Let $m$ be a positive integer, let $R$ be a (non-commutative) ring and let $I$ be a two-sided ideal of $R$ such that $x^m=0$ for all $x \in I$. Is there an integer $M$ such that $I^M=0$?
Thoughts:
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If $R$ is a finite dimensional algebra over a field, yes. See here.
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If we allow $m$ to depend on $x$, there are lots of counter-examples: Search this site for "nil" and "nilpotent". But I didn't find any counterexamples where the order of nilpotency was bounded.
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Let $G$ be a Tarski monster group for a prime $p$, let $R$ be the group algebra $\mathbb{F}_p[G]$ and let $I$ be the ideal generated by $1-g$ for $g \in G$. Then $(1-g)^p = 1-g^p=0$ for all $g \in G$. I thought this might be a good route to a counter-example, but $(g-h)^p$ is not $0$ for $g$ and $h$ non-commuting elements, so this needs more work.
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Let $S$ be the free noncommutative algebra over $\mathbb{F}_p$ on two generators $x_1$, $x_2$ and let $\mathfrak{m} = \langle x_1, x_2 \rangle \subset S$. Let $J$ be the ideal of $S$ generated by $f^p$ for $f \in \mathfrak{m}$. Take $R = S/J$ and $I = \mathfrak{m}/J$. By definition, $x^p=0$ for all $x \in I$. It looks like $R$ is infinite dimensional and $I$ is not nilpotent, but I haven't proved this.
Best Answer
For some prime $p$, let $R = \mathbb{F}_p[x_1, x_2, \ldots] / \langle x_1^p, x_2^p, \ldots \rangle$ and $I = \langle x_1, x_2, \ldots \rangle$. Then for every $f \in I$, $f^p = 0$; yet for each $n$, $x_1 x_2 \cdots x_n \ne 0$ so $I^n \ne 0$.