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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