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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?

edited Oct 13 at 16:47

asked Oct 13 at 16:42

Abramo
23●4

René Pannekoek · Answer 1 · 2012-10-13 17:00:36Z UTC

7

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!

edited Oct 13 at 17:00

answered Oct 13 at 16:45

René Pannekoek
1,092●3●12

	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 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 at 18:19

Is being principal a local property?

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