Lemma 93.12.1 (0DZ1)—The Stacks project

Lemma 93.12.1. Let $A' \to A$ be a surjection of rings with nilpotent kernel. Let $A' \to P'$ be a flat ring map. Set $P = P' \otimes _{A'} A$ . Let $M$ be an $A$ -flat $P$ -module. Then the following are equivalent

there is an $A'$ -flat $P'$ -module $M'$ with $M' \otimes _{P'} P = M$ , and
there is an object $K' \in D^-(P')$ with $K' \otimes _{P'}^\mathbf {L} P = M$ .

Proof. Suppose that $M'$ is as in (1). Then

$M = M' \otimes _ P P' = M' \otimes _{A'} A = M' \otimes _ A^\mathbf {L} A' = M' \otimes _{P'}^\mathbf {L} P$

The first two equalities are clear, the third holds because $M'$ is flat over $A'$ , and the fourth holds by More on Algebra, Lemma 15.61.2. Thus (2) holds. Conversely, suppose $K'$ is as in (2). We may and do assume $M$ is nonzero. Let $t$ be the largest integer such that $H^ t(K')$ is nonzero (exists because $M$ is nonzero). Then $H^ t(K') \otimes _{P'} P = H^ t(K' \otimes _{P'}^\mathbf {L} P)$ is zero if $t > 0$ . Since the kernel of $P' \to P$ is nilpotent this implies $H^ t(K') = 0$ by Nakayama's lemma a contradiction. Hence $t = 0$ (the case $t < 0$ is absurd as well). Then $M' = H^0(K')$ is a $P'$ -module such that $M = M' \otimes _{P'} P$ and the spectral sequence for Tor gives an injective map

$\text{Tor}_1^{P'}(M', P) \to H^{-1}(M' \otimes _{P'}^\mathbf {L} P) = 0$

By the reference on derived base change above $0 = \text{Tor}_1^{P'}(M', P) = \text{Tor}_1^{A'}(M', A)$ . We conclude that $M'$ is $A'$ -flat by Algebra, Lemma 10.99.8. $\square$

The Stacks project

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