TITLE: How far are the $p$-adic numbers from being algebraically closed?
QUESTION [48 upvotes]: A few days ago I was recalling some facts about the $p$-adic numbers, for example the fact that the $p$-adic metric is an ultrametric implies very strongly that there is no order on $\mathbb{Q}_p$, as any number in the interior of an open ball is in fact its center.
I know that if you take the completion of the algebraic closure of the $p$-adic completion you get something which is isomorphic to $\mathbb{C}$ (this result was very surprising until I studied model theory, then it became obvious).
Furthermore, if the algebraic closure is of an extension of dimension $2$ then the field is orderable, or even real closed. Either way, it implies that the $p$-adic numbers don't have this property.
So I was thinking, is there a $p$-adic number whose square equals $2$? $3$? $2011$? For which prime numbers $p$? How far down the rabbit hole of algebraic numbers can you go inside the $p$-adic numbers? Are there general results connecting the choice (or rather properties) of $p$ to the "amount" of algebraic closure it gives?

REPLY [14 votes]: Suppose that $K$ is an algebraic number field, i.e. a finite extension of $\mathbb Q$.  It has a ring of integers $\mathcal O_K$ (the integral closure of $\mathbb Z$ in $K$).  Suppose that there is a prime ideal $\wp \subset \mathcal O_K$ such that:

$p \in \wp,$ but $p \not\in \wp^2$.
The order of $\mathcal O_K/\wp = p.$  (Note that (1) implies in particular that
$\wp \cap \mathbb Z = p \mathbb Z$, so that $\mathcal O_K/\wp$ is an extension of $\mathbb Z/p\mathbb Z$.  We are now requiring that it in fact be the trivial extension.)

Then the number field $K$ embeds into $\mathbb Q_p$.  The converse also holds.
So if you want to know whether you can solve the equation $f(x) = 0$ in $\mathbb Q_p$ (where $f(x)$ is some irreducible polynomial in $\mathbb Q[x]$), then set
$K = \mathbb Q[x]/f(x)$ and apply this criterion.
This is easiest to do when
$f(x)$ has integral coefficients, and remains separable when reduced mod $p$
(something that you can check by computing the discriminant and seeing whether
or not it is divisible by $p$),
because in this case the criterion is equivalent to asking that $f(x)$ have a root mod $p$.
Incidentally, there are many $f(x)$ that satisfy this criterion
(because, among other
things, the algebraic closure of $\mathbb Q$ in $\mathbb Q_p$ has infinite degree over $\mathbb Q$), but there are also many $f(x)$ that don't.