Commutative J-trivial semigroup presentations

Commutative semigroups have presentations in the lattice of multisets, by their description as quotients of the free semimodule $F(S)$. In the case of commutative groups, there is no uniqueness of presentation even for the cyclic group $C_3$ on three elements. However, for the other fundamental type of commutative semigroups it is often possible to find a unique minimal generating set.

Theorem. let $S$ be a finite commutative J-trivial semigroup, then $S$ has a unique minimal generating set.

Proof. the semigroup $S$ is finite, so it satisfies the ACC on principal ideals. By the maximal condition, let $M$ be the set of all maximal principal ideals of $S$. Then by J-triviality each of these maximal principal ideals is associated with a unique irreducible element $x$. Let $X$ be the set of irreducible elements that generate $M$.

Let $I$ be the current set of irreducibles and $C$ the set of remaining elements of $S$. Then union $X$ to $I$ and remove $(X)$ from $C$. Then we have a new set $C$, which again satisfies the ACC so take its maximal elements and add them to the irreducibles and remove the elements they generate from $C$. Repeat this process a finite number of times to produce the unique minimal generating set $I$. $\square$

The basis of this is the ACC on principal ideals of the commutative J-trivial semigroup. This clearly holds when $S$ is finite. Suppose that $S$ is instead a finitely generated commutative J-trivial semigroup. Then $S$ satisfies the ACC on principal ideals as well.

Theorem. finitely generated commutative semigroups satisfy the ACC on principal ideals

Proof. define $X$ to be any finite set of elements and consider the map \[ f : F(X) \to S \] Mapping any word expressed in $X$ to $S$. Then $f$ is monotone (from the multiset ordering on $F(X)$ words to the increasing action preorder on $S$). Let $T$ be the ascending chain of principal ideals. Then form the ideal closure of $T$ to get an ideal $I$ in $S$ that contains every element of the ascending chain of principal ideals.

Then by the monotonicity of $F$ the ideal $I$ in $S$ reflects back to an ideal in $F(X)$ of the form $f^{-1}(I)$. By Dickson's lemma, this ideal $f^{-1}(X)$ is finitely generated. Consider the finite generators of $f^{-1}(X)$ then they reflect to a set of maximal principal ideals in $T$, these in turn in have a maximal represpentative (their join), so they do not generate the entire ascending chain. $\square$

We can now use this theorem with a generalization of the first theorem, to get a unique minimal generating set for any finitely generated commutative semigroup. More generally, this works for any commutative semigroup satisfying the ACC on principal ideals: for example $(\mathbb{Z}_+,*)$ satisfies the ACC on principal ideals and has a minimal generating set the set of prime numbers.

Theorem. let $S$ be a commutative J-trivial semigroup satisfying the ACC on principal ideals then it has a unique minimal generating set

Proof. (1) the base case: by the maximum condition $S$ has a set of maximum proper principal ideals, which by maximality are not generated by any element so they must be in any minimal generating set (2) we can remove from $S$ the generators of its maximum proper principal ideals and the elements generated by them to get another suborder satisfying the ACC. By transfinite induction this terminates after some ordinal number of steps to produce a minimal generating set. $\square$

One could speculate that the existence of unique generators is applicable to any commutative J-trivial semigroups, but this is not the case. A counter example is $(\mathbb{Q}_+,+)$. This doesn't satisfy the ACC on principal ideals (or any chain condition of any kind because it is dense), so it need not have a unique minimal generating set. A generating set for $(\mathbb{Q},+)$ is any lower interval, but by density there is no minimal lower interval so it doesn't have a unique minimal generating set.

Numerical semigroups are an example of a class of commutative finitely generated J-trivial semigroups. Naturally, it is known that any such numerical semigroup has a unique minimal generating set, which is finite because numerical semigroups are finitely generated. The size of the minimum generating set of a numerical semigroup is its embedding dimension.

Proposition. let $S$ be a commutative finitely generated semigroup, then $\frac{S}{H}$ is finitely generated.

Proof. let $X$ be a finite generating set for $S$ and $\pi : S \to \frac{S}{H}$. Then get the image $\pi(X)$, then this forms a finite generating set for $\frac{S}{H}$. Let $x \in \frac{S}{H}$ then choose a representative in $S$ for it and form its factorisation $t$ in terms of $X$ then $\pi(t)$ is a factorisation of $x$ in $\frac{S}{H}$ expressed in terms of $\pi(X)$, so every term in $\frac{S}{H}$ has a unique factorisation by $\pi(X)$. $\square$

In general, there is not a unique minimal generating set for a finite semigroup. The first counter example is the cyclic group $C_3$ on three elements: it has two different unique generators. In general, the number of generators for a cyclic group is the totient $\phi(n)$. We have a general idea of what finitely generated commutative groups are and what they look like, but no inherent notion of what their generating sets need to be.

The preceding theorem for finitely generated commutative semigroups, demonstrates that every finitely generated commutative semigroup is associated with a condensation $\frac{S}{H}$ which has a unique minimal generating set. This tells you the minimal set of J classes necessary to generate $S$, but still doesn't tell you how to generate each of the classes, or even if other J classes are necessary.

We can now speak about presentations. Suppose that we have a commutative semigroup $S$ with generating set $X$ then we have a map from the free commutative semigroup $F(X)$ to $S$. The free commutative semigroup $F(X)$ is a lattice ordered family of multisets, and the commutative J-trivial semigroup $S$ is naturally partially ordered. As a first step, I would like to relate the order on $F(X)$ to $S$.

Theorem. let $S$ be a commutative semigroup and $C$ a congruence of $S$ then the algebraic preorder on $\frac{S}{C}$ is the ordinary precedence preorder on $C$ with respect to the ordering of $S$.

Proof. suppose $A \subseteq B$ in $\frac{S}{C}$ then $\exists C : AC = B$. By an appropriate choice of representatives, this means that $ac = b$ for some $a \in A, c \in C, b \in B$ so that $a \subseteq b$ which implies that $a \frac{\subseteq}{C} b$. Thus, algebraic precedence in $\frac{S}{C}$ implies ordinary precedence.

Conversely, suppose that $a \frac{\subseteq}{C} b$. This means that there exists some $x$ such that $ax = b$. Consider the projection $pi : S \to \frac{S}{C}$. Then we can get the projection $\pi(x)$ so that $\pi(A)\pi(x) = \pi(B)$ so that $\pi(A) \subseteq \pi(B)$ in $\frac{S}{C}$. Thusly, ordinary precedence applies algebraic precedence in the quotient. $\square$

This implies that in order for the congruence of a commutative J-trivial semigroup to again be J-trivial, all of its equivalence classes must be convex. This generalizes the case for semilattice congruences, lattice congruences, etc. In those cases the congruence classes also need other conditions (like being a subsemilattice or a sublattice). In the finite case, this implies upper or lower bounds, or in the case of congruences of finite lattice that each congruence class is an interval.

Corollary. let $S$ be a commutative J-trivial semigroup and $C$ a congruence with J-trivial quotient. Then the congruence classes of $C$ are convex.

In the case of a semilattice congruence, we have that each congruence class is a subsemilattice. In a finite join semilattice, this implies the existence of upper bounds. This has a further implication: each congruence class has a unique representative. In the case of a finite semilattice congruence, these are the principal filters of irreducibles. It would be nice if this process could pass over to commutative J-trivial semigroups.

We see though, that it does not. The exceptional commutative J-trivial semigroup has neither upper or lower bounds for the congruence class of its intermediate element in its presentation determined by its unique minimal generating set. Thusly, it won't be as easy to characterize the representatives of congruence classes of commutative J-trivial semigroups as it is for semilattices.

The presentations of commutative J-trivial semigroups are interestingly linked to the combinatorics of multiset lattices. This is a major motivation for the study of commutative J-trivial semigroups. The following construction describes how to finite commutative J-trivial semigroups in terms of finite multiset lattices.

Definition. let $S$ be a finite commutative J-trivial semigroup with finite generating set $X$. Then define a map $m : X \to \mathbb{N}$ that assigns each $X$ element to the cardinality of its monogenic semigroup. Then this map $m: X \to \mathbb{N}$ defines a multiset, by considering its outputs to be multiplicities.

Then consider the power multiset $\wp(m)$ consisting of all submultisets of $m$. Then $\wp(m)$ forms a finite multiset lattice. The words of $S^1$ can now be presented by congruence classes in the finite multiset lattice $\wp(m)$. Define the power multiset partition of $S^1$ to be the family of all congruences of terms in $\wp(m)$ by the unique minimal generating set of $S$.

This associates to any element in a finite commutative J-trivial semigroup a finite convex family of multisets. As an application, we can consider any element that is maximally presentable to be any element whose generating set equivalence class has an upper bound. The maximally presentable elements form a subset of $S$.

That subset of commutative J-trivial monoids that have maximal representatives for each element, can be associated to closure operations on finite multiset lattices. This generalisations the representation of semilattices with identity by closures on finite boolean algebras.