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The Stacks project

Theorem 35.4.22. The following conditions are equivalent.

  1. The morphism f is a descent morphism for modules.

  2. The morphism f is an effective descent morphism for modules.

  3. The morphism f is universally injective.

Proof. It is clear that (b) implies (a). We now check that (a) implies (c). If f is not universally injective, we can find M \in \text{Mod}_ R such that the map 1_ M \otimes f: M \to M \otimes _ R S has nontrivial kernel N. The natural projection M \to M/N is not an isomorphism, but its image in DD_{S/R} is an isomorphism. Hence f^* is not fully faithful.

We finally check that (c) implies (b). By Lemmas 35.4.16 and 35.4.20, for (M, \theta ) \in DD_{S/R}, the natural map f^* f_*(M,\theta ) \to M is an isomorphism of S-modules. On the other hand, for M_0 \in \text{Mod}_ R, we may tensor (35.4.18.1) with M_0 over R to obtain an equalizer sequence, so M_0 \to f_* f^* M_0 is an isomorphism. Consequently, f_* and f^* are quasi-inverse functors, proving the claim. \square


Comments (2)

Comment #8763 by Dongryul Kim on

There seems to be at least a typo in the second-to-last sentence. We want to check that the unit map is an isomorphism. This can be checked after base-changing to thanks to univeral injectivity. By adjunction, the composition is the identity, while we have shown that the second map is an isomorphism. Hence becomes an isomorphism after applying .

Comment #9308 by on

OK, the typo is that a subscript was missing on the in the sentence, right? I fixed that. But I think it immediately follows from that sentence that is is an isomorphism. I did add another reference to the first sentence of the last paragraph to clarify. Changes here.

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  • 4 comment(s) on Section 35.4: Descent for universally injective morphisms

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