2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @EricWofsey Here actually a system of neighborhood of any point is give ($V_J$ is defined for any $\sigma$), so an open set should just be a union of the $V_J$'s right? Also, thanks for your answer! $\endgroup$ – Aaron Johnson Jan 7 '18 at 7:25
  • $\begingroup$ Oh, you're right, I misread. $\endgroup$ – Eric Wofsey Jan 7 '18 at 15:22
2
$\begingroup$

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 $J$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.