Is it possible for an Ideal I to be nil and yet $I^n \neq 0$ $\forall n \in \mathbb{N}$?

Question

Is it possible for an Ideal I to be nil and yet $I^n \neq 0$ $\forall n \in \mathbb{N}$?

I think it should be. For an ideal to be nil it means if $x \in I$, then there is a $n \in \mathbb{N}$ such that $x^n = 0$.

I am just a little curious.. Thanks in advance!

Answer 1

Set things up so you have elements $a_n$ in your ideal with $a_n^{n-1}\ne0$ and $a_n^n=0$. So let's take $$R=\Bbb Z[x_1,x_2,x_3\ldots]/\left\langle x_1,x_2^2,x_3^3,\ldots\right\rangle.$$ Write $a_i$ for the image of $x_i$ in $R$. These $a_i$ generate an ideal $I$ all of whose elements are nilpotent, but $I^n$ has the nonzero element $a_{n+1}^n$.

Answer 2

Take $R:=\mathbb{R}[X^\frac{1}{n} \ : \ n\in \mathbb{N}]$ and define $S:= R/(X)$. Then the ideal $I=(X^\frac{1}{n} \ : \ n\in \mathbb{N})$ is nil. However, for every $n\in \mathbb{N}$ holds $(X^\frac{1}{n+1})^n \neq 0$ (in $S$) and thus $I^n\neq 0$.