Restriction of automorphism to a fixed field an automorphism of the fixed field?

Question

If We assumed we had a Galois group of a finite field extension $Gal(K/k)$. and a fixed field $L = K^{H}$ where $H < Gal(K/k)$. why is it true that the restriction of an automorphism in $Gal(K/k)$ an automorphism of $L$?

Answer 1

This is not true in general.

Consider $k = \Bbb Q, K = \Bbb Q(\sqrt[3]{2}, \zeta_3)$. The extension $K/k$ is Galois, with $\mathrm{Gal}(K/k) \cong S_3$ (since the Galois group embeds in $S_3$ and has $6$ elements). Let $L = \Bbb Q(\sqrt[3]{2}) = K^H$ with $H=\mathrm{Gal}(K/L)$. The element $\sigma \in \mathrm{Gal}(K/k)$ such that $$\sigma(\sqrt[3]{2}) = \zeta_3 \sqrt[3]{2},\qquad \sigma(\zeta_3)=\zeta_3$$ doesn't even restrict to a morphism $\sigma\vert_L : L \to L$.

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However, if $\sigma(L) \subset L$, then $\sigma\vert_L : L \to L$ belongs to $\mathrm{Aut}_{k\mathrm{-alg.}}(L)$, since a field morphism is injective and $L$ is finite dimensional over $k$. The condition $\sigma(L) \subset L$ holds for any $\sigma \in \mathrm{Gal}(K/k)$ when $L/k$ is a normal extension, i.e. when $H$ is a normal subgroup of $\mathrm{Gal}(K/k)$.

Answer 2

This is only true if $H$ is a normal subgroup: In that case, $g^{-1}Hg=H$ for any $g\in G=\mathrm{Gal}(K/k)$, hence if $x\in L$, then $g^{-1}hg(x)=x$ or equivalently $g(x)=hg(x)$ for all $h\in H$, hence $g(x)\in L$ by definition of $L$. Therefore $g$ restricts to a $k$-algebra homomorphism of $L$ which is necessarily an automorphism as the same holds for $g^{-1}$.