How to describe the Galois group of the compositum of all quadratic extensions of Q?
This is Problem 1.7 from Gouvea's lecture notes on deformations of Galois representations. In particular, he asks you to show that it has many subgroups of finite index which are not closed. So here's what I've got so far, which may be wrong.
I can write the compositum as F = Q[√–1, √2, √3, √5, √7 ...] (can I?) and then the Galois group G = Gal(F/Q) is isomorphic to a direct product Πp (Z/2Z) where the product is taken over all primes p, as well as p=-1, and the pth component is generated by the conjugation σp defined by √p -> –√p.
An example of a subgroup which isn't closed would be the subgroup H consisting of finite products of conjugations, since for example, H contains the sequence σ2, σ2σ3, σ2σ3σ5, σ2σ3σ5σ7... which converges to the automorphism "conjugate everything", and this automorphism is not contained in H.
However this subgroup is nowhere near being finite index--it has the cardinality of the natural numbers, whereas G has the cardinality of the reals. The only finite index subgroups I can think of take are of the form Gal(F/K) where K is a finite extension of Q, but of course these are by definition all closed. So I guess I've stuffed up somewhere, and I'd be really grateful for any help?! In know this may seem a bit "homework questiony" but it's not, it's just something that's really bugging me!
Let's forget the Galois group structure: it is enough to consider the infinite product $G = \mathbb{Z}/(2)^\mathbb{N}$ with its profinite topology and exhibit non-closed subgroups of finite index.
An element of $G$ is naturally identified with a subset of $\mathbb{N}$: the string $(a_i)$ corresponds to the subset $A$ such that $n \in A$ if and only if $a_n = 1$. With this interpretation, the operation on $G$ just becomes the symmetric difference of sets.
Now consider the filter $F'$ of all cofinite subsets of $\mathbb{N}$, and let $F$ be any ultrafilter containing it. I show that we can associate to $F$ a subgroup of index $2$. The $F$ will be non closed since $F$ is not a principal ultrafilter (indeed any non-principal ultrafilter will do).
Consider the characteristic function of $F$: $$f = \chi_F \colon G = \mathcal{P}(\mathbb{N}) \to \mathbb{Z}/(2)$$. I claim that this is a homomorphism, and its kernel is the desired subgroup.
This is easy to verify directly. Let $A, B \in G$ and assume $f(A) = f(B) = 0$. This means that neither $A$ nor $B$ belong to $F$, so their complementary sets do. It follows that $A^c \cap B^c \in F$, so that $A \cup B \notin F$. A fortiori the symmetric difference $A \Delta B \notin F$, whic means $f(A \Delta B) = 0$. The other cases are similar.
One way to think of the situation is that the countable product of $\mathbf Z/2\mathbf Z$s that you are considering is a vector space $V$ over $\mathbf F_2$ that is uncountable, and thus has an (even more) uncountable dual space. If $f:V \to \mathbf F_2$ is a non-zero element of the dual space, its kernel is an index 2 subgroup of $V$.
Now the open subgroups that you know about correspond to (i.e. are the kernels of) the projections onto the various factors in the product, and these are countable in number. So there is a massive host of other index 2 subgroups, corresponding to all the other non-zero elements $f$ of the dual space.
As others have noted, you won't be able to write these down explicitly though.
The finite index subgroups that you say that you can think of are precisely the closed (equivalently open) finite index subgroups of $Gal(F/\mathbb{Q})$ with respect to the Krull topology on the group. These are really the only relevant ones in terms of Galois theory. But the group has non-closed subgroups too, but their existence requires some non-constructive principle such as Zorn's lemma. These subgroups won't be open or closed in the Krull topology.
Here is an easy point which is implicit in the other answers, but which is part of the question (as I read it).
Put $(p_1,p_2,p_3,p_4,\dots):=(-1,2,3,5,\dots)$, and define the fields $K_0,K_1,\dots$ by $K_0:=\mathbb Q$ and $K_n:=K_{n-1}(\sqrt{p_n})$ for $n\ge 1$. [It's clear that any quadratic field is contained into some $K_n$.] To make sure that the Galois group in question is the indicated one, we need to check that $\sqrt{p_n}$ is not in $K_{n-1}$. To do that we prove the following apparently stronger claim.
Let $n$ be a positive integer and $a$ the product of the $p_k^{e(k)}$, where $k$ runs over the integers $\ge n$ and $e(k)$ is an integer equal to $0$ for almost all $k$. Assume $\sqrt a\in K_{n-1}$. Then all the $e(k)$ are even.
This is easily proved by induction.
EDIT. Things can be described in this way: The obvious morphism from $\mathbb Q^{\times}/\mathbb Q^{\times 2}$ into our Galois group induces an isomorphism between the profinite completions. [It's nice to obtain a Galois group by applying a simple construction to the base field.]