I would like to understand the following part of this proof of the Main Theorem of Kummer Theory (source: James Milne's Fields and Galois Theory, page 73).
In particular, I would like to understand
- what the role/purpose of $H$ is in this proof, and
- why does $\forall \sigma \in H: \sigma(x) = x \: \: \forall x \in E = F[B^{1/n}]$ (this is how I understand the last underlined part) lead to $B = B(E)$?
I am grateful for any help!
I guess that your queries are only about the last paragraph, beginning with "Next consider a group B containing ${F^*}^n$ as a subgroup of finite index, and let $E=F(B^{1/n}$)...". According to the first paragraph, $E/F$ is an abelian extension with Galois group $G$ of exponent $n$ and, writing $B(E)=F^* \cap {E^*}^n$, there is an isomorphism $B(E)/{F^*}^n \to Hom(G,\mu_n)$ defined by $b \to (s \to s(b^{1/n})/b^{1/n})$ for all $s\in G$. Things get clearer when this iso. is interpreted as a $G$- duality, i.e. a non degenerate pairing $G \times B(E)/{F^*}^n \to \mathbf Z/n$ that is compatible with the action of $G$ because $F$ contains $\mu_n$. By construction, $B(E)$ contains $B$, so you can introduce the "orthogonal" $H$ of $B/{F^*}^n$ under the previous pairing and, practically by definition, check that $s\in H$ iff $s(b^{1/n})/b^{1/n}=1$ for all $b\in B$, iff $H=Gal(E/F(B^{1/n}))$, i.e. $H$ is trivial. By non-degeneracy, this means that $B/{F^*}^n = B(E)/{F^*}^n$, i.e. $B=B(E)$ .