I'm having trouble with a seemingly simple statement.
Suppose $N|K$ is a Galois extension with $L$ an additional field over $K$. Now for $\sigma \in \text{Gal}([N,L] | L)$, with $[N,L]$ being the composite of $N$ and $L$, I'd like to show that $\sigma(N) \subset N$.
I'm aware of the implication from the Fundamental Theorem of Galois Theory that says that for $K\leq E\leq F$ with $F|K$ and $E|K$ Galois extensions, $\sigma \in \text{Gal}(F|K): \sigma(E) \subset E$, which would solve my problem. But I can't seem to show that $[N,L]|K$ is Galois (I know an extension being Galois is not transitive).
Does it suffice to say that $[N,L]$ is also the splitting field over $K$ of a separable polynomial, just like $N$, although $[N,L]$ would eventually contain more elements?
This is my first question here, so let me know if I can improve the format of my question.
Best Answer
Let $M/K$ be a finite field extension, and $E,F$ two intermediate fields with $E/K$ Galois. Then $EF/F$ is Galois and there is an injective group homomorphism $$ \mathrm{Gal}(EF/F) \to \mathrm{Gal}(E/K), \quad \sigma \mapsto \sigma|_E $$ To see this, write $E$ as the splitting field of a separable polynomial $f\in K[X]$. Then $E/K$ is generated by the roots of $f$, so $EF=F(E)$ is generated over $F$ by the roots of $f$. Thus $EF/F$ is the splitting field of $f\in F[X]$, which is still separable, and hence $EF/F$ is Galois.
Now, if $\sigma\in\mathrm{Gal}(EF/F)$, then $\sigma$ permutes the roots of $f$, so restricts to an automorphism of $E$. It is easy to check that we have a group homomorphism between the Galois groups. Finally, if $\sigma$ is the identity on $E$, then it fixes all the roots of $f$, so $\sigma$ is the identity on $EF$, and the group homomorphism is injective.