# Subgroups of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$

If I'm understanding the main theorem of (infinite) Galois theory correctly, applied to $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, it gives us:

a) all its open subgroups are $\mathrm{Gal}(\overline{\mathbb{Q}}/K)$, with $K/\mathbb{Q}$ a finite extension.

b) all its closed (and not-open) subgroups are $\mathrm{Gal}(\overline{\mathbb{Q}}/K)$, with $K/\mathbb{Q}$ an infinite extension.

Is this correct?

I've also heard that for example $\mathrm{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ is a closed subgroup of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

How does this fit with the facts above? Does this mean that $\mathrm{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p) \cong \mathrm{Gal}(\overline{\mathbb{Q}}/K)$ for some $K$?

Also, what are its non-open subgroups? I assume this includes the Galois groups $\mathrm{Gal}(K/\mathbb{Q})$.

• what is an open subgroup ? it means that an infinite product of the elements of the group is not always in the group anymore ? May 28, 2016 at 13:28
• Galois groups are topological groups. A subgroup is open if an only if it contains a basic open subgroup. Here, a basic open subgroup is one that consists of all automorphisms leaving fixed all elements of a given finite extension $k\supset\Bbb Q$. May 28, 2016 at 13:45

Is this correct?

Yes. Infinite Galois theory provides a bijection between subfields of $\overline{\mathbb Q}$ and closed subgroups of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Since every open subgroup of a topological group is closed, and the open subgroups all have finite index in $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, both your statements are correct.

How does this fit with the facts above? Does this mean that $\mathrm{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p) \cong \mathrm{Gal}(\overline{\mathbb{Q}}/K)$ for some $K$?

The group $\mathrm{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)$ can be viewed as a subgroup of $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ as follows. Fix an embedding of $\overline{\mathbb Q}\hookrightarrow \overline{\mathbb Q}_p$. Then there is an inclusion $$\mathrm{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)\hookrightarrow\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)\\\sigma\mapsto\sigma|_{\overline{\mathbb Q}}$$

Choosing an embedding $\overline{\mathbb Q}\hookrightarrow \overline{\mathbb Q}_p$ is the same as picking a prime $\mathfrak p$ of $\overline{\mathbb Q}$ lying over $p$. In particular, we can identify $\mathrm{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)$ with the decomposition group $$D_\mathfrak p :=\{\sigma\in\ \mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q):\sigma(\mathfrak p) = \mathfrak p\} \subset\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q).$$

Analogously to the finite situation, the fixed field $\overline{\mathbb Q}^{D_\mathfrak p}$ can be identified with the maximal extension $K$ of $\mathbb Q$ such that $\mathfrak p\cap K$ is unramified and the residue field $\mathbb F_{\mathfrak p\cap K}=\mathbb F_p$.

The problem with this approach is that it is completely non-canonical and depends entirely on which choice of prime we chose of $\overline{\mathbb Q}$ lying above $p$. Practically, it is better to think of $\mathrm{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ as a subgroup of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ that is well-defined up to conjugation: analogously to the finite case, for different choices of primes $\mathfrak {p,p}'$, $D_\mathfrak p$ and $D_{\mathfrak p'}$ will be conjugate in $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

Also, what are its non-open subgroups? I assume this includes the Galois groups $\mathrm{Gal}(K/\mathbb{Q})$.

There are lots of non-closed and non-open subgroups, and they are not easy to describe. However the groups $\mathrm{Gal}(K/\mathbb{Q})$ should be seen as quotients of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, and not as subgroups.