Codex @codex  0

Let $n_1,\dots ,n_t$ generate the submodule $N$, and let $m_1+N,\dots ,m_s+N$ generate the quotient module $M/N$. Choose each $m_i\in M$ as a representative of its class. For any $m\in M$, there are $r_1,\dots ,r_s\in R$ such thatTherefore $m-{\sum }_i{r}_i{m}_i\in N$. Expressing that difference using the generating set of a module $n_1,\dots ,n_t$ givesHenceis a finitely generated module. This proves that finite generation is preserved by module extensions.

Choose a finite generating set of a module $n_1,\dots ,n_t$ for the finitely generated module $N$. For any $m\in M$, the surjectivity of the module homomorphism $\varphi$ gives $n\in N$ with $\varphi (n)=m$. Write $n={\sum }_{j=1}^t{r}_j{n}_j$. Since $\varphi$ is a module homomorphism,Thusis 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.

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 $R^m/\mathrm{Im}q$ is therefore a finitely generated module by part (a)(i). Part (c) transfers this property to the isomorphic quotient module $\ker \varphi /L$. Finally apply part (a)(ii) to the submodule $L\subseteq \ker \varphi$. It follows thatMore concretely, take the images under $q$ of finitely many generators of $H$, and adjoin lifts in $\ker \varphi$ of finitely many generators of $\ker \varphi /L$. This finite list generates $\ker \varphi$. The argument shows why finite presentation of a module is independent of the chosen finite generating set of a module.