Smallest Galois extension

Let $F$ be a finite dimensional Galois extension of $K$ and let $E$ be an intermediate field. Show that there is a unique smallest field $L$ such that $E\subset L\subset F$ and $L$ is Galois over $K$. Moreover, $\mathrm{Aut}_{L}F=\bigcap_{\sigma}\sigma(\mathrm{Aut}_{E}F)\sigma^{-1}$ where $\sigma$ runs over $\mathrm{Aut}_{K}F$.

I took $L=\bigcap\{F':E\subset F'\subset F,F' \;\text{Galois over}\; K\}$ but I could not prove the last statement.

• It seems like I should take $L$ to be the fixed field of $\bigcap_{\sigma}\sigma(\mathrm{Aut}_{E}F)\sigma^{-1}$ instead.
– cyc
Apr 28 '13 at 21:34

The field you are looking for is the Galois closure of $E$. That is, $L$ is the compositum of all the embeddings of $E$ in $F$. The assumptions on $F$ guarantee such a field will be unique. If you just started with a field $K$ and a finite extension $E$, then one would want to take the compositum over all embeddings of $E$ to get $L$, but this is not well-defined until you choose an algebraic closure. In this problem, $F$ sets the context.
The identity of automorphism groups is an application of the Galois correspondence to the compositum suggested above. Both of your characterizations of $L$ ought to be correct.