Lemma 15.90.19. Let $(R \to R', f)$ be a glueing pair. Let $M$ be an $R$-module which is not necessarily glueable for $(R \to R', f)$. Then $M$ is flat over $R$ if and only if $M \otimes _ R R'$ is flat over $R'$ and $M_ f$ is flat over $R_ f$.

**Proof.**
One direction of the lemma follows from Algebra, Lemma 10.39.7. For the other direction, assume $M \otimes _ R R'$ is flat over $R'$ and $M_ f$ is flat over $R_ f$. Let $\tilde M$ be as in Remark 15.90.18. If $\tilde M$ is flat over $R$, then applying Algebra, Lemma 10.39.12 to the short exact sequence $0 \to \mathop{\mathrm{Ker}}(M \to \tilde M) \to M \to \tilde M \to 0$ we find that $\mathop{\mathrm{Ker}}(M \to \tilde M) \otimes _ R (R' \oplus R_ f)$ is zero. Hence $M = \tilde M$ by Lemma 15.90.3 and we conclude. In other words, we may replace $M$ by $\tilde M$ and assume $M$ is glueable for $(R \to R', f)$. Let $N$ be a second $R$-module. It suffices to prove that $\text{Tor}_1^ R(M, N) = 0$, see Algebra, Lemma 10.75.8.

The long the exact sequence of Tors associated to the short exact sequence $0 \to R \to R' \oplus R_ f \to (R')_ f \to 0$ and $N$ gives an exact sequence

and isomorphisms $\text{Tor}_ i^ R(R', N) = \text{Tor}_ i^ R((R')_ f, N)$ for $i \geq 2$. Since $\text{Tor}_ i^ R((R')_ f, N) = \text{Tor}_ i^ R(R', N)_ f$ we conclude that $f$ is a nonzerodivisor on $\text{Tor}_1^ R(R', N)$ and invertible on $\text{Tor}_ i^ R(R', N)$ for $i \geq 2$. Since $M \otimes _ R R'$ is flat over $R'$ we have

by the spectral sequence of Example 15.62.2. Writing $M \otimes _ R R'$ as a filtered colimit of finite free $R'$-modules (Algebra, Theorem 10.81.4) we conclude that $f$ is a nonzerodivisor on $\text{Tor}_1^ R(M \otimes _ R R', N)$ and invertible on $\text{Tor}_ i^ R(M \otimes _ R R', N)$. Next, we consider the exact sequence $0 \to M \to M \otimes _ R R' \oplus M_ f \to M \otimes _ R (R')_ f \to 0$ coming from the fact that $M$ is glueable and the associated long exact sequence of $\text{Tor}$. The relevant part is

We conclude that $\text{Tor}_1^ R(M, N) = 0$ by our remarks above on the action on $f$ on $\text{Tor}_ i^ R(M \otimes _ R R', N)$. $\square$

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