Galois Cohomology - Elliptic Curves

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 $ℤ [G]$ -module.

Definition ( $H^{0}$ ). Define

H^{0} (G, A) = A^{G} = {a \in A | σ (a) = a \forall σ \in G} .

Definition (Cochains). Define

C^{1} (G, A) = {maps G \to A}

(called “cochains”).

Definition (Cocycles). Define

Z^{1} (G, A) = {{(a_{σ})}_{σ \in G} | a_{σ τ} = σ (a_{τ}) + a_{σ} \forall σ, τ \in G}

(called “cocycles”).

Definition (Coboundaries). Define

B^{1} (G, A) = {{(σ b - b)}_{σ \in G} | b \in A}

(called “coboundaries”).

Note. $C^{1} (G, A) \supset Z^{1} (G, A) \supset B^{1} (G, A)$ .

Then we can define:

Definition ( $H^{1}$ ). Define

H^{1} (G, A) = \frac{Z^{1} (G, A)}{B^{1} (G, A)} .

Remark. If $G$ acts trivially on $A$ , then

H^{1} (G, A) = Hom (G, A) .

Theorem 15.1. Assuming that:

we have a short exact sequence of $G$ -modules

$0 \to A \overset{ϕ}{\to} B \overset{ψ}{\to} C \to 0 .$

Then it gives rise to a long exact sequence of abelian groups:

0 \to A^{G} \overset{ϕ}{\to} B^{G} \overset{ψ}{\to} C^{G} \overset{δ}{\to} H^{1} (G, A) \overset{ϕ^{*}}{\to} H^{1} (G, B) \overset{ψ^{*}}{\to} H^{1} (G, C) .

Proof. Omitted. □

Definition ( $d e l t a$ ). Let $c \in C^{G}$ . Then $b \in B$ such that $ψ (b) = c$ . Then

ψ (σ b - b) = σ c - c = 0 \forall σ \in G

so $σ b - b = ϕ (a_{σ})$ for some $a_{σ} \in A$ .

Can check ${(a_{σ})}_{σ \in G} \in Z^{1} (G, A)$ .

Then $δ (c) =$ class of ${(a_{σ})}_{σ \in G}$ in $H^{1} (G, A)$ .

Theorem 15.2. Assuming that:

$A$ is a $G$ -module
$H ⊴ G$ a normal subgroup

Then there is an inflation restriction exact sequence

0 \to H^{1} (G ∕ H, A^{H}) \overset{\inf}{\to} H^{1} (G, A) \overset{res}{\to} H^{1} (H, A)

Proof. Omitted. □

Let $K$ be a perfect field. Then $Gal (\bar{K} ∕ K)$ is a topological group with basis of open subgroups being the $Gal (\bar{K} ∕ L)$ for $[L : K] < \infty$ .

If $G = Gal (\bar{K} ∕ K)$ then we modify the definition of $H^{1} (G, A)$ by insisting:

(1) The stabiliser of each $a \in A$ is an open subgroup of $G$ .
(2) All cochains $G \to A$ are continuous, where $A$ is given the discrete topology.

Then

H^{1} (Gal (\bar{K} ∕ K), A) = \lim_{\begin{array}{c} \to \\ L \\ L ∕ K finite Galois \end{array}} H^{1} (Gal (L ∕ K), A^{Gal (\bar{K} ∕ L)}) .

(direct limit is with respect to inflation maps).

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

$L ∕ K$ a finite Galois extension

Then

H^{1} (Gal (L ∕ K), L^{*}) = 0 .

Proof. Let $G = Gal (L ∕ K)$ . Let ${(a_{σ})}_{σ \in G} \in Z^{1} (G, L^{*})$ . Distinct automorphisms are linearly independent, so there exists $y \in L$ such that

\underset{= x}{\underset{⏟}{\sum_{τ \in G} a_{τ}^{- 1} τ (y)}} \neq 0 .

Then

\begin{array}{l} σ (x) & = \sum_{τ \in G} σ {(a_{τ})}^{- 1} σ τ (y) \\ = a_{σ} \underset{= x}{\underset{⏟}{\sum_{τ \in G} a_{σ τ}^{- 1} σ τ (y)}} \end{array}

Then $a_{σ} = \frac{σ (x)}{x}$ for all $σ \in G$ . ( $a_{σ} τ = σ (a_{τ}) a_{σ}$ $⟹$ $σ {(a_{τ})}^{- 1} = a_{σ} a_{σ τ}^{- 1}$ ).

So ${(a_{σ})}_{σ \in G} \in B^{1} (G, L^{*})$ . Therefore $H^{1} (G, L^{*}) = 0$ . □

Corollary. $H^{1} (Gal (\bar{K} ∕ K), {\bar{K}}^{*}) = 0$ .

Application: Assume $char K ∤ n$ . There is a short exact sequence of $Gal (\bar{K} ∕ K)$ -modules

\begin{array}{l} 0 \to μ_{n} \to {\bar{K}}^{*} & \to {\bar{K}}^{*} \to 0 \\ x & \mapsto x^{n} \end{array}

Long exact sequence:

\begin{array}{l} K^{*} & \to K^{*} \to H^{1} (Gal (\bar{K} ∕ K), μ_{n}) \to \overset{= 0 by Hilbert 90}{\overset{︷}{H^{1} (Gal (\bar{K} ∕ K), {\bar{K}}^{*})}} \\ x & \mapsto x^{n} \end{array}

Therefore $H^{1} (Gal (\bar{K} ∕ K), μ_{n}) ≅ K^{*} ∕ {(K^{*})}^{n}$ .

If $μ_{n} \subset K$ then

{Hom}_{cts} (Gal (\bar{K} ∕ K), μ_{n}) ≅ K^{*} ∕ {(K^{*})}^{n}

Finite subgroups of ${Hom}_{cts} (Gal (\bar{K} ∕ K), μ_{n})$ are of the form $Hom (Gal (L ∕ K), μ_{n})$ for $L ∕ K$ a finite abelian extension of $K$ of exponent dividing $n$ .

This gives another proof of Theorem 11.2.

Notation. $H^{1} (K, -)$ means $H^{1} (Gal (\bar{K} ∕ K), -)$ .

Let $ϕ : E \to E^{'}$ be an isogeny of elliptic curves over $K$ . Short exact sequence of $Gal (\bar{K} ∕ K)$ -modules

0 \to E [ϕ] \to E \overset{ϕ}{\to} E^{'} \to 0

has long exact seqeucne

E (K) \overset{ϕ}{\to} E^{'} (K) \overset{δ}{\to} H^{1} (K, E [ϕ]) \to H^{1} (K, E) \overset{ϕ_{*}}{\to} H^{1} (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 $\bar{K} \subset {\bar{K}}_{v}$ . Then $Gal ({\bar{K}}_{v} ∕ K_{v}) \subset Gal (\bar{K} ∕ K)$ .

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

\begin{array}{l} S^{(ϕ)} (E ∕ K) & = \ker (H^{1} (K, E [ϕ]) \to \prod_{v} H^{1} (K_{v}, E)) \\ = {α \in H^{1} (K, E [ϕ]) | {res}_{v} (α) \in Im (δ_{v}) \forall v} \end{array}

(the map $H^{1} (K, E [ϕ]) \to \prod_{v} H^{1} (K_{v}, E)$ is as in the commutative diagram above).

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

T S (E ∕ K) = \ker (H^{1} (K, E) \to \prod_{v} H^{1} (K_{v}, E)) .

We get a short exact sequence

0 \to \frac{E^{'} (K)}{ϕ E (K)} \to S^{(ϕ)} (E ∕ K) \to T S (E ∕ K) [ϕ_{K}] \to 0 .

Taking $ϕ = [n]$ gives

0 \to \frac{E (K)}{n E (K)} \to S^{(n)} (E ∕ K) \to T S (E ∕ K) [n] \to 0 .

Reorganising the proof of Mordell-Weil gives

Theorem 15.3. $S^{(n)} (E ∕ K)$ is finite.

Proof. For $L ∕ K$ 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] \subset E (K)

and hence by the Weil pairing

μ_{n} \subset K

Therefore $E [n] ≅ μ_{n} \times μ_{n}$ as a $Gal (\bar{K} ∕ K)$ -module. Then

\begin{array}{l} H^{1} (K, E [n]) & ≅ H^{1} (K, μ_{n}) \times H^{1} (K, μ_{n}) \\ ≅ K^{*} ∕ {(K^{*})}^{n} \times K^{*} ∕ {(K^{*})}^{n} \end{array}

Let

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

(a finite set of places).

Define the subgroup of $H^{1} (K, A)$ unramified outside of $S$ as

H^{1} (K, A; S) = \ker (H^{1} (K, A) \to \prod_{v \notin S} H^{1} (K_{v}^{nr}, 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

\times n

is surjective

\forall v \notin S

(see Theorem 9.8).

Therefore

\begin{array}{l} S^{(n)} (E ∕ K) & \subset H^{1} (K, E [n]; S) \\ ≅ H^{1} (K, μ_{n}; S) \times H^{1} (K, μ_{n}; S) \end{array}

But

\begin{array}{l} H^{1} (K, μ_{n}; S) & = \ker (\frac{K^{*}}{{(K^{*})}^{n}} \to \prod_{v \notin S} \frac{{(K_{v}^{nr})}^{*}}{{(K_{v}^{nr})}^{* n}}) \\ \subset K (S, n) \end{array}

which is finite by Lemma 11.4. □

Remark. $S^{(n)} (E ∕ K)$ is finite and effectively computable.

It is conjectured that $| T S (E ∕ K) | < \infty$ . This would imply that $rank E (K)$ is effectively computable.