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 $L$?

  • $\begingroup$ Ooops - I suppose you mean "restriction of an automorphism in $H$"? $\endgroup$ Feb 3, 2018 at 12:02
  • $\begingroup$ No I mean the restriction of any automorphism in Gal(K/k) to the fixed field L where L is a field fixed by the actions of a subgroup H of Gal(K/k) $\endgroup$ Feb 3, 2018 at 12:06
  • $\begingroup$ @Ecotistician If I understood correctly what you meant, then it is not true necessarily that the restriction of an automorphism of the whole Galois group to the middle field $\;L\;$ is an automorphism of $\;L\;$ ...in fact, unless $\;L/k\;$ is normal there will always be an element of the Galois group that is not an automorphism of $\;L/k\;$ ... $\endgroup$
    – DonAntonio
    Feb 3, 2018 at 12:14
  • $\begingroup$ @DonAntonio When will this be true? in our class we haven't learned the notion of normal extensions yet. I can see that if L itself is a splitting field then this is true (or that L is a galois over k). But does L being a fixed field of a subgroup of Gal(K/k) make it any closer for the restriction to be an automorphism in any way? $\endgroup$ Feb 3, 2018 at 12:18
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    $\begingroup$ @Ecotistician Sorry but I can't understand how can you be doing stuff with Galois extensions, one of which characteristics is being a normal extension, but you haven't yet learned what a normal extension is...You also should know by now what an algebraic extension is and what a sperable one is. In fact, "normal" enters also, pretty naturally, in the context of splitting fields... and you can see that both answers below mention this so important notion of "normal extension" as otherwise I really don't know how can this question be answered... $\endgroup$
    – DonAntonio
    Feb 3, 2018 at 13:39

2 Answers 2


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 $\sigma\vert_L : L \to L$.

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 $\mathrm{Gal}(K/k)$.


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 $g^{-1}$.


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