Local Class Field Theory

Pan Yan

Summer 2015

These are notes for a reading course with D. Wright on Local Class Field Theory in Summer 2015. The notes are prepared and written by Pan Yan ([email protected]). If you notice any mistakes or have any comments, please let me know.

Contents

1 Introduction 3 1.1 Outline of the course ...... 3 1.2 References ...... 3 1.3 Goals of the course ...... 3

2 Group Cohomology 4 2.1 Group rings ...... 4 2.2 G-modules ...... 4 2.3 Group cohomology via cochains ...... 6 2.4 Group cohomology via projective resolutions ...... 11 2.5 Homology ...... 14 2.6 Change of groups ...... 17 2.7 Tate cohomology ...... 23 2.8 Tate cohomology via complete resolutions ...... 29 2.9 Cup products ...... 29 2.10 Tate cohomology of cyclic groups ...... 32 2.11 Cohomological triviality ...... 36 2.12 Tate’s Theorem ...... 40

3 Profinite Groups 45 3.1 Inverse systems and inverse limits ...... 45 3.2 Topological structure of profinite groups ...... 46 3.3 Examples of profinite groups ...... 47 3.4 Direct systems and direct limits ...... 49

1 3.5 Discrete G-modules ...... 50 3.6 Cohomology of proﬁnite groups ...... 51 3.7 Galois cohomology ...... 52

4 Local Class Field Theory 55 4.1 Statements of the main theorems ...... 55 4.2 The fundamental class ...... 57 4.3 The local reciprocity map ...... 65 4.4 Lubin-Tate formal group law ...... 70

2 1 Introduction

1.1 Outline of the course Group Cohomology – Chapter IV of [CF67] (this will take about 2-3 weeks); Proﬁnite Groups – Chapter V of [CF67] (this will take about 2 weeks); Local Class Field Theory – Chapter VI of [CF67] (this will take the rest of the summer semester).

1.2 References The main reference is Algebraic Number Theory by Cassels & Fr¨ohlich [CF67]. Other references include [ANT], [Cas86], [FT91], [LT65], [Mil13], [Neu86], [Ser80], [Sha]. Note: Errata for Cassels & Fr¨ohlich [CF67] can be found at http://wwwf.imperial. ac.uk/~buzzard/errata.pdf (maintained by Kevin Buzzard).

1.3 Goals of the course Class Field Theory is the study of abelian extensions of (local or global) fields. In the case of Local Class Field Theory, we are mainly interested in abelian extensions of nonarchimedean local fields. Specifically, given a nonarchimedean local field K, we want to describe all finite abelian extensions of K totally in terms of the arithmetic of the base field K. To this end, we want to understand group cohomology (especially Tate cohomology for finite groups and Tate’s Theorem), inverse limits and profinite groups, direct limits, cohomology of profinite groups and Galois cohomology, the statements and proofs of Local Reciprocity Law and Local Existence Theorem.

3 2 Group Cohomology

2.1 Group rings Definition 2.1.1. Let G be a group, R a commutative ring with identity. The group ring P R[G] is the set of all formal finite sums g∈G agg with each ag ∈ R, i.e.,   X  R[G] = agg | ag ∈ R, almost all ag = 0 g∈G    X  = agg | ag ∈ R, only finitely many ag 6= 0 . g∈G 

The addition is deﬁned as     X X X  agg +  bgg = (ag + bg)g, g∈G g∈G g∈G and the multiplication is the involution     X X X X  agg  bgg = (agbhgh) = agh−1 bh g. g∈G g∈G g,h∈G g∈G

In this course we are mainly interested in the integral group ring   X  Z[G] = agg | ag ∈ Z, almost all ag = 0 . g∈G 

2.2 G-modules From now on G always means a group, unless otherwise indicated.

Definition 2.2.1. A (left) G-module is an abelian group A together with a G-action on A (i.e., a map G × A → A defined by (g, a) 7→ g · a) such that (i) 1 · a = a, ∀a ∈ A; (ii) g1 · (g2 · a) = (g1g2) · a, ∀a ∈ A, g1, g2 ∈ G; (iii) g · (a1 + a2) = g · a1 + g · a2, ∀a1, a2 ∈ A, g ∈ G. Remark 2.2.2. (i) A right G-module is defined similarly by replacing the above G-action with A × G → A defined by (a, g) 7→ g−1a. (ii) We will assume all G-modules are left G-modules unless otherwise indicated. (iii) G-modules are the same as Z[G]-modules.

4 Deﬁnition 2.2.3. A homomorphism ϕ : A → B of G-modules is a homomorphism of abelian groups such that ϕ(ga) = gϕ(a) for all a ∈ A, g ∈ G (hence it is compatible with the G-action). The group of G-module homomorphisms is denoted as HomG(A, B) = HomZ[G](A, B). A G-module homomorphism ϕ : A → B makes the following diagram commutative.

G−action A > A

ϕ ϕ ∨ G−action ∨ B > B If A, B are G-modules, we denote the group of all abelian group homomorphisms A → B as Hom(A, B). Note that Hom(A, B) actually has a G-module structure: if ϕ ∈ Hom(A, B), we can deﬁne the map

G × Hom(A, B) → Hom(A, B) A → B ! g · (ϕ : A → B) 7→ a 7→ gϕ(g−1a) which makes Hom(A, B) as a G-module.

Deﬁnition 2.2.4. A G-module A is trivial if g · a = a for all a ∈ A, g ∈ G.

Deﬁnition 2.2.5. Let A be a G-module. The group of G-invariants of A, denoted as AG, is AG = {a ∈ A | g · a = a, ∀g ∈ G, a ∈ A}.

Remark 2.2.6. (i) AG is the maximal trivial submodule of A. Indeed, AG is trivial, by deﬁnition. Now suppose B ⊂ A is a trivial submodule of A. Let b ∈ B, for any g ∈ G, we have g · b = b. So b ∈ AG. Hence B ⊂ AG. (ii) If A is a trivial G-module, then AG = A.

If A, B are G-modules, then

HomG(A, B) = {ϕ : A → B | ϕ(ga) = gϕ(a), ∀g ∈ G, a ∈ A}, (Hom(A, B))G = {ϕ : A → B | ϕ(a) = g · ϕ(a) = gϕ(g−1a), ∀g ∈ G, a ∈ A}.

Hence G HomG(A, B) = (Hom(A, B)) . So G ∼ G HomG(Z,A) = (Hom(Z,A)) = A .

5 Since the covariant functor Hom is left exact, it follows that AG is also a covariant left exact functor. More specifically, if 0 → A → B → C is an exact sequence of G-modules, then 0 → AG → BG → CG is an exact sequence of abelian groups. The i-th cohomology group Hi(G, A) of G with coefficients in A can be defined as the i-th derived functor on A of the functor of G-invariants. Alternatively, we give more specific definitions in the following two sections.

2.3 Group cohomology via cochains Definition 2.3.1. Let A be a G-module, and i ≥ 1. (i) The group of i-cochains of G with coefficients in A, denoted as Ci(G, A), is the set of functions from Gi to A, i.e., Ci(G, A) = {ϕ : Gi → A}. i i i i+1 (ii) The i-th differential d = dA : C (G, A) → C (G, A) is the map

i i X j d (ϕ)(g0, g1, ··· , gi) =g0ϕ(g1, ··· , gi) + (−1) ϕ(g0, ··· , gj−2, gj−1gj, ··· , gi) j=1 i+1 + (−1) ϕ(g0, ··· , gi−1).

Remark 2.3.2. C0(G, A) = {ϕ : G0 → A} = {ϕ : {p} → A} =∼ A.

Lemma 2.3.3. For any i ≥ 0, we have di+1 ◦ di = 0. So (Ci(G, A), di) is a cochain complex

d0 d1 0 > C0(G, A) > C1(G, A) > C2(G, A) > ··· .

Definition 2.3.4. Let i ≥ 0. (i) The group Zi(G, A) = ker di is the group of i-cocycles of G with coefficients in A. (ii) The group B0(G, A) = 0,Bi(G, A) = imdi−1 (i ≥ 1) is the group of i-coboundaries of G with coefficients in A.

Remark 2.3.5. Bi(G, A) ⊂ Zi(G, A) since di ◦ di−1 = 0.

Definition 2.3.6. The i-th cohomology group of G with coefficients in A is defined as

Zi(G, A) Hi(G, A) = . Bi(G, A)

Remark 2.3.7. The cohomology groups measure how far the cochain complex (Ci(G, A), di) is from being exact.

Here are some examples of cohomology groups of low degrees.

6 Lemma 2.3.8. (i) H0(G, A) = AG. (ii) Z1(G, A) = {ϕ : G → A | ϕ(gh) = gϕ(h) + ϕ(g), ∀g, h ∈ G}. B1(G, A) = {ϕ : G → A | ∃a ∈ A such that ϕ(g) = ga − a, ∀g ∈ G}.

(Elements in Z1(G, A) are called crossed homomorphism.) (iii) If A is a trivial G-module, then H1(G, A) = Hom(G, A). Proof. (i) H0(G, A) = Z0(G, A)/B0(G, A) = Z0(G, A) since B0(G, A) = 0 by definition. Z0(G, A) = ker d0, where d0 : C0(G, A) = A → C1(G, A) is defined by d0(a)(g) = ga − a. So ker d0 = {a ∈ A | ga − a = 0, ∀g ∈ G} = AG. Hence H0(G, A) = AG. (ii) By definition, B1(G, A) = im d0 = {ϕ : G → A | ∃a ∈ A such that ϕ(g) = ga − a, ∀g ∈ G}. Z1(G, A) = ker d1, where d1 : C1(G, A) → C2(G, A) is defined by d1(ϕ)(g, h) = gϕ(h) − ϕ(gh) + ϕ(g), where (g, h) ∈ G2. Hence Z1(G, A) = {ϕ : G → A | gϕ(h) − ϕ(gh) + ϕ(g) = 0, ∀g, h ∈ G} = {ϕ : G → A | ϕ(gh) = gϕ(h) + ϕ(g), ∀g, h ∈ G}. (iii) If A is a trivial G-module, then Z1(G, A) = {ϕ : G → A | ϕ(gh) = gϕ(h) + ϕ(g), ∀g, h ∈ G} = {ϕ : G → A | ϕ(gh) = ϕ(h) + ϕ(g), ∀g, h ∈ G} = Hom(G, A).

On the other hand, B1(G, A) = 0 since ga − a = 0 for any g ∈ G, a ∈ A. Hence, H1(G, A) = Z1(G, A)/B1(G, A) = Hom(G, A).

We can also compute 2-cocycle and 2-coboundary.

2 1 2 B (G, A) = im d = {ϕ : G → A | ∃ϕ0 : G → A such that

ϕ(g, h) = gϕ0(h) − ϕ0(gh) + ϕ0(g), ∀g, h ∈ G}. d2 : C2(G, A) → C3(G, A) is deﬁned by

2 d (ϕ)(g0, g1, g2) = g0ϕ(g1, g2) − ϕ(g0g1, g2) + ϕ(g0, g1g2) − ϕ(g0, g1). So

2 2 Z (G, A) = {ϕ : G → A | g0ϕ(g1, g2)−ϕ(g0g1, g2)+ϕ(g0, g1g2)−ϕ(g0, g1) = 0, ∀g0, g1, g2 ∈ G}. Such functions in Z2(G, A) are called factor systems.

7 Lemma 2.3.9. If α : A → B is a G-module homomorphism, then for each i ≥ 0, there is an induced homomorphism of groups

αi : Ci(G, A) → Ci(G, B) f 7→ α ◦ f which is compatible with the diﬀerentials, i.e.,

i i i+1 i dB ◦ α = α ◦ dA.

Proof. We only need to check the compatibility.

i i X i dB(α ◦ f)(g0, ··· , gi) =g0 · (α ◦ f)(g1, ··· , gi) + (−1) (α ◦ f)(g0, ··· , gj−1gj, ··· , gi) j=1 i+1 + (−1) (α ◦ f)(g0, ··· , gi−1) i X j =α(g0f(g1, ··· , gi) + (−1) f(g0, ··· , gj−1gj, ··· , gi) j=1 i+1 + (−1) f(g0, ··· , gi−1)) i =α ◦ dA(g0, ··· , gi).

Corollary 2.3.10. If α : A → B is a G-module homomorphism, then there are induced maps α∗ : Hi(G, A) → Hi(G, B) on cohomology. (Here we are omitting the superscripts.)

Actually Ci(G, ·) is an exact functor of G-modules.

Lemma 2.3.11. Suppose

ι π 0 > A > B > C > 0 is a short exact sequence of G-modules. Then the resulting sequence

ιi πi 0 > Ci(G, A) > Ci(G, B) > Ci(G, C) > 0 is exact.

8 i i Proof. First, we prove the injectivity of ι . Suppose ι (f) = ι◦f = 0, then ι(f(g1, ··· , gi)) = i i 0 for any (g1, ··· , gi) ∈ G . So f(g1, ··· , gi) = 0 for any (g1, ··· , gi) ∈ G since ι is injective. Hence f = 0. i i i i Second, we prove that im ι ⊂ ker π . Let f ∈ im ι , then there exists f0 ∈ C (G, A) i i such that f = ι ◦ f0. Then π (f) = π ◦ f = π ◦ (ι ◦ f0) = (π ◦ ι) ◦ f0 = 0. So f ∈ ker π and hence im ιi ⊂ ker πi. i i i i Third, we prove that ker π ⊂ im ι . Let f ∈ ker π ,(g1, ··· , gi) ∈ G . Then π ◦ f(g1, ··· , gi) = 0. So f(g1, ··· , gi) ∈ ker π = im ι. So there exists a ∈ A such that i f(g1, ··· , gi) = ι(a). For every (g1, ··· , gi) ∈ G and the corresponding a ∈ A, deﬁne i f0 ∈ C (G, A) such that f0(g1, ··· , gi) = a. Then

f(g1, ··· , gi) = ι(a) = ι ◦ f0(g1, ··· , gi)

i i and hence f = ι ◦ f0 = ι (f0). Hence f ∈ im ι . i i i Finally, we prove the surjectivity of π . Let f ∈ C (G, C), (g1, ··· , gi) ∈ G . Then f(g1, ··· , gi) ∈ C. So there exists b ∈ B such that f(g1, ··· , gi) = π(b) by the surjectivity i i of π. For every (g1, ··· , gi) ∈ G and the corresponding b ∈ B, deﬁne f0 ∈ C (G, B) such that f0(g1, ··· , gi) = b. Then

f(g1, ··· , gi) = π(b) = π ◦ f0(g1, ··· , gi)

i and hence f = π ◦ f0 = π (f0). Here is the main theorem of this section.

Theorem 2.3.12. Suppose that

ι π 0 > A > B > C > 0 is a short exact sequence of G-modules. Then there exists a long exact sequence of abelian groups

ι∗ π∗ δ0 0 > H0(G, A) > H0(G, B) > H0(G, C) > H1(G, A) > ··· .

Proof. Consider the following diagram with exact rows

ι π 0 > Ci(G, A) > Ci(G, B) > Ci(G, C) > 0

di di di ∨ A ∨ B ∨ C ι π 0 > Ci+1(G, A) > Ci+1(G, B) > Ci+1(G, C) > 0

9 for i ≥ 0. We claim that the following diagram is commutative with exact rows.

i i i C (G, A) ι C (G, B) π C (G, C) > > > 0 Bi(G, A) Bi(G, B) Bi(G, C) di di di ∨ A ∨ B ∨ C ι π 0 > Zi+1(G, A) > Zi+1(G, B) > Zi+1(G, C)

Here is the proof of the claim. First, the maps ι, π in the upper row are well-deﬁned. For example,

Ci(G, A) Ci(G, B) →ι Bi(G, A) Bi(G, B) f + Bi(G, A) 7→ ι ◦ f + Bi(G, B).

i Ci(G,B) i Suppose we have g + B (G, B) ∈ Bi(G,B) , then g ∈ C (G, C), and so there exists f ∈ Ci(G, B) such that π ◦f = g. Hence π ◦(f +Bi(G, B)) = π ◦f +Bi(G, C) = g +Bi(G, C). Hence the map π in the upper row is surjective. Suppose we have g + Bi(G, B) ∈ im ι, i Ci(G,A) i i then there exists f + B (G, A) ∈ Bi(G,A) such that ι(f + B (G, A)) = ι ◦ f + B (G, B) = g+Bi(G, B). Hence π(g+Bi(G, B)) = π◦ι(f+Bi(G, A)) = Bi(G, C) and so g+Bi(G, C) ∈ ker π. Hence im ι ⊂ ker π. Second, the maps in the lower row are also well-deﬁned. For example,

Zi+1(G, A) →ι Zi+1(G, B) f 7→ ι ◦ f.

i+1 i+1 i+1 i+1 Suppose f ∈ Z (G, A) = ker dA , then f ∈ C (G, A), So ι(f) ∈ C (G, B). Since i+1 i+1 i+1 i+1 dB (ι(f)) = ι ◦ dA (f) = 0 and so ι ◦ f ∈ ker dB = Z (G, B). Thus the map ι is well-deﬁned. Similarly, π is also well-deﬁned. Suppose f ∈ ker ι. Then f ∈ Ci+1(G, A) i+1 and ι(f)(g1, ··· , gi+1) = ι(f(g1, ··· , gi+1)) = 0 for any (g1, ··· , gi+1) ∈ G . Hence i+1 f(g1, ··· , gi+1) = 0 for any (g1, ··· , gi+1) ∈ G since ι : A → B is injective. So ι in the lower row is injective. Now suppose g ∈ im ι. Then there exists f ∈ Zi+1(G, A) such that g = ι(f). Then π(g) = π ◦ ι(f) = (π ◦ ι)(f) = 0. So f ∈ ker π. Therefore, im ι ⊂ ker π. Third, we need to check the column maps make sense. This is because

i i i+1 i+1 dA(C (G, A)) ⊂ ker dA = Z (G, A), and i i i i−1 dA(B (G, A)) = dA(im dA ) = 0. i Ci(G,A) Finally, we need to check the diagram is commutative. Let f + B (G, A) ∈ Bi(G,A) . Then i i i i i dB ◦ ι f + B (G, A) = dB ι ◦ f + B (G, B) = dB (ι(f)) ,

10 i i i ι ◦ dA f + B (G, A) = ι dA(f) . i i dB ◦ ι = ι ◦ dA by Lemma 2.3.9. Hence the diagram is commutative. So we have proved the claim. Now apply the Snake Lemma to get the exact sequence

i i i i i i ker dA → ker dB → ker dC → cokerdA → cokerdB → cokerdC for all i ≥ 0. Note that Zi(G, A) ker di = = Hi(G, A), A Bi(G, A) i+1 i+1 i Z (G, A) Z (G, A) i+1 cokerdA = i = i+1 = H (G, A). im dA B (G, A) The exactness of 0 → H0(G, A) → H0(G, B) a 7→ ι(a) is obvious. So we get the long exact sequence

ι∗ π∗ δ0 0 > H0(G, A) > H0(G, B) > H0(G, C) > H1(G, A) > ··· .

2.4 Group cohomology via projective resolutions The cohomology groups deﬁned in the last section can also be deﬁned in terms of projective resolutions. i+1 i+1 For i ≥ 0, let G denote the direct product of i + 1 copies of G. Z[G ] can be viewed as a G-module with the left action

g · (g0, g1, ··· , gi) = (gg0, gg1, ··· , ggi).

Deﬁnition 2.4.1. The standard resolution of Z by G-modules is a sequence of G-module homomorphisms

i+1 di−1 i 2 d0 ··· > Z[G ] > Z[G ] > ··· > Z[G ] > Z[G] > Z > 0 where : Z[G] → Z is the augmentation map deﬁned by   X X  agg = ag g∈G g∈G

11 and

i+2 i+1 di : Z[G ] → Z[G ] i+1 X j (g0, ··· , gi+1) 7→ (−1) (g0, ··· , gj−1, gj+1, ··· , gi+1) j=0 for each i ≥ 0.

i Remark 2.4.2. We may use (g0, ··· , gˆj, ··· , gi) ∈ G to denote the i-tuple excluding gj. Lemma 2.4.3. The standard resolution is exact.

Proof. Let d−1 = ε. For each i ≥ 0, we have

i+1 X j+k−s(j,k) di−1 ◦ di(g0, ··· , gi+1) = (−1) (g0, ··· , gˆj, ··· , gˆk, ··· , gi+1), k=0,k6=j where s(j, k) = 0 if k < j, s(j, k) = 1 if k > j. Each possible i-tuple appears twice in the term with opposite sign. Therefore, di−1 ◦ di = 0. Next, deﬁne

i+1 i+2 θi : Z[G ] → Z[G ] (g0, ··· , gi) 7→ (1, g0, ··· , gi).

Then

i X i di ◦ θi(g0, ··· , gi) = (g0, ··· , gi) − (−1) · (1, g0, ··· , gˆj, ··· , gi) j=0

= (g0, ··· , gi) − θi−1 ◦ di−1(g0, ··· , gi).

di ◦ θi + θi−1 ◦ di−1 = idZ[Gi+1].

If α ∈ ker di−1, then di(θi(α)) = α and so α ∈ im di. Hence ker di−1 ⊂ im di. To conclude, ker di−1 = im di. We want to consider the following complex

i i+1 D i+2 0 > HomG(Z[G],A) > ··· > HomG(Z[G ],A) > HomG(Z[G ],A) > ···

i i where D = DA is deﬁned by i D (ϕ) = ϕ ◦ di.

12 Theorem 2.4.4. The maps

i i+1 i ψ : HomG(Z[G ],A) → C (G, A) i ψ (ϕ)(g1, ··· , gi) = ϕ(1, g1, g1g2, ··· , g1g2 ··· gi) are isomorphisms for all i ≥ 0. Moreover, we have isomorphisms of complexes via

ψi+1 ◦ Di = di ◦ ψi.

Proof. First, we prove that ψi is injective. Suppose ψi(ϕ) = 0, then

ϕ(1, g1, g1g2, ··· , g1g2 ··· gi) = 0

−1 for any g1, ··· , gi ∈ G. Let h0, ··· , hi ∈ G and deﬁne gj = hj−1hj for all 1 ≤ j ≤ i. Then

−1 −1 ϕ(h0, h1, ··· , hi) = h0ϕ(1, h0 h1, ··· , h0 hi) = h0 · ϕ(1, g1, ··· , g1 ··· gi) = 0.

Hence ψi is injective. Second, we prove that ψi is also surjective. If f ∈ Ci(G, A), deﬁne

−1 −1 ϕ(h0, h1, ··· , hi) = h0f(h0 h1, ··· , hi−1hi).

Then

−1 −1 ϕ(gh0, gh1, ··· , ghi) = gh0f((gh0) gh1, ··· , (ghi−1) ghi) −1 −1 = gh0f(h0 h1, ··· , hi−1hi) = gϕ(h0, h1, ··· , hi) and hence

i ψ (ϕ)(h1, ··· , hi) = ϕ(1, h1, h1h2, ··· , h1h2 ··· hi) = f(h1, ··· , hi).

So ψi is surjective. Therefore, ψi is an isomorphism.

13 i+1 i+1 Let ϕ ∈ HomG(Z[G ],A), (g1, ··· , gi+1) ∈ G . Then

i+1 i ψ (D (ϕ))(g1, ··· , gi+1) i =D (ϕ)(1, g1, ··· , g1 ··· gi+1)

=ψ ◦ di(1, g1, ··· , g1 ··· gi+1) i+1 X j = (−1) ϕ(1, g1, ··· , g1\··· gj, ··· , g1 ··· gi+1) j=0 i X j =ϕ(g1, g1g2, ··· , g1 ··· gi+1) + (−1) ϕ(1, ··· , g1\··· gj, ··· , g1 ··· gi+1) j=1 i+1 + (−1) ϕ(1, g1, ··· , g1 ··· gi+1) i i X j i =g1ψ (ϕ)(g2, ··· , gi+1) + (−1) ψ (ϕ)(g1, ··· , gj−2, gj−1gj, gj+1, ··· , gi+1) j=1 i+1 i + (−1) ψ (ϕ)(g1, ··· , gi) i i =d (ψ (ϕ))(g1, ··· , gi+1).

i+1 i Corollary 2.4.5. The i-th cohomology group of the complex (HomG(Z[G ],A),DA) is isomorphic to Hi(G, A).

Remark 2.4.6. Actually the standard resolution is a projective resolution, this is because i+1 ∼ L [G ] i [G](1, g , ··· , g ) is free and the fact that every free module is Z = (g1,··· ,gi)∈G Z 1 i projective.

2.5 Homology

Let A, B be G-modules. Let A ⊗ B denote their tensor product over Z, and A ⊗G B = A⊗Z[G]B denote their tensor product over Z[G]. Note that A⊗B has a G-module structure via the action g(a ⊗ b) = (ga) ⊗ (gb).

Deﬁnition 2.5.1. The augmentation map is the homomorphism deﬁned by

: Z[G] → Z X X agg 7→ ag. g g

The augmentation ideal is IG = ker .

Lemma 2.5.2. IG is equal to the ideal of Z[G] generated by {g − 1 | g ∈ G}.

14 P P Proof. Clearly, g − 1 ∈ ker for any g ∈ G. Conversely, if g∈G ag = 0, then g∈G agg = P g∈G ag(g − 1). So ker is contained in the ideal generated by {g − 1 | g ∈ G}.

Deﬁnition 2.5.3. The group of G-coinvariants of A, denoted as AG, is

AG = A/IGA.

We have an exact sequence

ι 0 > IG > Z[G] > Z > 0. Since tensor product is right exact,

ι⊗id ⊗id IG ⊗G A > Z[G] ⊗G A > Z ⊗G A > 0 is exact. Hence,

∼ Z[G] ⊗G A ∼ A ∼ A ∼ A Z ⊗G A = im( ⊗ id) = = = = = AG. ker( ⊗ id) im(ι ⊗ id) IG ⊗G A IGA

Definition 2.5.4. The i-th homology group Hi(G, A) of G with coefficients in A is defined to be the i-th homology group of the sequence

3 d1 2 d0 d−1 ··· > Z[G ] ⊗G A > Z[G ] ⊗G A > Z[G] ⊗G A > 0 induced by the standard resolution. More speciﬁcally,

Zi(G, A) Hi(G, A) = , Bi(G, A) where Zi(G, A) = ker di−1 is the i-cycle of G with coeﬃcients in A, and Bi(G, A) = im di is the i-boundary of G with coeﬃcients in A.

Lemma 2.5.5. H0(G, A) = AG for any G-module A. Proof. Well, ∼ Z0(G, A) = ker d−1 = Z[G] ⊗G A = A.

The map d0 is deﬁned by d0 ((g0, g1) ⊗ a) = (g1 − g0)a. So B0(G, A) = im d0 = IGA. Hence, Z0(G, A) ∼ A H0(G, A) = = = AG. B0(G, A) IGA

15 If α : A → B is a G-module homomorphism, then there are induced maps α∗ : Hi(G, A) → Hi(G, B) on homology groups for each i ≥ 0. Theorem 2.5.6. Suppose that

ι π 0 > A > B > C > 0 is a short exact sequence of G-modules. Then there exists a long exact sequence of abelian groups

δ∗ ι∗ π∗ ··· > H1(G, C) > H0(G, A) > H0(G, B) > H0(G, C) > 0 where δ∗ : Hi(G, C) → Hi−1(G, A) is a connecting homomorphism. Proposition 2.5.7. Let G be a group, A be a G-module. Let [G, G] be the commutator subgroup of G so that Gab = G/[G, G] is the largest abelian quotient of G. Then ∼ ab H1(G, Z) = G .

Proof. Consider the exact sequence

0 > IG > Z[G] > Z > 0 of G-modules. First, we will show that H1(G, Z[G]) = 0. Suppose that (g0, g1) ⊗ g2 ∈ ker d0 = Z1(G, Z[G]) and g2 6= 0. Then

d0 ((g0, g1) ⊗ g2) = (g1 − g0) ⊗ g2 = 0 and hence g0 = g1. So

d1 ((x, g1, y) ⊗ g2) = ((g1, y) − (x, y) + (x, g1)) ⊗ g2 = (g1, g1) ⊗ g2 and hence (g0, g1) ⊗ g2 ∈ imd1 = B1(G, Z[G]). Hence H1(G, Z[G]) = 0. By Theorem 2.5.6, we have the following exact sequence

IG Z[G] 0 > H1(G, Z) > 2 > > H0(G, Z) > 0. IG IG

IG Z[G] Note that the map 2 → I is induced by IG → Z[G], hence is zero. Therefore, we have IG G

∼ IG H1(G, Z) = 2 . IG 2 Note that IG = IG · IG is the submodule of Z[G] generated by elements of the form (g − 1)(g0 − 1), g, g0 ∈ G.

16 The map

G IG → 2 [G, G] IG 2 g 7→ (g − 1) + IG is an isomorphism. Hence ∼ ab H1(G, Z) = G .

2.6 Change of groups Let H ⊂ G be a subgroup. Suppose B is a H-module. Then we can construct a G-module HomH (Z[G],B) by the G-action

(g · ϕ)(α) = ϕ(α · g).

This type of G-modules are coinduced from H to G.

Theorem 2.6.1 (Shapiro’s Lemma). For all i ≥ 0, we have

i ∼ i H (G, HomH (Z[G],B)) = H (H,B).

Proof. Let P be the standard resolution of Z by G-modules. Deﬁne the map

ψi : HomG (Pi, HomH (Z[G],B)) → HomH (Pi,B) ψi(θ)(x) 7→ θ(x)(1).

If θ ∈ ker ψi, then θ(x)(g) = (g · θ(x))(1) = θ(gx)(1) = 0 for all x ∈ Pi, g ∈ G. Hence θ = 0. So ψi is injective. Conversely, if ϕ ∈ HomH (Pi,B), then deﬁne θ by θ(x)(g) = ϕ(gx), then we have ψi(θ) = ϕ. So ψi is surjective. Hence ψi is an isomorphism. Then the result follows.

Remark 2.6.2. There is an analogous result for homology. We can form Z[G] ⊗H B as a G-module via the G-action g · (α ⊗ b) = (gα) ⊗ b. This type of G-modules are induced from H to G. Then we have ∼ Hi(G, Z[G] ⊗H B) = Hi(H,B) for any i ≥ 0.

17 Deﬁnition 2.6.3. We say that G-modules of the form

Z[G] ⊗Z X and Hom(Z[G],X) where X is an abelian group, are induced and coinduced G-modules, respectively.

Theorem 2.6.4. Suppose that A is a coinduce (resp. induced) G-module. Then we have i H (G, A) = 0 (resp. Hi(G, A) = 0) for all i ≥ 1.

Proof. Let X be an abelian group such that A = Hom(Z[G],X). By Theorem 2.6.1, we have

i i i ∼ i H (G, A) = H (G, Hom(Z[G],X)) = H (G, Hom{1}(Z[G],X)) = H ({1},X)

i for i ≥ 1. Since Z has a projective resolution of Z by itself, it follows that H ({1},X) = 0 for i ≥ 1. (Here is another way to show that. Suppose ϕ : {1} → X is a 1-cocycle, then ϕ(1) = ϕ(1 · 1) = ϕ(1) + ϕ(1) and hence ϕ(1) = 0. Let x ∈ X, then ϕ(1) = 1 · x − x. So ϕ is also a 1-coboundary. Hence H1({1},X) = 0. Now by the long exact sequence of i cohomology groups, H ({1},X) = 0 for all i ≥ 1.) Similarly, Hi(G, A) = 0 for i ≥ 1. Suppose f : G0 → G is a homomorphism of groups, we can regard a G-module A as a G0-module via the map f under the G0-action (g0, a) 7→ f(g0)a. There is an induced 0 homomorphism P → P of the standard resolution of Z, and hence homomorphisms of the cohomology groups f ∗ : Hi(G, A) → Hi(G0,A) for any G-module A, i ≥ 0. In particular, if G0 = H is a subgroup of G, and f : H → G is the embedding, then we have the restriction homomorphisms

Res : Hi(G, A) → Hi(H,A).

If H is a normal subgroup of G, consider f : G → G/H. For any G-module A, AH is a G/H-module via the G/H-action

G/H × AH → AH (g + H, a) 7→ (g + 1)a.

Hence we have the homomorphism

(2.1) Hi(G/H, AH ) → Hi(G, AH ).

The inﬂation homomorphisms are the composition of (2.1) with the homomorphism induced by AH → A, i.e., Inf : Hi(G/H, AH ) → Hi(G, A).

18 Remark 2.6.5. Similarly, for homology, if f : G0 → G is a homomorphism of groups, then we have homomorphisms of homology groups

0 f∗ : Hi(G ,A) → Hi(G, A) for any G-module A and i ≥ 0. In particular, if G0 = H is a subgroup of G, and f : H → G is the embedding, then we have the corestriction homomorphisms

Cor : Hi(H,A) → Hi(G, A).

Now back to cohomology, and consider the case G0 = G, f : G → G being the inner automorphism s 7→ tst−1. This turns A into another G-module, denoted as At, and gives the homomorphism

(2.2) Hi(G, A) → Hi(G, At).

Now the map ϕ : At → A deﬁned by a 7→ t−1a is an isomorphism, and hence gives the homomorphism

(2.3) Hi(G, At) → Hi(G, A).

Proposition 2.6.6. The composition of (2.2) with (2.3) is the identity map on Hi(G, A).

Before proving Proposition 2.6.6, we need a useful technique in group cohomology – dimension shifting. Let A be a G-module, and let G act on Hom(Z[G],A) by (g · ϕ)(a) = gϕ(g−1a).

We have an exact sequence

ι π ∗ (2.4) 0 → A → Hom(Z[G],A) → A → 0

∗ where ι is deﬁned by ι(a)(g) = a, ∀a ∈ A, g ∈ G, and A = coker(ι) = Hom(Z[G],A)/A. Then we have

Theorem 2.6.7 (Dimension Shifting). For any i ≥ 1, we have

Hi+1(G, A) =∼ Hi(G, A∗).

Proof. The exactness of (2.4) gives rise to the long exact sequence

i i ∗ δi i+1 ι∗ i+1 · · · → H (G, Hom(Z[G],A)) → H (G, A ) → H (G, A) → H (G, Hom(Z[G],A)) → · · ·

19 i for i ≥ 0. Since Hom(Z[G],A) is coinduced, H (G, Hom(Z[G],A)) = 0 for all i ≥ 1, by Theorem 2.6.4. So i 0 → Hi(G, A∗) →δ Hi+1(G, A) → 0 is exact, hence δi is both surjective and injective for all i ≥ 1. Therefore, δi is an isomorphism for all i ≥ 1 and hence

Hi(G, A∗) =∼ Hi+1(G, A) for all i ≥ 1.

Remark 2.6.8. There is an analogous result for homology. We regard Z[G] ⊗ A as a G-module via the G-action g · (α ⊗ a) = gα ⊗ ga. We have a short exact sequence

π 0 → A∗ → Z[G] ⊗ A → A → 0 where A∗ = ker π. Then we have ∼ Hi+1(G, A) = Hi(G, A∗) for any i ≥ 1.

Now we are ready to prove Proposition 2.6.6.

Proof of Proposition 2.6.6. We ﬁrst verify the case i = 0, then use dimension shifting to prove it by induction. For i = 0, we have

H0(G, At) = (At)G = {a ∈ At | g · a = a, ∀g ∈ G} = {a ∈ A | tgt−1a = a, ∀g ∈ G} = {a ∈ A | gt−1a = t−1a, ∀g ∈ G} = t · AG.

So (2.2) is just multiplication by t, and (2.3) is multiplication by t−1. Hence the composition is the identity. This proved the case i = 0. Now assume i ≥ 1 and the result holds for i. By the Dimension Shifting, we have the following commutative diagram

δ Hi(G, A∗) > Hi+1(G, A)

σ σ i∨ i+1∨ δ Hi(G, A∗) > Hi+1(G, A)

20 where σi, σi+1 are the composition of (2.2) with (2.3), δ is an isomorphism, σi is the i+1 identity map by the inductive hypothesis. Hence σi+1 is the identity map on H (G, A). This completes the proof.

Proposition 2.6.9 (The Restriction-Inﬂation Sequence). Let G be a group, A a G-module and H a normal subgroup of G. Then

Inf 0 → H1(G/H, AH ) → H1(G, A) →Res H1(H,A) is exact. Proof. Suppose f : G/H → AH ∈ Z1(G/H, AH ) is a 1-cocycle, then it induces a map Inf (f): G → A via f G → G/H → AH ,→ A.

Let g0, g1 ∈ G. Then

Inf (f)(g0g1) =f(g0g1H) = f((g0H)(g1H))

=(g0H)f(g1H) + f(g0H) H =g0f(g1H) + f(g0H) ( since f(g1H) ∈ A )

=g0Inf (f)(g1) + Inf (f)(g0).

So Inf (f) is an crossed homomorphism and hence Inf (f) ∈ Z1(G, A). Suppose f : G/H → AH ∈ B1(G/H, AH ) is a 1-coboundary, then there exists some a ∈ AH such that f(gH) = gH·a−a = ga−a for all g ∈ G. So Inf (f)(g) = f(gH) = ga−a for all g ∈ G. Hence Inf (f) ∈ B1(G, A). Suppose f : G → A ∈ Z1(G, A) is a 1-cocycle, then

Res (f)(h0h1) =f(h0h1)

=h0f(h1) + f(h0)

=h0Res (f)(h1) + Res (f)(h1)

1 for any h0, h1 ∈ H. So Res (f) ∈ Z (H,A). Suppose f : G → A ∈ B1(G, A) is a 1-coboundary, then there exists some a ∈ A such that Res (f)(h) = f(h) = ha − a for all h ∈ H. Hence Res (f) ∈ B1(H,A). Now we verify the exactness. Suppose Inf (f) ∈ B1(G, A), we need to show that f ∈ B1(G/H, AH ). There exists some a ∈ A such that Inf (f)(g) = ga − a = f(gH) for all g ∈ G. In particular, f(H) = Ha−a = 0 and hence a ∈ AH . Therefore, f ∈ B1(G/H, AH ). This proves that the inﬂation is injective. Suppose f : G/H → AH ∈ B1(G/H, AH ) is a 1-coboundary. Then for any h ∈ H, we have Res ◦ Inf (f)(h) = Res (f(H)) = Res (0) = 0. Hence Res ◦ Inf = 0.

21 Proposition 2.6.10. Let G be a group, A a G-module and H a normal subgroup of G. Let i ≥ 1 and suppose that Hj(H,A) = 0 for all 1 ≤ j ≤ i − 1. Then the sequence

Inf 0 → Hi(G/H, AH ) → Hi(G, A) →Res Hi(H,A) is exact. Proof. We argue by induction via dimension shifting. We already proved the case i = 1. Now suppose i ≥ 2 and suppose that the statement is true for i − 1. We have the exact sequence of G-modules

∗ 0 > A > Hom(Z[G],A) > A > 0

∗ where A = Hom(Z[G],A)/A. It is also an exact sequence of H-modules. So have have an exact sequence

0 0 0 ∗ 1 0 > H (H,A) > H (H, Hom(Z[G],A)) > H (H,A ) > H (H,A) in H-cohomology by Theorem 2.3.12. Since H1(H,A) = 0 and by Lemma 2.3.8, we have a short exact sequence

H H ∗ H 0 > A > (Hom(Z[G],A)) > (A ) > 0 in H-cohomology. Moreover, H ∼ ∼ H (Hom(Z[G],A)) = Hom(Z[G/H],A) = Hom(Z[G/H],A ) with the trivial H-action. Thus the connecting homomorphism

δi−1 : Hi−1(G/H, (A∗)H ) → Hi(G/H, AH ) is an isomorphism for i ≥ 2. Now consider the following commutative diagram

Inf Res 0 > Hi−1(G/H, (A∗)H ) > Hi−1(G, A∗) > Hi−1(H,A∗)

δi−1 δi−1 δi−1 ∨ ∨ ∨ Inf Res 0 > Hi(G/H, AH ) > Hi(G, A) > Hi(H,A) with vertical maps isomorphisms, and top row exact by inductive hypothesis. Hence the button row is also exact. This completes the proof.

Corollary 2.6.11. Let G be a group, A a G-module and H a normal subgroup of G. Let i ≥ 1 and suppose that Hj(H,A) = 0 for all 1 ≤ j ≤ i − 1. Then

Hj(G/H, AH ) =∼ Hj(G, A) for 1 ≤ j ≤ i − 1.

22 2.7 Tate cohomology

In this section we assume that G is a finite group. The norm element of Z[G] is defined P as NG = g∈G g, which defines a map

N = NA = NG,A : A → A X a 7→ g · a g∈G by left multiplication on any G-module A. The image X im (N) = N(A) = { g · a | a ∈ A} g∈G is the group of G-norms of A. ∗ G Lemma 2.7.1. The norm element induces a map N : AG → A .

Proof. For any g0 ∈ G, a ∈ A, we have X X X N((g0 − 1)a) = g · (g0 − 1)a = gg0a − ga = 0 g∈G g∈G g∈G and hence IGA ⊂ ker N. Moreover, X X X g0 · N(a) = g0 · ga = g0ga = ga = N(a) g∈G g∈G g∈G and hence N(A) ⊂ AG. Deﬁnition 2.7.2. We deﬁne AG Hˆ 0(G, A) = coker(N ∗) = , N(A)

∗ ker(N) Hˆ0(G, A) = ker(N ) = . IGA Since G is a ﬁnite group, the map Hom(Z[G],X) → Z[G] ⊗ X X ϕ 7→ g ⊗ ϕ(g) g∈G with its inverse Z[G] ⊗ X → Hom(Z[G],X) X g ⊗ xg 7→ (ϕ : g 7→ xg) g∈G where X is an abelian group, make these two G-modules an isomorphism. Hence the notion of induced and coinduced G-modules coincide for a ﬁnite group G.

23 Proposition 2.7.3. Let G be a ﬁnite group and A an induced G-module. Then Hˆ 0(G, A) = Hˆ0(G, A) = 0.

Proof. Let X be an abelian group such that A = Z[G] ⊗ X. Each element of A is uniquely P P G of the form g∈G g ⊗ xg where xg ∈ X. Suppose a = g∈G g ⊗ xg ∈ A , then X X X g0 · a = g0g ⊗ g0xg = g0g ⊗ xg = a = g ⊗ xg g∈G g∈G g∈G P for any g0 ∈ G. Hence all the xg are equal. Hence a = g∈G g ⊗ x = N(1 ⊗ x) and that a ∈ N(A). Therefore, AG = N(A) and hence Hˆ 0(G, A) = 0. P P Now suppose a = g∈G g ⊗ xg ∈ ker(N), then N g∈G g ⊗ xg = 0. Hence P g∈G xg = 0. So X X a = g ⊗ xg = (g − 1)(1 ⊗ xg) ∈ IGA. g∈G g∈G

So ker(N) ⊂ IGA and hence Hˆ0(G, A) = 0.

Definition 2.7.4. Let G be a finite group and A a G-module. Let i ∈ Z. We define the i-th Tate cohomology group to be

 i H (G, A), i ≥ 1,  Hˆ 0(G, A), i = 0, Hˆ i(G, A) = Hˆ (G, A), i = −1,  0  H−i−1(G, A), i ≤ −2.

Theorem 2.7.5. Let G be a finite group and A an induced G-module. Then Hˆ i(G, A) = 0 for all i ∈ Z. i Proof. By Theorem 2.6.4, H (G, A) = Hi(G, A) = 0 for all i ≥ 1. It suffices to check 0 Hˆ (G, A) and Hˆ0(G, A), which are proved in Proposition 2.7.3. Theorem 2.7.6 (Tate). Let G be a finite group. Suppose

ι π 0 > A > B > C > 0 is a short exact sequence of G-modules. Then there exists a long exact sequence of abelian groups

ι∗ π∗ δi ··· > Hˆ i(G, A) > Hˆ i(G, B) > Hˆ i(G, C) > Hˆ i+1(G, A) > ··· extending inﬁnitely in both directions.

24 Proof. Apply the Snake Lemma to the commutative diagram

··· > H1(G, C) > H0(G, A) > H0(G, B) > H0(G, C) > 0

NA NB NC ∨ ∨ ∨ ∨ ∨ 0 > H0(G, A) > H0(G, B) > H0(G, C) > H1(G, A) > ··· .

We also have dimension shifting for Tate cohomology groups. Actually the Tate cohomology groups can be shifted both up and down.

Theorem 2.7.7 (Dimension Shifting). Let G be a ﬁnite group and A a G-module. Then

i+1 ∼ i ∗ i−1 ∼ i Hˆ (G, A) = Hˆ (G, A ) and Hˆ (G, A) = Hˆ (G, A∗)

∗ for any i ∈ Z, where A = Hom(Z[G],A)/A, A∗ is the kernel of the map π : Z[G]⊗A → A deﬁned by π(g ⊗ a) = a for a ∈ A, g ∈ G.

Proof. This follows from Theorem 2.6.7 and Remark 2.6.8.

Let H be a subgroup of G. Recall that we have deﬁned the restriction homomorphisms

Res : Hi(G, A) → Hi(H,A) for all i ≥ 0. Hence it is deﬁned for the Tate cohomology groups Hˆ i(G, A) for all i ≥ 1. By dimension shifting via the following commutative diagram

Res Hˆ i−1(G, A) > Hˆ i−1(H,A)

∨ ∨ i Res i Hˆ (G, A∗) > Hˆ (H,A∗)

i it is deﬁned for Hˆ (G, A) for all i ∈ Z. Similarly, we have deﬁned the corestriction homomorphisms

Cor : Hi(H,A) → Hi(G, A) for all i ≥ 0. It is deﬁned for all the Tate cohomology groups Hˆ i(G, A) by dimension shifting.

Deﬁnition 2.7.8. We deﬁne

Res : H0(G, A) → H0(H,A) X a 7→ g−1 · a,˜ g∈G/H

25 H where a ∈ AG,a ˜ ∈ A is any lift of a, g is any coset representative of g ∈ G/H. We deﬁne

Cor : H0(H,A) → H0(G, A) X a 7→ g · a. g∈G/H

Proposition 2.7.9. Let G be a ﬁnite group, H a subgroup of G. There are maps

Res : Hi(G, A) → Hi(H,A) and Cor : Hi(H,A) → Hi(G, A) for all i ≥ 0 that provide morphisms of δ-functors.

Proof. We consider only the case of restriction since the case of corestriction is very similar. The exactness of

π 0 > A∗ > Z[G] ⊗ A > A > 0 where A∗ = ker π, gives rise to the following commutative diagram with exact rows

0 > H1(G, A) > H0(G, A∗) > H0(G, Z[G] ⊗ A) . . . Res Res Res ∨. ∨ ∨ 0 > H1(H,A) > H0(H,A∗) > H0(H, Z[G] ⊗ A).

The restriction Res : H0(G, Z[G] ⊗ A)) → H0(H, Z[G] ⊗ A)) is induced from Res : H0(G, A) → H0(H,A), hence it induces the restriction Res : H0(G, A∗) → H0(H,A∗) since A∗ = ker π ⊂ Z[G] ⊗ A. By the above diagram we are able to deﬁne Res : H1(G, A) → H1(H,A). Hence we can deﬁne Res : Hi(G, A) → Hi(H,A) for all i ≥ 2 as well via dimension shifting.

Corollary 2.7.10. Let G be a ﬁnite group, H a subgroup of G. There are maps

Res : Hˆ i(G, A) → Hˆ i(H,A) and Cor : Hˆ i(H,A) → Hˆ i(G, A) for all i ∈ Z that provide morphisms of δ-functors.

26 Proof. We only need to check the commutativity of one diagram in each case, and we check the case for restriction since the case for corestriction is very similar. Speciﬁcally, suppose

ι π 0 > A > B > C > 0 is an exact sequence of G-modules, then we need to check that the diagram

δ Hˆ −1(G, C) > Hˆ 0(G, A)

Res Res ∨ ∨ δ Hˆ −1(H,C) > Hˆ 0(H,A) is commutative. −1 ker(NG) Let c ∈ ker(NG : C → C) ⊂ C, and denote c to be its image in Hˆ (G, C) = . IGC Choose b ∈ B such that π(b) = c and consider N(b) ∈ BG. Note that X X X π(NG(b)) = π g · b = g · π(b) = g · c = 0, g∈G g∈G g∈G

G G G G G so NG(b) ∈ ker(π : B → C ) = im(ι : A → B ). So there exists some a ∈ A such 0 that ι(a) = NG(b). Then δ(c) is the image of a in Hˆ (G, A). Then Res (δ(c)) is the image of a in Hˆ 0(H,A). On the other hand, X Res(c) = g−1 · c,˜ g∈G/H

ˆ −1 P −1 wherec ˜ is the image of c in H (H,C). We lift c to g∈G/H g · c in the kernel of NH P −1 on C, then lift it to g∈G/H g · b ∈ B. Note that   X −1 NH  g · b = NG(b) = ι(a), g∈G/H hence δ(Res(c) is again the image of a in Hˆ 0(H,A).

Proposition 2.7.11. Let G be a ﬁnite group, H a subgroup of G. Suppose the index [G : H] = n is ﬁnite, then Cor ◦ Res is just the multiplication by n on Tate cohomology groups.

Proof. It suﬃces to show this on the zeroth cohomology and homology groups and then apply the dimension shifting. On cohomology we have

Res Cor AG > AH > AG

27 where Res is just the natural inclusion, and Cor is the map deﬁned in Deﬁnition 2.7.8. Let a ∈ AG, then g · a = a for any g ∈ G and hence X X Cor ◦ Res(a) = Cor(a) = g · a = a = na. g∈G/H g∈G/H

On homology, we have

Res Cor AG > AH > AG where Res is the map deﬁned in 2.7.8, and Cor is the natural quotient map. Let a ∈ AG anda ˜ ∈ AH be a lift of a, then X Cor ◦ Res(a) = Cor( g−1 · a˜) = na. g∈G/H

Remark 2.7.12. This also applies to homology groups and cohomology groups of an arbitrary group G (i.e., G need not be ﬁnite).

Corollary 2.7.13. Suppose G is a group of order n, A a G-module. Then all the Tate cohomology groups Hˆ i(G, A) vanishes by n.

Proof. Let H = {1}, then [G, {1}] = n. Corollary 2.7.11 implies that Cor ◦ Res = n on i i Hˆ (G, A) for all i ∈ Z. On the other hand, Hˆ (H,A) = 0 for all i ∈ Z (for the detail, see the Proof of Theorem 2.6.4). So Cor ◦ Res = 0.

Corollary 2.7.14. Suppose G is a finite group, A a finitely generated G-module. Then i Hˆ (G, A) is finite for all i ∈ Z. Proof. By the definitions of cohomology and homology groups in terms of cochains and chains, these groups are finitely generated. By Corollary 2.7.13, they are torsion, hence they are finite.

Let p be a prime and suppose pm is the highest power of p dividing the order |G| of a ﬁnite group G, i.e., |G| = pmq where (p, q) = 1. Sylow’s Theorem tells us that there exists subgroups of G having order pm and they are called Sylow p-subgroups of G.

Corollary 2.7.15. Suppose G is a ﬁnite group, and Gp is a Sylow p-subgroup of G for a prime p. Then for any G-module A and i ∈ Z, the kernel of

i i Res : Hˆ (G, A) → Hˆ (Gp,A) has no element of order p (alternatively, Res is injective on the p-primary component of Hˆ i(G, A)).

28 i n Proof. Let α ∈ Hˆ (G, A) with p α = 0 for some n ≥ 0. Then Cor ◦ Res(α) = [G : Gp]α. But [G : Gp] is prime to p, hence [G : Gp]α 6= 0 if α 6= 0. Hence α 6∈ ker(Res) if α 6= 0.

Corollary 2.7.16. Suppose G is a ﬁnite group, A any G-module, and Gp a Sylow p- subgroup of G for each prime p. Fix i ∈ Z, and suppose that

i i Res : Hˆ (G, A) → Hˆ (Gp,A) is trivial for all primes p. Then Hˆ i(G, A) = 0.

Proof. By Corollary 2.7.15, the p-primary components of Hˆ i(G, A) is trivial for all primes p. Hence Hˆ i(G, A) = 0.

2.8 Tate cohomology via complete resolutions Let G be a ﬁnite group. Tate cohomology can also be formed by complete resolution of G. Let P → Z be a resolution by ﬁnitely generated free G-modules (e.g. the standard resolution).

∗ Deﬁnition 2.8.1. The dual of P is deﬁned as P = HomZ(P, Z) with the G-module structure (g · ϕ)(x) = ϕ(g−1x).

We have two exact sequences

· · · → P1 → P0 → Z → 0, ∗ ∗ ∗ 0 → Z →→ P0 → P1 → · · · .

∗ We write P−n = Pn−1 for n ≥ 1 and gluing the above two resolutions together gives a complete resolution

L : · · · → P1 → P0 → P−1 → P−2 → · · · .

Proposition 2.8.2. Let G be a ﬁnite group and A be a G-module. Then Hˆ i(G, A) is isomorphic to the i-th cohomology group of the cochain HomG(L, A).

2.9 Cup products

Let G be a group and A, B any G-modules. We let G act on A⊗Z B by g·(a⊗b) = g·a⊗g·b. We consider the following maps on the standard complex

ϕi,j : Pi+j → Pi ⊗Z Pj (g0, ··· , gi+j) 7→ (g0, ··· , gi) ⊗ (gi, ··· , gi+j)

29 for i, j ≥ 0. There is an natural map

HomG(Pi,A) ⊗Z HomG(Pj,B) → HomG(Pi ⊗Z Pj,A ⊗Z B) ϕ ⊗ ϕ0 7→ (α ⊗ β 7→ ϕ(α) ⊗ ϕ0(β)).

These give rise to a map

HomG(Pi,A) ⊗Z HomG(Pj,B) → HomG(Pi+j,A ⊗Z B) ϕ ⊗ ϕ0 7→ ϕ ∪ ϕ0.

0 Deﬁnition 2.9.1. Let ϕ ∈ HomG(Pi,A), ϕ ∈ HomG(Pj,B), i, j ≥ 0. The cup product 0 ϕ ∪ ϕ ∈ HomG(Pi+j,A ⊗Z B) is deﬁned by 0 0 ϕ ∪ ϕ (g0, ··· , gi+j) = ϕ(g0, ··· , gi) ⊗ ϕ (gi, ··· , gi+j).

Remark 2.9.2. In cochains, we can deﬁne cup products

i j i+j C (G, A) ⊗Z C (G, B) → C (G, A ⊗Z B) by 0 0 f ∪ f (g1, ··· , gi+j) = f(g1, ··· , gi) ⊗ g1g2 ··· gif (gi+1, ··· , gi+j). Theorem 2.9.3. The cup products in Deﬁnition 2.9.1 induce unique maps which are also called cup products,

i j ∪ i+j H (G, A) ⊗Z H (G, B) → H (G, A ⊗Z B) for i, j ≥ 0 that are natural in A and B and satisfy the following properties. (i) For i = j = 0, the cup product

G G G A ⊗Z B → (A ⊗Z B) is induced by the identity on A ⊗Z B. (ii) If 0 → A1 → A → A2 → 0 is an exact sequence of G-modules, and

0 → A1 ⊗Z B → A ⊗Z B → A2 ⊗Z B → 0 is also exact, then

i+j+1 δ(α2 ∪ β) = δ(α2) ∪ β ∈ H (G, A ⊗Z B)

i j for all α2 ∈ H (G, A2), β ∈ H (G, B).

30 (iii) If 0 → B1 → B → B2 → 0 is an exact sequence of G-modules, and

0 → A ⊗Z B1 → A ⊗Z B → A ⊗Z B2 → 0 is also exact, then

i i+j+1 δ(α ∪ β2) = (−1) α ∪ δ(β) ∈ H (G, A ⊗Z B)

i j for all α ∈ H (G, A), β2 ∈ H (G, B2). Proposition 2.9.4. Let G be a group and A, B, C any G-modules. Let α ∈ Hi(G, A), β ∈ Hj(G, B), γ ∈ Hk(G, C). Then

i+j+k (α ∪ β) ∪ γ = α ∪ (β ∪ γ) ∈ H (G, A ⊗Z B ⊗Z C).

Proposition 2.9.5. Let G be a group and A, B any G-modules. Consider the natural isomorphism

sAB : A ⊗Z B → B ⊗Z A a ⊗ b 7→ b ⊗ a and the maps that it induces on cohomology. For all α ∈ Hi(G, A), β ∈ Hj(G, B), we have ∗ ij sAB(α ∪ β) = (−1) (β ∪ α).

Proposition 2.9.6. Let G be a group and A, B any G-modules. Let H be a subgroup of G of ﬁnite index. (i) If α ∈ Hi(G, A), β ∈ Hj(G, B), then

Res (α ∪ β) = Res (α) ∪ Res (β).

(ii) If α ∈ Hi(H,A), β ∈ Hj(G, B), then

Cor (α) ∪ β = Cor (α ∪ Res (β)) .

For ﬁnite groups, we also have cup products on Tate cohomology.

31 Theorem 2.9.7. Let G be a ﬁnite group and A, B any G-modules. There exists unique maps ˆ i ˆ j ∪ ˆ i+j H (G, A) ⊗Z H (G, B) → H (G, A ⊗Z B) for i, j ∈ Z that are natural in A and B and satisfy the following properties. (i) The diagram

0 0 ∪ 0 H (G, A) ⊗Z H (G, B) > H (G, A ⊗Z B)

∨ ∨ ˆ 0 ˆ 0 ∪ ˆ 0 H (G, A) ⊗Z H (G, B) > H (G, A ⊗Z B) commutes. (ii) If 0 → A1 → A → A2 → 0 is an exact sequence of G-modules, and

0 → A1 ⊗Z B → A ⊗Z B → A2 ⊗Z B → 0 is also exact, then

ˆ i+j+1 δ(α2 ∪ β) = δ(α2) ∪ β ∈ H (G, A ⊗Z B)

i j for all α2 ∈ Hˆ (G, A2), β ∈ Hˆ (G, B). (iii) If 0 → B1 → B → B2 → 0 is an exact sequence of G-modules, and

0 → A ⊗Z B1 → A ⊗Z B → A ⊗Z B2 → 0 is also exact, then

i ˆ i+j+1 δ(α ∪ β2) = (−1) α ∪ δ(β) ∈ H (G, A ⊗Z B)

i j for all α ∈ Hˆ (G, A), β2 ∈ Hˆ (G, B2).

2.10 Tate cohomology of cyclic groups P Pn−1 i Let G =< s > be a ﬁnite cyclic group of order n. Let D = s − 1, N = g∈G g = i=0 s . Then IG =< g − 1 | g ∈ G >=< D > as a Z[G]-module. Let A be any G-module. Notice that the action of G on A is determined by s, hence

a ∈ AG ⇔ s · a = a ⇔ (s − 1) · a = 0 ⇔ a ∈ ker(D : A → A),

32 IGA = im (D : A → A). G Therefore, A = ker(D),IGA = im (D). Hence,

AG ker(D) Hˆ 0(G, A) = = , N(A) im (N) ker(N) ker(N) Hˆ0(G, A) = = . IGA im (D) In particular, if A is induced, then ker(D) = im (N) and ker(N) = im (D). Now consider a complete resolution K for G. Let Ki = Z[G] for each i, and deﬁne maps d : Ki+1 → Ki by multiplication by D if i is even, and multiplication by N if i is odd. Then we obtain a complete resolution of Z

D N D · · · → Z[G] → Z[G] → Z[G] → Z[G] → · · · .

Then the complex HomG(K,A) is

· · · ← A ←N A ←D A ←N A ← · · · .

i ∼ i i+2 ∼ i Thus Hˆ (G, A) = H (HomG(K,A)). In particular, Hˆ (G, A) = Hˆ (G, A) since K is of periodic 2.

Proposition 2.10.1. For any i ∈ Z, we have

( G A , i ≡ 0 (mod 2), ˆ i N(A) H (G, A) = ker(N) D(A) , i ≡ 1 (mod 2).

2 G In particular, Hˆ (G, Z) = Z /N(Z) = Z/nZ. 2 Theorem 2.10.2. Cup product by a generator of Hˆ (G, Z) gives rise to an isomorphism

Hˆ i(G, A) → Hˆ i+2(G, A) for all i ∈ Z and any G-module A. Proof. Consider the two exact sequences of G-modules

0 > IG > Z[G] > Z > 0 and

N D 0 > Z > Z[G] > IG > 0.

33 i Since Hˆ (G, Z[G]) = 0 for all i ∈ Z, we have two isomorphisms

ˆ 0 δ ˆ 1 δ ˆ 2 H (G, Z) > H (G, IG) > H (G, Z).

0 Therefore, it reduces to show that cup product by a generator of Hˆ (G, Z) induces an automorphism of Hˆ i(G, A). By dimension shifting we reduce to the case i = 0. Note that

G 0 Z Hˆ (G, Z) = = Z/nZ, N(Z)

0 where n = |G|. A generator b of Hˆ (G, Z) is represented by an integer β which is prime to n, and cup product with b is multiplication by β. Since β is prime to n, there exists 0 an integer γ such that βγ ≡ 1 (mod n). Since Hˆ (G, Z) vanishes by multiplication of n, multiplication by β is therefore an automorphism of Hˆ 0(G, A).

Corollary 2.10.3. Suppose G is a ﬁnite cyclic group. A short exact sequence

0 → A → B → C → 0 of G-modules gives rise to an exact hexagon

Hˆ 0(G, A) > Hˆ 0(G, B) > > Hˆ 1(G, C) Hˆ 0(G, C) <

< Hˆ 1(G, B) < Hˆ 1(G, A) .

0 1 Definition 2.10.4. Suppose that Hˆ (G, A) and Hˆ (G, A) are finite groups, and let h0(A), h1(A) be their orders. We define the Herbrand quotient of A to be h (A) h(A) = 0 . h1(A) Proposition 2.10.5. Let G be a finite cyclic group. Let

0 → A → B → C → 0 be a short exact sequence of G-modules. If any two of h(A), h(B), h(C) are deﬁned, then the third one is also deﬁned, and

h(B) = h(A)h(C).

34 Proof. Without loss of generality, we assume that h0(A), h1(A), h0(B), h1(B) are finite. 0 0 We have an exact hexagon, and let M1 be the image of Hˆ (G, A) in Hˆ (G, B), M2 be the image of Hˆ 0(G, B) in Hˆ 0(G, C), and so on in clockwise direction round the hexagon. Then 0 0 → M2 → Hˆ (G, C) → M3 → 0 0 is exact. Now M2 is finite because it is a homomorphic image of Hˆ (G, B), and M3 is finite because it is a subgroup of Hˆ 1(G, A). Thus Hˆ 0(G, C) is also finite. Similarly, Hˆ 1(G, C) is finite. Thus h(C) is defined. Let mi be the order of Mi for 1 ≤ i ≤ 6. Consider

Proposition 2.10.6. Let G be a ﬁnite cyclic group. If A is a ﬁnite G-module, then h(A) = 1. Proof. The sequence

G ι D π 0 > A > A > A > AG > 0 is exact. Hence, ∼ AG = im(π) = A/ker(π) = A/im(D), ∼ ∼ A/AG = A/im(ι) = A/ker(D) = im(D), G and so |A | = |AG|. On the other hand,

∗ 1 N G 0 0 > Hˆ (G, A) > AG > A > Hˆ (G, A) > 0

1 −1 is exact because Hˆ (G, A) = Hˆ (G, A) = Hˆ0(G, A). The same technique implies that |Hˆ 1(G, A)| = |Hˆ 0(G, A)| and hence h(A) = 1.

Corollary 2.10.7. Let G be a finite cyclic group, A, B any G-modules. Suppose f : A → B is a G-homomorphism with finite kernel and cokernel. If either of h(A), h(B) is defined, then so is the other, and they are equal.

35 Proof. Without loss of generality, we assume that h(A) is deﬁned. The exactness of

0 > ker(f) > A > f(A) > 0 implies that h(f(A)) is deﬁned, and

h(A) = h(ker(f)) · h(f(A)) = h(f(A)) by Proposition 2.10.5 and Proposition 2.10.6. Similarly, the exactness of

0 > f(A) > B > coker(f) > 0 implies that h(B) is deﬁned, and

h(B) = h(f(A)) · h(coker(f)) = h(f(A)).

Therefore, h(A) = h(B).

2.11 Cohomological triviality We assume G to be a finite group. Definition 2.11.1. Let G be a finite group. A G module A is cohomologically trivial if i Hˆ (H,A) = 0 for all i ∈ Z and all subgroup H of G. Example 2.11.2. Induced G-modules are cohomologically trivial. This is because an induced G-module is also an induced H-module, and by Theorem 2.6.4. Example 2.11.3. Free G-modules are cohomologically trivial. To see this, let F be a free module over Z[G], hence over Z[H] on a generating set I for any subgroup H ⊂ G. Then ∼ L H ˆ 0 ˆ i F = i∈I Z[H] and so F = 0 and H (H,F ) = 0. Hence H (H,F ) = 0 for any i ∈ Z by the long exact sequence of Tate cohomology and dimension shifting. Let p be a prime number. Recall that a finite group G is called a p-group if its order |G| is a power of p. Lemma 2.11.4. Let G be a p-group where p is a prime number, and A be a G-module such that pA = 0. Then the following are equivalent. (i) A = 0. (ii) H0(G, A) = AG = 0. (iii) H0(G, A) = AG = 0. Proof. It is clear that (i) ⇒ (ii) and (i) ⇒ (iii). (ii) ⇒ (i): Let AG = 0 and a ∈ A. Then the submodule B of A generated by a is finite, of order a power of p, and BG = 0. The G-orbits in B are either {0} or a multiple of p since BG = 0. Since B has p-power order, the order of B has to be 1, and hence B = 0. Since A was arbitrary, A = 0.

36 (iii) ⇒ (i): Suppose AG = 0. Then X = HomZ(A, Fp) satisﬁes pX = 0, and

G G X = (HomZ(A, Fp)) = HomZ[G](A, Fp) = HomZ[G](AG, Fp) = 0. By the direction (ii) ⇒ (i) we just proved, X = 0, and hence A = 0.

Lemma 2.11.5. Let G be a p-group, and A be a G-module such that pA = 0. If H1(G, A) = 0, then A is a free module over Fp[G].

Proof. Since pA = 0, pAG = 0 and so AG is a vector space over Fp. Let eλ be a basis of AG over Fp, and lift it to aλ ∈ A. Let F be the free Fp[G]-module generated by aλ, then we have the canonical surjection π : F → A, and let R = ker(π). Then

π 0 > R > F > A > 0 is exact, and hence

π˜ 0 > RG > FG > AG > 0 is exact because H1(G, A) = 0 implies H1(G, R) = 0. The mapπ ˜ induced by π is an isomorphism, so RG = 0. Since pR = 0, we have R = 0 by Lemma 2.11.4. Hence π is an ∼ isomorphism. That is, A = F is a free module over Fp[G]. Theorem 2.11.6. Let G be a p-group, and A be a G-module such that pA = 0. The following are equivalent. (i) A is an induced G-module. (ii) A is cohomologically trivial. (iii) A is a free Fp[G]-module. i (iv) There exists some i ∈ Z such that Hˆ (G, A) = 0. Proof. It’s easy to see that (i) ⇒ (ii) ⇒ (iv). (iii) ⇒ (i): Suppose A is free over Fp[G] on a generating set I, then M ∼ M ∼ M ∼ M A = Fp[G] = Z[G]/pZ[G] = Z[G] ⊗ (Z/pZ) = Z[G] ⊗ ( Fp). i∈I i∈I i∈I i∈I Hence A is induced. ∗ (iv) ⇒ (iii): Since pA = 0, pA = pA∗ = 0. By dimension shifting, there is a G-module B such that pB = 0 and Hˆ j−2(G, B) =∼ Hˆ i+j(G, A) ˆ −2 ∼ ˆ i for all j ∈ Z. In particular, H1(G, B) = H (G, B) = H (G, A) is trivial. By Lemma 2.11.5, B is a free module over Fp[G]. By the case (iii) ⇒ (i) we just proved, B is cohomologically trivial, and hence A is also a cohomologically trivial G-module. Apply Lemma 2.11.5 again, and we obtain that A is a free Fp[G]-module.

37 Theorem 2.11.7. Let G be a p-group, and A be a G-module with no element of order p. Then the following are equivalent. (i) A is cohomologically trivial. i i+1 (ii) There exists some i ∈ Z such that Hˆ (G, A) = Hˆ (G, A) = 0. (iii) A/pA is a free Fp[G]-module. Proof. It is clear that (i) ⇒ (ii). (ii) ⇒ (iii): Since A has no p-torsion,

p 0 > A > A > A/pA > 0 is an exact sequence, and so we have an exact sequence

p p Hˆ i(G, A) > Hˆ i(G, A) > Hˆ i(G, A/pA) > Hˆ i+1(G, A) > Hˆ i+1(G, A).

Since Hˆ i(G, A) = Hˆ i+1(G, A) = 0, we have Hˆ i(G, A/pA) = 0 by the long exact sequence of Tate cohomology groups. By Theorem 2.11.6, A/pA is a free Fp[G]-module. (iii) ⇒ (i): Suppose A/pA is a free Fp[G]-module, then A/pA is cohomologically trivial by Theorem 2.11.6. Hence the map p : Hˆ i(G, A) → Hˆ i(G, A) of multiplication by p is i an isomorphism on Hˆ (G, A) for all i ∈ Z. For any subgroup H ⊂ G, consider the commutative diagram

p Hˆ i(H,A) > Hˆ i(H,A)

Cor Cor ∨ ∨ p Hˆ i(G, A) > Hˆ i(G, A).

It follows that the map p : Hˆ i(H,A) → Hˆ i(H,A) of multiplication by p is also an isomor- i i phism on Hˆ (H,A) for any subgroup H ⊂ G and any i ∈ Z. But Hˆ (H,A) is a p-group by Corollary 2.7.13, hence Hˆ i(H,A) = 0. Hence A is cohomologically trivial.

Corollary 2.11.8. Let G be a p-group, and A be a G-module that is free over Z and cohomologically trivial. If a G-module B is Z-torsion free, then HomZ(A, B) is cohomologically trivial.

Proof. Since B is Z-torsion free, the sequence p 0 > B > B > B/pB > 0 is exact, and hence so is the sequence

p 0 > HomZ(A, B) > HomZ(A, B) > HomZ(A, B/pB) > 0.

In particular, HomZ(A, B) has no p-torsion, and ∼ ∼ HomZ(A, B)/pHomZ(A, B) = HomZ(A, B/pB) = HomZ(A/pA, B/pB).

38 A/pA is free over Fp[G] by Theorem 2.11.7, hence it is an induced G-module, by Theorem 2.11.6. Let X be an abelian group such that A/pA = Z[G] ⊗Z X, then ∼ HomZ(A/pA, B/pB) = HomZ(Z[G] ⊗Z X, B/pB) = HomZ(Z[G], HomZ(X, B/pB)) by the Adjoint Isomorphism Theorem for Tensor and Hom. So HomZ(A/pA, B/pB), and hence HomZ(A, B)/pHomZ(A, B), are induced G-modules. Applying Theorem 2.11.6 again to conclude that HomZ(A, B) is cohomologically trivial.

Proposition 2.11.9. Let G be a ﬁnite group, and Gp be a Sylow p-subgroup of G for each prime p. Then A is cohomologically trivial as a G-module if and only if A is cohomologically trivial as a Gp-module for each prime p.

Proof. Suppose A is cohomologically trivial as a Gp-module for all p. Since all the Sylow p-subgroups are conjugate to each other, given a subgroup H ⊂ G, there exists some g ∈ G −1 i −1 such that Hp ⊂ gGpg . By the cohomological triviality of Gp, we have Hˆ (g Hpg, A) = 0 ˆ i for all i ∈ Z. Since conjugation by g is an isomorphism, we have H (Hp,A) = 0. Therefore, i i the restriction homomorphism Res:Hˆ (H,A) → Hˆ (Hp,A) is 0. Since this holds for all prime p, it follows that Hˆ i(H,A) = 0 by Corollary 2.7.16. Hence A is a cohomologically trivial G-module.

A G-module A is projective if HomG(A, ·) is an exact functor (or equivalent, A is a direct summand of a free G-module). Example 2.11.10. Projective G-modules are cohomologically trivial. Suppose P and Q are projective G-modules with F = P ⊕ Q being a free G-module. For any subgroup H ⊂ G, F is also free over Z[H]. Then Hˆ i(H,P ) ,→ Hˆ i(H,P ) ⊕ Hˆ i(H,Q) =∼ Hˆ i(H,P ⊕ Q) = Hˆ i(H,F ) = 0 for all i ∈ Z. Therefore, P is cohomologically trivial. We have the following theorem.

Theorem 2.11.11. Let G be a ﬁnite group, A a G-module that is free over Z. Then A is a cohomologically trivial G-module if and only if A is a projective G-module. Proof. We already showed that projective G-modules are cohomologically trivial. Suppose that A is a cohomologically trivial G-module. Since A is Z-free, it follows that Z[G] ⊗ A is a free G-module, A∗ is Z-torsion free, and the sequence

0 > HomZ(A, A∗) > HomZ(A, Z[G] ⊗ A) > HomZ(A, A) > 0 is exact. Moreover, HomZ(A, A∗) is a cohomologically trivial G-module by Corollary 2.11.8. Note that 0 G ∼ H (G, HomZ(A, A∗)) = (HomZ(A, A∗)) = HomZ[G](A, A∗),

39 0 G ∼ H (G, HomZ(A, Z[G] ⊗ A)) = (HomZ(A, Z[G] ⊗ A)) = HomZ[G](A, Z[G] ⊗ A). Hence the map

HomZ[G](A, Z[G] ⊗ A) → HomZ[G](A, A∗) is surjective by the long exact cohomology sequence attached to the above exact sequence. So the identity map of A lifts to a G-homomorphism A → Z[G] ⊗ A. Hence A is a direct summand of the free G-module Z[G] ⊗ A. Therefore, A is projective. Finally, we consider the general case. Theorem 2.11.12. Let G be a ﬁnite group, and A a G-module. Then the following are equivalent. (i) A is cohomologically trivial. ˆ i ˆ i+1 (ii) For each prime p, there exists some i ∈ Z such that H (Gp,A) = H (Gp,A) = 0. (iii) There is an exact sequence of G-modules

0 > P1 > P0 > A > 0 where P0,P1 are projective G-modules. Proof. It is clear that (i) ⇒ (ii). (iii) ⇒ (i) follows from the long exact sequence of Tate cohomology and the fact that projective G-modules are cohomologically trivial. (ii) ⇒ (iii): Choose a free (hence projective) G-module F which maps surjectively onto A, and let B be the kernel of the map. Then the sequence

0 > B > F > A > 0 is exact. Since F is cohomologically trivial, we have j−1 ∼ j Hˆ (Gp,A) = Hˆ (Gp,B) ˆ j for every j ∈ Z by the long exact sequence of Tate cohomology. It follows that H (Gp,B) vanishes for two consecutive values of j. B is Z-free because B is a subgroup of F . Then Theorem 2.11.7, Proposition 2.11.9, and Theorem 2.11.11 imply that B is projective.

2.12 Tate’s Theorem

Proposition 2.12.1. Let G be a ﬁnite group, f : A → B a G-homomorphism. Let Gp be a Sylow p-subgroup of G for each prime p. Suppose that, for each prime p, there exists some j ∈ Z such that ∗ ˆ i ˆ i fp : H (Gp,A) → H (Gp,B) is surjective for i = j − 1, an isomorphism for i = j, and injective for i = j + 1. Then

f ∗ : Hˆ i(H,A) → Hˆ i(H,B) is an isomorphism for all i ∈ Z and subgroups H ⊂ G.

40 Proof. For each prime p, consider the canonical injection

fp ⊗ ι : A → B ⊗ Hom(Z[Gp],A) of Gp-modules. Let C be the cokernel of the injection fp ⊗ ι. Then we have an exact sequence

0 > A > B ⊗ Hom(Z[Gp],A) > C > 0 of Gp-modules. Since Hom(Z[Gp],A) is Gp-cohomologically trivial, ˆ i ∼ ˆ i H (Gp,B ⊗ Hom(Z[Gp],A)) = H (Gp,B) for all i ∈ Z. So we have the long exact sequence

∗ ∗ i fp i i δ i+1 fp i+1 · · · → Hˆ (Gp,A) → Hˆ (Gp,B) → Hˆ (Gp,C) → Hˆ (Gp,A) → Hˆ (Gp,B) → · · · .

∗ ˆ j−1 ∗ Consider the case i = j − 1. The map fp being surjective on H (Gp,A) and fp being an j j−1 isomorphism and hence injective on Hˆ (Gp,A) implies that Hˆ (Gp,C) = 0. Similarly, ∗ ˆ j for i = j, fp being an isomorphism and hence surjective on H (Gp,A) and being injective j+1 j on Hˆ (Gp,A) implies that Hˆ (Gp,C) = 0. By Theorem 2.11.12, C is a cohomologically ∗ trivial Gp-module. Therefore, each f must be an isomorphism by the long exact sequence of Tate cohomology groups.

Theorem 2.12.2. Let G be a ﬁnite group, A, B, C be any G-modules, and

θ : A ⊗Z B → C

k be a G-module map. Let k ∈ Z and α ∈ Hˆ (G, A). For each subgroup H ⊂ G, deﬁne i ˆ i ˆ i+k ΘH,α : H (H,B) → H (H,C) β 7→ θ∗(Res(α) ∪ β).

For each prime p, suppose that there exists some j ∈ such that the map Θi is Z Gp,α i surjective for i = j − 1, an isomorphism for i = j, and injective for i = j + 1. Then ΘH,α is an isomorphism for all i ∈ Z and any subgroup H ⊂ G. Proof. First we consider the case k = 0. For α ∈ Hˆ 0(G, A), let a ∈ AG represent α. Consider the map ψ : B → C given by ψ(b) = θ(a ⊗ b). Note that for any g ∈ G,

ψ(gb) = θ(a ⊗ gb) = θ(ga ⊗ gb) = θ(g(a ⊗ b)) = gθ(a ⊗ b) = gψ(b).

So ψ is a G-homomorphism. We claim that the induced map on Tate cohomology

ψ∗ : Hˆ i(H,B) → Hˆ i(H,C)

41 i agrees with the map ΘH,α. To see the claim, we ﬁrst consider the case i = 0. Let b ∈ BH represent β ∈ Hˆ 0(H,B). Then ∗ ∗ i ψ (β) = θ(a ⊗ b) + N(C) = θ (Res(α) ∪ β) = ΘH,α(β). In general, we have a commutative diagram

0 > (A ⊗Z B)∗ > Z[G] ⊗Z (A ⊗Z B) > A ⊗Z B > 0

0 θ id [G]⊗θ θ ∨ ∨ Z ∨ 0 > C∗ > Z[G] ⊗ C > C > 0 with exact rows. Moreover, the top row of the above diagram is isomorphic to

0 > A ⊗Z B∗ > A ⊗Z (Z[G] ⊗Z B) > A ⊗Z B > 0 0 0 0 0 0 0 and we have a map ψ : B∗ → C∗ given by ψ (b ) = θ (a ⊗ b ) for all b ∈ B∗. Then we have a commutative diagram

0 > B∗ > Z[G] ⊗Z B > B > 0

0 ψ id [G]⊗ψ ψ ∨ ∨ Z ∨ 0 > C∗ > Z[G] ⊗Z C > C > 0 with exact rows. Then we have commutative diagrams

i−1 δ i Hˆ (H,B) > Hˆ (H,B∗)

ψ∗ (ψ0)∗ ∨ ∨ i−1 δ i Hˆ (H,C) > Hˆ (H,C∗) and

i−1 δ i Hˆ (H,B) > Hˆ (H,B∗)

Θi−1 (Θ0)i ∨ H,α ∨ H,α i−1 δ i Hˆ (H,C) > Hˆ (H,C∗)

0 i 0 ∗ where (Θ )H,α(β) = (θ ) (Res(α) ∪ β), and the connecting homomorphisms δ are isomorphisms since Z[G] ⊗Z B and Z[G] ⊗Z C are induced and hence cohomologically trivial. 0 ∗ 0 i ˆ i ∗ i−1 ˆ i−1 Now suppose (ψ ) = (Θ )H,α on H (H,B∗), then we have ψ = ΘH,α on H (H,B) as well. That is to say, if the statement is true for i, then it is also true for i − 1. A similar argument allows us to shift from i to i + 1. Hence we proved the claim. Now ψ∗ satisﬁes the condition of Proposition 2.12.1, hence ψ∗ is an isomorphism for all i ∈ Z. This proves Theorem 2.12.2 for k = 0.

42 The general case follows from another piece of dimension shifting. Let α ∈ Hˆ k−1(H,A), 0 k α = δ(α) ∈ Hˆ (H,A∗). Note that we also have an exact sequence

0 > A∗ ⊗Z B > (Z[G] ⊗Z A) ⊗Z B > A ⊗Z B > 0. Consider the diagram

Hˆ i(H,B) = Hˆ i(H,B)

Θi (Θ0)i ∨ H,α ∨ H,α0 i+k−1 δ i+k Hˆ (H,C) > Hˆ (H,C∗). It is commutative because

i 0 0 0 0 i δ ◦ ΘH,α(β) = δ ◦ θ(Res(α) ∪ β) = θ ◦ δ(Res(α) ∪ β) = θ ◦ (Res(α ) ∪ β) = (Θ )H,α0 (β).

i+k−1 i+k Also, the connecting homomorphism δ : Hˆ (H,C) → Hˆ (H,C∗) is an isomorphism. By assumption, there exists some j ∈ such that the map Θi is surjective for i = j −1, Z Gp,α an isomorphism for i = j, and injective for i = j + 1. By the commutativity of the above 0 i diagram, we have that the map (Θ )H,α0 is surjective for i = j, an isomorphism for i = j+1, and injective for i = j + 2. Assume Theorem 2.12.2 is true for k, then we have that all 0 i (Θ )H,α0 are isomorphisms. By the commutativity of the above diagram again, all the i maps ΘH,α are isomorphisms as well. This completes the proof. The following is a special case ﬁrst due to Tate.

Theorem 2.12.3 (Tate). Let G be a ﬁnite group, A be a G-module, and α ∈ H2(G, A). Let 1 Gp be a Sylow p-subgroup of G for each prime p. Suppose for each prime p, H (Gp,A) = 0 2 and H (Gp,A) is a cyclic group of order |Gp| generated by the restriction of α. Then the map

i i+2 Hˆ (H, Z) → Hˆ (H,A) β 7→ Res(α) ∪ β is an isomorphism for any i ∈ Z and any subgroup H ⊂ G. ˆ −1 ˆ 1 Proof. Take H = Gp. For i = −1, the map H (Gp, Z) → H (Gp,A) is surjective since ˆ 1 1 ˆ 1 ˆ 3 H (Gp,A) = H (Gp,A) = 0. For i = 1, the map H (Gp, Z) → H (Gp,A) is injective since ˆ 1 H (Gp, Z) =Hom(Gp, Z) (since Z is a trivial Gp-module, and by Lemma 2.3.8) =0.

43 For i = 0, we have Gp ˆ 0 Z Z H (Gp, Z) = = , NGp (Z) |Gp|Z ˆ 0 ˆ 2 ˆ 0 The map H (Gp, Z) → H (Gp,A) takes the image of n ∈ Z in H (Gp, Z) to nRes(α), hence 2 it is an isomorphism by the assumption on H (Gp,A). Then Theorem 2.12.2 implies that all the maps are isomorphisms.

44 3 Proﬁnite Groups

3.1 Inverse systems and inverse limits Definition 3.1.1. A directed partially ordered set I = (I, ≤) is a non-empty set I together with a binary relation ≤ which satisfies the following conditions: (i) (reflexive) i ≤ i for all i ∈ I. (ii) (transitive) If i ≤ j and j ≤ k, then i ≤ k for i, j, k ∈ I. (iii) (antisymmetric) If i ≤ j and j ≤ i, then i = j for i, j ∈ I. (iv) (directedness) If i, j ∈ I, then there exists some k ∈ I such that i ≤ k, j ≤ k. Definition 3.1.2. An inverse system (or projective system) of topological groups over I, is an object (Gi, ϕij,I) where for each i ∈ I, Gi is a topological group, and for each j ≤ i in I, there is a morphism (continuous group homomorphism) ϕij : Gi → Gj such that the diagram

ϕij Gi > Gj

ϕjk ϕik > ∨ Gk commutes whenever i, j, k ∈ I and k ≤ j ≤ i. In addition, we assume that ϕii is the identity map on Gi. We may simply write the inverse system as (Gi) if there is no confusion about the index set I.

0 0 0 Deﬁnition 3.1.3. Suppose (Gi, ϕij,I) and (Gi0 , ϕi0j0 ,I ) are two inverse systems. Let ψ : I0 → I be an order preserving map (i.e., if i0 ≤ j0, then ψ(i0) ≤ ψ(j0)). If for each 0 0 0 i ∈ I , we have a morphism fi0 : Gψ(i0) → Gi0 such that the diagram

fj0 0 Gψ(j0) > Gj0

0 ϕψ(j0)ψ(i0) ϕ 0 0 ∨ ∨ j i fi0 0 Gψ(i0) > Gi0

0 0 0 0 0 commutes whenever i ≤ j in I , then Ψ = (ψ; fi0 , i ∈ I ) is a morphism from (Gi, ϕij,I) 0 0 to (Gi0 , ϕi0j0 ,I ). Deﬁnition 3.1.4. The inverse limit (or projective limit)

G = lim G ←− i i Q of the inverse system (Gi, ϕij,I) is the subgroup of the direct product i∈I Gi of topological groups consisting of the tuples (gi) that satisfy the condition ϕij(gi) = gj if j ≤ i, Q and we assume that G has the topology induced by the product topology of i∈I Gi.

45 Lemma 3.1.5. If (Gi, ϕij,I) is an inverse system of Hausdorff topological groups, then lim G is a closed subgroup of Q G . ←− i i∈I i Proof. We show that Q G \ lim G is open in Q G . Let (g ) ∈ Q G \ lim G . i∈I i ←− i i∈I i i i∈I i ←− i Then there are some m, n ∈ I such that m ≤ n and ϕnm(gn) 6= gm. Choose disjoint neigh- 0 borhoods U and V of ϕnm(gn) and gm in Gm, respectively. Let U ⊂ Gn be an open neigh- 0 Q Q borhood of gn such that ϕnm(U ) ⊂ U. Consider the open subset W = i∈I Wi ⊂ i∈I Gi, 0 where Wn = U , Wm = V , and Wi = Gi for i 6= m, n. Then W is an open neighborhood of (g ) in Q G , disjoint from lim G . Hence Q G \ lim G is open in Q G i i∈I i ←− i i∈I i ←− i i∈I i and therefore, lim G is closed in Q G . ←− i i∈I i 3.2 Topological structure of profinite groups Definition 3.2.1. We say that G = lim G is a profinite group, pro-p group, or pronilpo- ←−i i tent group if all Gi are, respectively, finite groups, finite p-groups for a fixed prime p, or finite nilpotent groups, with the discrete topologies.

Example 3.2.2. Any ﬁnite group is trivially proﬁnite.

To characterize proﬁnite groups in terms of topological properties, we give some facts about topological groups.

Lemma 3.2.3. (i) An open subgroup H of a topological group G is closed. (ii) If G is a compact topological group, then a subgroup H is open if and only if H is closed of ﬁnite index.

Proof. (i) Let H be an open subgroup of G, then every left coset xH is an open set in G. Since H = G\ (∪x6∈H xH), H is closed. (ii) (⇒): Suppose H be an open subgroup of G, then H is closed in G by (i), and the left coset xH is also an open subset of G. Moreover, the left cosets of H in G form an open cover of G, hence H has finite index by the compactness of G. (⇐): Suppose H is a closed subgroup of G of finite index. To prove that H is open, we will prove that the complement of H, which is ∪x6∈H xH, is closed. Since H is of finite index, ∪x6∈H xH is a finite union of closed sets, hence ∪x6∈H xH is closed. Therefore, H is open.

Recall that a topological space is totally disconnected if the only non-empty connected subsets are the one-point sets, or equivalently, if for any two points there is an open and closed subset containing exactly one of them.

Lemma 3.2.4. Suppose G is a Hausdorff, compact, and totally disconnected topological group. If {Ni, i ∈ I} is a set of open normal subgroups of G, then ∩i∈I Ni = {1}. Theorem 3.2.5. If G is a topological group, then G is profinite if and only if G is Hausdorff, compact, and totally disconnected.

46 Corollary 3.2.6. If G is a proﬁnite group, then

G ∼ lim G/N = ←− N where N runs through all open normal subgroup of G.

Corollary 3.2.7. Let G be a proﬁnite group, and H ⊂ G be a closed subgroup of G, then

H ∼ lim H/H ∩ N = ←− N where N runs through all open normal subgroup of G.

Corollary 3.2.8. Let G be a proﬁnite group, and H ⊂ G be a closed normal subgroup of G, then G/H ∼ lim G/NH = ←− N where N runs through all open normal subgroup of G.

3.3 Examples of proﬁnite groups

Let G be a topological group. Consider the family N = {Ni, i ∈ I} of all normal subgroups of finite index in G. Note that N is not empty because G ∈ N . If H ∈ N ,K ∈ N , then H ∩ K ∈ N . We make N into a directed partially ordered set by defining j ≤ i if Ni is an open subgroup of Nj for Ni,Nj ∈ N . If Ni,Nj ∈ N and j ≤ i, define ϕij : G/Ni → G/Nj to be the natural homomorphism. Then (G/Ni, ϕij,I) is an inverse system of finite groups. We say that Gˆ = lim G/N ←− i i∈I is the profinite completion of G. For a fixed prime p, if N = {Ni, i ∈ I} is the family of all normal subgroups of G of index a power of p, then G = Gˆ = lim G/N pˆ p ←− i i∈I is called the pro-p completion of G.

Example 3.3.1. As a special example, consider the group of integers Z. Its proﬁnite completion is ˆ = lim /n . Z ←− Z Z n∈N This is isomorphic to the product of all p-adic integers:

ˆ ∼ Y Z = Zp. p

47 Its pro-p completion is ˆ = lim /pn = . Zp ←− Z Z Zp n∈N Profinite groups also arise from Galois theory. Let E/F be a Galois extension of fields. Let {Ki, i ∈ I} be the family of all finite intermediate Galois extensions: F ⊂ Ki ⊂ E and Ki/F is a finite Galois extension. Then [ E = Ki. i∈I From Galois theory, we know that (i) Gal(E/Ki) is a normal subgroup of Gal(E/F ). ∼ (ii) Gal(E/F )/Gal(E/Ki) = Gal(Ki/F ) is a finite group. (iii) If i, j ∈ I, then there is some n ∈ I such that Gal(E/Kn) is a subgroup of Gal(E/Ki)∩ Gal(E/Kj). (iv) ∩i∈I Gal(E/Ki) = {1}. There is a unique topology on Gal(E/F ), called the Krull topology, such that the collection {Gal(E/Ki), i ∈ I} is a fundamental system of neighborhoods of the identity element 1 of Gal(E/F ). If the Galois extension E/F is finite, then the Krull topology on Gal(E/F ) is the discrete topology. Consider the family of finite Galois groups {Gal(Ki/F ), i ∈ I}. We define the relation j ≤ i if Kj ⊂ Ki, or equivalently, if Gal(E/Ki) ⊂ Gal(E/Kj). Note that if Ki1 ,Ki2 are

finite Galois extensions of F contained in E, then so is the F -composite Ki1 Ki2 , which is some Kj, and i1 ≤ j, i2 ≤ j. Hence I is a directed partially ordered set. For j ≤ i, we define ϕij : Gal(Ki/F ) → Gal(Kj/F ) by restriction, that is, ϕij(σ) = σ|Kj , where σ ∈ Gal(Ki/F ). Then (Gal(Ki/F ), ϕij,I) forms an inverse system of finite Galois groups. Proposition 3.3.2. Gal(E/F ) =∼ lim Gal(K /F ). In particular, Gal(E/F ) is a profinite ←−i i group. Proof. Consider the homomorphism Y Ψ : Gal(E/F ) → lim Gal(K /F ) ⊂ Gal(K /F ) ←− i i i i∈I

σ 7→ (σ|Ki ). We will prove that Ψ is an isomorphism (algebraically). If σ 6= 1, then there exists x ∈ E∗ such that σ(x) 6= x, and there is some i ∈ I such that x ∈ Ki. Now σ|Ki 6= 1, so Ψ(σ) = (σ|Ki ) 6= 1. Hence Ψ is one-to-one. Suppose (σ ) ∈ lim Gal(K /F ), deﬁne σ : E → E by σ(x) = σ (x) for x ∈ K . Then i ←−i i i i σ ∈ Gal(E/F ) and Ψ(σ) = (σi). Hence Ψ is surjective. Finally we use the isomorphism Ψ to transfer the topology from lim Gal(K /F ) to ←−i i Gal(E/F ).

48 We showed that every Galois group of a Galois extension can be interpreted as a profinite group in Proposition 3.3.2. Actually the converse is also true. Every profinite group can be realized as the Galois group of some field extensions. Theorem 3.3.3. Let G be a profinite group. Then it is the Galois group of some field extension. Remark 3.3.4. Theorem 3.3.3 was proved in [Wat73] by generalizing Artin’s theorem that finite automorphism groups are Galois groups. However, it is still unknown whether any profinite group is the Galois group of some field extension over a fixed base field.

3.4 Direct systems and direct limits Deﬁnition 3.4.1. Let I = (I, ≤) be a directed partially ordered set. A direct (or inductive) system of abelian groups over I, is an object (Ai, ϕij,I) where for each i ∈ I, Ai is an abelian group, and for each i ≤ j in I, there is a group homomorphism ϕij : Ai → Aj such that the diagram

ϕij Ai > Aj

ϕjk ϕik > ∨ Ak commutes whenever i, j, k ∈ I and i ≤ j ≤ k. In addition, we assume that ϕii is the identity map on Ai. We may simply write the direct system as (Ai) or (Ai, ϕij) if there is no confusion about the index set I. 0 0 0 Deﬁnition 3.4.2. Suppose (Ai, ϕij,I) and (Ai0 , ϕi0j0 ,I ) are two direct systems. Let ψ : I → I0 be an order preserving map (i.e., if i ≤ j, then ψ(i) ≤ ψ(j)). If for each i ∈ I, 0 we have a homomorphism fi : Ai → Aψ(i) such that the diagram

fi 0 Ai > Aψ(i)

0 ϕij ϕψ(i)ψ(j) ∨ ∨ fj 0 Aj > Aψ(j) commutes whenever i ≤ j in I, then Ψ = (ψ; fi, i ∈ I) is a morphism from (Ai, ϕij,I) to 0 0 0 (Ai0 , ϕi0j0 ,I ).

Deﬁnition 3.4.3. The direct limit of the direct system (Ai, ϕij,I) is deﬁned as the disjoint union of the groups Ai module an equivalence relation lim A = t A / ∼ −→ i i∈I i where for x ∈ Ai, y ∈ Aj, x ∼ y if there exists some k ∈ I with i ≤ k, j ≤ k such that ϕik(x) = ϕjk(y).

49 3.5 Discrete G-modules Let G be a profinite group and A a (left) G-module. If H is an open subgroup of G, AH is the group of H-invariants in A.A G-module A satisfying [ A = AN N open normal subgroup of G is called a discrete G-module. Remark 3.5.1. In the definition, “normal” doesn’t matter so much. In fact, for a G- module A, [ [ AN = AH . N open normal subgroup of G H open subgroup of G S H The direction ⊆ is trivial. To see the other direction, let a ∈ H open subgroup of G A , then H T −1 there is some open subgroup H of G such that a ∈ A . Let N = g∈G g Hg. Since G is compact, N is a finite intersection of open sets, hence N is open. N is also a normal subgroup of G. Moreover, N ⊂ H. Hence a ∈ AH ⊂ AN . Proposition 3.5.2. Let G be a profinite group, and A be a G-module. The following are equivalent. (i) A is a discrete G-module. (ii) For every a ∈ A, the stabilizer StabG(a) = {g ∈ G : ga = a} is an open subgroup of G. (iii) The multiplication map G × A → A is continuous, where G has the usual topology for profinite groups and A is given the discrete topology. Proof. (i) ⇒ (iii): Let (g, a) ∈ G × A. Then there is some open normal subgroup N of G such that a ∈ AN . Then gN × {a} is an open neighborhood of (g, a) ∈ G × A mapping to ga. Hence the multiplication map G × A → A is continuous. (iii) ⇒ (ii): Let a ∈ A. Consider the restriction of the multiplication map to G × {a}. The preimage of a ∈ A is StabG(a) × {a}, which must be an open set. Hence StabG(a) is open. (ii) ⇒ (i): For any a ∈ A, there is an open normal subgroup

\ −1 N = g StabG(a)g g∈G

Stab (a) N of G that is contained in StabG(a), so a ∈ A G ⊂ A . Example 3.5.3. Let G be a profinite group. Then any G-module A with the trivial G-action is a discrete G-module. Less trivial examples of discrete G-modules come from Galois theory. Let E/F be a Galois extension of fields with G = Gal(E/F ). The following are discrete G-modules with action defined by g · a = g(a): (i) the additive group E; (ii) the multiplicative group E∗ = E\{0}; (iii) the multiplicative group of roots of 1 in E.

50 3.6 Cohomology of profinite groups Let G be a profinite group and A a discrete G-module. For n ≥ 0, we consider the group of n-th continuous cochains Cn(G, A) = {ϕ : Gn → A | ϕ is continuous} and define the n−th (continuous) differential dn : Cn(G, A) → Cn+1(G, A) by

n n X j d (ϕ)(g0, g1, ··· , gn) =g0ϕ(g1, ··· , gn) + (−1) ϕ(g0, ··· , gj−2, gj−1gj, ··· , gn) j=1 n+1 + (−1) ϕ(g0, ··· , gn−1).

Then (Cn(G, A), dn) is a cochain complex.

Definition 3.6.1. We define the n-th cohomology group of the profinite group G with coefficients in the discrete G-module A to be Zn(G, A) Hn(G, A) = , Bn(G, A) where Zn(G, A) = ker dn for n ≥ 0 are the (continuous) n-cocycles and B0(G, A) = 0, Bn(G, A) = im dn−1 for n ≥ 1 are the (continuous) n-coboundaries.

The cohomology group Hn(G, A) of a profinite group G with coefficients in a discrete G-module A are actually built up from the cohomology groups of finite groups. Let N = {Ni, i ∈ I} be the family of all open normal subgroups of G (hence G/Ni is finite, by Lemma 3.2.3). We define i ≤ j if Uj is an open subgroup of Ui. If Nj ⊂ Ni, then the projections n+1 n+1 n+1 G → (G/Nj) → (G/Ni) induce homomorphisms

n Ni n Nj n C (G/Ni,A ) → C (G/Nj,A ) → C (G, A).

Hence we have homomorphisms

Inf n Ni n Nj n H (G/Ni,A ) > H (G/Nj,A ) > H (G, A).

n N n N n N We deﬁne ϕij : H (G/Ni,A i ) → H (G/Nj,A j ). Then (H (G/Ni,A i ), ϕij,I) is a direct system.

Proposition 3.6.2. Let G be a proﬁnite group, and A be a discrete G-module. For any nonnegative integer n ≥ 0, we have

Hn(G, A) ∼ lim Hn(G/N ,ANi ), = −→ i i

n N where (H (G/Ni,A i ), ϕij,I) is a direct system describe above.

51 3.7 Galois cohomology Let E/F be a Galois extension with Galois group G = Gal(E/F ), which is a proﬁnite group by Proposition 3.3.2. Let {Ki, i ∈ I} be the family of all ﬁnite Galois extensions of F contained in E, and denote Ni = Gal(E/Ki), which are open normal subgroups of G. Then G = lim G/N ←− i i by Corollary 3.2.6. N Consider the additive group E, which is a discrete G-module. We have E i = Ki, ∼ each Ki is a Gal(Ki/F )-module and Gal(Ki/F ) = G/Ni. Let n ≥ 0 be an integer. Then

Hn(G, E) = lim Hn(G/N ,ENi ) ∼ lim Hn(Gal(K /F ),K ). −→ i = −→ i i i i Proposition 3.7.1. Let E/F be a Galois extension with Galois group G = Gal(E/F ). Then Hn(G, E) = 0 for all n ≥ 1.

Proof. This is a consequence of Proposition 3.7.2.

Proposition 3.7.2. Let E/F be a ﬁnite Galois extension with Galois group G = Gal(E/F ). n Then the Tate cohomology groups Hˆ (G, E) = 0 for all n ∈ Z. Proof. The Normal Basis Theorem says that there is an element α ∈ E such that {g(α), g ∈ ∼ G} is a basis of E/F . So E = Z[G] ⊗ F is an induced G-module, hence cohomologically trivial.

The cohomology of Gal(E/F ) with coeﬃcients in the additive group E is uninteresting. Now we look at the multiplicative group E∗ = E\{0} of E. Again, we have (E∗)Ni = ∗ ∗ ∼ Ki , each Ki is a Gal(Ki/F )-module and Gal(Ki/F ) = G/Ni. Let n ≥ 0 be an integer. Then Hn(G, E∗) = lim Hn(G/N , (E∗)Ni ) ∼ lim Hn(Gal(K /F ),K∗). −→ i = −→ i i i i Proposition 3.7.3 (Hilbert’s Theorem 90, Cohomology Version). Let E/F be a Galois extension with Galois group G = Gal(E/F ). Then H1(G, E∗) = 0.

Proof. Let ϕ : G → E∗ be a 1-cocycle (note that the multiplicative version of crossed homomorphism is ϕ(xy) = x(ϕ(y))ϕ(y), ∀x, y ∈ G). By the algebraic independence of automorphisms, there is some a ∈ E∗ such that X b = ϕ(x) · x(a) x∈G

52 is non-zero, i.e., b ∈ E∗. Let y ∈ G. Then X y(b) = [y (ϕ(x))] · [y ◦ x(a)] x∈G X = ϕ−1(y) · ϕ(yx) · [y ◦ x(a)] x∈G X =ϕ−1(y) ϕ(yx) · [y ◦ x(a)] x∈G X =ϕ−1(y) ϕ(z) · z(a) z∈G =ϕ−1(y) · b.

So ϕ(y) = b · y(b)−1. Replacing b with b−1, we get ϕ(y) = b−1 · y(b−1)−1 = y(b) · b−1. Hence ϕ is a 1-coboundary.

Proposition 3.7.4 (Hilbert’s Theorem 90). Let E/F be a ﬁnite cyclic extension with Galois group G = Gal(E/F ) generated by σ. If x ∈ E∗ has norm 1, then there exists ∗ y y ∈ E such that x = σ(y) . Proof. Since G is ﬁnite cyclic, we have

∗ ∗ 1 ∗ {x ∈ E : NE/F (x) = 1} {x ∈ E : NE/F (x) = 1} H (G, E ) = ∗ = y ∗ (1 − σ)(E ) { σ(y) , y ∈ E } by Proposition 2.10.1. The result follows since H1(G, E∗) = 0 by Proposition 3.7.3.

Definition 3.7.5. Let k be a field, let ksep be a separable closure of k. The Brauer group of k is defined to be Br (k) = H2(Gal(ksep/k), ksep∗). Remark 3.7.6. By definition, the Brauer group of a field k is the direct limit

Br (k) = lim H2(Gal(K /k),K∗) −→ i i i

sep where {Ki, i ∈ I} is the family of all ﬁnite Galois extension of k contained in k .

Example 3.7.7. The Brauer group of the ﬁeld R is ∼ Br (R) = Z/2Z.

sep This is because R = C, and Gal(C/R) is ﬁnite cyclic of order 2. Hence

∗Gal(C/R) ∗ 2 ∗ ˆ 0 ∗ C R ∼ Br (R) = H (Gal(C/R), C ) = H (Gal(C/R), C ) = ∗ = ∗ = Z/2Z. NC/R(C ) R+

53 Proposition 3.7.8. Let k ⊂ K ⊂ L be a tower of ﬁeld extensions with L/k, K/k Galois. Then there is an exact sequence

0 > H2(Gal(K/k),K∗) > H2(Gal(L/k),L∗) > H2(Gal(L/K),L∗).

Proof. Take G = Gal(L/k), H = Gal(L/K). Hilbert’s Theorem 90 implies that H1(H,L∗) = 0. The result follows by Proposition 2.6.10.

Corollary 3.7.9. Let K/k be a Galois extension. Then there is an exact sequence

0 > H2(Gal(K/k),K∗) > Br (k) > Br (K).

Corollary 3.7.10. Let {Ki, i ∈ I} be the family of all ﬁnite Galois extensions of a ﬁeld k in a separable closure ksep. Then

[ 2 ∗ Br (k) = H (Gal(Ki/k),Ki ). i∈I

54 4 Local Class Field Theory

4.1 Statements of the main theorems Recall that a field K is a nonarchimedean local field if it is complete with respect to the ∗ topology induced by a discrete valuation v : K Z with finite residue field. Let K be a local field. If K is archimedean, then K is either R or C. If K is nonarchimedean, then K is either a finite extension of the p-adic field Qp for some prime p, or a finite extension of the field of formal Laurent series Fp((T )) over a finite field Fp. We are interested in the case where K is nonarchimedean. Furthermore, for a nonarchimedean local field K, we denote: K∗ := K\{0} = the multiplicative group of K,

OK := {x ∈ K | v(x) ≥ 0} = the valuation ring of K,

mK := {x ∈ K | v(x) > 0} = the unique maximal ideal of OK ,

k := OK /mK = the residue ﬁeld of K,

UK := {x ∈ K | v(x) = 0} = the units of OK , (n) n UK := 1 + mK for n ≥ 1, π := a prime element (uniformizer) of K = a generator of mK

( such that mK = πOK ), K := the algebraic closure of K, Ksep := a fixed separable closure of K, Kunr := the maximal unramified extension of K in Ksep, Kab := the maximal abelian extension of K in Ksep. Let L be a finite unramified extension over a nonarchimedean local field K. Then L is Galois over K, and Gal(L/K) = Gal(l/k), where l is the residue field of L. Moreover, Gal(L/K) is a finite cyclic group, generated by the Frobenius element FrobL/K . Theorem 4.1.1 (Local Reciprocity Law). Let K be a nonarchimedean local field. Then there exists a unique homomorphism

∗ ab φK : K → Gal(K /K) satisfying the following properties: (i) For every prime element π ∈ K and every finite unramified extension L/K, φK (π) acts on L as FrobL/K . ∗ (ii) For every finite abelian extension L/K, NL/K (L ) is contained in the kernel of the map x 7→ φK (x)|L, and φK induces an isomophism ∗ ∗ φL/K : K /NL/K (L ) → Gal(L/K).

55 Remark 4.1.2. The maps φK , φL/K are usually called the local Artin maps, the local reciprocity maps, or the norm residue symbols.

Remark 4.1.3. Local Reciprocity Law says that the following diagram

φ K∗ K> Gal(Kab/K)

∨ ∗ ∨ K φL/K ∗ > Gal(L/K) NL/K (L ) is commutative. This implies that for all ﬁnite extensions K ⊂ E ⊂ L with L/K and E/K abelian, the diagram

∗ K φL/K ∗ > Gal(L/K) NL/K (L ) ∨ ∗ ∨ K φE/K ∗ > Gal(E/K) NE/K (E ) is commutative.

Deﬁnition 4.1.4. A subgroup H of K∗ is called a norm subgroup if there exists a ﬁnite ∗ abelian extension L/K such that H = NL/K (L ).

∗ Corollary 4.1.5. Let K be a nonarchimedean local ﬁeld. The map L 7→ NL/K (L ) is a bijection between the set of all ﬁnite abelian extensions and the set of norm subgroups of K∗.

Corollary 4.1.5 says that finite abelian extension of a nonarchimedean local field K corresponds bijectively to a norm subgroup of K∗. But this is unsatisfactory because to define norm subgroup we still need to know the extension. Local Existence Theorem (Theorem 4.1.6) gives a topological characterization of the norm subgroups of K∗. Hence we can characterize finite abelian extension of K totally in terms of the arithmetic of the ground field K.

Theorem 4.1.6 (Local Existence Theorem). A subgroup H in K∗ is a norm subgroup if and only if H is open of ﬁnite index in K∗.

Part (i) of Local Reciprocity Law will be proved by Proposition 4.3.8 and Remark 4.3.9. Part (ii) of Local Reciprocity Law will be proved by Theorem 4.3.2 and Remark 4.3.3. Local Existence Theorem will be proved by Proposition 4.4.25 and Proposition 4.4.26.

56 4.2 The fundamental class In this section, we will prove that for a ﬁnite Galois extension L of degree n over a nonarchimedean local ﬁeld K, Hˆ 2(Gal(L/K),L∗) is cyclic of order n, generated by the “fundamental class”.

Lemma 4.2.1. Let K be a nonarchimedean local field. Let L/K be a finite unramified extension. We have (1) ∼ ∗ UL/UL = l , and (i) (i+1) ∼ UL /UL = l as Gal(L/K)-modules for any i ≥ 1, where l is the residue field of L.

Proof. Note that since L/K is unramiﬁed, a prime element π of K is also a prime element of L. So (i) i i UL = 1 + mL = {1 + aπ | a ∈ OL}. The isomorphisms follow from the maps

∗ UL → l

u 7→ u mod mL and

(i) UL → l i 1 + aπ 7→ a mod mL

Lemma 4.2.2. Let K be a nonarchimedean local field. Let L/K be a finite unramified i ∗ extension. Then Hˆ (Gal(L/K), l ) = 0 for all i ∈ Z. Proof. Since L/K is finite unramified, Gal(L/K) =∼ Gal(l/k) and Gal(L/K) is finite cyclic. Moreover, l∗ is finite since it is the residue field of L. In particular, l∗ is a finite Gal(L/K)- module. Hence the Herbrand quotient

|Hˆ 0(Gal(L/K), l∗)| h(l∗) = = 1 |Hˆ 1(Gal(L/K), l∗)| by Proposition 2.10.6. Hilbert’s Theorem 90 implies that Hˆ 1(Gal(L/K), l∗) = 0. Hence 0 ∗ i ∗ i ∗ Hˆ (Gal(L/K), l ) = 0. Therefore, Hˆ (Gal(L/K), l ) = 0 for all i ∈ Z since Hˆ (Gal(L/K), l ) is 2-periodic.

57 ∗ ∗ Remark 4.2.3. Lemma 4.2.2 implies that the norm map Nl/k : l → k is surjective. Note ∗ 0 ∗ that the map N : H0(Gal(L/K), l ) → H (Gal(L/K), l ) induced by the norm element is ∗ ∗ exactly the norm Nl/k : l → k . Now ∗ Gal(l/k) ∗ ˆ 0 ∗ (l ) k H (Gal(l/k), l ) = ∗ = ∗ = 0, Nl/k(l ) Nl/k(l ) ∗ ∗ hence Nl/k(l ) = k . Lemma 4.2.4. Let K be a nonarchimedean local field. Let L/K be a finite unramified i extension. Then Hˆ (Gal(L/K), l) = 0 for all i ∈ Z. Proof. Since L/K is finite unramified, Gal(L/K) =∼ Gal(l/k) and Gal(L/K) is finite cyclic. The result follows from Proposition 3.7.2.

Remark 4.2.5. Similarly, Lemma 4.2.4 implies that the trace map Tl/k : l → k is surjective. Proposition 4.2.6. Let K be a nonarchimedean local field. Let L/K be a finite unramified extension. Then the norm NL/K : UL → UK is surjective. ∗ ∗ Proof. Let u ∈ UK . We know that the norm map N : l → k is surjective, by Remark 4.2.3. We also know that ∗ ∼ (1) l = UL/UL , and ∗ ∼ (1) k = UK /UK , (1) (1) by Lemma 4.2.1. So there exists v0 ∈ UL such that N(v0)UK ≡ u mod UK , i.e., (1) u/N(v0) ∈ UK . We know that the trace map T : l → k is surjective, by Remark 4.2.5. Also, by Lemma 4.2.1, we have ∼ (i) (i+1) l = UL /UL , and ∼ (i) (i+1) k = UK /UK , (1) (2) (1) (2) (1) for all i ≥ 1. The norm map N : UL /UL → UK /UK is induced by the map N : UL → (1) (2) UK /UK where Y N(1 + πx) = σ(1 + πx) σ∈Gal(L/K) Y = (1 + σ(πx)) σ∈Gal(L/K) X =1 + σ(πx) mod π2 σ∈Gal(L/K) =1 + T (πx) mod π2 (2) =T (πx) mod UK .

58 (1) (2) (1) (2) (1) So the norm map N : UL /UL → UK /UK is also surjective. So there exists v1 ∈ UL (2) (2) (2) such that N(v1)UK ≡ u/N(v0) mod UK , i.e., u/N(v0v1) ∈ UK . We can proceed (1) (2) (n) in this way to ﬁnd a sequence v0 ∈ UL, v1 ∈ UL , v2 ∈ UL , ··· , vn ∈ UL such that (n+1) Q∞ u/N(v0v1 ··· vn) ∈ UK . Let v = i=0 vi. Notice that

∞ \ (i) u/N(v) ∈ UK = {1}. i=1

Therefore, N(v) = u. Hence the norm map NL/K : UL → UK is surjective. Corollary 4.2.7. Let K be a nonarchimedean local field. Let L/K be an arbitrary unramified extension (possibly infinite). Then

i Hˆ (Gal(L/K),UL) = 0 for all i ∈ Z when Gal(L/K) is finite, and for all i ≥ 1 when Gal(L/K) is infinite. Proof. First consider the case where L/K is a finite unramified extension, then Gal(L/K) is finite cyclic. A prime element π ∈ K is also a prime element in L. So

∗ ∼ Z L = UL × π .

Hence, we have

i ∗ ∼ i i Z Hˆ (Gal(L/K),L ) = Hˆ (Gal(L/K),UL) ⊕ Hˆ (Gal(L/K), π )

1 ∗ 1 Hilbert’s Theorem 90 says that Hˆ (Gal(L/K),L ) = 0 and so Hˆ (Gal(L/K),UL) = 0 as well. By Proposition 4.2.6, we have

Gal(L/K) 0 UL UK Hˆ (Gal(L/K),UL) = = = 0. NL/K (UL) UK

ˆ i Hence H (Gal(L/K),UL) = 0 for all i ∈ Z since the Tate cohomology groups are 2- periodic. Now suppose L/K is unramified and possibly infinite. We no longer have well- defined Tate cohomology groups for i < 0, but we can work with the cohomology groups i H (Gal(L/K),UL) for i ≥ 1. In this case, we have

Hi(Gal(L/K),U ) = lim Hi(Gal(K /K),U ). L −→ i Ki Ki ﬁnite unramiﬁed over K

i Since each term in the direct limit is 0, we have H (Gal(L/K),UL) = 0. This completes the proof.

59 i unr unr Remark 4.2.8. Corollary 4.2.7 implies that H (Gal(K /K),UK ) = 0 for the maximal unramified extension Kunr/K for all i > 0. Corollary 4.2.9. Let K be a nonarchimedean local field. Let L/K be an arbitrary unramified extension (possibly infinite). Then i ∗ ∼ i Hˆ (Gal(L/K),L ) = Hˆ (Gal(L/K), Z) for all i ∈ Z when Gal(L/K) is finite, and for all i ≥ 1 when Gal(L/K) is infinite. Proof. We have an exact sequence

∗ ordL 0 > UL > L > Z > 0 of Gal(L/K)-modules where Gal(L/K) acts trivially on Z. The long exact sequence of Tate cohomology together with Corollary 4.2.7 give us the isomorphism.

Let L/K be unramified of finite degree n over a nonarchimedean local field K. Now consider the exact sequence

0 > Z > Q > Q/Z > 0 of Gal(L/K)-modules with trivial actions. Since multiplication by any positive integer (in particular, |Gal(L/K)|) on Q is an isomorphism Q → Q, hence an isomorphism i i i Hˆ (Gal(L/K), Q) → Hˆ (Gal(L/K), Q). But Gal(L/K) is ﬁnite, hence Hˆ (Gal(L/K), Q) = 0 for all i ∈ Z. By the long exact sequence of the Tate cohomology, we have 1 ∼ 2 Hˆ (Gal(L/K), Q/Z) = Hˆ (Gal(L/K), Z). By Corollary 4.2.9, we have the isomorphism 2 ∗ ∼ 1 Hˆ (Gal(L/K),L ) = Hˆ (Gal(L/K), Q/Z). Since Q/Z is a trivial Gal(L/K)-module, 1 ∼ Hˆ (Gal(L/K), Q/Z) = Hom(Gal(L/K), Q/Z) by Lemma 2.3.8. The map

Hom(Gal(L/K), Q/Z) → Q/Z f 7→ f(FrobL/K )

1 is an isomorphism from Hom(Gal(L/K), Q/Z) onto a subgroup n Z/Z of Q/Z (because L/K is unramiﬁed). We deﬁne the invariant map to be the composition ˆ 2 ∗ ∼ ˆ 2 ∼ invL/K : H (Gal(L/K),L ) = H (Gal(L/K), Z) = Hom(Gal(L/K), Q/Z) → Q/Z

f 7→ f(FrobL/K ). The invariant map is compatible with inﬂation in the following sense.

60 Proposition 4.2.10. Let K be a nonarchimedean local ﬁeld. Let K ⊂ L ⊂ E be a tower of ﬁelds with both L and E contained in Kunr. Then the diagram

2 ∗ invL/K Hˆ (Gal(L/K),L ) > Q/Z

Inf = ∨ ∨ 2 ∗ invE/K Hˆ (Gal(E/K),E ) > Q/Z is commutative. Proposition 4.2.11. Let K be a nonarchimedean local ﬁeld. There is a canonical isomorphism 2 unr unr∗ ∼ invK : H (Gal(K /K),K ) = Q/Z.

Proof. Let {Li, i ∈ I} be the family of all ﬁnite unramiﬁed extension of K. Then H2(Gal(Kunr/K),Kunr∗) ∼ lim H2(Gal(L /K),L∗). = −→ i i i

For every positive integer n, there exists an unramified extension Ln of degree n. Hence 1 the image of invK contains n for every n. Thus invK is an isomorphism. Proposition 4.2.12. Let K be a nonarchimedean local field. Let L/K be a finite extension of degree n. Then the diagram Res H2(Gal(Kunr/K),Kunr∗) > H2(Gal(Lunr/L),Lunr∗)

invK invL ∨ ∨ n Q/Z > Q/Z is commutative. Corollary 4.2.13. Let K be a nonarchimedean local ﬁeld. Let L/K be a ﬁnite extension 2 2 ∗ 2 unr unr∗ of degree n. Let H (L/K)unr = H (Gal(L/K),L ) ∩ H (Gal(K /K),K ). Then 2 2 unr unr∗ H (L/K)unr is cyclic of order n and is generated by the element uL/K ∈ H (Gal(K /K),K ) 1 with invK (uL/K ) = n . Proof. The sequence

2 2 unr unr∗ Res 2 unr unr∗ 0 > H (L/K)unr > H (Gal(K /K),K ) > H (Gal(L /L),L ) is exact. By Proposition 4.2.12, the diagram

2 2 unr unr∗ Res 2 unr unr∗ 0 > H (L/K)unr > H (Gal(K /K),K ) > H (Gal(L /L),L )

invK invL ∨ ∨ 1 n 0 > / > / > / nZ Z Q Z Q Z

61 is commutative with exact rows. Since invK and invL are isomorphisms by Proposition 2 ∼ 1 4.2.11, we obtain that H (L/K)unr = n Z/Z and hence the results follow. Corollary 4.2.14. Let K be a nonarchimedean local ﬁeld. Let L/K be a ﬁnite extension of degree n. Then n divides the order of H2(Gal(L/K),L∗).

Proof. Corollary 4.2.13 says that H2(Gal(L/K),L∗) contains a subgroup of order n. Gal(L/K) is a finite group, and L∗ is a finite generated Gal(L/K)-module by the Normal Basis Theorem. Then Corollary 2.7.14 implies that H2(Gal(L/K),L∗) is finite. Now La- grange’s Theorem in group theory implies that n divides the order of H2(Gal(L/K),L∗).

We have shown that for a finite unramified extension L over a nonarchimedean local i field K, we have Hˆ (Gal(L/K),UL) = 0 for all i, by Corollary 4.2.7. Such strong results do not hold for an arbitrary finite Galois extension. But we have the following results.

Lemma 4.2.15. Let K be a nonarchimedean local field. Let L/K be a finite Galois extension with Galois group Gal(L/K). There exists an open subgroup V of OL stable under the action and Hˆ i(Gal(L/K),V ) = 0 for all i ∈ Z. Proof. Normal Basis Theorem says that there is an element x ∈ L such that {σ(x) | σ ∈ Gal(L/K)} is a basis for L/K. Then there is a common denominator d in OK because Gal(L/K) is finite. Let yσ = dσ(x). Then {yσ | σ ∈ Gal(L/K)} is a basis of L/K with P elements in OL. Let V = σ∈Gal(L/K) OLyσ. Then ∼ ∼ V = OL[Gal(L/K)] = Z[Gal(L/K)] ⊗ OL,

i i.e., V is an induced Gal(L/K)-module. Hence Hˆ (Gal(L/K),V ) = 0 for all i ∈ Z. Proposition 4.2.16. Let K be a nonarchimedean local ﬁeld. Let L/K be a ﬁnite Galois extension with Galois group Gal(L/K). There exists an open subgroup V of UL stable under the action and Hˆ i(Gal(L/K),V ) = 0 for all i ∈ Z. Proof. We assume that K is of characteristic zero (for the characteristic p case, see page 134 of [CF67]). The power series

xn ex = 1 + x + ··· + + ··· n!

62 converges for ord(x) > ord(p)/(p − 1). It deﬁnes an isomorphism of an open neighborhood of 0 in L onto an open neighborhood of 1 in L∗, with the inverse map

x2 x3 log(1 + x) = x − + − · · · . 2 3

Both maps commute with the actions of G. Let V0 be an open neighborhood of 0 with ˆ i m H (Gal(L/K),V0) = 0 for all i ∈ Z, as in Lemma 4.2.15. Then π V0 will have the same m properties. Now take V = exp(π V0) and choose m suﬃciently large enough such that exp is a local isomorphism. Then the result follows.

Corollary 4.2.17. Let K be a nonarchimedean local ﬁeld. Let L/K be a cyclic extension ∗ of degree n with Galois group Gal(L/K). Then h(UL) = 1 and h(L ) = n.

i Proof. By Proposition 4.2.16, there exists an open subgroup V of UL such that Hˆ (Gal(L/K),V ) = 0 for all i ∈ Z. Hence h(V ) = 1. Since UL is a compact set, UL/V is ﬁnite. So h(UL/V ) = 1 by Proposition 2.10.6. The Herbrand quotient is multiplicative, hence

h(UL) = h(V )h(UL/V ) = 1.

∗ ∗ ∼ ∗ We know that L = UL × πZ, so L /UL = πZ = Z. So h(L ) = h(UL)h(Z) = h(Z). We have Gal(L/K) 0 Z Z Hˆ (Gal(L/K), Z) = = , NL/K (Z) nZ and ker(N ) ˆ 1 ˆ L/K H (Gal(L/K), Z) = H0(Gal(L/K), Z) = = 0. D(Z) Hence, ˆ 0 ∗ |H (Gal(L/K), Z)| n h(L ) = h(Z) = = = n. |Hˆ 1(Gal(L/K), Z)| 1

Corollary 4.2.18. Let K be a nonarchimedean local ﬁeld. Let L/K be a cyclic extension of degree n with Galois group Gal(L/K). Then Hˆ 2(Gal(L/K),L∗) is of order n.

We need the following lemma.

63 Lemma 4.2.19. Let G be a ﬁnite group and A be a G-module. Let m, j ≥ 0 be nonnegative integers. We assume that (i) Hˆ i(H,A) = 0 for all 0 < i < j and all subgroups H of G; (ii) if H1 ⊂ H2 ⊂ G, with H1 normal in H2 and H2/H1 is cyclic of prime order, then j H m the order of Hˆ (H2/H1,A 1 ) divides [H2 : H1] . Then the order of Hˆ j(G, A) divides [G : 1]m.

Proof. For a prime p, let Gp be a Sylow p-subgroup of G. Corollary 2.7.15 says that the j j restriction map Res : Hˆ (G, A) → Hˆ (Gp,A) is injective on the p-primary component of i i Hˆ (G, A). This allows us to reduce to the study of Hˆ (Gp,A), since we are interested in the order of Hˆ j(G, A). Thus, we may assume that G is a p-group. The strategy for the proof is by induction on the order of G. Assume that |G| > 1. Let H be a normal subgroup of G of order p. Apply the part j H (ii) of the hypothesis to H1 = H,H2 = G, and we obtain that the order of Hˆ (G/H, A ) divides [G : H]m = pm for all j > 0. By the induction hypothesis, we also obtain that the order of Hˆ j(H,A) divides [H : 1]m. Now part (i) of the hypothesis gives an exact sequence

0 → Hˆ j(G/H, AH ) →Inf Hˆ j(G, A) Res→ Hˆ j(H,A) by Proposition 2.6.10. Thus the order of Hˆ j(G, A) divides |Hˆ j(G/H, AH )| · |Hˆ j(H,A)|, which divides [G : H]m · [H : 1]m = [G : 1]m. This proves the case j > 0. For j = 0, note that we have an exact sequence

Theorem 4.2.20. Let K be a nonarchimedean local field. Let L/K be a finite Galois extension of degree n with Galois group Gal(L/K). Then Hˆ 2(Gal(L/K),L∗) is cyclic ˆ 2 unr unr∗ of order n. Moreover, there exists an element uL/K ∈ H (K /K, K ) generating ˆ 2 ∗ 1 H (Gal(L/K),L ) with invK (uL/K ) = n . Proof. We apply Lemma 4.2.19 to the case G = Gal(L/K), A = L∗, m = 1, j = 2. Note that part (i) of the hypothesis is satisfied by the Hilbert’s Theorem 90, and part (ii) of the hypothesis is satisfied by Corollary 4.2.18. Hence the order of Hˆ 2(Gal(L/K),L∗) divides n. However, by Corollary 4.2.13, Hˆ 2(Gal(L/K),L∗) contains a subgroup of order n generated ˆ 2 unr unr∗ 1 2 ∼ 1 by uL/K ∈ H (L /K, L ) such that invK (uL/K ) = n , namely H (L/K)unr = n Z/Z. ˆ 2 ∗ Therefore, H (Gal(L/K),L ) is cyclic of order n and generated by uL/K .

Remark 4.2.21. The generator uL/K is usually called the fundamental class. It will be used to deﬁne the local reciprocity map.

64 Corollary 4.2.22. Let K be a nonarchimedean local ﬁeld. Then we have

∗ H2(Gal(K/K), K ) = H2(Gal(Kunr/K),Kunr∗).

∗ Proof. Note that Kunr ⊂ K, hence H2(Gal(Kunr/K, Kunr∗) ⊂ H2(Gal(K/K, K ). For the other direction, Theorem 4.2.20 shows that for any ﬁnite Galois extension L/K of degree n, Hˆ 2(Gal(L/K),L∗) is cyclic of order n. Hence Hˆ 2(Gal(L/K),L∗) ⊂ H2(Gal(Kunr/K),Kunr∗). ∗ ∗ Since H2(Gal(K/K, K ) is the union of all such Hˆ 2(Gal(L/K),L∗), H2(Gal(K/K, K ) ⊂ H2(Gal(Kunr/K, Kunr∗).

Corollary 4.2.22 is important in the study of the Brauer groups. For now we use it (together with Corollary 4.2.13 and Theorem 4.2.20) to give the following theorem.

Theorem 4.2.23. Let K be a nonarchimedean local ﬁeld. There exists a canonical isomorphism 2 ∗ invK : H (Gal(K/K), K ) → Q/Z. Moreover, if L/K is a ﬁnite Galois extension of degree n with Galois group G = Gal(L/K), then the following diagram

∗ Res ∗ 0 > H2(G, L∗) > H2(Gal(K/K), K ) > H2(Gal(K/L), K )

invK invL ∨ ∨ 1 n 0 > / > / > / nZ Z Q Z Q Z

2 ∗ 1 is commutative, and therefore we recover the invariant map invL/K : H (G, L ) → n Z/Z.

4.3 The local reciprocity map First, we apply Tate’s Theorem (Theorem 2.12.3) to obtain the following theorem.

Theorem 4.3.1. Let K be a nonarchimedean local field. Let L/K be a finite Galois extension of degree n with Galois group Gal(L/K). Then cup product with uL/K defines an isomorphism

i i+2 ∗ Hˆ (Gal(L/K), Z) → Hˆ (Gal(L/K),L ) α 7→ α ∪ uL/K for all i ∈ Z. Now we apply Theorem 4.3.1 to the case i = −2 to get the following theorem.

65 Theorem 4.3.2. Let K be a nonarchimedean local field. Let L/K be a finite Galois extension of degree n with Galois group Gal(L/K). Then cup product with uL/K defines an isomorphism ∗ ab K Gal(L/K) → ∗ NL/K (L ) where Gab = G/[G, G] is the abelianization of G.

Proof. Note that

∗ Gal(L/K) ∗ ˆ 0 ∗ (L ) K H (Gal(L/K),L ) = ∗ = ∗ , NL/K (L ) NL/K (L ) and ˆ −2 H (Gal(L/K), Z) = H1(Gal(L/K), Z) = Gal(L/K)ab (by Proposition 2.5.7).

Remark 4.3.3. If L/K is ﬁnite abelian, then Gal(L/K)ab = Gal(L/K) and hence we get an isomorphism ∗ ∼ K Gal(L/K) = ∗ . NL/K (L ) Thus we prove part (ii) of Local Reciprocity Law (Theorem 4.1.1).

Definition 4.3.4. Let K be a nonarchimedean local field. Let L/K be a finite Galois extension. We define the local reciprocity map to be

∗ K ab φL/K : ∗ → Gal(L/K) NL/K (L )

∗ as in Theorem 4.3.2, which is the inverse of cup product by uL/K . Let x ∈ K , and let x K∗ be a representative of x in ∗ . We denote NL/K (L )

(x, L/K) := φL/K (x).

Lemma 4.3.5. Let K be a nonarchimedean local ﬁeld. Let K ⊂ E ⊂ L be a tower of ﬁnite Galois extensions. Then we have

Res(uL/K ) = uL/E,

Inf(uE/K ) = [L : E]uL/K .

66 Proof. Consider the following diagram

∗ Res ∗ Res ∗ H2(Gal(K/K), K ) > H2(Gal(K/E), K ) > H2(Gal(K/L), K )

invK invE invL ∨ ∨ ∨ [E:K] [L:E] Q/Z > Q/Z > Q/Z. Note that all vertical maps are isomorphisms by Theorem 4.2.23. Applying the kernel- cokernel lemma and we get the following commutative diagram

Inf Res 0 > H2(Gal(E/K),E∗) > H2(Gal(L/K),L∗) > H2(Gal(L/E),L∗)

∨invE/K ∨invL/K ∨invL/E 1 1 [E:K] 1 0 > / > / > / . [E : K]Z Z [L : K]Z Z [L : E]Z Z The second square commutes, implying that 1 1 [E : K] inv (u ) = [E : K] = = inv (u ) = inv Res(u ) L/K L/K [L : K] [L : E] L/E L/E L/E L/K and hence Res(uL/K ) = uL/E. Similarly, the commutativity of the ﬁrst square implies that

Inf(uE/K ) = [L : E]uL/K .

Proposition 4.3.6. Let K be a nonarchimedean local ﬁeld. Let K ⊂ E ⊂ L be a tower of ﬁnite Galois extensions with G = Gal(L/K), H = Gal(L/E). Then the diagrams

i uL/K i+2 ∗ Hˆ (G, Z) > Hˆ (G, L )

Res Res ∨ ∨ i uL/E i+2 ∗ Hˆ (H, Z) > Hˆ (H,L ) and

i uL/K i+2 ∗ Hˆ (G, Z) > Hˆ (G, L ) ∧ ∧ Cor Cor

i uL/E i+2 ∗ Hˆ (H, Z) > Hˆ (H,L ) are commutative for all i ∈ Z.

67 i Proof. Let a ∈ Hˆ (G, Z). Then

uL/E ∪ Res(a) = Res(uL/K ) ∪ Res(a) (by Lemma 4.3.5)

= Res(uL/K ∪ a) (by Proposition 2.9.6).

i Let b ∈ Hˆ (H, Z). Then

Cor(uL/E ∪ b) = Cor(Res(uL/K ) ∪ b) (by Lemma 4.3.5)

= uL/K ∪ Cor(b) (by Proposition 2.9.6).

Let L/K be a finite unramified extension of degree n with Galois group G = Gal(L/K). Let π be a prime element of K, then it is also a prime element of L, and defines a decomposition ∗ ∼ Z ∼ L = UL × π = UL × Z of G-modules. Thus ˆ i ∗ ∼ ˆ i ˆ i H (G, L ) = H (G, UL) ⊕ H (G, Z) ˆ i for all i ∈ Z. We already know that H (G, UL) = 0 by Corollary 4.2.7. Hence it remains i to consider Hˆ (G, Z). ˆ 1 First we determine a cocycle representing uL/K . Let f ∈ H (G, Q/Z) = Hom(G, Q/Z) i i ˆ 1 be the map such that f(FrobL/K ) = n mod Z for all i ∈ Z. Then f generates H (G, Q/Z) since FrobL/K generates G. Recall that we have an isomorphism 1 ∼ 2 δ : Hˆ (G, Q/Z) = Hˆ (G, Z). ˆ 2 ∗ Thus to determine the generator uL/K of H (G, L ) it is enough to determine δf. We choose a lifting of f to 1-cochain f˜ : G → Q. We choose the cochain f˜ to be the map i i FrobL/K 7→ n where 0 ≤ i ≤ n−1. Then using formulas for the connecting homomorphism δ we obtain ˜ i j i ˜ j ˜ i+j ˜ i δf(FrobL/K , FrobL/K ) = FrobL/K f(FrobL/K ) − f(FrobL/K ) + f(FrobL/K ) ( 0 if i + j ≤ n − 1, = 1 if i + j ≥ n.

Z ∗ ˆ 2 ∗ Recall that we can identify Z with π ⊂ L , so uL/K ∈ H (G, L ) is represented by the cocycle ϕ given by ( 0 if i + j ≤ n − 1, ϕ(Frobi , Frobj ) = L/K L/K π if i + j ≥ n.

ˆ 0 ∗ Note that since H (G, UL) = 0, we have UK ⊂ NL/K (L ) and so the class of π in ∗ ∗ K /NL/K (L ) is well-deﬁned.

68 Proposition 4.3.7. Let K be a nonarchimedean local ﬁeld. Let L/K be an unramiﬁed extension of degree n with Galois group G = Gal(L/K) generated by FrobL/K . Let x ∈ ∗ K∗ ˆ 0 ∗ ˆ 1 K , x ∈ ∗ = H (G, L ) be a representative of x, f ∈ Hom(G, / ) = H (G, / ). NL/K (L ) Q Z Q Z Then f φL/K (x) = invL/K (x ∪ δf).

ab ∼ Proof. Let φL/K (x) be the image of φL/K (x) under the isomorphism G = G = H1(G, Z) = ˆ −2 ˆ 0 ∗ H (G, Z). Then φL/K (x) ∪ uL/K = x ∈ H (G, L ). So

x ∪ δf = uL/K ∪ φL/K (x) ∪ δf = uL/K ∪ φL/K (x) ∪ δf (by Proposition 2.9.4) = uL/K ∪ δ φL/K (x) ∪ f (by Theorem 2.9.7)

ˆ −2 ˆ 1 ∼ ˆ −1 where φL/K (x) ∪ f is given by the cup product H (G, Z) ∪ H (G, Q/Z) = H (G, Q/Z), −1 ∼ 0 −1 represented by i/n since Hˆ (G, Q/Z) = Hˆ (G, Z) = Z/nZ and δ : Hˆ (G, Q/Z) → 0 Hˆ (G, Z) = Z/nZ is induced by the norm map, which is just multiplication by n since Q/Z and Z are trivial G-modules. Hence x ∪ δf = uL/K ∪ δ φL/K (x) ∪ f

= uL/K ∪ i. Note that f φL/K (x) = φL/K (x) ∪ f = i/n. Therefore,

invL/K (x ∪ δf) = invL/K (uL/K ∪ i) i = n = f φL/K (x) .

Proposition 4.3.8. Let K be a nonarchimedean local ﬁeld. Let L/K be an unramiﬁed extension of degree n with Galois group G = Gal(L/K) generated by FrobL/K . Then ord(x) ∗ φL/K (x) = (x, L/K) = FrobL/K for all x ∈ K .

ˆ 2 ∗ Proof. The invariant map invL/K : H (G, L ) → Q/Z is deﬁned by the composition

−1 ˆ 2 ∗ ord ˆ 2 δ ˆ 1 γ invL/K : H (G, L ) > H (G, ) > H (G, / ) > / =∼ Z =∼ Q Z Q Z

69 where γ(f) = f(FrobL/K ). Note that ord(x ∪ δf) = ord(x) ∪ δf. Hence f φL/K (x) = invL/K (x ∪ δf) (by Proposition 4.3.7) = γ ◦ δ−1 ◦ ord(x ∪ δf) = γ ◦ δ−1 (ord(x) ∪ δf) = ord(x) ∪ γ ◦ δ−1(δf) = ord(x) ∪ γ(f)

= ord(x)f(FrobL/K ) ord(x) = f FrobL/K .

ord(x) Since f was arbitrary, we conclude that φL/K (x) = FrobL/K . Remark 4.3.9. Proposition 4.3.8 proves part (i) of Local Reciprocity Law (Theorem 4.1.1) since ord(π) = 1.

4.4 Lubin-Tate formal group law In this section we study formal group laws and Lubin-Tate theory, which provides tools for a straightforward construction of totally ramiﬁed abelian extensions of a nonarchimedean local ﬁeld, and leads to a proof of the Local Existence Theorem.

Deﬁnition 4.4.1. Let A be a commutative ring with 1 and let F ∈ A[[X,Y ]]. We say that F is a commutative formal group law if (a) F (X,F (Y,Z)) = F (F (X,Y ),Z); (b) F (0,Y ) = Y and F (X, 0) = X; (c) there exists a unique G(X) such that F (X,G(X)) = 0; (d) F (X,Y ) = F (Y,X); (e) F (X,Y ) ≡ X + Y mod deg 2.

Remark 4.4.2. (b) implies that F (X,Y ) is of the form F (X,Y ) = X +Y +XYG(X,Y ), i.e., all the higher order terms are crossed terms. (e) implies that F (X,Y ) has no constant term.

Note that two formal power series are said to be congruent mod deg n if and only if they coincide in all terms of degree strictly less than n. Let K be a nonarchimedean local ﬁeld. Let A = OK and let F (X,Y ) be a commutative formal group law deﬁned over OK . If x, y ∈ mK , then F (x, y) converges to an element in mK . Under this composition mK becomes a group and we write F (mK ) to denote this group. For example, if we set F (X,Y ) = X + Y then F (mK ) = mK . If we set F (X,Y ) = X + Y + XY then we obtain the multiplicative group structure on 1 + mK .

70 Deﬁnition 4.4.3. Let F (X,Y ),G(X,Y ) be two commutative formal group laws over a commutative ring A.A homomorphism F → G is a power series h ∈ TA[[T ]] such that

h(F (X,Y )) = G(h(X), h(Y )).

If h has an inverse, i.e., there exists a homomorphism h0 : G → F such that

h ◦ h0 = T = h0 ◦ h, then h is said to be an isomorphism. A homomorphism f : F → F is called an endomorphism. Definition 4.4.4. Let F be a commutative formal group law over A. For any f, g ∈ TA[[T ]], we define (f +F g)(T ) = F (f(T ), g(T )). Definition 4.4.5. Let K be a nonarchimedean local field, q = |k| be the order of the residue field. Choose a prime element π ∈ K. We define Fπ to be the set of formal power series f ∈ OK [[X]] such that (i) f(X) ≡ πX mod deg 2; (ii) f(X) ≡ Xq mod π. Note that two formal power series are said to be congruent mod π if and only if the difference of coefficients of each degree is divisible by π.

p 2 p−1 p Example 4.4.6. K = Qp, π = p, f(X) = pX + 2 X + ··· pX + X is an example of such a power series.

Proposition 4.4.7. Let f, g ∈ Fπ, let n ∈ Z and let φ1(X1, ··· ,Xn) be a linear form in X1, ··· ,Xn with coeﬃcients in OK . Then there exists a unique φ ∈ OK [[X1, ··· ,Xn]] such that (i) φ ≡ φ1 mod deg 2; (ii) f(φ(X1, ··· ,Xn)) = φ(g(X1), ··· , g(Xn)).

Proof. Our approach is to construct such a φ by constructing a sequence {φj} with φj ∈ OK [[X1, ··· ,Xn]] such that φj is unique mod deg j + 1 and satisﬁes (i) and (ii) mod deg j + 1. Then we set φ = lim φj. First we consider φ1, which satisﬁes (i) and (ii) by assumption. Now suppose we have constructed φj for some positive integer j. Because φj is unique mod deg j +1, we must have φi ≡ φj mod deg j +1 for all i ≥ j. Thus φj+1 −φj contains only terms of degree j + 1. By assumption we have

f(φj(X1, ··· ,Xn)) = φj(g(X1), ··· , g(Xn)) mod deg j + 1.

Let Ej+1 = f(φj(X1, ··· ,Xn)) − φj(g(X1), ··· , g(Xn)) mod deg j + 2.

71 Let E φ = φ − j+1 . j+1 j π(1 − πj)

We need to show that φj+1 ∈ OK [[X1, ··· ,Xn]]. It suﬃces to show that π|Ej+1. To see this, note that f(X) ≡ g(X) ≡ Xq mod π, so

Ej+1 ≡ f(φj(X1, ··· ,Xn)) − φj(g(X1), ··· , g(Xn)) q q q ≡ φj(X1, ··· ,Xn) − φj(X1 , ··· ,Xn) ≡ 0 mod π.

Thus π|Ej+1 and hence φj+1 ∈ OK [[X1, ··· ,Xn]]. Since φj ≡ φ1 mod deg 2 by assumption, and Ej+1 has only terms of degree j + 1, we have φj+1 ≡ φj ≡ φ1 mod deg 2.

We have to show that φj+1 satisﬁes (ii) mod deg j + 2. Note that

E (g) f(φ ) − φ (g) ≡ f(φ ) − φ (g) + j+1 mod deg j + 2. j+1 j+1 j+1 j π(1 − πj)

By Taylor expansion, we have

f 00(φ ) f(φ ) = f(φ ) + f 0(φ )(φ − φ ) + j (φ − φ )2 + ··· j+1 j j j+1 j 2! j+1 j −E = f(φ ) + π j+1 + ··· j π(1 − πj) E ≡ f(φ ) − π j+1 mod deg j + 2. j π(1 − πj)

Similarly, if we consider Ej+1(g(X1), ··· , g(Xn)), we have

Ej+1(g(X1), ··· , g(Xn)) ≡ Ej+1(πX1, ··· , πXn) mod deg j + 2 j+1 ≡ π Ej+1(X1, ··· ,Xn) mod deg j + 2.

Therefore,

E (g) f(φ ) − φ (g) ≡ f(φ ) − φ (g) + j+1 mod deg j + 2 j+1 j+1 j+1 j π(1 − πj) E E ≡ f(φ ) − π j+1 − φ (g) + πj+1 j+1 mod deg j + 2 j π(1 − πj) j π(1 − πj)

≡ Ej+1 − Ej+1 mod deg j + 2 ≡ 0 mod deg j + 2.

72 j+1 It only remains to show the uniqueness of φj+1. We write φj+1 = φj + φ and suppose f(φj+1) − φj+1(g) ≡ 0 mod deg j + 2.

j+1 Ej+1 We will show that φ = − π(1−πj ) . As above, we use Taylor expansions to get

j+1 f(φj+1) ≡ f(φj) + πφ mod deg j + 2 and the fact that φj+1 only has terms of degree j + 1 to get

j+1 j+1 φj+1(g(X1), ··· , g(Xn)) ≡ π φ (X1, ··· ,Xn) mod deg j + 1.

Thus

0 ≡ f(φj+1) − φj+1(g) mod deg j + 2 j+1 j+1 ≡ f(φj) + πφ − φj(g) − φ (g) mod deg j + 2 j j+1 ≡ Ej+1 + π(1 − π )φ mod deg j + 2.

j+1 Ej+1 Therefore, φ = − π(1−πj ) . By induction, we are done.

Proposition 4.4.8. For every f ∈ Fπ, there is a unique commutative formal group law Ff with coeﬃcients in OK such that

f(Ff (X,Y )) = Ff (f(X), f(Y )), i.e., f is an endomorphism.

Proof. Let Ff be the unique solution to ( Ff (X,Y ) ≡ X + Y mod deg 2

f(Ff (X,Y )) = Ff (f(X), f(Y )) given by Proposition 4.4.7. We only need to check that Ff is a commutative formal group law. We can do this by uniqueness in Proposition4.4.7. Note that

Ff (X,Ff (Y,Z)) ≡ X + Ff (Y,Z) ≡ X + Y + Z mod deg 2,

Ff (Ff (X,Y ),Z) ≡ Ff (X,Y ) + Z ≡ X + Y + Z mod deg 2,

f(Ff (X,Ff (Y,Z))) = Ff (f(X), f(Ff (Y,Z))) = Ff (f(X),Ff (f(Y ), f(Z))),

f(Ff (Ff (X,Y ),Z)) = Ff (f(Ff (X,Y )), f(Z)) = Ff (Ff (f(X), f(Y )), f(Z)).

Hence Ff (X,Ff (Y,Z)) and Ff (Ff (X,Y ),Z) are both solutions to ( H(X,Y,Z) ≡ X + Y + Z mod deg 2 f(H(X,Y,Z)) = H(f(X), f(Y ), f(Z)).

73 By the uniqueness we have

Ff (X,Ff (Y,Z)) = Ff (Ff (X,Y ),Z).

Ff (0,Y ) and Y are both solutions to ( H(X,Y ) ≡ Y mod deg 2 f(H(X,Y )) = H(f(X), f(Y )).

Hence Ff (0,Y ) = Y . Ff (X, 0) and X are both solutions to ( H(X,Y ) ≡ X mod deg 2 f(H(X,Y )) = H(f(X), f(Y )).

Hence Ff (X, 0) = X. Similarly, we can verify that Ff is indeed a commutative formal group law.

Remark 4.4.9. The formal group laws Ff deﬁned by Proposition 4.4.8 are the Lubin-Tate formal group laws.

Proposition 4.4.10. Let f ∈ Fπ and Ff be the Lubin-Tate formal group law given by Proposition 4.4.8. Then for any a ∈ A = OK there exists a unique [a]f ∈ OK [[X]] such that (i) [a]f commutes with f; (ii) [a]f ≡ aX mod deg 2. Moreover, [a]f is an endomorphism of the group law Ff .

Proof. For any a ∈ OK and any f, g ∈ Fπ, let [a]f,g(T ) be the unique solution to ( [a]f,g(T ) ≡ aT mod deg 2

f([a]f,g(T )) = [a]f,g(g(T )) given by Proposition 4.4.7. Note that

Ff ([a]f,g(X), [a]f,g(Y )) ≡ [a]f,g(X) + [a]f,g(Y ) ≡ aX + aY mod deg 2,

[a]f,g(Fg(X,Y )) ≡ aFg(X,Y ) ≡ aX + aY mod deg 2,

Ff ([a]f,g(g(X)), [a]f,g(g(Y ))) = Ff (f([a]f,g(X)), f([a]f,g(Y ))) = f (Ff ([a]f,g(X), [a]f,g(Y ))) ,

[a]f,g(Fg(g(X), g(Y ))) = [a]f,g (g(Fg(X,Y )) = f ([a]f,g(Fg(X,Y ))) .

Hence both Ff ([a]f,g(X), [a]f,g(Y )) and [a]f,g(Fg(X,Y )) are solutions to ( H(X,Y ) ≡ aX + aY mod deg 2 f(H(X,Y )) = H(g(X), g(Y )).

74 By Proposition 4.4.7, we must have

Ff ([a]f,g(X), [a]f,g(Y )) = [a]f,g(Fg(X,Y )).

Now let f = g, and [a]f = [a]f,f , then we are done.

Remark 4.4.11. Using uniqueness of [a]f in Proposition 4.4.10, we have

[π]f (T ) = f(T ) and [1]f (T ) = T.

Proposition 4.4.12. Let f ∈ Fπ, a ∈ OK . The map a 7→ [a]f := [a]f,f is an injective ring homomorphism of A = OK to the ring EndOK (Ff ).

Proof. We ﬁrst show that [a]f ∈ EndOK (Ff ). Note that

[a]f (Ff (X,Y )) ≡ aFf (X,Y ) ≡ aX + aY mod deg 2, and Ff ([a]f (X), [a]f (Y )) ≡ [a]f (X) + [a]f (Y ) ≡ aX + aY mod deg 2.

Also, since both [a]f and Ff commutes with f, we have

f ([a]f (Ff (X,Y ))) = [a]f (f(Ff (X,Y ))) = [a]f (Ff (f(X), f(Y ))), and similarly

f (Ff ([a]f (X), [a]f (Y ))) = Ff (f([a]f (X)), f([a]f (Y ))) = Ff ([a]f (f(X)), [a]f (f(Y ))) .

By the uniqueness in Proposition 4.4.7 applied to g = f, φ1(X,Y ) = aX + aY , we have

[a]f (Ff (X,Y )) = Ff ([a]f (X), [a]f (Y )) .

Hence [a]f ∈ EndOK (Ff ). We also need to check the necessary properties for a ring homomorphism. The binary operations in the ring EndOK (Ff ) are +Ff and composition. So we need to verify that

[a]f +Ff [b]f = [a + b]f and [ab]f = [a]f ◦ [b]f . Both [a]f + [b]f and [a + b]f commute with f. Also,

[a]f +Ff [b]f ≡ aX + bX ≡ [a + b]f mod deg 2.

By the uniqueness in Proposition 4.4.10, we must have [a]f +Ff [b]f = [a + b]f . A similar argument shows that [ab]f = [a]f ◦ [b]f . Finally, the homomorphism is injective because [a]f ≡ aX mod deg 2, so a can be recovered as the coeﬃcient of the term of degree 1. Also note that [a]f has no constant term by Remark 4.4.2. Hence the map a 7→ [a]f := [a]f,f is injective.

75 ∼ Proposition 4.4.13. Let f, g ∈ Fπ. Then Ff = Fg.

−1 Proof. Let u ∈ UK . Then [u]f,g : Fg → Ff has an inverse [u ]g,f : Ff → Fg. Hence ∼ Ff = Fg. Let K be a nonarchimedean local field, q = |k| be the order of the residue field. We fix a prime element π of K. For f ∈ Fπ, let Ff be the corresponding Lubin-Tate formal group law given in Proposition 4.4.8. We write Mf for the group of points in mK equipped with the formal group law defined by Ff , i.e.,

Mf = Ff (mK ).

Then Mf has an OK -module structure given by

x +Ff y = Ff (x, y),

ax = [a]f (x). for any x, y ∈ Mf and a ∈ OK . Let

n n Ef = ker([π ]f ) and [ n Ef = Ef . n≥1

Then Ef is exactly the torsion submodule of Mf . Let

n n Kπ = K(Ef ),

[ n Kπ = K(Ef ), n≥1 and denote n Gπ,n = Gal(Kπ /K), then we have Gal(K /K) = lim G . π ←− π,n n n ∼ Proposition 4.4.14. The OK -module Ef is isomorphic to K/OK . Hence EndOK (Ef ) = n n ∼ n ∗ OK /(π ) and AutOK (Ef ) = (OK /(π )) .

Proof. (i) Proposition 4.4.13 implies that the choice of f ∈ Fπ is unimportant since the Lubin-Tate formal group laws are isomorphic for all elements of Fπ. Hence we may choose q f(X) = X + πX. Let a ∈ mK . Then f(X) − a = 0 has solutions in K, and in fact they belong to mK (we can prove this by using properties of nonarchimedean absolute values). Note that [π]f (T ) = f(T ), hence the map [π]f : Mf → Mf is surjective (given a ∈ Mf ,

76 there exists x ∈ Mf such that f(x) = a, hence [π]f (x) = f(x) = a). This implies that Mf is a divisible OK -module and hence a direct sum of copies of K/OK . Now consider 1 Ef = ker([πf ]) = {a ∈ Mf :[π]f (a) = 0 = f(a)}. 1 1 1 ∼ Hence Ef is exactly the set of roots of f, so |Ef | = q and hence Ef = OK /(π). Since [π]f is surjective, the sequence

1 n [π]f n−1 0 > Ef > Ef > Ef > 0 n n 1 n is exact. By induction, Ef has q elements. Moreover, since Ef is cyclic, Ef must be n n n ∼ n n cyclic. Therefore, Ef is cyclic of order q , and hence Ef = OK /(π ). Since Ef = ∪n≥1Ef , ∼ n it follows that Ef = K/OK . Also, the action of OK on Ef induces an isomorphism n ∼ n OK /(π ) = EndOK (Ef ).

n n−1 Theorem 4.4.15. (i) For each n ≥ 1, Kπ /K is totally ramified of degree (q − 1)q . n (ii) The action of OK on Ef defines an isomorphism n ∗ n (OK /(π )) → Gal(Kπ /K). n In particular, Kπ /K is abelian. n (iii) For each n ≥ 1, π is a norm from Kπ to K. q Proof. (i)-(ii) Again we choose f(X) = X + πX. Let x1 be a root of f(X), and define xn inductively by choosing xn to be a root of f(X) − xn−1. Consider the tower of Eisenstein extensions n K ⊂ K(x1) ⊂ K(x2) ⊂ · · · ⊂ K(xn) ⊂ Kπ where [K(x1): K] = q − 1 and [K(xn): K(xn−1)] = q for all n ≥ 2. This shows that n−1 K(xn) is totally ramified over K of degree (q − 1)q . n n (n) n Ef is the kernel of [π ]f , i.e., the set of roots of f ◦· · ·◦f = f . Hence Kπ is precisely (n) n the splitting field of f . So Gal(Kπ /K) can be identified with a subgroup of the group n n ∼ n ∗ of permutations of the set Ef , this subgroup is contained in AutOK (Ef ) = (OK /π ) , which has order (q − 1)qn−1. So n−1 n n n−1 (q − 1)q ≥ |Gal(Kπ /K)| = [Kπ : K] ≥ [K(xn): K] = (q − 1)q . n n ∼ n ∼ n ∗ This shows that Kπ = K(xn) and Gal(Kπ /K) = AutOK (Ef ) = (OK /(π )) . (iii) By construction, xn is a root of the polynomial

n−1 (f(X)/X) ◦ f (n−1) = X(q−1)q + ··· + π.

n−1 Since K(xn)/K = (q − 1)q , this polynomial must be the minimal polynomial of xn. Thus (q−1)qn−1 N n (x ) = (−1) π = π. Kπ /K n

77 unr The fact that Kπ is the union of totally ramified extensions of K and K is the union unr of unramified extensions of K implies that Kπ ∩ K = K. We define a homomorphism

∗ unr φπ : K → Gal(Kπ · K /K)

unr ∗ m ∗ by defining the restrictions of φπ(a) to Kπ and K for each a ∈ K . Let a = uπ ∈ K . Define m φπ(a)|Kunr = Frob and −1 φπ(a)|Kπ = [u ]f unr where Frob is the Frobenius element FrobKunr/K . Although K is complete, K is not complete, and we write K[unr for the completion of Kunr with respect to the unique extension of the valuation of K to Kunr. We also write Frob for the Frobenius element Frob . K\unr/K −1 Remark 4.4.16. We use u instead of u in the definition of the map φπ so that both unr Kπ · K and φπ are independent of the choice of the prime element π, as we will see in Theorem 4.4.20. Lemma 4.4.17. The homomorphisms O → O K\unr K\unr x 7→ Frob(x) − x and O → O K\unr K\unr x 7→ Frob(x)/x are surjective with kernels OK and UK respectively. Proposition 4.4.18. Let K be a nonarchimedan local field. Let π, π0 = uπ be two different primes of K. Let Ff and Fg be the Lubin-Tate formal group laws defined by f ∈ Fπ, g ∈ F 0 . Then there exists ε ∈ U such that Frob(ε) = εu and a power series h(T ) ∈ π K\unr O [[T ]] such that K\unr (i) h(T ) ≡ εT mod deg 2; (ii) Frob(h) = h ◦ [u]f ; (iii) h(Ff (X,Y )) = Fg(h(X), h(Y )); (iv) h ◦ [a]f = [a]g ◦ h for all a ∈ OK . Proof. We first show that there exists a h(T ) ∈ O [[T ]] that satisfies (i) and (ii). Let K\unr ε ∈ U such that Frob(ε) = εu (such ε exists by Lemma 4.4.17). Let h (T ) = εT , and K\unr 1 we construct a sequence of polynomials hr such that

r hr(T ) = hr−1(T ) + xT ,

78 Frob(hr) ≡ hr ◦ [u]f mod deg r + 1 for some x ∈ O . Note that h (T ) satisfies (i) and (ii). Suppose that h (T ) satisfies K\unr 1 r (i) and (ii). Let a ∈ O such that Frob(a) − a = c(εu)−r−1 where c is the coefficient K\unr r+1 r+1 of T in hr ◦ [u]f − Frob(hr). Note that such a exists by Lemma 4.4.17. Let b = aε , r+1 and we claim that hr+1(T ) = hr(T ) + bT satisfies (ii). Well, r+1 r+1 Frob(hr+1(T )) = Frob(hr(T )) + Frob(aε T ) r+1 = Frob(hr(T )) + Frob(a)Frob(εT ) c = Frob(h (T )) + a + Frob(εT )r+1 r (εu)r+1 c ≡ Frob(h (T )) + a + (Frob(ε))r+1T r+1 mod deg r + 2 r (εu)r+1 r+1 r+1 r+1 ≡ Frob(hr(T )) + a(Frob(ε)) T + cT mod deg r + 2 r+1 r+1 r+1 r+1 ≡ hr ◦ [u]f (T ) − cT + a(Frob(ε)) T + cT mod deg r + 2 r+1 r+1 ≡ hr ◦ [u]f (T ) + a(Frob(ε)) T mod deg r + 2 r+1 r+1 ≡ hr ◦ [u]f (T ) + a(εu) T mod deg r + 2 r+1 r+1 ≡ hr ◦ [u]f (T ) + aε T ◦ [u]f (T ) mod deg r + 2 r+1 ≡ hr ◦ [u]f (T ) + bT ◦ [u]f (T ) mod deg r + 2

≡ hr+1 ◦ [u]f (T ) mod deg r + 2.

Now take the limit of these hr to get the desired h so that it satisﬁes (i) and (ii). Next, we will show that h can be chosen so that g = Frob h ◦ f ◦ h−1. Deﬁne θ = Frob h ◦ f ◦ h−1. Note that −1 −1 −1 θ = Frob h ◦ f ◦ h = h ◦ [u]f ◦ f ◦ h = h ◦ f ◦ [u]f ◦ h . −1 By (ii) we have T = h ◦ [u]f ◦ (Frob h) (T ), hence −1 −1 h = [u]f ◦ (Frob h) .

Since f and [u]f have coeﬃcients in OK , we have −1 Frob θ = Frob h ◦ f ◦ [u]f ◦ (Frob h) = Frob h ◦ f ◦ h−1 = θ and so θ ∈ OK [[T ]]. Moreover, θ(T ) = Frob h ◦ f ◦ h−1(T ) ≡ Frob επε−1T mod T 2 ≡ εuπε−1T mod T 2 ≡ π0T mod T 2,

79 and θ(T ) = Frob h ◦ f ◦ h−1(T ) −1 q ≡ Frob h ◦ (h ) (T ) mod mK −1 q ≡ Frob h (Frob h (T )) mod mK q ≡ T mod mK .

0 0 Therefore, θ ∈ Fπ0 . Now let θ = [1]g,θ ◦ h. Then θ still satisﬁes (i) and (ii), and

0 0 −1 −1 Frob θ ◦ f ◦ (θ ) = [1]g,θ ◦ θ ◦ [1]g,θ = g. (iii) and (iv) follows from Proposition 4.4.7.

Lemma 4.4.19. Let K be a nonarchimedean local ﬁeld, L/K be an algebraic extension, and x ∈ Lˆ. If x is separable and algebraic over L, then x ∈ L. Proof. Let L0 = Lˆ ∩ K. Let σ ∈ Gal(K/L). We know that σ is continuous and it is the identity on L. Let x ∈ L0. Then x is the limit of elements in L and so the action of σ is trivial on x as well. Thus Gal(K/L) = Gal(K/L0). Then by Galois theory, we obtain L0 = L.

unr Theorem 4.4.20. The ﬁeld Kπ · K and the map φπ are independent of the choice of π.

0 Proof. Let π and π = uπ be two diﬀerent prime elements of K. Let f ∈ Fπ, g ∈ Fπ0 , and deﬁne h as in Proposition 4.4.18. Then

Frob ◦ h ◦ [π]f = (h ◦ [u]f ) ◦ [π]f

= h ◦ [uπ]f 0 = h ◦ [u ]f 0 = [π ]g ◦ h,

1 i.e., Frob ◦ h(f(T )) = g(h(T )). Therefore, for any a ∈ K, if a ∈ Ef , then f(a) = 0, so 1 1 −1 g(h(a)) = 0, and hence h(a) ∈ Eg . Similarly, if b ∈ Eg , then g(b) = 0, so f(h (b)) = 0, −1 1 1 1 and hence h (b) ∈ Ef . So h deﬁnes an isomorphism Ef → Eg . Thus

unr 1 unr 1 K[(Eg ) = K[(h(Ef )) unr 1 ⊂ K[(Ef ) unr −1 1 = K[(h (Eg )) unr 1 ⊂ K[(Eg ).

unr 1 unr 1 unr 1 Thus K[(Eg ) = K[(Ef ). Now combine Lemma 4.4.19 and we get K (Ef ) = unr 1 unr n unr n unr unr K (Eg ). Similarly, K (Ef ) = K (Eg ) for all n ≥ 1. Thus Kπ · K = Kπ0 · K .

80 We need to show that φπ = φπ0 as well. Note that both φπ(π) and φπ0 (π) act as Frob unr unr on K . Thus they agree on K . On Kπ, φπ(π) is the identity map. Our ﬁrst goal is to n n show that φπ0 (π) is also the identity map on Kπ. Note that Kπ is generated by Ef over n n K, and h gives an isomorphism Ef → Eg . It suﬃces that show that for all n ≥ 1 and n −1 0 x ∈ Ef , we have φπ0 (π)(h(x)) = h(x). Since π = u π , we have

−1 0 φπ0 (π) = φπ0 (u )φπ0 (π ) = σ1σ2 where ( Frob on Kunr σ = 1 n 1 on Ef and ( 1 on Kunr σ = 2 n [u]f on Ef . Since h has coeﬃcients in K[unr, we have

φπ0 (π)(h(x)) = σ1σ2(h(x))

= σ1(h(σ2(x)))

= σ1(h([u]f (x))) = h(x).

∗ Hence φπ0 (π) is identity on Kπ. Since π was arbitrary and prime elements generate K , we conclude that φπ does not depend on π.

Lemma 4.4.21. Let n, m ≥ 1. Let Km be the unique unramiﬁed extension of degree m ∗ (n) m n over K. Let Hn,m be the subgroup of K generated by UK and π . Then φπ(a)|Kπ ·Km = 1 for all a ∈ Hn,m.

∗ unr Lemma 4.4.22. Let a ∈ K . Then φK (a)|Kπ·K = φπ(a).

unr Proof. We know that both φK (π) and φπ(π) act as Frob on K . Hence they agree on unr n K . By part (iii) of Theorem 4.4.15, π is a norm from Kπ to K for every n ≥ 1. Hence n φK (π) acts trivially on Kπ for every n ≥ 1. Apply Lemma 4.4.21 with m = 1, then we n n unr get that φπ(π) acts trivially on Kπ for every n ≥ 1. So φK and φπ agree on Kπ · K S n unr unr for all n ≥ 1, so they must agree on the union n≥1 Kπ · K = Kπ · K . We are done since prime elements of K generate K∗ (every a ∈ K∗ can be written as a = uπi, and u = (uπ)π−1).

n ∗ (n) m Lemma 4.4.23. Let n, m ≥ 1. Let Kn,m = Kπ ·Km. Then NKn,m/K (Kn,m) = UK hπ i = Hn,m.

81 Proof. By Lemma 4.4.21, Hn,m is contained in the kernel of the local reciprocity map, i.e., ∗ Hn,m ⊂ NKn,m/K (Kn,m) by part (ii) of Theorem 4.1.1. On the other hand, we have

(n) m [K : Hn,m] = [UK : UK ][hπi : hπ i] = (q − 1)qnm n = [Kπ : K][Km : K] = [Kn,m : K].

By part (ii) of Theorem 4.1.1 we know that φK induces an isomorphism ∗ ∗ φKn,m/K : K /NKn,m/K (Kn,m) → Gal(Kn,m/K). Therefore, ∗ Hn,m = NKn,m/K (Kn,m).

ab unr Theorem 4.4.24. K = Kπ · K and φπ = φK . Now we are ready to prove the Local Existence Theorem. Proposition 4.4.25. Let K be a nonarchimedean local field. Let L/K be a finite abelian ∗ ∗ extension. Then NL/K (L ) is open subgroup of K of finite index. ∗ ∗ ∼ Proof. The local reciprocity map gives an isomorphism φL/K : K /NL/K (L ) = Gal(L/K), ∗ ∗ hence NL/K (L ) is of finite index. Now we need to show that NL/K (L ) is open. Note that UK is compact, hence NL/K (UL) is closed in UK (since UK is Hausdorff, and NL/K is ∗ ∗ ∗ continuous). UK is open in K , so NL/K (UK ) is open in K as well. Therefore, NL/K (L ) ∗ ∗ is open in K too, because it contains an open subgroup NL/K (UK ) of K . Proposition 4.4.26. Let K be a nonarchimedean local field. Suppose H is an open subgroup of K∗ with finite index, then H is a norm subgroup. n i Proof. Because H is open, there is some n ≥ 1 such that UK ⊂ H (since {UK } forms a neighborhood basis for the identity in K∗). Because H is of finite index in K∗, there are only a finite number of cosets of the form πjH, hence there exists some m ≥ 1 such m (n) m that π ∈ H. Hence H contains the subgroup Hn,m generated by UK and π . Let Km be the unique unramified extension of degree m over K, and consider the subfield n ab ∗ Kn,m = Kπ · Km of K . By Lemma 4.4.23, we have Hn,m = NKn,m/K (Kn,m). Let L be the subfield of Kn,m that is fixed by φKn,m/K (H). Then H is the kernel of the map ∗ φK : K → Gal(L/K), and by Theorem 4.1.1, we must have ∗ H = NL/K (L ).

Remark 4.4.27. Proposition 4.4.25 and Proposition 4.4.26 complete the proof of Local Existence Theorem (Theorem 4.1.6).

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