This is about an exercise on Fourier Analysis on Number Fields by Dinakar Ramakrishnan & Robert J. Valenza, exercise 1.14(a.i).
Let $K/F$ be a Galois extension with Galois group $G$.
Let $L$ be an intermediate field that is finite over $F$. For any $\sigma \in G$, define $N_L(\sigma)$ to be the set of $\tau \in G$ such that $\sigma$ and $\tau$ agree on $L$.The subsets $N_L(\sigma)$ constitute a subbase for a topology on $G$. Show (i) the topology remains unchanged if we restrict the subbase to normal intermediate fields that are finite over $F$ and (ii) that this topology is identical to the profinite topology on $G$.
This exercise says that the profinite topology can be induced by finite subextensions or more precisely normal finite subextensions through a subbase.
My questions:
a) Probably obvious, but did the authors mean "the subsets $N_L(\sigma)$" by $\{N_L(\sigma):F \subset L \subset K,[L:F]<\infty,\sigma \in G\}$?
b) While I cannot work out a solution of (i) yet, I suppose it is because of the existence of normal closure of finite extensions. Let $L$ be as above, we can find an intermediate $L' \supset L$ that is normal and finite over $F$, in which case $N_{L'}(\sigma) \subset N_L(\sigma)$. In other words, the subbase is obtained by "shrinking" those of $N_L(\sigma)$. How are we going to show that this "shrinking" does not damage the integrity of the subbase from this point? What property of normal & finite extension will be used here?
Thanks in advance and feel free to correct my mistakes if there is any.
a) Correct. When the authors say "the subsets $N_L(\sigma)$", they mean the collection $\{N_L(\sigma): F \subset L \subset K, [L: F] < \infty, \sigma \in G\}$, which is a family of subsets of $G$. For each intermediate field $L$ and each element $\sigma$ of $G$, we get one such subset.
b) To prove $(i)$, one needs to show that for any $L$ finite over $F$, and for any open set $U$ containing $N_L(\sigma)$, there exists a normal finite extension $L'$ over $F$ such that $N_{L'}(\sigma) \subset U$. By definition of the topology, it suffices to show that for any finite $L/F$ and any $\sigma \in G$, there exists a finite normal $L'/F$ with $N_L(\sigma) = N_{L'}(\sigma)$.
Now, if $L/F$ is finite, let $L'$ be the normal closure of $L/F$. Then any automorphism of $L'$ which fixes $F$ will restrict to an automorphism of $L$ which fixes $F$. Moreover, any automorphism of $L$ which fixes $F$ can be extended to an automorphism of $L'$ which fixes $F$, because $L'/F$ is normal. This implies that $N_L(\sigma) = N_{L'}(\sigma)$ for all $\sigma \in G$, and completes the proof of (i).
For $(ii)$, one needs to use the fact that the Galois group $G$ is a profinite group with respect to the topology where a base of neighborhoods of the identity is given by the kernels of the homomorphisms $G \rightarrow \text{Gal}(K/L)$ as $L$ ranges over all finite Galois subextensions of $K/F$. Then use the fact that for any finite Galois extension $L/F$ contained in $K/F$, $N_L(\sigma) = \sigma N_L(1) \sigma^{-1}$, so these subsets do indeed form a base for the topology of $G$, showing that this topology is the same as the one given by the subbase in the exercise.