The Stacks project

Proof. Assume (1) and choose an exact sequence $F_{-1} \to F_0 \to N \to 0$ with $F_ i$ finite free. Then we have an exact sequence

functorial in the $R$-module $M$. The functors $\mathop{\mathrm{Hom}}\nolimits _ R(F_ i, M)$ commute with filtered colimits as $\mathop{\mathrm{Hom}}\nolimits _ R(R^{\oplus n}, M) = M^{\oplus n}$. Since filtered colimits are exact (Lemma 10.8.8) we see that (2) holds.

Assume (2). By Lemma 10.11.3 we can write $N = \mathop{\mathrm{colim}}\nolimits N_ i$ as a filtered colimit such that $N_ i$ is of finite presentation for all $i$. Thus $\text{id}_ N$ factors through $N_ i$ for some $i$. This means that $N$ is a direct summand of a finitely presented $R$-module (namely $N_ i$) and hence finitely presented. $\square$