Infinite degree of the field of totally p-adic numbers 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!
 A: @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.
