . 15

11 Etale Algebras 18A Preparations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18B Separable Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19C Separable Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20D Separability Degree and Split Algebras . . . . . . . . . . . . . . . . . . . 21E Characterization of Etale Algebras . . . . . . . . . . . . . . . . . . . . . 23F Theorem of the Primitive Element . . . . . . . . . . . . . . . . . . . . . 26G Separable Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

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12 Galois Theory 29A Galois Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29B Main Theorem of Galois (Grothendieck version) . . . . . . . . . . . . . . 30C Proof of the Main Theorem of Galois Theory . . . . . . . . . . . . . . . 32

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8 Field Extensions

A Subfields

Remark 8.1. Let ϕ : K → A be a not necessarily commutative K-algebra 6= 0. Thenϕ is injective:

ϕ(1) = 1 6= 0⇒ 1 /∈ Ker(ϕ)⇒ Ker(ϕ) = 0

because Ker(ϕ) is an ideal of K, and the only ideals of a field K are K and 0.

Definition 8.2. (1) A K-algebra ϕ : K → L is called a field extension if L is a field.(2) A K-subalgebra of M of a field extension K → L is called a subextension if L is a

field.

Example 8.3. (1) Q ⊆ C is a field extension and R is a subextension of C.(2) K = Q[

√2] → C, a+ b

√2 7→ a− b

√2 is a field extension and R is a subextension

of C.

Remark and Definition 8.4. Let K → M be field extension and let (Li)i∈I be afamily of subextension of M . Then

⋂i∈I Li is a subextension of M .

Let S ⊆M be a subset. Then

K(S) :=⋂

L subextension of MS⊆L

L

is the smallest subextension of M containing S. It is called the subextension of Mgenerated by S or the subextension of M obtained by adjunction of S. One has

K(S) = f(s1, . . . , sm)

g(t1, . . . , tn)|n,m ∈ N0, f ∈ K[X1, . . . , Xm], g ∈ K[X1, . . . , Xn],

si, tj ∈ S, g(t1, . . . , tn) 6= 0.

For S = s1, . . . , sn we write K(s1, . . . , sn) instead of K(s1, . . . , sn). A field exten-sion K →M is called finitely generated if there exists a finite subset S ⊆M such thatM = K(S).One clearly has K[S] ⊆ K(S), where K[S] is the smallest K-subalgebra of M containingS.

Example 8.5. C = R(i) = R[i].

B Prime Fields

Definition 8.6. A field K is said to be a prime field if K ∼= Q or K ∼= Fp for someprime number p.

Proposition 8.7. Let R be a ring containing a field K. Then there exists a smallestsubfield K0 of R and this is a prime field. One has

char(R) = 0⇔ K0 = Q, char(R) = p⇔ K0 = Fp.

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Proof. Let ϕ : Z → R be the unique ring homomorphism. Then Im(ϕ) is containedin any subring of R and in particular in K. Hence Z/Ker(ϕ) = Im(ϕ) is an integraldomain and hence Ker(ϕ) is a prime ideal of Z (5.20). By (5.21) we have Ker(ϕ) = 0or Ker(ϕ) = pZ for some prime number p.(1) char(R) = p⇔ Ker(ϕ) = pZ⇔ Im(ϕ) ∼= Fp and in this case Im(ϕ) is the smallest

subfield of R.(2) char(R) = 0⇔ Ker(ϕ) = 0⇔ Im(ϕ) ∼= Z. Then Quot(Im(ϕ)) ∼= Q is a subfield K

(by the universal property of the quotient field), and it is the smallest subfield ofR.

Remark and Definition 8.8. Hence we say that a ring R that contains a field hascharacteristic p for a prime number p and write char(R) = p (resp. has characterstic 0and write char(R) = 0) if R contains Fp (resp. contains Q).If R is a ring of characteristic p one has p · 1 = 0 and hence for all a, b ∈ R

(a+ b)p = ap + bp

because for 1 ≤ k ≤ p− 1 the binomial coefficient(p

k

)=p(p− 1) · · · (p− k + 1)

(p− k)!

is divisible by p. This shows that

FrobR : R 7→ R, a 7→ ap,

is a ring endomorphism, called Frobenius endomorphism of R.

Definition 8.9. A ring R that contains a field is called perfect if either char(R) = 0or if char(R) = p > 0 and the Frobenius FrobR : a 7→ ap is bijective.

Example 8.10. (1) Every finite field F is perfect: FrobF is injective because Ker(FrobF) = a ∈ F ; ap = 0 = 0. Hence it is bijective because F is finite.

(2) Let K = Quot(Fp[T ]). As T is a prime element in Fp[T ] there exists no a ∈ K withap = T . Hence K is not perfect.

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9 Algebraic Extensions

A Finite Algebras

Definition 9.1. Let R be a ring and let A be an R-algebra. Then A is called a finite(resp. a free) R-algebra if A is a finitely generated (resp. a free) R-module.

Proposition 9.2. Let R → A → B be ring homomorphisms and suppose that A is afree R-algebra and that B is a free A-algebra. Then B is a free R-algebra and

rkR(B) = rkR(A) rkA(B).

Proof. We have isomorphisms of R-modules (the first one is even an isomorphism ofA-modules)

B ∼= A(I) ∼= (R(J))(I) = R(I×J).

Remark 9.3. Let R → R′ be a homomorphism of rings and let ϕ : R → A be anR-algebra. Then the scalar extension R′ ⊗R A is an R′-algebra via R′ → R′ ⊗R A,a′ 7→ a′ ⊗ 1. Morever: If A is free R-algebra, then R′ ⊗R A is a free R′-algebra andrkR′(R

′ ⊗R A) = rkR(A) by (6.33).

Definition and Remark 9.4. Let K be a field and let ϕ : K → A be a K-algebra.Then we call [A : K] := dimK(A) the degree of A over K.(1) One has A = K if and only if ϕ is an isomorphism.(2) If L is a field extension of K, then L⊗KA is an L-algebra and [L⊗KA : L] = [A : K]

by Remark 9.3.(3) If K → L is a field extension and L→ A is an L-algebra, then

[A : K] = [A : L][L : K]

by Proposition 9.2.

Definition 9.5. Let A be a ring.(1) An element a ∈ A is called nilpotent if there exists n ∈ N such that an = 0.(2) The ring A is called reduced if 0 is the only nilpotent element of A.

Remark 9.6. (1) Every integral domain is reduced.(2) Every subring of a reduced ring is reduced.(3) Let A1 and A2 be rings. Then A1 × A2 is reduced if and only if A1 and A2 are

reduced (an element (x1, x2) ∈ A1 × A2 is nilpotent if and only if x1 and x2 arenilpotent).

Proposition 9.7. Let A be a finite K-algebra.(1) A is an integral domain if and only if A is a field.(2) Every prime ideal of A is a maximal ideal.(3) There exist only finitely many maximal ideal m1, . . . ,mr in A with r ≤ [A : K].(4) A is reduced if and only if A ∼=

∏ri=1A/mi.

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Proof. (1). (Exercise 18) For x ∈ A let mx : A → A be the K-linear map a 7→ xa.Then

A integral domain ⇔ ∀ 0 6= x ∈ A : mx is injectivedimK(A) <∞⇔ ∀ 0 6= x ∈ A : mx is bijective

⇔ A is a field.

(2). Let p ⊂ A be a prime ideal. Then A/p is a finite K-algebra which is an integraldomain. Hence it is a field by (1). Therefore p is a maximal ideal (5.20).(3),(4). Any two maximal ideals m1 6= m2 of a ring are coprime. Hence if m1, . . . ,mt

are distinct maximal ideals of A, then

A→t∏i=1

A/mi

is surjective by the Chinese Remainder Theorem (5.16). Hence [A : K] ≥∑t

i=1[A/mi :K] which shows that there are only finitely many maximal ideals m1, . . . ,mr in A.Moreover

Ker(A→r∏i=1

A/mi) =

r⋂i=1

mi(2)=

⋂p⊂A

prime ideal

pEx.27

= a ∈ A ; a nilpotent,

which shows (4).

Example 9.8. Let f ∈ K[X] be a non-constant polynomial and let f = fe11 . . . ferr bean irreducible decomposition (i.e., the fi are irreducible and pairwise coprime, ei ∈ N).Let A := K[X]/(f) and for every g ∈ K[X] let g be its image in A.(1) A is a finite K-algebra with [A : K] = deg(f) (4.22).(2) One has

f irreducible⇔ A integral domain⇔ A field

(3) The maximal ideals of A (= prime ideals of A) are the ideals (fi) for i = 1, . . . , r(5.27).

(4) By the Chinese remainder theorem one has

A ∼=r∏i=1

K[X]/(feii )

and A is reduced if and only if e1 = e2 = · · · = er = 1 (A reduced ⇔ K[X]/(feii )reduced for all i by Remark 9.6 (3); ei = 1 ⇒ K[X]/(feii ) integral domain; ei > 1,then 0 6= fi ∈ K[X]/(feii ) with feii = 0).

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B Algebraic and Transcendental Elements

Definition 9.9. Let A be a K-algebra. An element a ∈ A is called algebraic over Kif it is integral over K (i.e. if there exists a monic polynomial f ∈ K[X] such thatf(a) = 0). If a is not algebraic, a is called transcendent.

Example 9.10. Let K = Q.(1) Let d ∈ Q, n ∈ N and let n

√d ∈ C. Then n

√d is algebraic because it is a root of the

polynomial Xn − d ∈ Q[X].(2) π, e ∈ C are transcendent over Q (usually proved with complex analysis), but they

are algebraic over R (as root of X − π or X − e ∈ R[X]).

Remark and Definition 9.11. Let A be a K-algebra and let a ∈ A. Consider theK-algebra homomorphism K[X]→ A, f 7→ f(a). Then

Im(ϕ) = K[a] = f(a) ; f ∈ K[X] .

The following assertions are equivalent.(i) a ∈ A is algebraic over K.(ii) ϕ is not injective.(iii) Ker(ϕ) = (µa,K) for a unique monic polynomial µa,K ∈ K[X].(iv) [K[a] : K] <∞.In this case, the polynomial µa,K ∈ K[X] is called minimal polynomial of a over K andone has degµa,K = [K[a] : K] (4.22).

Proof. The implications “(iv) ⇒ (ii) ⇒ (iii) ⇔ (i)” are clear. And (iii) shows thatK[a] ∼= K[X]/(µa,K) and hence [K[a] : K] = deg(µa,K) <∞.

Remark 9.12. Let A be a K-algebra, let a ∈ A, and let f ∈ K[X] with f(a) = 0. Thenµa,K divides f . One has f = µa,K if and only if f is monic and deg(f) = [K[a] : K].

Proposition 9.13. Let L be a field extension of K and let a ∈ L. Then:

a is algebraic over K ⇔ K[a] = K(a).

In this case µa,K is the unique irreducible monic polynomial f ∈ K[X] with f(a) = 0.

Proof. As K[a] ⊆ L, K[a] is an integral domain. Hence:

a algebraic⇔ K[a] finite K-algebra9.7⇒ K[a] field⇔ K[a] = K(a).

Moreover:a transcendent

9.11 (ii)⇔ K[a] ∼= K[X]⇒ K[a] ( K(a).

If a is algebraic, then (µa,K) is a maximal ideal of K[X] because K[a] ∼= K[X]/(µa,K)is a field. Hence µa,K is irreducible.Finally:

f(a) = 0⇒ µa,K | ff irred.⇒ (f) = (µa,K)

f monic⇒ f = µa,K .

Example 9.14. Let K = Q, L = C, let p ∈ Z be prime number > 0.

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(1) Let n ∈ N and n√p ∈ C some element with ( n

√p)n = p. Then n

√p is algebraic over

Q because it is a root of Xn− p. Moreover Xn− p is irreducible (Eisenstein (7.14))and monic. Hence µ n

√p,Q = Xn − p, Q( n

√p) = Q[ n

√p] and

[Q( n√p) : Q] = deg(µ n

√p,Q) = n.

(2) Let 1 6= ζ ∈ C an element with ζp = 1 (i.e., ζ = e2πik/p for some k = 1, . . . , p− 1).Then ζ is algebraic over Q because it is a root of Xp − 1 ∈ Q[X]. We have anirreducible decomposition (Exercise 41)

Xp − 1 = (X − 1)(Xp−1 +Xp−2 + · · ·+ 1).

As ζ 6= 1, we find µζ,Q = Xp−1 +Xp−2 + · · ·+X + 1 and hence [Q(ζ) : Q] = p− 1.

C Algebraic Extensions

Proposition 9.15. Let R be a ring, f1, . . . , fk ∈ R[X1, . . . , Xn], and let R → A bean R-algebra. For g ∈ R[X1, . . . , Xn] let g ∈ A[X1, . . . , Xn] be the image of g inA[X1, . . . , Xn]. Then we have an isomorphism of A-algebras

α : A⊗R R[X1, . . . , Xn]/(f1, . . . , fk)∼→ A[X1, . . . , Xn]/(f1, . . . , fk)

given by a⊗ g 7→ ag.

Proof. The map α0 : A⊗RR[X1, . . . , Xn]→ A[X1, . . . , Xn], a⊗g 7→ ag, is a well definedisomorphism of A-algebras with inverse given by Xi 7→ 1 ⊗ Xi. It sends 1 ⊗ fi to fi.Hence it induces the isomorphism α.

Example 9.16. We have C = R[i] = R[X]/(X2 + 1) because µi,R = X2 + 1. Hence

C⊗R C 9.15= C[X]/(X2 + 1) = C[X]/((X + i)(X − i))CRM

= C[X]/(X + i)× C[X]/(X − i) = C× C.

Corollary 9.17. Let R be a ring. Then one has an isomorphism of polynomial algebrasover R

R[X1, . . . , Xm]⊗R R[Y1, . . . , Yn]∼= (R[X1, . . . , Xm])[Y1, . . . , Yn] = R[X1, . . . , Xm, Y1, . . . , Yn]

given by f ⊗ g 7→ fg.

Remark 9.18. Let R be a ring and let A be an R-algebra. Let a1, a2 ∈ A. Then thehomomorphism of R-algebras

m : R[a1]⊗R R[a2] −→ R[a1, a2], x1 ⊗ x2 7→ x1x2

is surjective.

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Proof. The image of m is an R-subalgebra contained in R[a1, a2] and containing a1 anda2. Hence it is equal to R[a1, a2] because R[a1, a2] is the smallest R-subalgebra of Acontaining a1 and a2.

Definition 9.19. A K-algebra K → A is called algebraic if every element of A isalgebraic over K.

Proposition 9.20. Let K → A be a K-algebra. Then the following assertions areequivalent.(i) A is a finite K-algebra.(ii) A is an algebraic and finitely generated K-algebra.(iii) There exists a1, . . . , an ∈ A algebraic elements such that A = K[a1, . . . , an].

Proof. “(i) ⇒ (ii)” K → A finite ⇒ K → A finitely generated and [K[a] : K] <∞ forall a ∈ A 9.11⇒ a is algebraic over K.“(ii) ⇒ (iii)” Clear.“(iii) ⇒ (i)” K[ai] is finite over K for every i. Hence K[a1] ⊗K · · · ⊗K K[an] is finiteover K and therefore K[a1, . . . , an] is finite over K by Corollary 9.18.

Corollary 9.21. Let A be a K-algebra, S ⊆ A a set of algebraic elements such thatA = K[S]. Then A is an algebraic K-algebra.

Proof. Let a ∈ A. Choose T ⊆ S finite such that a ∈ K[T ]. Then Proposition 9.20shows that K[T ] is finite and a is algebraic over K.

Proposition 9.22. Let K → B be a K-algebra without zero-divisors. Then everyalgebraic K-subalgebra A of B is an algebraic field extension.

Proof. Let 0 6= a ∈ A. Then K[a] is a finite K-algebra by Proposition 9.20. It is anintegral domain because it is a subring of the integral domain. Hence it is a field byProposition 9.7 (2) and a is a unit in A.

Proposition 9.23. Let K → L be field extension and let L → A 6= 0 be an L-algebra.Then:

K → A is algebraic⇔ K → L and L → A are algebraic.

Proof. “⇒”. Obvious“⇐”. Let a ∈ A and f = Xn + bn−1X

n−1 + · · · + b0 ∈ L[X] with f(a) = 0. Hencea is algebraic over K(bn−1, . . . , b0) = K[bn−1, . . . , b0] (the equality holds because bi isalgebraic over K).

9.20 ⇒ K → K[b0, . . . , bn−1] → K[a, bn−1, . . . , b0] are finite9.4⇒ K → K[a, bn−1, . . . , b0] is finite9.20⇒ a algebraic over K.

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D Algebraic Closure

Proposition and Definition 9.24. A field K is called algebraically closed if it sat-isfies the following equivalent conditions.(i) Every f ∈ K[X] with deg(f) ≥ 1 has a root.(ii) Every f ∈ K[X] with deg(f) ≥ 1 splits in K[X] into a product of polynomials of

degree 1.(iii) Every irreducible f ∈ K[X] has degree 1.(iv) Every algebraic field extension of K is of degree 1 over K.

Proof. “(i) ⇔ (ii) ⇔ (iii)”: Linear Algebra (easy induction on deg(f)).“(iii)⇒ (iv)”: Let K → L be algebraic field extension, a ∈ L. Then (iii)⇒ deg(µa,K) =1 ⇒ µa,K = X − a ∈ K[X] ⇒ a ∈ K.“(iv) ⇒ (iii)”: Let f ∈ K[X] be irreducible. Then K → K[X]/(f) is a field extensionof degree deg(f)

9.20⇒ deg(f) = 1.

Example 9.25. C is algebraically closed (“Fundamental Theorem of Algebra”).

Definition 9.26. An algebraic closure of K is an algebraic field extension K → Ksuch that K is algebraically closed.

Theorem 9.27. Every field has an algebraic closure.

Proof. Let S be an uncountable set with K ⊂ S and such that card(S) > card(K). LetK → L be algebraic. Then we have L =

⋃n∈N Ln, where

Ln := a ∈ L ; a is root of a polynomial f ∈ K[X] with deg(f) ≤ n.

If K is finite, then Ln is finite and hence L is countable. Otherwise

card(K) ≤ card(Ln) ≤ ncard( f ∈ K[X] ; deg(f) ≤ n ) = card(K).

Therefore card(L) = card(K) if K is an infinite field. In particular card(L) < card(S).Let

X := K ⊆ L ⊆ S ; L is an algebraic extension of K.

It is partially ordered by inclusion. If Y ⊆ X is a totally ordered subset, then⋃L∈Y L

is an upper bound of Y in X. Hence there exists a maximal element L0 in X.Assume that L0 is not algebraically closed. By Proposition 9.24 there exists an algebraicextension L0 → M which is not bijective. As M is an algebraic extension of K(Proposition 9.23), we find

card(S \ L0) = card(S) > card(M \ L0).

Hence there exists an injective map i : M → S with i(x) = x for x ∈ L0. Identifying Mwith i(M) ⊆ S we obtain an element ofX that properly contains L0. Contradiction.

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E Extension of Field Homomorphisms

Definition and Remark 9.28. Let K → A1 and K → A2 be K-algebras. A K-algebra homomorphism A1 → A2 is also simply called a K-homomorphism. We denotethe set of K-homomorphisms A1 → A2 by

HomK-Alg(A1, A2).

Every K-homomorphism is K-linear. If L := A1 is a field extension of K and A :=A2 6= 0, then a K-homomorphism ϕ : L→ A is injective. In this case:(1) [L : K] ≤ [A : K](2) [L : K] = [A : K] ⇔ ϕ isomorphism of K-algebras.

Proposition 9.29. Let K → A and K → A′ be K-algebras. Let a ∈ A be algebraicover K.(1) Let ϕ : A→ A′ be a K-algebra homomorphism. Then ϕ(a) is algebraic over K and

µϕ(a),K divides µa,K (in particular µϕ(a),K = µa,K if µa,K is irreducible, for instanceif A is a field).

(2) The map

HomK-Alg(K[a], A′) −→ a′ ∈ A′ ; µa,K(a′) = 0 ,ϕ 7−→ ϕ(a)

is a bijection. In particular

# HomK-Alg(K[a], A′) ≤ degµa,K

if A′ is an integral domain.

Proof. (1). Let µa,K = Xn + an−1Xn−1 + · · ·+ a0 ∈ K[X]. Then

µa,K(ϕ(a)) = ϕ(a)n + an−1ϕ(a)n−1 + · · ·+ a0ϕ K-Alg-Hom

= ϕ(an + an−1an−1 + · · ·+ a0)

= ϕ(µa,K(a)) = ϕ(0) = 0.

This shows all claims.(2). Let a′ ∈ A′ with µa,K(a′) = 0. Then µa′,K |µa,K and hence (µa,K) ⊆ (µa′,K). Then

K[a]∼ // K[X]/(µa,K) // K[X]/(µa′,K)

∼ // K[a′] → A′,

a Xoo // X // a′

is a K-algebra homomorphism ϕ : K[a] → A with ϕ(a) = a′. It is unique because agenerates K[a] as a K-algebra.

Proposition 9.30. Let K → L be an algebraic extension and let K → Ω be analgebraically closed extension of K.

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(1) There exists a K-homomorphism ϕ : L→ Ω.(2) If L and Ω are algebraic closures of K, then ϕ is automatically a K-isomorphism.

Proof. (1). Consider

X := (Z, τ) ; K ⊆ Z ⊆ L subextension, τ : Z → Ω K-homomorphism.

Then X 6= ∅ because (K,K → Ω) ∈ X. Endow X with the following partial order:

(Z1, τ1) ≤ (Z2, τ2) :⇔ Z1 ⊆ Z2, τ2|Z1 = τ1.

Every totally ordered subset Y of X has an upper bound, namely

(⋃

(Z,τZ)∈Y

Z, τ) with τ |Z = τZ for all (Z, τZ) ∈ Y .

Hence X has a maximal element (Z0, τ0) be Zorn’s lemma. We claim that Z0 = L.Assume there existed a ∈ L \ Z0. We consider Ω as a Z0-algebra via τ0. Then µa,Z0 ∈Z0[X] has a root a′ ∈ Ω because Ω is algebraically closed. Hence Proposition 9.29implies that there exists a Z0-algebra homomorphism σ : Z0[a] → Ω with σ(a) = a′.This contradicts the maximimality of (Z0, τ0).(2). Now assume that L and Ω are algebraic closures of K. We have to show thatϕ is surjective. Its image ϕ(L) ∼= L is algebraically closed and ϕ(L) ⊆ Ω is algebraicbecause Ω is algebraic over K. Hence ϕ(L) = Ω by Proposition 9.24 (iv).

Corollary 9.31. Let K1 and K2 be algebraic closures of K. Then there exists a K-isomorphism K1

∼→ K2.

F Splitting Fields

Definition and Remark 9.32. Let F ⊂ K[X] be a set of non-constant polynomials.A splitting extension for F is a field extension E of K with the following proper-ties.(1) Every f ∈ F splits in E[X] into a product of polynomials of degree 1.(2) For each f ∈ F let Rf ⊆ E be the set of roots of f in E. Then E = K(

⋃f∈F Rf ).

As every element of Rf is algebraic, a splitting field is always an algebraic extension ofK by Corollary 9.21.

Example 9.33. Let f ∈ K[X] be a non-constant polynomial. Then a splitting fieldfor f is a field extension K → L such that f = u(X−a1) . . . (X−an) for a1, . . . , an ∈ Land u ∈ K× and such that L = K(a1, . . . , an) = K[a1, . . . , an]. In this case L is a finiteextension of K.

Proposition 9.34. Let F ⊂ K[X] be a set of non-constant polynomials.(1) There exists a splitting field of F .(2) Let L1 and L2 be two splitting fields of F , let Li be an algebraic closure of Li and let

ϕ : L1∼→ L2 be a K-isomorphism (as Li is also an algebraic closure of K, such an

isomorphism always exists by Corollary 9.31). Then ϕ induces a K-isomorphismL1

∼→ L2.In particular L1 and L2 are K-isomorphic.

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Proof. (1). Let K be an algebraic closure of K and let

S := a ∈ K ; a is root of some f ∈ F.

Then K(S) is a splitting field.(2). Let Si := a ∈ Li ; a is root of some f ∈ F. Let a1 ∈ S1 be root of f ∈ F . Asϕ is a K-algebra homomorphism, f(ϕ(a1)) = ϕ(f(a1)) = 0 and hence ϕ(a1) ∈ S2. Thesame argument shows: a2 ∈ S2 ⇒ ϕ−1(a2) ∈ S1. Hence ϕ(S1) = S2 which shows thatϕ induces a K-isomorphism

L1 = K(S1)∼→ L2 = K(S2).

Corollary 9.35. Let F ⊂ K[X] be a set of non-constant polynomials. Let Ω be a fieldextension of K and let L and L′ be splitting fields of F with L,L′ ⊆ Ω. Then L = L′.

Proof. We may replace Ω by an algebraic closure of Ω. Let K := a ∈ Ω ; a algebraic over K.Then K is an algebraic closure of K (Exercise 46) with L,L′ ⊆ K because L and L′

are algebraic extensions of K. Now apply Proposition 9.34 (2) to ϕ := idK .

Example 9.36. Let K = Q, p a prime number. Then f = X3 − p is irreducible inQ[X]. Let α := 3

√p ∈ R. The roots of f in C are α, ζα, ζ2α, where ζ := e2πi/3. Hence

the splitting field of f in C is

L := Q[α, ζα, ζ2α] = Q[α, ζ].

We have Q[α] ( L because Q[α] ⊆ R but L 6⊂ R. The irreducible decomposition of fover Q[α] is f = (X − α)g with g = (X − αζ)(X − αζ2) = X2 + αX + α2. Moreover:

[Q[ 3√p] : Q] = 3, [Q[ 3

√p, ζ] : Q[ 3

√p]] = 2.

Indeed, the first equality holds by Example 9.14. To show the second equality notefirst that [Q[ 3

√p, ζ] : Q[ 3

√p]] > 1 because ζ /∈ Q[ 3

√p]. But we also have

[Q[ 3√p, ζ] : Q[ 3

√p]] = deg(µζ,Q[ 3

√p]) ≤ deg(µζ,Q)

9.14= 2.

13

10 The Fundamental Theorem of Galois Theory for FiniteExtensions

A Normal extensions

Definition and Proposition 10.1. An algebraic field extension K → L is callednormal if it satisfies the following equivalent conditions.(i) L is the splitting field of a set F ⊂ K[X] of non-constant polynomials.(ii) Every irreducible polynomial in K[X] that has a root in L is a product of linear

polynomials in L[X].(iii) For every field extension L ⊆ L′ and every K-homomorphism σ : L→ L′ one has

σ(L) = L.(iv) For every algebraic closure L ⊆ L′ and every K-automorphism σ : L′

∼→ L′ onehas σ(L) = L.

Proof. (i) ⇒ (iii). If L is a splitting field of F , then σ(L) is also a splitting field of F .Hence σ(L) = L by Corollary 9.35.(iii) ⇔ (iv). “⇒” is clear. Conversely, let σ : L → L′ be a K-homomorphism. Re-placing L′ by an algebraic closure of L′ we may assume that L′ is algebraically closed.For every a ∈ L the image σ(a) is algebraic over K. Hence we may replace L′ by a′ ∈ L′ ; a′ is algebraic over K which is an algebraic closure of K (Exercise 46) con-taining L. By Proposition 9.30 we may extend σ to a K-homomorphism σ′ : L′ → L′

which is automatically an automorphism. Now we can apply (iv) to σ′ and obtainσ(L) = σ′(L) = L.(iii) ⇒ (ii). Let L → Ω be an algebraic closure of L, let f ∈ K[X] be irreducible anda ∈ L a root. We have to show that any root b ∈ Ω of f lies already in L. We have

f = µa,K = µb,K9.29 (2)⇒ ∃ϕ′ : K[a]→ Ω K-homomorphism with ϕ′(a) = b

9.30⇒ ∃ϕ : L→ Ω with ϕ|K[a] = ϕ′.

Now (iii) implies ϕ(L) = L and hence b = ϕ(a) ∈ L.(ii) ⇒ (i). Let S ⊆ L be any subset with L = K[S] and set F := µa,K ; a ∈ S ⊆K[X]. As µa,K is irreducible with root a in L, (ii) implies that µa,K is product ofpolynomials of degree 1 in L[X]. Hence L is a splitting field of F .

Proposition 10.2. Let K → L be a normal field extension and let M ⊆ L be asubextension.(1) For every K-homomorphism τ : M →M there exists a K-automorphism σ : L

∼→ Lsuch that τ = σ|M .

(2) K →M is normal ⇔ for every K-automorphism σ : L→ L one has σ(M) = M .

Proof. Let L be an algebraic closure of L. (1). The composition Mτ−→ M →

L extends to a K-automorphism σ of L by Proposition 9.30 and σ induces a K-automorphism σ of L by Proposition 10.1 (iv).

14

(2). Let M be normal and σ a K-automorphism of L. Then σ(M) = σ|M (M) = Mby Proposition 10.1 (iii).Conversely, every K-automorphism of L restricts to an automorphism of L (Proposi-tion 10.1). Hence σ(M) = M for all K-automorphisms σ of L. Hence M is normal byProposition 10.1 (iv).

Example 10.3. (1) Let K → L be a quadratic field extension (i.e., [L : K] = 2).Then K → L is normal.Indeed, let a ∈ L with deg µa,K = 2. Then we have µa,K = (X − a)g in L[X]. Asdeg(g) = 1, g = X − b for some b ∈ L and L = K[a, b] is the splitting field of µa,K .

(2) An algebraic closure is a normal extensions of K.(3) Q ⊆ Q[ 3

√p] (p prime number) is not normal but Q ⊆ Q[ 3

√p, e2πi/3] is normal

(Example 9.36).

B Fundamental Theorem of Galois Theory for Finite Extensions

Definition 10.4. A finite field extension K → L is called Galois extension if L is thesplitting field of a polynomial f ∈ K[X] such that f has only simple roots in L.

In particular, every finite Galois extension is normal.

Theorem 10.5. Let K → L be a finite Galois extension. Then the group

Gal(L/K) := AutK-Alg(L)

is finite and there is a bijective correspondence

(10.5.1)

M ⊆ L subextension of K → L ↔ H ⊆ Gal(L/K) subgroupM 7→ AutM -Alg(L)

LH ←7 H,

where LH := a ∈ L ; h(a) = a for all h ∈ H is the fixed subfield of H in L. Moreover,one has for every subgroup H of Gal(L/K):

[L : LH ] = |H|, [LH : K] = (Gal(L/K) : H).

Proof. Set G := Gal(L/K). (i). We first show that if M ⊆ L is a subextension ofK → L, then [L : M ] = |AutM -Alg(L)|. By hypothesis L = K[a1, . . . , an] with distinctelements a1, . . . , an ∈ L such that f := (X − a1) · · · (X − an) ∈ K[X]. For 1 ≤ i ≤ nset Mi−1 := M [a1, . . . , ai−1] and di := [Mi : Mi−1]. Then [L : M ] =

∏ni=1 di. Hence it

suffices to show that every K-homomorphism ϕ : Mi−1 → L has exactly di extensionsto a K-homomorphism Mi → L or, equivalently by Proposition 9.29, that µai,Mi−1

has

di distinct roots in L. But deg(µai,Mi−1) = di and has distinct roots because µai,Mi−1

divides f .(ii). Now we claim that if H is a subgroup of G, then |H| = [L : LH ]. We haveH ⊆ AutLH -Alg(L) and hence |H| ≤ [L : LH ] by (i). To show “≥” let k ∈ N with

15

k > |H| and b = (b1, . . . , bk) ∈ Lk. We have to show that b is linear dependent overLH , or equivalently that

(*) b⊥ ∩ (LH)k 6= 0.

Here we set for any subset S ⊆ Lk

S⊥ := c = (c1, . . . , ck) ∈ Lk ;k∑i=1

cisi = 0 ∀ s = (s1, . . . , sk) ∈ S .

For c ∈ b⊥ ∩ (LH)k and h ∈ H we also have c ∈ h(b)⊥ ∩ (LH)k and hence

(**) b⊥ ∩ (LH)k = (Hb)⊥ ∩ (LH)k,

where Hb := (h(b1), . . . , h(bk)) ; h ∈ H ⊆ Lk. Since the L-span of Hb in Lk hasL-dimension at most |H| < k, (Hb)⊥ 6= 0. Choose 0 6= x = (x1, . . . , xk) ∈ (Hb)⊥

such that the number of xi with xi = 0 is as large as possible. As x 6= 0 we find acoordinate xj with xj 6= 0. By multiplying x with x−1

j we may assume that xj = 1. As

the subspace (Hb)⊥ of Lk is stable under the action of H, we have h(x) − x ∈ (Hb)⊥

for all h ∈ H. Then the j-th coordinate of h(x)− x is zero and hence h(x) = x for allh ∈ H by the choice of x. Therefore

x ∈ (Hb)⊥ ∩ (LH)k(∗∗)= b⊥ ∩ (LH)k

which shows our claim.(iii). Proof of the bijectivity of (10.5.1). Let M ⊆ L be a subextension of K → L andH := AutM -Alg(L). Then M ⊆ LH and (i) and (ii) show that [L : M ] = [L : LH ] henceM = LH .Conversely let H ⊆ Gal(L/K) be a subgroup and H ′ := AutLH -Alg(L). Then H ⊆ H ′,and (i) and (ii) imply |H| = |H ′|. Hence H = H ′.

Proposition 10.6. Let K → L be a finite Galois extension. Via the bijection (10.5.1)a subgroup H of Gal(L/K) is a normal subgroup if and only if LH is a normal extensionof K.In this case one has an exact sequence of groups

1→ Gal(L/LH)︸ ︷︷ ︸=H

−→ Gal(L/K)︸ ︷︷ ︸=:G

σ 7→σ|LH−−−−→ Gal(LH/K)︸ ︷︷ ︸=G/H

→ 1

We will see in Remark 12.4 that K → LH ⊆ L are indeed Galois extensions of K(the first one only if LH is normal over K) justifying the notation Gal(L/LH) andGal(LH/K).

Proof. For σ ∈ Gal(L/K) we have

(*)

σ(LH) = σ(a) ; a ∈ L, γ(a) = a ∀γ ∈ H = a′ ∈ L ; γ(σ−1(a′)) = σ−1(a′) ∀γ ∈ H

= LσHσ−1.

16

As the map H 7→ LH is injective, we see that

LH normal extension of K10.2⇔ σ(LH) = LH for all σ ∈ Gal(L/K)(∗)⇔ H = σHσ−1 for all σ ∈ Gal(L/K).

Moreover, the group homomorphism σ 7→ σ|LH is well defined and surjective by Propo-

sition 10.2. Its kernel is clearly Gal(L/LH).

17

11 Etale Algebras

We continue to denote by K a field. All rings and all algebras are commutative (with1).

A Preparations

Lemma 11.1. Let R be a ring, let u : M → N be a homomorphism of R-modules andlet E 6= 0 be a free R-module. Then u is injective (resp. a surjective) if and only ifidE ⊗u : E ⊗RM → E ⊗R N is injective (resp. surjective).

Proof. Choose an isomorphism E ∼= R(I) of R-modules for some set I 6= ∅. By (6.29)we have a commutative diagram of R-linear maps

E ⊗RMidE ⊗u //

∼=

E ⊗R N∼=

M (I) (mi)i 7→(u(mi))i // N (I),

where the vertical maps are isomorphisms. This shows the claim.

Corollary 11.2. Let K a field, A a K-algebra. Let M and N be K-vector spaces. Letu : M → N be a K-linear map. Then u is injective (resp. surjective) if and only ifidA⊗u is injective (resp. surjective).In particular M = 0 ⇔ A⊗K M = 0.

Remark 11.3. Let K → L be a field extension and let V be an K-vector spce. ThenHomK(V,L), the set of K-linear maps V → L, is a L-subspace of the L-vector spaceLV = Map(V,L) of all maps V → L. One has an L-linear isomorphism

HomK(V,L)∼−→ HomL(L⊗K V,L), u 7→ (`⊗ v 7→ `v)

whose inverse is given by

HomL(L⊗K V,L) 3 u 7→ (v 7→ u(1⊗ v)).

Let V be a finite-dimensional K-vector space. Then

(11.3.1) dimK(V )9.3= dimL(L⊗K V ) = dimL((L⊗K V )∗) = dimL(HomK(V,L)).

Proposition 11.4. Let K → L be a field extension, let A be a K-algebra. Let

HomK-Alg(A,L) := A→ L homomorphism of K-algebras.

Then HomK-Alg(A,L) is a linear independent subset of the L-vector space HomK(A,L).In particular, if A is a finite K-algebra, then by (11.3.1) one has

(11.4.1) # HomK-Alg(A,L) ≤ [A : K].

18

Note that (11.4.1) can also be deduced from Proposition 9.7.

Proof. Let n ≥ 0 and show by induction on n that every tuple (ϕ1, . . . , ϕn) of distinctelements ϕi ∈ HomK-Alg(A,L) is linearly independent. n = 0 is trivial. Let n ≥ 1. Letx1, . . . , xn ∈ L such that

∑ni=1 xiϕi = 0. For a, b ∈ L we have

n−1∑i=1

xi(ϕi(a)− ϕn(a))ϕi(b) =n∑i=1

xiϕi(ab)− ϕn(a)n∑i=1

xiϕi(b) = 0 + ϕn(a) · 0 = 0

and hence∑n−1

i=1 xi(ϕi(a) − ϕn(a))ϕi = 0 for all a ∈ A. By induction hypothesis thisimplies that

xi(ϕi(a)− ϕn(a)) = 0, for all 1 ≤ i ≤ n− 1 and a ∈ A.

Since the ϕi are distinct, this implies xi = 0 for 1 ≤ i ≤ n− 1 and hence xnϕn = 0 andso xn = xnϕn(1) = 0.

B Separable Polynomials

Definition and Remark 11.5. Let R be a ring and let f = anXn + . . . a1X + a0 ∈

R[X]. Then we call

f ′ := nanXn−1 + (n− 1)an−1X

n−2 + · · ·+ 2a2X + a1 ∈ R[X]

the formal derivative of f .For all f, g ∈ R[X], a, b ∈ R one has

(af + bg)′ = af ′ + bg′ R-linearity,

(fg)′ = fg′ + f ′g Leibniz rule.

Indeed, linearity is immediate. By linearity it suffices to show the Leibnis rule only forf = Xn and for g = Xm which is also immediate.

Definition and Remark 11.6. Let R be an integral domain, f ∈ R[X] and a ∈ R.Then

orda(f) := supn ∈ N0 ; (X − a)n divides f ∈ N0 ∪ ∞

is called the order of the zero a of f . We say that a is a simple root of f if orda(f) = 1.One has

orda(f) =∞⇔ f = 0,

orda(f) = 0⇔ f(a) 6= 0,

orda(f) = 1⇔ f(a) = 0, f ′(a) 6= 0

Only the last equality requires an argument: orda(f) = 1 ⇔ f = (X − a)g withg ∈ R[X] such that g(a) 6= 0 ⇔ f(a) = 0 and f ′(a) 6= 0 (because f ′ = g + (X − a)g′

and hence f ′(a) = g(a)).

19

Definition 11.7. A polynomial 0 6= f ∈ K[X] is called separable if every root of f ina splitting field is simple.

Example: Let K = Q, f = (X3 − 2)(X + 1) is separable (roots in a splitting field are3√

2, ζ 3√

2, ζ2 3√

2,−1 with ζ = e2πi/3).f = (X − 1)2 is not separable.

Proposition 11.8. Let Ω be an algebraically closed extension of K. Let 0 6= f ∈ K[X].Then the following assertions are equivalent.(i) f is separable.(ii) Every root of f in Ω is simple.(iii) f and f ′ have no common root in Ω.(iv) f and f ′ are coprime in K[X].

Proof. We may assume that f is not constant.(i) ⇔ (ii). Let L be a splitting field of f . 9.30 ⇒ exists K-homomorphism L → Ω.(ii) ⇔ (iii). Remark 11.6.(iii) ⇔ (iv). As f and f ′ split completely over Ω, Hence:

(iii) ⇔ f and f ′ are coprime in Ω[X]

⇔ Ω⊗K K[X]/(f, f ′)9.15= Ω[X]/(f, f ′) = 0

11.2⇔ K[X]/(f, f ′) = 0

⇔ (iv).

Corollary 11.9. Let f ∈ K[X] be irreducible.(1) f separable ⇔ f ′ 6= 0.(2) Suppose that char(K) = 0. Then f is separable.

Proof. (1). f separable⇔ f and f ′ are coprimef irr.⇔ f ′ 6= 0 because any 0 6= g ∈ K[X]

with deg(g) < deg(f) is coprime to the irreducible polynomial f .(2). If char(K) = 0. Then deg(f ′) = deg(f)− 1 ≥ 0. Hence f ′ 6= 0 by (1).

C Separable Algebras

Definition 11.10. A K-algebra K → A is called separable or geometrically reduced ifK⊗K A is reduced for some (

9.31⇔ for all) algebraic closures K of K. A finite separableK-algebra is called etale K-algebra.

Proposition 11.11. Let K → A be a K-algebra.(1) If A is separable, then A is reduced.(2) If A is separable, then every K-subalgebra of A is separable.(3) If (Ai)1≤i≤k is a finite family of separable K-algebras, then

∏ki=1Ai is a separable

K-algebra.

20

Proof. Let K be an algebraic closure of K.(1). K → K injective

11.1⇒ A = K ⊗K A → K ⊗K A injective. Hence A is reduced ifK ⊗K A is reduced.(2). B ⊆ A subalgebra

11.1⇒ K ⊗K B → K ⊗K A injective.(3). By (6.29) we have K-linear isomorphism K ⊗K

∏ki=1Ai

∼→∏ki=1(K ⊗k Ai),

λ ⊗ (ai)i 7→ (λ ⊗ ai)i, and this is a homomorphism of K-algebras. Hence the claimfollows because products of reduced rings are again reduced (Remark 9.6 (3)).

Example 11.12. Let p be a prime number, K := Quot(Fp[T ]). Then f := Xp − T ∈K[X] is irreducible (Eisenstein with prime element T ). Hence L := K[X]/(f) is a fieldextension of degree p (in particular reduced). Let K be an algebraic closure of K andlet a ∈ K with ap = T (exists as f has a root in K). Then

K ⊗K L9.15= K[X]/(f) = K[X]/(Xp − ap) = K[X]/(X − a)p

is not reduced.Note that f is not separable because f ′ = pXp−1 = 0.

D Separability Degree and Split Algebras

Let A1 and A2 be K-algebras. We denote by HomK-Alg(A1, A2) the set of K-algebrahomomorphisms A1 → A2.

Remark 11.13. Let K → L be a field extension, let A be a K-algebra and let B bean L-algebra. Then K → L→ B is also a K-algebra and one has a bijection

HomK-Alg(A,B)↔ HomL-Alg(L⊗K A,B),

ϕ 7→ (`⊗ a 7→ `ϕ(a)),

(a 7→ ϕ(1⊗ a))←7 ϕ.

Definition 11.14. Let A be a finite K-algebra.(1) An element a ∈ A is called separable if µa,K is a separable polynomial.(2) Let Ω be an algebraically closed extension ofK. Then [A : K]s := # HomK-Alg(A,Ω)

is called the separability degree of A over K.

Proposition 11.15. Let A be a finite K-algebra. The separability degree [A : K]s isindependent of the choice of Ω. Moreover:(1) [A : K]s ≤ [A : K].(2) For every field extension K → L one has [L⊗K A : L]s = [A : K]s.(3) Let K ′ → K be a finite extension. Then [A : K ′]s = [A : K]s[K : K ′]s.

Proof. Let ϕ : A → Ω be a K-homomorphism. For all a ∈ A one has µa,K(ϕ(a)) = 0.Hence

ϕ(A) ⊆ Ωalg := w ∈ Ω ; w is algebraic over K.

But Ωalg is a an algebraic closure of K (Exercise 46) and hence we can always assumethat Ω is an algebraic closure. As two algebraic closures of K are K-isomorphic, itfollows that [A : K]s does not depend on the choice of Ω.

21

(1). This follows from Proposition 11.4.(2). This follows from Remark 11.13 applied to B = Ω an algebraic closure of L.(3). We may assume K ′ ⊆ K. Let Ω be an algebraic closure of K. For ϕ ∈HomK′-Alg(K,Ω) let

Tϕ := ψ ∈ HomK′-Alg(A,Ω) ; ψ|K = ϕ = HomK-Alg(A,Kϕ−→ Ω).

Then #Tϕ = [A : K]s. Hence

[A : K ′]s = # HomK′-Alg(A,Ω) =∑

ϕ∈HomK′-Alg(K,Ω)

#Tϕ = [K : K ′]s[A : K]s.

Example 11.16. Let n ∈ N. Let Kn = K × · · · ×K (n factors) be the product ring.It is a K-algebra via K → Kn, a 7→ (a, . . . , a). It is a finite reduced K-algebra with[Kn : K] = n. For n ≥ 2 this is not an integral domain: (1, 0, . . . , 0)(0, . . . , 0, 1) = 0.Every ideal of Kn is of the form a1 × · · · × an, where ai ⊆ K is an ideal and henceai = 0 or ai = K for all i. In particular, the maximal ideals of Kn are of the formK × · · · ×K × 0×K × · · · ×K.For every K-algebra K → A one has by (6.29) an isomorphism of A-modules

(11.16.1) A⊗K Kn ∼→ An, b⊗ (a1, . . . , an) 7→ (ba1, . . . , ban)

which is clearly even an isomorphism of A-algebras.Choosing for A an algebraic closure of K, we see that Kn is an etale K-algebra.

Remark 11.17. Let K → A be a K-algebra.(1) There is a bijection

HomK-Alg(A,K) = π : A→ K ring homomorphism ; K → Aπ−→ K is idK

↔ m ⊂ A maximal ideal ; K → Acan−→ A/m is an isomorphism

given by π 7→ Ker(π).(2) In particular we have for a finite K-algebra A with d := [A : K]

A ∼= Kd (isomorphism of K-algebras)9.7⇔ #m ⊂ A maximal ideal = d

⇔ # HomK-Alg(A,K) = d.

In this case set X := HomK-Alg(A,K). Then the isomorphism can be given by

(11.17.1) A∼−→ KX ∼= Kd, a 7→ (ϕ(a))ϕ∈X .

Definition 11.18. Let K → L be a field extension. A finite K-algebra A is calledL-split if L⊗K A ∼= Ld for some d ≥ 0.

One then has necessarily d = [A : K] = [L⊗K A : L].

22

Proposition 11.19. Let A be a finite K-algebra and let K → L be a field exten-sion.(1) Let L → L′ be a field extension. If A is L-split, then A is L′-split.(2) Let A be an L-split K-algbra and let A′ ⊆ A be a K-subalgebra. Then A′ is L-split.(3) Let K → K be an algebraic closure. The following implications hold

# HomK-Alg(A,L) = [A : K](a)⇔ A is L-split

(b)⇒ A is K-split(c)⇔ A is separable

Proof. (1).

L⊗K A ∼= Ld(11.16.1)⇒ L′ ⊗K A = L′ ⊗L (L⊗K A) ∼= L′ ⊗L Ld ∼= (L′)d.

(2). By Lemma 11.1, L ⊗K A′ is an L-subalgebra of L ⊗K A ∼= L[A:K]. Hence it isisomorphic to an L-algebra of the form Ld for some d (with d = [A′ : K]).(3). Let us show “

(a)⇔ ”. Remark 11.17 we find

L⊗K A ∼= L[A:K] ⇔ # HomL-Alg(L⊗K A,L) = [L⊗K A : L]

and # HomL-Alg(L⊗K A,L) = # HomK-Alg(A,L) by Remark 11.13 and [L⊗K A : L] =[A : K] by Remark 9.4 (2).Let us show “

(c)⇔ ”. By Proposition 9.7 we find that the finite K-algebra K ⊗K A isreduced if and only if K ⊗K A is a product of (necessarily finite) field extension of K.As K is algebraically closed, this means that A is K-split.Let us show “

(b)⇒ ”. Let L be an algebraic closure of L and let K ′ := a ∈ L ; a algebraic over K.Then K ′ is an algebraic closure of K and hence K-isomorphic to K. Hence:

A is L-split(1)⇒ A is L-split⇒ K ⊗K A is reduced.

E Characterization of Etale Algebras

Theorem 11.20. Let A be a finite K-algebra. Let K be an algebraic closure of K.Then the following assertions are equivalent.(i) A is a separable K-algebra (i.e. A is an etale K-algebra).(ii) There exists an isomorphism of K-algebras K ⊗K A 7→ K × · · · × K.(iii) Every a ∈ A is separable over K.(iv) There exists a1, . . . , an ∈ A separable elements such that A = K[a1, . . . , an].(v) [A : K]s = [A : K].(vi) There exists a separable polynomial f ∈ K[X] such that A ∼= K[X]/(f).

Partial proof of Theorem 11.20. We will prove

(vi)⇒ (ii)⇔ (v)⇔ (i)⇒ (iii)⇒ (iv)⇒ (ii)

The implication “(i) ⇒ (vi)” is called the theorem of the primitive element and provedin Theorem 11.26 below.(ii) ⇔ (v) ⇔ (i). Proposition 11.19 (3).

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(vi) ⇒ (ii). Let A = K[X]/(f) with f separable. We may assume that f is monic.Then f =

∏di=1(X − ai) in K[X] with a1, . . . , ad distinct. Hence

K ⊗K A9.15= K[X]/

∏i

(X − ai)9.8=

d∏i=1

K.

(i) ⇒ (iii). Let a ∈ A. Then K[a] ∼= K[X]/(µa,K) is a K-subalgebra of A and henceseparable (Proposition 11.11 (2)). Hence K ⊗K K[a] = K[X]/(µa,K) is reduced. As Kis algebraically closed, µa,K =

∏i(X − ai) ∈ K[X]. By Example 9.8, all roots ai have

to be distinct.(iii) ⇒ (iv). Clear.(iv) ⇒ (ii). For i = 1, . . . , n, K[ai] ∼= K[X]/(µai,K) with µai,K separable. Letdi := deg(µai,K). Then K ⊗K K[ai] ∼= Kdi follows from “(vi) ⇒ (ii)” which we alreadyproved. Hence

K ⊗K (K[a1]⊗K K[a2]⊗K · · · ⊗K K[an])∼= (K ⊗K K[a1])⊗K (K ⊗K K[a2])⊗K · · · ⊗K (K ⊗K K[an])

∼= Kd1 ⊗K · · · ⊗K Kdn

∼= Kd1···dn .

Now K[a1, . . . , an] is a quotient of the K-algebra K[a1] ⊗K K[a2] ⊗K · · · ⊗K K[an](Corollary 9.18), hence K⊗KK[a1, . . . , an] is isomorphic to a quotient of the K-algebraKd1···dn and hence isomorphic to Km for some m ≤

∏i di.

During the proof of the theorem we have seen the argument for the following corollary.

Corollary 11.21. Let A be a finite separable K-algebra and let a ⊆ A be an ideal.Then A/a is a finite separable K-algebra.

Corollary 11.22. Let A be a K-algebra. Then

A finite separable⇔ A is product of finite separable field extensions.

Proof. “⇐” by Proposition 11.11 (3).“⇒” IfA is separable, it is reduced and henceA ∼=

∏ri=1Ki for finite extensionsK → Ki

(Proposition 9.7). Each Ki is a quotient of A and hence separable by Corollary 11.21.

Corollary 11.23. Let A be an algebraic K-algebra. The following assertions are equiv-alent.(i) A is a separable K-algebra.(ii) Every element of A is separable over K.(iii) There exists a subset S ⊆ A of separable elements such that A = K[S].

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Proof. (i) ⇒ (ii). Let a ∈ A. Then K[a] is a finite K-subalgebra of A. As A is sepa-rable, K[a] is separable (Prop. 11.11 (1)). Hence a is separable by Theorem 11.20 (iii).(ii) ⇒ (iii). Obvious(iii) ⇒ (i). Let K be an algebraic closure of K. For every finite subset T ⊆ S theK-algebra K[T ] is finite (Proposition 9.20) and separable (Theorem 11.20) and henceK ⊗K K[T ] is reduced. One has

A = K[S] =⋃

T ⊆ S finite

K[T ]

and hence by Lemma 11.1

K ⊗K K[S] =⋃

T ⊆ S finite

(K ⊗K K[T ]).

Hence K ⊗K K[S] is reduced.

Corollary 11.24. Let K be a perfect field (e.g. if char(K) = 0).(1) A finite K-algebra A is separable if and only if A is reduced.(2) Every algebraic field extension of K is separable.

It is not difficult to see that conversely if every finite extension of a field K is separable,then K is perfect1.The first assertions holds more generally for arbitrary (not necessarily finite)K-algebras.This is usually proved using algebraic derivations (e.g. [BouA2] Chap. V, §15, Theorem3).

Proof. As every reduced finite algebra is a product of finite field extensions (Proposi-tion 9.7 (4)), it suffices to show the second assertion.Let K → L be an algebraic extension, a ∈ L. Assume that µa,K =:

∑ni=0 aiX

i is notseparable over K (Corollary 11.23) or – equivalently – that µ′a,K =

∑ni=1 iaiX

i−1 = 0(Corollary 11.9). This cannot happen if char(K) = 0. Hence suppose char(K) = p >0. Then ai = 0 if p - i. As K is perfect, we find bi ∈ K with bpi = ai for all i.Set g :=

∑p|i biX

i/p. Then gp =∑

p|i aiXi = µa,K . Contradiction because µa,K is

irreducible.

Proposition 11.25. Let K → L be an algebraic field extension and let L → A 6= 0 bean algebraic L-algebra. Then A is a separable K-algebra if and only if L is a separableextension of K and A is a separable L-algebra.

Proof. Let A be a separable K-algebra. Then L is isomorphic to a K-subalgebra of A,hence L is separable over K (Proposition 11.11). Let a ∈ A. Then µa,L divides the

1To see this, we may assume that char(K) = p > 0. Let K be an algebraic closure of K. Let a ∈ Kand let b ∈ K with bp = a. Then µb,K divides Xp − a = (X − b)p. By hypothesis µK,b is separable(because K[b] is separable over K), hence µb,K = X − b which shows b ∈ K.

A more careful argument shows that µb,K = Xp − a if b /∈ K. Hence a field of characteristic p isalready perfect if every extension of degree p is separable.

25

separable polynomial µa,K and hence is separable itself. Therefore every element a ofA is separable over L which shows that A is separable over L by Corollary 11.23.Conversely assume thatK → L and L → A are separable. We may assumeK ⊆ L ⊆ A.We have to show that every a ∈ A is separable over K. Let L′ = K[a0, . . . , an−1], wherethe ai ∈ L are the coefficients of the separable polynomial µa,L. This is a finite extensionof K contained in L, hence K → L′ is separable. As µa,L = µa,L′ , the finite L′-algebraK[a] is separable over L′ because it is generated by a separable element. Hence we mayassume that K → L and L → A are etale. Then

[A : K]9.4= [A : L][L : K]

11.20= [A : L]s[L : K]s

11.15= [A : K]s

which shows that A is an etale K-algebra

F Theorem of the Primitive Element

The implication “(i) ⇒ (vi)” in Theorem 11.20 follows from the following result.

Theorem 11.26. Let A be a finite separable K-algebra. Then there exists a ∈ A suchthat A = K[a].

Such an element a ∈ A is then sometimes called a primitive element of the K-algebraA. One then has A ∼= K[X]/(µa,K) and µa,K is separable because a is separable. Inparticular “(i) ⇒ (vi)” in Theorem 11.20 follows.

Proof. (i). We have A ∼=∏ri=1Ki for finite separable field extensions K → Ki by

Corollary 11.22. If Ki = K[ai], then A = K[(a1, . . . , ar)]. Hence we may assume thatA = L is a finite separable extension of K.(ii). Suppose that K is a finite field. Then L is a finite field and Exercise 16 showsthat there exists a ∈ L× with L× = 1 = a0, a, a2, . . . . Hence L = K[a]. Therefore wemay from now on assume that K is an infinite field.(iii). Let L = K[a1, . . . , an]. We prove the theorem by induction on n. For n = 1 we aredone. Let n > 1. The subextension K[a1, . . . , an−1] ⊆ L is finite separable and hencethere exists b ∈ K[a1, . . . , an−1] with K[b] = K[a1, . . . , an−1] by induction hypothesis.Hence we have L = K[b, an] and it suffices to consider the case that L = K[b, c] forb, c ∈ L.(iv). Let L = K[b, c]. Choose an algebraic closure K of K and write

HomK-Alg(L, K) = ϕ1, . . . , ϕm, m = [L : K]s = [L : K].

Considerg :=

∏1≤i<j≤m

[(ϕi(b)− ϕj(b)

)X +

(ϕi(c)− ϕj(c)

)]∈ K[X].

The ϕi are distinct hence ϕi(b) 6= ϕj(b) or ϕi(c) 6= ϕj(c) ∀i < j ⇒ g 6= 0.K infinite ⇒ ∃λ ∈ K with g(λ) 6= 0 and hence(

ϕi(b)− ϕj(b))λ+

(ϕi(c)− ϕj(c)

)6= 0 ∀i < j

⇒ ϕi(λb+ c) = λϕi(b) + ϕi(c) 6= λϕj(b) + ϕj(c) = ϕj(λb+ c).

26

Set a := λb + c. Then ϕi(a) 6= ϕj(a) ∀i 6= j. As ϕ1(a), . . . , ϕm(a) are the roots ofµa,K in K (Proposition 9.29), we find[

K[a] : K]

= deg(µa,K) = m = [L : K] =[L : K[a]

]·[K[a] : K

]which shows

[L : K[a]

]= 1. Therefore L = K[a].

G Separable Closure

Proposition and Definition 11.27. Let K → A be a K-algebra, and set

As := a ∈ A ; a is separable and algebraic over K.

Then As is the largest K-subalgebra of A that is algebraic and separable over K. If Ais a field, then As is a field.The K-algebra is called relative separable closure of K in A.

Proof. Let B ⊆ A be a K-subalgebra that is algebraic and separable over K. ThenB ⊆ As by Corollary 11.23 (ii). The K-algebra K[As] is algebraic (Corollary 9.21) andseparable (Corollary 11.23). Hence K[As] ⊆ As and so As = K[As].The last assertion follows from Proposition 9.22.

Definition and Remark 11.28. (1) A field K is called separably closed if the follow-ing equivalent conditions hold.(i) Every etale K-algebra A is K-split (i.e., isomorphic to K × · · · ×K).(ii) Every algebraic separable field extension of K has degree 1.(iii) Every separable polynomial f ∈ K[X] is a product of polynomials of degree 1.“(ii)⇒ (i)” follows from Corollary 11.22. “(i)⇒ (iii)⇒ (ii)” follows as in the proofof Proposition 9.24.

(2) A separable closure of a field K is an algebraic and separable field extension E ofK such that E is separably closed.

Proposition 11.29. Let K be a field.(1) Let Ω be an algebraically closed extension of K. Then

Ωs := a ∈ Ω ; a is separable and algebraic over K

is a separable closure of K. In particular separable closures of K exist.(2) Let K1 and K2 be separable closures of K. Then there exists a K-isomorphism

K1∼→ K2.

(3) Let Ksep be an algebraic closure of K and let K → L be a separable algebraic fieldextension. Then there exists a K-homomorphism ϕ : L → Ksep. If L is separablyclosed, then ϕ is automatically a K-isomorphism.

Proof. (1). 11.27 ⇒ Ωs is an algebraic and separable subextension of Ω. Let F bea finite separable extension of Ωs. As Ω is algebraically closed, there exists an Ωs-homomorphism ι : F → Ω (Proposition 9.30). Then the elements of ι(F ) ⊆ Ω areseparable and algebraic over K and hence ι(F ) ⊆ Ωs and hence [F : Ωs] = 1.

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(2). A separable closure is a splitting field of the set of all non-constant separablepolynomials in K[X]. Hence it is unique up to K-isomorphism by Proposition 9.34 (2).(3). Let K be an algebraic closure of Ksep. By Proposition 9.30 there exists a K-homomorphism ϕ : L → K. Every a ∈ L is separable, hence ϕ(a) ∈ K is separablebecause they have the same minimal polynomial (Proposition 9.29). Hence ϕ(L) ⊆Ks = Ksep.If L is separably closed, then L ∼= ϕ(L) is separably closed and ϕ(L) ⊆ Ksep is aseparable and algebraic extension. Hence ϕ(L) = Ksep by Definition 11.28 (ii).

Corollary 11.30. If K is perfect (e.g., if char(K) = 0), then every separable closureis an algebraic closure.

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12 Galois Theory

A Galois Extensions

Definition 12.1. An algebraic extension K → L is called Galois extension if it isnormal and separable. In this case we call

Gal(L/K) := AutK-Alg(L) = (σ : L→ L ; σ automorphism of K-algebras, )

the Galois group of L over K.

Proposition 12.2. For a finite field extension K → L the following assertions areequivalent.(i) K → L is normal and separable (i.e., a Galois extension in the sense of Defini-

tion 12.1).(ii) L is a splitting field of a separable polynomial f ∈ K[X] (i.e., K → L is a Galois

extension in the sense of Definition 10.4).(iii) # AutK-Alg(L) = [L : K].

Proof. (i) ⇔ (ii). If L is a splitting field of a polynomial f , then L is normal andL = K[a1, . . . , an], where ai are the roots of f in L. If f is separable, then all ai areseparable over K. Hence L is a separable extension of K by Theorem 11.20.Conversely, if L is separable over K, then L = K[a] with a ∈ L separable over K(Theorem 11.26). If L is also normal then L is the splitting field of the separablepolynomial µa,K .(i) ⇔ (iii). Let Ω be an algebraically closed extension of L. Then one always has

# AutK-Alg(L) ≤ # HomK-Alg(L,Ω) = [L : K]s ≤ [L : K].

The first inequality is an equality if and only if K → L is normal. The second inequalityis an equality if and only if K → L is separable.

Example 12.3. (1) A separable closure of a field K is a Galois extension of K. Indeedit is separable by definition and it is the splitting field of all separable polynomialsin K[X].

(2) Let K be a perfect field (e.g. if char(K) = 0). Then K → L is a Galois extensionif and only if it is a normal extension (Corollary 11.24).

Remark 12.4. Let K → M → L be a algebraic field extensions such that K → Lis a Galois extension. Then M → L is a Galois extension (it is separable by Proposi-tion 11.25 and it is normal because if L is the splitting field of a set F of polynomialsin K[X], we can view F as subset of M [X]).The extension K →M is a Galois extension if and only if it is normal.Note that if K → L is Galois extension, K → L is not necessarily normal (Exam-ple 9.36).Moreover, if K → M and M → L are Galois extensions, then K → L is in generalnot normal. Example: For a prime number p > 0 let

√p, 4√p ∈ R>0. The field

extensions Q ⊂ Q[√p] and Q[

√p] ⊂ Q[ 4

√p] have degree 2 because µ√p,Q = X2 − p and

29

µ 4√p,Q[√p] = X2 − √p. Hence they are normal (Example 10.3). They are separable

because char(Q) = 0. Therefore both extensions are Galois extensions. But µ 4√p,Q =

X4 − p has roots ± 4√p, ±i 4

√p and ±i 4

√p /∈ Q[ 4

√p] ⊆ R. Hence Q ⊂ Q[ 4

√p] is not

normal.

Remark and Definition 12.5. Let K → L be an algebraic extension, let S ⊆ L withL = K[S]. Let N be the splitting field of µa,K ; a ∈ S ⊆ K[X] in some algebraicclosure of L. Then N is a normal extension of K with L ⊆ N and there exists nonormal subextension N ′ ( N of K with L ⊆ N ′. Such a field extension is the called anormal hull of L over K.If K ⊆ L is finite (resp. separable), then we may choose S to be finite (resp. to consistof separable elements). Then K → N is finite (resp. a Galois extension).

Proposition 12.6. Let A be an etale K-algebra. Then there exists a finite Galoisextension K → L such that A is L-split (i.e., there exists an isomorphism of L-algebrasL⊗K A

∼→ Ld, where d = [A : K]).

Proof. By Theorem 11.26 there exists a ∈ A such that A = K[a]. As a is separableover K, µa,K is separable of degree d = [A : K]. Then a splitting field L of µa,K is afinite Galois extension (Proposition 12.2). Hence there exists d distinct roots of µa,Kin L and we have

# HomK-Alg(A,L)9.29= #b ∈ L root of µa,K = d.

This implies that A is L-split by Proposition 11.19 (3).

B Main Theorem of Galois (Grothendieck version)

Remark 12.7. Let K → L be a Galois extension and let K →M ⊆ L be a subexten-sion. Then Gal(L/M) is a subgroup of Gal(L/K). Suppose that K →M is normal (⇒K → M Galois extension). Then K → M ⊆ L are Galois extensions (Remark 12.4)and by Proposition 10.2 there is a well defined exact sequence of groups

(12.7.1) 1→ Gal(L/M) −→ Gal(L/K)σ 7→σ|M−−−−−→ Gal(M/K)→ 1

Definition 12.8. Let K → L be a Galois extension and let G := Gal(L/K).(1) We call a subgroup H ⊆ Gal(L/K) open, if there exists a subextension K →M →

L with [M : K] <∞ and Gal(L/M) ⊆ H.(2) Let X be a finite set. Then a G-action G × X → X is called continuous if there

exists an open subgroup H of G that acts trivially on X (i.e. σ ·x = x for all x ∈ Xand σ ∈ H).

Therefore a G-action on a finite set X is continuous if and only if there exists a finitesubextension K →M of L such that Gal(L/M) acts trivially on X.If K → L is finite, then any G-action on a finite set is continuous (take M = L).One can show that there exists a unique topology on Gal(L/K) making Gal(L/K) intoa topological group such that the open subgroups (as defined above) form a funda-mental system of open neighborhoods of the neutral element idL ∈ Gal(L/K). Then a

30

Gal(L/K)-action on a finite set X is continuous in the above sense if and only if themap G×X → X is a continuous map of topological spaces, where X is endowed withthe discrete topology.

From now on let Ksep be a separable closure of K, let K be an algebraic closure of Kand set GK := Gal(Ksep/K).

Remark 12.9. We will construct a contravariant functor2 from the category of etale(= finite separable) K-algebras to the category of finite sets endowed with a continuousGK-action.The functor on objects: Let A be an etale K-algebra and set

X := XA := HomK-Alg(A,Ksep) = HomK-Alg(A, K),

where the equality holds because for every ϕ ∈ HomK-Alg(A, K) one has ϕ(A) ⊆ Ks =Ksep. Then #X = [A : K]s = [A : K] =: d is finite. Define a GK-action on X by

GK ×X → X, (σ, ϕ) 7→ σ ϕ.

This action is continuous: First note that one has

(12.9.1)

A ∼= Kd (as K-algebras)11.17⇔ for all ϕ ∈ X one has ϕ(A) = K

⇔ σ ϕ = ϕ for all σ ∈ GKNow, if A is an etale K-algebra of degree d, then by Proposition 12.6 there exists afinite Galois extension K → L such that A is L-split. We can embed L into Ksep byProposition 11.29 (3) and hence may assume that L ⊆ Ksep. Applying (12.9.1) to theL-algebra L⊗K A ∼= Ld shows that GL acts trivially on XA.The functor on morphisms: Let u : A1 → A2 be a homomorphism of etale K-algebras,then

u∗ : XA2 −→ XA1 , (ϕ : A2 → Ksep) 7−→ ϕ uis a GK-equivariant map: For all σ ∈ GK and ϕ ∈ XA2 we have

u∗(σ · ϕ) = (σ ϕ) u = σ (ϕ u) = σ · u∗(ϕ).

Definition 12.10. Let C and D be categories. A contravariant functor F : C → D iscalled an equivalence of categories if for every object Y in D there exists an object X inC such that F (X) ∼= Y and if for all objects X1 and X2 in C the map HomC(X1, X2)→HomD(F (X2), F (X1)) is bijective.

Theorem 12.11. The above functor

(12.11.1) (category of etale K-algebras A) −→(

category of finite setswith continuous GK-action

)is an equivalence of categories. In other words:

2A contravariant functor F from a category C to a category D attaches to every object X in C anobject F (X) in D and to every morphism u ∈ HomC(X,Y ) a morphism F (u) ∈ HomD(F (Y ), F (X))such that F (idX) = idF (X) and F (u v) = F (v) F (u) for all composable morphisms u and v in C.

31

(1) If X is a finite set with continuous GK-action, then there exists an etale K-algebraA such that XA

∼= X (isomorphism of GK-sets).(2) For all etale K-algebra A1 and A2 the map

(12.11.2) HomK-Alg(A1, A2) −→ HomGK -sets(XA2 , XA1), u 7→ u∗

is bijective.

C Proof of the Main Theorem of Galois Theory

We prove the Main Theorem by reduction to more and more special situations. Thisalso has the advantage that we obtain more precise results in these situations.

Reduction to Sets with Action by a Finite Galois Group

Remark 12.12. Let G be a group acting on a set X. Let H be a normal subgroup ofG such that H acts trivially on X. Then the G-action on X induces a (well-defined)G/H-action

G/H ×X → X, (gH, x) 7→ gx.

If X is a finite set with continuous GK-action, then by definition there exists a finitesubextension K → L0 of Ksep (depending on X) such that GL0 ⊆ GK acts trivially onX. Let L be a normal hull of L0 in Ksep (i.e., if L0 = K[a], then L is the splitting fieldof µa,K in Ksep). Then K → L is a finite Galois extension such that GL acts triviallyon X. By Remark 12.7 GL is a normal subgroup of GK and GK/GL = Gal(L/K).Therefore the GK-action on X induces an action of the finite group Gal(L/K) on X.Moreover if X1 and X2 are two finite sets with continuous GK-action (resp. if A1 andA2 are two etale K-algebras) then there exists a finite Galois subextension K → L ofKsep such that GL acts trivially on X1 and X2 (resp. such that A1 and A2 are bothL-split): If Li = K[ai] (i = 1, 2) has this property for Xi (resp. for Ai), then we canchoose any finite Galois extensions K → L with L1, L2 ⊆ L ⊆ Ksep, for instance thesplitting field of µa1,K , µa2,K.Now fix a finite Galois subextension K → L of Ksep.The construction in Remark 12.9 yields a contravariant functor

(12.12.1)

(category of etale K-algebras A

that are L-split

)−→

(category of finite setswith Gal(L/K)-action

)A 7−→ HomK-Alg(A,L)

and the arguments above show that the categories in (12.11.1) are the union of thecategories in (12.12.1) if L runs through the set of all finite Galois subextension K → Lof Ksep. Hence it suffices to show the following result.

Theorem 12.13. For every finite Galois subextension K → L of Ksep, the functor(12.12.1) is a contravariant equivalence of categories.

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Reduction to Sets with a Transitive Action by a Finite Galois Group

Lemma 12.14. (1) The functor (12.12.1) induces a functor

(12.14.1)

category of finite separablefield extensions K → K ′

that are L-split

−→ (category of finite sets withtransitive Gal(L/K)-action

)A 7−→ HomK-Alg(A,L)

(2) To show Theorem 12.13 it suffices to show that the functor (12.14.1) is a con-travariant equivalence of categories.

Proof. We set G := Gal(L/K).(1). Let A be an etale K-algebra, A = K1 × · · · ×Kn, where Ki are finite separablefield extensions of K. If ϕ : A → L is a K-homomorphism, then ϕ(A) is a subfield ofL (Proposition 9.22), hence Ker(ϕ) is a maximal ideal and hence there exists i(ϕ) ∈1, . . . , n such that

Ker(ϕ) = mA,i(ϕ) := K1 × · · · ×Ki(ϕ)−1 ×Ki(ϕ)+1 × · · · ×Kn.

Therefore for all ϕ ∈ HomK-Alg(A,L) there is a unique i(ϕ) ∈ 1, . . . , n such that ϕinduces a K-homomorphism Ki(ϕ) → L. Hence we have

X =

n∐i=1

Xi, Xi := ϕ ∈ X ; i(ϕ) = i = HomK-Alg(Ki, L).

For σ ∈ G and ϕ ∈ Xi one has Ker(ϕσ) = Ker(ϕ) because σ is injective. In particular,ϕ σ ∈ Xi. In other words, Xi is a G-stable subset.We claim that G acts transitively on Xi, in other words, the Xi’s are the G-orbits.Indeed, let ϕ,ψ ∈ HomK-Alg(Ki, L). Let L ⊆ L be an algebraic closure of L. ByProposition 9.30 we can choose a K-homomorphism ι : Ki → L. Identifying Ki withι(Ki) we may assume that Ki ⊆ L. Again by Proposition 9.30 we may extend thecompositionsKi

ϕ−→ L → L andKiϕ−→ L → L toK-automorphisms ϕ, ψ : L→ L. Set

σ := ψ ϕ−1 ∈ AutK-Alg(L). By Proposition 10.1 (iii), σ := σ|L is a K-automorphism.

As ψ = σ ϕ, we obtain ψ = σ ϕ. This shows in particular (1).(2). Let X be a finite set with G-action and let X1, . . . , Xn be the G-orbits in X.Suppose that we have constructed a finite separable field extension Ki of K such thatXKi

∼= Xi (isomorphism of G-sets). Then A :=∏ni=1Ki is an etale K-algebra and we

have seen above that XA = X.Let A and A′ be etale K-algebra, A =

∏ni=1Ki and A′ =

∏jK′j for finite separable

field extensions Ki, K′j of K. Let u : A′ → A be a homomorphism of etale K-algebras.

We claim that for each i the composition A′u−→ A→ Ki induces a K-homomorphism

K ′j(i) → Ki for some j(i) and hence

(*) HomK-Alg(A,A′) =∐i

HomK-Alg(K ′j(i),Ki).

33

Let us show the claim: As the kernels of A′ → K ′j are exactly the maximal ideal ofA′ it suffices to show the following claim: Let m ⊂ A be a maximal ideal. We claimthat α−1(m) is also a maximal ideal. Indeed, α induces an injective homomorphismA′/α−1(m) → A/m this shows that A′/α−1(m) is an integral domain of finite K-dimension. Therefore it is a field by Proposition 9.7 and hence α−1(m) is a maximalideal.Similarly, if X := XA and X ′ := XA′ , then every G-equivariant map v : X → X ′ sendstwo elements x, x′ in the same orbit Xi := XKi ⊆ X into the same G-orbit X ′j(i) (if

x′ = gx for some g ∈ G, then v(x′) = v(gx) = gv(x)) and hence

(**) HomG-sets(X,X′) =

∐i

HomG-sets(Xi, X′j(i)).

Therefore if we have seen that the map HomK-Alg(A′, A) → HomG-sets(XA, XA′) isbijective for finite separable field extensions A and A′, then (*) and (**) show that themap is bijective for all etale K-algebras A and A′.

Proof for finite separable field extensions

It remains to prove that (12.14.1) is a contravariant equivalence of categories. SetG := Gal(L/K). We first remark that L itself is L-split. As [L : K] = # Gal(L/K) =# HomK-Alg(L,L) by Proposition 12.2, this follows from Proposition 11.19 (3). Nowlet X be a finite set with transitive Gal(L/K)-action and set

Θ(X) := f ∈ LX = Map(X,L) ; f(σx) = σ(f(x)) for all σ ∈ Gal(L/K).

This is a K-subalgebra of LX . As L is L-split, LX is L-split and hence Θ(X) is L-splitby Proposition 11.19 (2).Now the assertion that (12.14.1) is an equivalence of categories follows from the follow-ing two lemmas (the first lemma shows that Θ(X) is a finite separable field extensionof K with HomK-Alg(Θ(X), L) ∼= X).

Lemma 12.15. (1) Choose x0 ∈ X and let H := Gx0 = σ ∈ G ; σx0 = x0 be thestabilizer of X and set

M := a ∈ L ; τ(a) = a for all τ ∈ H.

Then M is a finite separable field extension of K of degree

(12.15.1) [M : K] = (G : H) = #X

and one has a K-isomorphism

(12.15.2) Θ(X)∼−→M, f 7→ f(x0).

(2) The map

(12.15.3) X −→ HomK-Alg(Θ(X), L), x 7→ (f 7→ f(x))

is an isomorphism of G-sets.

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Proof. (1). As L is a finite separable extension, M is a finite separable extension.Moreover [M : K] = (G : H) holds by Theorem 10.5 and G/H → X, σH 7→ σx0

is bijective. For f ∈ Θ(X) and σ ∈ H one has σ(f(x0)) = f(σx0) = f(x0). Hencef(x0) ∈M .Finally, if a ∈ M , then there is a unique map fa : X → L with fa(x0) = a: Everyx ∈ X is of the form x = σx0 for some σ ∈ G and we have to (and can) definefa(σx0) := σfa(x0) = σ(a) (this is independent of the choice of σ: If σ1x0 = σ2x0, then(σ−1

2 σ1)x0 = x0 and hence there exists τ ∈ H with σ2τ = σ1; therefore σ1(a) = σ2(a)because τ(a) = a). Hence we see that (12.15.2) is bijective.(2). For σ ∈ Gal(L/K) the map (12.15.3) sends σx to f 7→ f(σx) = σf(x) by thedefinition of Θ(X). Therefore (12.15.3) is G-equivariant.As Θ(X) is L-split, we find by Proposition 11.19 (3)

#X(12.15.1)

= [Θ(X) : K] = # HomK-Alg(Θ(X), L)

Hence it suffices to show that (12.15.3) is injective. Let x0, x1 ∈ X such that f(x0) =f(x1) for all f ∈ Θ(X). Choose σ ∈ G with σx0 = x1. Then

f(x0) = f(x1) = f(σx0) = σ(f(x0))

for all f ∈ Θ(X). As f 7→ f(x0) is bijection onto M by (1), we see that σ ∈ Gal(L/M).But Gal(L/M) = H by Theorem 10.5 and hence x1 = σx0 = x0.

Let X and Y be finite sets with G-action and let v : X → Y be a G-equivariant map.Define a K-homomorphism

Θ(v) : Θ(Y )→ Θ(X), g 7→ g v

Let M be an etale field extension of K. Then

Θ(HomK-Alg(M,L)) = f : HomK-Alg(M,L)→ L |σ(f(ϕ)) = f(σ ϕ)

∀ σ ∈ G,ϕ ∈ HomK-Alg(M,L)

and the map

(12.15.4)αM : M → Θ(HomK-Alg(M,L)),

a 7→ (ϕ 7→ ϕ(a))

is a homomorphism of K-algebras. In particular it is injective. As [M : K] =# HomK-Alg(M,L) = [Θ(HomK-Alg(M,L)) : K], it is an isomorphism of K-algebras.

Lemma 12.16. Let M and M ′ be etale field extensions of K that are L-split. Thenthe map

(12.16.1)HomK-Alg(M,M ′) −→ HomG-Sets(HomK-Alg(M ′, L),HomK-Alg(M ′, L)),

u 7−→ u∗.

is bijective.

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Proof. Let v ∈ HomG-Sets(HomK-Alg(M ′, L),HomK-Alg(M ′, L)) and define µ(v) as thecomposition

MαM−→ Θ(HomK-Alg(M,L))

Θ(v)−→ Θ(HomK-Alg(M ′, L))α−1M′−→M ′.

We claim that this defines an inverse map of (12.16.1). For u ∈ HomK-Alg(M,M ′) wehave a commutative diagram

MαM //

u

Θ(HomK-Alg(M,L))

Θ(u∗)

M ′αM′ // Θ(HomK-Alg(M ′, L)).

This shows that µ(u∗) = u.Conversely, let v ∈ HomG-Sets(HomK-Alg(M ′, L),HomK-Alg(M ′, L)). To show thatµ(v)∗ = v let ϕ′ ∈ HomK-Alg(M ′, L). Then we have to show

(*) v(ϕ′) = µ(v)∗(ϕ′) = ϕ′ α−1M ′ Θ(v) αM : M → L.

For a ∈M we have

Θ(v)(αM (a)) : HomK-Alg(M ′, L)→ L, ψ′ 7→ v(ψ′)(a).

Therefore α−1M ′(Θ(v)(αM (a))) is the unique element a′ ∈M ′ such that ψ′(a′) = v(ψ′)(a)

for all ψ′ ∈ HomK-Alg(M ′, L). This shows (*).

Relation to the Fundamental Theorem for finite Galois extensions

Finally, we will show how the equivalence (12.14.1)category of finite separablefield extensions K → K ′

that are L-split

−→ (category of finite sets withtransitive Gal(L/K)-action

)A 7−→ HomK-Alg(A,L)

and the Fundamental Theorem 10.5 for finite Galois extensions are related.We continue to fix a finite Galois extension K → L. One object in the category offinite separable field extensions of K that are L-split is L itself. It corresponds viathe equivalence (12.14.1) to the finite set XL = HomK-Alg(L,L) = Gal(L/K) withtransitive Gal(L/K)-action given by Gal(L/K)×XL → XL, (σ, ϕ) 7→ σ ϕ, i.e. by leftmultiplication.Consider the category FL of pairs (K ′, ι), where K ′ is a finite separable field extensionof K that is L-split and where ι ∈ HomK-Alg(K ′, L). A morphism (K ′1, ι1) → (K ′1, ι2)in this category is a K-homomorphism ϕ : K ′1 → K ′2 such that ι2 ϕ = ι1.Then the equivalence (12.14.1) yields an equivalence of this category with the follow-ing category SL: The category SL consists of pairs (X, v), where X is a finite set

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with transitive Gal(L/K)-action and where v : XL = Gal(L/K) → X is a Gal(L/K)-equivariant map. A morphism (X1, v1) → (X2, v2) in SL is a Gal(L/K)-equivariantmap w : X1 → X2 such that w v1 = v2.If (X, v) is an object in SL, then the map v is uniquely determined by v(idL) ∈ X asfor all σ ∈ Gal(L/K) one has v(σ) = σv(idL) because of the Gal(L/K)-equivariancyof v. Hence we can identify the category SL with the category of pairs (X,x), whereX is a finite set with transitive Gal(L/K)-action and where x ∈ X (then a morphism(X1, x1)→ (X2, x2) is a Gal(L/K)-equivariant map w : X1 → X2 with w(x1) = x2).Now (K ′, ι) 7→ ι(K ′) yields a bijection between isomorphism classes in FL and subexten-sions of L because any subextension of L is automatically separable (Proposition 11.25)and L-split (Proposition 11.19 (2)). On the other hand,

(X,x) 7→ Gal(L/K)x = σ ∈ Gal(L/K) ; σ(x) = x

yields a bijection between isomorphism classes in SL and subgroups H of Gal(L/K).This shows that the equivalence between FL and SL gives a bijection between subgroupsH of Gal(L/K) and subextensions of K → L. Moreover, Lemma 12.15 shows that thisbijection is indeed given by H 7→ a ∈ L ; tau(a) = a for all τ ∈ H.

References

[BouA2] N. Bourbaki, Algebra, Chapters IV–VII, Springer (2003).

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