# $R$ be an infinite commutative ring such that $R/I$ has only finitely many ideals for every non-zero ideal $I$ , what can we say about $R$?

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 )

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 $R$.
• @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. – Jeremy Rickard Jul 27 '16 at 8:56