Local Class Field Theory - Department of Mathematics · Local Class Field Theory Pan Yan Summer...

Local Class Field Theory

Pan Yan

Summer 2015

These are notes for a reading course with D. Wright on Local Class Field Theory inSummer 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 31.1 Outline of the course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Goals of the course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Group Cohomology 42.1 Group rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 G-modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 Group cohomology via cochains . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Group cohomology via projective resolutions . . . . . . . . . . . . . . . . . 112.5 Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.6 Change of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.7 Tate cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.8 Tate cohomology via complete resolutions . . . . . . . . . . . . . . . . . . . 292.9 Cup products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.10 Tate cohomology of cyclic groups . . . . . . . . . . . . . . . . . . . . . . . . 322.11 Cohomological triviality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.12 Tate’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 Profinite Groups 453.1 Inverse systems and inverse limits . . . . . . . . . . . . . . . . . . . . . . . . 453.2 Topological structure of profinite groups . . . . . . . . . . . . . . . . . . . . 463.3 Examples of profinite groups . . . . . . . . . . . . . . . . . . . . . . . . . . 473.4 Direct systems and direct limits . . . . . . . . . . . . . . . . . . . . . . . . . 49

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3.5 Discrete G-modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.6 Cohomology of profinite groups . . . . . . . . . . . . . . . . . . . . . . . . . 513.7 Galois cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4 Local Class Field Theory 554.1 Statements of the main theorems . . . . . . . . . . . . . . . . . . . . . . . . 554.2 The fundamental class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.3 The local reciprocity map . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.4 Lubin-Tate formal group law . . . . . . . . . . . . . . . . . . . . . . . . . . 70

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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 & Frohlich [CF67]. Otherreferences include [ANT], [Cas86], [FT91], [LT65], [Mil13], [Neu86], [Ser80], [Sha].

Note: Errata for Cassels & Frohlich [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 thecase of Local Class Field Theory, we are mainly interested in abelian extensions of nonar-chimedean local fields. Specifically, given a nonarchimedean local field K, we want todescribe all finite abelian extensions of K totally in terms of the arithmetic of the basefield K.

To this end, we want to understand group cohomology (especially Tate cohomologyfor 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 LocalReciprocity Law and Local Existence Theorem.

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http://wwwf.imperial.ac.uk/~buzzard/errata.pdf

http://wwwf.imperial.ac.uk/~buzzard/errata.pdf

2 Group Cohomology

2.1 Group rings

Definition 2.1.1. Let G be a group, R a commutative ring with identity. The group ringR[G] is the set of all formal finite sums

∑g∈G agg with each ag ∈ R, i.e.,

R[G] =

∑g∈G

agg | ag ∈ R, almost all ag = 0

=

∑g∈G

agg | ag ∈ R, only finitely many ag 6= 0

.

The addition is defined as∑g∈G

agg

+

∑g∈G

bgg

=∑g∈G

(ag + bg)g,

and the multiplication is the involution∑g∈G

agg

∑g∈G

bgg

=∑g,h∈G

(agbhgh) =∑g∈G

(agh−1bh

)g.

In this course we are mainly interested in the integral group ring

Z[G] =

∑g∈G

agg | ag ∈ Z, almost all ag = 0

.

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 onA (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-actionwith 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.

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Definition 2.2.3. A homomorphism ϕ : A → B of G-modules is a homomorphism ofabelian groups such that ϕ(ga) = gϕ(a) for all a ∈ A, g ∈ G (hence it is compatible withthe 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.

AG−action

> A

B

ϕ

∨G−action

> B

ϕ

∨

If A,B are G-modules, we denote the group of all abelian group homomorphismsA → B as Hom(A,B). Note that Hom(A,B) actually has a G-module structure: ifϕ ∈ Hom(A,B), we can define the map

G×Hom(A,B)→ Hom(A,B)

g · (ϕ : A→ B) 7→

(A→ B

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 asAG, 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, bydefinition. Now suppose B ⊂ A is a trivial submodule of A. Let b ∈ B, for any g ∈ G, wehave 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}.

HenceHomG(A,B) = (Hom(A,B))G.

SoHomG(Z, A) = (Hom(Z, A))G ∼= AG.

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Since the covariant functor Hom is left exact, it follows that AG is also a covariant leftexact 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 cohomologygroup H i(G,A) of G with coefficients in A can be defined as the i-th derived functor onA of the functor of G-invariants. Alternatively, we give more specific definitions in thefollowing 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}.(ii) The i-th differential di = diA : Ci(G,A)→ Ci+1(G,A) is the map

di(ϕ)(g0, g1, · · · , gi) =g0ϕ(g1, · · · , gi) +

i∑j=1

(−1)jϕ(g0, · · · , gj−2, gj−1gj , · · · , gi)

+ (−1)i+1ϕ(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 cochaincomplex

0 > C0(G,A)d0> C1(G,A)

d1> 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

H i(G,A) =Zi(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.

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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.

B2(G,A) = im d1 = {ϕ : G2 → 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

d2(ϕ)(g0, g1, g2) = g0ϕ(g1, g2)− ϕ(g0g1, g2) + ϕ(g0, g1g2)− ϕ(g0, g1).

So

Z2(G,A) = {ϕ : G2 → 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.

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

αi : Ci(G,A)→ Ci(G,B)

f 7→ α ◦ f

which is compatible with the differentials, i.e.,

diB ◦ αi = αi+1 ◦ diA.

Proof. We only need to check the compatibility.

diB(α ◦ f) (g0, · · · , gi) =g0 · (α ◦ f) (g1, · · · , gi) +i∑

j=1

(−1)i(α ◦ f) (g0, · · · , gj−1gj , · · · , gi)

+ (−1)i+1 (α ◦ f)(g0, · · · , gi−1)

=α(g0f(g1, · · · , gi) +

i∑j=1

(−1)jf(g0, · · · , gj−1gj , · · · , gi)

+ (−1)i+1f(g0, · · · , gi−1))

=α ◦ diA(g0, · · · , gi).

Corollary 2.3.10. If α : A → B is a G-module homomorphism, then there are inducedmaps

α∗ : H i(G,A)→ H i(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

0 > Aι> B

π> C > 0

is a short exact sequence of G-modules. Then the resulting sequence

0 > Ci(G,A)ιi> Ci(G,B)

πi> Ci(G,C) > 0

is exact.

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Proof. First, we prove the injectivity of ιi. Suppose ιi(f) = ι◦f = 0, then ι(f(g1, · · · , gi)) =0 for any (g1, · · · , gi) ∈ Gi. So f(g1, · · · , gi) = 0 for any (g1, · · · , gi) ∈ Gi since ι is injec-tive. Hence f = 0.

Second, we prove that im ιi ⊂ kerπi. Let f ∈ im ιi, then there exists f0 ∈ Ci(G,A)such that f = ι ◦ f0. Then πi(f) = π ◦ f = π ◦ (ι ◦ f0) = (π ◦ ι) ◦ f0 = 0. So f ∈ kerπi andhence im ιi ⊂ kerπi.

Third, we prove that kerπi ⊂ im ιi. Let f ∈ kerπi, (g1, · · · , gi) ∈ Gi. Then π ◦f(g1, · · · , gi) = 0. So f(g1, · · · , gi) ∈ kerπ = im ι. So there exists a ∈ A such thatf(g1, · · · , gi) = ι(a). For every (g1, · · · , gi) ∈ Gi and the corresponding a ∈ A, definef0 ∈ Ci(G,A) such that f0(g1, · · · , gi) = a. Then

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

and hence f = ι ◦ f0 = ιi(f0). Hence f ∈ im ιi.Finally, we prove the surjectivity of πi. Let f ∈ Ci(G,C), (g1, · · · , gi) ∈ Gi. Then

f(g1, · · · , gi) ∈ C. So there exists b ∈ B such that f(g1, · · · , gi) = π(b) by the surjectivityof π. For every (g1, · · · , gi) ∈ Gi and the corresponding b ∈ B, define f0 ∈ Ci(G,B) suchthat f0(g1, · · · , gi) = b. Then

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

and hence f = π ◦ f0 = πi(f0).

Here is the main theorem of this section.

Theorem 2.3.12. Suppose that

0 > Aι> B

π> C > 0

is a short exact sequence of G-modules. Then there exists a long exact sequence of abeliangroups

0 > H0(G,A)ι∗> H0(G,B)

π∗> H0(G,C)

δ0> H1(G,A) > · · · .

Proof. Consider the following diagram with exact rows

0 > Ci(G,A)ι> Ci(G,B)

π> Ci(G,C) > 0

0 > Ci+1(G,A)

diA∨ι> Ci+1(G,B)

diB∨π> Ci+1(G,C)

diC∨> 0

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

Ci(G,A)

Bi(G,A)

ι>Ci(G,B)

Bi(G,B)

π>Ci(G,C)

Bi(G,C)> 0

0 > Zi+1(G,A)

diA∨ι> Zi+1(G,B)

diB∨π> Zi+1(G,C)

diC∨

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

Ci(G,A)

Bi(G,A)

ι→ Ci(G,B)

Bi(G,B)

f +Bi(G,A) 7→ ι ◦ f +Bi(G,B).

Suppose we have g + Bi(G,B) ∈ Ci(G,B)Bi(G,B)

, then g ∈ Ci(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,

then there exists f + Bi(G,A) ∈ Ci(G,A)Bi(G,A)

such that ι(f + Bi(G,A)) = ι ◦ f + Bi(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) ∈kerπ. Hence im ι ⊂ kerπ.

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.

Suppose f ∈ Zi+1(G,A) = ker di+1A , then f ∈ Ci+1(G,A), So ι(f) ∈ Ci+1(G,B). Since

di+1B (ι(f)) = ι ◦ di+1

A (f) = 0 and so ι ◦ f ∈ ker di+1B = Zi+1(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)and ι(f)(g1, · · · , gi+1) = ι(f(g1, · · · , gi+1)) = 0 for any (g1, · · · , gi+1) ∈ Gi+1. Hencef(g1, · · · , gi+1) = 0 for any (g1, · · · , gi+1) ∈ Gi+1 since ι : A → B is injective. So ι in thelower row is injective. Now suppose g ∈ im ι. Then there exists f ∈ Zi+1(G,A) such thatg = ι(f). Then π(g) = π ◦ ι(f) = (π ◦ ι)(f) = 0. So f ∈ kerπ. Therefore, im ι ⊂ kerπ.

Third, we need to check the column maps make sense. This is because

diA(Ci(G,A)) ⊂ ker di+1A = Zi+1(G,A),

anddiA(Bi(G,A)) = diA(im di−1

A ) = 0.

Finally, we need to check the diagram is commutative. Let f + Bi(G,A) ∈ Ci(G,A)Bi(G,A)

.

ThendiB ◦ ι

(f +Bi(G,A)

)= diB

(ι ◦ f +Bi(G,B)

)= diB (ι(f)) ,

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ι ◦ diA(f +Bi(G,A)

)= ι(diA(f)

).

diB ◦ ι = ι ◦ diA by Lemma 2.3.9. Hence the diagram is commutative. So we have provedthe claim.

Now apply the Snake Lemma to get the exact sequence

ker diA → ker diB → ker diC → cokerdiA → cokerdiB → cokerdiC

for all i ≥ 0. Note that

ker diA =Zi(G,A)

Bi(G,A)= H i(G,A),

cokerdiA =Zi+1(G,A)

im diA=Zi+1(G,A)

Bi+1(G,A)= H i+1(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

0 > H0(G,A)ι∗> H0(G,B)

π∗> H0(G,C)

δ0> 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 projectiveresolutions.

For i ≥ 0, let Gi+1 denote the direct product of i + 1 copies of G. Z[Gi+1] can beviewed 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-modulehomomorphisms

· · · > Z[Gi+1]di−1> Z[Gi] > · · · > Z[G2]

d0> Z[G]

ε> Z > 0

where ε : Z[G]→ Z is the augmentation map defined by

ε

∑g∈G

agg

=∑g∈G

ag

11

and

di : Z[Gi+2]→ Z[Gi+1]

(g0, · · · , gi+1) 7→i+1∑j=0

(−1)j(g0, · · · , gj−1, gj+1, · · · , gi+1)

for each i ≥ 0.

Remark 2.4.2. We may use (g0, · · · , gj , · · · , gi) ∈ Gi 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

di−1 ◦ di(g0, · · · , gi+1) =i+1∑

k=0,k 6=j(−1)j+k−s(j,k)(g0, · · · , gj , · · · , gk, · · · , gi+1),

where s(j, k) = 0 if k < j, s(j, k) = 1 if k > j. Each possible i-tuple appears twice in theterm with opposite sign. Therefore, di−1 ◦ di = 0.

Next, define

θi : Z[Gi+1]→ Z[Gi+2]

(g0, · · · , gi) 7→ (1, g0, · · · , gi).

Then

di ◦ θi(g0, · · · , gi) = (g0, · · · , gi)−i∑

j=0

(−1)i · (1, g0, · · · , gj , · · · , gi)

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

Sodi ◦ 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

0 > HomG(Z[G], A) > · · · > HomG(Z[Gi+1], A)Di> HomG(Z[Gi+2], A) > · · ·

where Di = DiA is defined by

Di(ϕ) = ϕ ◦ di.

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Theorem 2.4.4. The maps

ψi : HomG(Z[Gi+1], A)→ Ci(G,A)

ψi(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

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

ϕ(h0, h1, · · · , hi) = h0ϕ(1, h−10 h1, · · · , h−1

0 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

ϕ(h0, h1, · · · , hi) = h0f(h−10 h1, · · · , h−1

i−1hi).

Then

ϕ(gh0, gh1, · · · , ghi) = gh0f((gh0)−1gh1, · · · , (ghi−1)−1ghi)

= gh0f(h−10 h1, · · · , h−1

i−1hi)

= gϕ(h0, h1, · · · , hi)

and hence

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

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

13

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

ψi+1(Di(ϕ))(g1, · · · , gi+1)

=Di(ϕ)(1, g1, · · · , g1 · · · gi+1)

=ψ ◦ di(1, g1, · · · , g1 · · · gi+1)

=i+1∑j=0

(−1)jϕ(1, g1, · · · , g1 · · · gj , · · · , g1 · · · gi+1)

=ϕ(g1, g1g2, · · · , g1 · · · gi+1) +

i∑j=1

(−1)jϕ(1, · · · , g1 · · · gj , · · · , g1 · · · gi+1)

+ (−1)i+1ϕ(1, g1, · · · , g1 · · · gi+1)

=g1ψi(ϕ)(g2, · · · , gi+1) +

i∑j=1

(−1)jψi(ϕ)(g1, · · · , gj−2, gj−1gj , gj+1, · · · , gi+1)

+ (−1)i+1ψi(ϕ)(g1, · · · , gi)=di(ψi(ϕ))(g1, · · · , gi+1).

Corollary 2.4.5. The i-th cohomology group of the complex (HomG(Z[Gi+1], A), DiA) is

isomorphic to H i(G,A).

Remark 2.4.6. Actually the standard resolution is a projective resolution, this is becauseZ[Gi+1] ∼=

⊕(g1,··· ,gi)∈Gi Z[G](1, g1, · · · , gi) is free and the fact that every free module is

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 structurevia the action g(a⊗ b) = (ga)⊗ (gb).

Definition 2.5.1. The augmentation map is the homomorphism defined by

ε : Z[G]→ Z∑g

agg 7→∑g

ag.

The augmentation ideal is IG = ker ε.

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

14

Proof. Clearly, g− 1 ∈ ker ε for any g ∈ G. Conversely, if∑

g∈G ag = 0, then∑

g∈G agg =∑g∈G ag(g − 1). So ker ε 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

0 > IGι> Z[G]

ε> Z > 0.

Since tensor product is right exact,

IG ⊗G Aι⊗id> Z[G]⊗G A

ε⊗id> Z⊗G A > 0

is exact. Hence,

Z⊗G A = im(ε⊗ id) ∼=Z[G]⊗G Aker(ε⊗ id)

∼=A

im(ι⊗ id)∼=

A

IG ⊗G A∼=

A

IGA= AG.

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

· · · > Z[G3]⊗G Ad1> Z[G2]⊗G A

d0> Z[G]⊗G A

d−1> 0

induced by the standard resolution. More specifically,

Hi(G,A) =Zi(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 diis 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 byd0 ((g0, g1)⊗ a) = (g1 − g0)a.

SoB0(G,A) = im d0 = IGA.

Hence,

H0(G,A) =Z0(G,A)

B0(G,A)∼=

A

IGA= AG.

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

0 > Aι> B

π> C > 0


· · · > H1(G,C)δ∗> H0(G,A)

ι∗> H0(G,B)

π∗> 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 commutatorsubgroup of G so that Gab = G/[G,G] is the largest abelian quotient of G. Then

H1(G,Z) ∼= Gab.

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

0 > H1(G,Z) >IGI2G

>Z[G]

IG> H0(G,Z) > 0.

Note that the map IGI2G→ Z[G]

IGis induced by IG → Z[G], hence is zero. Therefore, we have

H1(G,Z) ∼=IGI2G

.

Note that I2G = IG · IG is the submodule of Z[G] generated by elements of the form

(g − 1)(g′ − 1), g, g′ ∈ G.

16

The map

G

[G,G]→ IG

I2G

g 7→ (g − 1) + I2G

is an isomorphism. HenceH1(G,Z) ∼= Gab.

2.6 Change of groups

Let H ⊂ G be a subgroup. Suppose B is a H-module. Then we can construct a G-moduleHomH(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

H i(G,HomH(Z[G], B)) ∼= H i(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 ∈ kerψi, thenθ(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 ψiis an isomorphism. Then the result follows.

Remark 2.6.2. There is an analogous result for homology. We can form Z[G] ⊗H B asa G-module via the G-action g · (α ⊗ b) = (gα) ⊗ b. This type of G-modules are inducedfrom 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 haveH i(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, wehave

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

for i ≥ 1. Since Z has a projective resolution of Z by itself, it follows that H i({1}, X) = 0for i ≥ 1. (Here is another way to show that. Suppose ϕ : {1} → X is a 1-cocycle, thenϕ(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 ofcohomology groups, H i({1}, X) = 0 for all i ≥ 1.) Similarly, Hi(G,A) = 0 for i ≥ 1.

Suppose f : G′ → G is a homomorphism of groups, we can regard a G-module A asa G′-module via the map f under the G′-action (g′, a) 7→ f(g′)a. There is an inducedhomomorphism P ′ → P of the standard resolution of Z, and hence homomorphisms of thecohomology groups

f∗ : H i(G,A)→ H i(G′, A)

for any G-module A, i ≥ 0. In particular, if G′ = H is a subgroup of G, and f : H → Gis the embedding, then we have the restriction homomorphisms

Res : H i(G,A)→ H i(H,A).

If H is a normal subgroup of G, consider f : G→ G/H. For any G-module A, AH isa 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) H i(G/H,AH)→ H i(G,AH).

The inflation homomorphisms are the composition of (2.1) with the homomorphism in-duced by AH → A, i.e.,

Inf : H i(G/H,AH)→ H i(G,A).

18

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

f∗ : Hi(G′, A)→ Hi(G,A)

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

Cor : Hi(H,A)→ Hi(G,A).

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

(2.2) H i(G,A)→ H i(G,At).

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

(2.3) H i(G,At)→ H i(G,A).

Proposition 2.6.6. The composition of (2.2) with (2.3) is the identity map on H i(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

(2.4) 0→ Aι→ Hom(Z[G], A)

π→ 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

H i+1(G,A) ∼= H i(G,A∗).

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

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

ι∗→ H i+1(G,Hom(Z[G], A))→ · · ·

19

for i ≥ 0. Since Hom(Z[G], A) is coinduced, H i(G,Hom(Z[G], A)) = 0 for all i ≥ 1, byTheorem 2.6.4. So

0→ H i(G,A∗)δi→ H i+1(G,A)→ 0

is exact, hence δi is both surjective and injective for all i ≥ 1. Therefore, δi is an isomor-phism for all i ≥ 1 and hence

H i(G,A∗) ∼= H i+1(G,A)

for all i ≥ 1.

Remark 2.6.8. There is an analogous result for homology. We regard Z[G] ⊗ A as aG-module via the G-action

g · (α⊗ a) = gα⊗ ga.

We have a short exact sequence

0→ A∗ → Z[G]⊗A π→ A→ 0

where A∗ = kerπ. 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 toprove 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 compo-sition 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 thefollowing commutative diagram

H i(G,A∗)δ> H i+1(G,A)

H i(G,A∗)

σi∨δ> H i+1(G,A)

σi+1∨

20

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

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

0→ H1(G/H,AH)Inf→ 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 mapInf (f) : G→ A via

G→ G/Hf→ AH ↪→ A.

Let g0, g1 ∈ G. Then

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

=(g0H)f(g1H) + f(g0H)

=g0f(g1H) + f(g0H) ( since f(g1H) ∈ AH)

=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−afor 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)

for any h0, h1 ∈ H. So Res (f) ∈ Z1(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 allg ∈ 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, wehave

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

0→ H i(G/H,AH)Inf→ H i(G,A)

Res→ H i(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 exactsequence 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 havean exact sequence

0 > H0(H,A) > H0(H,Hom(Z[G], A)) > H0(H,A∗) > H1(H,A)

in H-cohomology by Theorem 2.3.12. Since H1(H,A) = 0 and by Lemma 2.3.8, we havea short exact sequence

0 > AH > (Hom(Z[G], A))H > (A∗)H > 0

in H-cohomology. Moreover,

(Hom(Z[G], A))H ∼= Hom(Z[G/H], A) ∼= Hom(Z[G/H], AH)

with the trivial H-action. Thus the connecting homomorphism

δi−1 : H i−1(G/H, (A∗)H)→ H i(G/H,AH)

is an isomorphism for i ≥ 2. Now consider the following commutative diagram

0 > H i−1(G/H, (A∗)H)Inf> H i−1(G,A∗)

Res> H i−1(H,A∗)

0 > H i(G/H,AH)

δi−1

∨Inf> H i(G,A)

δi−1

∨Res

> H i(H,A)

δi−1

∨

with vertical maps isomorphisms, and top row exact by inductive hypothesis. Hence thebutton 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. Leti ≥ 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 definedas NG =

∑g∈G g, which defines a map

N = NA = NG,A : A→ A

a 7→∑g∈G

g · a

by left multiplication on any G-module A. The image

im (N) = N(A) = {∑g∈G

g · a | a ∈ A}

is the group of G-norms of A.

Lemma 2.7.1. The norm element induces a map N∗ : AG → AG.

Proof. For any g0 ∈ G, a ∈ A, we have

N((g0 − 1)a) =∑g∈G

g · (g0 − 1)a =∑g∈G

gg0a−∑g∈G

ga = 0

and hence IGA ⊂ kerN . Moreover,

g0 ·N(a) = g0 ·∑g∈G

ga =∑g∈G

g0ga =∑g∈G

ga = N(a)

and hence N(A) ⊂ AG.

Definition 2.7.2. We define

H0(G,A) = coker(N∗) =AG

N(A),

H0(G,A) = ker(N∗) =ker(N)

IGA.

Since G is a finite group, the map

Hom(Z[G], X)→ Z[G]⊗X

ϕ 7→∑g∈G

g ⊗ ϕ(g)

with its inverse

Z[G]⊗X → Hom(Z[G], X)∑g∈G

g ⊗ xg 7→ (ϕ : g 7→ xg)

where X is an abelian group, make these two G-modules an isomorphism. Hence thenotion 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 H0(G,A) =H0(G,A) = 0.

Proof. Let X be an abelian group such that A = Z[G]⊗X. Each element of A is uniquelyof the form

∑g∈G g ⊗ xg where xg ∈ X. Suppose a =

∑g∈G g ⊗ xg ∈ AG, then

g0 · a =∑g∈G

g0g ⊗ g0xg =∑g∈G

g0g ⊗ xg = a =∑g∈G

g ⊗ xg

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 H0(G,A) = 0.

Now suppose a =∑

g∈G g ⊗ xg ∈ ker(N), then N(∑

g∈G g ⊗ xg)

= 0. Hence∑g∈G xg = 0. So

a =∑g∈G

g ⊗ xg =∑g∈G

(g − 1)(1⊗ xg) ∈ IGA.

So ker(N) ⊂ IGA and hence H0(G,A) = 0.

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

H i(G,A) =

H i(G,A), i ≥ 1,

H0(G,A), i = 0,

H0(G,A), i = −1,

H−i−1(G,A), i ≤ −2.

Theorem 2.7.5. Let G be a finite group and A an induced G-module. Then H i(G,A) = 0for all i ∈ Z.

Proof. By Theorem 2.6.4, H i(G,A) = Hi(G,A) = 0 for all i ≥ 1. It suffices to checkH0(G,A) and H0(G,A), which are proved in Proposition 2.7.3.

Theorem 2.7.6 (Tate). Let G be a finite group. Suppose

0 > Aι> B

π> C > 0


· · · > H i(G,A)ι∗> H i(G,B)

π∗> H i(G,C)

δi> H 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

0∨

> H0(G,A)

NA∨> H0(G,B)

NB∨> H0(G,C)

NC∨> H1(G,A)

∨> · · · .

We also have dimension shifting for Tate cohomology groups. Actually the Tate coho-mology 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

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

for any i ∈ Z, where A∗ = Hom(Z[G], A)/A, A∗ is the kernel of the map π : Z[G]⊗A→ Adefined 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 : H i(G,A)→ H i(H,A)

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

H i−1(G,A)Res> H i−1(H,A)

H i(G,A∗)

∨Res> H i(H,A∗)

∨

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

Cor : H i(H,A)→ H i(G,A)

for all i ≥ 0. It is defined for all the Tate cohomology groups H i(G,A) by dimensionshifting.

Definition 2.7.8. We define

Res : H0(G,A)→ H0(H,A)

a 7→∑

g∈G/H

g−1 · a,

25

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

Cor : H0(H,A)→ H0(G,A)

a 7→∑

g∈G/H

g · a.

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)

andCor : H i(H,A)→ H i(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

0 > A∗ > Z[G]⊗A π> A > 0

where A∗ = kerπ, gives rise to the following commutative diagram with exact rows

0 > H1(G,A) > H0(G,A∗) > H0(G,Z[G]⊗A)

0 > H1(H,A)

Res∨

........> H0(H,A∗)

Res∨

> H0(H,Z[G]⊗A).

Res∨

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∗) sinceA∗ = kerπ ⊂ 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 viadimension shifting.

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

Res : H i(G,A)→ H i(H,A)

andCor : H i(H,A)→ H 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 wecheck the case for restriction since the case for corestriction is very similar. Specifically,suppose

0 > Aι> B

π> C > 0

is an exact sequence of G-modules, then we need to check that the diagram

H−1(G,C)δ> H0(G,A)

H−1(H,C)

Res∨

δ> H0(H,A)

Res∨

is commutative.Let c ∈ ker(NG : C → C) ⊂ C, and denote c to be its image in H−1(G,C) = ker(NG)

IGC.

Choose b ∈ B such that π(b) = c and consider N(b) ∈ BG. Note that

π(NG(b)) = π∑g∈G

g · b =∑g∈G

g · π(b) =∑g∈G

g · c = 0,

so NG(b) ∈ ker(π : BG → CG) = im(ι : AG → BG). So there exists some a ∈ AG suchthat ι(a) = NG(b). Then δ(c) is the image of a in H0(G,A). Then Res (δ(c)) is the imageof a in H0(H,A). On the other hand,

Res(c) =∑

g∈G/H

g−1 · c,

where c is the image of c in H−1(H,C). We lift c to∑

g∈G/H g−1 · c in the kernel of NH

on C, then lift it to∑

g∈G/H g−1 · b ∈ B. Note that

NH

∑g∈G/H

g−1 · b

= NG(b) = ι(a),

hence δ(Res(c) is again the image of a in H0(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 cohomologygroups.

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

AGRes> AH

Cor> AG

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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

Cor ◦ Res(a) = Cor(a) =∑

g∈G/H

g · a =∑

g∈G/H

a = na.

On homology, we have

AGRes> AH

Cor> AG

where Res is the map defined in 2.7.8, and Cor is the natural quotient map. Let a ∈ AGand a ∈ AH be a lift of a, then

Cor ◦ Res(a) = Cor(∑

g∈G/H

g−1 · a) = na.

Remark 2.7.12. This also applies to homology groups and cohomology groups of an ar-bitrary 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 Tatecohomology groups H i(G,A) vanishes by n.

Proof. Let H = {1}, then [G, {1}] = n. Corollary 2.7.11 implies that Cor ◦ Res = n onH i(G,A) for all i ∈ Z. On the other hand, H i(H,A) = 0 for all i ∈ Z (for the detail, seethe 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. ThenH i(G,A) is finite for all i ∈ Z.

Proof. By the definitions of cohomology and homology groups in terms of cochains andchains, these groups are finitely generated. By Corollary 2.7.13, they are torsion, hencethey are finite.

Let p be a prime and suppose pm is the highest power of p dividing the order |G| of afinite group G, i.e., |G| = pmq where (p, q) = 1. Sylow’s Theorem tells us that there existssubgroups 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 aprime p. Then for any G-module A and i ∈ Z, the kernel of

Res : H i(G,A)→ H i(Gp, A)

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

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Proof. Let α ∈ H i(G,A) with pnα = 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

Res : H i(G,A)→ H i(Gp, A)

is trivial for all primes p. Then H i(G,A) = 0.

Proof. By Corollary 2.7.15, the p-primary components of H i(G,A) is trivial for all primesp. Hence H 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 ofG. Let P → Z be a resolution by finitely generated free G-modules (e.g. the standardresolution).

Definition 2.8.1. The dual of P is defined as P ∗ = HomZ(P,Z) with the G-modulestructure

(g · ϕ)(x) = ϕ(g−1x).

We have two exact sequences

· · · → P1 → P0ε→ Z→ 0,

0→ Z ε∗→→ P ∗0 → P ∗1 → · · · .

We write P−n = P ∗n−1 for n ≥ 1 and gluing the above two resolutions together gives acomplete 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 H i(G,A) isisomorphic 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⊗ZB 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)

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for i, j ≥ 0. There is an natural map

HomG(Pi, A)⊗Z HomG(Pj , B)→ HomG(Pi ⊗Z Pj , A⊗Z B)

ϕ⊗ ϕ′ 7→ (α⊗ β 7→ ϕ(α)⊗ ϕ′(β)).

These give rise to a map

HomG(Pi, A)⊗Z HomG(Pj , B)→ HomG(Pi+j , A⊗Z B)

ϕ⊗ ϕ′ 7→ ϕ ∪ ϕ′.

Definition 2.9.1. Let ϕ ∈ HomG(Pi, A), ϕ′ ∈ HomG(Pj , B), i, j ≥ 0. The cup productϕ ∪ ϕ′ ∈ HomG(Pi+j , A⊗Z B) is defined by(

ϕ ∪ ϕ′)

(g0, · · · , gi+j) = ϕ(g0, · · · , gi)⊗ ϕ′(gi, · · · , gi+j).

Remark 2.9.2. In cochains, we can define cup products

Ci(G,A)⊗Z Cj(G,B)→ Ci+j(G,A⊗Z B)

by (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 alsocalled cup products,

H i(G,A)⊗Z Hj(G,B)

∪→ H i+j(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

AG ⊗Z BG → (A⊗Z B)G

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

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

for all α2 ∈ H i(G,A2), β ∈ Hj(G,B).

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(iii) If0→ B1 → B → B2 → 0


0→ A⊗Z B1 → A⊗Z B → A⊗Z B2 → 0

is also exact, then

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

for all α ∈ H i(G,A), β2 ∈ Hj(G,B2).

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

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

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

sAB : A⊗Z B → B ⊗Z A

a⊗ b 7→ b⊗ a

and the maps that it induces on cohomology. For all α ∈ H i(G,A), β ∈ Hj(G,B), wehave

s∗AB(α ∪ β) = (−1)ij(β ∪ α).

Proposition 2.9.6. Let G be a group and A,B any G-modules. Let H be a subgroup ofG of finite index.

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

Res (α ∪ β) = Res (α) ∪ Res (β).

(ii) If α ∈ H i(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.

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Theorem 2.9.7. Let G be a finite group and A,B any G-modules. There exists uniquemaps

H i(G,A)⊗Z Hj(G,B)

∪→ H i+j(G,A⊗Z B)

for i, j ∈ Z that are natural in A and B and satisfy the following properties.(i) The diagram

H0(G,A)⊗Z H0(G,B)

∪> H0(G,A⊗Z B)

H0(G,A)⊗Z H0(G,B)

∨∪> H0(G,A⊗Z B)

∨

commutes.(ii) If

0→ A1 → A→ A2 → 0


0→ A1 ⊗Z B → A⊗Z B → A2 ⊗Z B → 0

is also exact, then

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

for all α2 ∈ H i(G,A2), β ∈ Hj(G,B).(iii) If

0→ B1 → B → B2 → 0


0→ A⊗Z B1 → A⊗Z B → A⊗Z B2 → 0

is also exact, then

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

for all α ∈ H i(G,A), β2 ∈ Hj(G,B2).

2.10 Tate cohomology of cyclic groups

Let G =< s > be a finite cyclic group of order n. Let D = s− 1, N =∑

g∈G g =∑n−1

i=0 si.

Then IG =< g − 1 | g ∈ G >=< D > as a Z[G]-module. Let A be any G-module. Noticethat 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).

Therefore, AG = ker(D), IGA = im (D). Hence,

H0(G,A) =AG

N(A)=

ker(D)

im (N),

H0(G,A) =ker(N)

IGA=

ker(N)

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 isodd. Then we obtain a complete resolution of Z

· · · → Z[G]D→ Z[G]

N→ Z[G]D→ Z[G]→ · · · .

Then the complex HomG(K,A) is

· · · ← AN← A

D← AN← A← · · · .

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

Proposition 2.10.1. For any i ∈ Z, we have

H i(G,A) =

{AG

N(A) , i ≡ 0 (mod 2),ker(N)D(A) , i ≡ 1 (mod 2).

In particular, H2(G,Z) = ZG/N(Z) = Z/nZ.

Theorem 2.10.2. Cup product by a generator of H2(G,Z) gives rise to an isomorphism

H i(G,A)→ H 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]ε> Z > 0

and

0 > Z N> Z[G]

D> IG > 0.

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

H0(G,Z)δ> H1(G, IG)

δ> H2(G,Z).

Therefore, it reduces to show that cup product by a generator of H0(G,Z) induces anautomorphism of H i(G,A). By dimension shifting we reduce to the case i = 0. Note that

H0(G,Z) =ZG

N(Z)= Z/nZ,

where n = |G|. A generator b of H0(G,Z) is represented by an integer β which is primeto n, and cup product with b is multiplication by β. Since β is prime to n, there existsan integer γ such that βγ ≡ 1 (mod n). Since H0(G,Z) vanishes by multiplication of n,multiplication by β is therefore an automorphism of H0(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

H0(G,A) > H0(G,B)

H1(G,C)

>

H0(G,C)

>

H1(G,B) <

<

H1(G,A)<

.

Definition 2.10.4. Suppose that H0(G,A) and H1(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) =h0(A)

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, thenthe third one is also defined, and

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

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Proof. Without loss of generality, we assume that h0(A), h1(A), h0(B), h1(B) are finite.We have an exact hexagon, and let M1 be the image of H0(G,A) in H0(G,B), M2 bethe image of H0(G,B) in H0(G,C), and so on in clockwise direction round the hexagon.Then

0→M2 → H0(G,C)→M3 → 0

is exact. Now M2 is finite because it is a homomorphic image of H0(G,B), and M3 is finitebecause it is a subgroup of H1(G,A). Thus H0(G,C) is also finite. Similarly, H1(G,C)is finite. Thus h(C) is defined.

Let mi be the order of Mi for 1 ≤ i ≤ 6. Consider

H1(G,C)δ→ H0(G,A)

ι→ H0(G,B),

we have

M1 = im (ι) ∼=H0(G,A)

ker(ι)=H0(G,A)

im (δ)=H0(G,A)

M6.

Hence |H0(G,A)| = m1m6. Similarly, |H0(G,B)| = m1m2, |H0(G,C)| = m2m3, |H1(G,A)| =m3m4, |H1(G,B)| = m4m5, |H1(G,C)| = m5m6. Therefore,

h(B) =h0(B)

h1(B)=m1m2

m4m5=m1m6

m3m4· m2m3

m5m6= h(A)h(C).

Proposition 2.10.6. Let G be a finite cyclic group. If A is a finite G-module, thenh(A) = 1.

Proof. The sequence

0 > AGι> A

D> A

π> AG > 0

is exact. Hence,AG = im(π) ∼= A/ker(π) = A/im(D),

A/AG ∼= A/im(ι) = A/ker(D) ∼= im(D),

and so |AG| = |AG|. On the other hand,

0 > H1(G,A) > AGN∗> AG > H0(G,A) > 0

is exact because H1(G,A) = H−1(G,A) = H0(G,A). The same technique implies that|H1(G,A)| = |H0(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→ Bis 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 ifH i(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 aninduced 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 freemodule over Z[G], hence over Z[H] on a generating set I for any subgroup H ⊂ G. ThenF ∼=

⊕i∈I Z[H] and so FH = 0 and H0(H,F ) = 0. Hence H i(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-modulesuch 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 psince 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

XG = (HomZ(A,Fp))G = 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. IfH1(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 ofAG over Fp, and lift it to aλ ∈ A. Let F be the free Fp[G]-module generated by aλ, thenwe have the canonical surjection π : F → A, and let R = ker(π). Then

0 > R > Fπ> A > 0

is exact, and hence

0 > RG > FGπ> AG > 0

is exact because H1(G,A) = 0 implies H1(G,R) = 0. The map π induced by π is anisomorphism, so RG = 0. Since pR = 0, we have R = 0 by Lemma 2.11.4. Hence π is anisomorphism. 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. Thefollowing are equivalent.(i) A is an induced G-module.(ii) A is cohomologically trivial.(iii) A is a free Fp[G]-module.(iv) There exists some i ∈ Z such that H i(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

A =⊕i∈I

Fp[G] ∼=⊕i∈I

Z[G]/pZ[G] ∼=⊕i∈I

Z[G]⊗ (Z/pZ) ∼= Z[G]⊗ (⊕i∈I

Fp).

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 andHj−2(G,B) ∼= H i+j(G,A)

for all j ∈ Z. In particular, H1(G,B) = H−2(G,B) ∼= H i(G,A) is trivial. By Lemma2.11.5, B is a free module over Fp[G]. By the case (iii) ⇒ (i) we just proved, B iscohomologically trivial, and hence A is also a cohomologically trivial G-module. ApplyLemma 2.11.5 again, and we obtain that A is a free Fp[G]-module.

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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.(ii) There exists some i ∈ Z such that H i(G,A) = H i+1(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,

0 > Ap> A > A/pA > 0

is an exact sequence, and so we have an exact sequence

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

p> H i+1(G,A).

Since H i(G,A) = H i+1(G,A) = 0, we have H i(G,A/pA) = 0 by the long exact sequenceof 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 trivialby Theorem 2.11.6. Hence the map p : H i(G,A) → H i(G,A) of multiplication by p isan isomorphism on H i(G,A) for all i ∈ Z. For any subgroup H ⊂ G, consider thecommutative diagram

H i(H,A)p> H i(H,A)

H i(G,A)

Cor∨

p> H i(G,A).

Cor∨

It follows that the map p : H i(H,A)→ H i(H,A) of multiplication by p is also an isomor-phism on H i(H,A) for any subgroup H ⊂ G and any i ∈ Z. But H i(H,A) is a p-groupby Corollary 2.7.13, hence H 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 coho-mologically trivial. If a G-module B is Z-torsion free, then HomZ(A,B) is cohomologicallytrivial.

Proof. Since B is Z-torsion free, the sequence

0 > Bp> B > B/pB > 0

is exact, and hence so is the sequence

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

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A/pA is free over Fp[G] by Theorem 2.11.7, hence it is an induced G-module, by Theorem2.11.6. 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 2.11.6again 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 eachprime p. Then A is cohomologically trivial as a G-module if and only if A is cohomologi-cally 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 Sylowp-subgroups are conjugate to each other, given a subgroup H ⊂ G, there exists some g ∈ Gsuch thatHp ⊂ gGpg−1. By the cohomological triviality ofGp, we have H i(g−1Hpg,A) = 0for all i ∈ Z. Since conjugation by g is an isomorphism, we have H i(Hp, A) = 0. Therefore,the restriction homomorphism Res:H i(H,A) → H i(Hp, A) is 0. Since this holds for allprime p, it follows that H i(H,A) = 0 by Corollary 2.7.16. Hence A is a cohomologicallytrivial G-module.

A G-module A is projective if HomG(A, ·) is an exact functor (or equivalent, A is adirect summand of a free G-module).

Example 2.11.10. Projective G-modules are cohomologically trivial. Suppose P and Qare projective G-modules with F = P ⊕ Q being a free G-module. For any subgroupH ⊂ G, F is also free over Z[H]. Then

H i(H,P ) ↪→ H i(H,P )⊕ H i(H,Q) ∼= H i(H,P ⊕Q) = H 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 isa 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. Supposethat A is a cohomologically trivial G-module. Since A is Z-free, it follows that Z[G] ⊗ Ais 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 Corollary2.11.8. Note that

H0 (G,HomZ(A,A∗)) = (HomZ(A,A∗))G ∼= HomZ[G](A,A∗),

39

H0 (G,HomZ(A,Z[G]⊗A)) = (HomZ(A,Z[G]⊗A))G ∼= HomZ[G](A,Z[G]⊗A).

Hence the mapHomZ[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 directsummand 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 areequivalent.(i) A is cohomologically trivial.(ii) For each prime p, there exists some i ∈ Z such that H i(Gp, A) = H i+1(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 Tatecohomology and the fact that projective G-modules are cohomologically trivial.

(ii)⇒ (iii): Choose a free (hence projective) G-module F which maps surjectively ontoA, 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

Hj−1(Gp, A) ∼= Hj(Gp, B)

for every j ∈ Z by the long exact sequence of Tate cohomology. It follows that Hj(Gp, B)vanishes for two consecutive values of j. B is Z-free because B is a subgroup of F . ThenTheorem 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 Gpbe a Sylow p-subgroup of G for each prime p. Suppose that, for each prime p, there existssome j ∈ Z such that

f∗p : H i(Gp, A)→ H i(Gp, B)

is surjective for i = j − 1, an isomorphism for i = j, and injective for i = j + 1. Then

f∗ : H i(H,A)→ H 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 exactsequence

0 > A > B ⊗Hom(Z[Gp], A) > C > 0

of Gp-modules. Since Hom(Z[Gp], A) is Gp-cohomologically trivial,

H i(Gp, B ⊗Hom(Z[Gp], A)) ∼= H i(Gp, B)

for all i ∈ Z. So we have the long exact sequence

· · · → H i(Gp, A)f∗p→ H i(Gp, B)→ H i(Gp, C)

δ→ H i+1(Gp, A)f∗p→ H i+1(Gp, B)→ · · · .

Consider the case i = j−1. The map f∗p being surjective on Hj−1(Gp, A) and f∗p being an

isomorphism and hence injective on Hj(Gp, A) implies that Hj−1(Gp, C) = 0. Similarly,for i = j, f∗p being an isomorphism and hence surjective on Hj(Gp, A) and being injective

on Hj+1(Gp, A) implies that Hj(Gp, C) = 0. By Theorem 2.11.12, C is a cohomologicallytrivial Gp-module. Therefore, each f∗ must be an isomorphism by the long exact sequenceof 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

be a G-module map. Let k ∈ Z and α ∈ Hk(G,A). For each subgroup H ⊂ G, define

ΘiH,α : H i(H,B)→ H i+k(H,C)

β 7→ θ∗(Res(α) ∪ β).

For each prime p, suppose that there exists some j ∈ Z such that the map ΘiGp,α

is

surjective for i = j − 1, an isomorphism for i = j, and injective for i = j + 1. Then ΘiH,α

is an isomorphism for all i ∈ Z and any subgroup H ⊂ G.

Proof. First we consider the case k = 0. For α ∈ H0(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∗ : H i(H,B)→ H i(H,C)

41

agrees with the map ΘiH,α.

To see the claim, we first consider the case i = 0. Let b ∈ BH represent β ∈ H0(H,B).Then

ψ∗(β) = θ(a⊗ b) +N(C) = θ∗(Res(α) ∪ β) = ΘiH,α(β).

In general, we have a commutative diagram

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

0 > C∗

θ′

∨> Z[G]⊗ C

idZ[G]⊗θ∨

> C

θ

∨> 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

and we have a map ψ′ : B∗ → C∗ given by ψ′(b′) = θ′(a⊗ b′) for all b′ ∈ B∗. Then we havea commutative diagram

0 > B∗ > Z[G]⊗Z B > B > 0

0 > C∗

ψ′

∨> Z[G]⊗Z C

idZ[G]⊗ψ∨

> C

ψ

∨> 0

with exact rows. Then we have commutative diagrams

H i−1(H,B)δ> H i(H,B∗)

H i−1(H,C)

ψ∗

∨δ> H i(H,C∗)

(ψ′)∗

∨

and

H i−1(H,B)δ> H i(H,B∗)

H i−1(H,C)

Θi−1H,α∨

δ> H i(H,C∗)

(Θ′)iH,α∨

where (Θ′)iH,α(β) = (θ′)∗(Res(α) ∪ β), and the connecting homomorphisms δ are isomor-phisms since Z[G] ⊗Z B and Z[G] ⊗Z C are induced and hence cohomologically trivial.Now suppose (ψ′)∗ = (Θ′)iH,α on H i(H,B∗), then we have ψ∗ = Θi−1

H,α on H i−1(H,B) aswell. That is to say, if the statement is true for i, then it is also true for i− 1. A similarargument 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 forall i ∈ Z. This proves Theorem 2.12.2 for k = 0.

42

The general case follows from another piece of dimension shifting. Let α ∈ Hk−1(H,A),α′ = δ(α) ∈ Hk(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

H i(H,B) = H i(H,B)

H i+k−1(H,C)

ΘiH,α∨δ> H i+k(H,C∗).

(Θ′)iH,α′∨

It is commutative because

δ ◦ΘiH,α(β) = δ ◦ θ(Res(α) ∪ β) = θ′ ◦ δ(Res(α) ∪ β) = θ′ ◦ (Res(α′) ∪ β) = (Θ′)iH,α′(β).

Also, the connecting homomorphism δ : H i+k−1(H,C)→ H i+k(H,C∗) is an isomorphism.By assumption, there exists some j ∈ Z such that the map Θi

Gp,αis surjective for i = j−1,

an isomorphism for i = j, and injective for i = j + 1. By the commutativity of the abovediagram, we have that the map (Θ′)iH,α′ 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(Θ′)iH,α′ are isomorphisms. By the commutativity of the above diagram again, all the

maps ΘiH,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). LetGp be a Sylow p-subgroup of G for each prime p. Suppose for each prime p, H1(Gp, A) = 0and H2(Gp, A) is a cyclic group of order |Gp| generated by the restriction of α. Then themap

H i(H,Z)→ H i+2(H,A)

β 7→ Res(α) ∪ β

is an isomorphism for any i ∈ Z and any subgroup H ⊂ G.

Proof. Take H = Gp. For i = −1, the map H−1(Gp,Z) → H1(Gp, A) is surjective sinceH1(Gp, A) = H1(Gp, A) = 0. For i = 1, the map H1(Gp,Z) → H3(Gp, A) is injectivesince

H1(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

H0(Gp,Z) =ZGp

NGp(Z)=

Z|Gp|Z

,

The map H0(Gp,Z)→ H2(Gp, A) takes the image of n ∈ Z in H0(Gp,Z) to nRes(α), henceit is an isomorphism by the assumption on H2(Gp, A). Then Theorem 2.12.2 implies thatall 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 togetherwith 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 ≤ iin I, there is a morphism (continuous group homomorphism) ϕij : Gi → Gj such that thediagram

Giϕij> Gj

Gk

ϕjk∨ϕik >

commutes whenever i, j, k ∈ I and k ≤ j ≤ i. In addition, we assume that ϕii is the identitymap on Gi. We may simply write the inverse system as (Gi) if there is no confusion aboutthe index set I.

Definition 3.1.3. Suppose (Gi, ϕij , I) and (G′i′ , ϕ′i′j′ , I

′) are two inverse systems. Letψ : I ′ → I be an order preserving map (i.e., if i′ ≤ j′, then ψ(i′) ≤ ψ(j′)). If for eachi′ ∈ I ′, we have a morphism fi′ : Gψ(i′) → G′i′ such that the diagram

Gψ(j′)

fj′> G′j′

Gψ(i′)

ϕψ(j′)ψ(i′)∨

fi′> G′i′

ϕ′j′i′∨

commutes whenever i′ ≤ j′ in I ′, then Ψ = (ψ; fi′ , i′ ∈ I ′) is a morphism from (Gi, ϕij , I)

to (G′i′ , ϕi′j′ , I′).

Definition 3.1.4. The inverse limit (or projective limit)

G = lim←−i

Gi

of the inverse system (Gi, ϕij , I) is the subgroup of the direct product∏i∈I Gi of topo-

logical groups consisting of the tuples (gi) that satisfy the condition ϕij(gi) = gj if j ≤ i,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, thenlim←−Gi is a closed subgroup of

∏i∈I Gi.

Proof. We show that(∏

i∈I Gi)\(lim←−Gi

)is open in

∏i∈I Gi. Let (gi) ∈

(∏i∈I Gi

)\(lim←−Gi

).

Then there are some m,n ∈ I such that m ≤ n and ϕnm(gn) 6= gm. Choose disjoint neigh-borhoods U and V of ϕnm(gn) and gm in Gm, respectively. Let U ′ ⊂ Gn be an open neigh-borhood of gn such that ϕnm(U ′) ⊂ U . Consider the open subset W =

∏i∈IWi ⊂

∏i∈I Gi,

where Wn = U ′, Wm = V , and Wi = Gi for i 6= m,n. Then W is an open neighborhoodof (gi) in

(∏i∈I Gi

), disjoint from lim←−Gi. Hence

(∏i∈I Gi

)\(lim←−Gi

)is open in

∏i∈I Gi

and therefore, lim←−Gi is closed in∏i∈I Gi.

3.2 Topological structure of profinite groups

Definition 3.2.1. We say that G = lim←−iGi is a profinite group, pro-p group, or pronilpo-tent group if all Gi are, respectively, finite groups, finite p-groups for a fixed prime p, orfinite 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 factsabout 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\ (∪x 6∈HxH), H is closed.

(ii) (⇒): Suppose H be an open subgroup of G, then H is closed in G by (i), and theleft coset xH is also an open subset of G. Moreover, the left cosets of H in G form anopen 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 ∪x 6∈HxH, is closed. Since H is of finiteindex, ∪x 6∈HxH is a finite union of closed sets, hence ∪x 6∈HxH is closed. Therefore, H isopen.

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

Lemma 3.2.4. Suppose G is a Hausdorff, compact, and totally disconnected topologicalgroup. If {Ni, i ∈ I} is a set of open normal subgroups of G, then ∩i∈INi = {1}.

Theorem 3.2.5. If G is a topological group, then G is profinite if and only if G isHausdorff, compact, and totally disconnected.

46

Corollary 3.2.6. If G is a profinite group, then

G ∼= lim←−N

G/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←−N

H/H ∩N


Corollary 3.2.8. Let G be a profinite group, and H ⊂ G be a closed normal subgroup ofG, then

G/H ∼= lim←−N

G/NH


3.3 Examples of profinite groups

Let G be a topological group. Consider the family N = {Ni, i ∈ I} of all normal subgroupsof finite index in G. Note that N is not empty because G ∈ N . If H ∈ N ,K ∈ N , thenH ∩K ∈ N . We make N into a directed partially ordered set by defining j ≤ i if Ni is anopen 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

G = lim←−i∈I

G/Ni

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, thenGp = Gp = lim←−

i∈IG/Ni

is called the pro-p completion of G.

Example 3.3.1. As a special example, consider the group of integers Z. Its profinitecompletion is

Z = lim←−n∈N

Z/nZ.

This is isomorphic to the product of all p-adic integers:

Z ∼=∏p

Zp.

47

Its pro-p completion isZp = lim←−

n∈NZ/pnZ = Zp.

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 andKi/F is a finite Galois extension. Then

E =⋃i∈I

Ki.

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∈IGal(E/Ki) = {1}.

There is a unique topology on Gal(E/F ), called the Krull topology , such that thecollection {Gal(E/Ki), i ∈ I} is a fundamental system of neighborhoods of the identityelement 1 of Gal(E/F ). If the Galois extension E/F is finite, then the Krull topology onGal(E/F ) is the discrete topology.

Consider the family of finite Galois groups {Gal(Ki/F ), i ∈ I}. We define the relationj ≤ i if Kj ⊂ Ki, or equivalently, if Gal(E/Ki) ⊂ Gal(E/Kj). Note that if Ki1 ,Ki2 arefinite Galois extensions of F contained in E, then so is the F -composite Ki1Ki2 , which issome Kj , and i1 ≤ j, i2 ≤ j. Hence I is a directed partially ordered set. For j ≤ i, wedefine

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←−i Gal(Ki/F ). In particular, Gal(E/F ) is a profinitegroup.

Proof. Consider the homomorphism

Ψ : Gal(E/F )→ lim←−i

Gal(Ki/F ) ⊂∏i∈I

Gal(Ki/F )

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 (σi) ∈ lim←−i Gal(Ki/F ), define σ : E → E by σ(x) = σi(x) for x ∈ Ki. Then

σ ∈ Gal(E/F ) and Ψ(σ) = (σi). Hence Ψ is surjective.Finally we use the isomorphism Ψ to transfer the topology from lim←−i Gal(Ki/F ) to

Gal(E/F ).

48

We showed that every Galois group of a Galois extension can be interpreted as aprofinite group in Proposition 3.3.2. Actually the converse is also true. Every profinitegroup 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 fieldextension.

Remark 3.3.4. Theorem 3.3.3 was proved in [Wat73] by generalizing Artin’s theoremthat finite automorphism groups are Galois groups. However, it is still unknown whetherany 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 induc-tive) system of abelian groups over I, is an object (Ai, ϕij , I) where for each i ∈ I, Ai isan abelian group, and for each i ≤ j in I, there is a group homomorphism ϕij : Ai → Ajsuch that the diagram

Aiϕij> Aj

Ak

ϕjk∨ϕik >

commutes whenever i, j, k ∈ I and i ≤ j ≤ k. In addition, we assume that ϕii is theidentity map on Ai. We may simply write the direct system as (Ai) or (Ai, ϕij) if there isno confusion about the index set I.

Definition 3.4.2. Suppose (Ai, ϕij , I) and (A′i′ , ϕ′i′j′ , I

′) are two direct systems. Letψ : I → I ′ be an order preserving map (i.e., if i ≤ j, then ψ(i) ≤ ψ(j)). If for each i ∈ I,we have a homomorphism fi : Ai → A′ψ(i) such that the diagram

Aifi> A′ψ(i)

Aj

ϕij∨

fj> A′ψ(j)

ϕ′ψ(i)ψ(j)∨

commutes whenever i ≤ j in I, then Ψ = (ψ; fi, i ∈ I) is a morphism from (Ai, ϕij , I) to(A′i′ , ϕ

′i′j′ , I

′).

Definition 3.4.3. The direct limit of the direct system (Ai, ϕij , I) is defined as the disjointunion of the groups Ai module an equivalence relation

lim−→Ai = ti∈IAi/ ∼

where for x ∈ Ai, y ∈ Aj , x ∼ y if there exists some k ∈ I with i ≤ k, j ≤ k such thatϕik(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 =⋃

N open normal subgroup of G

AN

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, ⋃

N open normal subgroup of G

AN =⋃

H open subgroup of G

AH .

The direction ⊆ is trivial. To see the other direction, let a ∈⋃H open subgroup of GA

H , then

there is some open subgroup H of G such that a ∈ AH . Let N =⋂g∈G g

−1Hg. Since Gis compact, N is a finite intersection of open sets, hence N is open. N is also a normalsubgroup 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 areequivalent.(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 forprofinite 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 Gsuch that a ∈ AN . Then gN ×{a} is an open neighborhood of (g, a) ∈ G×A mapping toga. 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) isopen.

(ii) ⇒ (i): For any a ∈ A, there is an open normal subgroup

N =⋂g∈G

g−1StabG(a)g

of G that is contained in StabG(a), so a ∈ AStabG(a) ⊂ AN .

Example 3.5.3. Let G be a profinite group. Then any G-module A with the trivialG-action is a discrete G-module. Less trivial examples of discrete G-modules come fromGalois theory. Let E/F be a Galois extension of fields with G = Gal(E/F ). The followingare 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 groupof n-th continuous cochains Cn(G,A) = {ϕ : Gn → A | ϕ is continuous} and define then−th (continuous) differential dn : Cn(G,A)→ Cn+1(G,A) by

dn(ϕ)(g0, g1, · · · , gn) =g0ϕ(g1, · · · , gn) +n∑j=1

(−1)jϕ(g0, · · · , gj−2, gj−1gj , · · · , gn)

+ (−1)n+1ϕ(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 withcoefficients in the discrete G-module A to be

Hn(G,A) =Zn(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 discreteG-module A are actually built up from the cohomology groups of finite groups. LetN = {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 theprojections

Gn+1 → (G/Nj)n+1 → (G/Ni)

n+1

induce homomorphisms

Cn(G/Ni, ANi)→ Cn(G/Nj , A

Nj )→ Cn(G,A).

Hence we have homomorphisms

Hn(G/Ni, ANi) > Hn(G/Nj , A

Nj )Inf> Hn(G,A).

We define ϕij : Hn(G/Ni, ANi) → Hn(G/Nj , A

Nj ). Then (Hn(G/Ni, ANi), ϕij , I) is a

direct system.

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

Hn(G,A) ∼= lim−→i

Hn(G/Ni, ANi),

where (Hn(G/Ni, ANi), ϕij , I) is a direct system describe above.

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3.7 Galois cohomology

Let E/F be a Galois extension with Galois group G = Gal(E/F ), which is a profinitegroup by Proposition 3.3.2. Let {Ki, i ∈ I} be the family of all finite Galois extensions ofF contained in E, and denote Ni = Gal(E/Ki), which are open normal subgroups of G.Then

G = lim←−i

G/Ni

by Corollary 3.2.6.Consider the additive group E, which is a discrete G-module. We have ENi = 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−→i

Hn(G/Ni, ENi) ∼= lim−→

i

Hn(Gal(Ki/F ),Ki).

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 ).Then the Tate cohomology groups Hn(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 cohomologicallytrivial.

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 =

K∗i , each K∗i is a Gal(Ki/F )-module and Gal(Ki/F ) ∼= G/Ni. Let n ≥ 0 be an integer.Then

Hn(G,E∗) = lim−→i

Hn(G/Ni, (E∗)Ni) ∼= lim−→

i

Hn(Gal(Ki/F ),K∗i ).

Proposition 3.7.3 (Hilbert’s Theorem 90, Cohomology Version). Let E/F be a Galoisextension 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 crossedhomomorphism is ϕ(xy) = x(ϕ(y))ϕ(y),∀x, y ∈ G). By the algebraic independence ofautomorphisms, there is some a ∈ E∗ such that

b =∑x∈G

ϕ(x) · x(a)

52

is non-zero, i.e., b ∈ E∗. Let y ∈ G. Then

y(b) =∑x∈G

[y (ϕ(x))] · [y ◦ x(a)]

=∑x∈G

ϕ−1(y) · ϕ(yx) · [y ◦ x(a)]

=ϕ−1(y)∑x∈G

ϕ(yx) · [y ◦ x(a)]

=ϕ−1(y)∑z∈G

ϕ(z) · z(a)

=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 withGalois group G = Gal(E/F ) generated by σ. If x ∈ E∗ has norm 1, then there existsy ∈ E∗ such that x = y

σ(y) .

Proof. Since G is finite cyclic, we have

H1(G,E∗) ={x ∈ E∗ : NE/F (x) = 1}

(1− σ)(E∗)={x ∈ E∗ : NE/F (x) = 1}

{ yσ(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 groupof 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−→i

H2(Gal(Ki/k),K∗i )

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

Example 3.7.7. The Brauer group of the field R is

Br (R) ∼= Z/2Z.

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

Br (R) = H2(Gal(C/R),C∗) = H0(Gal(C/R),C∗) =C∗Gal(C/R)

NC/R(C∗)=

R∗

R∗+∼= Z/2Z.

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. TakeG = Gal(L/k), H = Gal(L/K). Hilbert’s Theorem 90 implies thatH1(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 fieldk in a separable closure ksep. Then

Br (k) =⋃i∈I

H2(Gal(Ki/k),K∗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 thetopology 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 nonar-chimedean, then K is either a finite extension of the p-adic field Qp for some prime p, or afinite extension of the field of formal Laurent series Fp((T )) over a finite field Fp. We areinterested 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 ,

U(n)K := 1 + mn

K 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 Lis 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. Thenthere exists a unique homomorphism

φK : K∗ → Gal(Kab/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 localreciprocity maps, or the norm residue symbols.

Remark 4.1.3. Local Reciprocity Law says that the following diagram

K∗φK> Gal(Kab/K)

K∗

NL/K(L∗)

∨φL/K

> Gal(L/K)∨

is commutative. This implies that for all finite extensions K ⊂ E ⊂ L with L/K andE/K abelian, the diagram

K∗

NL/K(L∗)

φL/K> Gal(L/K)

K∗

NE/K(E∗)

∨φE/K

> Gal(E/K)∨

is commutative.

Definition 4.1.4. A subgroup H of K∗ is called a norm subgroup if there exists a finiteabelian 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 abijection between the set of all finite abelian extensions and the set of norm subgroups ofK∗.

Corollary 4.1.5 says that finite abelian extension of a nonarchimedean local field Kcorresponds bijectively to a norm subgroup of K∗. But this is unsatisfactory becauseto 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∗. Hencewe can characterize finite abelian extension of K totally in terms of the arithmetic of theground field K.

Theorem 4.1.6 (Local Existence Theorem). A subgroup H in K∗ is a norm subgroup ifand 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 Remark4.3.9. Part (ii) of Local Reciprocity Law will be proved by Theorem 4.3.2 and Remark4.3.3. Local Existence Theorem will be proved by Proposition 4.4.25 and Proposition4.4.26.

56

4.2 The fundamental class

In this section, we will prove that for a finite Galois extension L of degree n over anonarchimedean local field K, H2(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 unramifiedextension. We have

UL/U(1)L∼= l∗,

andU

(i)L /U

(i+1)L

∼= 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 elementof L. So

U(i)L = 1 + mi

L = {1 + aπi | a ∈ OL}.

The isomorphisms follow from the maps

UL → l∗

u 7→ u mod mL

and

U(i)L → l

1 + aπi 7→ a mod mL

Lemma 4.2.2. Let K be a nonarchimedean local field. Let L/K be a finite unramifiedextension. Then H i(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

h(l∗) =|H0(Gal(L/K), l∗)||H1(Gal(L/K), l∗)|

= 1

by Proposition 2.10.6. Hilbert’s Theorem 90 implies that H1(Gal(L/K), l∗) = 0. HenceH0(Gal(L/K), l∗) = 0. Therefore, H i(Gal(L/K), l∗) = 0 for all i ∈ Z since H i(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. Notethat the map N : H0(Gal(L/K), l∗) → H0(Gal(L/K), l∗) induced by the norm element isexactly the norm Nl/k : l∗ → k∗. Now

H0(Gal(l/k), l∗) =(l∗)Gal(l/k)

Nl/k(l∗)=

k∗

Nl/k(l∗)= 0,

hence Nl/k(l∗) = k∗.

Lemma 4.2.4. Let K be a nonarchimedean local field. Let L/K be a finite unramifiedextension. Then H i(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 surjec-tive.

Proposition 4.2.6. Let K be a nonarchimedean local field. Let L/K be a finite unramifiedextension. 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 Remark4.2.3. We also know that

l∗ ∼= UL/U(1)L ,

andk∗ ∼= UK/U

(1)K ,

by Lemma 4.2.1. So there exists v0 ∈ UL such that N(v0)U(1)K ≡ u mod U

(1)K , i.e.,

u/N(v0) ∈ U (1)K . We know that the trace map T : l → k is surjective, by Remark 4.2.5.

Also, by Lemma 4.2.1, we have

l ∼= U(i)L /U

(i+1)L ,

andk ∼= U

(i)K /U

(i+1)K ,

for all i ≥ 1. The norm map N : U(1)L /U

(2)L → U

(1)K /U

(2)K is induced by the map N : U

(1)L →

U(1)K /U

(2)K where

N(1 + πx) =∏

σ∈Gal(L/K)

σ(1 + πx)

=∏

σ∈Gal(L/K)

(1 + σ(πx))

=1 +∑

σ∈Gal(L/K)

σ(πx) mod π2

=1 + T (πx) mod π2

=T (πx) mod U(2)K .

58

So the norm map N : U(1)L /U

(2)L → U

(1)K /U

(2)K is also surjective. So there exists v1 ∈ U (1)

L

such that N(v1)U(2)K ≡ u/N(v0) mod U

(2)K , i.e., u/N(v0v1) ∈ U

(2)K . We can proceed

in this way to find a sequence v0 ∈ UL, v1 ∈ U(1)L , v2 ∈ U

(2)L , · · · , vn ∈ U

(n)L such that

u/N(v0v1 · · · vn) ∈ U (n+1)K . Let v =

∏∞i=0 vi. Notice that

u/N(v) ∈∞⋂i=1

U(i)K = {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 un-ramified extension (possibly infinite). Then

H i(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

L∗ ∼= UL × πZ.

Hence, we have

H i(Gal(L/K), L∗) ∼= H i(Gal(L/K), UL)⊕ H i(Gal(L/K), πZ)

Hilbert’s Theorem 90 says that H1(Gal(L/K), L∗) = 0 and so H1(Gal(L/K), UL) = 0 aswell. By Proposition 4.2.6, we have

H0(Gal(L/K), UL) =UL

Gal(L/K)

NL/K(UL)=UKUK

= 0.

Hence H i(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 groupsH i(Gal(L/K), UL) for i ≥ 1. In this case, we have

H i(Gal(L/K), UL) = lim−→Ki finite unramified over K

H i(Gal(Ki/K), UKi).

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

59

Remark 4.2.8. Corollary 4.2.7 implies that H i(Gal(Kunr/K), UunrK ) = 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 un-ramified extension (possibly infinite). Then

H i(Gal(L/K), L∗) ∼= H i(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

0 > UL > L∗ordL

> Z > 0

of Gal(L/K)-modules where Gal(L/K) acts trivially on Z. The long exact sequence ofTate 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. Nowconsider 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 isomorphismH i(Gal(L/K),Q)→ H i(Gal(L/K),Q). But Gal(L/K) is finite, hence H i(Gal(L/K),Q) =0 for all i ∈ Z. By the long exact sequence of the Tate cohomology, we have

H1(Gal(L/K),Q/Z) ∼= H2(Gal(L/K),Z).

By Corollary 4.2.9, we have the isomorphism

H2(Gal(L/K), L∗) ∼= H1(Gal(L/K),Q/Z).

Since Q/Z is a trivial Gal(L/K)-module,

H1(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/Zf 7→ f(FrobL/K)

is an isomorphism from Hom(Gal(L/K),Q/Z) onto a subgroup 1nZ/Z of Q/Z (because

L/K is unramified). We define the invariant map to be the composition

invL/K : H2(Gal(L/K), L∗) ∼= H2(Gal(L/K),Z) ∼= Hom(Gal(L/K),Q/Z)→ Q/Zf 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 towerof fields with both L and E contained in Kunr. Then the diagram

H2(Gal(L/K), L∗)invL/K

> Q/Z

H2(Gal(E/K), E∗)

Inf∨

invE/K> Q/Z

=∨

is commutative.

Proposition 4.2.11. Let K be a nonarchimedean local field. There is a canonical iso-morphism

invK : H2(Gal(Kunr/K),Kunr∗) ∼= Q/Z.

Proof. Let {Li, i ∈ I} be the family of all finite unramified extension of K. Then

H2(Gal(Kunr/K),Kunr∗) ∼= lim−→i

H2(Gal(Li/K), L∗i ).

For every positive integer n, there exists an unramified extension Ln of degree n. Hencethe image of invK contains 1

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 extensionof degree n. Then the diagram

H2(Gal(Kunr/K),Kunr∗)Res> H2(Gal(Lunr/L), Lunr∗)

Q/Z

invK∨

n> Q/Z

invL∨

is commutative.

Corollary 4.2.13. Let K be a nonarchimedean local field. Let L/K be a finite extensionof degree n. Let H2(L/K)unr = H2(Gal(L/K), L∗) ∩ H2(Gal(Kunr/K),Kunr∗). ThenH2(L/K)unr is cyclic of order n and is generated by the element uL/K ∈ H2(Gal(Kunr/K),Kunr∗)

with invK(uL/K) = 1n .

Proof. The sequence

0 > H2(L/K)unr > H2(Gal(Kunr/K),Kunr∗)Res> H2(Gal(Lunr/L), Lunr∗)

is exact. By Proposition 4.2.12, the diagram

0 > H2(L/K)unr > H2(Gal(Kunr/K),Kunr∗)Res> H2(Gal(Lunr/L), Lunr∗)

0 >1

nZ/Z > Q/Z

invK∨

n> Q/Z

invL∨

61

is commutative with exact rows. Since invK and invL are isomorphisms by Proposition4.2.11, we obtain that H2(L/K)unr

∼= 1nZ/Z and hence the results follow.

Corollary 4.2.14. Let K be a nonarchimedean local field. Let L/K be a finite extensionof 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 NormalBasis 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 localfield K, we have H i(Gal(L/K), UL) = 0 for all i, by Corollary 4.2.7. Such strong resultsdo 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 Galoisextension with Galois group Gal(L/K). There exists an open subgroup V of OL stableunder the action and

H 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 becauseGal(L/K) is finite. Let yσ = dσ(x). Then {yσ | σ ∈ Gal(L/K)} is a basis of L/K withelements in OL. Let V =

∑σ∈Gal(L/K)OLyσ. Then

V ∼= OL[Gal(L/K)] ∼= Z[Gal(L/K)]⊗OL,

i.e., V is an induced Gal(L/K)-module. Hence H i(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 Galoisextension with Galois group Gal(L/K). There exists an open subgroup V of UL stableunder the action and

H 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 page134 of [CF67]). The power series

ex = 1 + x+ · · ·+ xn

n!+ · · ·

62

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

log(1 + x) = x− x2

2+x3

3− · · · .

Both maps commute with the actions of G. Let V0 be an open neighborhood of 0 withH i(Gal(L/K), V0) = 0 for all i ∈ Z, as in Lemma 4.2.15. Then πmV0 will have the sameproperties. Now take V = exp(πmV0) and choose m sufficiently large enough such thatexp 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 extensionof degree n with Galois group Gal(L/K). Then h(UL) = 1 and h(L∗) = n.

Proof. By Proposition 4.2.16, there exists an open subgroup V of UL such that H i(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 ) = 1by 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). Wehave

H0(Gal(L/K),Z) =ZGal(L/K)

NL/K(Z)=

ZnZ

,

and

H1(Gal(L/K),Z) = H0(Gal(L/K),Z) =ker(NL/K)

D(Z)= 0.

Hence,

h(L∗) = h(Z) =|H0(Gal(L/K),Z)||H1(Gal(L/K),Z)|

=n

1= n.

Corollary 4.2.18. Let K be a nonarchimedean local field. Let L/K be a cyclic extensionof degree n with Galois group Gal(L/K). Then H2(Gal(L/K), L∗) is of order n.

Proof. We have

h(L∗) =|H2(Gal(L/K), L∗)||H1(Gal(L/K), L∗)|

= n

by Corollary 4.2.17. Moreover, Hilbert’s Theorem 90 says that H1(Gal(L/K), L∗) = 0and so |H1(Gal(L/K), L∗)| = 1. Hence |H2(Gal(L/K), L∗)| = 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 nonnegativeintegers. We assume that

(i) H 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

the order of Hj(H2/H1, AH1) divides [H2 : H1]m.

Then the order of Hj(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 therestriction map Res : Hj(G,A) → Hj(Gp, A) is injective on the p-primary component ofH i(G,A). This allows us to reduce to the study of H i(Gp, A), since we are interested inthe order of Hj(G,A). Thus, we may assume that G is a p-group. The strategy for theproof 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(ii) of the hypothesis to H1 = H,H2 = G, and we obtain that the order of Hj(G/H,AH)divides [G : H]m = pm for all j > 0. By the induction hypothesis, we also obtain that theorder of Hj(H,A) divides [H : 1]m. Now part (i) of the hypothesis gives an exact sequence

0→ Hj(G/H,AH)Inf→ Hj(G,A)

Res→ Hj(H,A)

by Proposition 2.6.10. Thus the order of Hj(G,A) divides |Hj(G/H,AH)| · |Hj(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

AH

NH(A)

Cor>

AG

NG(A)>

(AH)G/H

NG/H(AH)

where H is a normal subgroup of G of order p. Similarly, we have |H0(G,A)| divides|H0(H,A)|·|H0(G/H,AH)|, where |H0(H,A)| divides [H : 1]m and |H0(G/H,AH)| divides[G : H]m by hypothesis. Hence |H0(G,A)| divides [G : 1]m. This completes the proof.

Theorem 4.2.20. Let K be a nonarchimedean local field. Let L/K be a finite Galoisextension of degree n with Galois group Gal(L/K). Then H2(Gal(L/K), L∗) is cyclicof order n. Moreover, there exists an element uL/K ∈ H2(Kunr/K,Kunr∗) generating

H2(Gal(L/K), L∗) with invK(uL/K) = 1n .

Proof. We apply Lemma 4.2.19 to the case G = Gal(L/K), A = L∗, m = 1, j = 2. Notethat part (i) of the hypothesis is satisfied by the Hilbert’s Theorem 90, and part (ii) of thehypothesis is satisfied by Corollary 4.2.18. Hence the order of H2(Gal(L/K), L∗) divides n.However, by Corollary 4.2.13, H2(Gal(L/K), L∗) contains a subgroup of order n generatedby uL/K ∈ H2(Lunr/K,Lunr∗) such that invK(uL/K) = 1

n , namely H2(L/K)unr∼= 1

nZ/Z.

Therefore, H2(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 beused to define the local reciprocity map.

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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,H2(Gal(L/K), L∗) is cyclic of order n. Hence H2(Gal(L/K), L∗) ⊂ H2(Gal(Kunr/K),Kunr∗).Since H2(Gal(K/K,K

∗) is the union of all such H2(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 iso-morphism

invK : H2(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

0 > H2(G,L∗) > H2(Gal(K/K),K∗)

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

∗)

0 >1

nZ/Z > Q/Z

invK∨

n> Q/Z

invL∨

is commutative, and therefore we recover the invariant map invL/K : H2(G,L∗)→ 1nZ/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 Galoisextension of degree n with Galois group Gal(L/K). Then cup product with uL/K definesan isomorphism

H i(Gal(L/K),Z)→ H i+2(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.

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Theorem 4.3.2. Let K be a nonarchimedean local field. Let L/K be a finite Galoisextension of degree n with Galois group Gal(L/K). Then cup product with uL/K definesan isomorphism

Gal(L/K)ab → K∗

NL/K(L∗)

where Gab = G/[G,G] is the abelianization of G.

Proof. Note that

H0(Gal(L/K), L∗) =(L∗)Gal(L/K)

NL/K(L∗)=

K∗

NL/K(L∗),

and

H−2(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 getan isomorphism

Gal(L/K) ∼=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 Galoisextension. We define the local reciprocity map to be

φL/K :K∗

NL/K(L∗)→ Gal(L/K)ab

as in Theorem 4.3.2, which is the inverse of cup product by uL/K . Let x ∈ K∗, and let x

be a representative of x in K∗

NL/K(L∗) . We denote

(x, L/K) := φL/K(x).

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

Res(uL/K) = uL/E ,

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

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Proof. Consider the following diagram

H2(Gal(K/K),K∗)

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

∗)

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

∗)

Q/Z

invK∨

[E:K]> Q/Z

invE∨

[L:E]> Q/Z.

invL∨

Note that all vertical maps are isomorphisms by Theorem 4.2.23. Applying the kernel-cokernel lemma and we get the following commutative diagram

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

Res> H2(Gal(L/E), L∗)

0 >1

[E : K]Z/Z

invE/K∨>

1

[L : K]Z/Z

invL/K∨[E:K]

>1

[L : E]Z/Z.

invL/E∨

The second square commutes, implying that

[E : K](invL/K(uL/K)

)= [E : K]

1

[L : K]=

1

[L : E]= invL/E(uL/E) = invL/E

(Res(uL/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 towerof finite Galois extensions with G = Gal(L/K), H = Gal(L/E). Then the diagrams

H i(G,Z)uL/K

> H i+2(G,L∗)

H i(H,Z)

Res∨

uL/E> H i+2(H,L∗)

Res∨

and

H i(G,Z)uL/K

> H i+2(G,L∗)

H i(H,Z)

Cor∧

uL/E> H i+2(H,L∗)

Cor∧

are commutative for all i ∈ Z.

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Proof. Let a ∈ H i(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).

Let b ∈ H i(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 adecomposition

L∗ ∼= UL × πZ ∼= UL × Z

of G-modules. ThusH i(G,L∗) ∼= H i(G,UL)⊕ H i(G,Z)

for all i ∈ Z. We already know that H i(G,UL) = 0 by Corollary 4.2.7. Hence it remainsto consider H i(G,Z).

First we determine a cocycle representing uL/K . Let f ∈ H1(G,Q/Z) = Hom(G,Q/Z)

be the map such that f(FrobiL/K) = in mod Z for all i ∈ Z. Then f generates H1(G,Q/Z)

since FrobL/K generates G. Recall that we have an isomorphism

δ : H1(G,Q/Z) ∼= H2(G,Z).

Thus to determine the generator uL/K of H2(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 mapFrobiL/K 7→

in where 0 ≤ i ≤ n−1. Then using formulas for the connecting homomorphism

δ we obtain

δf(FrobiL/K ,FrobjL/K) = FrobiL/K f(FrobjL/K)− f(Frobi+jL/K) + f(FrobiL/K)

=

{0 if i+ j ≤ n− 1,

1 if i+ j ≥ n.

Recall that we can identify Z with πZ ⊂ L∗, so uL/K ∈ H2(G,L∗) is represented by thecocycle ϕ given by

ϕ(FrobiL/K ,FrobjL/K) =

{0 if i+ j ≤ n− 1,

π if i+ j ≥ n.

Note that since H0(G,UL) = 0, we have UK ⊂ NL/K(L∗) and so the class of π inK∗/NL/K(L∗) is well-defined.

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Proposition 4.3.7. Let K be a nonarchimedean local field. Let L/K be an unramifiedextension of degree n with Galois group G = Gal(L/K) generated by FrobL/K . Let x ∈K∗, x ∈ K∗

NL/K(L∗) = H0(G,L∗) be a representative of x, f ∈ Hom(G,Q/Z) = H1(G,Q/Z).

Thenf(φL/K(x)

)= invL/K(x ∪ δf).

Proof. Let φL/K(x) be the image of φL/K(x) under the isomorphismG = Gab ∼= H1(G,Z) =

H−2(G,Z). Then φL/K(x) ∪ uL/K = x ∈ H0(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)

where φL/K(x)∪ f is given by the cup product H−2(G,Z)∪ H1(G,Q/Z) ∼= H−1(G,Q/Z),

represented by i/n since H−1(G,Q/Z) ∼= H0(G,Z) = Z/nZ and δ : H−1(G,Q/Z) →H0(G,Z) = Z/nZ is induced by the norm map, which is just multiplication by n sinceQ/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 unramifiedextension of degree n with Galois group G = Gal(L/K) generated by FrobL/K . Then

φL/K(x) = (x, L/K) = Frobord(x)L/K for all x ∈ K∗.

Proof. The invariant map invL/K : H2(G,L∗)→ Q/Z is defined by the composition

invL/K : H2(G,L∗)ord

∼=> H2(G,Z)

δ−1

∼=> H1(G,Q/Z)

γ> 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)

= f(

Frobord(x)L/K

).

Since f was arbitrary, we conclude that φL/K(x) = Frobord(x)L/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 fora straightforward construction of totally ramified abelian extensions of a nonarchimedeanlocal 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 saythat 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 ifthey 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 commutativeformal group law defined over OK . If x, y ∈ mK , then F (x, y) converges to an elementin mK . Under this composition mK becomes a group and we write F (mK) to denotethis group. For example, if we set F (X,Y ) = X + Y then F (mK) = mK . If we setF (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 acommutative 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 h′ : G→ F such that

h ◦ h′ = T = h′ ◦ h,

then h is said to be an isomorphism. A homomorphism f : F → F is called an endomor-phism.

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 theresidue field. Choose a prime element π ∈ K. We define Fπ to be the set of formal powerseries 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 thedifference of coefficients of each degree is divisible by π.

Example 4.4.6. K = Qp, π = p, f(X) = pX +(p2

)X2 + · · · pXp−1 +Xp 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 formin 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} withφj ∈ 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 = limφj .

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 containsonly terms of degree j + 1. By assumption we have

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

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

71

Let

φj+1 = φj −Ej+1

π(1− πj).

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

Ej+1 ≡ f(φj(X1, · · · , Xn))− φj(g(X1), · · · , g(Xn))

≡ φj(X1, · · · , Xn)q − φj(Xq1 , · · · , X

qn)

≡ 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

haveφj+1 ≡ φj ≡ φ1 mod deg 2.

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

f(φj+1)− φj+1(g) ≡ f(φj+1)− φj(g) +Ej+1(g)

π(1− πj)mod deg j + 2.

By Taylor expansion, we have

f(φj+1) = f(φj) + f ′(φj)(φj+1 − φj) +f ′′(φj)

2!(φj+1 − φj)2 + · · ·

= f(φj) + π

(−Ej+1

π(1− πj)

)+ · · ·

≡ f(φj)− π(

Ej+1

π(1− πj)

)mod deg j + 2.

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

≡ πj+1Ej+1(X1, · · · , Xn) mod deg j + 2.

Therefore,

f(φj+1)− φj+1(g) ≡ f(φj+1)− φj(g) +Ej+1(g)

π(1− πj)mod deg j + 2

≡ f(φj)− π(

Ej+1

π(1− πj)

)− φj(g) + πj+1

(Ej+1

π(1− πj)

)mod deg j + 2

≡ Ej+1 − Ej+1 mod deg j + 2

≡ 0 mod deg j + 2.

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It only remains to show the uniqueness of φj+1. We write φj+1 = φj + φj+1 andsuppose

f(φj+1)− φj+1(g) ≡ 0 mod deg j + 2.

We will show that φj+1 = − Ej+1

π(1−πj) . As above, we use Taylor expansions to get

f(φj+1) ≡ f(φj) + πφj+1 mod deg j + 2

and the fact that φj+1 only has terms of degree j + 1 to get

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

Thus

0 ≡ f(φj+1)− φj+1(g) mod deg j + 2

≡ f(φj) + πφj+1 − φj(g)− φj+1(g) mod deg j + 2

≡ Ej+1 + π(1− πj)φj+1 mod deg j + 2.

Therefore, φj+1 = − Ej+1

π(1−πj) . By induction, we are done.

Proposition 4.4.8. For every f ∈ Fπ, there is a unique commutative formal group lawFf 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 grouplaw. 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)).

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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 formalgroup law.

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

Proposition 4.4.10. Let f ∈ Fπ and Ff be the Lubin-Tate formal group law given byProposition 4.4.8. Then for any a ∈ A = OK there exists a unique [a]f ∈ OK [[X]] suchthat(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 injectivering 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,

andFf ([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 withf . 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 similarargument shows that [ab]f = [a]f ◦ [b]f .

Finally, the homomorphism is injective because [a]f ≡ aX mod deg 2, so a can berecovered as the coefficient of the term of degree 1. Also note that [a]f has no constantterm by Remark 4.4.2. Hence the map a 7→ [a]f := [a]f,f is injective.

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

Proof. Let u ∈ UK . Then [u]f,g : Fg → Ff has an inverse [u−1]g,f : Ff → Fg. HenceFf ∼= Fg.

Let K be a nonarchimedean local field, q = |k| be the order of the residue field. Wefix a prime element π of K. For f ∈ Fπ, let Ff be the corresponding Lubin-Tate formalgroup law given in Proposition 4.4.8. We write Mf for the group of points in mK equippedwith 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

Enf = ker([πn]f )

andEf =

⋃n≥1

Enf .

Then Ef is exactly the torsion submodule of Mf . Let

Knπ = K(Enf ),

Kπ =⋃n≥1

K(Enf ),

and denoteGπ,n = Gal(Kn

π/K),

then we haveGal(Kπ/K) = lim←−

n

Gπ,n.

Proposition 4.4.14. The OK-module Ef is isomorphic to K/OK . Hence EndOK (Enf ) ∼=OK/(πn) and AutOK (Enf ) ∼= (OK/(πn))∗.

Proof. (i) Proposition 4.4.13 implies that the choice of f ∈ Fπ is unimportant since theLubin-Tate formal group laws are isomorphic for all elements of Fπ. Hence we may choosef(X) = Xq + πX. Let a ∈ mK . Then f(X)− a = 0 has solutions in K, and in fact theybelong 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 ,

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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

E1f = ker([πf ]) = {a ∈Mf : [π]f (a) = 0 = f(a)}.

Hence E1f is exactly the set of roots of f , so |E1

f | = q and hence E1f∼= OK/(π). Since [π]f

is surjective, the sequence

0 > E1f > Enf

[π]f> En−1

f > 0

is exact. By induction, Enf has qn elements. Moreover, since E1f is cyclic, Enf must be

cyclic. Therefore, Enf is cyclic of order qn, and hence Enf∼= OK/(πn). Since Ef = ∪n≥1E

nf ,

it follows that Ef ∼= K/OK . Also, the action of OK on Enf induces an isomorphismOK/(πn) ∼= EndOK (Enf ).

Theorem 4.4.15. (i) For each n ≥ 1, Knπ/K is totally ramified of degree (q − 1)qn−1.

(ii) The action of OK on Enf defines an isomorphism

(OK/(πn))∗ → Gal(Knπ/K).

In particular, Knπ/K is abelian.

(iii) For each n ≥ 1, π is a norm from Knπ to K.

Proof. (i)-(ii) Again we choose f(X) = Xq +πX. Let x1 be a root of f(X), and define xninductively by choosing xn to be a root of f(X)− xn−1. Consider the tower of Eisensteinextensions

K ⊂ K(x1) ⊂ K(x2) ⊂ · · · ⊂ K(xn) ⊂ Knπ

where [K(x1) : K] = q − 1 and [K(xn) : K(xn−1)] = q for all n ≥ 2. This shows thatK(xn) is totally ramified over K of degree (q − 1)qn−1.

Enf is the kernel of [πn]f , i.e., the set of roots of f ◦· · ·◦f = f (n). Hence Knπ is precisely

the splitting field of f (n). So Gal(Knπ/K) can be identified with a subgroup of the group

of permutations of the set Enf , this subgroup is contained in AutOK (Enf ) ∼= (OK/πn)∗,

which has order (q − 1)qn−1. So

(q − 1)qn−1 ≥ |Gal(Knπ/K)| = [Kn

π : K] ≥ [K(xn) : K] = (q − 1)qn−1.

This shows that Knπ = K(xn) and Gal(Kn

π/K) ∼= AutOK (Enf ) ∼= (OK/(πn))∗.(iii) By construction, xn is a root of the polynomial

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

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

NKnπ /K

(xn) = (−1)(q−1)qn−1π = π.

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

φπ : K∗ → Gal(Kπ ·Kunr/K)

by defining the restrictions of φπ(a) to Kπ and Kunr for each a ∈ K∗. Let a = uπm ∈ K∗.Define

φπ(a)|Kunr = Frobm

andφπ(a)|Kπ = [u−1]f

where Frob is the Frobenius element FrobKunr/K . Although K is complete, Kunr is not

complete, and we write Kunr for the completion of Kunr with respect to the unique ex-tension of the valuation of K to Kunr. We also write Frob for the Frobenius elementFrob

Kunr/K.

Remark 4.4.16. We use u−1 instead of u in the definition of the map φπ so that bothKπ ·Kunr and φπ are independent of the choice of the prime element π, as we will see inTheorem 4.4.20.

Lemma 4.4.17. The homomorphisms

OKunr → OKunr

x 7→ Frob(x)− x

and

OKunr → OKunr

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, π′ = uπ be two differentprimes of K. Let Ff and Fg be the Lubin-Tate formal group laws defined by f ∈ Fπ, g ∈Fπ′. Then there exists ε ∈ U

Kunr such that Frob(ε) = εu and a power series h(T ) ∈OKunr [[T ]] such that

(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 ) ∈ OKunr [[T ]] that satisfies (i) and (ii). Let

ε ∈ UKunr such that Frob(ε) = εu (such ε exists by Lemma 4.4.17). Let h1(T ) = εT , and

we construct a sequence of polynomials hr such that

hr(T ) = hr−1(T ) + xT r,

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Frob(hr) ≡ hr ◦ [u]f mod deg r + 1

for some x ∈ OKunr . Note that h1(T ) satisfies (i) and (ii). Suppose that hr(T ) satisfies

(i) and (ii). Let a ∈ OKunr such that Frob(a) − a = c(εu)−r−1 where c is the coefficient

of T r+1 in hr ◦ [u]f − Frob(hr). Note that such a exists by Lemma 4.4.17. Let b = aεr+1,and we claim that hr+1(T ) = hr(T ) + bT r+1 satisfies (ii). Well,

Frob(hr+1(T )) = Frob(hr(T )) + Frob(aεr+1T r+1)

= Frob(hr(T )) + Frob(a)Frob(εT )r+1

= Frob(hr(T )) +

(a+

c

(εu)r+1

)Frob(εT )r+1

≡ Frob(hr(T )) +

(a+

c

(εu)r+1

)(Frob(ε))r+1T r+1 mod deg r + 2

≡ Frob(hr(T )) + a(Frob(ε))r+1T r+1 + cT r+1 mod deg r + 2

≡ hr ◦ [u]f (T )− cT r+1 + a(Frob(ε))r+1T r+1 + cT r+1 mod deg r + 2

≡ hr ◦ [u]f (T ) + a(Frob(ε))r+1T r+1 mod deg r + 2

≡ hr ◦ [u]f (T ) + a(εu)r+1T r+1 mod deg r + 2

≡ hr ◦ [u]f (T ) + aεr+1T r+1 ◦ [u]f (T ) mod deg r + 2

≡ hr ◦ [u]f (T ) + bT r+1 ◦ [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

θ = Frob h ◦ f ◦ h−1 = h ◦ [u]f ◦ f ◦ h−1 = h ◦ f ◦ [u]f ◦ h−1.

By (ii) we have T = h ◦ [u]f ◦ (Frob h)−1(T ), hence

h−1 = [u]f ◦ (Frob h)−1.

Since f and [u]f have coefficients in OK , we have

Frob θ = Frob h ◦ f ◦ [u]f ◦ (Frob h)−1

= 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

≡ π′T mod T 2,

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and

θ(T ) = Frob h ◦ f ◦ h−1(T )

≡ Frob h ◦ (h−1)q(T ) mod mK

≡ Frob h (Frob h−1(T q)) mod mK

≡ T q mod mK .

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

Frob θ′ ◦ f ◦ (θ′)−1 = [1]g,θ ◦ θ ◦ [1]−1g,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 ∈ L. If x is separable and algebraic over L, then x ∈ L.

Proof. Let L′ = L ∩K. Let σ ∈ Gal(K/L). We know that σ is continuous and it is theidentity on L. Let x ∈ L′. Then x is the limit of elements in L and so the action of σis trivial on x as well. Thus Gal(K/L) = Gal(K/L′). Then by Galois theory, we obtainL′ = L.

Theorem 4.4.20. The field Kπ ·Kunr and the map φπ are independent of the choice ofπ.

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

Frob ◦ h ◦ [π]f = (h ◦ [u]f ) ◦ [π]f

= h ◦ [uπ]f

= h ◦ [u′]f

= [π′]g ◦ h,

i.e., Frob ◦ h(f(T )) = g(h(T )). Therefore, for any a ∈ K, if a ∈ E1f , then f(a) = 0, so

g(h(a)) = 0, and hence h(a) ∈ E1g . Similarly, if b ∈ E1

g , then g(b) = 0, so f(h−1(b)) = 0,and hence h−1(b) ∈ E1

f . So h defines an isomorphism E1f → E1

g . Thus

Kunr(E1g ) = Kunr(h(E1

f ))

⊂ Kunr(E1f )

= Kunr(h−1(E1g ))

⊂ Kunr(E1g ).

Thus Kunr(E1g ) = Kunr(E1

f ). Now combine Lemma 4.4.19 and we get Kunr(E1f ) =

Kunr(E1g ). Similarly, Kunr(Enf ) = Kunr(Eng ) for all n ≥ 1. Thus Kπ ·Kunr = Kπ′ ·Kunr.

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We need to show that φπ = φπ′ as well. Note that both φπ(π) and φπ′(π) act as Frobon Kunr. Thus they agree on Kunr. On Kπ, φπ(π) is the identity map. Our first goal is toshow that φπ′(π) is also the identity map on Kπ. Note that Kn

π is generated by Enf overK, and h gives an isomorphism Enf → Eng . It suffices that show that for all n ≥ 1 and

x ∈ Enf , we have φπ′(π)(h(x)) = h(x). Since π = u−1π′, we have

φπ′(π) = φπ′(u−1)φπ′(π

′) = σ1σ2

where

σ1 =

{Frob on Kunr

1 on Enf

and

σ2 =

{1 on Kunr

[u]f on Enf .

Since h has coefficients in Kunr, we have

φπ′(π)(h(x)) = σ1σ2(h(x))

= σ1(h(σ2(x)))

= σ1(h([u]f (x)))

= h(x).

Hence φπ′(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

over K. Let Hn,m be the subgroup of K∗ generated by U(n)K and πm. Then φπ(a)|Kn

π ·Km = 1for all a ∈ Hn,m.

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

Proof. We know that both φK(π) and φπ(π) act as Frob on Kunr. Hence they agree onKunr. By part (iii) of Theorem 4.4.15, π is a norm from Kn

π to K for every n ≥ 1. HenceφK(π) acts trivially on Kn

π for every n ≥ 1. Apply Lemma 4.4.21 with m = 1, then weget that φπ(π) acts trivially on Kn

π for every n ≥ 1. So φK and φπ agree on Knπ · Kunr

for all n ≥ 1, so they must agree on the union⋃n≥1K

nπ ·Kunr = Kπ ·Kunr. We are done

since prime elements of K generate K∗ (every a ∈ K∗ can be written as a = uπi, andu = (uπ)π−1).

Lemma 4.4.23. Let n,m ≥ 1. Let Kn,m = Knπ ·Km. Then NKn,m/K(K∗n,m) = U

(n)K 〈πm〉 =

Hn,m.

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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(K∗n,m) by part (ii) of Theorem 4.1.1. On the other hand, we have

[K : Hn,m] = [UK : U(n)K ][〈π〉 : 〈πm〉]

= (q − 1)qnm

= [Knπ : 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(K∗n,m)→ Gal(Kn,m/K).

Therefore,Hn,m = NKn,m/K(K∗n,m).

Theorem 4.4.24. Kab = Kπ ·Kunr 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 abelianextension. 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. Notethat UK is compact, hence NL/K(UL) is closed in UK (since UK is Hausdorff, and NL/K iscontinuous). 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 opensubgroup of K∗ with finite index, then H is a norm subgroup.

Proof. Because H is open, there is some n ≥ 1 such that UnK ⊂ H (since {U iK} formsa neighborhood basis for the identity in K∗). Because H is of finite index in K∗, thereare only a finite number of cosets of the form πjH, hence there exists some m ≥ 1 such

that πm ∈ H. Hence H contains the subgroup Hn,m generated by U(n)K and πm. Let

Km be the unique unramified extension of degree m over K, and consider the subfieldKn,m = Kn

π · Km of Kab. By Lemma 4.4.23, we have Hn,m = NKn,m/K(K∗n,m). Let Lbe the subfield of Kn,m that is fixed by φKn,m/K(H). Then H is the kernel of the mapφK : 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 LocalExistence Theorem (Theorem 4.1.6).

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References

[ANT] Algebraic Number Theory lecture notes. Available at http://math.okstate.edu/people/pyan/AlgebraicNumberTheory.pdf.

[Cas86] Cassels, J.W.S., Local Fields, Cambridge University Press, Cambridge, 1986.

[CF67] Cassels, J.W.S., Frohlich, A., Algebraic Number Theory, Washington DC, 1967.

[FT91] Frohlich, A., Taylor, M., Algebraic Number Theory, Cambridge University Press,1991.

[LT65] Lubin, J., Tate, J., Formal complex multiplication in local fields, Annals of Math-ematics (1965): 380-387.

[Mil13] Milne J.S., Class Field Theory (v4.02), 2013. Available at http://www.jmilne.

org/math/.

[Neu86] Neukirch, J., Class Field Theory, Springer, 1986.

[Ser80] Serre, J.P., Local Fields, Springer-Verlag, 1980.

[Sha] Sharifi, R., An Introduction to Group Cohomology, Lecture notes, available at http://math.arizona.edu/~sharifi/groupcoh.pdf.

[Wat73] Waterhouse, W.C., Profinite groups are Galois groups, Proceedings of the Amer-ican Mathematical Society 42 (1973), 639-640.

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http://math.okstate.edu/people/pyan/AlgebraicNumberTheory.pdf

http://math.okstate.edu/people/pyan/AlgebraicNumberTheory.pdf

http://www.jmilne.org/math/

http://www.jmilne.org/math/

http://math.arizona.edu/~sharifi/groupcoh.pdf

http://math.arizona.edu/~sharifi/groupcoh.pdf

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