Localisation and completion of $\mathbb{Z}$

Question

This question might seem a little vague, but any information will be helpful.

Fix a prime $p$ in $\mathbb{Z}$. It is easy to see as rings that the localisation at $(p)$ is contained in the $(p)$-adic completion (i.e. $\mathbb{Z}_{(p)}\subset\mathbb{Z}_p$), but what other elements can we claim is in the completion? Can we say something like $\sqrt[n]{q} \in\mathbb{Z}_p$ for any prime $q$ other than $p$ (probably by using Hensel's lemma)?

Edit: As suggested in the comments, this older question gives a good insight into the answer.

The question came up because someone told me that the maximal ideal $m_{O_\mathbb{C_p}}$ (the maximal ideal of the ring of integers of $\mathbb{C}_p$) is generated by $\{\sqrt[n]{p}: n\in\mathbb{N}\}$. Does that mean that $\mathbb{Q}_p[\{\sqrt[n]{p}: n\in\mathbb{N}\}]=\mathbb{Q}_p^{alg}$ (where $\mathbb{Q}_p^{alg}$ is the algebraic closure of $\mathbb{Q}_p$)?

Again, I know the question is pretty vague, but thanks in advance for any help!

Does this answer your question? Square roots in the $p$-adics — Steven Stadnicki, Apr 25 '22 at 15:10
(The question I've pointed to isn't a precise duplicate but it very specifically answers the question about square roots and points to how to use Hensel's lemma to find the algebraic numbers in $\mathbb{Z}_p$.) — Steven Stadnicki, Apr 25 '22 at 15:11
This definitely helps with the first half of my questions. Thanks for that! — Coherent Sheaf, Apr 25 '22 at 15:20

reuns · Accepted Answer · 2022-04-25T16:13:15.817

$\Bbb{Z}_p$ is uncountable, it contains a lot of elements non-algebraic over $\Bbb{Q}$.

For the elements algebraic over $\Bbb{Q}$ they are a bit complicated to describe. What we can say

A number field $L$ embeds into $\Bbb{Z}_p$ iff its ring of integers has an unramified prime ideal above $p$ and whose redidue field is $\Bbb{F}_p$.
Also $\Bbb{Q}_p\cap \overline{\Bbb{Q}}$ is generated over $\Bbb{Q}$ by some elements for which Hensel lemma holds. More precisely $f\in \Bbb{Q}[x]$ irreducible has a root in $\Bbb{Z}_p$ iff there is $g\in \Bbb{Z}[x]$ monic irreducible with a simple root in $\Bbb{Z}/(p)$ and such that $f$ has a root in $\Bbb{Q}[x]/(g)$.

For your last question, no. $\mathbb{Q}_p^{alg} = K_\infty$ where $F=\Bbb{Q}_p[\zeta_{p^\infty-1}], K_0=F[p^{1/(p^\infty-1)}], K_{n+1}= K_n[K_n^{1/p}]$.

$F$ is the maximal unramified extension, $K_0$ is the maximal tamely ramified extension, and the solvability of the ramification group plus Kummer theory gives that $K_\infty$ is algebraically closed.

The maximal ideal of $O_{\Bbb{C}_p}$ is generated by the $p^{1/n}$ just because it is the valuation ring of a $\Bbb{Q}$-valued valuation.

Thanks a lot! This is incredibly helpful! – Coherent Sheaf Apr 25 '22 at 16:11 — Coherent Sheaf, Apr 25 '22 at 16:11

Localisation and completion of $\mathbb{Z}$

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