= Solution
{c}
Choose a finite <generating set of a module> $n_1,\ldots,n_t$ for the <finitely generated module> $N$. For any $m\in M$, the <surjectivity> of the <module homomorphism> $\phi$ gives $n\in N$ with $\phi(n)=m$. Write $n=\sum_{j=1}^t r_jn_j$. Since $\phi$ is a <module homomorphism>,
$$
m=\phi\left(\sum_{j=1}^t r_jn_j\right)=\sum_{j=1}^t r_j\phi(n_j).
$$
Thus
$$
\boxed{M=R\phi(n_1)+\cdots+R\phi(n_t)}
$$
is a <finitely generated module>. The argument includes $N=0$, in which case <surjectivity> forces $M=0$ and the empty <generating set of a module> suffices.
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