1. ## Field extension question

I'm trying to answer this question:

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

Does anyone have any ideas?

2. Hi. Can you verify if I'm right?

$\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathb b{Q}(\sqrt[4]{2})$ (normal/normal)
$\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{2})\subset\mathbb{Q}(\sqrt[6]{2})$ (not normal/normal)
$\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathb b{Q}(\sqrt[6]{2})$ (normal/not normal)
$\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{2})\subset\mathbb{Q}(\sqrt[9]{2})$ (not normal/not normal)

Now assume $M$ is a normal extention over $K,$ and $K\subseteq L\subseteq M$
$M$ is normal over $K$ iff given an algebraic closure $C$ of $M,$ any $K$-isomorphism from $M$ to another subfield of $C$ is a $K$-automorphism of $M.$ Since a $L$-isomorphism from $M$ to another subfield of $C$ is also a $K$-isomorphism, then it is a $L$-automorphism of $M$ and $M$ is a normal extension over $L.$

Finally, consider:

$\mathbb{Q}\subset\mathbb{Q}(j)\subset\mathbb{Q}(\s qrt[3]{2},j)$
$\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{2})\subset\mathbb{Q}(\sqrt[3]{2},j)$

$\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{2},j)$ is a normal extension, but what about $\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}\subset\mathbb{Q}(j)$?

3. Originally Posted by KSM08
I'm trying to answer this question:

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

Does anyone have any ideas?

$K=\mathbb{Q}\leq L=\mathbb{Q}(\sqrt{2})\leq M=\mathbb{Q}(2^{1\slash 4})$ . $M\slash K$ isn't normal, but $M\slash L\,,\,L\slash K$ are.

$K=\mathbb{Q}\leq L=\mathbb{Q}(2^{1\slash 4})\leq M=\mathbb{Q}(2^{1\slash 8})$ . $M\slash K\,,\,L/K$ aren't normal, but $M\slash L$ is.

Now you try other cases.

Tonio