I am currently studying the separable degree of an algebraic extension. I have come across the following result: if $K \subset L \subset M$ are algebraic extensions and $M$ is normal over $K$, show that $[L:K]_{s} = |\operatorname{Hom}_{K}(L,M)|$. I want to prove it.
My attempt:
By definition, the separable degree of $L$ over $K$, $[L:K]_{s}$, equals $|\operatorname{Hom}_{K}(L,K')|$, where $K'$ is an algebraic closure of $K$. Therefore I should try to find a one-to-one relation between homomorphisms $L \rightarrow K'$ and homomorphisms $L \rightarrow M$ (fixing $K$). But I do not know what the relation is.