I am trying to teach myself Galois theory. Is it true that every for a field extension K->L, that every normal subgroup of Gal(L:K) is of the form Gal(L:M) for some intermediate field M, ie K->M->L?
[Math] Normal subgroups of the Galois group
galois-theory
Related Solutions
You might introduce a Grothendieck topology on the Galois subextensions of $L/K$; then the Galois group is indeed a sheaf.
I just want to remark the following: There is a natural homeomorphism between $Gal(L/K)$ and $Spec(L \otimes_K \overline{K})$. Namely, if $\sigma \in Gal(L/K)$, then the kernel of $L \otimes_K \overline{K} \to L \otimes_K \overline{K} \subseteq \overline{K} \otimes_K \overline{K} \to \overline{K}$ is a prime ideal of $L \otimes_K \overline{K}$. This is easily checked to be a homeomorphism if $L/K$ is finite, and then also in general. In particular, you can endow the profinite space $Gal(L/K)$ with a sheaf so that we get an affine scheme. If $L'/K$ is a Galois subextension of $L/K$, then the restriction $Gal(L/K) \to Gal(L'/K)$ comes from the inclusion $L \otimes_K \overline{K} \subseteq L' \otimes_K \overline{K}$. This provides another reason why $Gal(-/K)$ is a "sheaf", namely because $Spec$ commutes with filtered direct limits.
It is always a good idea to look at the local question first. What is the maximal pro-$p$ extension of $\mathbf{Q}_p$ ? There is a vast literature on the subject, starting with Demushkin whose results are exposed in an old Bourbaki talk by Serre on the Structure de certains pro-p-groupes (d'après Demuškin), (http://www.numdam.org/item?id=SB_1962-1964__8__145_0). They turn out to be extremely interesting groups which satisfy a kind "Poincaré duality" and have a striking presentation mysteriously similar to the presentation of the fundamental group of a compact orientable surface. Now I have to go for my yoga class; I hope you can find more on the web by yourself.
Addendum. Another good idea which I record here although I'm sure you don't need to be reminded of it is that in the local situation one should first ask what happens at primes $l\neq p$. What is the maximal pro-$l$ extension of $\mathbf{Q}_p$ ? This is much easier to answer because the only possible ramification is tame. You have the maximal unramified $l$-extension, which is a $\mathbf{Z}_l$-extension, and then a totally ramified extension of group $\mathbf{Z}_l(1)$, the projective limit of the roots of $1$ of $l$-power order. As a pro-$l$ group it admits the presentation
$\langle\tau,\sigma\mid\sigma\tau\sigma^{-1}=\tau^p\rangle$.
All this can be found in Chapter 16 of Hasse's Zahlentherie and in a paper by Albert in the Annals in the late 30s.
Addendum 2. A third good idea is to look at the function field analogues, wherein you replace $\mathbf{Q}$ by a function field $k(X)$ over a finite field $k$ of characterisitc $p$, where $X$ is a smooth projective absolutely irreducible curve over $k$. There would again be two cases, according as you are looking at the maximal pro-$l$ extension for a prime $l\neq p$ or for the prime $l=p$; the former should be much easier. You can simplify the problem by replacing $k$ by an algebraic closure. We are then in the setting of Abhyankar's conjecture (1957) which was very useful to Grothendieck as a first test for his theory of schemes. In any case, the conjecture was settled by Raynaud and Harbatter in the early 90s.
Best Answer
Let $L/K$ be an arbitrary extension of fields, and put $G = \operatorname{Aut}(L/K)$.
For a subgroup $H$ of $G$, put $L^H = \{x \in L \ | \ \forall \sigma \in H, \ \sigma(x) = x\}$.
We say that a subgroup $H$ of $G$ is closed if $H = \operatorname{Aut}(L/L^H)$.
Theorem (Artin-Kaplansky): Every finite subgroup $H$ of $G$ is closed.
Moreover, a closed subgroup $H$ of $G$ is normal iff for all $\sigma \in G$, $\sigma L^H = L^H$. If $L/K$ is algebraic, this says that $L^H/K$ is normal.
Also, if $[L:K]$ is finite, then $|G| \leq [L:K]$. So for finite extensions, every subgroup is closed.
If $[L:K]$ is infinite, it need not be the case that every subgroup $H$ of $G$ is closed. In fact, if $L/K$ is algebraic and Galois of infinite degree, then there are always nonclosed subgroups, since any closed subgroup $H$ of $G$ is profinite, hence finite or uncountably infinite. But the group $G$ itself is uncountably infinite, so has countably infinite subgroups.
To see a particular example of a nonclosed subgroup which is moreover normal, take $K = \mathbb{F}_p$ and $L = \overline{\mathbb{F}_p}$, an algebraic closure. Then $\operatorname{Aut}(L/K) \cong \widehat{\mathbb{Z}}$, and the closed subgroups are those of the form $n \widehat{\mathbb{Z}}$: they are all open, and in particular uncountably infinite. Then the subgroup of $G$ generated (not topologically generated!) by the Frobenius automorphism $\operatorname{Fr}: x \mapsto x^p$ is isomorphic to $\mathbb{Z}$, and is a nonclosed normal subgroup whose closure is $\widehat{\mathbb{Z}}$.