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A ring $R$ is said to have a bounded index (of nilpotency) if there is a positive integer $m$ such that $x^m = 0$ for every nilpotent $x\in R$. I wonder whether this property transfers to the ring of matrices, that is,

If $R$ has bounded index, then $M_n(R)$ has bounded index?

If $R$ is a field, then the answer is positive and extends obviously to integral domains. Noetherian rings have also the bounded index property, and it would be nice to have an answer in this case. (When $R$ is noetherian and reduced the answer is also positive.)

Remark. This question was suggested by Bound on nilpotency index of endomorphisms.

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up vote 2 down vote accepted

Edit Suppose the nilradical of $R$ is nilpotent, then the question has a positive answer.

First we reduce to the case $R$ is reduced. Let $I$ be the nilradical of $R$. Let $r\ge 1$ be such that $I^r=0$. Suppose we proved that $M_n(R/I)$ has index bounded by some $m$. Then any nilpotent element $T\in M_n(R)$ satisfies $T^m(R^n)\subseteq I R^n$. So $T^{mr}(R^n)\subseteq I^r R^n=0$ and $T^{mr}=0$.

So suppose $R$ is reduced. Consider the canonical injection $$ R\to \prod_{\mathfrak p} (R/\mathfrak p)$$ where $\mathfrak p$ runs the minimal prime ideals of $R$. Then $$ M_n(R)\to \prod_{\mathfrak p} M_n(R/\mathfrak p)$$ is injective. Now observe that over a domain $R/\mathfrak p$, the index of $M_n(R/\mathfrak p)$ is bounded by $n$. So $M_n(R)$ has index bounded by $n$.

In the general case ($I$ not necessarily nilpotent, but $x^r=0$ for all $x\in I$), but $R$ commutative, we still can reduce to the reduced case. We have $T^{m}(R^n)\subseteq IR^n$ as above. In fact, let $J_T$ be the ideal of $R$ generated by the entries of $T^m\in M_n(R)$. Then $J_T$ is a sub-module of $I$ generated by $n^2$ elements. So $J_T^{n^2r}=0$ and $T^{mn^2r}(R^n)\subseteq J_T^{n^2r}R^n=0$. To be more precise, we saw above that we can take $m=n$ for $R/I$, so we get $$T^{n^3r}=0$$ for all nilpotent $T\in M_n(R)$ if $x^r=0$ for all nilpotent $x\in R$.

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