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 Lemma 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 $M_0 \to f_* f^* M$ is an isomorphism. Consequently, $f_*$ and $f^*$ are quasi-inverse functors, proving the claim. $\square$

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 $M_0 \to f_\ast f^\ast M_0$ is an isomorphism. This can be checked after base-changing to $S$ thanks to univeral injectivity. By adjunction, the composition $f^\ast M_0 \to f^\ast f_\ast f^\ast M_0 \to f^\ast M_0$ is the identity, while we have shown that the second map is an isomorphism. Hence $M_0 \to f_\ast f^\ast M_0$ becomes an isomorphism after applying $f^\ast$.

Comment #9308 by on

OK, the typo is that a subscript $0$ was missing on the $M$ in the sentence, right? I fixed that. But I think it immediately follows from that sentence that $\text{id} \to f_* f^*$ 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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