Combining two normal field extensions

Question

Suppose $M/N$ and $N/K$ are both normal field extensions, does that necessarily imply that $M/K$ is also a normal extension?

I have the counterexample that $\mathbb {Q}[{^4}\sqrt{-5}] / \mathbb {Q}[\sqrt{-5}]$ and $\mathbb {Q}[\sqrt{-5}]/\mathbb {Q}$ are both normal extensions (can easily check as both extensions have degree 2).

But $\mathbb {Q}[{^4}\sqrt{-5}]/\mathbb {Q}$ isn't a normal extension, demonstrated by the $\mathbb {Q}$-conjugates of ${^4}\sqrt{-5}$.

Is this a correct counterexample, or have I made a mistake somewhere.

Answer 1

I think that the argument doesn't work with ${^4}\sqrt{-5}$ because all the Q-conjugate of ${^4}\sqrt{-5}$ are in $L$. (To see this, you can write ${^4}\sqrt{-5} = \xi_4 {^4}\sqrt{5}$ where $ \xi_4$ is the primitive 4th root of unity). By definition of the primitive root, you will be able to generate the other Q-conjugate.

Your argument should work with $5$ instead of $-5$.

As a general remark : you should avoid using notations such as $\sqrt{-5}$ and rater use $i\sqrt{5}$ (same for the $4^{th}$ power as I wrote it above).

Answer 2

Your counter-example is correct, see e.g. https://math.stackexchange.com/a/3437967/300700. In general, if $M/N$ and $N/K$ are both normal, let us consider that all these extensions are contained in an algebraic closure $\bar K$. Then it is easy to see that $M/K$ is normal iff $M$ is stable under the action of all the prolongations to $M$ of the $K$-embeddings of $N$ into $\bar K$.