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The Stacks project

Lemma 10.127.5. Let A be a ring and let M, N be A-modules. Suppose that R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i is a directed colimit of A-algebras.

  1. If M is a finite A-module, and u, u' : M \to N are A-module maps such that u \otimes 1 = u' \otimes 1 : M \otimes _ A R \to N \otimes _ A R then for some i we have u \otimes 1 = u' \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i.

  2. If N is a finite A-module and u : M \to N is an A-module map such that u \otimes 1 : M \otimes _ A R \to N \otimes _ A R is surjective, then for some i the map u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i is surjective.

  3. If N is a finitely presented A-module, and v : N \otimes _ A R \to M \otimes _ A R is an R-module map, then there exists an i and an R_ i-module map v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i such that v = v_ i \otimes 1.

  4. If M is a finite A-module, N is a finitely presented A-module, and u : M \to N is an A-module map such that u \otimes 1 : M \otimes _ A R \to N \otimes _ A R is an isomorphism, then for some i the map u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i is an isomorphism.

Proof. To prove (1) assume u is as in (1) and let x_1, \ldots , x_ m \in M be generators. Since N \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i N \otimes _ A R_ i we may pick an i \in I such that u(x_ j) \otimes 1 = u'(x_ j) \otimes 1 in M \otimes _ A R_ i, j = 1, \ldots , m. For such an i we have u \otimes 1 = u' \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i.

To prove (2) assume u \otimes 1 surjective and let y_1, \ldots , y_ m \in N be generators. Since N \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i N \otimes _ A R_ i we may pick an i \in I and z_ j \in M \otimes _ A R_ i, j = 1, \ldots , m whose images in N \otimes _ A R equal y_ j \otimes 1. For such an i the map u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i is surjective.

To prove (3) let y_1, \ldots , y_ m \in N be generators. Let K = \mathop{\mathrm{Ker}}(A^{\oplus m} \to N) where the map is given by the rule (a_1, \ldots , a_ m) \mapsto \sum a_ j x_ j. Let k_1, \ldots , k_ t be generators for K. Say k_ s = (k_{s1}, \ldots , k_{sm}). Since M \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i M \otimes _ A R_ i we may pick an i \in I and z_ j \in M \otimes _ A R_ i, j = 1, \ldots , m whose images in M \otimes _ A R equal v(y_ j \otimes 1). We want to use the z_ j to define the map v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i. Since K \otimes _ A R_ i \to R_ i^{\oplus m} \to N \otimes _ A R_ i \to 0 is a presentation, it suffices to check that \xi _ s = \sum _ j k_{sj}z_ j is zero in M \otimes _ A R_ i for each s = 1, \ldots , t. This may not be the case, but since the image of \xi _ s in M \otimes _ A R is zero we see that it will be the case after increasing i a bit.

To prove (4) assume u \otimes 1 is an isomorphism, that M is finite, and that N is finitely presented. Let v : N \otimes _ A R \to M \otimes _ A R be an inverse to u \otimes 1. Apply part (3) to get a map v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i for some i. Apply part (1) to see that, after increasing i we have v_ i \circ (u \otimes 1) = \text{id}_{M \otimes _ R R_ i} and (u \otimes 1) \circ v_ i = \text{id}_{N \otimes _ R R_ i}. \square


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