Lemma 15.61.6. Let $R$ be a ring. Let $A$, $B$ be $R$-algebras. The following are equivalent

1. $A$ and $B$ are Tor independent over $R$,

2. for every pair of primes $\mathfrak p \subset A$ and $\mathfrak q \subset B$ lying over the same prime $\mathfrak r \subset R$ the rings $A_\mathfrak p$ and $B_\mathfrak q$ are Tor independent over $R_\mathfrak r$, and

3. For every prime $\mathfrak s$ of $A \otimes _ R B$ the module

$\text{Tor}_ i^ R(A, B)_\mathfrak s = \text{Tor}_ i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q)_\mathfrak s$

(where $\mathfrak p = A \cap \mathfrak s$, $\mathfrak q = B \cap \mathfrak s$ and $\mathfrak r = R \cap \mathfrak s$) is zero.

Proof. Let $\mathfrak s$ be a prime of $A \otimes _ R B$ as in (3). The equality

$\text{Tor}_ i^ R(A, B)_\mathfrak s = \text{Tor}_ i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q)_\mathfrak s$

where $\mathfrak p = A \cap \mathfrak s$, $\mathfrak q = B \cap \mathfrak s$ and $\mathfrak r = R \cap \mathfrak s$ follows from Lemma 15.61.3. Hence (2) implies (3). Since we can test the vanishing of modules by localizing at primes (Algebra, Lemma 10.23.1) we conclude that (3) implies (1). For (1) $\Rightarrow$ (2) we use that

$\text{Tor}_ i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q) = \text{Tor}_ i^ R(A, B) \otimes _{(A \otimes _ R B)} (A_\mathfrak p \otimes _{R_{\mathfrak r}} B_\mathfrak q)$

again by Lemma 15.61.3. $\square$

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