nLab tensor product of modules

Context

Monoidal categories

monoidal categories

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,

$A\otimes_R B = A\otimes B / (a,r\cdot b) \sim (a\cdot r,b).$
Definition

The tensor product $A \otimes_R B$ is the coequalizer of the two maps

$A\otimes R \otimes B \;\rightrightarrows\; A\otimes B$

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 $(R Mod, \otimes)$ is equivalently an $R$-algebra.

Proposition

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

$A \otimes \left(\oplus_{s \in S} B_s\right) \simeq \oplus_{s \in S} ( A \otimes B_c ) \,.$

Exactness properties

Let $R$ be a commutative ring.

Proposition

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

$(-) \otimes_R N \colon R Mod \to R Mod$
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 $(-)\otimes_R N$ is a left adjoint functor, its right adjoint being the internal hom $[N,-]$ (see at Mod). By the discussion at adjoint functor this means that $(-) \otimes_R N$ even preserves all colimits, in particular the finite colimits.

Remark

The interpretation of this statement in higher category theoryis that $Mod_A$ is a 2-abelian group? (see also the discussion at 2-ring).

Remark

The functor $(-)\otimes_R N$ is not a left exact functor (hence not an exact functor) for all choices of $R$ and $N$.

Example

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

$0 \to \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0 \,.$

The functor $(-) \otimes \mathbb{Z}/2\mathbb{Z}$ sends this to

$0 \to \mathbb{Z}/2\mathbb{Z} \stackrel{0}{\to} \mathbb{Z}/2\mathbb{Z} \stackrel{id}{\to} \mathbb{Z}/2\mathbb{Z} \to 0 \,.$

Here the morphism on the left is the 0-morphism: in components it is given for all $n_1, n_2 \in \mathbb{Z}$ by

\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 $(-)\otimes_R N$ is always exact are free modules.

Example

Let $i \colon N_1 \hookrightarrow N_2$ be an inclusion of a submodule. For $S \in$ Set write $R^{\oplus {\vert S\vert}} = R[S]$ for the free module on $S$. Then

$i \otimes_R N \colon N_1 \otimes_R R^{\oplus {\vert S\vert}} \to N_2 \otimes_R R^{\oplus {\vert S\vert}}$

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

$i^{\oplus {\vert S\vert}} \colon N_1^{\oplus {\vert S\vert}} \hookrightarrow N_2^{\oplus {\vert S\vert}} \,.$

There are more modules $N$ than the free ones for which $(-)\otimes_R N$ is exact. One says

Definition

If $N \in R Mod$ is such that $(-)\otimes_R N \colon R Mod \to R Mod$ 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 $(-)\otimes_R N$ to be exact is given by the Tor-functor $Tor^1(-,N)$. 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)