In Page 279 of Algebriac Number Theory by Neukirch, it states that every field extension is composite of its finite subextensions. I know how to prove this in Galois Theory, but I don't see how to translate the proof to abstract Galois Theory.
In Neukrich, we take a profinite group $G$, and denote $G_K$ to be a closed subgroup of $G$ where we call $K$ as fix field of $G_K$. We say $L/K$ is a field extension if $G_L\subset G_K$. If $G_L$ is an open subgroup (equivalent to $G_L$ has finite index) in $G_K$, then we we say $L/K$ is a finite extension. The composite of $K_i$, denoted as $\prod_i K_i$ is the fix field of $\bigcap_i G_{K_i}$. So we need to prove that if $G_L\subset G_K$, then $$G_L = \bigcap_{G_L\subset G_M\subset G_K, [G_K: G_M]<\infty} G_M$$
I feel that this should be related to the topology on $G$ that there is open neighborhood basis of $1$ consisting of open subgroups of $G$.
Best Answer
I think I got an answer.
Since $G_K\supset G_L$ are closed subgroup of a profnite $G$, then we know that $G_K/G_L$ is again a profinite group. Then it suffice to show that the intersection of all open subgroup of a profinite group is $\{1\}$. This follows from open normal subgroups form a clopen neighborhood basis of $\{1\}$, and a profinite group is compact, Hausdorff and totally disconnected. Then the intersection of those open normal subgroups is the same as intersection of all clopen subsets containing $1$, and this equals the connected component containing $1$ which is $\{1\}$.