Condensation of commutative semigroups

A commutative semigroup is naturally preordered by the factorisation preorder $ x \leq y \Leftrightarrow \exists z : xz = y$. This generalizes the divisibility ordering of the natural numbers, which is the factorisation preorder of multiplication. As is natural, in any preorder there are symmetric components which partition the relation. Further, for any preorder we can get the condensation by reducing all symmetric components to individual elements. I will show that an analagous condensation is possible for commutative semigroups.

Theorem. the symmetric components of a commutative semigroup form a congruence

Proof. let H denote the symmetric components, and now suppose that $a = c [H]$ and $b = d [H]$. That means that there exists $x_1,x_2,y_1,y_2$ such that $ax_1=c$, $a = cy_1$, $bx_2 = d$, and $b = dy_2$. We can now get that $ab = (cy_1)(dy_2)$ which means that $ab = (y_1y_2)cd$ and dually $(ax_1)(bx_2) = cd$ which means that $(x_1x_2)ab = cd$. Therefore, $ab$ and $cd$ are both factors of one another, so they belong to the same symmetric component. Which means that the symmetric components $H$ form a congruence.

Theorem. the quotient of the commutative semigroup by its symmetric components is antisymmetric.

Proof. let $H_1$, $H_2$ be symmetric components, suppose that $H_1 = H_2$ then there exists $I_1, I_2$ such that $I_1H_1 = H_2$ and $H_1 = H_2I_2$. Let $x$ be an element of $H_1$ and $y$ be an element of $H_2$ then there exists an element $a_1 \in I_1$ such that $ax$ is in $H_2$ then since $a_1x$ and $y$ are both in $H_2$ there exists some $b_1$ such that $a_1b_1x = y$. Dually, there exists some $a_2$ such that $a_2y$ is in $H_1$ and then since $a_2y$ and $x$ are both in $H_1$ there exists some $b_2$ such that $a_2b_2y = x$. This means that $x$ and $y$ both belong to the same symmetric component, which is a contradiction because we supposed that they were in different symmetric components at the start. Therefore, $\frac{S}{H}$ is antisymmetric.

From now on we can refer to $\frac{S}{H}$ as the condensation by analogy with preorders. As with how we can condense the symmetric components of any preorder to get a poset, we can condense the symmetric components of any commutative semigroup to get a posetal commutative semigroup. This quotient structure fully determines the order theory of commutative semigroups.

Commutative algebra:
Ideal multiplication forms a commutative posetal semigroup, even when its multiplicative semigroup is not posetal. I will show here that in the special case of PIDs, ideal multiplication is simply the condensation of multiplicative semigroup.

Theorem. in a PID the ideal multiplication is isomorphic to the condensation of the multiplicative semigroup.

Proof. every multiplicative ideal in a PID is sum closed, to see this notice that every PID has identity, so the sum of any number of elements in $aX$ is of the form $(a + a + ... + ax + ay + ...)$ which is simply $(1+1+...+x+y+...)a$. Two multiplicative ideals are equal if they are in the same symmetric component, so we have a natural one to one mapping $f : \frac{(R,*)}{H} \to (Ideals(R),*)$ between the condensation of the multiplicative semigroup and the multiplicative semigroup of ideals.

Let $H_1,H_2$ be elements of the condensation semigroup. Then let $a \in H_1$ and $b \in H_2$ and now since the condensation is a quotient semigroup $ab \in H_1H_2$. Therefore, $f(ab) = (ab) = f(a)*f(b) = (a)*(b) = (ab) $ because the product of two principal ideals in a PID is the product of their generators. Dually, in the other direction if we are given two principal ideals $(a)$ and $(b)$ their product is $(ab)$ so $f^{-1}( (a)(b) ) = f^{-1}((a))*f^{-1}((b)) = H_a * H_b = H_{ab}$.

Examples:
1. The condensation of a commutative semigroup is a semilattice iff it is a commutative clifford semigroup.

2. In the multiplicative semigroup of a field the condensation is a two element semilattice consisting of units and the zero element. The multiplicative semigroups of fields are clifford.

3. Let $(\mathbb{Z},*)$ be the multiplicative group of the integers. Then the condensation $\frac{(\mathbb{Z},*)}{H}$ is isomorphic to $(\mathbb{N},*)$ because the absolute value of two integers is determined by the absolute value of its arguments. Therefore, by modding out signs we get the condensation.

4. The condensation of a finite monogenic semigroup is a finite commutative aperiodic semigroup.

Notes:
The theory of $\frac{S}{H}$ was first explored by Kolibiarova in the late fifties. It is mentioned in chapter five of Grillet's commutative semigroups. It is the first thing anyone should know about commutative semigroups.