tensor product of modules


Monoidal categories



The tensor product of modules.



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:


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.


If the ring RR happens to be a field, then RR-modules are vector spaces and the tensor product of RR-modules becomes the tensor product of vector spaces.


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.


The tensor product of modules can be generalized to the tensor product of functors.


Monoidal category structure

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


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


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.


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.


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.


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


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.


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


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.


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.


Detailed discussion specifically for tensor products of modules is in

  • Keith Conrad, Tensor products (pdf)

Revised on November 16, 2016 06:49:57 by Urs Schreiber (