Lemma 37.69.1. Consider a commutative diagram of schemes

\[ \xymatrix{ Z' \ar[d] \ar[r] & Y' \ar[d] \\ X' \ar[r] & S' } \]

Let $S \to S'$ be a morphism. Denote by $X$ and $Y$ the base changes of $X'$ and $Y'$ to $S$. Assume $Y' \to S'$ and $Z' \to X'$ are flat. Then $X \times _ S Y$ and $Z'$ are Tor independent over $X' \times _{S'} Y'$.

**Proof.**
The question is local, hence we may assume all schemes are affine (some details omitted). Observe that

\[ \xymatrix{ X \times _ S Y \ar[r] \ar[d] & X' \times _{S'} Y' \ar[d] \\ X \ar[r] & X' } \]

is cartesian with flat vertical arrows. Write $X = \mathop{\mathrm{Spec}}(A)$, $X' = \mathop{\mathrm{Spec}}(A')$, $X' \times _{S'} Y' = \mathop{\mathrm{Spec}}(B')$. Then $X \times _ S Y = \mathop{\mathrm{Spec}}(A \otimes _{A'} B')$. Write $Z' = \mathop{\mathrm{Spec}}(C')$. We have to show

\[ \text{Tor}_ p^{B'}(A \otimes _{A'} B', C') = 0, \quad \text{for } p > 0 \]

Since $A' \to B'$ is flat we have $A \otimes _{A'} B' = A \otimes _{A'}^\mathbf {L} B'$. Hence

\[ (A \otimes _{A'} B') \otimes _{B'}^\mathbf {L} C' = (A \otimes _{A'}^\mathbf {L} B') \otimes _{B'}^\mathbf {L} C' = A \otimes _{A'}^\mathbf {L} C' = A \otimes _{A'} C' \]

The second equality by More on Algebra, Lemma 15.60.5. The last equality because $A' \to C'$ is flat. This proves the lemma.
$\square$

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