Let $R$ be a commutative ring. How to prove the following:
If $\chi_A(t) \equiv t^n \bmod\operatorname{nil}(R)$ then $A \in M_n(R)$ is nilpotent.
Note $\chi_A$ is the characteristic polynomial of $A$.
This is in reference to "Strongly clean matrix rings over commutative local rings" (Borooah)
Let $\chi_A(T)\in R[T]$ be the characteristic polynomial of $A\in M_n(R)$. Say $\chi_A(T)\equiv T^n\bmod{\rm nil}(R)$.
By the Cayley-Hamilton theorem we know $\chi_A(A)=0$. But $\chi_A(A)=A^n+f(A)$ where $f(\cdot)$ is some polynomial with nilpotent coefficients from ${\rm nil}(R)$, by the hypothesis. Sums and products of nilpotent elements are nilpotent in commutative ring, so since all the scalars involved in $f(A)$ are nilpotent, so too must $f(A)$ be. Thus $A^n=-f(A)$ is nilpotent. If a power of something is nilpotent, then it is too.