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)?


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    $\begingroup$ 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). $\endgroup$ Dec 25, 2019 at 22:13
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    $\begingroup$ 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. $\endgroup$
    – rschwieb
    Dec 26, 2019 at 0:14

1 Answer 1


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, $$f^2 = \sum_{i=1}^n g_i^2x_i^2 + 2\sum_{1\le i<j\le n}g_ig_jx_ix_j = 0$$ 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 $k$.


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