The Stacks project

Lemma 15.88.16. Let $\varphi : R \to S$ be a flat ring map and let $I \subset R$ be a finitely generated ideal such that $R/I \to S/IS$ is an isomorphism.

  1. Given an $R$-module $N$, an $S$-module $M'$ and an $S$-module map $\varphi : M' \to N \otimes _ R S$ whose kernel and cokernel are $I$-power torsion, there exists an $R$-module map $\psi : M \to N$ and an isomorphism $M \otimes _ R S = M'$ compatible with $\varphi $ and $\psi $.

  2. Given an $R$-module $M$, an $S$-module $N'$ and an $S$-module map $\varphi : M \otimes _ R S \to N'$ whose kernel and cokernel are $I$-power torsion, there exists an $R$-module map $\psi : M \to N$ and an isomorphism $N \otimes _ R S = N'$ compatible with $\varphi $ and $\psi $.

In both cases we have $\mathop{\mathrm{Ker}}(\varphi ) \cong \mathop{\mathrm{Ker}}(\psi )$ and $\mathop{\mathrm{Coker}}(\varphi ) \cong \mathop{\mathrm{Coker}}(\psi )$.

Proof. Proof of (1). Say $I = (f_1, \ldots , f_ t)$. It is clear that the localization $\varphi _{f_ i}$ is an isomorphism. Thus we see that $(M', N_{f_ i}, \varphi _{f_ i}, can_{ij})$ is an object of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$, see Remark 15.88.10. By Proposition 15.88.15 we conclude that there exists an $R$-module $M$ such that $M' = M \otimes _ R S$ and $N_{f_ i} = M_{f_ i}$ compatibly with the isomorphisms $\varphi _{f_ i}$ and $can_{ij}$. There is a morphism

\[ (M \otimes _ R S, M_{f_ i}, can_ i, can_{ij}) = (M', N_{f_ i}, \varphi _{f_ i}, can_{ij}) \to (N \otimes _ R S, N_{f_ i}, can_ i, can_{ij}) \]

of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ which uses $\varphi $ in the first component. This corresponds to an $R$-module map $\psi : M \to N$ (by the equivalence of categories of Proposition 15.88.15). The composition of the base change of $M \to N$ with the isomorphism $M' \cong M \otimes _ R S$ is $\varphi $, in other words $M \to N$ is compatible with $\varphi $.

Proof of (2). This is just the dual of the argument above. Namely, the localization $\varphi _{f_ i}$ is an isomorphism. Thus we see that $(N', M_{f_ i}, \varphi _{f_ i}^{-1}, can_{ij})$ is an object of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$, see Remark 15.88.10. By Proposition 15.88.15 we conclude that there exists an $R$-module $N$ such that $N' = N \otimes _ R S$ and $N_{f_ i} = M_{f_ i}$ compatibly with the isomorphisms $\varphi _{f_ i}^{-1}$ and $can_{ij}$. There is a morphism

\[ (M \otimes _ R S, M_{f_ i}, can_ i, can_{ij}) \to (N', M_{f_ i}, \varphi _{f_ i}, can_{ij}) = (N \otimes _ R S, N_{f_ i}, can_ i, can_{ij}) \]

of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ which uses $\varphi $ in the first component. This corresponds to an $R$-module map $\psi : M \to N$ (by the equivalence of categories of Proposition 15.88.15). The composition of the base change of $M \to N$ with the isomorphism $N' \cong N \otimes _ R S$ is $\varphi $, in other words $M \to N$ is compatible with $\varphi $.

The final statement follows for example from Lemma 15.88.3. $\square$


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