Let $R$ be a commutative ring with identity and let $A \in M_n(R)$ be such that $$\mbox{tr}A = \mbox{tr}A^2 = \cdots = \mbox{tr}A^n = 0 .$$ I want to show that $n!A^n= 0$.
Any suggestion or reference would be helpful.
P.S.: When $R$ is a field of characteristic zero I can prove that $A^n=0$ but I have no idea for the general case.
Let $$\chi(\lambda) = \det(\lambda I_n - A ) = \lambda^n - e_1 \lambda^{n-1} + e_2 \lambda^{n-2} + \cdots + (-1)^n e_n$$ be the characteristic polynomial of $A$. By Cayley–Hamilton theorem (which is valid for any commutative ring), we have
$$\chi(A) = A^n - e_1 A^{n-1} + e_2 A^{n-2} + \cdots + (-1)^n e_n I_n = 0\tag{*1}$$
Let $p_k = \text{tr}A^k$ for $k = 1, \ldots, n$. The $e_k$ are elementary symmetric polynomials in the roots of $\chi(\lambda)$ and they are related to $p_k$ through the Newton identities (again valid for any commutative ring):
$$\begin{align} e_1 = & p_1,\\ 2 e_2 = & e_1 p_1 - p_2,\\ 3 e_3 = & e_2 p_1 - e_1 p_2 + p_3,\\ 4 e_4 = & e_3 p_1 - e_2 p_2 + e_1 p_3 - p_4,\\ \vdots \end{align}$$ As a result, $$p_1 = p_2 = \cdots = p_n = 0\quad\implies\quad e_1 = 2 e_2 = 3 e_3 = \cdots = n e_n = 0$$
Multiply $(*1)$ by $n!$ then give us $\;n!A^n = 0\;$.
The following argument also works in prime characteristic. The coefficients of the characteristic polynomial $$ \chi(t)=\sum^n_{j=0} (-1)^j \omega_j (A)\:t^{n-j}\; $$ of $A$ satisfy the following identities: $$ \sum^j_{i=1} (-1)^{i+1} {\rm tr}(A^i)\:\omega_{j-i} (A) =j\cdot \omega_j (A) \hbox{ with } \omega_0 (A)=1,\; \omega_{n+j}(A) = 0 \quad \forall \; j\in \mathbb{N} $$ We have either $p\mid n!$, or we have $p>n$ so that $\omega_j (A) = 0 \quad\forall \, j \ge 1$. In this case $A$ has characteristic polynomial $t^n$, so that $A$ is nilpotent with $A^n=0$. Together this means $n!A^n=0$.
