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Consider $Q_p$ and $R$ where $Q_p$ is $p-$adic numbers with $p-$adic topology. If $Q_p\to R$ is a ring homomorphism, then certainly it must be trivial as $Z\to Z$ inducing $Q\to Q$ but infinite series of p-adic element does not converge in $R$ in general.

$\textbf{Q:}$ It is natural to ask whether one can classify all continuous maps $Hom_{Cts}(Q_p,R)$ besides constant maps. It is certainly possible to get just set theoretical embedding by axiom of choice. Does $Hom_{Cts}(Q_p,R)$ have non-constant functions? I do not see an obvious choice of candidate for non-constant maps.

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There are many continuous, non-constant functions; for example, the $p$-adic norm is one such (as is any positive power of the $p$-adic norm).

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  • $\begingroup$ (Topologically, $\mathbb{Q}_p$ is a countable disjoint union of $\mathbb{Z}_p$ which looks like the Cantor set, so there are of continuous maps.) $\endgroup$ – hunter Sep 11 at 18:15
  • $\begingroup$ Oh, in that case, I was dumb. I forgot the obvious $p-$adic norms. Why cantor sets are homeomorphic to $Z_p$ here? It looks like $Z_p$ is countable but cantor set is not. Do I need axiom of choice? $\endgroup$ – user45765 Sep 11 at 18:16
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    $\begingroup$ @user45765 In fact, $\mathbb{Z}_p$ is uncountable. For the homeomorphism to the Cantor set, no AC is needed. You can see a nice explicit proof here: personal.psu.edu/axk29/TOPOLOGY/p-adic.pdf $\endgroup$ – hunter Sep 11 at 18:19
  • $\begingroup$ Thanks a lot for clarification. $\endgroup$ – user45765 Sep 11 at 18:19

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