Lemma 10.92.1. Let M be an R-module. Write M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i where (M_ i, f_{ij}) is a directed system of finitely presented R-modules. If M is Mittag-Leffler and countably generated, then there is a directed countable subset I' \subset I such that M \cong \mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i.
Proof. Let x_1, x_2, \ldots be a countable set of generators for M. For each x_ n choose i \in I such that x_ n is in the image of the canonical map f_ i: M_ i \to M; let I'_{0} \subset I be the set of all these i. Now since M is Mittag-Leffler, for each i \in I'_{0} we can choose j \in I such that j \geq i and f_{ij}: M_ i \to M_ j factors through f_{ik}: M_ i \to M_ k for all k \geq i (condition (3) of Proposition 10.88.6); let I'_1 be the union of I'_0 with all of these j. Since I'_1 is a countable, we can enlarge it to a countable directed set I'_{2} \subset I. Now we can apply the same procedure to I'_{2} as we did to I'_{0} to get a new countable set I'_{3} \subset I. Then we enlarge I'_{3} to a countable directed set I'_{4}. Continuing in this way—adding in a j as in Proposition 10.88.6 (3) for each i \in I'_{\ell } if \ell is odd and enlarging I'_{\ell } to a directed set if \ell is even—we get a sequence of subsets I'_{\ell } \subset I for \ell \geq 0. The union I' = \bigcup I'_{\ell } satisfies:
I' is countable and directed;
each x_ n is in the image of f_ i: M_ i \to M for some i \in I';
if i \in I', then there is j \in I' such that j \geq i and f_{ij}: M_ i \to M_ j factors through f_{ik}: M_ i \to M_ k for all k \in I with k \geq i. In particular \mathop{\mathrm{Ker}}(f_{ik}) \subset \mathop{\mathrm{Ker}}(f_{ij}) for k \geq i.
We claim that the canonical map \mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i \to \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i = M is an isomorphism. By (2) it is surjective. For injectivity, suppose x \in \mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i maps to 0 in \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i. Representing x by an element \tilde{x} \in M_ i for some i \in I', this means that f_{ik}(\tilde{x}) = 0 for some k \in I, k \geq i. But then by (3) there is j \in I', j \geq i, such that f_{ij}(\tilde{x}) = 0. Hence x = 0 in \mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i. \square
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