Lemma 15.103.7. Let $R$ be a ring and $f \in R$. Let $M \in D(R)$. Assume
$H^ i(M) = 0$ for $i > 0$,
$M \otimes _ R^\mathbf {L} R_ f$ is isomorphic to a flat $R_ f$-module placed in degree $0$,
$M \otimes _ R^\mathbf {L} R/fR$ is isomorphic to a flat $R/fR$-module placed in degree $0$.
Then $M$ is isomorphic to a flat $R$-module placed in degree $0$.
Proof. Let $K$ be an $R$-module. It suffices to show that $N = M \otimes _ R^\mathbf {L} K$ is isomorphic to an $R$-module placed in degree $0$, see Section 15.67. By assumption (1) we see that $N$ only has nonvanishing cohomology in degrees $\leq 0$. Thus, by Lemma 15.103.6, it suffices to show that $H^ i(N)_ f = 0$ for $i < 0$ and that $H^ i(C) = 0$ for $i < -1$ where $C$ is the cone of $f : N \to N$. For the first, we note that
By our assumption (2) this is zero, except if $i = 0$ and then one gets $H^0(M)_ f \otimes _ R K$. For the second, let $C'$ be the cone of $f : K \to K$. Then
Now $C'$ only has nonzero cohomology in degrees $0$ and $-1$ equal to $K/fK$ and $K[f]$; in other words, there is a distinguished triangle
On the other hand, for any $R$-module $K'$ annihilated by $f$ we have
By our assumption (3) this is equal to the module $H^0(M \otimes _ R^\mathbf {L} R/fR) \otimes _{R/fR} K'$ placed in degree $0$. Combining the above we conclude that $C$ only has nonzero cohomology in degrees $0$ and $-1$ and the proof is complete. $\square$
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