$L \vert K$ normal extension with $K \subset M \subset L$. Show that $L \vert M$ is normal

Question

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?

Answer 1

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.

Answer 2

$L/K$ need not be finite for this to he true. Your proof is unclear though. For one, you should justify that you can write $M$ in this way. It may be true, but it doesn't seem obvious to me. Also, I don't think the $f$ you defined has any reason to be in $M[x]$. I'll provide a solution to the result below under a spoiler tag, since I think you're not far from a solution.

Let $L/M/K$ with $L/K$ normal. Then by definition, there is some subset $S \subseteq K[x]$ such that $L$ is the splitting field of $S$. Then as $K \subseteq M$, $K[x] \subseteq M[x]$ so $S \subseteq M[x]$. I claim then that $L$ is the splitting field of $S$ over $M$, which will show that $L/M$ is normal. There isn't anything happening in the proof of this, but I'll write the details. Indeed, as $L$ is the splitting field of $S$ over $K$, everything in $S$ splits in $L$. Furthermore, if there was some $L/E/M$ with $L/E$ proper such that $S$ splits in $E$, then we'd have that for $L/E/K$ contradicting the fact that $L/K$ is the splitting field of $S$. Thus, $L/M$ is the splitting field of $S$ so $L/M$ is normal.