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A commutative ring is called zero-dimensional if all its prime ideals are maximal, and a ring is said to have a bounded index of nilpotency if there is a positive integer $n$ such that $x^n = 0$ for every nilpotent $x$ in the ring. Does every zero-dimensional commutative ring have a bounded index of nilpotency?

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up vote 2 down vote accepted

What ahout $k[x_2,x_3,\dots]/(x_2^2,x_3^3,x_4^4,\dots)$?

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A very nice example again, Mariano! – Georges Elencwajg Oct 27 '12 at 7:41

Here is a kind of explanation and generalization of Mariano's example.

It is known that a ring $R$ is zero-dimensional if and only if for all $x \in R$ there is some $n \geq 1$ such that $x^{n+1} | x^n$. It follows that zero-dimensional rings are closed unter directed colimits. Alternatively, this follows from the fact that $\mathrm{Spec} : \mathsf{Ring} \to \mathsf{Top}^{\mathrm{op}}$ preserves directed colimits.

In particular it is quite easy to construct arbitrary "large" zero-dimensional rings! But of course directed colimits can destroy bounded index of nilpotency.

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