Galois Correspondence - Surjectivity of Map Between Groups under Restriction

Question

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 $\lambda(x)$??

Answer 1

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 $L[x]$

Answer 2

I'll show a different way to solve it. Note that the kernel of $\pi$ is exactly $Gal(E/L)$, so from the first isomorphism theorem $Im(\pi)\cong Gal(E/K)/Gal(E/L)$, and hence:

$|Im(\pi)|=\frac{|Gal(E/K)|}{|Gal(E/L)|}=\frac{[E:K]}{[E:L]}=[L:K]$

Now, if we can prove that $L/K$ is a Galois extension then we would get $|Im(\pi)|=[L:K]=|Gal(L/K)|$ and since everything is finite $Im(\pi)=Gal(L/K)$. So let's prove $L/K$ is Galois. It is separable because $E/K$ is separable, so we only need to show it is a normal extension. Suppose it isn't. Then there is an element $\alpha\in L$ with a minimal polynomial $m_\alpha$ over $K$ such that $m_\alpha$ has a root $\beta\notin L$. Since $E/K$ is normal we know that $\beta\in E$, so it is in $E\setminus L$. It is well known that there is a natural isomorphism $\varphi:K(\alpha)\to K(\beta)$ such that $\varphi(\alpha)=\beta$. It is also known that it must have at least once extension to a homomorphism $\bar{\varphi}:E\to\overline{E}$. But $E/K$ is normal and hence $\bar{\varphi}$ is actually an automorphism of $E$, which by the definition of $\varphi$ must act on $K$ as the identity. So $\bar{\varphi}\in Gal(E/K)$. However $\bar{\varphi}(\alpha)=\beta\notin L$ which contradicts the assumption that $\sigma(L)\subseteq L$ for all $\sigma\in Gal(E/K)$.

Note: it might be a bit unclear what is the "natural" $\varphi$. Well, we know that $K(\alpha)\cong E/(m_\alpha)\cong K(\beta)$, since $m\alpha$ is the minimal polynomial of both $\alpha$ and $\beta$ over $K$. So $\varphi$ is just the composition of the corresponding isomorphisms.