If $K/k$ is a finite Galois extension with a galois abelian group, and $ L/k$ any field extension

Question

any hints would be greatly appreciated to help me prove the following:

If $K/k$ is a finite Galois extension with a galois abelian group, and $ L/k$ any field extension then $KL/L$ is a finite galois extension with an abelian group.

So far I have proved that $KL/L$ is finite, and attempted to prove that its a separable and normal extension by trying show that $L$ is the fixed field of $Aut(KL/L)$ but failed, $L\subset L^{Aut(KL/L)}$ was trivial i and tried doing the other side by contradiction but got nowhere.

i had also previously briefly considered trying to show that $KL$ is the splitting field of a separable polynomial in $L[x]$ but disregarded that ianidea immediately since i don't think its even possible to think of an explicit polynomial.

Answer 1

Assume $K$ is the splitting field of $f(X) \in k[X]$. Then $LK$ is the splitting field of $f(X)$ regarded as an element of $L[X]$.

Answer 2

Let $\sigma$ be an embedding of $KL$ in a given algebraic closure over $L$. Then $\sigma(KL)=\sigma(K)\sigma(L)=KL$ since $K/k$ is Galois and in particular normal. This shows that $KL/L$ is normal. Since $KL/L$ is finite and $K/k$ is separable, $KL=L(a_1,...a_n)$ where each $a_i\in K$ are separable over $k$ and thus separable over $L$. Then $KL$ is separable over $L$.

Hence, $KL/L$ is finite Galois, and we denote its Galois group by $G(KL/L)$, and similarly for $K/k$. We define a restriction map $f:G(KL/L)\rightarrow G(K/K\cap L)$. $f$ is clearly a homomorphism. The identity automorphism in $G(K/K\cap L)$ fixes $K$ by definition and thus any of its extension to $KL/L$ must fix all of $KL$, i.e. it is the identity in $G(KL/L)$. On the other hand, any $\sigma\in G(K/K\cap L)$ may be extended to an automorphism on $KL$. We contend that the fixed field $K^{G(KL/L)}$ is precisely $K\cap L$.

Let $x\in K\cap L$. Then $x$ is fixed by any automorphism in $G(KL/L)$ and so it lies in $K^{G(KL/L)}$. Conversely, let $x\in K^{G(KL/L)}$. Then $\tau x=x$ for each $\tau\in G(KL/L)$ and thus $x\in L$. But $K^{G(KL/L)}$ is a subfield of $L$, so we must have $x\in L$. So $x\in K\cap L$. This prove that $K^{G(KL/L)}=K\cap L$.

Therefore $f$ is in fact an isomorphism, and we have $G(KL/L)\cong G(K/K\cap L)$. Since any embedding $\tau$ satisfies $\tau(K\cap L)=\tau(K)\cap\tau(L), K\cap L/k$ is Galois, and we have a homomorphism defined by restriction $g: G(K/k)\rightarrow G(K\cap L/k)$. One sees at once it is surjective, and using the same method as before, we see it has kernel $G(K/K\cap L)$, which is isomorphic to $G(KL/L)$. Then it is a subgroup of $G(K/k)$ which is abelian, so it is abelian also. This concludes our proof.