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

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.

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 $$L[X]$$.
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.