Suppose $K \subset L \subset E$ are finite field extensions and $E$ is galois over $K$. I wish to show that if $L$ is stable w.r.t $K\subset E$, that is:
$$ \forall \, Gal(E/K), \; \sigma(L) \subseteq L$$
Then the following map is a surjective homomorphism of groups:
$$ \pi \, : Gal(E/K) \rightarrow Gal(L/K) \, : \sigma \mapsto \sigma|_{L} $$
The homomorphism part is fine. To show surjectivity, the hint is to use the Second Isomorphism theorem for fields:
Thm:Second Isomorphism Theorem for Fields
Let $K,K'$ be fields, $f(x) = \sum_{i=0}^{N} \alpha^{i}x^{i} \in K[x]$ then given an isomorphism $\xi\,: K\rightarrow K'$, and denoting the polynomial $f^{\xi}(x) = \sum_{i=0}^{N} \xi(\alpha_{i})x^{i}$, there is an isomorphism:
$$ \psi\, : E \rightarrow E' \quad \psi\big|_{K} = \xi$$
Where $E$ is the splitting field of $f(x)$ and $E'$ the splitting field of $f^{\xi}(x)$.
I try to apply this by picking $\sigma \in Gal(L/K)$, and settng $ \xi = \sigma \, : L \rightarrow L$. Then since $E$ is galois over $K$, $E$ is galois over $L$.
This means equivalently that $E$ is the splitting field of a separable polynomial $\lambda(x) \in L[x]$ over $L$.
IF the splitting field of $\lambda^{\sigma}(x)$ is the SAME as that of $\lambda(x)$ – $E$ then we can immeadiately apply the second isomorphism theorem and find a morphism:
$$ \hat{\sigma} \, : E \rightarrow E \quad \hat{\sigma}\big|_{L} = \sigma$$
i.e for every element in $\sigma \in Gal(L/K)$ there is some element in $\hat{\sigma} \in Gal(E/K)$ such that the restriction of $\hat{\sigma}$ is $\sigma$. So the morphism is surjective.
However – why is it true that the splitting field of $\lambda^{\sigma}(x)$ is the same as $\lambda(x)$??
Best Answer
HINT: Take $f$ to be a polyonimal in $K[x]$, s.t. $E$ is the splitting field of it over $K$. Consider it as an element of $L[x]$