**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

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(N, M) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_0, M) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_{-1}, M) \]

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 $M = \mathop{\mathrm{colim}}\nolimits M_ i$ as a filtered colimit such that $M_ i$ is of finite presentation for all $i$. Thus $\text{id}_ M$ factors through $M_ i$ for some $i$. This means that $M$ is a direct summand of a finitely presented $R$-module (namely $M_ i$) and hence finitely presented.
$\square$

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