$\xymatrix{ S \ar[r] & S' \\ R \ar[r] \ar[u] & R' \ar[u] }$

be a commutative diagram of local homomorphisms of local Noetherian rings. Let $I \subset R$ be a proper ideal. Let $M$ be a finite $S$-module. Denote $I' = IR'$ and $M' = M \otimes _ S S'$. Assume that

1. $S'$ is a localization of the tensor product $S \otimes _ R R'$,

2. $M/IM$ is flat over $R/I$,

3. $\text{Tor}_1^ R(M, R/I) \to \text{Tor}_1^{R'}(M', R'/I')$ is zero.

Then $M'$ is flat over $R'$.

Proof. Since $S'$ is a localization of $S \otimes _ R R'$ we see that $M'$ is a localization of $M \otimes _ R R'$. Note that by Lemma 10.39.7 the module $M/IM \otimes _{R/I} R'/I' = M \otimes _ R R' /I'(M \otimes _ R R')$ is flat over $R'/I'$. Hence also $M'/I'M'$ is flat over $R'/I'$ as the localization of a flat module is flat. By Lemma 10.99.10 it suffices to show that $\text{Tor}_1^{R'}(M', R'/I')$ is zero. Since $M'$ is a localization of $M \otimes _ R R'$, the last assumption implies that it suffices to show that $\text{Tor}_1^ R(M, R/I) \otimes _ R R' \to \text{Tor}_1^{R'}(M \otimes _ R R', R'/I')$ is surjective.

By Lemma 10.99.13 we see that $\text{Tor}_1^ R(M, R'/I') \to \text{Tor}_1^{R'}(M \otimes _ R R', R'/I')$ is surjective. So now it suffices to show that $\text{Tor}_1^ R(M, R/I) \otimes _ R R' \to \text{Tor}_1^ R(M, R'/I')$ is surjective. This follows from Lemma 10.99.12 by looking at the ring maps $R \to R/I \to R'/I'$ and the module $M$. $\square$

Comment #1218 by JuanPablo on

In the statement here $I$ should be a proper ideal of $R$, and $M'$ a finite $S'$-module (for lemma 10.95.10, tag 00ML, to apply).

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).