Let $R$ be a ring. For each $n>0$ I want to find an ideal $I$ of $R$ such that $I^n=0$ but $I^{n-1}\not= 0$. Clearly this won't work for $R=\Bbb{Z}$ or $\Bbb{Z}/n\Bbb{Z}$. And I ran out of examples...
2026-04-03 03:55:48.1775188548
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Example of ideals such that $I^n=0$ but $I^{n-1}\not= 0$
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Another example of how you can engineer a single ring for a given $n$: let $T_n(F)$ be the ring of $n\times n$ upper triangular matrices over a field, and let $I$ be the ideal of strictly upper triangular matrices. This ideal has the property you seek.
Then you can look at $\prod_{n=1}^\infty T_n(F)$ to find a ring with an ideal for each $n$ with this property.
One can do the same with Chris Dugale's example, or even $\Bbb Z/(p^n)$ where $p$ is a fixed prime of your choice.
Look at $k[x]/\langle x^n \rangle$, and consider the principal ideal generated by $x+\langle x^n \rangle$.