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

By the assumption that $M \otimes _ R R'$ is flat over $R'$, the sequence

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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