Galois automorphism and normal (extension) of normal is normal

abstract-algebragalois-theorynormal-extension

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 $L$.

  1. $K$ galois over $L$ means it takes certain number of these factors and its automorphisms permute roots WITHIN each of them.
  2. 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 $\mathrm{Aut}(K/F)\to \mathrm{Gal}(L/F)$.

  • Since $L/F$ is Galois, this map is well-defined. Indeed, any $F$-automorphism $\sigma\colon K\to K$ restricts to a map $\sigma|_L\colon L\to K$ and, since $L/F$ is Galois, $\sigma|_L$ must have image in $L$.
  • The kernel is exactly $\mathrm{Gal}(K/L)$.
  • By assumption, the map is surjective.

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.

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