## 15.63 Products and Tor

The simplest example of the product maps comes from the following situation. Suppose that $K^\bullet , L^\bullet \in D(R)$. Then there are maps

15.63.0.1
\begin{equation} \label{more-algebra-equation-simple-tor-product} H^ i(K^\bullet ) \otimes _ R H^ j(L^\bullet ) \longrightarrow H^{i + j}(K^\bullet \otimes _ R^{\mathbf{L}} L^\bullet ) \end{equation}

Namely, to define these maps we may assume that one of $K^\bullet , L^\bullet$ is a K-flat complex of $R$-modules (for example a bounded above complex of free or projective $R$-modules). In that case $K^\bullet \otimes _ R^{\mathbf{L}} L^\bullet$ is represented by the complex $\text{Tot}(K^\bullet \otimes _ R L^\bullet )$, see Section 15.59 (or Section 15.57). Next, suppose that $\xi \in H^ i(K^\bullet )$ and $\zeta \in H^ j(L^\bullet )$. Choose $k \in \mathop{\mathrm{Ker}}(K^ i \to K^{i + 1})$ and $l \in \mathop{\mathrm{Ker}}(L^ j \to L^{j + 1})$ representing $\xi$ and $\zeta$. Then we set

$\xi \cup \zeta = \text{class of }k \otimes l\text{ in } H^{i + j}(\text{Tot}(K^\bullet \otimes _ R L^\bullet )).$

This make sense because the formula (see Homology, Definition 12.18.3) for the differential $\text{d}$ on the total complex shows that $k \otimes l$ is a cocycle. Moreover, if $k' = d_ K(k'')$ for some $k'' \in K^{i - 1}$, then $k' \otimes l = \text{d}(k'' \otimes l)$ because $l$ is a cocycle. Similarly, altering the choice of $l$ representing $\zeta$ does not change the class of $k \otimes l$. It is equally clear that $\cup$ is bilinear, and hence to a general element of $H^ i(K^\bullet ) \otimes _ R H^ j(L^\bullet )$ we assign

$\sum \xi _ i \otimes \zeta _ i \longmapsto \sum \xi _ i \cup \zeta _ i$

in $H^{i + j}(\text{Tot}(K^\bullet \otimes _ R L^\bullet ))$.

Let $R \to A$ be a ring map. Let $K^\bullet , L^\bullet \in D(R)$. Then we have a canonical identification

15.63.0.2
\begin{equation} \label{more-algebra-equation-pullback-derived-tensor-product} (K^\bullet \otimes _ R^{\mathbf{L}} A) \otimes _ A^{\mathbf{L}} (L^\bullet \otimes _ R^{\mathbf{L}} A) = (K^\bullet \otimes _ R^{\mathbf{L}} L^\bullet ) \otimes _ R^{\mathbf{L}} A \end{equation}

in $D(A)$. It is constructed as follows. First, choose K-flat resolutions $P^\bullet \to K^\bullet$ and $Q^\bullet \to L^\bullet$ over $R$. Then the left hand side is represented by the complex $\text{Tot}((P^\bullet \otimes _ R A) \otimes _ A (Q^\bullet \otimes _ R A))$ and the right hand side by the complex $\text{Tot}(P^\bullet \otimes _ R Q^\bullet ) \otimes _ R A$. These complexes are canonically isomorphic. Thus the construction above induces products

$\text{Tor}^ R_ n(K^\bullet , A) \otimes _ A \text{Tor}^ R_ m(L^\bullet , A) \longrightarrow \text{Tor}_{n + m}^ R(K^\bullet \otimes _ R^\mathbf {L} L^\bullet , A)$

which are occasionally useful.

Let $M$, $N$ be $R$-modules. Using the general construction above, the canonical map $M \otimes _ R^\mathbf {L} N \to M \otimes _ R N$ and functoriality of $\text{Tor}$ we obtain canonical maps

15.63.0.3
\begin{equation} \label{more-algebra-equation-tor-product} \text{Tor}^ R_ n(M, A) \otimes _ A \text{Tor}^ R_ m(N, A) \longrightarrow \text{Tor}_{n + m}^ R(M \otimes _ R N, A) \end{equation}

Here is a direct construction using projective resolutions. First, choose projective resolutions

$P_\bullet \to M, \quad Q_\bullet \to N, \quad T_\bullet \to M \otimes _ R N$

over $R$. We have $H_0(\text{Tot}(P_\bullet \otimes _ R Q_\bullet )) = M \otimes _ R N$ by right exactness of $\otimes _ R$. Hence Derived Categories, Lemmas 13.19.6 and 13.19.7 guarantee the existence and uniqueness of a map of complexes $\mu : \text{Tot}(P_\bullet \otimes _ R Q_\bullet ) \to T_\bullet$ such that $H_0(\mu ) = \text{id}_{M \otimes _ R N}$. This induces a canonical map

\begin{align*} (M \otimes _ R^{\mathbf{L}} A) \otimes _ A^{\mathbf{L}} (N \otimes _ R^{\mathbf{L}} A) & = \text{Tot}((P_\bullet \otimes _ R A) \otimes _ A (Q_\bullet \otimes _ R A)) \\ & = \text{Tot}(P_\bullet \otimes _ R Q_\bullet ) \otimes _ R A \\ & \to T_\bullet \otimes _ R A \\ & = (M \otimes _ R N) \otimes _ R^{\mathbf{L}} A \end{align*}

in $D(A)$. Hence the products (15.63.0.3) above are constructed using (15.63.0.1) over $A$ to construct

$\text{Tor}^ R_ n(M, A) \otimes _ A \text{Tor}^ R_ m(N, A) \to H^{-n-m}((M \otimes _ R^{\mathbf{L}} A) \otimes _ A^{\mathbf{L}} (N \otimes _ R^{\mathbf{L}} A))$

and then composing by the displayed map above to end up in $\text{Tor}_{n + m}^ R(M \otimes _ R N, A)$.

An interesting special case of the above occurs when $M = N = B$ where $B$ is an $R$-algebra. In this case we obtain maps

$\text{Tor}_ n^ R(B, A) \otimes _ A \text{Tor}_ m^ R(B, A) \longrightarrow \text{Tor}_{n + m}^ R(B \otimes _ R B, A) \longrightarrow \text{Tor}_{n + m}^ R(B, A)$

the second arrow being induced by the multiplication map $B \otimes _ R B \to B$ via functoriality for $\text{Tor}$. In other words we obtain an $A$-algebra structure on $\text{Tor}^ R_{\star }(B, A)$. This algebra structure has many intriguing properties (associativity, graded commutative, $B$-algebra structure, divided powers in some case, etc) which we will discuss elsewhere (insert future reference here).

Lemma 15.63.1. Let $R$ be a ring. Let $A, B, C$ be $R$-algebras and let $B \to C$ be an $R$-algebra map. Then the induced map

$\text{Tor}^ R_{\star }(B, A) \longrightarrow \text{Tor}^ R_{\star }(C, A)$

is an $A$-algebra homomorphism.

Proof. Omitted. Hint: You can prove this by working through the definitions, writing all the complexes explicitly. $\square$

Comment #5792 by Jinhyun Park on

It appears to me that there is a typo. In the displayed equation in the paragraph just above Lemma 068K, the Tor's subscrip indices of the 2nd and the 3rd term should be n+m, not n.

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