1

It is well-known that the localization (at any multiplicative set) of any principal ideal ring is again a principal ideal ring (PIR). A Dedekind domain localized at any prime ideal is a DVR, so is again a PIR. These domains show that a commutative Noetherian ring need not be a PIR to be locally a PIR. Do we have any characterizing conditions for a general commutative Noetherian ring to be locally a PIR? It seems to me that such conditions should be quite restrictive.

Chris Leary
  • 2,993
  • You essentially gave all examples already: these rings are exactly the finite products of Dedekind domains and Artinian PIRs – math54321 Apr 19 '23 at 07:10
  • @math54321- Thank you so much. I was unaware of such a characterization. Indeed, it is pretty restrictive. Best regards. – Chris Leary Apr 19 '23 at 15:49
  • @math54321-Did you mean Noetherian PIRs? A Noetherian local ring with principal maximal ideal is a PIR. – Chris Leary Apr 19 '23 at 16:01
  • I did mean Artinian. There is a classic result (maybe due to Hungerford) that a PIR is a finite product of quotients of PIDs. In particular a local non-Artinian PIR is necessarily a domain, i.e. a DVR – math54321 Apr 19 '23 at 16:09
  • @math54321-Right. I should probably go back and reread Hungerford's paper a bit more carefully. In any event, if you care to write this up as an answer I would be pleased to accept it. Thanks again. – Chris Leary Apr 19 '23 at 17:05
  • I've written an answer with an elementary argument, without using Hungerford's result. By the way, the Noetherian hypothesis is crucial: there exist domains that are locally DVRs which are not Dedekind. – math54321 Apr 19 '23 at 18:08
  • @math54321 - That' good to know. Thanks. – Chris Leary Apr 19 '23 at 21:53

1 Answers1

1

Let $R$ be a Noetherian ring which is locally a PIR. Then $\dim R \le 1$, and we can write $R = \prod_{i=1}^n R_i$ as a finite product where each $\operatorname{Spec}(R_i)$ is connected, i.e. $R_i$ has no nontrivial idempotents. There are $2$ cases to consider:

(1) $\dim R_i = 0$: then $R_i$ is Artinian, hence a product of Artinian local rings, hence is Artinian local (since $R_i$ has no idempotents), so $R_i$ is an Artinian PIR.

(2) $\dim R_i = 1$: in this case $R_i$ must be a domain. It suffices to show that $R_i$ is locally a domain, since a Noetherian ring that is locally a domain is a finite product of domains, and as before since $R_i$ has no idempotents, the product can only have one factor. Now localizations of $R_i$ are localizations of $R$, so this reduces to the statement that a $1$-dimensional local PIR is a domain. This can be seen by Nakayama's lemma, or a direct argument: if $S$ is a local PIR with maximal ideal $(x)$, and $a, b \in (x)$ with $ab = 0$, then writing $a = ux^n$, $b = vx^m$ for some units $u, v \in S \setminus (x)$ (which is possible since $\bigcap_{k \ge 0} (x)^k = 0$) gives $x^{n+m} = 0$, i.e. $\dim S = 0$.

So each $1$-dimensional factor $R_i$ is a Noetherian domain of dimension $1$ which is locally a PIR, hence locally a DVR, i.e. $R_i$ is a Dedekind domain. Thus $R$ is a finite product of Artinian PIRs and Dedekind domains.

math54321
  • 902
  • 4
  • 9