Question: $K \subset E \subset L$ finite field extensions and $L$ normal over $K$.
(a) Is $L$ normal over $E$?
(b) Is $E$ normal over $K$?
So far, I believe I have done (b) (see below) but am stuck on (a).
My answer for (a): Take $l \in L$ and let $P_K(X) \in K[X]$ and $P_E(X) \in E[X]$ be its minimal polynomials over $K$ and $E$ respectively.
$L/K$ is normal $\implies$ $P_K(X)$ splits in $L[X]$. $P_K(l)=0$ and $P_E(X)$ generates the ideal of all polynomials in $E[X]$ such that $P_E(l)=0$, therefore $P_E(X)$ divides $P_K(X)$.
Therefore, $P_E(X)$ splits completely in $L[X]$ so $L/E$ is a normal extension.
My idea for (b): If this were false, and $E/K$ is not a normal extension, then I could prove this by finding a counter-example. e.g: $L=\mathbb{Q}(A, B)$, $E=\mathbb{Q}(A)$ and $K=\mathbb{Q}$.
I have not made any progress with this method and would appreciate some help.
Thank you in advance
You seem to have answered part (a) in what you state is your answer to part (b), anyway, for part (b), ie. is E normal over K:
let K = $\mathbb{Q}$, L = The splitting field of $x^3 - 2$ and E = $\mathbb{Q}(\sqrt[3]{2})$ (ie. $\mathbb{Q}/(x^2-3)$ or equivalently $\mathbb{Q}$ with one of the roots of $x^3 - 2$ adjoined)
Now K $\subset$ E $\subset$ L and L is normal over K (as it is a splitting field)
The 3 roots of $x^3 - 2$ are $\sqrt[3]{2}$, $\sqrt[3]{2}\omega$ and $\sqrt[3]{2}\omega^2$ where $\omega$ is the complex cube root of unity ($\omega^3 = 1$)
but $\sqrt[3]{2}$ is the only one which is in $\mathbb{Q}(\sqrt[3]{2})$ (can you see why?)
Thus $\mathbb{Q}(\sqrt[3]{2})$ is not normal over $\mathbb{Q}$
and so E is not normal over K in general