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I understand the somewhat default example with $k[X_1…]/(x_1,x_2^2…)$.

My question is, would $k[X_1…]/(x_1^2,x_2^2,x_3^2,…)$ work as well? Since for any $n$ I could just take the product of $x_1\cdot…\cdot x_n\neq 0$? Or is my understanding of ideal powers wrong?

average math enjoyer
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    That ring would work but the example you give doesn’t work because it always squares to $0$. For any given $n$, study the $n$th and $(n+1)$th powers of the element $x_1 + \ldots + x_n$. – Arkady Sep 23 '23 at 10:53
  • @Arkady Hmm, I do not understand, why is "squares to 0" a problem? – average math enjoyer Sep 23 '23 at 11:47

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Arkady is incorrect and you are correct. $\text{nil}(A)^n$ is spanned by all products of $n$ elements of $\text{nil}(A)$ so it does in fact contain $x_1 x_2 \dots x_n$ which is not zero, and this is already enough to show $\text{nil}(A)^n \neq 0$. (I think Arkady interpreted you to be considering $x_1 x_2 \dots x_n$ as an element of $\text{nil}(A)$ rather than as an element of $\text{nil}(A)^n$.)

Qiaochu Yuan
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