tensor product of commutative monoids in nLab

Context

Algebra

Monoidal categories

Monoid theory

• Idea
• Definition
• Properties
• Symmetric monoidal category structure
• Monoids
• In (∞,1)(\infty,1)-category theory
• Related concepts
• References

Definition

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

• (a 1,b)+(a 2,b)∼(a 1+a 2,b)(a_1,b)+(a_2,b)\sim (a_1+a_2,b)

• (a,b 1)+(a,b 2)∼(a,b 1+b 2)(a,b_1)+(a,b_2)\sim (a,b_1+b_2)

• (0,b)∼0(0,b)\sim 0

• (a,0)∼0(a,0)\sim 0

for all a,a 1,a 2∈Aa, a_1, a_2 \in A and b,b 1,b 2∈Bb, b_1, b_2 \in B.

Definition/Proposition

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

f:A×B→⊗A⊗B→ϕC. f : A \times B \stackrel{\otimes}{\to} A \otimes B \stackrel{\phi}{\to} C \,.

Proposition

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

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

Proof

To see that ℕ\mathbb{N} is the unit object, consider for any commutative monoid AA the map

A⊗ℕ→A A \otimes \mathbb{N} \to A

which sends for n∈ℕn \in \mathbb{N}

(a,n)↦n⋅a≔a+a+⋯+a⏟ nsummands. (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,1+1⋯+1⏟ nsummands)=(a,1)+(a,1)+⋯+(a,1)⏟ nsummands. (a, n) = (a, \underbrace{1 + 1 \cdots + 1}_{n\; summands}) = \underbrace{ (a,1) + (a,1) + \cdots + (a,1) }_{n\; summands} \,.

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

Showing that σ A,B\sigma_{A, B} is natural in A,BA, B is trivial, so σ\sigma is a braiding. σ 2\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⊗⊕ s∈SB s≃⊕ s∈S(A⊗B s). A \otimes \oplus_{s \in S} B_s \simeq \oplus_{s \in S} ( A \otimes B_s ) \,.

Proposition

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

Proof

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

⋅:A⊗A→A \cdot : A \otimes A \to A

is bilinear means by the above that for all a 1,a 2,b∈Aa_1, a_2, b \in A we have

(a 1+a 2)⋅b=a 1⋅b+a 2⋅b (a_1 + a_2) \cdot b = a_1 \cdot b + a_2 \cdot b

b⋅(a 1+a 2)=b⋅a 1+b⋅a 2. b \cdot (a_1 + a_2) = b \cdot a_1 + b \cdot a_2 \,.

0⋅b=0. 0 \cdot b = 0 \,.

b⋅0=0. b \cdot 0 = 0 \,.

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