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The Stacks project

Lemma 10.92.2. Let R be a ring. Let M be an R-module. Assume M is Mittag-Leffler and countably generated. For any R-module map f : P \to M with P finitely generated there exists an endomorphism \alpha : M \to M such that

  1. \alpha : M \to M factors through a finitely presented R-module, and

  2. \alpha \circ f = f.

Proof. Write M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i as a directed colimit of finitely presented R-modules with I countable, see Lemma 10.92.1. The transition maps are denoted f_{ij} and we use f_ i : M_ i \to M to denote the canonical maps into M. Set N = \prod _{s \in I} M_ s. Denote

M_ i^* = \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N) = \prod \nolimits _{s \in I} \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, M_ s)

so that (M_ i^*) is an inverse system of R-modules over I. Note that \mathop{\mathrm{Hom}}\nolimits _ R(M, N) = \mathop{\mathrm{lim}}\nolimits M_ i^*. As M is Mittag-Leffler, we find for every i \in I an index k(i) \geq i such that

E_ i := \bigcap \nolimits _{i' \geq i} \mathop{\mathrm{Im}}(M_{i'}^* \to M_ i^*) = \mathop{\mathrm{Im}}(M_{k(i)}^* \to M_ i^*)

Choose and fix j \in I such that \mathop{\mathrm{Im}}(P \to M) \subset \mathop{\mathrm{Im}}(M_ j \to M). This is possible as P is finitely generated. Set k = k(j). Let x = (0, \ldots , 0, \text{id}_{M_ k}, 0, \ldots , 0) \in M_ k^* and note that this maps to y = (0, \ldots , 0, f_{jk}, 0, \ldots , 0) \in M_ j^*. By our choice of k we see that y \in E_ j. By Example 10.86.2 the transition maps E_ i \to E_ j are surjective for each i \geq j and \mathop{\mathrm{lim}}\nolimits E_ i = \mathop{\mathrm{lim}}\nolimits M_ i^* = \mathop{\mathrm{Hom}}\nolimits _ R(M, N). Hence Lemma 10.86.3 guarantees there exists an element z \in \mathop{\mathrm{Hom}}\nolimits _ R(M, N) which maps to y in E_ j \subset M_ j^*. Let z_ k be the kth component of z. Then z_ k : M \to M_ k is a homomorphism such that

\xymatrix{ M \ar[r]_{z_ k} & M_ k \\ M_ j \ar[ru]_{f_{jk}} \ar[u]^{f_ j} }

commutes. Let \alpha : M \to M be the composition f_ k \circ z_ k : M \to M_ k \to M. Then \alpha factors through a finitely presented module by construction and \alpha \circ f_ j = f_ j. Since the image of f is contained in the image of f_ j this also implies that \alpha \circ f = f. \square


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