If $R$ is a unital ring, it is well-known that its Jacobson radical $J(R)$ contains no non-zero idempotent element of $R$. My question:

Is there a ring $R$ such that $J(R)$ contains a non-zero idempotent ideal of $R$?

If $R$ is Noetherian, then each ideal is finitely generated. So, in case $I\subseteq J(R)$ with $I=I^2$ we would have $I=I^2\subseteq IJ(R)\subseteq I$. Now, by Nakayama's lemma we get $I=0$.

But, I do not have any idea in general case. Thanks for any help!


There are (non-noetherian) local rings $(R,\mathfrak m)$ such that $\mathfrak m^2=\mathfrak m$ and $\mathfrak m\ne0$.

For instance, $R=K[X_1,\dots,X_n,\dots]/(X_1,X_2^2-X_1\dots,X^2_{n+1}-X_n,\dots)$.

If you are looking for $R$ an integral domain, then there are valuations rings with the above property.

  • $\begingroup$ Thanks for the answer. Would you please introduce to me the valuation rings. $\endgroup$ – karparvar Jan 2 '16 at 15:16
  • $\begingroup$ @karparvar What exactly do you want? $\endgroup$ – user26857 Jan 2 '16 at 15:21
  • $\begingroup$ With the assumption that $R$ is an integral domain, what are the valuation rings with the aforementioned property? $\endgroup$ – karparvar Jan 2 '16 at 15:29
  • $\begingroup$ A non-discrete valuation ring of rank one. $\endgroup$ – user26857 Jan 2 '16 at 15:30

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