22.29 Bimodules and tensor product
Let R be a ring. Let A and B be R-algebras. Let M be a right A-module. Let N be a (A, B)-bimodule. Then M \otimes _ A N is a right B-module.
If in the situation of the previous paragraph A and B are \mathbf{Z}-graded algebras, M is a graded A-module, and N is a graded (A, B)-bimodule, then M \otimes _ A N is a right graded B-module. The construction is functorial in M and defines a functor
- \otimes _ A N : \text{Mod}^{gr}_ A \longrightarrow \text{Mod}^{gr}_ B
of graded categories as in Example 22.25.6. Namely, if M and M' are graded A-modules and f : M \to M' is an A-module homomorphism homogeneous of degree n, then f \otimes \text{id}_ N : M \otimes _ A N \to M' \otimes _ A N is a B-module homomorphism homogeneous of degree n.
If in the situation of the previous paragraph (A, \text{d}) and (B, \text{d}) are differential graded algebras, M is a differential graded A-module, and N is a differential graded (A, B)-bimodule, then M \otimes _ A N is a right differential graded B-module.
Lemma 22.29.1. 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. Then M \mapsto M \otimes _ A N defines a functor
- \otimes _ A N : \text{Mod}^{dg}_{(A, \text{d})} \longrightarrow \text{Mod}^{dg}_{(B, \text{d})}
of differential graded categories. This functor induces functors
\text{Mod}_{(A, \text{d})} \to \text{Mod}_{(B, \text{d})} \quad \text{and}\quad K(\text{Mod}_{(A, \text{d})}) \to K(\text{Mod}_{(B, \text{d})})
by an application of Lemma 22.26.5.
Proof.
Above we have seen how the construction defines a functor of underlying graded categories. Thus it suffices to show that the construction is compatible with differentials. Let M and M' be differential graded A-modules and let f : M \to M' be an A-module homomorphism which is homogeneous of degree n. Then we have
\text{d}(f) = \text{d}_{M'} \circ f - (-1)^ n f \circ \text{d}_ M
On the other hand, we have
\text{d}(f \otimes \text{id}_ N) = \text{d}_{M' \otimes _ A N} \circ (f \otimes \text{id}_ N) - (-1)^ n (f \otimes \text{id}_ N) \circ \text{d}_{M \otimes _ A N}
Applying this to an element x \otimes y with x \in M and y \in N homogeneous we get
\begin{align*} \text{d}(f \otimes \text{id}_ N)(x \otimes y) = & \text{d}_{M'}(f(x)) \otimes y + (-1)^{n + \deg (x)}f(x) \otimes \text{d}_ N(y) \\ & - (-1)^ n f(\text{d}_ M(x)) \otimes y - (-1)^{n + \deg (x)}f(x) \otimes \text{d}_ N(y) \\ = & \text{d}(f) (x \otimes y) \end{align*}
Thus we see that \text{d}(f) \otimes \text{id}_ N = \text{d}(f \otimes \text{id}_ N) and the proof is complete.
\square
If we have a ring R and R-algebras A, B, and C, a right A-module M, an (A, B)-bimodule N, and a (B, C)-bimodule N', then N \otimes _ B N' is a (A, C)-bimodule and we have
(M \otimes _ A N) \otimes _ B N' = M \otimes _ A (N \otimes _ B N')
This equality continuous to hold in the graded and in the differential graded case. See More on Algebra, Section 15.72 for sign rules.
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