Suppose that there exists a finitely generated free module F, and an A-module homo. $\phi : F \to M$ which is onto. Then prove that M is finitely generated.
As F is free so F $\approx A^n$ for some $n\in \mathbb{N}$. Also, $\phi$ is onto so for every m there exists f $\in F$ such that $\phi(f)=m$. Now, F is finitely generated means that there exists a finite set {$f_1,…,f_n$} such that any f $\in F$ can be represented as $f=a_1f_1+…+a_nf_n$ , but I am unable to use it to prove that M is finitely generated. I think I should not try this question by method of contradiction.
Best Answer
The freeness assumption is not necessary. Any homomorphic image of a finitely generated module is finitely generated.
This follows from Noether's isomorphism theorem: if $f:A\to B$ is a surjective homomorphism, then $B\simeq A/\ker(f),$ so any set of generators for $A$ yields a set of generators for $B$ - if $A$ is generated by $a_1,\dots,a_n$ then $A/\ker(f)$ is generated by $\bar{a}_1,\dots,\bar{a}_n$.
In your notation, the generators of $M$ will be $\phi(f_1),\dots \phi(f_n)$ and $m=a_1\phi(f_1)+\dots+a_n\phi(f_n)$.