Lemma 22.33.1. The functor (22.33.0.1) defines an exact functor of triangulated categories K(\text{Mod}_{(A, \text{d})}) \to K(\text{Mod}_{(B, \text{d})}).
22.33 Derived tensor product
This section is analogous to More on Algebra, Section 15.60.
Let R be a ring. Let (A, \text{d}) and (B, \text{d}) be differential graded algebras over R. Let N be a differential graded (A, B)-bimodule. Consider the functor
22.33.0.1
\begin{equation} \label{dga-equation-bc} \text{Mod}_{(A, \text{d})} \longrightarrow \text{Mod}_{(B, \text{d})},\quad M \longmapsto M \otimes _ A N \end{equation}
defined in Section 22.29.
Proof. Via Lemma 22.29.1 and Remark 22.29.2 this follows from the general principle of Lemma 22.27.17. \square
At this point we can consider the diagram
\xymatrix{ K(\text{Mod}_{(A, \text{d})}) \ar[d] \ar[rr]_{- \otimes _ A N} \ar[rrd]_ F & & K(\text{Mod}_{(B, \text{d})}) \ar[d] \\ D(A, \text{d}) \ar@{..>}[rr] & & D(B, \text{d}) }
The dotted arrow that we will construct below will be the left derived functor of the composition F. (Warning: the diagram will not commute.) Namely, in the general setting of Derived Categories, Section 13.14 we want to compute the left derived functor of F with respect to the multiplicative system of quasi-isomorphisms in K(\text{Mod}_{(A, \text{d})}).
Lemma 22.33.2. In the situation above, the left derived functor of F exists. We denote it - \otimes _ A^\mathbf {L} N : D(A, \text{d}) \to D(B, \text{d}).
Proof. We will use Derived Categories, Lemma 13.14.15 to prove this. As our collection \mathcal{P} of objects we will use the objects with property (P). Property (1) was shown in Lemma 22.20.4. Property (2) holds because if s : P \to P' is a quasi-isomorphism of modules with property (P), then s is a homotopy equivalence by Lemma 22.22.3. \square
Lemma 22.33.3. Let R be a ring. Let (A, \text{d}) and (B, \text{d}) be differential graded R-algebras. Let f : N \to N' be a homomorphism of differential graded (A, B)-bimodules. Then f induces a morphism of functors
1\otimes f : - \otimes _ A^\mathbf {L} N \longrightarrow - \otimes _ A^\mathbf {L} N'
If f is a quasi-isomorphism, then 1 \otimes f is an isomorphism of functors.
Proof. Let M be a differential graded A-module with property (P). Then 1 \otimes f : M \otimes _ A N \to M \otimes _ A N' is a map of differential graded B-modules. Moreover, this is functorial with respect to M. Since the functors - \otimes _ A^\mathbf {L} N and - \otimes _ A^\mathbf {L} N' are computed by tensoring on objects with property (P) (Lemma 22.33.2) we obtain a transformation of functors as indicated.
Assume that f is a quasi-isomorphism. Let F_\bullet be the given filtration on M. Observe that M \otimes _ A N = \mathop{\mathrm{colim}}\nolimits F_ i(M) \otimes _ A N and M \otimes _ A N' = \mathop{\mathrm{colim}}\nolimits F_ i(M) \otimes _ A N'. Hence it suffices to show that F_ n(M) \otimes _ A N \to F_ n(M) \otimes _ A N' is a quasi-isomorphism (filtered colimits are exact, see Algebra, Lemma 10.8.8). Since the inclusions F_ n(M) \to F_{n + 1}(M) are split as maps of graded A-modules we see that
0 \to F_ n(M) \otimes _ A N \to F_{n + 1}(M) \otimes _ A N \to F_{n + 1}(M)/F_ n(M) \otimes _ A N \to 0
is a short exact sequence of differential graded B-modules. There is a similar sequence for N' and f induces a map of short exact sequences. Hence by induction on n (starting with n = -1 when F_{-1}(M) = 0) we conclude that it suffices to show that the map F_{n + 1}(M)/F_ n(M) \otimes _ A N \to F_{n + 1}(M)/F_ n(M) \otimes _ A N' is a quasi-isomorphism. This is true because F_{n + 1}(M)/F_ n(M) is a direct sum of shifts of A and the result is true for A[k] as f : N \to N' is a quasi-isomorphism. \square
Lemma 22.33.4. Let R be a ring. Let (A, \text{d}) and (B, \text{d}) be differential graded R-algebras. Let N be a differential graded (A, B)-bimodule which has property (P) as a left differential graded A-module. Then M \otimes _ A^\mathbf {L} N is computed by M \otimes _ A N for all differential graded A-modules M.
Proof. Let f : M \to M' be a homomorphism of differential graded A-modules which is a quasi-isomorphism. We claim that f \otimes \text{id} : M \otimes _ A N \to M' \otimes _ A N is a quasi-isomorphism. If this is true, then by the construction of the derived tensor product in the proof of Lemma 22.33.2 we obtain the desired result. The construction of the map f \otimes \text{id} only depends on the left differential graded A-module structure on N. Moreover, we have M \otimes _ A N = N \otimes _{A^{opp}} M = N \otimes _{A^{opp}}^\mathbf {L} M because N has property (P) as a differential graded A^{opp}-module. Hence the claim follows from Lemma 22.33.3. \square
Lemma 22.33.5. Let R be a ring. Let (A, \text{d}) and (B, \text{d}) be differential graded R-algebras. Let N be a differential graded (A, B)-bimodule. Then the functor
- \otimes _ A^\mathbf {L} N : D(A, \text{d}) \longrightarrow D(B, \text{d})
of Lemma 22.33.2 is a left adjoint to the functor
R\mathop{\mathrm{Hom}}\nolimits (N, -) : D(B, \text{d}) \longrightarrow D(A, \text{d})
of Lemma 22.31.2.
Proof. This follows from Derived Categories, Lemma 13.30.1 and the fact that - \otimes _ A N and \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, -) are adjoint by Lemma 22.30.3. \square
Example 22.33.6. Let R be a ring. Let (A, \text{d}) \to (B, \text{d}) be a homomorphism of differential graded R-algebras. Then we can view B as a differential graded (A, B)-bimodule and we get a functor
- \otimes _ A B : D(A, \text{d}) \longrightarrow D(B, \text{d})
By Lemma 22.33.5 the left adjoint of this is the functor R\mathop{\mathrm{Hom}}\nolimits (B, -). For a differential graded B-module let us denote N_ A the differential graded A-module obtained from N by restriction via A \to B. Then we clearly have a canonical isomorphism
\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(B, N) \longrightarrow N_ A,\quad f \longmapsto f(1)
functorial in the B-module N. Thus we see that R\mathop{\mathrm{Hom}}\nolimits (B, -) is the restriction functor and we obtain
\mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(M, N_ A) = \mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(M \otimes ^\mathbf {L}_ A B, N)
bifunctorially in M and N exactly as in the case of commutative rings. Finally, observe that restriction is a tensor functor as well, since N_ A = N \otimes _ B {}_ BB_ A = N \otimes _ B^\mathbf {L} {}_ BB_ A where {}_ BB_ A is B viewed as a differential graded (B, A)-bimodule.
Lemma 22.33.7. With notation and assumptions as in Lemma 22.33.5. Assume
N defines a compact object of D(B, \text{d}), and
the map H^ k(A) \to \mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(N, N[k]) is an isomorphism for all k \in \mathbf{Z}.
Then the functor -\otimes _ A^\mathbf {L} N is fully faithful.
Proof. Our functor has a left adjoint given by R\mathop{\mathrm{Hom}}\nolimits (N, -) by Lemma 22.33.5. By Categories, Lemma 4.24.4 it suffices to show that for a differential graded A-module M the map
M \longrightarrow R\mathop{\mathrm{Hom}}\nolimits (N, M \otimes _ A^\mathbf {L} N)
is an isomorphism in D(A, \text{d}). For this it suffices to show that
H^ n(M) \longrightarrow \text{Ext}^ n_{D(B, \text{d})}(N, M \otimes _ A^\mathbf {L} N)
is an isomorphism, see Lemma 22.31.4. Since N is a compact object the right hand side commutes with direct sums. Thus by Remark 22.22.5 it suffices to prove this map is an isomorphism for M = A[k]. Since (A[k] \otimes _ A^\mathbf {L} N) = N[k] by Remark 22.29.2, assumption (2) on N is that the result holds for these. \square
Lemma 22.33.8. Let R \to R' be a ring map. Let (A, \text{d}) be a differential graded R-algebra. Let (A', \text{d}) be the base change, i.e., A' = A \otimes _ R R'. If A is K-flat as a complex of R-modules, then
- \otimes _ A^\mathbf {L} A' : D(A, \text{d}) \to D(A', \text{d}) is equal to the right derived functor of
K(A, \text{d}) \longrightarrow K(A', \text{d}),\quad M \longmapsto M \otimes _ R R'the diagram
\xymatrix{ D(A, \text{d}) \ar[r]_{- \otimes _ A^\mathbf {L} A'} \ar[d]_{restriction} & D(A', \text{d}) \ar[d]^{restriction} \\ D(R) \ar[r]^{- \otimes _ R^\mathbf {L} R'} & D(R') }commutes, and
if M is K-flat as a complex of R-modules, then the differential graded A'-module M \otimes _ R R' represents M \otimes _ A^\mathbf {L} A'.
Proof. For any differential graded A-module M there is a canonical map
c_ M : M \otimes _ R R' \longrightarrow M \otimes _ A A'
Let P be a differential graded A-module with property (P). We claim that c_ P is an isomorphism and that P is K-flat as a complex of R-modules. This will prove all the results stated in the lemma by formal arguments using the definition of derived tensor product in Lemma 22.33.2 and More on Algebra, Section 15.59.
Let F_\bullet be the filtration on P showing that P has property (P). Note that c_ A is an isomorphism and A is K-flat as a complex of R-modules by assumption. Hence the same is true for direct sums of shifts of A (you can use More on Algebra, Lemma 15.59.8 to deal with direct sums if you like). Hence this holds for the complexes F_{p + 1}P/F_ pP. Since the short exact sequences
0 \to F_ pP \to F_{p + 1}P \to F_{p + 1}P/F_ pP \to 0
are split exact as sequences of graded modules, we can argue by induction that c_{F_ pP} is an isomorphism for all p and that F_ pP is K-flat as a complex of R-modules (use More on Algebra, Lemma 15.59.5). Finally, using that P = \mathop{\mathrm{colim}}\nolimits F_ pP we conclude that c_ P is an isomorphism and that P is K-flat as a complex of R-modules (use More on Algebra, Lemma 15.59.8). \square
Lemma 22.33.9. Let R be a ring. Let (A, \text{d}) and (B, \text{d}) be differential graded R-algebras. Let T be a differential graded (A, B)-bimodule. Assume
T defines a compact object of D(B, \text{d}), and
S = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(T, B) represents R\mathop{\mathrm{Hom}}\nolimits (T, B) in D(A, \text{d}).
Then S has a structure of a differential graded (B, A)-bimodule and there is an isomorphism
N \otimes _ B^\mathbf {L} S \longrightarrow R\mathop{\mathrm{Hom}}\nolimits (T, N)
functorial in N in D(B, \text{d}).
Proof. Write \mathcal{B} = \text{Mod}^{dg}_{(B, \text{d})}. The right A-module structure on S comes from the map A \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, T) and the composition \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, B) \otimes \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, T) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, B) defined in Example 22.26.8. Using this multiplication a second time there is a map
c_ N : N \otimes _ B S = \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(B, N) \otimes _ B \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, B) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, N)
functorial in N. Given N we can choose quasi-isomorphisms P \to N \to I where P, resp. I is a differential graded B-module with property (P), resp. (I). Then using c_ N we obtain a map P \otimes _ B S \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(T, I) between the objects representing S \otimes _ B^\mathbf {L} N and R\mathop{\mathrm{Hom}}\nolimits (T, N). Clearly this defines a transformation of functors c as in the lemma.
To prove that c is an isomorphism of functors, we may assume N is a differential graded B-module which has property (P). Since T defines a compact object in D(B, \text{d}) and since both sides of the arrow define exact functors of triangulated categories, we reduce using Lemma 22.20.1 to the case where N has a finite filtration whose graded pieces are direct sums of B[k]. Using again that both sides of the arrow are exact functors of triangulated categories and compactness of T we reduce to the case N = B[k]. Assumption (2) is exactly the assumption that c is an isomorphism in this case. \square
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