$L \vert K$ normal extension with $K \subset M \subset L$. Show that $L \vert M$ is normal Let $L \vert K$ be a field extension and let $K \subset M \subset L$ be a subextension. Show that if $L \vert K$ is normal, so is $L \vert M$.
So at this point I'm asking myself if $L \vert K$ must be a finite extension.
If not, I would argue using the fact that in this case $L$ is a splitting field of a polynomial of degree $n$ with roots $\alpha_1, \dots, \alpha_n \in L$ which implies we can write $M = K(\alpha_1, \dots, \alpha_i)$ with $i \in \{1, \dots, n\}$.
Now, let $f(X) \in  M[X]$ be defined as $f(X):=\prod_{j=i}^n (X- \alpha_j)$
It follows that $L$ is a splitting field of $f(X) \in M[X]$ and therefore $L \vert M$ a normal extension what was to be shown. 
My questions: Has $L \vert K$ to be finite? If yes, what would be the proof in this case? In the first case, is my proof correct?

Let $p$ be an irreducible element of $M[X]$ suppose that $P$ has a root $u\in L$, since $L|K$ is algebraic ? $K(u)|K$ is finite and denote by $f$ the minimal polynomial of $u$ over $K$, since $L|K$ is normal, all the roots of $f$ are in $L$ and remark that $p$ divides $f$ since $p$ is irreducible, we deduce that all the roots of $p$ are in $L$ and $L|M$ is normal.
