Lemma 10.98.14. Let

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

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

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

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

$\text{Tor}_1^ R(M, R/I) \to \text{Tor}_1^{R'}(M', R'/I')$ 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 10.38.7 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 10.98.10 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 10.98.13 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 10.98.12 by looking at the ring maps $R \to R/I \to R'/I'$ and the module $M$.
$\square$

## Comments (2)

Comment #1218 by JuanPablo on

Comment #1246 by Johan on