1. Algebraic Equations 1.1

LECTURE NOTES FOR MATH 512

JIAN-SHU LI

1. Algebraic Equations 1.1. Algebraic Extensions. • Simple extensions, algebracity • algebraic versus finite (principle: algebraic extensions are finite in nature. • degree of extensions: [E : k] = [E : F ][F : k]. • compositum EF is the field of fractions of ring of elements of the form

a1b1 + ··· + anbn, (aj ∈ E, bj ∈ F ) 1.2. Algebraic Closure. We want to show: Theorem 1.1. Let k be any field. Then there is an algebraic closure of k which is unique up to isomorphism Step 1: Existence • embeddings: any non-trivial homomorphism between two fields is an embedding. • extension of embedding, embedding over a subfield • f σ Lemma 1.2. We have • E/k algebraic, σ : E −→ E an embedding. Then σ is an automorphism. • σ(EF ) = σ(E)σ(F ) Definition 1.3. Algebracally closed, algebraic closure.

Lemma 1.4. Let f1, ··· , fn ∈ k[X]. Then there is an extension E of k in which every fj has a root. Here is the key step, whose proof is not completely obvious: Theorem 1.5. There is an algebraically closed ﬁeld containg k.

Proof. First, we show that there is an extension E1 in which every polynomial f ∈ k[X] of degree ≥ 1 has a root in E1. Remember what we did for a single polynomial: we got k[X]/(f(X)). In the case of two polynomials may be we want to try k[X1,X2]/(f1(X1), f2(X2)). We want to enlarge the denominator to a maximal ideal, and we want to do this for all polynomials simutaneouly ..... Formally: to each polynomial f ∈ k[X] with degree ≥ 1 we associate a symbol Xf . Let S be the set of all such letters, and consider the polynomial ring k[S].

Date: Spring 2003. 1 2 JIAN-SHU LI

Claim: the ideal generated by all f(Xf ) is not the unit ideal. Now let m be a maximal ideal containing all f(Xf ), and consider

k[S] −→ k[S]/m = E1 Inductively,, we get E1 ⊆ E2 ⊆ · · · We take E = ∪En. Corollary 1.6. There is an algebraic closure ka of k.

a Proof. Take k = union of all algebraic subextensions of E, with E as above. Step 2: Uniqueness. Let E/k be algebraic. Let σ : k −→ L be an embedding of k into an algebraically closed field L. • If α ∈ E then the number of extensions of σ to k(α) is equal to the number of distinct roots of the irreducible polynomial of α over k. Theorem 1.7. There is an extension of σ to E. If E is algebraically closed and L is algebraic over k then any such extension is an isomorphism onto L. Proof. Use Zorn’s lemma. Consider the family of pairs (F, τ) with F = subfield of E containing k, τ an extension of σ to F . Define (F, τ) ≤ (F 0, τ 0) if F ⊆ F 0 and τ 0 is an extension of τ..... 1.3. Normal extensions. A splitting field of a polynomial f over k is an extension E/k such that • f splits into linear factors in E • E is generated by the roots of f Theorem 1.8. Let E,K be splitting field of the same polynomial f. Then any embedding E −→ Ka induces an isomorphism of E onto K (in particular, splitting fields of the same f are isomorphic to each other). The proof is obvious. We can also consider a family of polynomials and consider the corresponding splitting field. We have a result similar to the above. The next result describes normality. Theorem 1.9. Let k ⊆ K ⊆ ka. TFAE • every embedding of K in ka over k induces an automorphism of K • K is the splitting field in ka of a family of polynomials in k[X] • Every irreducible polynomial in k[X] which has a root in K splits completely in K. Some properties: • Normal extensions remain normal under lifting • if k ⊆ E ⊆ K and K is normal over k, then it is normal over E. • If K1,K2 are normal over k then K1K2 and K1 ∩ K2 are also normal over k. LECTURE NOTES FOR MATH 512 3

Let E be a ﬁnite extension of k. The smallest normal extension (in Ea) of k containing E is called the normal closure of E. It is equal to the intersection of all normal extensions in Ea containig E. It is also equal to the compositum

(σ1E) ··· (σnE) a where σ1, ··· , σn are the different embeddings of of E into E . If E is separable over k then so is its normal closure. If σ is an embedding of E into ka over k, we say σE is a conjugate of E in ka. a If α is algebraic over k and σ1, ··· , σn are the distinct embeddings of k(α) into k , a we call σ1α, ··· , σnα the conjugates of α in k . (Source of term: 1 − i is a conjugate of 1 + i). 1.4. Separable extensions. Let E be algebraic over F , σ : F −→ L an embedding of F into an algebraically closed field L. We may assume L to be algebraic over σF . Let Sσ be the set of extensions of σ to E. We draw a diagram to show that if (σ, L) is replaced by some (τ, L0) then there is a bijection between Sσ and Sτ . This enables us to define the separable degree of E/F to be

[E : F ]s = #(Sσ) Theorem 1.10. Suppose k ⊆ F ⊆ E. We have

[E : k]s = [E : F ]s[F : k]s

[E : k]s ≤ [E : k]

For a finite extension E we say it is separable if [E : k]s = [E : k]. Note that k(α)/k is separablle iff the irreducible polynomial of α has no multiple roots. In such case we say α is separable. Lemma 1.11. A finite extension is separable iff each of its elements is separable Definition 1.12. Let E be an arbitrary algebraic extension of k. We say E/k is separable if every element in E is separable. Corollary 1.13. E is separable iff it is generated by a family of separable elements Theorem 1.14. (Primitive Element Theorem) Let E be a finite extension of k. Then E = k(α) iff there exists only finitely many intermediate fields k ⊆ F ⊆ E. If E is separable over k then such an element α exists Proof. If k is finite then so is E, and we know E× is a cyclic group, hence E is generated by one element. (As we shall see later, all algebraic extensions of a finite field are separable). Assume k is infinite. Suppose there are only finitely many field between k and E. Consider α, β ∈ E. As c ranges over k, only finitely many fields k(α + cβ) could be distinct. So there exists c1 6= c2 with k(α + c1β) = k(α + c1β) = F

Then (c1 − c2)β ∈ F , etc. Now suppose E = k(α). Let f be the irred. poly of α over k. For any intermediate F let gF be the irred. poly of α over F . Then gF divides f so there are 4 JIAN-SHU LI only finitely many gF . We want o show that F is determined by gF . In fact F is equal to the field F0 generated over k by the coefficients of gF (this we show by comparing degrees: [E : F ] = [E : F0] = deg gF ). Suppose E is separable. May assume E = k(α, β). Let σ1, ··· , σn be the distinct embeddings. Want to show that σj(α + cβ) distinct for some c ∈ k. Look at Y P (X) = (σiα + σiβX − σjα − σjβX)

1.5. Finite Fields. Any field must contain a primitive field isomorphic to Q (char. 0), or Fp (char. p). r Suppose F is finite field, then it contains some Fp. So F has q = p elements, is the splitting field of Xq − X, etc. a • In Fp there is unique extension of Fq of degree n • Frobenius • Aut(Fq) is cyclic of order n, generated by Frob.

• extensions of Fq are Fqm . All extensions are normal and separable.

1.6. Inseparable extensions.

[E : k] = [E : k]s[E : k]i etc.

• Given any algebraic ext E/k there is a maximal separable subextension E0. If E/k is normal then so is E0/k.

2. Galois Theory 2.1. Galois Extensions. Definition 2.1. Galois = normal + separable Theorem 2.2. (Main Theorem of Galois Theory) Let K/k be finite Galois, G = Gal(K/k). • There is a bijection K ⊇ E ←→ H ⊆ G, E = KH ,H = Gal(K/E) • H is normal iff E/k is Galois • If E/k is Galois then

Gal(K/k)/Gal(K/E) ' Gal(E/k), σ 7→ σ|E Theorem 2.3. Let K/k be Galois (may be inﬁnite), and G = Gal(K/k). Then • k = KG • If k ⊆ F ⊆ K then K/F is Galois • The map F 7→ Gal(K/F ) is injective.

H Proof. We have F = K if H = Gal(K/F ). LECTURE NOTES FOR MATH 512 5

Terminology: Gal(K/F ) is associated with F , and H = Gal(K/F ) belongs to F . Note: If λ : K ' λK then λK/λk is Galois, and Gal(λK/λk) = λGal(K/k)λ−1 etc. In particular, suppose K/k is Galois and k ⊆ F ⊆ K. Let λ : F −→ K be any embedding of F (extended to K). Then Gal(K/λF ) = λGal(K/F )λ−1 Theorem 2.4. Suppose K/k is Galois, G = Gal(K/k). In the correspondence F −→ H we have F/k normal iff H is normal in G. In such case, restriction gives an isomorphism G/H ' Gal(F/k) Corollary 2.5. • H ∩ H0 ←→ FF 0 • F ∩ F 0 = fixed field of the smallest subgroup of Gal(K/k) containing H,H0. • F ⊆ F 0 ⇐⇒ H ⊇ H0 • If K/k is finite then there are only finitely many intermediate fields between k and K. Corollary 2.6. E/k separable, K = smallest normal extension of k containing E. Then K/k is Galois. There are only finitely many intermediate fields between k and E Lemma 2.7. E/k separable. Assume that there is an integer n such that every element of E is of degree ≤ n over k. Then [E : k] ≤ n

Proof. Clear from the Primitive Element Theorem Theorem 2.8. (Artin) K = field, G = finite group of automorphisms of K, of order n, and k = KG. Then K/k is finite Galois with Galois group G.

Proof. Let α ∈ K. Let σ1, ··· , σn be the elements of G. Then Y f(X) = (X − σjα) is a polynomial with coefficients in k. So α is separable, of degree ≤ n, etc. Corollary 2.9. K/k finite Galois. Then every subgroup of Gal(K/k) belong to some intermediate field F . Infinite Case: Review: Inverse limits Let K/k be Galois, with G = Gal(K/k). For any finite Galois subextension F we set H = Gal(K/F ). Then H is of finite index in G (equal to [F : k]). We have G −→ G/H for each such H. We get G −→ lim G/H This is an isomorphism (proof easy). Krull topology • Basis of top: cosets of Gal(K/F ), with F/k finite Galois • Basis of top: cosets of Gal(K/F ), with F/k finite 6 JIAN-SHU LI

• lim G/H as closed subset Y lim G/H −→ G/H ←− the right hand side being compact by Tychonoﬀ. • every closed subgroup of ﬁnite index in G is open

Proof. [ H = G \ σjH

with σj run through a set of reps of non-identity cosets. • Let H be a subgroup of G. Let F = KH . Then Gal(K/F ) = closure of H

Proof. Clearly H ⊆ Gal(K/F ). Let E be any ﬁnite Galois extension of F . Since EH = F we see

H|E = Gal(E/F ) ' Gal(K/F )/Gal(K/E) Thus for any σ ∈ Gal(K/F ) and any E as above, the neighborhood σGal(K/E) contains a point of H. Thus Gal(K/F ) is the closure of H. • closed subgroups in G are those of the form Gal(K/F ), F any intermediate ﬁeld

Proof. We show any Gal(K/F ) is closed. Take any σ 6∈ Gal(K/F ). Then there is an α ∈ F with σ(α) 6= α. Let E be the normal closure of k(α). Then σGal(K/E) is open and disjoint from Gal(K/F ). Conversely suppose H is closed. Set F = KH . We know Gal(K/F ) is the closure of H. Since H is closed, we see it is equal to Gal(K/F ). Corollary 2.10. In a ablelian extension every subextension is Galois and abelian. In a cyclic extension ext. every subext is cyclic Theorem 2.11. K/k Galois, E/k arbitrary. Suppose both are contained in some field. Then Gal(KE/E) ' Gal(K/K ∩ E) In particular, [KF : F ] divides [K : k]. Proof. The restriction map is obviously injetive. Let G be the image. In the finite case we only need to show that KG = K ∩ E, which is also obvious. In the infinite case we show the map is continuous. Let F/E ∩K be finite Galois and contained in K. Then the preimage of Gal(K/F ) is simply Gal(KE/F E), and F E/E is finite Galois. So the map is indeed continuous. Then as the image of a compact set, G must be closed. So G = Gal(K/K ∩ E) as claimed. Corollary 2.12. ** • In the above suppose both K,E Galois. Then the map

Gal(KE/k) −→ Gal(K/k) × Gal(E/k), σ 7→ (σ|K , σ|E) is an injective homomorphism, and is an isomorphism if K ∩ E = k. LECTURE NOTES FOR MATH 512 7

Proof. The map is clearly an injective homomorphism. Suppose K ∩E = k. By the previous result we see that Gal(K/k)×1 and 1×Gal(E/k) are both in the image So the product is in the image. In the inﬁnite case we observe that the map is clearly continuous. It takes closed sets to closed sets (because closed sets are compact). Hence also take open sets to open sets. Hence (in case K ∩ E = k) the inverse is also continuous.

• Case of n Galois extensions, Ki/k, assuming Ki+1 ∩ (K1 ··· Ki−1) = k. • When G = G1 × · · · × Gn. • If K,L abelian over k then so is KL. Proof. Consider the map

Gal(KL/k) −→ Gal(K/k) × Gal(L/k), σ 7→ (σ|K , σ|L) 2.2. Examples. Example 2.13. • If K is the splitting ﬁeld of a separable polynomial of degree n then Gal(K/k) ⊆ Sn. In particular, [K : k] ≤ n! • Quadratic extensions • Cubic extensions: 2 δ = (α1 − α2)(α2 − α3)(α1 − α3), ∆ = δ Even permutations preserve δ. If f(X) = X3 + aX + b then ∆ = −4a3 − 27b2.

Gal(K/k) ' A3 ⇐⇒ ∆ is a sqaure

Example 2.14. K = k(t1, ··· , tn), G = Sn. n Y n n f(X) = (X − tj) = X − s1X + ··· + (−1) sn j=1 Then Sn k(t1, ··· , tn) = k(s1, ··· , sn) Indeed let G F = K ⊇ k(s1, ··· , sn) Then [K : F ] = n! while [K : k(s1, ··· , sn)] ≤ n!

Example 2.15. It is known that Sn is the Galois group of some extension of Q. It is not known whether for any given finite group G there is a Galois extension of Q with Galois group isomorphic to G. Example 2.16. (Fundamental Theorem of Algebra) • Every element of C = R(i) has a square root • every finite extension of R(i) is contained in some finite Galois extension K of R. 8 JIAN-SHU LI

• Let G = Gal(K/R),H = a 2-Sylow subgroup of G, F = KH . Then [F : R] is odd. By the primitive element theorem, F = R(α), and α is a root of an irreducible polynomial of odd degree over R. This degree must be 1. So G = H is a 2-group, hence solvable. • Let G1 = Gal(K/R(i)). If G1 is not trivial then it contains a subgroup G2 of index 2, necessarily normal. Let F = KG2 . Then [F : R(i)] = 2. But we have seen that C could have no quadratic extension. Hence G1 must be trivial, K = C. So C is algebraically closed. 2.3. Roots of Unity. • Xn − 1 is separable iff n is not divisible by the characteristic of the field. Assume this to be the case. • splitting field = k(ζ) with ζ a primitive n-th root of unity. • one has × Gal(k(ζ)) ⊆ (Z/nZ) , and hence [k(ζ): k] divides ϕ(n) Lemma 2.17. We have × [Q(ζ): Q] = ϕ(n), Gal(Q(ζ)/Q) ' (Z/nZ) Proof. Let f be the minimal polynomial of ζ. It suffices to show: ζp is a root of f for any p not dividing n. Suppose not. Write Xn − 1 = f(X)g(X) Then ζp is a root of g, and ζ is a root of g(Xp). So f(X) divides g(Xp). Write g(Xp) = f(X)h(X) Then g(X)p ≡ f(X)g(X) ( mod p) n Hence X − 1 has a multiple root mod p. Contradiction. Corollary 2.18. • If n, m are relatively prime then

Q(ζn) ∩ Q(ζm) = Q • over Q the irreducible poly of ζp is Xp−1 + ··· + 1 Let Y Φd(X) = (X − ζ) period ζ=d Then n Y X − 1 = Φd(X) d|n and Xn − 1 Φn(X) = Q d|n,d

• pr−1 Φpr (X) = Φp(X )

r1 rs • If n = p1 ··· ps then

r1−1 rs−1 p1 ···ps Φn(X) = Φp1···ps (X )

• if n is odd then Φ2n(X) = Φn(−X) • If p is a prime not dividing n then

p Φn(X ) Φpn(X) = Φn(X) If p|n then p Φpn(X) = Φn(X ) • Let µ be the M¨obius function deﬁned by  0, if p2|n  r µ(n) = (−1) , if n = p1 ··· pr  1, if n = 1 Then X 1, if n = 1 µ(d) = 0, if n > 1 d|n

Proof. Let r be the number of the distinct primes dividing n. Then

r X X r µ(d) = (−1)j = (1 − 1)r j d|n j=0

• Y n/d µ(d) Φn(X) = (X − 1) d|n A Theorem of Kronecker asserts that every abelian extension is contained in some Q(ζ). Here we verify this for quadratic extensions. We have (1 + i)2 2 = i and if p is an odd prime and X x S = ζx p x then −1 S2 = p p So our assertion follows. 10 JIAN-SHU LI

2.4. Linear Independence of characters. Let G be a monoid. • def of character • examples • linear independence of characters (Artin) Proof. Starting from a minimla expression for linear independence. What 0 it means to have χ 6= χ ?

Corollary 2.20. Let E/k be ﬁnite. Let σ1, ··· , σr be distinct embeddings of E into a k . If w1, ··· , wn is a basis of E over k then the vectors

ξ1 = (σ1w1, ··· , σ1wn) ···

ξr = (σrw1, ··· , σrwn) are linearly independent. 2.5. Norm and Trace.

Definition 2.21. Let E/k be a finite extension. σ1, ··· , σr different embeddings into ka. r E Y [E:k]i N(α) = NE/k(α) = Nk (α) = ( σjα) j=1 r E X T (α) = T rE/k(α) = Tk (α) = [E : k]i · σjα j=1 Lemma 2.22. • N is a homomorphism • T is a linear map E F E E F E • Nk = Nk ◦ NF ,Tk = Tk ◦ TF • If α is a root of the irred polynomial n n−1 f(X) = X + a1X + ··· + an then k(α) n k(α) Nk (α) = (−1) an,Tk (α) = −a1 Theorem 2.23. Suppose E/k is finite separable. Then E • Tk is nonzero Proof. indepence of characters • (x, y) 7→ T (xy) is non-degenerate • existence of dual basis Let E/k be any finite extension. Each α determines a k-linear map

mα : E −→ E, x 7→ αx Proposition 2.24. One has

det(mα) = NE/k(α), Tr(mα) = TE/k(α)

Proof. Let F = k(α). Then NF/k(α) is determined by the minimal polynomial of α, and [E:F ] NE/k(α) = NF/k(α) ,TE/k(α) = [E : F ] · TF/k(α) i Let w1, ··· , wr be a basis of E/F . Then {α wj} is a basis of E/k, and we calculate the determinant and trace by means of this basis. LECTURE NOTES FOR MATH 512 11

2.6. Cyclic Extensions. Theorem 2.25. (Hilbert 90) Let K/k be cyclic with G = Gal(K/k) =< σ >. Let β ∈ K. Then N(β) = 1 iff β = α/σα for some α ∈ K. Proof. BY Artin, the map given by n−2 f = id + βσ + β1+σσ2 + ··· + β1+σ+···+σ σn−1 σ is non-zero. So there exist θ ∈ K with α = f(θ) 6= 0. One verifies βα = α. Theorem 2.26. Let k be a field, n > 0 an integer prime to the characteristic of k. Assume that k contains the n-th roots of unity. (1) Let K/k be cyclic of degree n. Then there is α ∈ K such that K = k(α), with α satisfying an equation Xn − a = 0 for some a ∈ k. (2) Conversely, let a ∈ k. Let α be a root of Xn − a. Then k(α) is cyclic of degree d|n, and αd is an element of k. Proof. To prove the first part apply Hilbert’s Theorem 90 with β = ζ−1, where ζ ∈ k is a primitive n-th root of unity. Theorem 2.27. (Hilbert 90, Additive Form) Let K/k be cyclic with G = Gal(K/k) =< σ >. Let β ∈ K. Then T (β) = 0 iff β = α − σα for some α ∈ K. Proof. Suppose T (β) = 0. Choose θ ∈ K with T (θ) 6= 0, and put

1 2 n−1 α = [βθσ + (β + σβ)βσ + ··· + (β + σβ + ··· + σn−2β)θσ ] T (θ) Theorem 2.28. (Artin-Schreier) Let k be a field of characteristic p. (1) Let K be a cyclic ext of degree p. Then there is an α ∈ K such that K = k(α), and α satisfies an equation Xp − X − a = 0 with a ∈ k. (2) Conversely, given a ∈ k, the poly f(X) = Xp − X − a either has one root in k, in which case all its roots are in k, or it is irreducible. In the latter case, if α is a root then k(α) is cyclic of degree p over k. Proof. For (1) we apply Hilbert 90 with β = −1. Then there is an α with σα = α + 1 It follows that [k(α): k] ≥ p and a = αp − α ∈ k, etc. For (2) let α be a root of f. Then α + j is also a root for 1 ≤ j ≤ p. Assume no root lies in k. We claim that f is irreducible. Suppose not, and let f(X) = g(X)h(X) be a non-trivial factorization with d = deg g, 1 ≤ d < p. Since Y f(X) = (X − α − i) we see the coefficient of Xd−1 in g(X) is sum of d terms of the form − α − i = −dα − j 12 JIAN-SHU LI

Since d 6= 0 in k we see α ∈ k. Contradiction. Now we see k(α) is the splitting ﬁeld of f, and there is σ ∈ Gal(k(α)/k) with σ(α) = α + 1, etc

2.7. Solvable and Radical Extensions. Definition 2.29. A (separable) extension E/k is said to be solvable if the following equivalent conditions are satisfied • the Galois group of the normal closure of E is solvable • there is an extension L/k containing E with solvable Galois group. Definition 2.30. A finite extension F/k is said to be solvable by radicals if it is separable, and there is a finite ext E of k containing F , with a tower of extensions

k = E0 ⊆ E1 ⊆ · · · ⊆ Em = E such that each Ej+1/Ej is one of the following types • adjoining a root of unity n • adjoining a root of X − a with a ∈ Ej and n prime to the characteristic • if the characteristic is p > 0, adjoining a root of Xp − X − a. Theorem 2.31. Let E be a separable extension of k. Then E is solvable by radicals if and only if E/k is solvable. Proof. Suppose E/k is solvable. Let K be a ﬁnite Galois ext of k containing E. Let m = product of all primes dividing [K : k] and diﬀerent from the characteristic of k

Let F = k(ζm). It suffices to show that KF/F is solvable by radicals. Now there is a tower of subfields between F and KF , each cyclic of prime order p, necessarily dividing [K : k]. By the two theorems in the previous section, we see that KF/F is solvable by radicals. Conversely, assume E/k is solvable by radicals. Let K be the Galois closure of E in Ea. Then K/k is solvable by radicals because K is the compositum of conjugates of E. Let m be as above, and let F = k(ζm). It suffices to show that KF/F is solvable. But this follows from the definition of “solvable by radicals”.

2.8. Abelian Kummer Theory (Kummer Pairing). Deﬁnition 2.32. Let k be a ﬁeld and m a positive integer. A Galois extension K/k is said to be of exponent m if σm = 1 for all σ ∈ Gal(K/k).

a Consider all extensions in a ﬁxed k . Let µm be the group of m-th roots of unity. Assume • m is prime to the characteristic of k • µm ⊆ k For a ∈ k the symbol k1/m is not well deﬁned, yet k(a1/m) has a clear meaning. Let • B= a subgroup of k× containing (k×)m. 1/m 1/m • KB = k(B ) = compositum of all k(a ). This is Galois over k. LECTURE NOTES FOR MATH 512 13

Let a ∈ B and α an m-th root of a. Then for any σ ∈ Gal(KB/k) one has σα = ωσα for some m-th root of unity ωσ. We may write

ωσ = σα/α

This is independent of the choice of α, and the map σ 7→ ωσ is a homomorphism. Write

ωσ =< σ, a >

This is a bilinear pairing (what do we mean by this before we prove that KB/k is abelian ?).

Gal(KB/k) × B −→ µm which obviously factors through B/(k×)m. Theorem 2.33. In the above setting we have • The left kernel is 1, the right kernel is (k×)m. • KB/k is abelian of exponent m. • We have × m [KB : k] = [B :(k ) ] If this is ﬁnite then B/(k×)m ' Gˆ

Theorem 2.34. The map B 7→ KB gives a bijection between: • subgroups of k× containg (k×)m • ablelian extensions of k of exponent m.

Proof. First we show that the map is injective. Suppose B1 ⊆ B2 then obvi- 1/m ously KB1 ⊆ KB2 . Conversely assume KB1 ⊆ KB2 . Let b ∈ B1. Then k(b ) 1/m is contained in a finite subextension of k(B2 ). We may therefore assume that × m × B2/(k ) is finitely generated, hence finite. Let B3 be the subgroup of k gener- 1/m 1/m ated by b and B2. Then k(B2 ) = k(B3 ). Hence × m × m (B2 :(k ) ) = (B3 :(k ) )

Hence B2 = B3 so b ∈ B2. Hence B1 ⊆ B2. Now we show surjectivity. Assume that K is an abelian extension of k of exponent m. Any ﬁnite subextension of K/k is of exponent m, and is therefore a compositum of cyclic extension of exponent m. But every such cyclic extension can be obtained be obtained by joining an m-th root. Thus K can be obtained by joining a family of m-th roots, say m-th roots of {bj}. Let B be the subgroup of × × m k generated by all bj and (k ) . Additive Theory: Suppose that k has characteristic p > 0. Then in place of x 7→ xm we have the operator p deﬁned by p(x) = xp − x Exercise 2.35. Develope the additive theory Note that p is an additive homomorphism. A root of Xp − X − a is denoted p−1a. If B is a subgroup of k containg pk we let −1 KB = k(p B) 14 JIAN-SHU LI

Theorem 2.36. The map B 7→ k(p−1B) is a bijection from the set of subgroups of k containing pk onto the set of abelian extensions of k of exponent p. Let G = Gal(KB/k). If σ ∈ G and pα = a, let < σ, a >= σα − α. The bilinear map G × B −→ Z/pZ, (σ, a) 7→< σ, a > has left kernel 1 and right kernel pk. We have

[KB : k] = [B : pk]

Proof. Exercise Witt vectors

2.9. Galois Cohomology. • A = abelian group with group law written additively. • G = group acting on A via automorphisms. Thus G −→ Aut(A) • a 1-cocycle is a family {ασ}σ∈G with ασ ∈ A, such that

ασ + σατ = αστ • a 1-coboundary is a family for which there is a β ∈ A such that

ασ = σβ − β We define the first cohomology group by H1(G, A) = Z1(G, A)/B1(G, A) Theorem 2.37. Let K/k be finite Galois, with G = Gal(K/k). Then H1(G, K×) = 1,H1(G, K) = 0 Proof. A multiplicative cocycle satisfies σ ασατ = αστ There is a θ ∈ K such that X β = ατ τ(θ) 6= 0 −1 −1 One checks that σβ = ασ β for all σ ∈ G. Now replace β by β . For the additive part take 1 X β = − α τ(θ) T (θ) τ Example 2.38. (Hilbert’s Theorem 90) Suppose G =< γ > is cyclic of order n. Define X T : A −→ A, T (a) = σ(a)

If {ασ} is a cocycle then by deﬁnition

ασ + σατ = αστ

Taking σ = τ = 1, we see that α1 = 0. Let αγ = a. Taking σ = τ = γ in the γ cocycle relation we ﬁnd αγ2 = a + a . Inductively we ﬁnd

γ γr−1 αγr = a + a + ··· + a , LECTURE NOTES FOR MATH 512 15 and by setting r = n we see that T (a) = 0. Thus the map {ασ} 7→ αγ = a identiﬁes Z1(G, A) with the kernel of T , and the image of B1(G, A) is precisely (γ − 1)A. So we obtain H1(G, A) ' KerT/(γ − 1)A Thus the above Theorem gives Hilbert’s Theorem 90 when G is cyclic. Note that H1(•, •) is a bi-functor, contravariant in the ﬁrst variable, and conva- raiant in the second. Thus in particular, any G-modules homomorphism E −→ E0 gives rise to a homomorphism H1(G, E) −→ H1(G, E0). Lemma 2.39. Let G be a group and E a G-module. Let τ be in the center of G,. Then H1(G, E) is annihilated by the map x 7→ τx − x on E.

Proof. Let {ασ} be a 1-cocycle of G in E. Write f(σ) = ασ. Since τ is in the center of G, we have (using the cocycle condition) τf(σ) − f(σ) = f(τσ) − f(τ) − f(σ) = f(στ) − f(σ) − f(τ) = σf(τ) − f(τ) The right hand side is a coboundary Comments on the rest of this Chapter: The preceeding lemma is used in §11 on Non-abelian Kummer Extension, which we will not discuss. We shall also skip §12, 13. On the other hand §14 has already been discussed. The last section §15 try to set the connection with theory of modular forms. While this a fascinating subject, I don’t feel that we have enough time to give any reasonable introductory treatment. The interested student may try to read [?] or the ﬁrst six chapters of [?]. Suggested Exercises As always, I suggest that you do ALL of the exercises in the book (in this case all the exercise for Ch. VI, on pp. 320-332). The point is that you’ll need to do this to truely understand the material of this chapter, and if you truely understand, then doing all these exercises will not take that much time. However, in case you are still not convinced, I suggest that you at least do the following: p. 320-323: 1-(a),(e), (i),(j),(k),(m). 2-(b),(d),(f), 5. 8, 10, 13, 14 p.323-326: 17, 18, 20, 23, 24, 27, 31,32 p. 326-327: 33, 35, 36 p. 327-330: 38, 41, 44, 45 p. 330-332: 46, 47, 48, 49, 50, 51. •

References 1. S. Gelbart, Automorphic Forms on Adele groups, Annals of Math Studies, Vol. 83, Princeton University Press, (1975). 2. G. Shimura, Introduction to the Arithmetic Theory of Automorphic Forms, Princeton Uni- versity Press (1971)