Let $K$ be a finite extension of $\Bbb{Q}_p$. Let $E/K$ be an elliptic curve over $K$.
Let $E(K)$ be the Mordell-Weil group of $E/K$.
Let $\widehat{E(K)}$ be the profinite completion of $E(K)$, that is, its completion with respect to finite index subgroups.
Is it true that $E(K)=\widehat{E(K)}$ ?
If we admit the result $E(K)\cong {\Bbb{Z}_p}^{[K:\Bbb{Q}_p]}\times (\text{finite group})$, it is true.
But I'm seeking a more elementary way without heavy machinery. Thank you for your help.
cf. $E(K)\cong {\Bbb{Z}_p}^{[K:\Bbb{Q}_p]}\times (\text{finite group})$ is in Theorem 5.2(a) of [Mat] A. Mattuck, Abelian varieties over p-adic ground fields, Ann. of Math. (2) 62 (1955), 92– 119.
You can easily check that $E(K)$ is a topological group that continuously embeds as a closed subspace $\mathbb{P}^2(K)$, hence it is a compact and totally disconnected topological group. That it is profinite (and thus isomorphic to its profinite completion) follows from the relevant Bourbaki volume (referred to in the beginning of Serre’s Galois Cohomology).
You can also avoid having to refer to Bourbaki by the following method:
Lemma: Every subgroup $nE(K)$ for $n \geq 1$ is open.
Proof: since it’s a compact subgroup, it’s enough to show that it has finite index, ie that $E(K)/nE(K)$ is finite.
Now, $E(K)/nE(K) \rightarrow H^1(K,E[n])$ is injective, so it’s enough to show that $H^1(K,E[n])$ is finite.
There is a finite Galois extension $L/K$ such that $E[n] \cong \mu_n \oplus \mu_n$ as $G_L$-modules, hence $H^1(L,E[n])=H^1(L,\mu_n)^{\oplus 2}=(L^{\times}/L^{\times n})^{\oplus 2}$ which is finite.
The kernel of the restriction map $H^1(K,E[n]) \rightarrow H^1(L,E[n])$ is exactly $H^1(\mathrm{Gal}(L/K), E[n](L))$: it’s the cohomology of a finite group in a finite module, hence is finite.
Finally $H^1(K,E[n])$ is finite, QED.
Corollary: every finite index subgroup of $E(K)$ is open.
Proof: a subgroup of index $n$ contains $nE(K)$.
Thus the map $E(K) \rightarrow \widehat{E(K)}$ is continuous between compact spaces with dense image, hence it is onto. To show it is an isomorphism (hence an homeomorphism since both spaces are Hausdorff compact), all we need to do is prove that it is injective, or, to put it differently, that the intersection of all subgroups of finite index of $E(K)$ is trivial.
A family of subgroups of $E(K)$ with finite index with trivial intersection is given, if I’m not mistaken, by the theory of the formal group (see the homonymous chapter of Silverman).