## 15.57 Derived tensor product

We can construct the derived tensor product in greater generality. In fact, it turns out that the boundedness assumptions are not necessary, provided we choose K-flat resolutions. In this section we use Homology, Example 12.22.2 and Homology, Definition 12.22.3 to turn a pair of complexes of modules into a double complex and its associated total complex.

Lemma 15.57.1. Let $R$ be a ring. Let $P^\bullet $ be a complex of $R$-modules. Let $\alpha , \beta : L^\bullet \to M^\bullet $ be homotopy equivalent maps of complexes. Then $\alpha $ and $\beta $ induce homotopy equivalent maps

\[ \text{Tot}(\alpha \otimes \text{id}_ P), \text{Tot}(\beta \otimes \text{id}_ P) : \text{Tot}(L^\bullet \otimes _ R P^\bullet ) \longrightarrow \text{Tot}(M^\bullet \otimes _ R P^\bullet ). \]

In particular the construction $L^\bullet \mapsto \text{Tot}(L^\bullet \otimes _ R P^\bullet )$ defines an endo-functor of the homotopy category of complexes.

**Proof.**
Say $\alpha = \beta + dh + hd$ for some homotopy $h$ defined by $h^ n : L^ n \to M^{n - 1}$. Set

\[ H^ n = \bigoplus \nolimits _{a + b = n} h^ a \otimes \text{id}_{P^ b} : \bigoplus \nolimits _{a + b = n} L^ a \otimes _ R P^ b \longrightarrow \bigoplus \nolimits _{a + b = n} M^{a - 1} \otimes _ R P^ b \]

Then a straightforward computation shows that

\[ \text{Tot}(\alpha \otimes \text{id}_ P) = \text{Tot}(\beta \otimes \text{id}_ P) + dH + Hd \]

as maps $\text{Tot}(L^\bullet \otimes _ R P^\bullet ) \to \text{Tot}(M^\bullet \otimes _ R P^\bullet )$.
$\square$

Lemma 15.57.2. Let $R$ be a ring. Let $P^\bullet $ be a complex of $R$-modules. The functor

\[ K(\text{Mod}_ R) \longrightarrow K(\text{Mod}_ R), \quad L^\bullet \longmapsto \text{Tot}(L^\bullet \otimes _ R P^\bullet ) \]

is an exact functor of triangulated categories.

**Proof.**
By our definition of the triangulated structure on $K(\text{Mod}_ R)$ we have to check that our functor maps a termwise split short exact sequence of complexes to a termwise split short exact sequence of complexes. As the terms of $\text{Tot}(L^\bullet \otimes _ R P^\bullet )$ are direct sums of the tensor products $L^ a \otimes _ R P^ b$ this is clear.
$\square$

The following definition will allow us to think intelligently about derived tensor products of unbounded complexes.

Definition 15.57.3. Let $R$ be a ring. A complex $K^\bullet $ is called *K-flat* if for every acyclic complex $M^\bullet $ the total complex $\text{Tot}(M^\bullet \otimes _ R K^\bullet )$ is acyclic.

Lemma 15.57.4. Let $R$ be a ring. Let $K^\bullet $ be a K-flat complex. Then the functor

\[ K(\text{Mod}_ R) \longrightarrow K(\text{Mod}_ R), \quad L^\bullet \longmapsto \text{Tot}(L^\bullet \otimes _ R K^\bullet ) \]

transforms quasi-isomorphisms into quasi-isomorphisms.

**Proof.**
Follows from Lemma 15.57.2 and the fact that quasi-isomorphisms in $K(\text{Mod}_ R)$ and $K(\text{Mod}_ A)$ are characterized by having acyclic cones.
$\square$

Lemma 15.57.5. Let $R \to R'$ be a ring map. If $K^\bullet $ is a K-flat complex of $R$-modules, then $K^\bullet \otimes _ R R'$ is a K-flat complex of $R'$-modules.

**Proof.**
Follows from the definitions and the fact that $(K^\bullet \otimes _ R R') \otimes _{R'} L^\bullet = K^\bullet \otimes _ R L^\bullet $ for any complex $L^\bullet $ of $R'$-modules.
$\square$

Lemma 15.57.6. Let $R$ be a ring. If $K^\bullet $, $L^\bullet $ are K-flat complexes of $R$-modules, then $\text{Tot}(K^\bullet \otimes _ R L^\bullet )$ is a K-flat complex of $R$-modules.

**Proof.**
Follows from the isomorphism

\[ \text{Tot}(M^\bullet \otimes _ R \text{Tot}(K^\bullet \otimes _ R L^\bullet )) = \text{Tot}(\text{Tot}(M^\bullet \otimes _ R K^\bullet ) \otimes _ R L^\bullet ) \]

and the definition.
$\square$

Lemma 15.57.7. Let $R$ be a ring. Let $(K_1^\bullet , K_2^\bullet , K_3^\bullet )$ be a distinguished triangle in $K(\text{Mod}_ R)$. If two out of three of $K_ i^\bullet $ are K-flat, so is the third.

**Proof.**
Follows from Lemma 15.57.2 and the fact that in a distinguished triangle in $K(\text{Mod}_ A)$ if two out of three are acyclic, so is the third.
$\square$

Lemma 15.57.8. Let $R$ be a ring. Let $0 \to K_1^\bullet \to K_2^\bullet \to K_3^\bullet \to 0$ be a short exact sequence of complexes. If $K_3^ n$ is flat for all $n \in \mathbf{Z}$ and two out of three of $K_ i^\bullet $ are K-flat, so is the third.

**Proof.**
Let $L^\bullet $ be a complex of $R$-modules. Then

\[ 0 \to \text{Tot}(L^\bullet \otimes _ R K_1^\bullet ) \to \text{Tot}(L^\bullet \otimes _ R K_2^\bullet ) \to \text{Tot}(L^\bullet \otimes _ R K_3^\bullet ) \to 0 \]

is a short exact sequence of complexes. Namely, for each $n, m$ the sequence of modules $0 \to L^ n \otimes _ R K_1^ m \to L^ n \otimes _ R K_2^ m \to L^ n \otimes _ R K_3^ m \to 0$ is exact by Algebra, Lemma 10.38.12 and the sequence of complexes is a direct sum of these. Thus the lemma follows from this and the fact that in a short exact sequence of complexes if two out of three are acyclic, so is the third.
$\square$

Lemma 15.57.9. Let $R$ be a ring. Let $P^\bullet $ be a bounded above complex of flat $R$-modules. Then $P^\bullet $ is K-flat.

**Proof.**
Let $L^\bullet $ be an acyclic complex of $R$-modules. Let $\xi \in H^ n(\text{Tot}(L^\bullet \otimes _ R P^\bullet ))$. We have to show that $\xi = 0$. Since $\text{Tot}^ n(L^\bullet \otimes _ R P^\bullet )$ is a direct sum with terms $L^ a \otimes _ R P^ b$ we see that $\xi $ comes from an element in $H^ n(\text{Tot}(\tau _{\leq m}L^\bullet \otimes _ R P^\bullet ))$ for some $m \in \mathbf{Z}$. Since $\tau _{\leq m}L^\bullet $ is also acyclic we may replace $L^\bullet $ by $\tau _{\leq m}L^\bullet $. Hence we may assume that $L^\bullet $ is bounded above. In this case the spectral sequence of Homology, Lemma 12.22.6 has

\[ {}'E_1^{p, q} = H^ p(L^\bullet \otimes _ R P^ q) \]

which is zero as $P^ q$ is flat and $L^\bullet $ acyclic. Hence $H^*(\text{Tot}(L^\bullet \otimes _ R P^\bullet )) = 0$.
$\square$

In the following lemma by a colimit of a system of complexes we mean the termwise colimit.

Lemma 15.57.10. Let $R$ be a ring. Let $K_1^\bullet \to K_2^\bullet \to \ldots $ be a system of K-flat complexes. Then $\mathop{\mathrm{colim}}\nolimits _ i K_ i^\bullet $ is K-flat. More generally any filtered colimit of K-flat complexes is K-flat.

**Proof.**
Because we are taking termwise colimits we have

\[ \mathop{\mathrm{colim}}\nolimits _ i \text{Tot}(M^\bullet \otimes _ R K_ i^\bullet ) = \text{Tot}(M^\bullet \otimes _ R \mathop{\mathrm{colim}}\nolimits _ i K_ i^\bullet ) \]

by Algebra, Lemma 10.11.9. Hence the lemma follows from the fact that filtered colimits are exact, see Algebra, Lemma 10.8.8.
$\square$

Lemma 15.57.11. Let $R$ be a ring. Let $K^\bullet $ be a complex of $R$-modules. If $K^\bullet \otimes _ R M$ is acyclic for all finitely presented $R$-modules $M$, then $K^\bullet $ is K-flat.

**Proof.**
We will use repeatedly that tensor product commute with colimits (Algebra, Lemma 10.11.9). Thus we see that $K^\bullet \otimes _ R M$ is acyclic for any $R$-module $M$, because any $R$-module is a filtered colimit of finitely presented $R$-modules $M$, see Algebra, Lemma 10.8.12. Let $M^\bullet $ be an acyclic complex of $R$-modules. We have to show that $\text{Tot}(M^\bullet \otimes _ R K^\bullet )$ is acyclic. Since $M^\bullet = \mathop{\mathrm{colim}}\nolimits \tau _{\leq n} M^\bullet $ (termwise colimit) we have

\[ \text{Tot}(M^\bullet \otimes _ R K^\bullet ) = \mathop{\mathrm{colim}}\nolimits \text{Tot}(\tau _{\leq n} M^\bullet \otimes _ R K^\bullet ) \]

with truncations as in Homology, Section 12.14. As filtered colimits are exact (Algebra, Lemma 10.8.8) we may replace $M^\bullet $ by $\tau _{\leq n}M^\bullet $ and assume that $M^\bullet $ is bounded above. In the bounded above case, we can write $M^\bullet = \mathop{\mathrm{colim}}\nolimits \sigma _{\geq -n} M^\bullet $ where the complexes $\sigma _{\geq -n} M^\bullet $ are bounded but possibly no longer acyclic. Arguing as above we reduce to the case where $M^\bullet $ is a bounded complex. Finally, for a bounded complex $M^ a \to \ldots \to M^ b$ we can argue by induction on the length $b - a$ of the complex. The case $b - a = 1$ we have seen above. For $b - a > 1$ we consider the split short exact sequence of complexes

\[ 0 \to \sigma _{\geq a + 1}M^\bullet \to M^\bullet \to M^ a[-a] \to 0 \]

and we apply Lemma 15.57.2 to do the induction step. Some details omitted.
$\square$

Lemma 15.57.12. Let $R$ be a ring. For any complex $M^\bullet $ there exists a K-flat complex $K^\bullet $ and a quasi-isomorphism $K^\bullet \to M^\bullet $. Moreover each $K^ n$ is a flat $R$-module.

**Proof.**
Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\text{Mod}_ R)$ be the class of flat $R$-modules. By Derived Categories, Lemma 13.28.1 there exists a system $K_1^\bullet \to K_2^\bullet \to \ldots $ and a diagram

\[ \xymatrix{ K_1^\bullet \ar[d] \ar[r] & K_2^\bullet \ar[d] \ar[r] & \ldots \\ \tau _{\leq 1}M^\bullet \ar[r] & \tau _{\leq 2}M^\bullet \ar[r] & \ldots } \]

with the properties (1), (2), (3) listed in that lemma. These properties imply each complex $K_ i^\bullet $ is a bounded above complex of flat modules. Hence $K_ i^\bullet $ is K-flat by Lemma 15.57.9. The induced map $\mathop{\mathrm{colim}}\nolimits _ i K_ i^\bullet \to M^\bullet $ is a quasi-isomorphism by construction. The complex $\mathop{\mathrm{colim}}\nolimits _ i K_ i^\bullet $ is K-flat by Lemma 15.57.10. The final assertion of the lemma is true because the colimit of a system of flat modules is flat, see Algebra, Lemma 10.38.3.
$\square$

Lemma 15.57.14. Let $R$ be a ring. Let $\alpha : P^\bullet \to Q^\bullet $ be a quasi-isomorphism of K-flat complexes of $R$-modules. For every complex $L^\bullet $ of $R$-modules the induced map

\[ \text{Tot}(\text{id}_ L \otimes \alpha ) : \text{Tot}(L^\bullet \otimes _ R P^\bullet ) \longrightarrow \text{Tot}(L^\bullet \otimes _ R Q^\bullet ) \]

is a quasi-isomorphism.

**Proof.**
Choose a quasi-isomorphism $K^\bullet \to L^\bullet $ with $K^\bullet $ a K-flat complex, see Lemma 15.57.12. Consider the commutative diagram

\[ \xymatrix{ \text{Tot}(K^\bullet \otimes _ R P^\bullet ) \ar[r] \ar[d] & \text{Tot}(K^\bullet \otimes _ R Q^\bullet ) \ar[d] \\ \text{Tot}(L^\bullet \otimes _ R P^\bullet ) \ar[r] & \text{Tot}(L^\bullet \otimes _ R Q^\bullet ) } \]

The result follows as by Lemma 15.57.4 the vertical arrows and the top horizontal arrow are quasi-isomorphisms.
$\square$

Let $R$ be a ring. Let $M^\bullet $ be an object of $D(R)$. Choose a K-flat resolution $K^\bullet \to M^\bullet $, see Lemma 15.57.12. By Lemmas 15.57.1 and 15.57.2 we obtain an exact functor of triangulated categories

\[ K(\text{Mod}_ R) \longrightarrow K(\text{Mod}_ R), \quad L^\bullet \longmapsto \text{Tot}(L^\bullet \otimes _ R K^\bullet ) \]

By Lemma 15.57.4 this functor induces a functor $D(R) \to D(R)$ simply because $D(R)$ is the localization of $K(\text{Mod}_ R)$ at quasi-isomorphism. By Lemma 15.57.14 the resulting functor (up to isomorphism) does not depend on the choice of the K-flat resolution.

Definition 15.57.15. Let $R$ be a ring. Let $M^\bullet $ be an object of $D(R)$. The *derived tensor product*

\[ - \otimes _ R^{\mathbf{L}} M^\bullet : D(R) \longrightarrow D(R) \]

is the exact functor of triangulated categories described above.

This functor extends the functor (15.56.0.1). It is clear from our explicit constructions that there is an isomorphism (involving a choice of signs, see below)

\[ M^\bullet \otimes _ R^{\mathbf{L}} L^\bullet \cong L^\bullet \otimes _ R^{\mathbf{L}} M^\bullet \]

whenever both $L^\bullet $ and $M^\bullet $ are in $D(R)$. Hence when we write $M^\bullet \otimes _ R^{\mathbf{L}} L^\bullet $ we will usually be agnostic about which variable we are using to define the derived tensor product with.

Lemma 15.57.16. Let $R$ be a ring. Let $K^\bullet , L^\bullet $ be complexes of $R$-modules. There is a canonical isomorphism

\[ K^\bullet \otimes _ R^\mathbf {L} L^\bullet \longrightarrow L^\bullet \otimes _ R^\mathbf {L} K^\bullet \]

functorial in both complexes which uses a sign of $(-1)^{pq}$ for the map $K^ p \otimes _ R L^ q \to L^ q \otimes _ R K^ p$ (see proof for explanation).

**Proof.**
Replace the complexes by K-flat complexes $K^\bullet , L^\bullet $. Then we consider the map

\[ \text{Tot}(K^\bullet \otimes _ R L^\bullet ) \longrightarrow \text{Tot}(L^\bullet \otimes _ R K^\bullet ) \]

given by using $(-1)^{pq}$ times the canonical map $K^ p \otimes _ R L^ q \to L^ q \otimes _ R K^ p$. This is an isomorphism. To see that it is a map of complexes we compute for $x \in K^ p$ and $y \in L^ q$ that

\[ \text{d}(x \otimes y) = \text{d}_ K(x) \otimes y + (-1)^ px \otimes \text{d}_ L(y) \]

Our rule says the right hand side is mapped to

\[ (-1)^{(p + 1)q}y \otimes \text{d}_ K(x) + (-1)^{p + p(q + 1)} \text{d}_ L(y) \otimes x \]

On the other hand, we see that

\[ \text{d}((-1)^{pq}y \otimes x) = (-1)^{pq} \text{d}_ L(y) \otimes x + (-1)^{pq + q} y \otimes \text{d}_ K(x) \]

These two expressions agree by inspection and the lemma is proved.
$\square$

Lemma 15.57.17. Let $R$ be a ring. Let $K^\bullet , L^\bullet , M^\bullet $ be complexes of $R$-modules. There is a canonical isomorphism

\[ (K^\bullet \otimes _ R^\mathbf {L} L^\bullet ) \otimes _ R^\mathbf {L} M^\bullet = K^\bullet \otimes _ R^\mathbf {L} (L^\bullet \otimes _ R^\mathbf {L} M^\bullet ) \]

functorial in all three complexes.

**Proof.**
Replace the complexes by K-flat complexes and apply Homology, Remark 12.22.8.
$\square$

## Comments (2)

Comment #3061 by shanbei on

Comment #3165 by Johan on