# nLab tensor product of modules

### Context

#### Monoidal categories

monoidal categories

## With traces

• trace

• traced monoidal category?

# Contents

## Idea

The tensor product of modules.

## Definition

###### Definition

Let $R$ be a commutative ring and consider the multicategory $R$Mod of $R$-modules and $R$-multilinear maps. In this case the tensor product of modules $A{\otimes }_{R}B$ of $R$-modules $A$ and $B$ can be constructed as the quotient of the tensor product of abelian groups $A\otimes B$ underlying them by the action of $R$; that is,

$A{\otimes }_{R}B=A\otimes B/\left(a,r\cdot b\right)\sim \left(a\cdot r,b\right).$A\otimes_R B = A\otimes B / (a,r\cdot b) \sim (a\cdot r,b).
###### Definition

The tensor product $A{\otimes }_{R}B$ is the coequalizer of the two maps

$A\otimes R\otimes B\phantom{\rule{thickmathspace}{0ex}}⇉\phantom{\rule{thickmathspace}{0ex}}A\otimes B$A\otimes R \otimes B \;\rightrightarrows\; A\otimes B

given by the action of $R$ on $A$ and on $B$.

###### Definition

If $R$ is a field, then $R$-modules are vector spaces; this gives probably the most familiar case of a tensor product spaces, which is also probably the situation where the concept was first conceived.

###### Remark

This tensor product can be generalized to the case when $R$ is not commutative, as long as $A$ is a right $R$-module and $B$ is a left $R$-module. More generally yet, if $R$ is a monoid in any monoidal category (a ring being a monoid in Ab with its tensor product), we can define the tensor product of a left and a right $R$-module in an analogous way. If $R$ is a commutative monoid in a symmetric monoidal category, so that left and right $R$-modules coincide, then $A{\otimes }_{R}B$ is again an $R$-module, while if $R$ is not commutative then $A{\otimes }_{R}B$ will no longer be an $R$-module of any sort.

## Properties

### Monoidal category structure

The category $R$Mod equipped with the tensor product of modules ${\otimes }_{R}$ becomes a monoidal category.

###### Proposition

A monoid in $\left(R\mathrm{Mod},\otimes \right)$ is equivalently an $R$-algebra.

###### Proposition

The tensor product of modules distributes over the direct sum of modules:

$A\otimes \left({\oplus }_{s\in S}{B}_{s}\right)\simeq {\oplus }_{s\in S}\left(A\otimes {B}_{c}\right)\phantom{\rule{thinmathspace}{0ex}}.$A \otimes \left(\oplus_{s \in S} B_s\right) \simeq \oplus_{s \in S} ( A \otimes B_c ) \,.

### Exactness properties

Let $R$ be a commutative ring.

###### Proposition

For $N\in R\mathrm{Mod}$ a module, the functor of tensoring with this module

$\left(-\right){\otimes }_{R}N:R\mathrm{Mod}\to R\mathrm{Mod}$(-) \otimes_R N \colon R Mod \to R Mod
###### Proof

The functor is additive by the distributivity of tensor products over direct sums, prop. 2.

A general abstract way of seeing that the functor is right exact is to notice that $\left(-\right){\otimes }_{R}N$ is a left adjoint functor, its right adjoint being the internal hom $\left[N,-\right]$ (see at Mod). By the discussion at adjoint functor this means that $\left(-\right){\otimes }_{R}N$ even preserves all colimits, in particular the finite colimits.

###### Remark

The interpretation of this statement in higher category theoryis that ${\mathrm{Mod}}_{A}$ is a 2-abelian group? (see also the discussion at 2-ring).

###### Remark

The functor $\left(-\right){\otimes }_{R}N$ is not a left exact functor (hence not an exact functor) for all choices of $R$ and $N$.

###### Example

Let $R≔ℤ$, hence $R\mathrm{Mod}\simeq$ Ab and let $N≔ℤ/2ℤ$ the cyclic group or order 2. Moreover, consider the inclusion $ℤ\stackrel{\cdot 2}{↪}𝕋$ sitting in the short exact sequence

$0\to ℤ\stackrel{\cdot 2}{\to }ℤ\to ℤ/2ℤ\to 0\phantom{\rule{thinmathspace}{0ex}}.$0 \to \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0 \,.

The functor $\left(-\right)\otimes ℤ/2ℤ$ sends this to

$0\to ℤ/2ℤ\stackrel{0}{\to }ℤ/2ℤ\stackrel{\mathrm{id}}{\to }ℤ/2ℤ\to 0\phantom{\rule{thinmathspace}{0ex}}.$0 \to \mathbb{Z}/2\mathbb{Z} \stackrel{0}{\to} \mathbb{Z}/2\mathbb{Z} \stackrel{id}{\to} \mathbb{Z}/2\mathbb{Z} \to 0 \,.

Here the morphism on the left is the 0-morphism: in components it is given for all ${n}_{1},{n}_{2}\in ℤ$ by

$\begin{array}{rl}\left({n}_{1},{n}_{2}\mathrm{mod}2\right)& ↦\left(2{n}_{1},{n}_{2}\mathrm{mod}2\right)\\ & \simeq 2\left({n}_{1},{n}_{2}\mathrm{mod}2\right)\\ & \simeq \left({n}_{1},2{n}_{2}\mathrm{mod}2\right)\\ & \simeq \left({n}_{1},0\right)\\ & \simeq 0\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\begin{aligned} (n_1, n_2 mod 2) & \mapsto (2 n_1, n_2 mod 2) \\ & \simeq 2 (n_1, n_2 mod 2) \\ & \simeq (n_1, 2 n_2 mod 2) \\ & \simeq (n_1, 0) \\ & \simeq 0 \end{aligned} \,.

Hence this is not a short exact sequence anymore.

One kind of module $N$ for which $\left(-\right){\otimes }_{R}N$ is always exact are free modules.

###### Example

Let $i:{N}_{1}↪{N}_{2}$ be an inclusion of a submodule. For $S\in$ Set write ${R}^{\oplus \mid S\mid }=R\left[S\right]$ for the free module on $S$. Then

$i{\otimes }_{R}N:{N}_{1}{\otimes }_{R}{R}^{\oplus \mid S\mid }\to {N}_{2}{\otimes }_{R}{R}^{\oplus \mid S\mid }$i \otimes_R N \colon N_1 \otimes_R R^{\oplus {\vert S\vert}} \to N_2 \otimes_R R^{\oplus {\vert S\vert}}

is again a monomorphism. Indeed, due to the distributivity of the tensor product over the direct sum and using that $R\in R\mathrm{Mod}$ is the tensor unit, this is

${i}^{\oplus \mid S\mid }:{N}_{1}^{\oplus \mid S\mid }↪{N}_{2}^{\oplus \mid S\mid }\phantom{\rule{thinmathspace}{0ex}}.$i^{\oplus {\vert S\vert}} \colon N_1^{\oplus {\vert S\vert}} \hookrightarrow N_2^{\oplus {\vert S\vert}} \,.

There are more modules $N$ than the free ones for which $\left(-\right){\otimes }_{R}N$ is exact. One says

###### Definition

If $N\in R\mathrm{Mod}$ is such that $\left(-\right){\otimes }_{R}N:R\mathrm{Mod}\to R\mathrm{Mod}$ is a left exact functor (hence an exact functor), $N$ is called a flat module.

###### Remark

For a general module, a measure of the failure of $\left(-\right){\otimes }_{R}N$ to be exact is given by the Tor-functor ${\mathrm{Tor}}^{1}\left(-,N\right)$. See there for more details.

## References

An general exposition is in

• Collin Roberts, Introduction to the tensor product (pdf)

Detailed discussion specifically for tensor products of modules is in

• Keith Conrad, Tensor products (pdf)

Revised on February 11, 2013 20:57:13 by Urs Schreiber (89.204.138.151)