Let $p$ be an odd prime number. It is known that $\mathbb{Z}_p^{\times}$ is topologically cyclic.

Now let $\chi_{\mathrm{cyclo}} : Gal(\bar{\mathbb{Q}}_p / \mathbb{Q}_p) \to \mathbb{Z}_p^{\times}$ be the $p$-adic cyclotomic character. This character factors through $Gal(\mathbb{Q}_p(\mu_{p^{\infty}} / \mathbb{Q}_p))$ which is also topologically cyclic.

Can I find a topological generator $g$ of $Gal(\mathbb{Q}_p(\mu_{p^{\infty}} / \mathbb{Q}_p))$ such that the image $\chi_{\mathrm{cyclo}}(g) \in \mathbb{N}$ ?

  • $\begingroup$ I am probably confused, but isn't $\chi_\text{cyclo}$ surjective? $\endgroup$ – user8268 Sep 19 '13 at 20:00

The cyclotomic characters induce an isomorphism between $Gal(\mathbb Q_p(\mu_{p^{\infty}})/\mathbb Q_p)$ and $\mathbb Z_p^{\times}$. So your question is equivalent to asking whether there is a topological generator of $\mathbb Z_p^{\times}$ which lies in $\mathbb N$.

The answer is yes: choose any natural number that is a primitive root modulo $p^2$ (and hence modulo $p^n$ for all $n$).

[This was already pointed out in an answer by user8268 that is now deleted.]

  • $\begingroup$ Nice answer. Is it still true if we replace $Gal(\bar{\mathbb{Q}}_p / \mathbb{Q}_p)$ by $Gal(\bar{\mathbb{Q}}_p / K)$ where $K/\mathbb{Q}_p$ is finite ? $\endgroup$ – user65490 Sep 19 '13 at 20:15
  • $\begingroup$ My guess is that the answer is still yes, because the image of Gal(\bar{\mathbb{Q}}_p / K) by the cyclo. character is an open subgroup of $\mathbb{Z}_p^{\times}$ of the form $\mu \times 1 + p^n \mathbb{Z}_p$. $\endgroup$ – user65490 Sep 19 '13 at 20:56
  • $\begingroup$ @user65490: Dear user, The answer is still yes. We are just looking at an open subgroup of the procyclic group $\mathbb Z_p^{\times}$, so it is again procycic, and a generator can be taken to be a power of our chosen generator of $\mathbb Z_p^{\times}$. Now just note that a power of an element of $\mathbb N$ is again in $\mathbb N$. Regards, $\endgroup$ – Matt E Sep 20 '13 at 1:29

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