Normality and Field Extensions

Let M : L : K be finite field extensions. When M is not normal over K, give four examples to show that this gives no information about the normality of M over L or of L over K. What are the possibilities if M is normal over K?

-
Can you give just any example of a normal and a not normal extension? – Hagen von Eitzen Nov 6 '12 at 16:39
@Hagen so $A=\mathbb{Q}(2^{1/2})$ is a splitting field for $a(x)=x^2 - 2$ because all roots of $a$ are in $A$ and since $a(x)\in\mathbb{Q}[x]$ I know that $A$ is a normal extension of $\mathbb{Q}$. Alternatively, let $B=\mathbb{Q}(2^{1/3})$ and $b(x)=x^3 - 2$ then not all roots of $b$ are in $B$ so $B$ is not a normal extension of $\mathbb{Q}$ – Dexter Nov 6 '12 at 16:49
Now for the first part of the problem you might take $M=\mathbb Q(2^{1/3})$, $K=\mathbb Q$ and cheat a little by letting $L=M$ or $L=K$, respectively. For the second part, waht can you sa about $M:L$ if $M:K$ is normal? – Hagen von Eitzen Nov 6 '12 at 17:54
Given that extensions of finite fields are always normal, I judged that for the OP finite only refers to the extension as opposed to field, so I removed the finite-fields tag. – Jyrki Lahtonen Nov 7 '12 at 8:06

The question being 4 months old, I guess it's safe to assume the homework assignment was due a while ago, so we can give answers.

In all cases, take $K={\bf Q}$.

Let $M={\bf Q}(2^{1/16})$, so $M$ is not normal over $K$.

Take $L_1={\bf Q}(2^{1/8})$, $L_2={\bf Q}(2^{1/4})$, $L_3={\bf Q}(2^{1/2})$.

Then: $M/L_1$ is normal, $L_1/K$ is not. $M/L_2$ is not normal, neither is $L_2/K$. $M/L_3$ is not normal, but $L_3/K$ is. $L_2/K$ is not normal, but $L_2/L_3$ and $L_3/K$ both are.

Still with $K={\bf Q}$:

If $M/K$ is normal, then $M/L$ is normal, but $L/K$ could go either way. E.g., if $M={\bf Q}(\root4\of2,i)$, then $L_1={\bf Q}(\root4\of2)$ is not normal over $K$, but $L_2={\bf Q}(i)$ is.

-

Here's a couple examples. Let $\mathbb{F}$ have characteristic $p$ and consider

1. $\mathbb{F}_p(x):\mathbb{F}_p(x^p):\mathbb{F}$,

2. $\mathbb{F}_p(x,\alpha):\mathbb{F}_p(x):\mathbb{F}(x^p)$

where $\alpha$ is algebraic.

-