Problem Statement: Given $F \subseteq L \subseteq K$ field extensions such that $K/L$ and $L/F$ are galois, and any automorphism of $L/F$ extends to an automorphism of $K$, show that $K/F$ is galois.
From the (counter) example $\mathbb{Q},\mathbb{Q}(\sqrt{2}),\mathbb{Q}(2^{1/4})$, I get the vague idea given $f$ irreducible over $F$, it may split into factors over $L$.
- $K$ galois over $L$ means it takes certain number of these factors and its automorphisms permute roots WITHIN each of them.
- But since automorphism of $L/F$ permutes one factor to another, the ability to extend to an automorphism of $K$ means that auto of $K$ can also permute BETWEEN these factors.
Hence $K$ does not miss roots. But how do I turn this into a solid proof? I was also thinking galois group argument…?
Best Answer
It's always helpful to think in terms of automorphisms.
Consider the restriction map $\mathrm{Aut}(K/F)\to \mathrm{Gal}(L/F)$.
Putting this all together with the first isomorphism theorem, we find that $|\mathrm{Aut}(K/F)| = |\mathrm{Gal}(K/L)||\mathrm{Gal}(L/F)| = [K:F]$ and the result follows.