Finitely presented module

Remarks

Equivalently, $M$ is finitely generated and the kernel of some surjection $R^n \to M$ is finitely generated. By Schanuel's lemma, the kernel of every surjection $R^n \to M$ is then finitely generated. The map $\varphi$ is given by an $n \times m$ matrix over $R$, so $M$ is the quotient of $R^n$ by the column span of a matrix. Over a Noetherian ring, every finitely generated module is finitely presented, since submodules of $R^n$ are finitely generated.

Example: over $R = \Z$, the module $\Z/6\Z$ is finitely presented. The sequence $\Z \xrightarrow{6} \Z \to \Z/6\Z \to 0$, with the second map the reduction modulo $6$, is exact, because the image of multiplication by $6$ is $6\Z$, which is the kernel of the reduction.

Non-example: let $k$ be a field, $R = k[x_1, x_2, x_3, \ldots]$ the polynomial ring in infinitely many variables, and $M = R/(x_1, x_2, \ldots) \cong k$. Then $M$ is finitely generated, by the class of $1$. The kernel of $R \to M$ is the ideal $(x_1, x_2, \ldots)$, which is not finitely generated, so $M$ is not finitely presented.