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 If $f$ is a quasi-isomorphism, then $1 \otimes f$ is an isomorphism of functors.
\[ 1\otimes f : - \otimes _ A^\mathbf {L} N \longrightarrow - \otimes _ A^\mathbf {L} N' \]
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 of Lemma 22.33.2 is a left adjoint to the functor of Lemma 22.31.2.
\[ - \otimes _ A^\mathbf {L} N : D(A, \text{d}) \longrightarrow D(B, \text{d}) \]
\[ R\mathop{\mathrm{Hom}}\nolimits (N, -) : D(B, \text{d}) \longrightarrow D(A, \text{d}) \]
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 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 functorial in the $B$-module $N$. Thus we see that $R\mathop{\mathrm{Hom}}\nolimits (B, -)$ is the restriction functor and we obtain 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.
\[ - \otimes _ A B : D(A, \text{d}) \longrightarrow D(B, \text{d}) \]
\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(B, N) \longrightarrow N_ A,\quad f \longmapsto f(1) \]
\[ \mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(M, N_ A) = \mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(M \otimes ^\mathbf {L}_ A B, N) \]
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
the diagram
commutes, and
if $M$ is K-flat as a complex of $R$-modules, then the differential graded $A'$-module $M \otimes _ R R'$ represents $M \otimes _ A^\mathbf {L} A'$.
\[ K(A, \text{d}) \longrightarrow K(A', \text{d}),\quad M \longmapsto M \otimes _ R R' \]
\[ \xymatrix{ D(A, \text{d}) \ar[r]_{- \otimes _ A^\mathbf {L} A'} \ar[d]_{restriction} & D(A', \text{d}) \ar[d]^{restriction} \\ D(R) \ar[r]^{- \otimes _ R^\mathbf {L} R'} & D(R') } \]
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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