nLab tensor product of commutative monoids

Idea

For $A$ and $B$ two commutative monoids, their tensor product $A \otimes B$ is a new commutative monoid such that a monoid homomorphism $A \otimes B \to C$ is equivalently a bilinear map out of $A$ and $B$.

Definition

Definition

Let CMon be the collection of commutative monoids, regarded as a multicategory whose multimorphisms are the multilinear maps $A_1, \cdots, A_n \to B$.

The tensor product $A, B \mapsto A \otimes B$ in this multicategory is the tensor product of commutative monoids.

Equivalently this means explicitly:

Definition

For $A, B$ two commutative monoids, their tensor product of commutative monoids is the commutative monoid $A \otimes B$ which is the quotient of the free commutative monoid on the product of their underlying sets $A \times B$ by the relations

• $(a_1,b)+(a_2,b)\sim (a_1+a_2,b)$

• $(a,b_1)+(a,b_2)\sim (a,b_1+b_2)$

• $(0,b)\sim 0$

• $(a,0)\sim 0$

for all $a, a_1, a_2 \in A$ and $b, b_1, b_2 \in B$.

Remark

By definition of the free construction and the quotient there is a canonical function of the underlying sets

$$p_{A,B} \;\colon\; U(A) \times U(B) \overset{}{\longrightarrow} U(A \otimes B)$$

(where $U \colon CMon \to Set$ is the forgetful functor).

On elements this sends $(a,b)$ to the equivalence class that it represents under the above equivalence relations.

The following relates the tensor product to bilinear functions. It is a definition or a proposition dependening on whether one takes the notion of bilinear function to be defined before or after that of tensor product of commutative monoids.

Definition/Proposition

A function of underlying sets $f : A \times B \to C$ is a bilinear function precisely if it factors by the morphism of through a monoid homomorphism $\phi : A \otimes B \to C$ out of the tensor product:

$$f : A \times B \stackrel{\otimes}{\to} A \otimes B \stackrel{\phi}{\to} C \,.$$

Properties

Symmetric monoidal category structure

Proposition

Equipped with the tensor product $\otimes$ of def. and the exchange map $\sigma_{A, B}: A\otimes B \to B \otimes A$ generated by $\sigma_{A, B}(a, b) = (b, a)$, CMon becomes a symmetric monoidal category.

The unit object in $(CMon, \otimes)$ is the additive monoid of natural numbers $\mathbb{N}$.

Proof

To see that $\mathbb{N}$ is the unit object, consider for any commitative momoid $A$ the map

$$A \otimes \mathbb{N} \to A$$

which sends for $n \in \mathbb{N}$

$$(a, n) \mapsto n \cdot a \coloneqq \underbrace{a + a + \cdots + a}_{n\;summands} \,.$$

Due to the quotient relation defining the tensor product, the element on the left is also equal to

$$(a, n) = (a, \underbrace{1 + 1 \cdots + 1}_{n\; summands}) = \underbrace{ (a,1) + (a,1) + \cdots + (a,1) }_{n\; summands} \,.$$

This shows that $A \otimes \mathbb{N} \to A$ is in fact an isomorphism.

Showing that $\sigma_{A, B}$ is natural in $A, B$ is trivial, so $\sigma$ is a braiding. $\sigma^2$ is identity, so it gives CMon a symmetric monoidal structure.

Proposition

The tensor product of commutative monoids distributes over the biproduct of commutative monoids

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

Monoids

Proposition

A monoid in $(CMon, \otimes)$ is equivalently a rig.

Proof

Let $(A, \cdot)$ be a monoid in $(CMon, \otimes)$. The fact that the multiplication

$$\cdot : A \otimes A \to A$$

is bilinear means by the above that for all $a_1, a_2, b \in A$ we have

$$(a_1 + a_2) \cdot b = a_1 \cdot b + a_2 \cdot b$$

$$b \cdot (a_1 + a_2) = b \cdot a_1 + b \cdot a_2 \,.$$

$$0 \cdot b = 0 \,.$$

$$b \cdot 0 = 0 \,.$$

This is precisely the distributivity law and absorption law? of the rig.

Related concepts

• tensor product of abelian groups

• tensor product of chain complexes

• tensor product of modules

• tensor product of algebras over a commutative monad