
## 22.23 Tensor product

This section should be moved somewhere else. Let $R$ be a ring. Let $A$ be an $R$-algebra (see Section 22.2). Given a right $A$-module $M$ and a left $A$-module $N$ there is a tensor product

$M \otimes _ A N$

This tensor product is a module over $R$. As an $R$-module $M \otimes _ A N$ is generated by symbols $x \otimes y$ with $x \in M$ and $y \in N$ subject to the relations

$\begin{matrix} (x_1 + x_2) \otimes y - x_1 \otimes y - x_2 \otimes y, \\ x \otimes (y_1 + y_2) - x \otimes y_1 - x \otimes y_2, \\ xa \otimes y - x \otimes ay \end{matrix}$

for $a \in R$, $x, x_1, x_2 \in M$ and $y, y_1, y_2 \in N$. Thus $M \otimes _ A N$ is the receptacle of the universal $A$-bilinear map $M \times N \to M \otimes _ A N$, $(x, y) \mapsto x \otimes y$.

We list some properties of the tensor product

1. In each variable the tensor product is right exact, in fact commutes with direct sums and arbitrary colimits.

2. If $A$, $M$, $N$ are graded and the module structures are compatible with gradings then $M \otimes _ A N$ is graded as well. Then $n$th graded piece $(M \otimes _ A N)^ n$ of $M \otimes _ A N$ is the quotient of $\bigoplus _{p + q = n} M^ p \otimes _{A^0} N^ q$ by the submodule generated by $x \otimes ay - xa \otimes y$ where $x \in M^ p$, $y \in N^ q$, and $a \in A^{n - p - q}$.

3. If $(A, \text{d})$ is a differential graded algebra, and $M$ and $N$ are (left and right) differential graded $A$-modules, then $M \otimes _ A N$ is a differential graded $R$-module with differential

$\text{d}(x \otimes y) = \text{d}(x) \otimes y + (-1)^ ix \otimes \text{d}(y)$

for $x \in M^ i$ and $y \in N$.

4. If we have a second $R$-algebra $B$ and $N$ is a $(A, B)$-bimodule (this means: the induced $R$-module structures are the same and the $A$ and $B$ actions commute) then $M \otimes _ A N$ is a right $B$-module.

5. If $A$ and $B$ are graded algebras, $M$ is a graded $A$-module, and $N$ is an $(A, B)$-bimodule which comes with a grading such that it is both a left graded $A$-module and a right graded $B$-module, then $M \otimes _ A N$ is a graded $B$-module.

6. If $(A, \text{d})$ and $(B, \text{d})$ are differential graded algebras, $M$ is a differential graded $A$-module, and $N$ is an $(A, B)$-bimodule which comes with a grading and a differential such that it is both a left differential graded $A$-module and a right differential graded $B$-module, then $M \otimes _ A N$ is a differential graded $B$-module.

7. If we have $R$-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 (compare with Homology, Remark 12.22.8 for the sign rules).

In (6), the condition may be more hevalleysuccinctly stated by saying that $N$ is a differential graded module over $A^{opp} \otimes _ R B$. We state the following as a lemma.

Lemma 22.23.1. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras, and let $N$ be an $(A, B)$-bimodule which comes with a grading and a differential such that it is both a left differential graded $A$-module and a right differential graded $B$-module. 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 22.19.5.

Proof. This follows from the discussion above. $\square$

If $A$ is an algebra and $M$, $M'$ are right $A$-modules, then we define

$\mathop{\mathrm{Hom}}\nolimits _ A(M, M') = \{ f : M \to M'\mid f \text{ is }A\text{-linear}\}$

as usual. If $A$ is graded and $M$ and $M'$ are graded $A$-modules, then we recall (Example 22.18.6) that

$\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ A^{gr}}(M, M') = \bigoplus \nolimits _{n \in \mathbf{Z}} \mathop{\mathrm{Hom}}\nolimits ^ n(M, M')$

where $\mathop{\mathrm{Hom}}\nolimits ^ n(M, M')$ is the collection of all $A$-module maps $M \to M'$ which are homogeneous of degree $n$.

Lemma 22.23.2. Let $A$ and $B$ be algebras. Let $M$ be a right $A$-module, $N$ an $(A, B)$-bimodule, and $N'$ a right $B$-module. Then we have

$\mathop{\mathrm{Hom}}\nolimits _ B(M \otimes _ A N, N') = \mathop{\mathrm{Hom}}\nolimits _ A(M, \mathop{\mathrm{Hom}}\nolimits _ B(N, N'))$

If $A$, $B$, $M$, $N$, $N'$ are compatibly graded, then we have

$\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ B^{gr}}(M \otimes _ A N, N') = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ A^{gr}}(M, \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ B^{gr}}(N, N'))$

Proof. This follows by interpreting both sides as $A$-bilinear maps $\psi : M \times N \to N'$ which are $B$-linear on the right. $\square$

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