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Definition: A p-adic field is a finite extension of $Q_p$.

Question: Let $E$ be a p-adic field, $G$ is a nontrivial additive compact subgroup of $E$, how to prove: E is isomorphic to $Z_p^n$ for some positive integer $n$. This isomorphism is not only a topological group isomorphism but also a $Z_p$ module isomorphism.

I guess we can prove it using non-archimedean analysis, but I still don't know how to prove it.

Thanks for any answers!

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$G$ is compact, thus closed in $E$. It is stable by multiplication by an element of $\mathbb{Z}$ (dense in $\mathbb{Z}_p$), thus is a $\mathbb{Z}_p$-submodule of $E$.

Note that there exists a finite $\mathbb{Q}_p$-base of the vector subspace $V$ spanned by $G$, with vectors $a_1, \ldots, a_n$.

Now, for each $v \in V$, denote $v_i \in \mathbb{Q}_p$ to be the coordinate of $v$ in the direction $a_i$.

Then $v \longmapsto v_i$ is a linear form, thus is continuous, hence $G_i=\{g_i,\,g \in G\} \subset \mathbb{Q}_p$ is compact. Therefore, there is a $N>0$ such that for each $i$, $p^NG_i \subset \mathbb{Z}_p$.

Now, let $$G’=\bigoplus_{i=1}^n{\frac{a_i}{p^N}\mathbb{Z}_p}.$$

$G’$ is a finitely generated $\mathbb{Z}_p$-module, thus is Noetherian, and since $G$ is a submodule of $G’$, $G$ is finitely generated over $\mathbb{Z}_p$.

Since $\mathbb{Z}_p$ is principal and $G$ has no torsion, and all the nontrivial quotients of $\mathbb{Z}_p$ are finite, $G$ is a power of $\mathbb{Z}_p$.

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    $\begingroup$ Thanks for your perfect answer. But it seems to me that we don't need the fact that all the nontrivial quotients of $Z_p$ are finite, because any finitely generated torsion-free module over a PID is a free module of finite rank. This is corollary 2.4 in Conrad's paper. $\endgroup$ – Sssss Jul 8 at 12:29
  • $\begingroup$ You are right, that part is indeed pointless. I was (wrongly) concerned with the idea that some quotients of $\mathbb{Z}_p$ would still be torsion-free, but that’s obviously impossible. $\endgroup$ – Mindlack Jul 8 at 12:34

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