# Examples for commutative rings with many nilpotent elements yet the nilradical is not

What is an example (if exists) for a commutative ring that contains many nilpotent elements (preferably even with the same exponent, i.e., there are $$a_1,...,a_r\in \mathbb{R}$$ with $$a_i^k=0$$), such that the ideal of nilpotent elements is not nilpotent by itself (i.e., $$a_1...a_{k-1}a_k$$ should not be $$0$$, for example)?

Thanks!

• Something like $\mathbb{Z}[X_{1}, X_{2}, X_{3}, \ldots]/\langle X_{1}^{2}, X_{2}^{3}, X_{3}^{4}, \ldots \rangle$ should work, I think. Many nilpotents, but the nilradical should be the ideal generated by the variables, and no power of this is zero (choose a variable of sufficiently large index). Dec 25, 2019 at 22:13
• to get this from the Database of ring theory: search for Jacobson rings that are not reduced and don’t have a nilpotent Jacobson radical. You could throw in commutative too, but in this case the results were already commutative. Dec 26, 2019 at 0:14

You need an infinite number of generators for the ideal in the commutative setting. Perhaps the simplest example would be $$R=k[x_1,\dots,x_n,\dots]/(x_1^2,\dots,x_n^2,\dots)$$, where $$k$$ is field of characteristic 2 (see note).
We observe that $$I=(x_1,\dots,x_n,\dots)$$ is not nilpotent, since $$0\ne x_1\cdots x_n\in I^n$$ for each $$n\ge 1$$. On the other hand, the square of any element of $$I$$ is $$0$$, as we shall show. Let $$f\in I$$. Expressing $$f$$ as a linear combination of the generators of $$I$$, we have $$f = g_1x_1+\dots+g_nx_n$$ for some $$n>0$$ and some $$g_1,\dots,g_n\in R$$. Then, $$f^2 = \sum_{i=1}^n g_i^2x_i^2 + 2\sum_{1\le i as desired.
Thus, all elements of $$I$$ (and in fact all nilpotent elements of $$R$$) have the property that their square is $$0$$. So, an exponent of $$2$$ is attainable.
Note: One cannot decrease the exponent in $$(x_1+\dots+x_n)^{n+1}=0$$ for an arbitrary coefficient field $$k$$.