Explicit description of $\Bbb Q_p \cap \bar{\Bbb Q}$

Question

Note that we can embed $\Bbb Q_p$ into $\Bbb C$, as it is discussed here. But as far as I understand, this embedding sends the power series to transcendental elements, so we can't certainly embed $\Bbb Q_p$ into $\bar{\Bbb Q}$ (I love typing barBbbQ). And when we complete $\Bbb Q$ with the $p$-adic norm, then we get, for example, $(p-1)$-th roots of unity (to see this, just note that $x^{p-1} - 1$ factors in $\Bbb Z / p$, so by Hensel's lemma it factors in $\Bbb Q_p$).

So the question is: do we get anything else? What?

Answer 1

Tautologically, you want to describe the number fields $K \subset \overline{\mathbf{Q}}$ which embed in $\mathbf{Q}_p$. It's certainly not going to be possible to describe such fields completely. For example, consider the polynomial:

$$f(x) = x(x-1)^{n-1} + p g(x),$$

where $g(x)$ is an arbitrary polynomial. For a random choice of $g(x)$, the polynomial will be irreducible and (generically, by Hilbert irreducibility) have splitting field $S_n$. On the other hand, Hensel's Lemma will ensure that $K = \mathbf{Q}[x]/f(x)$ embeds into $\mathbf{Q}_p$.

For a Galois extension $K/\mathbf{Q}$, the condition is equivalent to asking that $K$ is unramified at $p$, and that the Frobenius element (conjugacy class) $\mathrm{Frob}_p \in G = \mathrm{Gal}(K/\mathbf{Q})$ is the identity.

For example, if $K/\mathbf{Q}$ is abelian, then the ramification condition implies that the conductor $N$ is prime to $p$, that is, $K \subset \mathbf{Q}(\zeta_N)$. The largest subfield of $\mathbf{Q}(\zeta_N)$ which embeds into $\mathbf{Q}_p$ corresponds, by Galois theory, to the largest quotient on which $\mathrm{Frob}_p$ is trivial. Yet

$$\mathrm{Frob}_p = [p] \in (\mathbf{Z}/N \mathbf{Z})^{\times} = G = \mathrm{Gal}(\mathbf{Q}(\zeta_N)/\mathbf{Q}).$$

The corresponding extension has Galois group $(\mathbf{Z}/N \mathbf{Z})^{\times}/[p].$

For example:

1. $\mathbf{Q}(\zeta_N) \hookrightarrow \mathbf{Q}_p$ if and only if $[p]$ is trivial in $G$, or equivalently, if $p \equiv 1 \mod N$.

2. There will be infinitely many abelian fields of any degree inside $\mathbf{Q}_p$ --- to generate extensions of degree $M$, for example, choose $N$ to be divisible by at least two distinct primes of the form $1 \mod M$. Then $G$ surjects onto $(\mathbf{Z}/M \mathbf{Z})^2$, and hence $G/[p]$ surjects onto $\mathbf{Z}/M \mathbf{Z}.$

3. A version of the inverse Galois problem predicts that there exists (arbitary many) fields $K$ with Galois group any finite group $G$ in which any (fixed) prime $p$ splits completely.

Answer 2

Let $K$ be a number field unramified at $p$, and $f$ be the minimal polynomial of a generator of $\mathcal O_K$. Then any root of $f$ modulo $p$ is simple (since $\mathrm{disc} f \not\equiv 0 \pmod{p}$), and hence (by Hensel) lifts to a root in $\mathrm Z_p$. Thus, if such a root exists, we have an embedding $K \subset \mathrm Q_p$.

On the other hand, assume that $K$ is totally ramified at $p$ with ramification index $e$. Then any root of $f$ in $\overline{\mathbb Q}_p$ has valuation $1/f$, and therefore cannot belong to $\mathbb Q_p$. This means that $K \cap \mathbb Q_p = \mathbb Q$.

More generally, we may write the localization $\widehat{K}_p = K \otimes_{\mathbb Q} \mathbb Q_p$ as a direct product of finite extensions of $\mathbb Q_p$, depending on the factorization of the ideal $p \mathcal O_K$ in $\mathcal O_K$. We immediately see that there exists a morphism $K \hookrightarrow \mathbb Q_p$ iff one of these extensions is trivial.

For example in the case of quadratic extensions: the quadratic number fields included in $\mathbb Q_p$ are exactly those which are split (not ramified, not inert) at $p$.

Answer 3

The set $\mathbb Q_p\cap \overline{\mathbb Q}$ is simply the set of $p$-adic numbers which satisfy a polynomial with coefficients in $\mathbb Q$. In fact you can see the two sets living inside the same one, namely $\overline{\mathbb Q_p}$.