The subject of this post will be the ideals theory of commutative semigroups. A quantale of ideals will be constructed for commutative semigroups, similar to commutative rings. Ideal multiplication in the context of semigroups will be briefly discussed. Finaly, the ideals of commutative semigroups will be related to the two most important structural decompositions of commutative semigroups: condensation and semilattice decompositions.
Algebraic operations on subsets:
Lattice ordered magmas:Let $M$ be a magma with a Moore family of closed sets $F$ formed by a closure operator $cl_F$. Then we can form a lattice ordered magma on $F$ by \[ ST = cl_F(\{st : s \in S, t \in T\}) \] A lattice ordered magma, is a magma that is internal to the category of partial orders and monotone maps. Therefore, it only remains to show that $(F,*)$ is order compatible.
Proposition. (F,*) is a lattice ordered magma
Proof. Let $S_1,T_1,S_2,T_2 \in F$ such that $S_1 \subseteq S_2$ and $T_1 \subseteq T_2$. Then $S_1T_1 = \{st : s \in S_1, t \in T_1 \}$. Then by $s \in S_2$ and $t \in T_2$ so $st \in S_2T_2$. Therefore, $S_1T_1 \subseteq S_2T_2$, now by the fact that closure operators are monotone we have $cl_F(S_1T_1) \subseteq cl_F(S_2T_2)$. $\square$
Properties preserved by subset multiplication:
Let $M$ be a magma then $\wp(M)$ is a lattice ordered magma. In order to show it is a lattice ordered semigroup, all it remains is to preserve associativity. In fact, associativity is just one of the properties preserved by $\wp(M)$.
Proposition. let $M$ be a magma then $\wp(M)$ preserves the associativity and commutativity of $M$.
Proof. (1) let $A,B,C \subseteq S$ then it remains to show that $(AB)C = A(BC)$. Let $x \in (AB)C$ then $x$ can be factored as $(a_1b_1)c_1$. By associativity this equals $a_1(b_1c_1)$ which is in $A(BC)$. The reverse cases follow in the same way.
(2) let $A,B \subseteq S$ then if $ab \in AB$ by commutativity $ab = ba \in BA$ Likewise, if $ba \in BA$ then $ba = ab \in AB$. Therefore, $AB = BA$ so subset multiplication commutes.$\square$
It follows from this that if $S$ is a commutative semigroup, then $\wp(S)$ is a commutative semigroup as well.
Subsemigroup multiplication in the commutative case:
Let $S$ be a semigroup, then $Sub(S)$ always forms a lattice ordered magma. It need not be the case that $Sub(S)$ is a lattice ordered semigroup, which would follow if $Sub(S)$ were a subsemigroup of $\wp(S)$. The product of subsemigroups is a subsemigroup if they permute, so we do have that $Sub(S)$ is a lattice ordered semigroup in the commutative case.
Theorem. let $S$ be a commutative semigroup, then $(Sub(S),*)$ is a subsemigroup of $(\wp(S),*)$.
Proof. let $A,B \in Sub(S)$ then $AB = \{ab : a \in A, b \in B \}$. Let $a_1b_1, a_2b_2 \in AB$ then it remains to show that their product is in $AB$. The product of $a_1b_1$ and $a_2b_2$ is $a_1b_2a_2b_2$. Then $A$ and $B$ commute by the fact that $S$ is a commutative semigroup, so we can rewrite this as $(a_1a_2)(b_1b_2)$. By the fact that $A,B$ are subsemigroups we have $a_1a_2 \in A$ and $b_1b_2 \in B$ so $(a_1a_2)(b_1b_2) \in AB$. It follows that $Sub(S)$ is a subsemigroup of $\wp(S)$.
Subalgebras form more then just lattices:
Let $A$ be an any algebraic structure, then $Sub(A)$ is always more then just a lattice, because the algebraic operations on elements of a set always pass on to their subalgebras. It may be that $Sub(S)$ is merely a magma, and so it lacks desired properties like associativity, but it is always the case that an algebraic structure is there. Therefore, $Sub(S)$ must always be considered an ordered algebraic structure.
Quantales:
Quantale of subsets:Let $S$ be a commutative semigroup, then we saw that $\wp(S)$ is a lattice ordered semigroup. In order to show that it is a quantale, it only remains to show that the product distributes over unions. Then $(\wp(S),\vee,*)$ will form a commutative semiring, and $\wp(S)$ will be a quantale.
Theorem. let $S$ be a commutative semiring and $A,B,C \in \wp(S)$. Then $A(B \cup C) = AB \cup AC$.
Proof. the expression $A(B \cup C)$ can be broken down in the following manner: \[ A(B \cup C) \] \[ = \{ax : a \in A, x \in B \lor x \in C \}\] \[ = \{ax : a \in A, x \in B\} \cup \{ax : a \in A, x \in C \} \] \[ = AB \cup AC \] The first and last expressions combined yield $A(B \cup C) = AB \cup AC$. $\square$
The quantale of subsets is associated with a residual operation $A:B$ which is the largest subset that whose product with a given set $B$ is less then $A$. \[ A:B = \{ c : aB \subseteq A \} \] This can clearly be seen to be a residual so that if $C = A:B$ then $AD \subseteq B$ implies $D \subseteq C$. The residual property of this expression clearly follows from the residual properties of quantales.
Join decompositions of quantales:
The distributivity property of quantales, allows you to express any product of elements of a quantale in terms of products of join components of each element. Let $a,b \in Q$ and suppose we have join decompositions $a = x_1 \vee ... \vee x^n$ and $b = y_1 \vee ... \vee y^m$. Then we can express the product $ab$ as: \[ ab \] \[ = (x_1 \vee ... \vee x_n)(y_1 \vee ... \vee y_m) \] \[ = \bigvee_{1 \leq i \leq n, 1 \leq j \leq m} x_i y_j\] It follows that in a quantale, a product can always be reduced to a combination of the join irreducible components of each element. This generalizes how the product of sets is determined by the elements of each set. Quantales are therefore generalizations of operations like products of sets.
Quantale of ideals:
Let $S$ be a commutative semigroup, then the ideals of $S$ are all two sided. The product of ideals of $S$ is always again an ideal, so $Ideals(S)$ forms a subsemigroup.
Theorem. let $I,J$ be ideals of $S$ then the product of $I$ and $J$ as sets $IJ$ is an ideal.
Proof. let $ij \in IJ$, and let $x \in S$. Then the product $x(ij)$ is equal to $(xi)j$. By the fact that $I$ is an ideal, $xi \in I$. Therefore, $(xi)j \in IJ$. $\square$
It is a basic fact that ideals of a semigroup are closed under unions and intersections. The preceding theorem demonstrates that they are also closed under the operation of taking products. Taken together, this means that $Ideals(S)$ is a sub-quantale of the quantale of subset multiplication.
Corollary. $Ideals(S)$ is a subquantale of $\wp(S)$
A similar sort of construction applies for left and right ideals of a semigroup, but for now we can use this to understand ideals of commutative semigroups. Most constructions including the residual are inherited from the quantale of sets. The residual of ideals is the same as the residual used to define the subobject classifier in the topos of monoid actions.
Rees semigroup congruences:
The ideals of a commutative ring form a quantale, which also has a quotient operation that produces ring objects. In order to make commutative semigroups more analagous to rings, we also need a quotient operation. That is where Rees semigroup congruences come in. By a Rees semigroup congruence, we can associate a semigroup object to any semigroup ideal. \[ q : Ideals(S) \to Ob(Sgrp) \] The addition of this final component makes the ideals of commutative semigroups analogous to those of commutative rings. They are quantales with associated quotient objects.
Ideals theory:
Ideals and condensation:Let $S$ be a semigroup, then we can form a map $\pi : S \to \frac{S}{H}$ that maps the elements of any given semigroup to its corresponding $H$ class. We want to show that this maps both reflects and preserves ideals, which would mean that ideals can be fully determined by the condensation.
Lemma. let $f : S \to T$ be a commutative semigroup epimorphism. Then $f$ preserves ideals.
Proof. let $I$ be an ideal of $S$ and $f(I)$ its image in $T$. Let $x \in f(I)$. This means that there exists $i \in I$ such that $f(i) = x$. Let $t \in T$. By the fact that $f$ is an epimorphism, there exists $s$ such that $t = f(s)$. Then the product $xt$ is equal to $f(i)f(s)$. By the fact that $f$ is a morphism of semigroups, we have $f(i)f(s) = f(is)$. $I$ is an ideal and $i \in I$, so $is \in I$. Finally, $is \in I$ and $xt = f(is)$ implies that $xt \in f(I)$. This holds for all $x \in f(I)$ and $t \in T$ so $f(I)$ is an ideal. $\square$
Lemma. let $f : S \to T$ be a commutative semigroup epimorphism. Then $f$ reflects ideals.
Proof. let $I \subset T$ be an ideal. Then consider $f^{-1}(I) \subseteq S$. Let $x \in f^{-1}(I)$ then $f(x) \in I$. Let $y \in S$. Then by the fact that $f$ is a morphism of semigroups $f(xy) = f(x)f(y)$. We have that $f(x) \in I$ and $I$ is an ideal so $f(x)f(y) \in I$. Since $f(x)f(y) = f(xy)$ this means $f(xy) \in I$. It follows that $xy \in f^{-1}(I)$. Finally, this means that $f^{-1}(I)$ is an ideal. $\square$
Lemma. the induced map $\pi: Ideals(S) \to Ideals(\frac{S}{H})$ is one to one and onto.
Proof. (1) let $I$ be an ideal, then $I$ is clearly $H$ closed. Therefore, any two semigroup ideals must have different sets of $H$ classes, which means the map to their $H$ classes $\pi$ is injective.
(2) the map $\pi$ both preserves and reflects ideals, so for any ideal in $I \in Ideals(\frac{S}{H})$, we have a corresponding ideal $f^{-1}(I)$ with image $I$. $\square$
Lemma. $\pi : S \to \frac{S}{H}$ induces a semigroup homomorphism from $(Ideals(S),*)$ to $(Ideals(\frac{S}{H}),*)$.
Proof. let $I,J$ be ideals in $Ideals(S)$. Then their product is the set of products of elements of both, and $\pi(IJ)$ is the H classes of the products of each element. Then by the fact that $H$ is a congruence, this can be decomposed into a product of the $H$ classes of $i$ and $j$ which makes this a semigroup homomorphism. \[ \pi(IJ) \] \[ = \pi(\{ij: i I, j \in J\}) \] \[ = \{(ij)_H : i \in I, j \in J\} \] \[ = \{i_H j_H : i \in I, j \in J \} \] \[ \pi(I)\pi(J) \] Then by the fact that for all ideals $I,J$ we have $\pi(IJ) = \pi(I)\pi(J)$ we have that $\pi$ is a semigroup homomorphism of ideal multiplication semigroups. $\square$
We can clearly combine these four lemmas to get a quantale isomorphism from $S$ to $\frac{S}{H}$. The first three lemmas produce a lattice isomorphism, and the final one relates the product of the two quantales.
Theorem. $\pi : S \to \frac{S}{H}$ induces a quantale isomorphism: \[Im(\pi) : Ideals(S) \leftrightarrow Ideals(\frac{S}{H}) \] The special case of the ideal quantale of a PID is essentially dealt with by using the condensation, and by noting that the quantale of ideals of a PID can be completely recovered from its multiplication semigroup. This allows us to create a general ideal theory of commutative semigroups based upon the condensation $\frac{S}{H}$.
Radical ideals and semilattice decompositions:
The ideals of a commutative semigroup are determined by its condensation. It would be nice if the radical ideals would be determined by its semilattice decomposition, which is the second most important quotient associated to any commutative semigroup. We will show that this is the case, thereby creating a full theory of both ideals and radical ideals of commutative semigroups.
Lemma. radical ideals are Archidemean closed
Proof. let $S$ be a semigroup with radical ideal $I$. Let $x \in I$ and suppose $x,y$ are in the same archimedean component, then $\exists n \in \mathbb{Z}_+, z \in S: y^n = xz$. By the fact that $x \in I$ and $I$ is an ideal, $xz \in I$. By the fact that $I$ is radical, $y$ is in $I$, so $I$ is Archimedean closed. $\square$
Lemma. Archimedean closed ideals are radical
Proof. let $I$ be an Archimedean closed ideal and suppose there exists $x \not\in I$ and $y \in I$ such that $x^n \in I$. Then by the fact that $I$ is Archimedean closed, $x$ belongs in a separate Archimedean component $C$ from $y$, but then since Archimedean components form a congruence with semilattice quotient, $C^2$ should equal $C$, but we have an element $x \in C$ such that $x^n \not\in C$, which contradicts semilattice decompositions. So, the Archimedean closed ideals must be radical. $\square$
Theorem. $\pi : S \to \frac{S}{\gamma}$ induces a one to one and onto map of radical ideals
Proof. (1) by the fact that radical ideals are Archimedean closed, $\pi$ always produces different image radical ideals from different input ideals. Therefore, $\pi$ is one to one.
(2) By the fact that Archidemean closed ideals are radical, the inverse image of a radical ideal $I$ is a radical ideal $f^{-1}(I)$ which has $I$ as an image. As every radical ideal of $\frac{S}{\gamma}$ is the image of some radical ideal of $S$, we have that $\pi$ is onto. $\square$
The radical ideals of a commutative semigroup are isomorphic to the ideals of its semilattice decomposition. Ideal lattices are always distributive, because they are lattices of subobjects of a topos. So this implies that the lattice of radical ideals of a commutative semigroup is distributive.
Corollary. the lattice of radical ideals of a commutative semigroup is distributive
The induced map of ideals of the condensation was one to one. In the case of semilattice condensation, it is merely a monotone Galois connection, whose closed sets are the radical ideals of the semigroup and whose adjoint is a Galois insertion that inserts a radical ideal into the ideals lattice of the semigroup.
Overview and comparison:
It was known by Dedekind that the lattice of ideals of a commutative ring is modular. The lattice of radical ideals is not only distributive, it is dual to a spatial locale which is why we can construct the spectrum $Spec(R)$. The lattice of ideals of a commutative semigroup is clearly distributive, because it is a subobject lattice of a topos. The lattice of radical ideals of a commutative semigroup is distributive by basically the same reason.
| Commutative ring | Commutative semigroup | |
|---|---|---|
| Ideals: | Modular | Distributive |
| Radical ideals: | Distributive | Distributive |
References:
The Algebraic Theory of Semigroups
By A.H Clifford
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