Lemma 35.4.16. For (M,\theta ) \in DD_{S/R}, the diagram
is a split equalizer.
Lemma 35.4.16. For (M,\theta ) \in DD_{S/R}, the diagram
is a split equalizer.
Proof. Define the ring homomorphisms \sigma ^0_0: S_2 \to S_1 and \sigma _0^1, \sigma _1^1: S_3 \to S_2 by the formulas
We then take the auxiliary morphisms to be 1_ M \otimes \sigma _0^0: M \otimes _{S, \delta _1^1} S_2 \to M and 1_ M \otimes \sigma _0^1: M \otimes _{S,\delta _{12}^1} S_3 \to M \otimes _{S, \delta _1^1} S_2. Of the compatibilities required in (35.4.2.1), the first follows from tensoring the cocycle condition (35.4.14.1) with \sigma _1^1 and the others are immediate. \square
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