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I have a question on the degree of the field of totally $p$-adic numbers over $\mathbb{Q}$.

Fix $p$ a prime in $\mathbb{Z}$. We say that $\alpha \in \overline{\mathbb{Q}}$ is totally $p$-adic if $p$ splits completely in the ring of integers of the number field $\mathbb{Q}(\alpha)$. This is equivalent to say that for all valuation $v$ on $\mathbb{Q}(\alpha)$ above $p$, the completion $\mathbb{Q}(\alpha)_v$ is trivial, i.e., $\mathbb{Q}(\alpha)_v = \mathbb{Q}_p$.

Note the analogy with the concept of 'totally real': we say that $\alpha$ is totally real if every conjugate of $\alpha$ lies in $\mathbb{R}$, which is the same as saying that the completion of $\mathbb{Q}(\alpha)$ under any archimidean absolute value is just $\mathbb{R}$ (and $\mathbb{Q}_p$ is similar to $\mathbb{R}$ in a lot of senses).

Write $\mathbb{Q}^{tp}$ the field of totally $p$-adic numbers and $\mathbb{Q}^{tr}$ for the field of totally real ones. We know that the degree $[\mathbb{Q}^{tr}:\mathbb{Q}]$ is infinite, and so i would like to know if the degree $[\mathbb{Q}^{tp}:\mathbb{Q}]$ is also infinite or not (and why of course).

My attempt: The argument of proving that the degree $[\mathbb{Q}^{tr}:\mathbb{Q}]$ is infinite is very simple: take $K = \mathbb{Q}(\sqrt2, \sqrt3, \sqrt5,..., \sqrt{p_n})$. Then $K \subset \mathbb{Q}^{tr}$ and $[K:\mathbb{Q}] = 2^n$. So we have subfields of $\mathbb{Q}^{tr}$ with arbitrarely large degree.

I would like to export this idea to the $p$-adic world, but for example, $\sqrt{2}$ is not necessarily totally $p$-adic; it actually depends on the conditions of $p$, and so i'm stuck....

Thanks in advance!

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    $\begingroup$ What about considering quadratic reciprocity? If $p \equiv 1 \mod{4}$, then there exist arbitrarily many primes $q_1, \dots , q_n$ which are squares modulo $p$ (these are all other primes $q$ satisfying $q \equiv 1 \mod{4}$), thus $\Bbb{Q}(\sqrt{q_1}, \dots , \sqrt{q_n}) \subset \Bbb{Q}^{tp}$. $\endgroup$
    – Crostul
    Apr 10 '18 at 8:36
  • $\begingroup$ Sorry but i don't see why $p$ splits completely in $\mathbb{Q}(\sqrt{q})$ when $q$ is a square mod $p$. $\endgroup$ Apr 11 '18 at 9:17
  • $\begingroup$ This is a consequence of Hensel lemma. Since $x^2-q=0$ has two solutions $\mod{p}$, you can lift them to $p-$adic integers. $\endgroup$
    – Crostul
    Apr 11 '18 at 13:17
  • $\begingroup$ Yes that's right, thanks! $\endgroup$ Apr 11 '18 at 15:14
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@Crostul’s comment is the most direct route to a proof, but here is a more conceptual approach that shows where this field appears naturally.

Let $G_\mathbb Q=\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ be the absolute Galois group of $\mathbb Q$. Fix a place $v$ of $\overline{\mathbb Q}$ above $p$, and use it to define the decomposition group $D_v$ of $G_\mathbb Q$: $$D_v = \{\sigma\in G_\mathbb Q:\sigma\cdot v = v\}.$$ Whilst $D_v$ depends on the choice of prime, if $v,w$ are places above $p$, then $D_v$ is conjugate to $D_w$.

Then $\mathbb Q^{tp} = \overline{\mathbb Q}^{D_v}$ -- i.e. $\mathbb Q^{tp}$ is the fixed field of $D_v$. In particular, $\mathbb Q^{tp}$ is infinite and $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q^{tp}) \cong D_v$.

Before proving this, it's worth noting that $D_v\cong\mathrm{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)$, so this is completely analogous to the archimedean case, where $\mathrm{Gal}(\overline{\mathbb Q}/\overline{\mathbb Q}\cap\mathbb R)\cong \mathrm{Gal}(\overline{\mathbb R}/\mathbb R)$.


We can see this is true as follows. First note that $\mathbb Q^{tp}$ is compositum of all extensions of $\mathbb Q$ in which $p$ splits completely. If $K$ is a finite Galois extension of $\mathbb Q$, $v$ is a place of $K$ above $p$, and $D_v$ is the corresponding decomposition group, then $p$ splits completely in $L = K^{D_v}$. Extending this using infinite Galois theory gives the result.

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  • $\begingroup$ Sorry but i cannot see why $p$ splits completely in the ring of integers of $K^{D_v}$. $\endgroup$ Apr 11 '18 at 8:52
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    $\begingroup$ This is a standard result - see here, Prop 14.1.4. $\endgroup$
    – Mathmo123
    Apr 11 '18 at 16:06
  • $\begingroup$ Yes you are right. But i still have an issue with the infinity of the extension. I mean, we then can construct number fields $K$ with arbitrarily large degree which are Galois over $\mathbb{Q}$ and such that $K^D$ is in $\mathbb{Q}^{tp}$ (with $D$ the decomposition group) and $[K^D:\mathbb{Q}]$ is the number of distinct primes on $K$ above $p$. Why does this number of primes can also be arbitrarily large? $\endgroup$ Apr 12 '18 at 8:37
  • $\begingroup$ If $K, L$ are two finite extensions in which $p$ splits completely, then $p$ will also split completely in the compositum $K\cdot L$. For well chosen fields, the number of primes above $p$ in $K\cdot L$ will be $[L:\mathbb Q][K:\mathbb Q]$. Continuing this, you'd expect this number to become arbitrarily large. $\endgroup$
    – Mathmo123
    Apr 12 '18 at 13:26
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    $\begingroup$ In an infinite extension $K/\mathbb Q$, there will often be infinitely many primes of $K$ lying over $p$. $\endgroup$
    – Mathmo123
    Apr 12 '18 at 13:27

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