Definition 15.59.1. 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.

## 15.59 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.

Lemma 15.59.2. Let $R$ be a ring. Let $K^\bullet $ be a K-flat complex. Then the functor

transforms quasi-isomorphisms into quasi-isomorphisms.

**Proof.**
Follows from Lemma 15.58.4 and the fact that quasi-isomorphisms in $K(R)$ are characterized by having acyclic cones.
$\square$

Lemma 15.59.3. 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 15.59.4. 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

and the definition. $\square$

Lemma 15.59.5. Let $R$ be a ring. Let $(K_1^\bullet , K_2^\bullet , K_3^\bullet )$ be a distinguished triangle in $K(R)$. If two out of three of $K_ i^\bullet $ are K-flat, so is the third.

**Proof.**
Follows from Lemma 15.58.4 and the fact that in a distinguished triangle in $K(R)$ if two out of three are acyclic, so is the third.
$\square$

Lemma 15.59.6. 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

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 15.59.7. 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

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 15.59.8. 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

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 15.59.9. 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

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

and we apply Lemma 15.58.4 to do the induction step. Some details omitted. $\square$

Lemma 15.59.10. 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

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 15.59.7. 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 15.59.8. 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 15.59.11. In fact, we can do better than Lemma 15.59.10. 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

by subcomplexes such that

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

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

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

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

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

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,

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

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

Lemma 15.59.12. 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

is a quasi-isomorphism.

**Proof.**
Choose a quasi-isomorphism $K^\bullet \to L^\bullet $ with $K^\bullet $ a K-flat complex, see Lemma 15.59.10. Consider the commutative diagram

The result follows as by Lemma 15.59.2 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 15.59.10. By Lemmas 15.58.2 and 15.58.4 we obtain an exact functor of triangulated categories

By Lemma 15.59.2 this functor induces a functor $D(R) \to D(R)$ simply because $D(R)$ is the localization of $K(R)$ at quasi-isomorphism. By Lemma 15.59.12 the resulting functor (up to isomorphism) does not depend on the choice of the K-flat resolution.

Definition 15.59.13. Let $R$ be a ring. Let $M^\bullet $ be an object of $D(R)$. The *derived tensor product*

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)

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

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.**
We may and do replace the complexes by K-flat complexes $K^\bullet $ and $L^\bullet $ and then we use the commutativity constraint discussed in Section 15.58.
$\square$

Lemma 15.59.15. Let $R$ be a ring. Let $K^\bullet , L^\bullet , M^\bullet $ be complexes of $R$-modules. There is a canonical isomorphism

functorial in all three complexes.

**Proof.**
Replace the complexes by K-flat complexes and use the associativity constraint in Section 15.58.
$\square$

Lemma 15.59.16. 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

$N^\bullet $ is K-flat,

$c$ is a quasi-isomorphism,

$a$ 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 $K(R)$ 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 15.59.10. 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 15.59.5 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

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (4)

Comment #3061 by shanbei on

Comment #3165 by Johan on

Comment #4341 by Manuel Hoff on

Comment #4491 by Johan on