0
$\begingroup$

The following is my question: Let $K/F$ be a Galois extension with Galois group $G = Gal(K/F)$, with intermediate field $L: F \subseteq L \subseteq K$ which corresponds to subgroup $H \leq G$ by the fundamental theorem of Galois theory (FTGT); $L: H = Gal(K/L), L = K^H$, where $K^H$ is the subfield of $K$ fixed by $H$. Let $\sigma \in G$. Prove that $\sigma(L)$ is the intermediate field which corresponds with the subgroup $\sigma H \sigma^{-1} \leq G$ by the FTGT.

I am expecting normalcy to come up but beyond that intuition, I am stumped.

$\endgroup$
1
  • $\begingroup$ I see that this question was marked down. I would be happy to improve it, but I am not sure what is wrong with it. Can you let me know? (I realize that my own work on the problem is rather small, but I am just too confused to get very far with it. I am trying, however, and will update the question with any insights that I have.) $\endgroup$
    – kevin
    Apr 2, 2014 at 0:44

1 Answer 1

1
$\begingroup$

An element $x$ of $K$ is fixed by $\sigma H\sigma^{-1}$ iff, for all $h\in H$, we have $\sigma (h(\sigma^{-1}(x)))=x$, which is equivalent to $h(\sigma^{-1}(x))=\sigma^{-1}(x)$. That last equation says that $\sigma^{-1}(x)$ is fixed by $h$, so the equation holds for all $h\in H$ iff $\sigma^{-1}(x)\in L$, i.e., iff $x\in\sigma(L)$.

$\endgroup$
1
  • $\begingroup$ Oh wow! That makes a lot of sense now. Good job and thank you! $\endgroup$
    – kevin
    Apr 2, 2014 at 1:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .