Definition 15.61.1. Let $R$ be a ring. Let $A$, $B$ be $R$-algebras. We say $A$ and $B$ are *Tor independent over $R$* if $\text{Tor}_ p^ R(A, B) = 0$ for all $p > 0$.

## 15.61 Tor independence

Consider a commutative diagram

of rings. Given an object $K$ of $D(A)$ we can consider its restriction to an object of $D(R)$. We can then consider take the derived change of rings of $K$ to an object of $D(A')$ and $D(R')$. We claim there is a functorial comparison map

in $D(R')$. To construct this comparison map choose a K-flat complex $K^\bullet $ of $A$-modules representing $K$. Next, choose a quasi-isomorphism $E^\bullet \to K^\bullet $ where $E^\bullet $ is a K-flat complex of $R$-modules. The map above is the map

In general there is no chance that this map is an isomorphism.

However, we often encounter the situation where the diagram above is a “base change” diagram of rings, i.e., $A' = A \otimes _ R R'$. In this situation, for any $A$-module $M$ we have $M \otimes _ A A' = M \otimes _ R R'$. Thus $- \otimes _ R R'$ is equal to $- \otimes _ A A'$ as a functor $\text{Mod}_ A \to \text{Mod}_{A'}$. In general this equality **does not extend** to derived tensor products. In other words, the comparison map is not an isomorphism. A simple example is to take $R = k[x]$, $A = R' = A' = k[x]/(x) = k$ and $K^\bullet = A[0]$. Clearly, a necessary condition is that $\text{Tor}_ p^ R(A, R') = 0$ for all $p > 0$.

Lemma 15.61.2. The comparison map (15.61.0.1) is an isomorphism if $A' = A \otimes _ R R'$ and $A$ and $R'$ are Tor independent over $R$.

**Proof.**
To prove this we choose a free resolution $F^\bullet \to R'$ of $R'$ as an $R$-module. Because $A$ and $R'$ are Tor independent over $R$ we see that $F^\bullet \otimes _ R A$ is a free $A$-module resolution of $A'$ over $A$. By our general construction of the derived tensor product above we see that

as desired. $\square$

Lemma 15.61.3. 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

**Proof.**
By Algebra, Section 10.76 there are canonical maps

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 10.81.4) 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

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

Lemma 15.61.4. Let $R \to A$ and $R \to B$ be ring maps. Let $R \to R'$ be a ring map and set $A' = A \otimes _ R R'$ and $B' = B \otimes _ R R'$. If $A$ and $B$ are tor independent over $R$ and $R \to R'$ is flat, then $A'$ and $B'$ are tor independent over $R'$.

**Proof.**
Follows immediately from Lemma 15.61.3 and Definition 15.61.1.
$\square$

Lemma 15.61.5. Assumptions as in Lemma 15.61.3. For $M \in D(A)$ there are canonical isomorphisms

of $A' \otimes _{R'} B'$-modules.

**Proof.**
Let us elucidate the two sides of the equation. On the left hand side we have the composition of the functors $D(A) \to D(A') \to D(R') \to D(B')$ with the functor $H^ i : D(B') \to \text{Mod}_{B'}$. Since there is a map from $A'$ to the endomorphisms of the object $(M \otimes _ A^\mathbf {L} A') \otimes _{R'}^\mathbf {L} B'$ in $D(B')$, we see that the left hand side is indeed an $A' \otimes _{R'} B'$-module. By the same arguments we see that $H^ i(M \otimes _ R^\mathbf {L} B)$ has an $A \otimes _ R B$-module structure.

We first prove the result in case $B' = R' \otimes _ R B$. In this case we choose a resolution $F^\bullet \to B$ by free $R$-modules. We also choose a K-flat complex $M^\bullet $ of $A$-modules representing $M$. Then the left hand side is represented by

The final equality because $A \to A'$ is flat. The final module is the desired module because $A' \otimes _{R'} B' = A' \otimes _ R B$ since we've assumed $B' = R' \otimes _ R B$ in this paragraph.

General case. Suppose that $B' \to B''$ is a flat ring map. Then it is easy to see that

and

Thus the result for $B'$ implies the result for $B''$. Since we've proven the result for $R' \otimes _ R B$ in the previous paragraph, this implies the result in general. $\square$

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

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

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

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

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

again by Lemma 15.61.3. $\square$

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Comment #7765 by Andrea Panontin on