# A prime in a separable extension splits completly iff it does so in the galois closure.

I just did the following exercise out of Neukirch's Algebraic Number Theory:

"A prime ideal p of K is totally split in the separable extension L|K iff it is totally split in the Galois closure N|K of L|K"

Now, I managed to solve this using a complicated argument using decomposition groups and it was not at all too pretty. Considering Neukirch introduces the decomposition group in the pages following this exercise, I suspect that there is a nicer way to prove this. So, my question is:

Are there any slick ways to show this? If so, how?

-

If a prime ideal $\mathfrak p$ of $K$ is totally split in two separable extensions $L|K$ and $L'|K$, then it is also totally split in the composite extension.
The galois closure $N$ of $L|K$ is the composite field of the subfields $\sigma L$ where $\sigma$ ranges over $G(N|L)$. Since $\mathfrak p$ is totally split in $L$ it is also totally split in $\sigma L$ (if $\mathfrak p\mathcal O_L = \mathfrak P_1\cdots\mathfrak P_r$ in $L$ then $\mathfrak p \mathcal O_{\sigma L} = (\sigma \mathfrak P_1)\cdots(\sigma \mathfrak P_r)$ in $\sigma L$). So $\mathfrak p$ is also totally split in $\prod_{\sigma} \sigma L = N$ by the quoted exercise.