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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?

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    $\begingroup$ Over $L$, $\sqrt 2$ will also bring its conjugate $-\sqrt 2$ into the field extension, as over $K$. Rather try $\sqrt[3]3$. $\endgroup$ – Berci Feb 16 '13 at 21:23
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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.

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  • $\begingroup$ This was the most helpful answer for me. Thanks! $\endgroup$ – Jonas Feb 16 '13 at 22:26
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It's a bit of an overkill, but you can take a Galois extension of $\mathbf{Q}$ which has $S_5$ as the Galois group. Then choose $K = \mathbf{Q}$, $L$ to correspond to $S_4$ in the Galois correspondence, and $M$ to correspond to $S_3$. None of the subgroups is normal in any other, so no field is normal over any other.

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Your $M$ couldn't possibly work because $[M:L\textbf{]}=2$ and any extension of degree $2$ is normal.

Let $M=\Bbb Q(\root \large 3\of 2, \root \large 5 \of 3), L=\Bbb Q(\root \large 3\of 2)$ and $K=\Bbb Q$.

You already know that $L:K$ isn't normal.

To see that $M:L$ isn't normal consider the polynomial $p(x):=x^5-3\in K[x\textbf{]}\subseteq L[x\textbf{]}$.

If $p(x)$ is irreducible over $L$ we're done because $M$ is a real field and $p(x)$ has complex roots.

Let $L'=\Bbb Q(\root \large 5\of 3)$.

We have $[L:K\textbf{]}=3$ and $[L':K\textbf{]}=5$. Therefore $[M:K\textbf{]}\ge 15$.

Since $\root \large 5\of 3$ is a root of $p(x)$ it follows that $[M:L\textbf{]}\leq 5$.

So we have $15\leq[M:K\textbf{]}=[M:L\textbf{]}[L:K\textbf{]}=[M:L\textbf{]}\cdot 3\leq 5\cdot 3=15$. This implies that $[M:L\textbf{]}=5$ and therefore $p(x)$ is irreducible over $L$.

It follows that $M:L$ isn't normal.

The important thing here is that $3$ and $5$ are coprime and they're both greater than $2$.

The same argument could have be done with $\Bbb Q(\root \large 3\of 2, \root \large 5 \of 2)$, for instance.

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