Given a module N over a ring R, you can characterize whether or not N is a finite module or a finitely presented module in terms of the functor \mathop{\mathrm{Hom}}\nolimits _ R(N, -).
Lemma 10.11.1. Let R be a ring. Let N be an R-module. The following are equivalent
N is a finite R-module,
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
For purposes of reference, we define what it means to have a relation between elements of a module.
Definition 10.11.2. Let R be a ring. Let M be an R-module. Let n \geq 0 and x_ i \in M for i = 1, \ldots , n. A relation between x_1, \ldots , x_ n in M is a sequence of elements f_1, \ldots , f_ n \in R such that \sum _{i = 1, \ldots , n} f_ i x_ i = 0.
Lemma 10.11.3. Let R be a ring and let M be an R-module. Then M is the colimit of a directed system (M_ i, \mu _{ij}) of R-modules with all M_ i finitely presented R-modules.
Proof. Consider any finite subset S \subset M and any finite collection of relations E among the elements of S. So each s \in S corresponds to x_ s \in M and each e \in E consists of a vector of elements f_{e, s} \in R such that \sum f_{e, s} x_ s = 0. Let M_{S, E} be the cokernel of the map
There are canonical maps M_{S, E} \to M. If S \subset S' and if the elements of E correspond, via this map, to relations in E', then there is an obvious map M_{S, E} \to M_{S', E'} commuting with the maps to M. Let I be the set of pairs (S, E) with ordering by inclusion as above. It is clear that the colimit of this directed system is M. \square
Lemma 10.11.4. Let R be a ring. Let N be an R-module. The following are equivalent
N is a finitely presented R-module,
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 bijective.
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
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