# Since $K/F$ is Galois, every embedding of $K$ fixing $F$ is an automorphism of $K$?

I'm trying to read a proof in Dummit and Foote of the statement

Suppose $K/F$ is a Galois extension and $F'/F$ is any extension. Then $KF'/F'$ is a Galois extension, and $Gal(KF'/F') \cong Gal(K/K \cap F')$.

One line I am confused about is

Since $K/F$ is Galois, every embedding of $K$ fixing $F$ is an automorphism of $K$, so the map $\varphi: Gal(KF'/F') \to Gal(K/F$), $\sigma \mapsto \sigma\vert_K$ defined by restricting an automorphism $\sigma$ to the subfield $K$ is well-defined.

I take it this means automorphisms of $KF'$ fixing $F'$ (thus fixing $F$) also send $K$ to $K$? If that's what it means, why is it true?

Thanks!

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Yes, you are right in your guess. Every $\sigma_K$ defines an embedding of $K$ to the bigger field field $KF'$. To show that $\varphi$ is well-defined, you need to show $\sigma(K)=K$ (as sets). That's the meaning the first sentence in your quotation. As for why, $K/F$ is Galois, in particular, it is normal. Take any $\alpha \in K$, any embedding fixing $F$ must sends it to a conjugate of $\alpha$. Being normal, $K$ contains all such conjugates. –  Jiangwei Xue May 8 '11 at 16:45
You might find useful reading this simplified version of this theorem I asked about in the past. The proof carries through to your generalized version. –  quanta May 8 '11 at 16:52
Clever, thanks :) –  badatmath May 8 '11 at 18:42