nLab
tensor product of modules

Contents

Idea

The tensor product of modules.

Definition

Definition

Let RR be a commutative ring and consider the multicategory RRMod of RR-modules and RR-multilinear maps. In this case the tensor product of modules A RBA\otimes_R B of RR-modules AA and BB can be constructed as the quotient of the tensor product of abelian groups ABA\otimes B underlying them by the action of RR; that is,

A RB=AB/(a,rb)(ar,b). A\otimes_R B = A\otimes B / (a,r\cdot b) \sim (a\cdot r,b).

More category-theoretically:

Definition

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

ARBAB A\otimes R \otimes B \;\rightrightarrows\; A\otimes B

given by the action of RR on AA and on BB.

Definition

If RR is a field, then RR-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 RR is not commutative, as long as AA is a right RR-module and BB is a left RR-module. More generally yet, if RR 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 RR-module in an analogous way. If RR is a commutative monoid in a symmetric monoidal category, so that left and right RR-modules coincide, then A RBA\otimes_R B is again an RR-module, while if RR is not commutative then A RBA\otimes_R B will no longer be an RR-module of any sort.

Properties

Monoidal category structure

The category RRMod equipped with the tensor product of modules R\otimes_R becomes a monoidal category.

Proposition

A monoid in (RMod,)(R Mod, \otimes) is equivalently an RR-algebra.

Proposition

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

A( sSB s) sS(AB c). A \otimes \left(\oplus_{s \in S} B_s\right) \simeq \oplus_{s \in S} ( A \otimes B_c ) \,.

Characterization by exact additive functors

See Eilenberg-Watts theorem.

Exactness properties

Let RR be a commutative ring.

Proposition

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

() RN:RModRMod (-) \otimes_R N \colon R Mod \to R Mod

is an additve right exact functor.

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

Remark

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

Remark

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

Example

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

02/20. 0 \to \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0 \,.

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

0/20/2id/20. 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 2n_1, n_2 \in \mathbb{Z} by

(n 1,n 2mod2) (2n 1,n 2mod2) 2(n 1,n 2mod2) (n 1,2n 2mod2) (n 1,0) 0. \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 NN for which () RN(-)\otimes_R N is always exact are free modules.

Example

Let i:N 1N 2i \colon N_1 \hookrightarrow N_2 be an inclusion of a submodule. For SS \in Set write R |S|=R[S]R^{\oplus {\vert S\vert}} = R[S] for the free module on SS. Then

i RN:N 1 RR |S|N 2 RR |S| 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 RRModR \in R Mod is the tensor unit, this is

i |S|:N 1 |S|N 2 |S|. i^{\oplus {\vert S\vert}} \colon N_1^{\oplus {\vert S\vert}} \hookrightarrow N_2^{\oplus {\vert S\vert}} \,.

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

Definition

If NRModN \in R Mod is such that () RN:RModRMod(-)\otimes_R N \colon R Mod \to R Mod is a left exact functor (hence an exact functor), NN is called a flat module.

Remark

For a general module, a measure of the failure of () RN(-)\otimes_R N to be exact is given by the Tor-functor Tor 1(,N)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)