## 15.58 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 12.18.2 and Homology, Definition 12.18.3 to turn a pair of complexes of modules into a double complex and its associated total complex.

Lemma 15.58.1. 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 15.58.2. Let $R$ be a ring. Let $P^\bullet$ be a complex of $R$-modules. The functors

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

and

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

are exact functors of triangulated categories.

Proof. This follows from Derived Categories, Remark 13.10.9. $\square$

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

Definition 15.58.3. 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 15.58.4. 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 15.58.2 and the fact that quasi-isomorphisms in $K(\text{Mod}_ R)$ and $K(\text{Mod}_ A)$ are characterized by having acyclic cones. $\square$

Lemma 15.58.5. 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 15.58.6. 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 15.58.7. 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 15.58.2 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 15.58.8. 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 10.39.12 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 15.58.9. 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 12.25.3 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 15.58.10. 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 10.12.9. Hence the lemma follows from the fact that filtered colimits are exact, see Algebra, Lemma 10.8.8. $\square$

Lemma 15.58.11. 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 10.12.9). 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 10.11.3. 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 12.15. 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 15.58.2 to do the induction step. Some details omitted. $\square$

Lemma 15.58.12. Let $R$ be a ring. For any complex $M^\bullet$ there exists a K-flat complex $K^\bullet$ whose terms are flat $R$-modules and a quasi-isomorphism $K^\bullet \to M^\bullet$ which is termwise surjective.

Proof. Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\text{Mod}_ R)$ be the class of flat $R$-modules. By Derived Categories, Lemma 13.29.1 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 15.58.9. The induced map $\mathop{\mathrm{colim}}\nolimits _ i K_ i^\bullet \to M^\bullet$ is a quasi-isomorphism and termwise surjective by construction. The complex $\mathop{\mathrm{colim}}\nolimits _ i K_ i^\bullet$ is K-flat by Lemma 15.58.10. The terms $\mathop{\mathrm{colim}}\nolimits K_ i^ n$ are flat because filtered colimits of flat modules are flat, see Algebra, Lemma 10.39.3. $\square$

Remark 15.58.13. In fact, we can do better than Lemma 15.58.12. Namely, we can find a quasi-isomorphism $P^\bullet \to M^\bullet$ where $P^\bullet$ is a complex of $R$-modules endowed with a filtration

$0 = F_{-1}P^\bullet \subset F_0P^\bullet \subset F_1P^\bullet \subset \ldots \subset P^\bullet$

by subcomplexes such that

1. $P^\bullet = \bigcup F_ pP^\bullet$,

2. the inclusions $F_ iP^\bullet \to F_{i + 1}P^\bullet$ are termwise split injections,

3. the quotients $F_{i + 1}P^\bullet /F_ iP^\bullet$ are isomorphic to direct sums of shifts $R[k]$ (as complexes, so differentials are zero).

This will be shown in Differential Graded Algebra, Lemma 22.20.4. Moreover, given such a complex we obtain a distinguished triangle

$\bigoplus F_ iP^\bullet \to \bigoplus F_ iP^\bullet \to M^\bullet \to \bigoplus F_ iP^\bullet [1]$

in $D(R)$. Using this we can sometimes reduce statements about general complexes to statements about $R[k]$ (this of course only works if the statement is preserved under taking direct sums). More precisely, let $T$ be a property of objects of $D(R)$. Suppose that

1. if $K_ i \in D(R)$, $i \in I$ is a family of objects with $T(K_ i)$ for all $i \in I$, then $T(\bigoplus K_ i)$,

2. if $K \to L \to M \to K[1]$ is a distinguished triangle and $T$ holds for two, then $T$ holds for the third object,

3. $T(R[k])$ holds for all $k$.

Then $T$ holds for all objects of $D(R)$.

Lemma 15.58.14. 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 15.58.12. 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 15.58.4 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 15.58.12. By Lemmas 15.58.1 and 15.58.2 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 15.58.4 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 15.58.14 the resulting functor (up to isomorphism) does not depend on the choice of the K-flat resolution.

Definition 15.58.15. 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 (15.57.0.1). 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 15.58.16. 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 15.58.17. 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 12.18.4. $\square$

Lemma 15.58.18. Let $R$ be a ring. Let $a : K^\bullet \to L^\bullet$ be a map of complexes of $R$-modules. If $K^\bullet$ is K-flat, then there exist a complex $N^\bullet$ and maps of complexes $b : K^\bullet \to N^\bullet$ and $c : N^\bullet \to L^\bullet$ such that

1. $N^\bullet$ is K-flat,

2. $c$ is a quasi-isomorphism,

3. $a$ is homotopic to $c \circ b$.

If the terms of $K^\bullet$ are flat, then we may choose $N^\bullet$, $b$, and $c$ such that the same is true for $N^\bullet$.

Proof. We will use that the homotopy category $K(\text{Mod}_ R)$ is a triangulated category, see Derived Categories, Proposition 13.10.3. Choose a distinguished triangle $K^\bullet \to L^\bullet \to C^\bullet \to K^\bullet [1]$. Choose a quasi-isomorphism $M^\bullet \to C^\bullet$ with $M^\bullet$ K-flat with flat terms, see Lemma 15.58.12. By the axioms of triangulated categories, we may fit the composition $M^\bullet \to C^\bullet \to K^\bullet [1]$ into a distinguished triangle $K^\bullet \to N^\bullet \to M^\bullet \to K^\bullet [1]$. By Lemma 15.58.7 we see that $N^\bullet$ is K-flat. Again using the axioms of triangulated categories, we can choose a map $N^\bullet \to L^\bullet$ fitting into the following morphism of distinghuised triangles

$\xymatrix{ K^\bullet \ar[r] \ar[d] & N^\bullet \ar[r] \ar[d] & M^\bullet \ar[r] \ar[d] & K^\bullet [1] \ar[d] \\ K^\bullet \ar[r] & L^\bullet \ar[r] & C^\bullet \ar[r] & K^\bullet [1] }$

Since two out of three of the arrows are quasi-isomorphisms, so is the third arrow $N^\bullet \to L^\bullet$ by the long exact sequences of cohomology associated to these distinguished triangles (or you can look at the image of this diagram in $D(R)$ and use Derived Categories, Lemma 13.4.3 if you like). This finishes the proof of (1), (2), and (3). To prove the final assertion, we may choose $N^\bullet$ such that $N^ n \cong M^ n \oplus K^ n$, see Derived Categories, Lemma 13.10.7. Hence we get the desired flatness if the terms of $K^\bullet$ are flat. $\square$

Comment #3061 by shanbei on

This might be a silly question: is there a way to choose K-flat resolutions functorially for complexes of sheaves of O_X modules where (X,O_X) is some ringed site.

For K-injective resolutions this is done in tag 079P, so I'm wondering if something more general is true (and perhaps is already in the SP).

Comment #3165 by on

OK, I think there is a way to do this. First you prove that you can do it for bounded above complexes by using the fact that there is a functorial way to get a surjection from a flat module to a given module. Then you do the construction in Lemma 21.17.10. Not sure this would be useful?

Comment #4341 by Manuel Hoff on

I have a question concerning the material in this section. For a ring $R$ and a complex of $R$-modules $M^\bullet$ one defines the derived tensor product $- \otimes _ R^{\mathbf{L}} M^\bullet$. But I can't find any reference to a relation with the functor $F \colon K(\text{Mod}_ R) \to K(\text{Mod}_ R)$ sending $N^\bullet \mapsto \text{Tot}(N^\bullet \otimes _ R M^\bullet )$. Is it true that $LF$ is defined everywhere on $K(\text{Mod}_ R)$ in the sense of Section 13.15 (Tag 05S7) and that we have $LF = - \otimes _ R^{\mathbf{L}} M^\bullet$ as endofunctors of $D(R)$?

Comment #4491 by on

@#4341. Yes, this is true. Namely, we apply the second part of Lemma 13.14.15 where the collection $\mathcal{P}$ is the K-flat complexes. This is allowed by Lemmas 15.58.12 and 15.58.14.

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