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
\[ \xymatrix{ A \ar[r] & A' \\ R \ar[r] \ar[u] & R' \ar[u] } \]
of rings. Given an object $K$ of $D(A)$ we can consider its derived base change $K \otimes _ A^\mathbf {L} A'$ to an object of $D(A')$. Or we can take the restriction of $K$ to an object of $D(R)$ and consider the derived base change of this to an object of $D(R')$, denoted $K \otimes _ R^\mathbf {L} R'$ We claim there is a functorial comparison map
15.61.0.1
\begin{equation} \label{more-algebra-equation-comparison-map} K \otimes _ R^{\mathbf{L}} R' \longrightarrow K \otimes _ A^{\mathbf{L}} A' \end{equation}
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
\[ K \otimes _ R^{\mathbf{L}} R' = E^\bullet \otimes _ R R' \longrightarrow K^\bullet \otimes _ A A' = K \otimes _ A^{\mathbf{L}} A' \]
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
\[ K^\bullet \otimes _ A A' \cong \text{Tot}(K^\bullet \otimes _ A (F^\bullet \otimes _ R A)) = \text{Tot}(K^\bullet \otimes _ R F^\bullet ) \cong \text{Tot}(E^\bullet \otimes _ R F^\bullet ) \cong E^\bullet \otimes _ R R' \]
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
\[ \xymatrix{ A' & R' \ar[r] \ar[l] & B' \\ A \ar[u] & R \ar[l] \ar[u] \ar[r] & B \ar[u] } \]
\[ \text{Tor}_ i^ R(A, B) \otimes _{(A \otimes _ R B)} (A' \otimes _{R'} B') = \text{Tor}_ i^{R'}(A', B') \]
Proof. By Algebra, Section 10.76 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 10.81.4) 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
\[ \text{Tor}_ i^{R'}(M_ i, N_ j) = \text{Tor}_ i^ R(A, B) \otimes _{A \otimes _ R B} (M_ i \otimes _{R'} N_ j) \]
as one can see by writing everything out in terms of bases. Taking the colimit we get the result of the lemma. $\square$
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.
\[ H^ i((M \otimes _ A^\mathbf {L} A') \otimes _{R'}^\mathbf {L} B') = H^ i(M \otimes _ R^\mathbf {L} B) \otimes _{(A \otimes _ R B)} (A' \otimes _{R'} B') \]
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
\begin{align*} H^ i(\text{Tot}((M^\bullet \otimes _ A A') \otimes _{R'} (R' \otimes _ R F^\bullet ))) & = H^ i(\text{Tot}(M^\bullet \otimes _ A A' \otimes _ R F^\bullet )) \\ & = H^ i(\text{Tot}(M^\bullet \otimes _ R F^\bullet ) \otimes _ A A') \\ & = H^ i(M \otimes _ R^\mathbf {L} B) \otimes _ A A' \end{align*}
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
\[ H^ i((M \otimes _ A^\mathbf {L} A') \otimes _{R'}^\mathbf {L} B'') = H^ i((M \otimes _ A^\mathbf {L} A') \otimes _{R'}^\mathbf {L} B') \otimes _{B'} B'' \]
and
\[ H^ i(M \otimes _ R^\mathbf {L} B) \otimes _{(A \otimes _ R B)} (A' \otimes _{R'} B'') = \left( H^ i(M \otimes _ R^\mathbf {L} B) \otimes _{(A \otimes _ R B)} (A' \otimes _{R'} B') \right) \otimes _{B'} B'' \]
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
(where $\mathfrak p = A \cap \mathfrak s$, $\mathfrak q = B \cap \mathfrak s$ and $\mathfrak r = R \cap \mathfrak s$) is zero.
\[ \text{Tor}_ i^ R(A, B)_\mathfrak s = \text{Tor}_ i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q)_\mathfrak s \]
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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