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
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
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. $\square$
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