Topology on the infinite Galois group

Question

I was reading the statement of the local class field theory from Kato's "Number Theory 2", and got confused on the topology of infinite Galois groups.

1. Given a Galois extension $L/K$, the topology of $Gal(L/K)$ is defined by the following fundamental system of neighborhood: $$V_J = \{\tau \in Gal(L/K) \mid \tau(x) = \sigma (x) \forall x \in J\}$$ where $J$ is a finite subset of $L$. Then the main theorem of the infinite Galois theory says there's a bijection $$\{\textrm{fields $M$ such that $K \subset M \subset L$}\} \leftrightarrow \{\textrm{closed subgroups $H$ of $G$}\}$$ where $M$ corresponds to $H = \{\sigma \in G \mid \sigma(x) = x \forall x \in M\}$. Now my question is why is $H$ a closed subgroup? It seems like $H$ is in the fundamental system of neighborhood of 1. (My understanding of a fundamental system of neighborhood is any open set is a union of elements in the system, is that correct?)

2. When stating LCFT, the book claims that there's a bijection $$\{\textrm{finite abelian extensions of $K$}\}\leftrightarrow \{\textrm{open subgroups of $Gal(K^{ab}/K)$}\}$$ given by Galois theory, where $L/K$ corresponds to the kernel of $Gal(K^{ab}/K) \to Gal(L/K)$. I think this is exactly the bijection I wrote in 1, but why is this kernel an open subgroup now, rather than closed?

Answer 1

The key to what is going on here is that if $G$ is a topological group and $H\subseteq G$ is an open subgroup, then $H$ is also closed. Indeed, $G$ can be written as the disjoint union of the cosets of $H$. The union of all the cosets except $H$ itself is thus open, and so its complement, namely $H$, is closed.

So in particular, the $V_J$ are by definition open, but since they are subgroups, this implies they are also closed. The subgroup $H = \{\sigma \in G \mid \sigma(x) = x \forall x \in M\}$ is then closed, being the intersection of all the $V_J$ where $J$ ranges over all finite subsets of $M$.

For your second question, if $L$ is a finite extension of $K$, then you can take $J$ to be a finite set that generates $L$ over $K$ and then $V_J$ is equal to the subgroup corresponding to $L$. So when $L$ is finite, the corresponding subgroup is open as well as closed (since the $V_J$ are open). Conversely, any open subgroup contains $V_J$ for some finite set $J$, and then the corresponding subfield must be contained in the extension of $K$ generated by $J$.