Lemma 10.99.13. Let R \to R' be a ring map. Let I \subset R be an ideal and I' = IR'. Let M be an R-module and set M' = M \otimes _ R R'. The natural map \text{Tor}_1^ R(R'/I', M) \to \text{Tor}_1^{R'}(R'/I', M') is surjective.
Proof. Let F_2 \to F_1 \to F_0 \to M \to 0 be a free resolution of M over R. Set F_ i' = F_ i \otimes _ R R'. The sequence F_2' \to F_1' \to F_0' \to M' \to 0 may no longer be exact at F_1'. A free resolution of M' over R' therefore looks like
for a suitable free module F_2'' over R'. Next, note that F_ i \otimes _ R R'/I' = F_ i' / IF_ i' = F_ i'/I'F_ i'. So the complex F_2'/I'F_2' \to F_1'/I'F_1' \to F_0'/I'F_0' computes \text{Tor}_1^ R(M, R'/I'). On the other hand F_ i' \otimes _{R'} R'/I' = F_ i'/I'F_ i' and similarly for F_2''. Thus the complex F_2'/I'F_2' \oplus F_2''/I'F_2'' \to F_1'/I'F_1' \to F_0'/I'F_0' computes \text{Tor}_1^{R'}(M', R'/I'). Since the vertical map on complexes
clearly induces a surjection on cohomology we win. \square
Comments (0)