composition of field extensions

Question

Let fields $K\subseteq L\subseteq M$. Then we know that if $L$ is a finite extension of $K$ and $M$ a finite extension of $L$, then $M$ is a finite extension of $K$. Can we generalize this property?

if $L$ is an algebraic extension (not assumed to be simple extension) of $K$ and $M$ an algebraic extension of $L$, then $M$ is an algebraic extension of $K$.

if $L$ is a separable extension of $K$ and $M$ a separable extension of $L$, then $M$ is a separable extension of $K$.

if $L$ is a normal extension of $K$ and $M$ a normal extension of $L$, then $M$ is a normal extension of $K$.

if $L$ is a Galois extension of $K$ and $M$ a Galois extension of $L$, then $M$ is a Galois extension of $K$.

I really have no idea to prove or disprove...

Answer 1

This one holds true

if $L$ is an algebraic extension $K$ and $M$ an algebraic extension of $L$, then $M$ is an algebraic extension of $K$

You argue by reducing it to the finite case.

Take $a \in M$. There will be a polynomial $b = b_{0} + b_{1} x + \dots + b_{n-1} x^{n-1} + x^{n} \in L[x]$ of which $a$ is a root. Now all $b_{i} \in L$ are algebraic over $K$, so $E = K(b_{0}, \dots, b_{n-1})$ is a finite extension of $K$. But $a$ is algebraic over $E$, as $b \in E[x]$, so $E(a)$ is also finite over $E$, which implies that $a \in E(a)$, a finite extension of $K$, so $a$ is algebraic over $K$.

Answer 2

Hint: let $ K=\mathbb Q, L=\mathbb Q (\sqrt 2), M=\mathbb Q (\sqrt {\sqrt {2}}) $. Now both $ L/K $ and $ M/L $ are normal. What can you say about $ M/K $?

From this you can deduce the same result about Galois extensions.