The closure of a number field under the $p$-adic topology

Question

We fix an inclusion $\overline{\mathbb{Q}}\hookrightarrow\overline{\mathbb{Q}_p}$.

Given $K$ a finite extension of $\mathbb{Q}_p$, can we always find $K_0$ a finite extension of $\mathbb{Q}$ such that the closure of $K_0$ in $\overline{\mathbb{Q}_p}$ under the $p$-adic topology is $K$?

Another question is how to determine all elements of $\overline{\mathbb{Q}}\cap\mathbb{Q}_p$?

Answer 1

Both of your questions have answers named after famous lemmas.

For your first question: yes, this follows from Krasner's lemma. Basically any polynomial sufficiently close to the minimal polynomial for $K$ but with rational coefficients will cut out the right extension for you.

EDIT: this is not complete, see the comments. For your second: if $\alpha \in \overline{\mathbb{Q}}$ and $p$ is not ramified in the extension $\mathbb{Q}(\alpha)$ you can appeal to Hensel's lemma. If $p$ is ramified and the extension is Galois then $\alpha$ is not in $\mathbb{Q}_p$; otherwise I am not sure.

Answer 2

Looks to me as if your second question is: “What is the algebraic closure of $\Bbb Q$ in $\Bbb Q_p$?”

I can’t imagine how one would describe all elements of that field, but at least I can point out that for every $n\ge1$ this field has a subfield (in fact many) that is of degree $n$ over $\Bbb Q$.

Start with the polynomial $g(X)=X^n+X+p$, which has a root in $\Bbb Q_p$. For all I know, it’s $\Bbb Q$-irreducible already, but if it isn’t, I’ll jiggle it a little so it becomes $\Bbb Q$-irreducible.

Let $q$ be any other prime than $p$, and let $a_0\in\Bbb Z$ with $a_0\equiv0\pmod{p^2}$ but $a_0\equiv q-p\pmod{q^2}$, using Chinese Remainder. Also, let $a_1\in\Bbb Z$ with $a_1\equiv0\pmod p$ and $a_1\equiv-1\pmod q$.

Now let $f(X)=g(X)+a_1X+a_0=X^n+(a_1+1)X+(a_0+p)$, our jiggled polynomial. Looking through the $p$-lens, we see that $v_p(a_1+1)=0$ and $v_p(a_0+p)=1$, so Hensel still gives us a $\Bbb Q_p$-root of $f$. On the other hand, the $q$-view gives us $v_q(a_1+1)\ge1$ and $v_q(a_0+p)=1$, which is enough to bring in Eisenstein (using the prime $q$), and $\Bbb Q$-irreducibility of $f$.