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tensor product of modules

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,

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

More category-theoretically:

Definition

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

$$A\otimes R\otimes B\rightrightarrows A\otimes B$$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 $(R\mathrm{Mod},\otimes )$ is equivalently an $R$-algebra.

Proposition

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

$$A\otimes \left({\oplus }_{s\in S}{B}_s\right)\simeq {\oplus }_{s\in S}(A\otimes B_c).$$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 $R$ be a commutative ring.

Proposition

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

$$(-){\otimes }_{R}N:R\mathrm{Mod}\to R\mathrm{Mod}$$(-) \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 $(-){\otimes }_{R}N$ 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 ${\mathrm{Mod}}_A$ 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 $R≔\mathbb{Z}$, hence $R\mathrm{Mod}\simeq$ Ab and let $N≔\mathbb{Z}/2\mathbb{Z}$ the cyclic group or order 2. Moreover, consider the inclusion $\mathbb{Z}\stackrel{\cdot 2}{\hookrightarrow }𝕋$ sitting in the short exact sequence

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

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

$$\begin{array}{rl}(n_1,n_2\mathrm{mod}2)& \mapsto (2n_1,n_2\mathrm{mod}2)\\ & \simeq 2(n_1,n_2\mathrm{mod}2)\\ & \simeq (n_1,2n_2\mathrm{mod}2)\\ & \simeq (n_1,0)\\ & \simeq 0\end{array}.$$\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 $i:N_1\hookrightarrow N_2$ be an inclusion of a submodule. For $S\in$ Set write $R^{\oplus \mid S\mid }=R[S]$ 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

$$i^{\oplus \mid S\mid }:N_1^{\oplus \mid S\mid }\hookrightarrow N_2^{\oplus \mid S\mid }.$$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 $N\in R\mathrm{Mod}$ is such that $(-){\otimes }_{R}N:R\mathrm{Mod}\to R\mathrm{Mod}$ 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 $(-){\otimes }_{R}N$ to be exact is given by the Tor-functor ${\mathrm{Tor}}^1(-,N)$. See there for more details.

Related concepts

• tensor product of abelian groups

• tensor product of algebras

• tensor product of chain complexes

• tensor product of ∞-modules

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)