The Stacks project

Proposition 15.89.15. Assume $\varphi : R \to S$ is a flat ring map and $I = (f_1, \ldots , f_ t) \subset R$ is an ideal such that $R/I \to S/IS$ is an isomorphism. Then $\text{Can}$ and $H^0$ are quasi-inverse equivalences of categories

\[ \text{Mod}_ R = \text{Glue}(R \to S, f_1, \ldots , f_ t) \]

Proof. We have already seen that $H^0 \circ \text{Can}$ is isomorphic to the identity functor, see Lemma 15.89.11. Consider an object $\mathbf{M} = (M', M_ i, \alpha _ i, \alpha _{ij})$ of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$. We get a natural morphism

\[ \Psi : (H^0(\mathbf{M}) \otimes _ R S, H^0(\mathbf{M})_{f_ i}, \text{can}_ i, \text{can}_{ij}) \longrightarrow (M', M_ i, \alpha _ i, \alpha _{ij}). \]

Namely, by definition $H^0(\mathbf{M})$ comes equipped with compatible $R$-module maps $H^0(\mathbf{M}) \to M'$ and $H^0(\mathbf{M}) \to M_ i$. We have to show that this map is an isomorphism.

Pick an index $i$ and set $R' = R_{f_ i}$. Combining Lemmas 15.89.14 and 15.89.13 we see that $\Psi \otimes _ R R'$ is an isomorphism. Hence the kernel, resp. cokernel of $\Psi $ is a system of the form $(K, 0, 0, 0)$, resp. $(Q, 0, 0, 0)$. Note that $H^0((K, 0, 0, 0)) = K$, that $H^0$ is left exact, and that by construction $H^0(\Psi )$ is bijective. Hence we see $K = 0$, i.e., the kernel of $\Psi $ is zero.

The conclusion of the above is that we obtain a short exact sequence

\[ 0 \to H^0(\mathbf{M}) \otimes _ R S \to M' \to Q \to 0 \]

and that $M_ i = H^0(\mathbf{M})_{f_ i}$. Note that we may think of $Q$ as an $R$-module which is $I$-power torsion so that $Q = Q \otimes _ R S$. By Lemma 15.89.8 we see that there exists a commutative diagram

\[ \xymatrix{ 0 \ar[r] & H^0(\mathbf{M}) \ar[r] \ar[d] & E \ar[r] \ar[d] & Q \ar[r] \ar[d] & 0 \\ 0 \ar[r] & H^0(\mathbf{M}) \otimes _ R S \ar[r] & M' \ar[r] & Q \ar[r] & 0 } \]

with exact rows. This clearly determines an isomorphism $\text{Can}(E) \to (M', M_ i, \alpha _ i, \alpha _{ij})$ in the category $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ and we win. (Of course, a posteriori we have $Q = 0$.) $\square$

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