I'm looking for an example of a commutative ring $R$ with zero Krull dimension such that $R$ satisfies one of the following conditions:
(1) $R$ has no nonzero proper idempotent ideals, but has a nonzero non-nilpotent ideal.
(2) $R$ has no nonzero nilpotent ideals, but has a nonzero idempotent ideal.
As for (2), I only know that $R$ would be von Neumann regular by a classical theorem of commutative algebra.
Thanks for any help!
An example for part (1) may be the ring $R=F[X_2,X_3,\dots]/(X_2^2,X_3^3,\dots)$, where $F$ is a field. At the outset, note that any prime ideal of $R$ is of the form $P/J$, where $J=(X_2^2,X_3^3,\dots)$ and $P$ is a prime ideal of $F[X_2,X_3,\dots]$ containing $J$, so that $P$ contains all the $X_i$'s, whence it is equal to the maximal ideal $M=(X_2,X_3,\dots)$ of $F[X_2,X_3,\dots]$. Thus, $R$ is a local ring of zero Krull dimension with the maximal ideal $\overline M=(\overline X_2,\overline X_3,\dots)$, where the bars refer to modulo $J$. The ideal $\overline M$, having elements of arbitrary high nilpotency indices, is not nilpotent. On the other hand, since any non-zero polynomial of degree $k$ in $\overline M$ could not belong to $\overline M^{k+1}$, we deduce that $\cap_{i\ge 1}\overline M^i=(0)$. Now, if $I$ is a proper idempotent ideal of $R$, we get the conclusion that $I=\cap_{i\ge1}I^i\subseteq\cap_{i\ge 1}\overline M^i=(0)$.