abstract algebra - Field extensions that are not normal

I am trying to come up with field extensions $M : L : K$ such that none of the three extensions $M:L, L:K, M:K$ are normal.

So far, I have tried letting $K = \mathbb{Q}, L = \mathbb{Q}(\sqrt[3]{2})$. I know that $L$ is not normal over $K$ since $x^3 - 2$ is an irreducible polynomial over $K$ with a root in $L$ but does not split in $L$, due to having complex roots.

Now I am not sure what a suitable choice of $M$ would be. I am using $M = \mathbb{Q}(\sqrt[3]{2},\sqrt{2})$, which is not normal over $K$ again by using $x^3 - 2$ as the non-splitting irreducible polynomial over $K$. To show that $M$ is not normal over $L$, I am trying to use the polynomial $x^6 - 32$: it has a root $\sqrt[6]{32} = \sqrt[3]{2} \cdot \sqrt{2}$ in $M$, and does not split in $M$ since it has complex roots, but how can we show this polynomial is irreducible over $L$, if indeed it is irreducible?

Answer

For your choice of $M$ the extension $M/L$ will be normal, because any field extension of degree $2$ is normal see here. Rather than extend $L$ by $\sqrt{2}$ let's try $\sqrt[9]{2}$. The polynomial $x^9-2$ is irreducible over $\mathbb Q$ by Eisenstein's criterion so $\sqrt[9]{2}$ has degree $9$ and can't be contained in $L=\mathbb Q(\sqrt[3]{2})$ which only has degree $3$. So $[\mathbb Q(\sqrt[9]{2}): \mathbb Q(\sqrt[3]{2})]=3$ and in particular the minimum polynomial of $\sqrt[9]{2}$ over $\mathbb Q(\sqrt[3]{2})$ is forced to be $x^3-\sqrt[3]{2}$. Then it's easily verified that each of the extensions in the chain $\mathbb Q \subset \mathbb Q(\sqrt[3]{2}) \subset \mathbb Q(\sqrt[9]{2})$ is not normal by arguing that $\mathbb Q(\sqrt[9]{2})$ is a purely real field.