Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

It becomes easy if you use the preceding exercise in Neukirch's book:

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.

share|improve this answer
    
Right, yeah, I thought of that one. I guess tghere might not be nice arguments avoiding arguments of that sort. –  Dedalus Sep 2 '12 at 12:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.