I want to find an ideal which is nil but not nilpotent.
For instance, in the (commutative) ring $$R= \mathbb{Z}[x_1,x_2,x_3,…]/(x_1^2,x_2^3,x_3^4,…),$$
the ideal $I$ generated by $\bar{x_1}, \bar{x_2}, \bar{x_3},…$ is nil, but is not nilpotent.
How can I write the proof explicitly.
Best Answer
$I$ is nil because every element is in an ideal generated by finitely many nilpotent elements. (Every polynomial in $\mathbb{Z}[x_1,x_2,x_3,...]$ uses only finitely many variables.)
$I$ is not nilpotent because the nilpotency index of $\bar x_n$ is $n+1$ and so there can't be a fixed $N$ such that $I^N=0$.