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Let $R$ be a commutative ring such that for every prime ideal $P$ of $R$, the ring $R/P$ is a PID. Do you know how these rings are called or another characterization of them?

I know there are a lot of examples of this kind of ring, for example every commutative Artinian ring. But I am trying to characterize these rings.

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    $\begingroup$ Do you actually mean the quotient or rather the localisation? Also, you may want $P$ to be non-zero otherwise at least for a domains $R$ ought to be a PID. $\endgroup$
    – user9072
    Commented Feb 15, 2016 at 13:38
  • $\begingroup$ I am going to characterize commutative rings R s.t. for every prime ideal P of R, R?P is a PID. I know there are a lot of examples of there kind of rings. For example every commutative Artinian ring. But I am trying to characterize these rings. Thank you $\endgroup$
    – Najmeh Dehghani
    Commented Feb 17, 2016 at 13:48

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Certainly, if R is a Dedekind domain, then all ideals are one-and-a-half generated. So, if P is a (non-zero!) prime ideal, the quotient ring R/P must be a PID.

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  • $\begingroup$ Thanks. I know there are a lot of rings with this condition. As you said Dedekind domain, for each nonzero prime ideal P of R, R/ P is a PID. Or as another example if R is a commutative Artinian ring, then for every prime ideal P we have R/P is a field. But I need to obtain a {\bf characterization} for commutative rings s.t. for every prime P, R?P is a PID . $\endgroup$
    – Najmeh Dehghani
    Commented Feb 17, 2016 at 13:46
  • $\begingroup$ What does one-and-a-half generated mean? $\endgroup$
    – Steven Landsburg
    Commented Feb 17, 2016 at 15:09
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    $\begingroup$ @StevenLandsburg it means that you can complete every non-zero $i \in I$ to some two element generating set of $I$. $\endgroup$
    – user9072
    Commented Feb 17, 2016 at 15:27
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    $\begingroup$ @quid: Thanks! $\phantom{xxxxxxxxxxxxxx}$ $\endgroup$
    – Steven Landsburg
    Commented Feb 17, 2016 at 17:10
  • $\begingroup$ Note that the parenthetical "(non-zero!)" is actually important, as it rules out most Dedekind domains having the property Najmeh wants. $\endgroup$
    – Pace Nielsen
    Commented Feb 17, 2016 at 21:23

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