## 10.11 Characterizing finite and finitely presented modules

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 $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ R(N, M_ i) \to \mathop{\mathrm{Hom}}\nolimits _ R(K, M)$ 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

\[ R^{\# E} \longrightarrow R^{\# S}, \quad (g_ e)_{e\in E} \longmapsto (\sum g_ e f_{e, s})_{s\in S}. \]

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 $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ R(N, M_ i) \to \mathop{\mathrm{Hom}}\nolimits _ R(K, M)$ 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

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

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.
$\square$

## Comments (1)

Comment #6040 by Shurui Liu on