Trace of nilpotent matrix over a ring Let $R$ be a commutative ring with unity, and $n$ a positive integer.
Let $A\in \mathfrak{M}_n(R)$ such that there exists $m\in \mathbb N$, for which $A^m=0$.
Is it true that there exists $\ell\in \mathbb N$, such that $\bigl(\text{tr}(A)\bigr)^\ell=0$ ?
Remark : It is true for $n=1$ and $n=2$.

If $\mathfrak{p}$ is any prime ideal of $R$, then $A/\mathfrak{p}$ is a domain. We can embed it in its fraction field, then in an algebraic closure. Then your matrix mod $\mathfrak{p}$ is triangularisable, with of course eigenvalues $0$.
Hence, the reduction mod $\mathfrak{p}$ of your matrix has vanishing trace. Since this is true for every prime ideal, this shows that the trace of $A$ is in the intersection of all prime ideals of $R$, that is in the nilradical of $R$, which is the ideal of nilpotent elements of $R$.
This concludes the proof.
