Determinant modulo $m$ for a matrix Let $A$ be a matrix having entries in $\mathbb{Z}_n$, i.e. the ring of intergers modulo $n$. Suppose $$A^m \equiv 0 \pmod{n}$$ for some positive integer $m$, then can we say that $$(\det(A))^m \equiv 0 \pmod n,$$ i.e., $\det A$ is also a nipotent modulo $n$?
According to me it is right. As $$\det(A^m) = (\det A)^m,$$ and thus modulo $m$, we must get $$(\det A)^m \equiv 0 \pmod{n}.$$ Am i right?

Yes. You are right. You can think about the determinant as a polynomial in the entries of $A$. Now, if you assume $A^m=0$ over $\mathbb{Z}_n$ then obviously a polynomial in its entries will be divisible by $n$ as well. From the equality $\det(A)^m=\det(A^m)$ you conclude what you want
