Is there a ring $R$ with zero Krull-dimension such that $R$ has neither nonzero idempotent nor nilpotent ideals?

I know that this ring could not be Noetherian, because if this is so, then $R$ would be Artinian and hence, the Jacobson radical would be nilpotent. (Of course, this is a contradiction if the Jacobson radical is non-zero.)

  • $\begingroup$ Presumably you also want $R$ to not be a field? $\endgroup$ – Eric Wofsey Oct 11 '16 at 7:46
  • $\begingroup$ $R$ is always nonzero idempotent, so I guess you mean to exclude that? $\endgroup$ – rschwieb Oct 11 '16 at 19:57

If $R$ is zero-dimensional then $\operatorname{Spec} R$ is totally disconnected. Since idempotent elements of $R$ correspond to clopen subsets of $\operatorname{Spec} R$, this means that if $R$ has no nontrivial idempotent ideals, $\operatorname{Spec} R$ is a point. This means that $R$ has a unique prime ideal, which is equal to its nilradical. If $R$ has no nontrivial nilpotent ideals, this means the nilradical is $0$. Thus $0$ is the unique prime ideal of $R$ and in particular is maximal, which means $R$ is a field.

  • $\begingroup$ How do we deduce from "$R$ has no nilpotent ideals" that $R$ has zero nilradical? $\endgroup$ – karparvar Oct 13 '16 at 4:14
  • $\begingroup$ If $r\in R$ is a nonzero nilpotent, then $(r)$ is a nonzero nilpotent ideal. $\endgroup$ – Eric Wofsey Oct 13 '16 at 4:16

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