Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a number ring and a Dedekind domain. We have the following result:

For every ideal $I\subset R$ $$ I = \bigcap_P I_P $$ where $I_P$ denotes the localization of $I$ at $P$ and the intersection is taken over all the prime ideals $P$ of $R$.

My question is: Can we deduce from this that if $I_P$ is principal in $R_P$ for every prime ideal $P$ of $R$, then $I$ itself is principal? Or stated differently, is being principal a local property?

share|improve this question
add comment

1 Answer

Only in principal ideal domains (PIDs). If by number ring you mean Dedekind domain, then all its localizations at prime ideals are discrete valuation rings (except the one at 0 which is a field), which are principal ideal domains. So every ideal in a Dedekind Domain is locally principal. But of course there are Dedekind domains that are not PIDs!

share|improve this answer
    
Yes I forgot to write Dedekind domain. I need to prove that a Dedekind domain which is also a UFD is a PID... it is driving me crazy! By the way of course is $R$ is a PID then $I$ is principal! ;-) –  Abramo Oct 13 '12 at 16:51
    
To help you with your problem: There are two steps one needs to consider (whose proof I leave to you). 1.) In a UFD any prime ideal of height one is principal. 2.) An ideal maximal among non-principal ideals is automatically a prime ideal. –  Lennart Galinat Oct 13 '12 at 18:19
add comment

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.