Is there a commutative ring $R$ with zero Krull dimension such that its Jacobson radical is nil but not nilpotent?
Of course, in Noetherian case (which leads to Artinian case) for $R$ each nil ideal is nilpotent, so the ring, if it exists, should not be Noetherian.
Does any body have an answer to the raised question? Thanks!
Let $K$ be a field, let $A = K[X_1, X_2, ...]/(X_1, X_2^2, X_3^3, ...)$. The ring $A$ has only one prime ideal, namely $\mathfrak{p}$ generated by the images of $X_1, X_2, ...$ under the quotient map. Note that the generators of the ideal $\mathfrak{p}$ are nilpotent, hence $\mathfrak{p}$ is contained in and equal to the nilradical.
But $\mathfrak{p}$ is not nilpotent, for $\mathfrak{p}^n $, $ n \in \mathbf{N}$, contains $X_{n + 1}^n$ which is non-zero.
