How can I find an idempotent element in a matrix ring?

Question

I have an $r \times r$ matrix $M$ with rational entries and suppose $M$ has $\operatorname{rank} = n-k, k < n$. Then I proceed as follows : Let $m(x)$ be the minimal polynomial of $M$, and make sure it is irreducible. Then $m(x) = m'(x)x$, where $m'(x)$ has a nonzero constant term. Then, if we set $N = m'(M)$ we have that the minimal polynomial of $N$ is of the form $x^2 + ax$, so that $-N/a$ is an idempotent matrix of rank $k$. Moreover if $E =$ the Concatenation of the nullspace of $M$ and the nullspace of $N$ then all the matrices $A$ diagonally splits in blocks of $(n-k)\times (n-k)$ and $k \times k$ matrices. It's not difficult to prove this if we diagonalize $M$ using its roots in $\Bbb C$, but I would like a pure linear algebra proof that I can generalize to other rings than $\Bbb Q$.