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The Stacks project

Proposition 10.88.6. Let M be an R-module. Let (M_ i, f_{ij}) be a directed system of finitely presented R-modules, indexed by I, such that M = \mathop{\mathrm{colim}}\nolimits M_ i. Let f_ i: M_ i \to M be the canonical map. The following are equivalent:

  1. For every finitely presented R-module P and module map f: P \to M, there exists a finitely presented R-module Q and a module map g: P \to Q such that g and f dominate each other, i.e., \mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ N) = \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ N) for every R-module N.

  2. For each i \in I, there exists j \geq i such that f_{ij}: M_ i \to M_ j dominates f_ i: M_ i \to M.

  3. For each i \in I, there exists j \geq i such that f_{ij}: M_ i \to M_ j factors through f_{ik}: M_ i \to M_ k for all k \geq i.

  4. For every R-module N, the inverse system (\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N), \mathop{\mathrm{Hom}}\nolimits _ R(f_{ij}, N)) is Mittag-Leffler.

  5. For N = \prod _{s \in I} M_ s, the inverse system (\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N), \mathop{\mathrm{Hom}}\nolimits _ R(f_{ij}, N)) is Mittag-Leffler.

Proof. First we prove the equivalence of (1) and (2). Suppose (1) holds and let i \in I. Corresponding to the map f_ i: M_ i \to M, we can choose g: M_ i \to Q as in (1). Since M_ i and Q are of finite presentation, so is \mathop{\mathrm{Coker}}(g). Then by Lemma 10.88.5, f_ i : M_ i \to M factors through g: M_ i \to Q, say f_ i = h \circ g for some h: Q \to M. Then since Q is finitely presented, h factors through M_ j \to M for some j \geq i, say h = f_ j \circ h' for some h': Q \to M_ j. In total we have a commutative diagram

\xymatrix{ & M & \\ M_ i \ar[dr]_ g \ar[ur]^{f_ i} \ar[rr]^{f_{ij}} & & M_ j \ar[ul]_{f_ j} \\ & Q \ar[ur]_{h'} & }

Thus f_{ij} dominates g. But g dominates f_ i, so f_{ij} dominates f_ i.

Conversely, suppose (2) holds. Let P be of finite presentation and f: P \to M a module map. Then f factors through f_ i: M_ i \to M for some i \in I, say f = f_ i \circ g' for some g': P \to M_ i. Choose by (2) a j \geq i such that f_{ij} dominates f_ i. We have a commutative diagram

\xymatrix{ P \ar[d]_{g'} \ar[r]^{f} & M \\ M_ i \ar[ur]^{f_ i} \ar[r]_{f_{ij}} & M_ j \ar[u]_{f_ j} }

From the diagram and the fact that f_{ij} dominates f_ i, we find that f and f_{ij} \circ g' dominate each other. Hence taking g = f_{ij} \circ g' : P \to M_ j works.

Next we prove (2) is equivalent to (3). Let i \in I. It is always true that f_ i dominates f_{ik} for k \geq i, since f_ i factors through f_{ik}. If (2) holds, choose j \geq i such that f_{ij} dominates f_ i. Then since domination is a transitive relation, f_{ij} dominates f_{ik} for k \geq i. All M_ i are of finite presentation, so \mathop{\mathrm{Coker}}(f_{ik}) is of finite presentation for k \geq i. By Lemma 10.88.5, f_{ij} factors through f_{ik} for all k \geq i. Thus (2) implies (3). On the other hand, if (3) holds then for any R-module N, f_{ij} \otimes _ R \text{id}_ N factors through f_{ik} \otimes _ R \text{id}_ N for k \geq i. So \mathop{\mathrm{Ker}}(f_{ik} \otimes _ R \text{id}_ N) \subset \mathop{\mathrm{Ker}}(f_{ij} \otimes _ R \text{id}_ N) for k \geq i. But \mathop{\mathrm{Ker}}(f_ i \otimes _ R \text{id}_ N: M_ i \otimes _ R N \to M \otimes _ R N) is the union of \mathop{\mathrm{Ker}}(f_{ik} \otimes _ R \text{id}_ N) for k \geq i. Thus \mathop{\mathrm{Ker}}(f_ i \otimes _ R \text{id}_ N) \subset \mathop{\mathrm{Ker}}(f_{ij} \otimes _ R \text{id}_ N) for any R-module N, which by definition means f_{ij} dominates f_ i.

It is trivial that (3) implies (4) implies (5). We show (5) implies (3). Let N = \prod _{s \in I} M_ s. If (5) holds, then given i \in I choose j \geq i such that

\mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ j, N) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, N)) = \mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ k, N) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, N))

for all k \geq j. Passing the product over s \in I outside of the \mathop{\mathrm{Hom}}\nolimits 's and looking at the maps on each component of the product, this says

\mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ j, M_ s) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ s)) = \mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ k, M_ s) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ s))

for all k \geq j and s \in I. Taking s = j we have

\mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ j, M_ j) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ j)) = \mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ k, M_ j) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ j))

for all k \geq j. Since f_{ij} is the image of \text{id} \in \mathop{\mathrm{Hom}}\nolimits (M_ j, M_ j) under \mathop{\mathrm{Hom}}\nolimits (M_ j, M_ j) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ j), this shows that for any k \geq j there is h \in \mathop{\mathrm{Hom}}\nolimits (M_ k, M_ j) such that f_{ij} = h \circ f_{ik}. If j \geq k then we can take h = f_{kj}. Hence (3) holds. \square


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