= Solution
{c}
Assume $H=\ker\psi$ is a <finitely generated module>. The restriction $q:H\to L$ is a <surjective> <module homomorphism>, so part (a)(i) makes $L$ a <finitely generated module>.
The <free module> $R^m$ has its finite standard <basis of a module>. Its <quotient module> $R^m/\operatorname{Im}q$ is therefore a <finitely generated module> by part (a)(i). Part (c) transfers this property to the isomorphic <quotient module> $\ker\phi/L$. Finally apply part (a)(ii) to the <submodule> $L\subseteq\ker\phi$. It follows that
$$
\boxed{\ker\psi\text{ finitely generated}\ \Longrightarrow\ \ker\phi\text{ finitely generated}}.
$$
More concretely, take the images under $q$ of finitely many generators of $H$, and adjoin lifts in $\ker\phi$ of finitely many generators of $\ker\phi/L$. This finite list generates $\ker\phi$. The argument shows why <finite presentation of a module> is independent of the chosen finite <generating set of a module>.
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