I was thinking about how to prove that if $K/F$ is Galois of Galois group $G$ and $L$ is the fixed field of $H This apparently requires $H$ to be normal to be true, but what bothers me is that I have an argument that only seems to require that $K$ be finite over $F$, without even needing to be Galois and without needing $F$ to be a fixfield. Let $f$ be some automorphism of $F$, and let $K=F(\alpha)$ for algebraic $\alpha$, then the mapping $\sum a_i \alpha^i\to\sum f(a_i)\alpha^i$ is surely an automorphism of $K$, no?
Why do automorphisms of a fixedfield extend to automorphisms of the whole (Galois) field?
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3This isn't true unless $L$ is also Galois, the quotient map only works for normal extensions otherwise $G/H$ is not a group. – 2017-01-06
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0@AdamHughes Is that not built into the fact that $L$ is a fixed field? – 2017-01-06
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1No, take for example $\Bbb Q(\zeta_3, \sqrt[3]{2})$ the subfield $\Bbb Q(\sqrt[3]{2})$ is the fixed field of complex conjugation (it is the maximal, real subfield) but clearly it is not Galois. – 2017-01-06
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0Here I go riding my hobby horse, but an extensive examination of examples would have told you where you went wrong. – 2017-01-07
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0Wait, well then why is it true that the Galois group of $L/F$ is isomorphic to $G/H$? What assumption am I missing? – 2017-01-07
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0@JackM, $L/F$ is Galois iff $H$ is normal in $G$. – 2017-01-08
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0I've updated the question. – 2017-01-08