Processing math: 100%

The Stacks project

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:

  1. I' is countable and directed;

  2. each x_ n is in the image of f_ i: M_ i \to M for some i \in I';

  3. 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


Comments (0)


Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.