nilpotent endomorphism on finitely generated modules over a domain 
If $R$ is a domain and $f: R^n \to R^n$ is an $R$-module endomorphism. Suppose $f^m = 0$ for some $m> 0$. Show that $f^n = 0$.

The cases $ m \le n$ is trivial. When $m>n$, I don't have much idea how to start. I tried to apply the Cayley-Hamilton theorem but doesn't seem to help much.
 A: Let $Q$ be the field of fractions of $R$, and let $F:Q^n\to Q^n$ be the morphism of $Q$-modules which has the same matrix as $f$ (with respect to the standard bases of $R^n$ in the case of $f$ and of $Q^n$ in the case of $F$) 


*

*Show that $F$ is nilpotent, and that therefore $F^n=0$.

*If $i:R^n\to Q^n$ is the obvious inclusion, then we have $F\circ i=i\circ f$. Use this, and the fact that $i$ is an injective function, to show that $f^n=0$.
A: It's convenient but unnecessary to pass to the fraction field. Your hunch that you should apply Cayley-Hamilton works. By Cayley-Hamilton, we have
$$f^n = a_{n-1} f^{n-1} + \dots + a_i f^i$$
where $a_k \in R$ and $a_i \neq 0$ (of course if all the $a_i$ are zero then we are done). Multiplying by $f^{m-i-1}$ gives
$$a_i f^{m-1} = 0$$
and hence, since $R$ is an integral domain, that
$$f^{m-1} = 0.$$
Now you can multiply by $f^{m-i-2}$, etc. Eventually you conclude that $f^n = 0$ as desired. 
A: View the matix as lying in a matrix ring of the same size, but having entries in the algebraic closure of the field of factions for $R$.
It therefore has a Jordan Normal Form, and since it is nilpotent, it must have all zeros on the diagonal. The normal form is then a strictly upper triangular.
Finally, it is easy to prove that $M^n=0$ for any strictly upper triangular matrix in the $n\times n$ matrix ring. The same is true for the original matrix, which is a conjugate of this normal form.
