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!


$G$ 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 $\mathbb{Z}_p$.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.