Commutative monoid

A commutative monoid is a monoid whose operation is commutative. The natural numbers with addition form a commutative monoid that is not an abelian group, because positive elements have no additive inverses.

The free commutative monoid on $d$ generators consists of the nonnegative integral sums of those generators, with unique coefficients. Its semigroup algebra over a commutative ring is the polynomial ring in $d$ variables. In a smooth toric chart, the dual-cone lattice points form such a monoid, yielding affine-space coordinates.

A commutative monoid is cancellative when $a+c=b+c$ implies $a=b$. Cancellation makes the natural map into its Grothendieck group injective.

For an integer $m\ge 2$, the positive integers congruent to one modulo $m$ form a cancellative commutative monoid under multiplication. The identity is one, and it is the only unit. Proper factors outside this congruence class do not witness reducibility inside the monoid. Hence irreducible elements of a commutative monoid need not coincide with ordinary prime numbers.

The congruence submonoid of the positive integers with modulus ten has four distinct irreducible elements of a commutative monoid $21,221,51,91$ with equal pairwise products. Their ordinary prime factors are respectively $(3,7),(13,17),(3,17),(7,13)$. None of these proper prime factors is congruent to one modulo ten. The Euclid lemma proves that a product of two distinct primes has no other proper positive divisors, so all four are irreducible in this monoid. This gives nonunique factorization without assuming the fundamental theorem of arithmetic.

A nonunit element of a commutative monoid is irreducible if each factorization has a unit factor. In a submonoid of the positive integers under multiplication, the only unit is one. Irreducibility is relative to the allowed factors: a composite integer can be irreducible in a smaller monoid even though it is not a prime number.

A cancellative commutative monoid is torsion-free in this sense when $na=nb$ implies $a=b$ for every positive integer $n$. Its Grothendieck group is then a torsion-free abelian group, because $n(a-b)=0$ forces $na=nb$. This is the condition inherited by a submonoid of a torsion-free abelian group.

This category has commutative monoids and their homomorphisms. Pointwise addition makes each hom-set a commutative monoid, and composition is additive in both variables. Finite cartesian products are also coproducts, using the coordinate injections with zero in the other coordinates and summing images. Therefore it is a semi-additive category, but it is not an additive category. The reflexive pair from the submonoid $\{(m,n):m\le n\le 2m\}$ of ${\mathbb{N}}^2$ to $\mathbb{N}$, using the two projections and section $m\mapsto (m,m)$, has an arrow from $1$ to $2$ and none from $2$ to $1$ in its represented graph. Thus a reflexive pair here need not yield a groupoid.