The Stacks project

Lemma 35.37.1. Let $S$ be a scheme. Let $\{ X_ i \to S\} _{i\in I}$ be an fpqc covering, see Topologies, Definition 34.9.1. Let $(V_ i/X_ i, \varphi _{ij})$ be a descent datum relative to $\{ X_ i \to S\} $. If each morphism $V_ i \to X_ i$ is affine, then the descent datum is effective.

Proof. Being affine is a property of morphisms of schemes which is local on the base and preserved under any base change, see Morphisms, Lemmas 29.11.3 and 29.11.8. Hence Lemma 35.36.2 applies and it suffices to prove the statement of the lemma in case the fpqc-covering is given by a single $\{ X \to S\} $ flat surjective morphism of affines. Say $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$ so that $R \to A$ is a faithfully flat ring map. Let $(V, \varphi )$ be a descent datum relative to $X$ over $S$ and assume that $V \to X$ is affine. Then $V \to X$ being affine implies that $V = \mathop{\mathrm{Spec}}(B)$ for some $A$-algebra $B$ (see Morphisms, Definition 29.11.1). The isomorphism $\varphi $ corresponds to an isomorphism of rings

\[ \varphi ^\sharp : B \otimes _ R A \longleftarrow A \otimes _ R B \]

as $A \otimes _ R A$-algebras. The cocycle condition on $\varphi $ says that

\[ \xymatrix{ B \otimes _ R A \otimes _ R A & & A \otimes _ R A \otimes _ R B \ar[ll] \ar[ld]\\ & A \otimes _ R B \otimes _ R A \ar[lu] & } \]

is commutative. Inverting these arrows we see that we have a descent datum for modules with respect to $R \to A$ as in Definition 35.3.1. Hence we may apply Proposition 35.3.9 to obtain an $R$-module $C = \mathop{\mathrm{Ker}}(B \to A \otimes _ R B)$ and an isomorphism $A \otimes _ R C \cong B$ respecting descent data. Given any pair $c, c' \in C$ the product $cc'$ in $B$ lies in $C$ since the map $\varphi $ is an algebra homomorphism. Hence $C$ is an $R$-algebra whose base change to $A$ is isomorphic to $B$ compatibly with descent data. Applying $\mathop{\mathrm{Spec}}$ we obtain a scheme $U$ over $S$ such that $(V, \varphi ) \cong (X \times _ S U, can)$ as desired. $\square$

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