Lemma 10.99.14. 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 10.39.7 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
Comments (2)
Comment #1218 by JuanPablo on
Comment #1246 by Johan on