A Short Course on Galois Cohomology - William Stein - University of ...

The following proposition is proved using formal groups; it’s Proposition
VII.6.3 in [Sil92].
Proposition 21.3. Suppose E is an elliptic curve over a finite extension k
of Qp and let R be the ring of integers of k. Then E(k) contains a subgroup
of finite index that is isomorphic to (R, +).
For simplicity, assume that R = Zp. Then we have an exact sequence
0 → Zp → E(Qp) → M → 0,
where M is a finite group. Applying the Hom(−, Q/Z) (continuous homomorphisms)
Pontryagin duality gives an exact sequence
We have
0 → Hom(M, Q/Z) → Hom(E(Qp), Q/Z) → Hom(Zp, Q/Z) → 0.
Hom(Zp, Q/Z) ∼ = Qp/Zp,
where the isomorphism sends ϕ : Zp → Q/Z to ϕ(1) ∈ Qp/Zp ⊂ Q/Z. Thus
we have an exact sequence
0 → F → H 1 (Qp, E) → Qp/Zp → 0,
where F is a finite group. So that’s what cohomology of an elliptic curve
over a local field looks like:
The group H 1 (Qp, E) has a lot of elements of order a power of
p, and not much else.
21.2 Duality over a Finite Field
Let k be a finite field. Then the dualizing module is Q/Z with trivial action
(not ¯ k ∗ as above), and we define
M ∗ = Hom(M, Q/Z).
Theorem 21.4. For every finite module M and r = 0, 1, cup product induces
a nondegenerate pairing
H r (k, M) × H 1−r (k, M ∗ ) → Q/Z.
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