Let $R$ be a commutative ring and let $M$ be an $R$-module. The following is equivalent to the usual notion of finitely generated:
$M$ is a finitely generated $R$-module if $\text{Hom}_R(M,{-})$ preserves filtered colimits of monomorphism.
As an exercise in understanding this definition, I'd like to try prove Nakayama's lemma using this definition of a finitely generated $R$-module.
Suppose $I$ is an ideal in $R$ such that $IM=M$. Our goal is to show that there is $a=1+x$ with $x\in I$ such that $aM=0$.
With the assumption $IM=M$ we have $\text{Hom}_R(M/IM,N)=0$. For any $R$-module $N$ we have $\text{Hom}_R(M/IM,N)\cong \text{Hom}_R(M,\text{Hom}_R(R/I, N))$. I don't see anywhere to go from here.