15 Galois Cohomology

Let G be a group and A be a G-module (an abelian group with an action of G via group homomorphisms). G-module means exactly the same thing as [G]-module.

Definition (H0). Define

H0(G,A)=AG={aA|σ(a)=a σG}.

Definition (Cochains). Define

C1(G,A)={maps GA}

(called “cochains”).

Definition (Cocycles). Define

Z1(G,A)={(aσ)σG|aστ=σ(aτ)+aσ σ,τG}

(called “cocycles”).

Definition (Coboundaries). Define

B1(G,A)={(σbb)σG|bA}

(called “coboundaries”).

Note. C1(G,A)Z1(G,A)B1(G,A).

Then we can define:

Definition (H1). Define

H1(G,A)=Z1(G,A)B1(G,A).

Remark. If G acts trivially on A, then

H1(G,A)=Hom(G,A).

Theorem 15.1. Assuming that:

  • we have a short exact sequence of G-modules

    0AϕBψC0.

Then it gives rise to a long exact sequence of abelian groups:
0AGϕBGψCGδH1(G,A)ϕH1(G,B)ψH1(G,C).

Proof. Omitted.

Definition (delta). Let cCG. Then bB such that ψ(b)=c. Then

ψ(σbb)=σcc=0σG

so σbb=ϕ(aσ) for some aσA.

Can check (aσ)σGZ1(G,A).

Then δ(c)= class of (aσ)σG in H1(G,A).

Theorem 15.2. Assuming that:

  • A is a G-module

  • HG a normal subgroup

Then there is an inflation restriction exact sequence
0H1(GH,AH)infH1(G,A)resH1(H,A)

Proof. Omitted.

Let K be a perfect field. Then Gal(K¯K) is a topological group with basis of open subgroups being the Gal(K¯L) for [L:K]<.

If G=Gal(K¯K) then we modify the definition of H1(G,A) by insisting:

Then

H1(Gal(K¯K),A)=limLLK finite GaloisH1(Gal(LK),AGal(K¯L)).

(direct limit is with respect to inflation maps).

Theorem (Hilbert’s Theorem 90). Assuming that:

  • LK a finite Galois extension

Then
H1(Gal(LK),L)=0.

Proof. Let G=Gal(LK). Let (aσ)σGZ1(G,L). Distinct automorphisms are linearly independent, so there exists yL such that

τGaτ1τ(y)=x0.

Then

σ(x)=τGσ(aτ)1στ(y)=aστGaστ1στ(y)=x

Then aσ=σ(x)x for all σG. (aστ=σ(aτ)aσ σ(aτ)1=aσaστ1).

So (aσ)σGB1(G,L). Therefore H1(G,L)=0.

Corollary. H1(Gal(K¯K),K¯)=0.

Application: Assume charKn. There is a short exact sequence of Gal(K¯K)-modules

0μnK¯K¯0xxn

Long exact sequence:

KKH1(Gal(K¯K),μn)H1(Gal(K¯K),K¯)=0 by Hilbert 90xxn

Therefore H1(Gal(K¯K),μn)K(K)n.

If μnK then

Homcts(Gal(K¯K),μn)K(K)n

Finite subgroups of Homcts(Gal(K¯K),μn) are of the form Hom(Gal(LK),μn) for LK a finite abelian extension of K of exponent dividing n.

This gives another proof of Theorem 11.2.

Notation. H1(K,) means H1(Gal(K¯K),).

Let ϕ:EE be an isogeny of elliptic curves over K. Short exact sequence of Gal(K¯K)-modules

0E[ϕ]EϕE0

has long exact seqeucne

E(K)ϕE(K)δH1(K,E[ϕ])H1(K,E)ϕH1(K,E).

We get a short exact sequence

              ′
 0           EϕE(K(K-))            H1 (K,E [ϕ])          H1 (K, E)[ϕ∗]  0


            ∏   E′(Kv)-        ∏    1               ∏    1        ∗
δrrδ0esesVvv           v ϕE(Kv)           vH  (Kv,E[ϕ])        vH  (Kv, E)[ϕ0]
Now take K a number field.

For each place v, fix an embedding K¯K¯v. Then Gal(K¯vKv)Gal(K¯K).

Definition (Selmer-group). We define the ϕ-Selmer group

S(ϕ)(EK)=ker(H1(K,E[ϕ])vH1(Kv,E))={αH1(K,E[ϕ])|resv(α)Im(δv) v}

(the map H1(K,E[ϕ])vH1(Kv,E) is as in the commutative diagram above).

Definition (Tate-Shafarevich group). The Tate-Shafarevich group is

TS(EK)=ker(H1(K,E)vH1(Kv,E)).

We get a short exact sequence

0E(K)ϕE(K)S(ϕ)(EK)TS(EK)[ϕK]0.

Taking ϕ=[n] gives

0E(K)nE(K)S(n)(EK)TS(EK)[n]0.

Reorganising the proof of Mordell-Weil gives

Theorem 15.3. S(n)(EK) is finite.

Proof. For LK a finite Galois extension,

                  ◜-------fin◞it◟e-------◝
 0                H1 (Gal(L∕K ),E(L)[n ]) H1 (K,E [n])      H1(L,E [n])


inr⊃⊃fes                                     S (n)(E ∕K )       S(n)(E∕L )
Therefore by extending our field, we may assume E[n]E(K) and hence by the Weil pairing μnK.

Therefore E[n]μn×μn as a Gal(K¯K)-module. Then

H1(K,E[n])H1(K,μn)×H1(K,μn)K(K)n×K(K)n

Let

S={primes of bad reduction for E}{v|n}

(a finite set of places).

Define the subgroup of H1(K,A) unramified outside of S as

H1(K,A;S)=ker(H1(K,A)vSH1(Kvnr,A)).

There is a commutative diagram with exact rows:

    E (Kv )       E(Kv )          H1(Kv,E [n])


×δr×0nvesn  E (Knr )      E(Kvnr )        H1(Knr,E [n])
        v                             v
The bottom map ×n is surjective vS (see Theorem 9.8).

Therefore

S(n)(EK)H1(K,E[n];S)H1(K,μn;S)×H1(K,μn;S)

But

H1(K,μn;S)=ker(K(K)nvS(Kvnr)(Kvnr)n)K(S,n)

which is finite by Lemma 11.4.

Remark. S(n)(EK) is finite and effectively computable.

It is conjectured that |TS(EK)|<. This would imply that rankE(K) is effectively computable.