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've been looking at Janusz book Algebraic Number Fields. I can't seem to figure out the following correspondence:

Let $\mathfrak{p}$ be a finite prime of a field K and L/K a finite Galois extension. Then we get a valuation ring R from $\mathfrak{p}$ by picking one valuation. Janusz then writes (p. 123) "Identify $\mathfrak{p}$ as the maximal ideal of R". The problem is that given the maximal ideal in the valuation ring R, I can't see how we get back a valuation on K unless we assume that the ring R is a DVR?

I must be missing something here? Finally if we then let R' be the integral closure of R in L and $\mathfrak{P}$ some prime containing the maximal ideal of R should this somehow define a prime on L?

share|improve this question
1  
I believe in your last line you mean "valuation on $L$". –  Zev Chonoles Feb 5 '11 at 3:21
1  
Dear dstt, not every prime (e.g. the zero prime) corresponds to a valuation. However, every nonzero prime in a Dedekind domain does produce a valuation (since the localization at that prime is a DVR, and the ring and the localization have the same quotient field). –  Akhil Mathew Feb 5 '11 at 3:50
    
Is the process like this? Since these are algebraic number fields, the conditions in the comment by @Akhil Mathew are satisfied, and you get a valuation from a prime ideal, and then get the valuation ring of the valuation which coincides with the localization at the prime ideal, and the prime ideal is the same as the maximal ideal as the localization or simply the definition shows, so we see that the maximal ideals correspond to valuations in K this way. In any case, thanks. –  awllower Feb 27 '11 at 14:53
    
Besides, the ring R is always a discrete valuation ring as explained by @Akhil Mathew. –  awllower Feb 27 '11 at 14:54

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.