Assume the field extension $L:K$ is Galois (i.e. finite, normal and separable), with $M$ an intermediate field.
- Show that $L:M$ and $M:K$ are finite separable field extensions.
Attempt: Both $M:K$ and $L:M$ are algebraic ($M:K$ since $M \subset L$, so every $\alpha \in M$ is algebraic over $K$ and $L:M$ since any $\alpha \in L$ is algebraic over $K$, so also over $M$).
Then $M:K$ is separable again since $M \subset L$ (every element of $L$ so in particular every element of $M$ is separable over $K$).
For $L:M$, take any $\alpha \in L$, then min$_M(\alpha)$ divides min$_K(\alpha)$ in $M[X]$. Since min$_K(\alpha)$ has no multiple zeros in a splitting field, neither does min$_M(\alpha)$, i.e. $\alpha$ is separable over $M$.
I'm not quite about the proof for finite, would it involve the Tower Law?
- Show that $L:M$ is a normal extension.
I'm a bit stuck here, any help for this one?
Yes, to show that $L/M$ and $M/K$ are finite extensions use the Tower Law, $$ [L:K] = [L:M][M:K]. $$
To show that $L/M$ is a normal extension, let $f \in M[x]$ be an irreducible polynomial which has a root $\alpha$ in $L$. We want to show that $f$ splits over $L$. Let $g \in K[x]$ be the minimal polynomial of $\alpha$ over $K$. Then, $f$ divides $g$. Since $L/K$ is normal, $g$ splits over $L$ and therefore so does $f$.