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It is known that if $R$ is an infinite commutative ring such that for every non-zero ideal $I$ , $R/I$ is finite then $R$ is a Noetheian domain . It is also known that if $R$ is a PID then for every non-zero ideal $I$ of $R$ , $R/I$ has only finitely many ideals . Now I want to ask the following question :

Let $R$ be an infinite commutative ring such that for every non-zero ideal $I$ of $R$ , $R/I$ has only finitely many ideals ; then can we say $R$ is an integral domain ? or that $R$ is a PIR ? ( I am only able to see that $R$ is Noetherian ) Please help . Thanks in advance

( NOTE : all considered rings are with unity )

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If $k$ is any infinite field then the ring $R=k[x]/(x^2)$ is a counterexample, since the ring $R$ has only the ideals 0, $(x)$, and $R$.

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  • $\begingroup$ okay so as the ring itself has finitely many ideals only so will be the case for any quotient ring quotiented by any non-zero ideal . And the ring isn't an I.D . Thanks . But it is a PIR .... right ? $\endgroup$ – user228168 Jul 27 '16 at 6:07
  • $\begingroup$ @SaunDev Yes this is a PIR. It's only a counterexample to the claim that it's an integral domain. $\endgroup$ – Ted Jul 27 '16 at 6:31
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    $\begingroup$ @SaunDev A slight adaptation of Ted's example gives a ring that is not a PIR. Take $R=k[x,y]/(x^2,xy,y^2)$. For every non-zero ideal $I$, $R/I$ is isomorphic to one of $0$, $k$ or $k[t]/(t^2)$, all of which have only finitely many ideals. $\endgroup$ – Jeremy Rickard Jul 27 '16 at 8:56

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