Lemma 10.11.1. Let $R$ be a ring. Let $N$ be an $R$-module. The following are equivalent

1. $N$ is a finite $R$-module,

2. for any filtered colimit $M = \mathop{\mathrm{colim}}\nolimits M_ i$ of $R$-modules the map $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ R(N, M_ i) \to \mathop{\mathrm{Hom}}\nolimits _ R(N, M)$ is injective.

Proof. Assume (1) and choose generators $x_1, \ldots , x_ m$ for $N$. If $N \to M_ i$ is a module map and the composition $N \to M_ i \to M$ is zero, then because $M = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} M_{i'}$ for each $j \in \{ 1, \ldots , m\}$ we can find a $i' \geq i$ such that $x_ j$ maps to zero in $M_{i'}$. Since there are finitely many $x_ j$ we can find a single $i'$ which works for all of them. Then the composition $N \to M_ i \to M_{i'}$ is zero and we conclude the map is injective, i.e., part (2) holds.

Assume (2). For a finite subset $E \subset N$ denote $N_ E \subset N$ the $R$-submodule generated by the elements of $E$. Then $0 = \mathop{\mathrm{colim}}\nolimits N/N_ E$ is a filtered colimit. Hence we see that $\text{id} : N \to N$ maps into $N_ E$ for some $E$, i.e., $N$ is finitely generated. $\square$

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