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. $\square$

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