Galois Theory. Torsten Wedhorn. July 13, These lecture notes contain the final third of my lecture on abstract algebra.

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1 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. The symbol K always denotes a field. Contents 8 Field Extensions 3 A Subfields B Prime Fields Algebraic Extensions 5 A Finite Algebras B Algebraic and Transcendental Elements C Algebraic Extensions D Algebraic Closure E Extension of Field Homomorphisms F Splitting Fields The Fundamental Theorem of Galois Theory for Finite Extensions 14 A Normal extensions B Fundamental Theorem of Galois Theory for Finite Extensions Étale Algebras 18 A Preparations B Separable Polynomials C Separable Algebras D Separability Degree and Split Algebras E Characterization of Étale Algebras F Theorem of the Primitive Element G Separable Closure

2 12 Galois Theory 29 A Galois Extensions B Main Theorem of Galois (Grothendieck version) C Proof of the Main Theorem of Galois Theory

3 8 Field Extensions A Subfields Remark 8.1. Let ϕ: K A be a not necessarily commutative K-algebra 0. Then ϕ is injective: ϕ(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 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 (L i ) i I be a family of subextension of M. Then i I L i is a subextension of M. Let S M be a subset. Then K(S) := L L subextension of M S L is the smallest subextension of M containing S. It is called the subextension of M generated by S or the subextension of M obtained by adjunction of S. One has K(S) = { f(s 1,..., s m ) g(t 1,..., t n ) n, m N 0, f K[X 1,..., X m ], g K[X 1,..., X n ], s i, t j S, g(t 1,..., t n ) 0}. For S = {s 1,..., s n } we write K(s 1,..., s n ) instead of K({s 1,..., s n }). A field extension K M is called finitely generated if there exists a finite subset S M such that M = K(S). One clearly has K[S] K(S), where K[S] is the smallest K-subalgebra of M containing S. 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 = F p for some prime number p. Proposition 8.7. Let R be a ring containing a field K. Then there exists a smallest subfield K 0 of R and this is a prime field. One has char(r) = 0 K 0 = Q, char(r) = p K 0 = F p. 3

4 Proof. Let ϕ: Z R be the unique ring homomorphism. Then Im(ϕ) is contained in any subring of R and in particular in K. Hence Z/ Ker(ϕ) = Im(ϕ) is an integral domain and hence Ker(ϕ) is a prime ideal of Z (5.20). By (5.21) we have Ker(ϕ) = 0 or Ker(ϕ) = pz for some prime number p. (1) char(r) = p Ker(ϕ) = pz Im(ϕ) = F p 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 of R. Remark and Definition 8.8. Hence we say that a ring R that contains a field has characteristic p for a prime number p and write char(r) = p (resp. has characterstic 0 and write char(r) = 0) if R contains F p (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 = a p + b p because for 1 k p 1 the binomial coefficient ( ) p p(p 1) (p k + 1) = k (p k)! is divisible by p. This shows that Frob R : R R, a a p, 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) = 0 or if char(r) = p > 0 and the Frobenius Frob R : a a p is bijective. Example (1) Every finite field F is perfect: Frob F is injective because Ker(Frob F ) = { a F ; a p = 0 } = 0. Hence it is bijective because F is finite. (2) Let K = Quot(F p [T ]). As T is a prime element in F p [T ] there exists no a K with a p = T. Hence K is not perfect. 4

5 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 a free R-algebra and that B is a free A-algebra. Then B is a free R-algebra and rk R (B) = rk R (A) rk A (B). Proof. We have isomorphisms of R-modules (the first one is even an isomorphism of A-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 an R-algebra. Then the scalar extension R R A is an R -algebra via R R R A, a a 1. Morever: If A is free R-algebra, then R R A is a free R -algebra and rk R (R R A) = rk R (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] := dim K (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 K A is an L-algebra and [L K A : L] = [A : K] by Remark 9.3. (3) If K L is a field extension and L A is an L-algebra, then by Proposition 9.2. [A : K] = [A : L][L : K] Definition 9.5. Let A be a ring. (1) An element a A is called nilpotent if there exists n N such that a n = 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 A 1 and A 2 be rings. Then A 1 A 2 is reduced if and only if A 1 and A 2 are reduced (an element (x 1, x 2 ) A 1 A 2 is nilpotent if and only if x 1 and x 2 are nilpotent). 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 m 1,..., m r in A with r [A : K]. (4) A is reduced if and only if A = r i=1 A/m i. 5

6 Proof. (1). Then (Exercise 18) For x A let m x : A A be the K-linear map a xa. A integral domain 0 x A: m x is injective dim K (A) < 0 x A: m x 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 integral domain. Hence it is a field by (1). Therefore p is a maximal ideal (5.20). (3),(4). Any two maximal ideals m 1 m 2 of a ring are coprime. Hence if m 1,..., m t are distinct maximal ideals of A, then A t A/m i i=1 is surjective by the Chinese Remainder Theorem (5.16). Hence [A : K] t i=1 [A/m i : K] which shows that there are only finitely many maximal ideals m 1,..., m r in A. Moreover Ker(A which shows (4). r A/m i ) = i=1 r i=1 m i (2) = p A prime ideal p Ex.27 = { a A ; a nilpotent}, Example 9.8. Let f K[X] be a non-constant polynomial and let f = f e f r er be an irreducible decomposition (i.e., the f i are irreducible and pairwise coprime, e i N). Let A := K[X]/(f) and for every g K[X] let ḡ 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 ( f i ) for i = 1,..., r (5.27). (4) By the Chinese remainder theorem one has A = r i=1 K[X]/(f e i i ) and A is reduced if and only if e 1 = e 2 = = e r = 1 (A reduced K[X]/(f e i reduced for all i by Remark 9.6 (3); e i = 1 K[X]/(f e i i ) integral domain; e i > 1, then 0 f i K[X]/(f e i i ) with f e i i = 0). i ) 6

7 B Algebraic and Transcendental Elements Definition 9.9. Let A be a K-algebra. An element a A is called algebraic over K if it is integral over K (i.e. if there exists a monic polynomial f K[X] such that f(a) = 0). If a is not algebraic, a is called transcendent. Example 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 X n 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 Let A be a K-algebra and let a A. Consider the K-algebra homomorphism K[X] A, f 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) = (µ a,k ) for a unique monic polynomial µ a,k K[X]. (iv) [K[a] : K] <. In this case, the polynomial µ a,k K[X] is called minimal polynomial of a over K and one has deg µ a,k = [K[a] : K] (4.22). Proof. The implications (iv) (ii) (iii) (i) are clear. K[a] = K[X]/(µ a,k ) and hence [K[a] : K] = deg(µ a,k ) <. And (iii) shows that Remark Let A be a K-algebra, let a A, and let f K[X] with f(a) = 0. Then µ a,k divides f. One has f = µ a,k if and only if f is monic and deg(f) = [K[a] : K]. Proposition 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 µ a,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: Moreover: a algebraic K[a] finite K-algebra 9.7 K[a] field K[a] = K(a). a transcendent 9.11 (ii) K[a] = K[X] K[a] K(a). If a is algebraic, then (µ a,k ) is a maximal ideal of K[X] because K[a] = K[X]/(µ a,k ) is a field. Hence µ a,k is irreducible. Finally: f(a) = 0 µ a,k f f irred. (f) = (µ a,k ) f monic f = µ a,k. Example Let K = Q, L = C, let p Z be prime number > 0. 7

8 (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 X n p. Moreover X n p is irreducible (Eisenstein (7.14)) and monic. Hence µ n p,q = X n p, Q( n p) = Q[ n p] and [Q( n p) : Q] = deg(µ n p,q) = n. (2) Let 1 ζ C an element with ζ p = 1 (i.e., ζ = e 2πik/p for some k = 1,..., p 1). Then ζ is algebraic over Q because it is a root of X p 1 Q[X]. We have an irreducible decomposition (Exercise 41) C X p 1 = (X 1)(X p 1 + X p ). As ζ 1, we find µ ζ,q = X p 1 + X p X + 1 and hence [Q(ζ) : Q] = p 1. Algebraic Extensions Proposition Let R be a ring, f 1,..., f k R[X 1,..., X n ], and let R A be an R-algebra. For g R[X 1,..., X n ] let g A[X 1,..., X n ] be the image of g in A[X 1,..., X n ]. Then we have an isomorphism of A-algebras given by a g a g. α: A R R[X 1,..., X n ]/(f 1,..., f k ) A[X 1,..., X n ]/( f 1,..., f k ) Proof. The map α 0 : A R R[X 1,..., X n ] A[X 1,..., X n ], a g a g, is a well defined isomorphism of A-algebras with inverse given by X i 1 X i. It sends 1 f i to f i. Hence it induces the isomorphism α. Example We have C = R[i] = R[X]/(X 2 + 1) because µ i,r = X Hence C R C 9.15 = C[X]/(X 2 + 1) = C[X]/((X + i)(x i)) CRM = C[X]/(X + i) C[X]/(X i) = C C. Corollary Let R be a ring. Then one has an isomorphism of polynomial algebras over R given by f g fg. R[X 1,..., X m ] R R[Y 1,..., Y n ] = (R[X 1,..., X m ])[Y 1,..., Y n ] = R[X 1,..., X m, Y 1,..., Y n ] Remark Let R be a ring and let A be an R-algebra. Let a 1, a 2 A. Then the homomorphism of R-algebras is surjective. m: R[a 1 ] R R[a 2 ] R[a 1, a 2 ], x 1 x 2 x 1 x 2 8

9 Proof. The image of m is an R-subalgebra contained in R[a 1, a 2 ] and containing a 1 and a 2. Hence it is equal to R[a 1, a 2 ] because R[a 1, a 2 ] is the smallest R-subalgebra of A containing a 1 and a 2. Definition A K-algebra K A is called algebraic if every element of A is algebraic over K. Proposition Let K A be a K-algebra. Then the following assertions are equivalent. (i) A is a finite K-algebra. (ii) A is an algebraic and finitely generated K-algebra. (iii) There exists a 1,..., a n A algebraic elements such that A = K[a 1,..., a n ]. Proof. (i) (ii) K A finite K A finitely generated and [K[a] : K] < for all a A 9.11 a is algebraic over K. (ii) (iii) Clear. (iii) (i) K[a i ] is finite over K for every i. Hence K[a 1 ] K K K[a n ] is finite over K and therefore K[a 1,..., a n ] is finite over K by Corollary Corollary Let A be a K-algebra, S A a set of algebraic elements such that A = K[S]. Then A is an algebraic K-algebra. Proof. Let a A. Choose T S finite such that a K[T ]. Then Proposition 9.20 shows that K[T ] is finite and a is algebraic over K. Proposition Let K B be a K-algebra without zero-divisors. algebraic K-subalgebra A of B is an algebraic field extension. Then every Proof. Let 0 a A. Then K[a] is a finite K-algebra by Proposition It is an integral domain because it is a subring of the integral domain. Hence it is a field by Proposition 9.7 (2) and a is a unit in A. Proposition Let K L be field extension and let L A 0 be an L-algebra. Then: K A is algebraic K L and L A are algebraic. Proof.. Obvious. Let a A and f = X n + b n 1 X n b 0 L[X] with f(a) = 0. Hence a is algebraic over K(b n 1,..., b 0 ) = K[b n 1,..., b 0 ] (the equality holds because b i is algebraic over K) K K[b 0,..., b n 1 ] K[a, b n 1,..., b 0 ] are finite 9.4 K K[a, b n 1,..., b 0 ] is finite 9.20 a algebraic over K. 9

10 D Algebraic Closure Proposition and Definition A field K is called algebraically closed if it satisfies 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(µ a,k ) = 1 µ a,k = X a K[X] a K. (iv) (iii) : Let f K[X] be irreducible. Then K K[X]/(f) is a field extension of degree deg(f) 9.20 deg(f) = 1. Example C is algebraically closed ( Fundamental Theorem of Algebra ). Definition An algebraic closure of K is an algebraic field extension K K such that K is algebraically closed. Theorem Every field has an algebraic closure. Proof. Let S be an uncountable set with K S and such that card(s) > card(k). Let K L be algebraic. Then we have L = n N L n, where L n := { a L ; a is root of a polynomial f K[X] with deg(f) n}. If K is finite, then L n is finite and hence L is countable. Otherwise card(k) card(l n ) 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 L 0 in X. Assume that L 0 is not algebraically closed. By Proposition 9.24 there exists an algebraic extension L 0 M which is not bijective. As M is an algebraic extension of K (Proposition 9.23), we find card(s \ L 0 ) = card(s) > card(m \ L 0 ). Hence there exists an injective map i: M S with i(x) = x for x L 0. Identifying M with i(m) S we obtain an element of X that properly contains L 0. Contradiction. 10

11 E Extension of Field Homomorphisms Definition and Remark Let K A 1 and K A 2 be K-algebras. A K- algebra homomorphism A 1 A 2 is also simply called a K-homomorphism. We denote the set of K-homomorphisms A 1 A 2 by Hom K-Alg (A 1, A 2 ). Every K-homomorphism is K-linear. If L := A 1 is a field extension of K and A := A 2 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 Let K A and K A be K-algebras. Let a A be algebraic over K. (1) Let ϕ: A A be a K-algebra homomorphism. Then ϕ(a) is algebraic over K and µ ϕ(a),k divides µ a,k (in particular µ ϕ(a),k = µ a,k if µ a,k is irreducible, for instance if A is a field). (2) The map is a bijection. In particular Hom K-Alg (K[a], A ) { a A ; µ a,k (a ) = 0 }, ϕ ϕ(a) if A is an integral domain. # Hom K-Alg (K[a], A ) deg µ a,k Proof. (1). Let µ a,k = X n + a n 1 X n a 0 K[X]. Then µ a,k (ϕ(a)) = ϕ(a) n + a n 1 ϕ(a) n a 0 ϕ K-Alg-Hom = ϕ(a n + a n 1 a n a 0 ) = ϕ(µ a,k (a)) = ϕ(0) = 0. This shows all claims. (2). Let a A with µ a,k (a ) = 0. Then µ a,k µ a,k and hence (µ a,k ) (µ a,k). Then K[a] K[X]/(µ a,k ) K[X]/(µ a,k) K[a ] A, a X X a is a K-algebra homomorphism ϕ: K[a] A with ϕ(a) = a. It is unique because a generates K[a] as a K-algebra. Proposition Let K L be an algebraic extension and let K Ω be an algebraically closed extension of K. 11

12 (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 because (K, K Ω) X. Endow X with the following partial order: (Z 1, τ 1 ) (Z 2, τ 2 ) : Z 1 Z 2, τ 2 Z1 = τ 1. Every totally ordered subset Y of X has an upper bound, namely ( Z, τ) with τ Z = τ Z for all (Z, τ Z ) Y. (Z,τ Z ) Y Hence X has a maximal element (Z 0, τ 0 ) be Zorn s lemma. We claim that Z 0 = L. Assume there existed a L \ Z 0. We consider Ω as a Z 0 -algebra via τ 0. Then µ a,z0 Z 0 [X] has a root a Ω because Ω is algebraically closed. Hence Proposition 9.29 implies that there exists a Z 0 -algebra homomorphism σ : Z 0 [a] Ω with σ(a) = a. This contradicts the maximimality of (Z 0, τ 0 ). (2). Now assume that L and Ω are algebraic closures of K. We have to show that ϕ is surjective. Its image ϕ(l) = L is algebraically closed and ϕ(l) Ω is algebraic because Ω is algebraic over K. Hence ϕ(l) = Ω by Proposition 9.24 (iv). Corollary Let K1 and K 2 be algebraic closures of K. Then there exists a K- isomorphism K 1 K2. F Splitting Fields Definition and Remark 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 properties. (1) Every f F splits in E[X] into a product of polynomials of degree 1. (2) For each f F let R f E be the set of roots of f in E. Then E = K( f F R f ). As every element of R f is algebraic, a splitting field is always an algebraic extension of K by Corollary Example Let f K[X] be a non-constant polynomial. Then a splitting field for f is a field extension K L such that f = u(x a 1 )... (X a n ) for a 1,..., a n L and u K and such that L = K(a 1,..., a n ) = K[a 1,..., a n ]. In this case L is a finite extension of K. Proposition Let F K[X] be a set of non-constant polynomials. (1) There exists a splitting field of F. (2) Let L 1 and L 2 be two splitting fields of F, let L i be an algebraic closure of L i and let ϕ: L 1 L2 be a K-isomorphism (as L i is also an algebraic closure of K, such an isomorphism always exists by Corollary 9.31). Then ϕ induces a K-isomorphism L 1 L2. In particular L 1 and L 2 are K-isomorphic. 12

13 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 S i := { a L i ; a is root of some f F}. Let a 1 S 1 be root of f F. As ϕ is a K-algebra homomorphism, f( ϕ(a 1 )) = ϕ(f(a 1 )) = 0 and hence ϕ(a 1 ) S 2. The same argument shows: a 2 S 2 ϕ 1 (a 2 ) S 1. Hence ϕ(s 1 ) = S 2 which shows that ϕ induces a K-isomorphism L 1 = K(S 1 ) L 2 = K(S 2 ). Corollary Let F K[X] be a set of non-constant polynomials. Let Ω be a field extension 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 := id K. Example Let K = Q, p a prime number. Then f = X 3 p is irreducible in Q[X]. Let α := 3 p R. The roots of f in C are α, ζα, ζ 2 α, where ζ := e 2πi/3. Hence the splitting field of f in C is L := Q[α, ζα, ζ 2 α] = Q[α, ζ]. We have Q[α] L because Q[α] R but L R. The irreducible decomposition of f over Q[α] is f = (X α)g with g = (X αζ)(x αζ 2 ) = X 2 + αx + α 2. Moreover: [Q[ 3 p] : Q] = 3, [Q[ 3 p, ζ] : Q[ 3 p]] = 2. Indeed, the first equality holds by Example To show the second equality note first 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(µ ζ,q[ 3 p]) deg(µ ζ,q ) 9.14 = 2. 13

14 10 The Fundamental Theorem of Galois Theory for Finite Extensions A Normal extensions Definition and Proposition An algebraic field extension K L is called normal 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 one has 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 (iii) (iv). is clear. Conversely, let σ : L L be a K-homomorphism. Replacing 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) containing 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 obtain σ(l) = σ (L) = L. (iii) (ii). Let L Ω be an algebraic closure of L, let f K[X] be irreducible and a L a root. We have to show that any root b Ω of f lies already in L. We have f = µ a,k = µ b,k ϕ : K[a] Ω K-homomorphism with ϕ (a) = b 9.30 ϕ: L Ω with ϕ K[a] = ϕ (2) 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 := { µ a,k ; a S } K[X]. As µ a,k is irreducible with root a in L, (ii) implies that µ a,k is product of polynomials of degree 1 in L[X]. Hence L is a splitting field of F. Proposition Let K L be a normal field extension and let M L be a subextension. (1) For every K-homomorphism τ : M M there exists a K-automorphism σ : L L such 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 M τ 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

15 (2). Let M be normal and σ a K-automorphism of L. Then σ(m) = σ M (M) = M by Proposition 10.1 (iii). Conversely, every K-automorphism of L restricts to an automorphism of L (Proposition 10.1). Hence σ(m) = M for all K-automorphisms σ of L. Hence M is normal by Proposition 10.1 (iv). Example (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 µ a,k = 2. Then we have µ a,k = (X a)g in L[X]. As deg(g) = 1, g = X b for some b L and L = K[a, b] is the splitting field of µ a,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, e 2πi/3 ] is normal (Example 9.36). B Fundamental Theorem of Galois Theory for Finite Extensions Definition A finite field extension K L is called Galois extension if L is the splitting 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 Let K L be a finite Galois extension. Then the group Gal(L/K) := Aut K-Alg (L) is finite and there is a bijective correspondence (10.5.1) {M L subextension of K L} {H Gal(L/K) subgroup} M Aut M-Alg (L) L H H, where L H := { 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 : L H ] = H, [L H : K] = (Gal(L/K) : H). Proof. Set G := Gal(L/K). (i). We first show that if M L is a subextension of K L, then [L : M] = Aut M-Alg (L). By hypothesis L = K[a 1,..., a n ] with distinct elements a 1,..., a n L such that f := (X a 1 ) (X a n ) K[X]. For 1 i n set M i 1 := M[a 1,..., a i 1 ] and d i := [ M i : Mi 1 ]. Then [L : M] = n i=1 d i. Hence it suffices to show that every K-homomorphism ϕ: Mi 1 L has exactly d i extensions to a K-homomorphism M i L or, equivalently by Proposition 9.29, that µ ai, M i 1 has d i distinct roots in L. But deg(µ ai, M i 1 ) = d i and has distinct roots because µ ai, M i 1 divides f. (ii). Now we claim that if H is a subgroup of G, then H = [L : L H ]. We have H Aut L H -Alg (L) and hence H [L : LH ] by (i). To show let k N with 15

16 k > H and b = (b 1,..., b k ) L k. We have to show that b is linear dependent over L H, or equivalently that (*) b (L H ) k 0. Here we set for any subset S L k S := { c = (c 1,..., c k ) L k ; k c i s i = 0 s = (s 1,..., s k ) S }. i=1 For c b (L H ) k and h H we also have c h(b) (L H ) k and hence (**) b (L H ) k = (Hb) (L H ) k, where Hb := { (h(b 1 ),..., h(b k )) ; h H } L k. Since the L-span of Hb in L k has L-dimension at most H < k, (Hb) 0. Choose 0 x = (x 1,..., x k ) (Hb) such that the number of x i with x i = 0 is as large as possible. As x 0 we find a coordinate x j with x j 0. By multiplying x with x 1 j we may assume that x j = 1. As the subspace (Hb) of L k 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 all h H by the choice of x. Therefore x (Hb) (L H ) k ( ) = b (L H ) k which shows our claim. (iii). Proof of the bijectivity of (10.5.1). Let M L be a subextension of K L and H := Aut M-Alg (L). Then M L H and (i) and (ii) show that [L : M] = [L : L H ] hence M = L H. Conversely let H Gal(L/K) be a subgroup and H := Aut L H -Alg (L). Then H H, and (i) and (ii) imply H = H. Hence H = H. Proposition 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 L H is a normal extension of K. In this case one has an exact sequence of groups 1 Gal(L/L H ) }{{} =H σ σ L Gal(L/K) H Gal(L H /K) 1 }{{}}{{} =:G =G/H We will see in Remark 12.4 that K L H L are indeed Galois extensions of K (the first one only if L H is normal over K) justifying the notation Gal(L/L H ) and Gal(L H /K). Proof. For σ Gal(L/K) we have (*) σ(l H ) = { σ(a) ; a L, γ(a) = a γ H } = { a L ; γ(σ 1 (a )) = σ 1 (a ) γ H } = L σhσ 1. 16

17 As the map H L H is injective, we see that 10.2 L H normal extension of K σ(l H ) = L H for all σ Gal(L/K) ( ) H = σhσ 1 for all σ Gal(L/K). is well defined and surjective by Propo- Moreover, the group homomorphism σ σ L H sition Its kernel is clearly Gal(L/L H ). 17

18 11 Étale Algebras We continue to denote by K a field. All rings and all algebras are commutative (with 1). A Preparations Lemma Let R be a ring, let u: M N be a homomorphism of R-modules and let E 0 be a free R-module. Then u is injective (resp. a surjective) if and only if id E u: E R M E R N is injective (resp. surjective). Proof. Choose an isomorphism E = R (I) of R-modules for some set I. By (6.29) we have a commutative diagram of R-linear maps E R M id E u E R N = = M (I) (m i ) i (u(m i )) i N (I), where the vertical maps are isomorphisms. This shows the claim. Corollary Let K a field, A a K-algebra. Let M and N be K-vector spaces. Let u: M N be a K-linear map. Then u is injective (resp. surjective) if and only if id A u is injective (resp. surjective). In particular M = 0 A K M = 0. Remark Let K L be a field extension and let V be an K-vector spce. Then Hom K (V, L), the set of K-linear maps V L, is a L-subspace of the L-vector space L V = Map(V, L) of all maps V L. One has an L-linear isomorphism Hom K (V, L) Hom L (L K V, L), u (l v lv) whose inverse is given by Hom L (L K V, L) ũ (v ũ(1 v)). Let V be a finite-dimensional K-vector space. Then (11.3.1) dim K (V ) 9.3 = dim L (L K V ) = dim L ((L K V ) ) = dim L (Hom K (V, L)). Proposition Let K L be a field extension, let A be a K-algebra. Let Hom K-Alg (A, L) := {A L homomorphism of K-algebras}. Then Hom K-Alg (A, L) is a linear independent subset of the L-vector space Hom K (A, L). In particular, if A is a finite K-algebra, then by (11.3.1) one has (11.4.1) # Hom K-Alg (A, L) [A : K]. 18

19 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 (ϕ 1,..., ϕ n ) of distinct elements ϕ i Hom K-Alg (A, L) is linearly independent. n = 0 is trivial. Let n 1. Let x 1,..., x n L such that n i=1 x iϕ i = 0. For a, b L we have n 1 x i (ϕ i (a) ϕ n (a))ϕ i (b) = i=1 n n x i ϕ i (ab) ϕ n (a) x i ϕ i (b) = 0 + ϕ n (a) 0 = 0 i=1 i=1 and hence n 1 i=1 x i(ϕ i (a) ϕ n (a))ϕ i = 0 for all a A. By induction hypothesis this implies that x i (ϕ i (a) ϕ n (a)) = 0, for all 1 i n 1 and a A. Since the ϕ i are distinct, this implies x i = 0 for 1 i n 1 and hence x n ϕ n = 0 and so x n = x n ϕ n (1) = 0. B Separable Polynomials Definition and Remark Let R be a ring and let f = a n X n +... a 1 X + a 0 R[X]. Then we call f := na n X n 1 + (n 1)a n 1 X n a 2 X + a 1 R[X] the formal derivative of f. For all f, g R[X], a, b R one has (af + bg) = af + bg (fg) = fg + f g R-linearity, Leibniz rule. Indeed, linearity is immediate. By linearity it suffices to show the Leibnis rule only for f = X n and for g = X m which is also immediate. Definition and Remark Let R be an integral domain, f R[X] and a R. Then ord a (f) := sup{ n N 0 ; (X a) n divides f} N 0 { } is called the order of the zero a of f. We say that a is a simple root of f if ord a (f) = 1. One has ord a (f) = f = 0, ord a (f) = 0 f(a) 0, ord a (f) = 1 f(a) = 0, f (a) 0 Only the last equality requires an argument: ord a (f) = 1 f = (X a)g with g R[X] such that g(a) 0 f(a) = 0 and f (a) 0 (because f = g + (X a)g and hence f (a) = g(a)). 19

20 Definition A polynomial 0 f K[X] is called separable if every root of f in a splitting field is simple. Example: Let K = Q, f = (X 3 2)(X + 1) is separable (roots in a splitting field are 3 2, ζ 3 2, ζ 2 3 2, 1 with ζ = e 2πi/3 ). f = (X 1) 2 is not separable. Proposition Let Ω be an algebraically closed extension of K. Let 0 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 exists K-homomorphism L Ω. (ii) (iii). Remark (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 ) = K[X]/(f, f ) = 0 (iv). Corollary Let f K[X] be irreducible. (1) f separable f 0. (2) Suppose that char(k) = 0. Then f is separable. Proof. (1). f separable f and f are coprime f irr. f 0 because any 0 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 0 by (1). C Separable Algebras Definition A K-algebra K A is called separable or geometrically reduced if K K A is reduced for some ( 9.31 for all) algebraic closures K of K. A finite separable K-algebra is called étale K-algebra. Proposition 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 (A i ) 1 i k is a finite family of separable K-algebras, then k i=1 A i is a separable K-algebra. 20

21 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 if K 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 K i=1 A i k i=1 ( K k A i ), λ (a i ) i (λ a i ) i, and this is a homomorphism of K-algebras. Hence the claim follows because products of reduced rings are again reduced (Remark 9.6 (3)). Example Let p be a prime number, K := Quot(F p [T ]). Then f := X p T K[X] is irreducible (Eisenstein with prime element T ). Hence L := K[X]/(f) is a field extension of degree p (in particular reduced). Let K be an algebraic closure of K and let a K with a p = T (exists as f has a root in K). Then K K L 9.15 = K[X]/(f) = K[X]/(X p a p ) = K[X]/(X a) p is not reduced. Note that f is not separable because f = px p 1 = 0. D Separability Degree and Split Algebras Let A 1 and A 2 be K-algebras. We denote by Hom K-Alg (A 1, A 2 ) the set of K-algebra homomorphisms A 1 A 2. Remark Let K L be a field extension, let A be a K-algebra and let B be an L-algebra. Then K L B is also a K-algebra and one has a bijection Hom K-Alg (A, B) Hom L-Alg (L K A, B), (a ϕ(1 a)) ϕ. ϕ (l a lϕ(a)), Definition Let A be a finite K-algebra. (1) An element a A is called separable if µ a,k is a separable polynomial. (2) Let Ω be an algebraically closed extension of K. Then [A : K] s := # Hom K-Alg (A, Ω) is called the separability degree of A over K. Proposition Let A be a finite K-algebra. The separability degree [A : K] s is independent 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 µ a,k (ϕ(a)) = 0. Hence ϕ(a) Ω alg := { w Ω ; w is algebraic over K}. But Ω alg is a an algebraic closure of K (Exercise 46) and hence we can always assume that Ω is an algebraic closure. As two algebraic closures of K are K-isomorphic, it follows that [A : K] s does not depend on the choice of Ω. 21

22 (1). This follows from Proposition (2). This follows from Remark 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 Hom K -Alg(K, Ω) let T ϕ := { ψ Hom K -Alg(A, Ω) ; ψ K = ϕ } = Hom K-Alg (A, K Then #T ϕ = [A : K] s. Hence [A : K ] s = # Hom K -Alg(A, Ω) = ϕ Hom K -Alg (K,Ω) ϕ Ω). #T ϕ = [K : K ] s [A : K] s. Example Let n N. Let K n = K K (n factors) be the product ring. It is a K-algebra via K K n, a (a,..., a). It is a finite reduced K-algebra with [K n : K] = n. For n 2 this is not an integral domain: (1, 0,..., 0)(0,..., 0, 1) = 0. Every ideal of K n is of the form a 1 a n, where a i K is an ideal and hence a i = 0 or a i = K for all i. In particular, the maximal ideals of K n are of the form K K 0 K K. For every K-algebra K A one has by (6.29) an isomorphism of A-modules ( ) A K K n A n, b (a 1,..., a n ) (ba 1,..., ba n ) which is clearly even an isomorphism of A-algebras. Choosing for A an algebraic closure of K, we see that K n is an étale K-algebra. Remark Let K A be a K-algebra. (1) There is a bijection Hom K-Alg (A, K) = { π : A K ring homomorphism ; K A { m A maximal ideal ; K A can A/m is an isomorphism} given by π Ker(π). (2) In particular we have for a finite K-algebra A with d := [A : K] A = K d (isomorphism of K-algebras) 9.7 #{m A maximal ideal} = d # Hom K-Alg (A, K) = d. π K is id K } In this case set X := Hom K-Alg (A, K). Then the isomorphism can be given by ( ) A K X = K d, a (ϕ(a)) ϕ X. Definition Let K L be a field extension. A finite K-algebra A is called L-split if L K A = L d for some d 0. One then has necessarily d = [A : K] = [L K A : L]. 22

23 Proposition Let A be a finite K-algebra and let K L be a field extension. (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 # Hom K-Alg (A, L) = [A : K] Proof. (1). (a) A is L-split (b) A is K-split (c) A is separable L K A = L d ( ) L K A = L L (L K A) = L L L d = (L ) d. (2). By Lemma 11.1, L K A is an L-subalgebra of L K A = L [A:K]. Hence it is isomorphic to an L-algebra of the form L d for some d (with d = [A : K]). (3). Let us show (a). Remark we find L K A = L [A:K] # Hom L-Alg (L K A, L) = [L K A : L] and # Hom L-Alg (L K A, L) = # Hom K-Alg (A, L) by Remark 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 is reduced 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 Étale Algebras Theorem 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 étale K-algebra). (ii) There exists an isomorphism of K-algebras K K A K K. (iii) Every a A is separable over K. (iv) There exists a 1,..., a n A separable elements such that A = K[a 1,..., a n ]. (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 We will prove (vi) (ii) (v) (i) (iii) (iv) (ii) The implication (i) (vi) is called the theorem of the primitive element and proved in Theorem below. (ii) (v) (i). Proposition (3). 23

24 (vi) (ii). Let A = K[X]/(f) with f separable. We may assume that f is monic. Then f = d i=1 (X a i) in K[X] with a 1,..., a d distinct. Hence K K A 9.15 = K[X]/ i (X a i ) 9.8 = d i=1 K. (i) (iii). Let a A. Then K[a] = K[X]/(µ a,k ) is a K-subalgebra of A and hence separable (Proposition (2)). Hence K K K[a] = K[X]/(µ a,k ) is reduced. As K is algebraically closed, µ a,k = i (X a i) K[X]. By Example 9.8, all roots a i have to be distinct. (iii) (iv). Clear. (iv) (ii). For i = 1,..., n, K[a i ] = K[X]/(µ ai,k) with µ ai,k separable. Let d i := deg(µ ai,k). Then K K K[a i ] = K d i follows from (vi) (ii) which we already proved. Hence K K (K[a 1 ] K K[a 2 ] K K K[a n ]) = ( K K K[a 1 ]) K ( K K K[a 2 ]) K K ( K K K[a n ]) = K d 1 K K Kd n = K d 1 d n. Now K[a 1,..., a n ] is a quotient of the K-algebra K[a 1 ] K K[a 2 ] K K K[a n ] (Corollary 9.18), hence K K K[a 1,..., a n ] is isomorphic to a quotient of the K-algebra K d 1 d n and hence isomorphic to K m for some m i d i. During the proof of the theorem we have seen the argument for the following corollary. Corollary 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 Let A be a K-algebra. Then A finite separable A is product of finite separable field extensions. Proof. by Proposition (3). If A is separable, it is reduced and hence A = r i=1 K i for finite extensions K K i (Proposition 9.7). Each K i is a quotient of A and hence separable by Corollary Corollary Let A be an algebraic K-algebra. The following assertions are equivalent. (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]. 24

25 Proof. (i) (ii). Let a A. Then K[a] is a finite K-subalgebra of A. As A is separable, K[a] is separable (Prop (1)). Hence a is separable by Theorem (iii). (ii) (iii). Obvious (iii) (i). Let K be an algebraic closure of K. For every finite subset T S the K-algebra K[T ] is finite (Proposition 9.20) and separable (Theorem 11.20) and hence K K K[T ] is reduced. One has and hence by Lemma 11.1 Hence K K K[S] is reduced. A = K[S] = K K K[S] = T S finite T S finite K[T ] ( K K K[T ]). Corollary 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 perfect 1. 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, Theorem 3). Proof. As every reduced finite algebra is a product of finite field extensions (Proposition 9.7 (4)), it suffices to show the second assertion. Let K L be an algebraic extension, a L. Assume that µ a,k =: n i=0 a ix i is not separable over K (Corollary 11.23) or equivalently that µ a,k = n i=1 ia ix i 1 = 0 (Corollary 11.9). This cannot happen if char(k) = 0. Hence suppose char(k) = p > 0. Then a i = 0 if p i. As K is perfect, we find b i K with b p i = a i for all i. Set g := p i b ix i/p. Then g p = p i a ix i = µ a,k. Contradiction because µ a,k is irreducible. Proposition Let K L be an algebraic field extension and let L A 0 be an algebraic L-algebra. Then A is a separable K-algebra if and only if L is a separable extension 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 µ a,l divides the 1 To see this, we may assume that char(k) = p > 0. Let K be an algebraic closure of K. Let a K and let b K with b p = a. Then µ b,k divides X p a = (X b) p. By hypothesis µ K,b is separable (because K[b] is separable over K), hence µ b,k = X b which shows b K. A more careful argument shows that µ b,k = X p a if b / K. Hence a field of characteristic p is already perfect if every extension of degree p is separable. 25

26 separable polynomial µ a,k and hence is separable itself. Therefore every element a of A is separable over L which shows that A is separable over L by Corollary Conversely assume that K L and L A are separable. We may assume K L A. We have to show that every a A is separable over K. Let L = K[a 0,..., a n 1 ], where the a i L are the coefficients of the separable polynomial µ a,l. This is a finite extension of K contained in L, hence K L is separable. As µ a,l = µ a,l, the finite L -algebra K[a] is separable over L because it is generated by a separable element. Hence we may assume that K L and L A are étale. Then [A : K] 9.4 = [A : L][L : K] = [A : L] s [L : K] s = [A : K] s which shows that A is an étale K-algebra F Theorem of the Primitive Element The implication (i) (vi) in Theorem follows from the following result. Theorem Let A be a finite separable K-algebra. Then there exists a A such that A = K[a]. Such an element a A is then sometimes called a primitive element of the K-algebra A. One then has A = K[X]/(µ a,k ) and µ a,k is separable because a is separable. In particular (i) (vi) in Theorem follows. Proof. (i). We have A = r i=1 K i for finite separable field extensions K K i by Corollary If K i = K[a i ], then A = K[(a 1,..., a r )]. Hence we may assume that A = 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 shows that there exists a L with L = {1 = a 0, a, a 2,... }. Hence L = K[a]. Therefore we may from now on assume that K is an infinite field. (iii). Let L = K[a 1,..., a n ]. We prove the theorem by induction on n. For n = 1 we are done. Let n > 1. The subextension K[a 1,..., a n 1 ] L is finite separable and hence there exists b K[a 1,..., a n 1 ] with K[b] = K[a 1,..., a n 1 ] by induction hypothesis. Hence we have L = K[b, a n ] and it suffices to consider the case that L = K[b, c] for b, c L. (iv). Let L = K[b, c]. Choose an algebraic closure K of K and write Hom K-Alg (L, K) = {ϕ 1,..., ϕ m }, m = [L : K] s = [L : K]. Consider g := 1 i<j m [ (ϕi (b) ϕ j (b) ) X + ( ϕ i (c) ϕ j (c) )] K[X]. The ϕ i are distinct hence ϕ i (b) ϕ j (b) or ϕ i (c) ϕ j (c) i < j g 0. K infinite λ K with g(λ) 0 and hence ( ϕi (b) ϕ j (b) ) λ + ( ϕ i (c) ϕ j (c) ) 0 i < j ϕ i (λb + c) = λϕ i (b) + ϕ i (c) λϕ j (b) + ϕ j (c) = ϕ j (λb + c). 26

27 Set a := λb + c. Then ϕ i (a) ϕ j (a) i j. As {ϕ 1 (a),..., ϕ m (a)} are the roots of µ a,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 Let K A be a K-algebra, and set A s := { a A ; a is separable and algebraic over K}. Then A s is the largest K-subalgebra of A that is algebraic and separable over K. If A is a field, then A s 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. Then B A s by Corollary (ii). The K-algebra K[A s ] is algebraic (Corollary 9.21) and separable (Corollary 11.23). Hence K[A s ] A s and so A s = K[A s ]. The last assertion follows from Proposition Definition and Remark (1) A field K is called separably closed if the following equivalent conditions hold. (i) Every étale 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 (i) (iii) (ii) follows as in the proof of Proposition (2) A separable closure of a field K is an algebraic and separable field extension E of K such that E is separably closed. Proposition Let K be a field. (1) Let Ω be an algebraically closed extension of K. Then Ω s := { a Ω ; a is separable and algebraic over K} is a separable closure of K. In particular separable closures of K exist. (2) Let K 1 and K 2 be separable closures of K. Then there exists a K-isomorphism K 1 K2. (3) Let K sep be an algebraic closure of K and let K L be a separable algebraic field extension. Then there exists a K-homomorphism ϕ: L K sep. If L is separably closed, then ϕ is automatically a K-isomorphism. Proof. (1) Ω s is an algebraic and separable subextension of Ω. Let F be a finite separable extension of Ω s. As Ω is algebraically closed, there exists an Ω s - homomorphism ι: F Ω (Proposition 9.30). Then the elements of ι(f ) Ω are separable and algebraic over K and hence ι(f ) Ω s and hence [F : Ω s ] = 1. 27

28 (2). A separable closure is a splitting field of the set of all non-constant separable polynomials in K[X]. Hence it is unique up to K-isomorphism by Proposition 9.34 (2). (3). Let K be an algebraic closure of K sep. By Proposition 9.30 there exists a K- homomorphism ϕ: L K. Every a L is separable, hence ϕ(a) K is separable because they have the same minimal polynomial (Proposition 9.29). Hence ϕ(l) K s = K sep. If L is separably closed, then L = ϕ(l) is separably closed and ϕ(l) K sep is a separable and algebraic extension. Hence ϕ(l) = K sep by Definition (ii). Corollary If K is perfect (e.g., if char(k) = 0), then every separable closure is an algebraic closure. 28

29 12 Galois Theory A Galois Extensions Definition An algebraic extension K L is called Galois extension if it is normal and separable. In this case we call Gal(L/K) := Aut K-Alg (L) = ({ σ : L L ; σ automorphism of K-algebras}, ) the Galois group of L over K. Proposition For a finite field extension K L the following assertions are equivalent. (i) K L is normal and separable (i.e., a Galois extension in the sense of Definition 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) # Aut K-Alg (L) = [L : K]. Proof. (i) (ii). If L is a splitting field of a polynomial f, then L is normal and L = K[a 1,..., a n ], where a i are the roots of f in L. If f is separable, then all a i are separable over K. Hence L is a separable extension of K by Theorem 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 separable polynomial µ a,k. (i) (iii). Let Ω be an algebraically closed extension of L. Then one always has # Aut K-Alg (L) # Hom K-Alg (L, Ω) = [L : K] s [L : K]. The first inequality is an equality if and only if K L is normal. The second inequality is an equality if and only if K L is separable. Example (1) A separable closure of a field K is a Galois extension of K. Indeed it is separable by definition and it is the splitting field of all separable polynomials in K[X]. (2) Let K be a perfect field (e.g. if char(k) = 0). Then K L is a Galois extension if and only if it is a normal extension (Corollary 11.24). Remark Let K M L be a algebraic field extensions such that K L is a Galois extension. Then M L is a Galois extension (it is separable by Proposition and it is normal because if L is the splitting field of a set F of polynomials in 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 (Example 9.36). Moreover, if K M and M L are Galois extensions, then K L is in general not 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 = X 2 p and 29

30 µ 4 p,q[ p] = X 2 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 = X 4 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 Let K L be an algebraic extension, let S L with L = K[S]. Let N be the splitting field of { µ a,k ; a S } K[X] in some algebraic closure of L. Then N is a normal extension of K with L N and there exists no normal subextension N N of K with L N. Such a field extension is the called a normal hull of L over K. If K L is finite (resp. separable), then we may choose S to be finite (resp. to consist of separable elements). Then K N is finite (resp. a Galois extension). Proposition Let A be an étale K-algebra. Then there exists a finite Galois extension K L such that A is L-split (i.e., there exists an isomorphism of L-algebras L K A L d, where d = [A : K]). Proof. By Theorem there exists a A such that A = K[a]. As a is separable over K, µ a,k is separable of degree d = [A : K]. Then a splitting field L of µ a,k is a finite Galois extension (Proposition 12.2). Hence there exists d distinct roots of µ a,k in L and we have # Hom K-Alg (A, L) 9.29 = #{b L root of µ a,k } = d. This implies that A is L-split by Proposition (3). B Main Theorem of Galois (Grothendieck version) Remark Let K L be a Galois extension and let K M L be a subextension. 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 σ M Gal(M/K) 1 Definition 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 X and σ H). Therefore a G-action on a finite set X is continuous if and only if there exists a finite subextension 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) into a topological group such that the open subgroups (as defined above) form a fundamental system of open neighborhoods of the neutral element id L Gal(L/K). Then a 30