Lemma 10.99.12. Let R \to R' \to R'' be ring maps. Let M be an R-module. Suppose that M \otimes _ R R' is flat over R'. Then the natural map \text{Tor}_1^ R(M, R') \otimes _{R'} R'' \to \text{Tor}_1^ R(M, R'') is onto.
Proof. Let F_\bullet be a free resolution of M over R. The complex F_2 \otimes _ R R' \to F_1\otimes _ R R' \to F_0 \otimes _ R R' computes \text{Tor}_1^ R(M, R'). The complex F_2 \otimes _ R R'' \to F_1\otimes _ R R'' \to F_0 \otimes _ R R'' computes \text{Tor}_1^ R(M, R''). Note that F_ i \otimes _ R R' \otimes _{R'} R'' = F_ i \otimes _ R R''. Let K' = \mathop{\mathrm{Ker}}(F_1\otimes _ R R' \to F_0 \otimes _ R R') and similarly K'' = \mathop{\mathrm{Ker}}(F_1\otimes _ R R'' \to F_0 \otimes _ R R''). Thus we have an exact sequence
0 \to K' \to F_1\otimes _ R R' \to F_0 \otimes _ R R' \to M \otimes _ R R' \to 0.
By the assumption that M \otimes _ R R' is flat over R', the sequence
K' \otimes _{R'} R'' \to F_1 \otimes _ R R'' \to F_0 \otimes _ R R'' \to M \otimes _ R R'' \to 0
is still exact. This means that K' \otimes _{R'} R'' \to K'' is surjective. Since \text{Tor}_1^ R(M, R') is a quotient of K' and \text{Tor}_1^ R(M, R'') is a quotient of K'' we win. \square
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