Lemma 15.60.3. Consider a commutative diagram of rings

Assume that $R'$ is flat over $R$ and $A'$ is flat over $A \otimes _ R R'$ and $B'$ is flat over $R' \otimes _ R B$. Then

Lemma 15.60.3. Consider a commutative diagram of rings

\[ \xymatrix{ A' & R' \ar[r] \ar[l] & B' \\ A \ar[u] & R \ar[l] \ar[u] \ar[r] & B \ar[u] } \]

Assume that $R'$ is flat over $R$ and $A'$ is flat over $A \otimes _ R R'$ and $B'$ is flat over $R' \otimes _ R B$. Then

\[ \text{Tor}_ i^ R(A, B) \otimes _{(A \otimes _ R B)} (A' \otimes _{R'} B') = \text{Tor}_ i^{R'}(A', B') \]

**Proof.**
By Algebra, Section 10.75 there are canonical maps

\[ \text{Tor}_ i^ R(A, B) \longrightarrow \text{Tor}_ i^{R'}(A \otimes _ R R', B \otimes _ R R') \longrightarrow \text{Tor}_ i^{R'}(A', B') \]

These induce a map from left to right in the formula of the lemma.

Take a free resolution $F_\bullet \to A$ of $A$ as an $R$-module. Then we see that $F_\bullet \otimes _ R R'$ is a resolution of $A \otimes _ R R'$. Hence $\text{Tor}_ i^{R'}(A \otimes _ R R', B \otimes _ R R')$ is computed by $F_\bullet \otimes _ R B \otimes _ R R'$. By our assumption that $R'$ is flat over $R$, this computes $\text{Tor}_ i^ R(A, B) \otimes _ R R'$. Thus $\text{Tor}_ i^{R'}(A \otimes _ R R', B \otimes _ R R') = \text{Tor}_ i^ R(A, B) \otimes _ R R'$ (uses only flatness of $R'$ over $R$).

By Lazard's theorem (Algebra, Theorem 10.80.4) we can write $A'$, resp. $B'$ as a filtered colimit of finite free $A \otimes _ R R'$, resp. $B \otimes _ R R'$-modules. Say $A' = \mathop{\mathrm{colim}}\nolimits M_ i$ and $B' = \mathop{\mathrm{colim}}\nolimits N_ j$. The result above gives

\[ \text{Tor}_ i^{R'}(M_ i, N_ j) = \text{Tor}_ i^ R(A, B) \otimes _{A \otimes _ R B} (M_ i \otimes _{R'} N_ j) \]

as one can see by writing everything out in terms of bases. Taking the colimit we get the result of the lemma. $\square$

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