Embedding the absolute Galois group of $ \mathbf Q_p $ into that of $ \mathbf Q $

Question

It is mentioned, in passing, in many sources that the restriction map $ \textrm{Gal}(\bar{\mathbf Q_p}/\mathbf Q_p) \to \textrm{Gal}(\bar{\mathbf Q}/\mathbf Q) $ is an embedding, that is, an element of the absolute Galois group of $ \mathbf Q_p $ is completely determined by its action on the algebraic numbers. I have been trying to prove this statement, to no avail...

I suspect that one might take a monic polynomial $ f \in \mathbf Q_p[X] $, and approximate it closely by a sequence of monic polynomials $ f_n \in \mathbf Q[X] $. Hopefully, as the sup-norm difference between the polynomials gets smaller, so do their roots get closer together - however, I have not been able to turn this into a rigorous argument. How would you go about proving this fact?

Related: (1)

Answer 1

R. Raylor gives an outline as follows: the absolute value $|\cdot |_p$ on $\mathbb{Q}$ has a unique extension to an absolute value on $\overline{\mathbb{Q}_p}$ and $G_p:=\textrm{Gal}(\bar{\mathbb Q_p}/\mathbb Q_p)$ is identified with the group of automorphisms of $\overline{\mathbb{Q}_p}$ which preserve $|\cdot |_p$, and it easy to see that this is the group of continuous automorphisms of $\overline{\mathbb{Q}_p}$. Thus we obtain, for each embedding $\overline{Q}\hookrightarrow \overline{\mathbb{Q}_p}$ a closed embedding $G_p\hookrightarrow \textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

Answer 2

If $P \in \Bbb Q_p[X]$ is a degree $n$ polynomial with $n$ roots $x_i$, consider the map $\phi_P$ that takes a degree $n-1$ polynomial $Q$ and outputs the degree $n-1$ polynomial $\prod_i (Y - Q(x_i)) - Y^n$.

It turns out that the Jacobian of $\phi_P$ at $Q=X$ is the discriminant of $P$ (the discriminant has to divide it, then you can check that the degrees coincide).

If this discriminant is nonzero (so if $P$ has distinct roots) this entails that there are neighbourhoods of $Q = X$ and of $P$ where $\phi_P$ is a local homeomorphism.

In particular, if $P$ is the minimal polynomial of some $x \in \overline {\Bbb Q_p}$ then every polynomial close enough to $P$ is a $\phi_P(Q)$ for some $Q$, which is an annihilating polynomial of the $Q(x_i) \in \Bbb Q_p(x_i)$.

In fact it would be great if $\phi_P(Q)$ was the minimal polynomial of $Q(x_1)$, because then the inclusion $\Bbb Q_p(Q(x_1)) \subset \Bbb Q_p(x_1)$ must be an equality and then we can just pick $Q$ such that $\phi_P(Q)$ is a rational approximation of $P$.

If it's not the minimal polynomial of $Q(x_1)$ then we must have a link between $\{1 ; Q(x_1) ; Q(x_1^2) ; \ldots Q(x_1^{n-1})\}$, so some big determinant whose entries are polynomials in the coefficients of $P$ and $Q$ is zero.

Since the determinant is continuous and is nonzero for $Q = X$, it stays nonzero for $Q$ close enough to $X$.

Then you have a small neighbourhood of $P$ where all the splitting fields are the same. In particular you can always get a rational appoximation of $P$ that has the same splitting field as $P$.

This shows that $\overline{\Bbb Q_p} = \Bbb Q_p \overline {\Bbb Q}$