Let $F$ be a field and let $\zeta$ be a primitive $n$-th root of unity in $F$. Also, let $E/F$ be a finite Galois extension with Galois group $G$.
Now I am trying to understand the following Theorem from Milne's Fields and Galois Theory (page 73):
Question What exactly does this map in this theorem do?
In the sections before the theorem, there were some more maps which seemed to play a role in understanding the map in the theorem (and I think this was the author's intention). The previous sections are these (on page 72):
However, I have not figured out yet how the map in the theorem and the previously discussed maps/sequences are related. Could you please help me explaining that?
As you are aware, this is the main theorem of Kummer theory. I keep you notations and assumptions, but beware that you forgot the hypothesis that the characteristic of $F$ does not divide $n$. Besides, it will be more convenient to replace a subgroup $B$ containing ${F^\times}^n$ as a subgroup of finite index, by the quotient $\bar B=B.{F^\times}^n/{F^\times}^n$ considered as a finite subgroup of $F^\times/{F^\times}^n$. The reason is that, thanks to the hypotheses, $F(b^{1/n})/F$ depends only on the class $\bar b$ of $b$ mod ${F^\times}^n$. We'll denote this extension by $F({\bar b}^{1/n})/F$. Analogously, $F(B^{1/n})$ will be written $F({\bar B}^{1/n})$. Now the map that you ask about can be rewritten as $E\in (a) \to \bar B_E \in (F^\times\cap {E^\times}^n)/{E^\times}^n \in (b)$ (with an obvious rewriting of (b)). By definition, $E=F(\bar B^{1/n})$, which is easily shown to be a galois extension, with group $G_{E/F}$ finite abelian of exponent $n$. Usually, $ \bar B_E$ is called the radical of $E$, and Kummer's main theorem states thatit is canonically isomorphic to Hom ($G_{E/F}, \mu_n)$ (you already gave thje cohomological proof !). It follows that $ \bar B_E$ and $G_{E/F}$ have the same order, and the correspondence between (a) and (b) is bijective.