Divisibility Theory of Commutative Rings and Ideal Distributivity

TL;DR

This paper explores a special class of commutative rings where elements have unique divisibility properties, establishing their algebraic structure and linking them to Bézout monoids through a representation theorem.

Contribution

It characterizes a finitely axiomatizable class of rings with unique divisibility and connects their ideal structure to Bézout monoids via a new representation theorem.

Findings

Class of rings is a finitely axiomatizable, ideal distributive quasivariety.

This class is generated by integral domains with trivial units.

Provides evidence supporting the conjecture linking Bézout monoids to ideal lattices.

Abstract

We begin by investigating the class of commutative unital rings in which no two distinct elements divide the same elements. We prove that this class forms a finitely axiomatizable, relatively ideal distributive quasivariety, and it equals the quasivariety generated by the class of integral domains with trivial unit group. We end the paper by proving a representation theorem that provides more evidence to the conjecture that B\'ezout monoids describe exactly the monoids of finitely generated ideals of commutative unital rings with distributive ideal lattice.

Equations

$$(s_1=t_1)\wedge \cdots \wedge (s_n=t_n)\to (s_0=t_0)$$

$$I\wedge (J\vee K)=(I\wedge J)\vee (I\wedge K).$$

$$⟨X,Y,Z\mid XYZ=Z⟩.$$

$$s=1s=(e\wedge f)s=es\wedge fs=es\wedge 0=es\wedge ef=e(s\wedge f),$$

$$S_1(B\setminus P)$$

$$(S_a\cup \{0\})\wedge P$$

$$(c_1{f}_1)(c_2{f}_2)$$

$$(d_1\wedge e_1)(d_2\wedge e_2)$$

$$(c_1{f}_1)(d_2\wedge e_2)$$

$$e_1{e}_2$$

$$e\in P$$

$$cf=c(f^{\prime }\wedge s{e}^{\prime })=c{f}^{\prime }\wedge cs{e}^{\prime }=s{f}^{\prime }\wedge s{e}^{\prime }=s(f^{\prime }\wedge e^{\prime })=s1=s,$$

$$s=s1=s(f^{\prime }\wedge e^{\prime })=s{f}^{\prime }\wedge s{e}^{\prime }=c{f}^{\prime }\wedge s{e}^{\prime }$$

$$d_1^2=d_1^2\wedge 0=d_1^2\wedge d_1{e}_1=d_1(d_1\wedge e_1)=d_1(d_2\wedge e_2)=d_1{d}_2\wedge d_1{e}_2=d_1{d}_2\wedge 0=d_1{d}_2.$$

$$e_1^2=0\wedge e_1^2=d_1{e}_1\wedge e_1^2=(d_1\wedge e_1)e_1=(d_2\wedge e_2)e_1=d_2{e}_1\wedge e_2{e}_1=0\wedge e_2{e}_1=e_1{e}_2.$$

$$\biggl{\{}\sum_{i=1}^{n}k_{i}s_{i}:0\neq k_{i}\in L,\ s_{i}\in C,\ \bigwedge_{i=1}^{n}s_{i}=1,\,\,n\in\mathbb{N}\biggr{\}},$$

$$hi:=(\hat{h}{\chi }_1)i\text{for all i∈I}.$$

$$(h_1,i_1)(h_2,i_2)=(h_1{h}_2,h_1{i}_2+h_2{i}_1+i_1{i}_2)$$

$$rR=(c,{\chi }_e)R=cR\oplus {\chi }_{e}R=cR\oplus gR\text{ for some e∈P,if c∈Sa∪{0}}.$$

$$rR\ni r\chi_{e_{\ell}}=(c,g)(0,\chi_{e_{\ell}})=\biggl{(}c,\sum_{j=0}^{n-1}a_{j}\chi_{e_{j}}\biggr{)}(0,\chi_{e_{\ell}})=\bigl{(}0,(a_{\ell}+c)\chi_{e_{\ell}}\bigr{)}=(a_{\ell}+c)\chi_{e_{\ell}}.$$

$$e_{=}=\underset{\begin{array}{c}1\le j\le n-1\\ b_j=0\end{array}}{\bigwedge }{e}_j\text{and}{e}_{\ne }=\underset{\begin{array}{c}1\le j\le n-1\\ b_j\ne 0\end{array}}{\bigwedge }{e}_j.$$

$${\chi }_{e_{=}}=\sum _{\begin{array}{c}1\le j\le n-1\\ b_j=0\end{array}}{\chi }_{e_j},{\chi }_{e_{\ne }}=\sum _{\begin{array}{c}1\le j\le n-1\\ b_j\ne 0\end{array}}{\chi }_{e_j},{\chi }_{e_{=}}{\chi }_{e_{\ne }}={\chi }_0=0,\text{and}{\chi }_{e_{=}}+{\chi }_{e_{\ne }}={\chi }_e.$$

$$\qquad rR\ni r-i=\biggl{(}c,\sum_{j=0}^{n-1}a_{j}\chi_{e_{j}}\biggr{)}-\biggl{(}0,\sum_{j=1}^{n-1}(a_{j}+c)\chi_{e_{j}}\biggr{)}=\biggl{(}c,\sum_{j=1}^{n-1}(-c)\chi_{e_{j}}\biggr{)}\\ =c-c\chi_{e}=c(1,-\chi_{e}).\qquad$$

$$rR=c(1,-{\chi }_e)R\oplus {\chi }_{e_{\ne }}R=c(1,-{\chi }_{e_{=}})R.$$

$${\mathcal{I}}_{\text{princ}}(R)=\{c(1,-{\chi }_e)R:c\in S_1,e\in P\}\cup \{(d,{\chi }_e)R:d\in S_a\cup \{0\},e\in P\}.$$

$$c(1,-\chi_{e_{1}e_{2}})=h^{-1}hc(1,-\chi_{e_{1}e_{2}})=h^{-1}\bigl{(}c_{1}(1,-\chi_{e_{1}e_{2}})+c_{2}(1,-\chi_{e_{1}e_{2}})\bigr{)}\in J.$$

$$c(1,-\chi_{e_{1}})=h^{-1}hc(1,-\chi_{e_{1}})=h^{-1}\bigl{(}c_{1}(1,-\chi_{e_{1}})+d_{2}(1,-\chi_{e_{1}})\bigr{)}\\ =h^{-1}\bigl{(}c_{1}(1,-\chi_{e_{1}})+(d_{2},0)\bigr{)}\in J_{1}+J_{2}=J,$$

$$c(1,-{\chi }_e)=c(1,-{\chi }_{e_1})+c{\chi }_{e_1{e}_2}=c(1,-{\chi }_{e_1})+{\chi }_{e_2}c{\chi }_{e_1}\in J+J_2=J,$$

$$J=(d_{1}R+d_{2}R)\oplus ({\chi }_{e_1}R+{\chi }_{e_2}R)=(d_1\wedge d_2)R\oplus {\chi }_{e_1\vee e_2}R=(d_1\wedge d_2,{\chi }_{e_1\vee e_2})R,$$

$$\Phi :S$$

$$cf$$

$$d\wedge e$$

$$\Phi(c_{1}f_{1})\Phi(c_{2}f_{2})=c_{1}(1,-\chi_{e_{1}})Rc_{2}(1,-\chi_{e_{2}})R=c_{1}c_{2}(1,-\chi_{e_{1}\wedge e_{2}})R\\ =\Phi\bigl{(}c_{1}c_{2}(f_{1}\vee f_{2})\bigr{)}=\Phi\bigl{(}(c_{1}f_{1})(c_{2}f_{2})\bigr{)}.$$

$$\Phi(d_{1}\wedge e_{1})\Phi(d_{2}\wedge e_{2})=(d_{1},\chi_{e_{1}})R(d_{2},\chi_{e_{2}})R=(d_{1}d_{2},d_{1}\chi_{e_{2}}+d_{2}\chi_{e_{1}}+\chi_{e_{1}}\chi_{e_{2}})R\\ =(d_{1}d_{2},\chi_{e_{1}\vee e_{2}})R=\Phi\bigl{(}d_{1}d_{2}\wedge(e_{1}\vee e_{2})\bigr{)}=\Phi\bigl{(}(d_{1}\wedge e_{1})(d_{2}\wedge e_{2})\bigr{)}.$$

Full text

This material is based upon work supported by the National Research, Development and Innovation Office NKFIH K119934, the Vietnam Institute for Advanced Study in Mathematics (VIASM), the Vietnamese Institute of Mathematics, the National Science Foundation grant no. DMS 1500254, the Hungarian National Foundation for Scientific Research (OTKA) grant no. K115518, and the National Research, Development and Innovation Fund of Hungary (NKFI) grant no. K128042.

1. Introduction

In this paper we discuss four perspectives for studying the divisibility theory of commutative unital rings.

The first perspective is that of multiplicative monoids. If $R$ is a commutative unital ring, then in the multiplicative monoid $\langle R;\cdot\rangle$ we say that “$a$ divides $b$” (written $a\mid b$) provided there exists $c$ such that $ac=b$. This relation of divisibility is a reflexive, transitive relation on $R$, i.e. a quasiorder on $R$, and we call the structure $R$, under the relation of divisibility, the divisibility quasiorder of $R$.

The second perspective identifies two elements of $\langle R;\cdot\rangle$ if they divide the same elements. This identification means that we are passing from the divisibility quasiorder to its quotient partial order. Since $a$ and $b$ divide the same elements exactly when the principal ideals $(a)$ and $(b)$ are equal, this second perspective involves studying the poset ${\mathcal{I}}_{\textrm{princ}}(R)$ of principal ideals of $R$. (We order the elements of ${\mathcal{I}}_{\textrm{princ}}(R)$ by inclusion, thereby reversing the quotient order.)

The third perspective involves enlarging the poset ${\mathcal{I}}_{\textrm{princ}}(R)$ to the join semilattice ${\mathcal{I}}_{\textrm{f.g.}}(R)$ of finitely generated ideals of $R$.

The fourth perspective involves enlarging ${\mathcal{I}}_{\textrm{f.g.}}(R)$ still further to the lattice ${\mathcal{I}}(R)$ of all ideals.

The passage from $\langle R;\cdot\rangle$ to ${\mathcal{I}}_{\textrm{princ}}(R)$, involving the formation of a quotient, results in a loss of information. One sees this clearly by noting that when $\mathbb{E}$ and $\mathbb{F}$ are fields we must have ${\mathcal{I}}_{\textrm{princ}}(\mathbb{E})\cong{\mathcal{I}}_{\textrm{princ}}(\mathbb{F})$, yet we needn’t have $\langle\mathbb{E};\cdot\rangle\cong\langle\mathbb{F};\cdot\rangle$. (These latter monoids needn’t even have the same size.) Our first main goal is to investigate the situation where the passage from $\langle R;\cdot\rangle$ to ${\mathcal{I}}_{\textrm{princ}}(R)$ results in no loss of information. Equivalently, we examine and characterize the situation where the divisibility quasiorder of $R$ is already a partial order. We will find that the class of commutative unital rings whose divisibility quasiorder is a partial order is a relatively ideal distributive quasivariety. Moreover, we show that this quasivariety is axiomatized relative to the class of commutative rings by the single quasi-identity $xyz=z\to yz=z$. Finally, this quasivariety is exactly the quasivariety generated by the class of integral domains which have trivial unit group.

Our second main objective is a contribution to the representation of Bézout monoids as divisibility theories $\mathcal{I}_{\rm princ}(R)=\mathcal{I}_{\rm f.g}(R)$ of Bézout rings (i.e., commutative unital rings whose finitely generated ideals are principal). Bézout monoids (see Definition 4.1) were introduced in [5] (see also [2]) as an abstract framework for studying divisibility in Bézout rings, and more generally, in arithmetical rings (i.e., commutative unital rings $R$ with distributive ideal lattice $\mathcal{I}(R)$). It is known that the divisibility theory $\mathcal{I}_{\rm f.g}(R)$ of every arithmetical ring $R$ is a Bézout monoid [2], and it is conjectured that, conversely, every Bézout monoid can be represented as the divisibility theory $\mathcal{I}_{\rm f.g}(R)$ of an arithmetical ring $R$. By a result of Anderson [1], if this is the case, then in fact, every Bézout monoid can be represented as the divisibility theory $\mathcal{I}_{\rm princ}(R)=\mathcal{I}_{\rm f.g}(R)$ of a Bézout ring $R$. The conjecture has been proved for semihereditary Bézout monoids [3] and Bézout monoids which have only one minimal $m$-prime filter [4]. In this paper we will prove the conjecture for another class of Bézout monoids, which we call Vasconcelos monoids, due to the fact that the divisibility theory of the Bézout ring constructed by Vasconcelos [12, Example 3.2] belongs to this class.

2. The quasivariety of

rings whose principal ideals have

unique generators

Our goal in this section is to describe the class of commutative rings whose principal ideals have unique generators. The main result is that this class is a relatively ideal distributive quasivariety, so let us explain now what that means. (For more details about relatively congruence distributive/modular quasivarieties, we refer to [8, 9, 10].)

A quasi-identity in the language of commutative rings is a universally quantified implication of the form

[TABLE]

where $s_{i}$ and $t_{i}$ are ring terms (= “words”, or “polynomials”). We allow $n=0$, in which case the quasi-identity reduces to an identity: $s_{0}=t_{0}$ (universally quantified). To emphasize this last point: identities are special quasi-identities.

A variety is a class axiomatized by identities. A quasi-variety is a class axiomatized by quasi-identities. For an example of the former, the class of commutative rings is a variety. For an example of the latter, the class of rings axiomatized by the identities defining commutative rings together with the quasi-identity $(x^{2}=0)\to(x=0)$ is the quasivariety of reduced commutative rings (rings with no nonzero nilpotent elements).

If $\mathcal{Q}$ is a quasivariety of commutative rings, $R\in\mathcal{Q}$, and $I\lhd R$ is an ideal of $R$, then $I$ is a $\mathcal{Q}$-ideal (or a relative ideal) if $R/I\in{\mathcal{Q}}$. For example, if $\mathcal{Q}$ is the quasivariety of commutative reduced rings and $R\in\mathcal{Q}$, then $I$ is a relative ideal of $R$ exactly when $I$ is a semiprime ideal of $R$.

The collection of $\mathcal{Q}$-ideals of some $R\in\mathcal{Q}$, when ordered by inclusion, forms an algebraic lattice. It is not a sublattice of the ordinary ideal lattice, but it is a subset of the ordinary ideal lattice that is closed under arbitrary meet.

A quasivariety $\mathcal{Q}$ of commutative rings is relatively ideal distributive if the $\mathcal{Q}$-ideal lattice of any member of $\mathbb{Q}$ satisfies the distributive law:

[TABLE]

Here, the meet operation is just intersection ($I\wedge J=I\cap J$) while the join operation depends on $\mathcal{Q}$; all that can be said is that $I\vee J$ is the least $\mathcal{Q}$-ideal that contains $I\cup J$ (or, equivalently, contains $I+J$).

It is interesting to recognize that a quasivariety of rings is relatively ideal distributive, because any distributive algebraic lattice is isomorphic to the lattice of open sets of a topology defined on the set of meet irreducible lattice elements. This means that, if $\mathcal{Q}$ is relatively ideal distributive, then to each member of $\mathcal{Q}$ there is a naturally associated topological space, its $\mathcal{Q}$-spectrum. It is possible to treat a member $R\in\mathcal{Q}$ as a ring of functions defined over the spectrum of $R$. It turns out that the quasivariety of commutative reduced rings, mentioned earlier as an example, is relatively ideal distributive, and for this $\mathcal{Q}$ the $\mathcal{Q}$-spectrum of any $R\in\mathcal{Q}$ is just the ordinary prime spectrum of $R$.

The main result of this section is that the class of commutative rings whose principal ideals have unique generators is a relatively ideal distributive quasivariety, and for which we provide an axiomatization.

Theorem 2.1**.**

Let $\mathcal{Q}$ be the class of all commutative rings $R$ (with $1$) having the property that each principal ideal has a unique generator. Let $\mathcal{D}$ be the class of domains in $\mathcal{Q}$.

1. (1)

$\mathcal{Q}$* is a quasivariety. It is exactly the class of rings axiomatized by the quasi-identity $(xyz=z)\to(yz=z)$ along with the identities defining the variety of all commutative rings. All rings in $\mathcal{Q}$ have trivial unit group and are reduced. Such rings are $\mathbb{F}_{2}$-algebras.* 2. (2)

$\mathcal{D}$* is exactly the class of domains with trivial unit group.* 3. (3)

$\mathcal{Q}$* consist of the subrings of products of members of $\mathcal{D}$ (we write ${\mathcal{Q}}=\mathsf{S}\mathsf{P}({\mathcal{D}})$).* 4. (4)

${\mathcal{Q}}$* is a relatively ideal distributive quasivariety.* 5. (5)

The class of locally finite algebras in ${\mathcal{Q}}$ is the class of Boolean rings. This class is the largest subvariety of $\mathcal{Q}$.

Proof.

We argue the first two claims of Item (1) together. Namely, we show that $R\in\mathcal{Q}$ if and only if $R$ belongs to the quasivariety of commutative rings satisfying $(xyz=z)\to(yz=z)$.

For the “if” part, let $R$ be a commutative ring satisfying $(xyz=z)\to(yz=z)$. Choose $r\in R$ and assume that $(r)=(s)$ for some $s$. Then $s=qr$ and $r=ps$ for some $p,q\in R$. Since $pqr=r$, the quasi-identity yields $qr=r$, or $s=r$. Thus, $(r)=(s)$ implies $r=s$, showing that $R$ satisfies the unique generator property for principal ideals. Conversely, for “only if”, suppose that $R$ does not satisfy $(xyz=z)\to(yz=z)$. $R$ must have elements $p,q,r$ such that $pqr=r$ and $qr\neq r$, Then $(r)=(qr)$ and $qr\neq r$, so $R$ does not have the unique generator property.

For the second to last statement of Item (1), suppose that $R\in{\mathcal{Q}}$ and that $u$ is a unit in $R$. Then $(u)=R=(1)$, so by the unique generator property $u=1$. Also, to see that $R$ is reduced, assume that $n\in R$ satisfies $n^{2}=0$. Then $1+n$ is a unit (with inverse $1-n$), so $1+n=1$, so $n=0$.

For the final statement of Item (1), the fact that any $R\in{\mathcal{Q}}$ is an $\mathbb{F}_{2}$-algebra follows from the fact that $-1$ is a unit, so $1=-1$. Then the prime subring of $R$ is isomorphic to $\mathbb{F}_{2}$, which is enough to establish that $R$ is an $\mathbb{F}_{2}$-algebra.

For Item (2), if $D\in{\mathcal{D}}$, then $D$ is a domain by definition, and it has trivial unit group by Item (1). Conversely, suppose that $D$ is a domain with trivial unit group. If $(a)=(b)$ in $D$, then $a$ and $b$ must differ by a unit, hence $a=b$, showing that $D$ has the unique generator property, so $D$ is a domain in $\mathcal{Q}$, yielding $D\in\mathcal{D}$.

In order to establish Item (3) we first prove a claim.

Claim 2.2**.**

If $R\in{\mathcal{Q}}$ and $S\subseteq R$ is a subset, then the annihilator $A=\mbox{\rm ann}(S)$ is a $\mathcal{Q}$-ideal (meaning that $R/A\in{\mathcal{Q}}$).

Proof of claim. For this we must verify that $R/A$ satisfies the quasi-identity $(xyz=z)\to(yz=z)$. Equivalently, we must show that if $x,y,z\in R$ and $xyz\equiv z\pmod{A}$, then $yz\equiv z\pmod{A}$. We begin: If $xyz\equiv z\pmod{A}$, then $(xyz-z)\in A$, so $(xyz-z)s=0$ for any $s\in S$. This means that $xy(zs)=(zs)$ for any $s\in S$. Applying the quasi-identity from Item (1) with $zs$ in place of $z$ we derive that $yzs=zs$, or $(yz-z)s=0$ for any $r\in I$. Hence $yz\equiv z\pmod{A}$, as desired.

Next we argue that if $R\in{\mathcal{Q}}$ is not a domain, then $R$ has disjoint nonzero $\mathcal{Q}$-ideals $I$ and $J$. If $R$ is not a domain, then there exist nonzero $r$ and $s$ such that $rs=0$. Take $I=\mbox{\rm ann}(r)$ and $J=\mbox{\rm ann}(I)$. $I$ is nonzero since it contains $s$, and $J$ is nonzero since it contains $r$. Both $I$ and $J$ are $\mathcal{Q}$-ideals by Claim 2.2. If $t\in I\cap J$, then $t^{2}\in IJ=\{0\}$, so $t$ is nilpotent. According to Item (1), any $R\in{\mathcal{Q}}$ is reduced, so $t=0$. Thus $I$ and $J$ are indeed disjoint nonzero $\mathcal{Q}$-ideals.

The argument for Item (3) is completed by noting that any quasi-variety $\mathcal{Q}$ is expressible as $\mathsf{S}\mathsf{P}(\mathcal{K})$ where $\mathcal{K}$ is the subclass of relatively subdirectly irreducible members of $\mathcal{Q}$. This is a version of Birkhoff’s Subdirect Representation Theorem, stated for quasivarieties, and it holds for quasivarieties because relative ideal/congruence lattices are algebraic. The previous paragraph shows that the only members of ${\mathcal{Q}}$ that could possibly be relatively subdirectly irreducible are the domains. (That is, $R$ not a domain $\Rightarrow$ $R$ has disjoint nonzero $\mathcal{Q}$-ideals $\Rightarrow$ $R$ is not relatively subdirectly irreducible.)

To prove Item (4), we refer to general criteria from [10] for proving that a quasivariety is relatively congruence distributive. Specifically we will use Theorems 4.1 and 4.3 of that paper, along with some of the remarks between those theorems.

Here is a summary of what we are citing. From Theorem 4.1 of [10], a quasivariety is relatively congruence modular if and only if it satisfies the “extension principle” and the “relative shifting lemma”. From remarks following the proof of Theorem 4.1, the “extension principle” can be replaced by the “weak extension principle”. From Theorem 2.1 of that paper, the “relative shifting lemma” can be replaced by the “existence of quasi-Day terms”. Finally, from Theorem 4.3 of that paper, a quasivariety is relatively congruence distributive if and only if it is relatively congruence distributive and no member has a nonzero abelian congruence.

What this reduces to in our setting is this: to prove that our quasivariety $\mathcal{Q}$ is relatively ideal distributive (Item (4)) it suffices to show that $\mathcal{Q}$ (i) has “quasi-Day terms”, (ii) satisfies the “weak extension principle”, and (iii) has no member with a nontrivial abelian congruence (i.e., a nonzero ideal $A$ satisfying $A^{2}=0$).

Condition (i) holds since $\mathcal{Q}$ has ordinary Day terms, in fact a Maltsev term. (More explicitly, the singleton set $\Sigma_{s}:=\{(p(w,x,y,z),q(w,x,y,z)\}$ where $p(w,x,y,z):=w-x+y$ and $q(w,x,y,z):=z$ meets the defining conditions from Theorem 2.1(2) of [10] for “quasi-Day terms”.)

Condition (iii) holds since if $A\lhd R\in{\mathcal{Q}}$ and $A^{2}=0$, then the elements of $A$ are nilpotent. As argued in the proof of Item (1), the only nilpotent element in $R$ is [math], hence $A=0$.

Condition (ii) means that if $R\in{\mathcal{Q}}$ has disjoint ideals $I$ and $J$, then $I$ and $J$ can be extended to $\mathcal{Q}$-ideals $\overline{I}\supseteq I$ and $\overline{J}\supseteq J$ that are also disjoint. To prove that Condition (ii) holds we modify an argument from above: If $R$ has ideals $I$ and $J$ such that $I\cap J=0$, then $IJ=0$. The ${\mathcal{Q}}$-ideal $\overline{J}=\mbox{\rm ann}(I)$ contains $J$, the ${\mathcal{Q}}$-ideal $\overline{I}=\mbox{\rm ann}(\overline{J})$ contains $I$, both are $\mathcal{Q}$-congruences, and $\overline{I}\cap\overline{J}=0$ (since the elements in this intersection square to zero and $R$ is reduced). This shows that disjoint ideals $I$ and $J$ may be extended to disjoint $\mathcal{Q}$-ideals.

For Item (5), to show that a locally finite ring in $\mathcal{Q}$ is a Boolean ring it suffices to show that any finite ring $F\in\mathcal{Q}$ is Boolean. (The reason this reduction is permitted is that the property of being a Boolean ring is expressible by the identity $x^{2}=x$, and a locally finite structure satisfies a universal sentence if and only if its finite substructures satisfy the sentence.)

So choose a finite $F\in\mathcal{Q}$. As $F$ has trivial unit group, and $1+{\rm rad}(F)\subseteq U(F)$, we get that $F$ must be semiprimitive. Since $F$ is finite it must be a product of fields. Since $F$ has only trivial units, each factor field must have size $2$, so $F$ is Boolean.

Conversely, if $B$ is any Boolean ring, then multiplication is a semilattice operation, so $xyz\leq yz\leq z$ in the semilattice order for any $x,y,z\in B$. If, in $B$, we have first = last ($xyz=z$), then we must have middle = last ($yz=z$). Hence $B\in{\mathcal{Q}}$.

To complete the proof of Item (5) we must show that if $\mathcal{V}$ is a variety and ${\mathcal{V}}\subseteq{\mathcal{Q}}$, then $\mathcal{V}$ consists of Boolean rings. For this it suffices to show that if $R\in{\mathcal{Q}}$ is not Boolean (i.e., $R$ has an element $r$ satisfying $r\neq r^{2}$), then $R\notin{\mathcal{V}}$. This holds because $\langle r^{2}\rangle\subsetneq\langle r\rangle$ by the unique generator property, so $r/\langle r^{2}\rangle$ is a nonzero nilpotent element of $R/\langle r^{2}\rangle$, establishing that some homomorphic image of $R$ is not in $\mathcal{Q}$. ∎

By substituting $z=1$ in the quasi-identity $(xyz=z)\to(yz=z)$ we obtain the consequence $(xy=1)\to(y=1)$, which expresses that the unit group is trivial. Since a consequence can be no stronger than the original statement, this is enough to deduce that the quasivariety of commutative rings with trivial unit group contains the quasivariety of commutative rings whose principal ideals have unique generators. This containment is proper, and the following example describes a commutative ring satisfying $(xy=1)\to(y=1)$ but not $(xyz=z)\to(yz=z)$.

Example 2.3**.**

Let $R$ be the commutative $\mathbb{F}_{2}$-algebra presented by

[TABLE]

That is, $R$ is the quotient of the polynomial ring $\mathbb{F}_{2}[X,Y,Z]$ by the ideal $(XYZ-Z)$.

We may view the relation $XYZ-Z=0$ as a reduction rule $XYZ\to Z$ to produce a normal form for elements of $R$. This single rule is applied as follows: choose a monomial of the form $XYZW$ ($W$ is a product of variables) of an element in a coset of $(XYZ-Z)\subseteq\mathbb{F}_{2}[X,Y,Z]$ and replace $XYZW$ by $ZW$ That is, if each of $X,Y,Z$ appear in a monomial, we delete one instance of $X$ and one instance of $Y$ from that monomial.

The Diamond Lemma applies to show that there is a normal form for elements of $R$, and the elements in normal form are exactly the polynomials over $\mathbb{F}_{2}$ in the generators $X,Y,Z$ where no monomial is divisible by each of $X,Y$, and $Z$.

Note that each application of the reduction rule reduces the $X$-degree and the $Y$-degree of some monomial, but does not alter the $Z$-degree of any monomial. This is enough to prove that the unit group of $R$ is trivial. For if $R$ had a unit $u$ with inverse $v$, then the $Z$-degree of the product $uv=1$ is zero, but it is also the sum of the $Z$-degrees of $u$ and $v$. Hence the normal form of a unit must be $Z$-free. But then $u$ and $v$ would then be inverse units in the subring $\mathbb{F}_{2}[X,Y]$, where all elements are in normal form. Now one can argue in this subring, using $X$-degree and $Y$-degree, to conclude that none of $X,Y,Z$ appear in the normal form of a unit. We are left with $u=v=1$ as the only possibility.

Notice also that $YZ-Z$ is in normal form, so $YZ-Z\neq 0$ in $R$. This shows that $R$ fails to satisfy $(xyz=z)\to(yz=z)$, but does satisfy $(xy=1)\to(y=1)$. In particular, the fact that $XYZ=Z$ while $YZ\neq Z$ means that $(YZ)=(Z)$, while $YZ\neq Z$, so the principal ideal $(Z)$ does not have a unique generator.

3. Some related quasivarieties

Recall from the introduction that one of our goals was to identify the situation when the passage from the monoid $\langle R;\cdot\rangle$ to the poset of principal ideals results in no loss of information. This led to the investigation of rings whose principal ideals have unique generators. For these rings we progressed from (1) to (2) to (3) on this list of equivalent conditions:

1. (1)

Principal ideals of $R$ have unique generators. 2. (2)

$R$ satisfies $(xyz=z)\to(yz=z)$. 3. (3)

$R$ belongs to the quasivariety generated by domains with trivial unit group, which is the class of domains satisfying $(xy=1)\to(y=1)$.

If we had progressed in the reverse direction, taking the property of “trivial unit group” as the object of interest, the results would have been different. The quasivariety of commutative rings with trivial unit group is not generated by its subclass of domains, and it is not relatively ideal distributive.

Notice also that progressing in the reverse direction and trying to obtain the same results requires one to strengthen $(xy=1)\to(y=1)$ to $(xyz=z)\to(yz=z)$.

In this section, we will progress in the reverse direction starting with a generalization of the quasi-identity $(xy=1)\to(y=1)$. Namely, we will investigate the quasivariety of commutative rings generated by domains satisfying $(xy=1)\to(y^{n}=1)$, which asserts that every unit has exponent dividing $n$. (The previous section was entirely concerned with the case $n=1$.)

If $D$ is a domain satisfying $(xy=1)\to(y^{n}=1)$, then its group of units is a cyclic group of order dividing $n$. The reason for this is that unit group $U(D)$ is an abelian group satisfying $x^{n}=1$, hence $U(D)$ is a locally finite abelian group. If $U(D)$ is not cyclic, then it contains a finite noncyclic subgroup $G\subseteq U(D)$. But now $G$ is a finite noncyclic subgroup of the field of fractions of $D$, and we all know that the multiplicative group of a field contains no finite noncyclic subgroup.

So let ${\mathcal{Q}}_{|n}$ denote the quasivariety of commutative rings generated by the domains $D$ for which $U(D)$ is cyclic of order dividing $n$. We shall show that this quasivariety is relatively ideal distributive and we shall provide an axiomatization for ${\mathcal{Q}}_{|n}$.

Write ${\mathcal{D}}_{|n}$ for the class of domains whose unit group is cyclic of order dividing $n$.

Theorem 3.1**.**

By definition, we have that ${\mathcal{Q}}_{|n}$ is the quasivariety generated by ${\mathcal{D}}_{|n}$.

1. (1)

${\mathcal{Q}}_{|n}$* is axiomatized by*

1. (a)

the identities defining commutative rings, 2. (b)

the quasi-identity $(x^{2}=0)\to(x=0)$, which expresses that the only nilpotent element is [math], and 3. (c)

the quasi-identity $(xyz=z)\to(y^{n}z=z)$. 2. (2)

${\mathcal{Q}}_{|n}$* is a relatively ideal distributive quasivariety.*

Proof.

To prove Item (1), let $\mathcal{K}$ be the quasivariety axiomatized by the sentences in (a), (b), and (c). It is easy to see that ${\mathcal{D}}_{|n}$ satisfies the quasi-identities in (a), (b) and (c), so ${\mathcal{D}}_{|n}\subseteq{\mathcal{K}}$, and therefore ${\mathcal{Q}}_{|n}\subseteq{\mathcal{K}}$.

Conversely, we must show that ${\mathcal{K}}\subseteq{\mathcal{Q}}_{|n}$ For this, we need the analogue of Claim 2.2 for ${\mathcal{K}}$:

Claim 3.2**.**

If $R\in{\mathcal{K}}$ and $S\subseteq R$ is a subset, then the annihilator $A=\mbox{\rm ann}(S)$ is a ${\mathcal{K}}$-ideal.

Proof of claim. Our goal is to prove that $R/A\in{\mathcal{K}}$, so we must prove that $R/A$ is a commutative ring satisfying $(x^{2}=0)\to(x=0)$ and $(xyz=z)\to(y^{n}z=z)$. It is clear that $R/A$ is a commutative ring (identities are preserved under quotients), so we only need to verify that $R/A$ satisfies $(x^{2}=0)\to(x=0)$ and $(xyz=z)\to(y^{n}z=z)$. For the second of these, the proof is exactly like the proof of Claim 2.2, while for the first there is an extra idea. We prove the first only.

To prove that $R/A$ satisfies $(x^{2}=0)\to(x=0)$, we must show that $R$ satisfies $x^{2}\equiv 0\pmod{A}$ implies $x\equiv 0\pmod{A}$. If $x^{2}\equiv 0\pmod{A}$, or $x^{2}\in A$, then $x^{2}s=0$ for all $s\in S$. This implies $(xs)^{2}=(x^{2}s)s=0$ for all $s\in S$. (This is the “extra idea”.) But $R$ satisfies $(x^{2}=0)\to(x=0)$, so from $(xs)^{2}=0$ we deduce $xs=0$ for all $s\in S$. This proves that $x\in A$ or $x\equiv 0\pmod{A}$.

We will use Claim 3.2 the same way we used Claim 2.2 in the proof of Theorem 2.1. If $R\in\mathcal{K}$ is not a domain, then there exist nonzero $r$ and $s$ such that $rs=0$. Take $I=\mbox{\rm ann}(r)$ and $J=\mbox{\rm ann}(I)$. $I$ is nonzero since it contains $s$, and $J$ is nonzero since it contains $r$. By Claim 3.2, $I$ and $J$ are $\mathcal{K}$-ideals. Any element in $I\cap J$ must square to zero, so since $\mathcal{K}$ satisfies axiom (b) we get $I\cap J=\{0\}$. Thus, if $R$ is not a domain, then it has a pair of nonzero, disjoint, ${\mathcal{K}}$-ideals. This is enough to guarantee that $R$ is not subdirectly irreducible relative to $\mathcal{K}$.

In the contrapositive form, we have shown that any relatively subdirectly irreducible member of $\mathcal{K}$ is a domain. Hence ${\mathcal{K}}$ is generated by its subclass of domains.

But if $D\in{\mathcal{K}}$ is a domain, then by (1)(c) we derive that $D$ satisfies $(xyz=z)\to(y^{n}z=z)$. For domains we can cancel $z$ to obtain that $D$ satisfies $(xy=1)\to(y^{n}=1)$. This shows that the domains in ${\mathcal{K}}$ lie in ${\mathcal{D}}_{|n}$, so ${\mathcal{K}}$ is contained the quasivariety generated by ${\mathcal{D}}_{|n}$, which is ${\mathcal{Q}}_{|n}$.

Item (2) of this theorem is proved exactly like Item (4) of Theorem 2.1. ∎

Observation 3.3**.**

A quick test to rule out that some nonzero ring $R$ belongs to some quasivariety $\mathcal{Q}_{|n}$ is to show that the prime subring of $R$ does not belong to $\mathcal{Q}_{|n}$. Since the prime subring of $R$ is isomorphic to $\mathbb{Z}_{k}$ for some $k>1$, and since the units of $\mathbb{Z}_{k}$ are easy to calculate, it is not hard to derive some consequences.

In fact, it is not hard to write down the exact relationship between $k$ and $n$ which guarantees that $\mathbb{Z}_{k}\notin{\mathcal{Q}}_{|n}$, but here we only mention that if $n$ is odd, then $\mathbb{Z}_{k}\notin{\mathcal{Q}}_{|n}$ unless $k=2$. Also, $\mathbb{Z}\notin{\mathcal{Q}}_{|n}$.

To see this, note that if $R=\mathbb{Z}_{k}$ for $k\neq 2$ or $R=\mathbb{Z}$, then $1\neq-1$, so $U(R)$ contains an element $-1$ of order $2$. But if $n$ is odd, then the rings in ${\mathbb{Q}}_{|n}$ have unit groups of odd exponent, hence do not have units of order $2$.

In particular, this shows that if $n$ is odd, then each ring in ${\mathbb{Q}}_{|n}$ may be thought of as a $\mathbb{F}_{2}$-algebra.

We can use this observation to show that not all the quasivarieties ${\mathcal{Q}}_{|n}$ are distinct, in particular

Theorem 3.4**.**

If $p$ is an odd prime, then ${\mathcal{Q}}_{|p}={\mathcal{Q}}_{|1}$ unless $p$ is a Mersenne prime.

Proof.

To prove that ${\mathcal{Q}}_{|p}={\mathcal{Q}}_{|1}$ when $p$ is an odd non-Mersenne prime, it will suffice to show that these quasivarieties contain the same domains. We always have ${\mathcal{Q}}_{|m}\subseteq{\mathcal{Q}}_{|n}$ when $m\mid n$, from the definition of these quasivarieties, so we must show that any domain $D\in{\mathcal{Q}}_{|p}$ is contained in ${\mathcal{Q}}_{|1}$ (i.e. has a trivial unit group).

Choose $D\in{\mathcal{Q}}_{|p}$. From Observation 3.3, we know (since $p$ is odd) that $D$ is an $\mathbb{F}_{2}$-algebra. Suppose that $\theta\in D$ is a nontrivial unit. Since $\theta$ has finite multiplicative order, and the prime subring of $D$ is finite, the subring $S\subseteq D$ generated by $\theta$ is finite. $S$ is a subring of a domain itself, hence it is a field, and $U(S)=S^{\times}$. $S$ belongs to ${\mathcal{Q}}_{|p}$, so $S^{\times}$ has order dividing $p$, and it must therefore be that $|S^{\times}|=p$. This shows that $S$ is a finite field of characteristic $2$ and of cardinality $|S|=p+1$. We derive that $p+1=2^{s}$ for some $s$, or $p=2^{s}-1$. This completes the proof that ${\mathcal{Q}}_{|p}={\mathcal{Q}}_{|1}$ unless $p$ is a Mersenne prime. ∎

The primality of $p$ did not play a big role in the proof. The same argument shows that if $n$ is any odd number, then ${\mathcal{Q}}_{|n}={\mathcal{Q}}_{|1}$ unless $n$ is divisible by some number $x>1$ of the form $x=2^{s}-1$. So, for example, ${\mathcal{Q}}_{|55}={\mathcal{Q}}_{|25}={\mathcal{Q}}_{|1}$. But if $n$ is divisible by some number $x>1$ of the form $x=2^{s}-1$, then ${\mathcal{Q}}_{|n}$ will contain some finite fields that are not in ${\mathcal{Q}}_{|1}$.

4. A representation theorem for Vasconcelos monoids

In this section we will focus on the structure ${\mathcal{I}}_{\textrm{f.g.}}(R)$ formed by the finitely generated ideals of a commutative unital ring $R$. We will refer to ${\mathcal{I}}_{\textrm{f.g.}}(R)$ as the divisibility theory of $R$. Traditionally, ${\mathcal{I}}_{\textrm{f.g.}}(R)$ is considered partially ordered under reverse inclusion: $I\leq_{l}J$ $\Leftrightarrow$ $I\supseteq J$. Thus $\langle{\mathcal{I}}_{\textrm{f.g.}}(R);\wedge\rangle$ is a meet subsemilattice of the dual ${\mathcal{I}}^{d}(R)$ of the ideal lattice of $R$ where $I\wedge J=I+J$ for all $I,J\in{\mathcal{I}}_{\textrm{f.g.}}(R)$. In general, ${\mathcal{I}}_{\textrm{f.g.}}(R)$ is not a sublattice of ${\mathcal{I}}^{d}(R)$, because the intersection of finitely generated ideals of $R$ need not be finitely generated.

Ideal multiplication induces another monoid structure $\langle{\mathcal{I}}_{\textrm{f.g.}}(R);\cdot\rangle$ on ${\mathcal{I}}_{\textrm{f.g.}}(R)$. Divisibility with respect to this multiplication yields another partial order on ${\mathcal{I}}_{\textrm{f.g.}}(R)$: for any $I,J\in{\mathcal{I}}_{\textrm{f.g.}}(R)$, we let $I\leq_{m}J$ $\Leftrightarrow$ $J=IK$ for some $K\in{\mathcal{I}}_{\textrm{f.g.}}(R)$. In general, the partial orders $\leq_{l}$ and $\leq_{m}$ are different.

However, if $R$ is arithmetical [6], that is, if the ideal lattice of $R$ is distributive, then

• •

the partial orders $\leq_{l}$ and $\leq_{m}$ coincide (see [7, Theorem 2] and [2, p. 3972]), and will be denoted by $\leq$; furthermore,

• •

the intersection of any two finitely generated ideals is finitely generated [2, Proposition 1.7].

Consequently,

• •

$\langle{\mathcal{I}}_{\textrm{f.g.}}(R);\wedge,\vee\rangle$ is a sublattice of ${\mathcal{I}}^{d}(R)$; namely, $I\wedge J=I+J$ and $I\vee J=I\cap J=I_{1}J$ for all $I,J\in{\mathcal{I}}_{\textrm{f.g.}}(R)$, where $I_{1}\in{\mathcal{I}}_{\textrm{f.g.}}(R)$ satisfies $I=(I\wedge J)I_{1}$ (see [2, Proposition 1.2 and its proof]).

In fact, it follows that

• •

$\langle{\mathcal{I}}_{\textrm{f.g.}}(R);\cdot,\leq,\wedge,\vee\rangle$ is a Bézout monoid as defined below (see [2, Proposition 1.6]).

Definition 4.1**.**

(See [5, Section 4], [2, Definition 1.1].)

A Bézout monoid $S$ is a commutative monoid $\langle S;\cdot\rangle$ with [math] such that

1. (1)

the divisibility relation $a\mid b\Leftrightarrow aS\supseteq bS$ is a partial order $\leq$ on $S$, called the natural partial order of $S$, 2. (2)

$\leq$ is a lattice order, and the induced lattice $\langle S;\wedge,\vee\rangle$ is distributive; 3. (3)

$\cdot$ is distributive over $\wedge$ and $\vee$; and 4. (4)

$S$ is hyper-normal, that is, for any elements $a,b\in S$, if $d=a\wedge b$ and $a_{1}\in S$ is such that $a=da_{1}$, then there exists an element $b_{1}\in S$ satisfying $a_{1}\wedge b_{1}=1$ and $b=db_{1}$.

As we mentioned right before this definition, the divisibility theory ${\mathcal{I}}_{\textrm{f.g.}}(R)$ of an arithmetical ring $R$ is a Bézout monoid. An major open problem is whether the converse of this statement is also true; that is: Is every Bézout monoid isomorphic to the divisibility theory of an arithmetical ring? Primary examples of arithmetical rings are Bézout rings, that is, commutative unital rings in which all finitely generated ideals are principal. However, there are many other classes of arithmetical rings, e.g., semihereditary rings and rings of weak dimension $1$. Nevertheless, by a result of Anderson [1], the divisibility theory ${\mathcal{I}}_{\textrm{f.g.}}(R)$ of every arithmetical ring $R$ is isomorphic to the divisibility theory ${\mathcal{I}}_{\textrm{f.g.}}(\sf{R})={\mathcal{I}}_{\textrm{princ}}(\sf{R})$ of a Bézout ring $\sf{R}$. Therefore, the question above is equivalent to the following:

Is every Bézout monoid isomorphic to the divisibility theory of a Bézout ring?

The answer is known to be YES for Bézout monoids that are either semihereditary [3] or have only one minimal $m$-prime filter [4] (for the definitions, see the forthcoming paragraphs).

Our aim in this section is to prove a similar representation theorem for a class of Bézout monoids that we will call Vasconcelos monoids. The name is motivated by the fact that a well-known example of a Bézout ring constructed by Vasconcelos [12, Example 3.2] (see also [11, Example 3]) has its divisibility theory in this class.

Before defining Vasconcelos monoids we recall some definitions and basic facts on Bézout monoids. Let $S$ be a Bézout monoid. For any element $a\in S$ we let $a^{\perp}$ denote the annihilator $\{x\in S:ax=0\}$ of $a$. By a filter of $S$ we mean a lattice filter of $S$, that is, a subset $F$ of $S$ such that $F$ is up-closed (i.e., $b\in F$ whenever $S\ni b\geq a\in F$) and is closed under $\wedge$. Clearly, every filter $F$ of $S$ is a monoid ideal of $S$ as well, that is, $F$ satisfies $FS\subseteq F$. Furthermore, properties (2) and (3) in the definition of a Bézout monoid imply that $a^{\perp}$ is a filter of $S$ for every element $a\in S$. A filter $F$ of $S$ is said to be multiplication prime (briefly, $m$-prime) if $F\not=S$ and $ab\in F$ implies $a\in F$ or $b\in F$ for all $a,b\in S$. If $F$ is a filter of $S$, we define $S/F$ to be the factor of $S=\langle S;\cdot,\leq,\wedge,\vee\rangle$ by the congruence $\equiv_{F}$ defined for all $a,b\in S$ by $a\equiv_{F}b$ iff $a\wedge c=b\wedge c$ for some $c\in F$. It is not hard to show (see [2, Theorem 2.15]) that $S/F$ is a Bézout monoid. The following two basic properties of Bézout monoids will also be important later on.

Proposition 4.2**.**

If $S$ is Bézout monoid, then

1. (1)

$1$* is the unique invertible element of $S$; and* 2. (2)

the idempotent elements of $S$ form a Boolean algebra, and for any two idempotent elements $e,f$ of $S$ we have $e\vee f=ef$. **[2, Propositions 1.14, 1.16]**

It follows from statement (2) above that if $S$ is a Bézout monoid, then for any idempotent $e^{2}=e\in S$ there exists a unique idempotent $f^{2}=f\in S$ such that $e\wedge f=1$ and $e\vee f=ef=0$. We will refer to $f$ as the orthogonal complement of $e$.

For a Bézout monoid $S$, we say that

• •

$S$ is semihereditary if for each $a\in S$ there exists an idempotent $e^{2}=e\in S$ such that $a^{\perp}=eS$; and

• •

$S$ is semiprime if it has no non-zero nilpotent elements, or equivalently, if the intersection of its $m$-prime filters is $\{0\}$.

It is known that semihereditary Bézout monoids are semiprime [3, Corollary 2.2].

For the rest of this section we will focus on Bézout monoids $S$ that are semiprime. For each $a\in S$ let $S_{a}=\{b\in S:b^{\perp}=a^{\perp}\}$. In particular, $S_{1}$ is the set of all elements $b\in S$ satisfying $b^{\perp}=1^{\perp}=\{0\}$; that is, $S_{1}$ is the set of all elements of $S$ that are not zero divisors (in $S$).

Proposition 4.3**.**

[2, Proposition 3.3, Theorem 3.5]** Let $S$ be a semiprime Bézout monoid, and let $B$ be the Boolean algebra of all idempotent elements of $S$.

1. (1)

$(st)^{\perp}=(s\vee t)^{\perp}$* and $(s\wedge t)^{\perp}=s^{\perp}\cap t^{\perp}$ hold for arbitrary elements $s,t\in S$.* 2. (2)

For every element $a\in S$,

1. (i)

$S_{a}$* is a cancellative subsemigroup of $S$, which is closed under $\wedge$ and $\vee$, and* 2. (ii)

the map $S_{a}\to S/a^{\perp}$, $x\mapsto x/{\equiv_{a^{\perp}}}$ embeds $S_{a}$ into the subsemigroup $(S/a^{\perp})_{1}$ of $S/a^{\perp}$; moreover, $S_{a}$ and $(S/a^{\perp})_{1}$ have the same group of quotients. 3. (3)

$S$* is the disjoint union of its distinct (cancellative) submonoids $S_{a}$, where $S_{0}=\{0\}$ and $S_{1}$ consists of all elements of $S$ that are not zero divisors in $S$. In particular, $S_{e_{1}}\cap S_{e_{2}}=\emptyset$ whenever $e_{1},e_{2}$ are different elements in $B$.* 4. (4)

$S_{e}=eS_{1}$* holds for every $e\in B$.*

Proof.

For a proof of (1), (2), and the first statement in (3) we refer the reader to [2, Proposition 3.3] and [2, Theorem 3.5]. For any $a,b\in S$ the definitions of $S_{a}$ and $S_{b}$ imply that $S_{a}\not=S_{b}\ \Leftrightarrow\ a^{\perp}\not=b^{\perp}\ \Leftrightarrow\ S_{a}\cap S_{b}=\emptyset$. Therefore the second statement in (3) is an immediate consequence of the fact that for $e_{1},e_{2}\in B$ we have $e_{1}^{\perp}\cap B\not=e_{2}^{\perp}\cap B$ whenever $e_{1}\not=e_{2}$.

In (4) the inclusion $S_{e}\supseteq eS_{1}$ holds, because $(ec)^{\perp}=e^{\perp}$ for all $c\in S_{1}$. To prove the reverse inclusion $S_{e}\subseteq eS_{1}$, let $s\in S_{e}$, and let $f$ be the orthogonal complement of $e$. Then $fs=0$, because $s\in S_{e}$ implies that $f\in e^{\perp}=s^{\perp}$. Hence,

[TABLE]

where $s\wedge f\in S_{1}$, because $(s\wedge f)^{\perp}=s^{\perp}\cap f^{\perp}=e^{\perp}\cap f^{\perp}=(e\wedge f)^{\perp}=1^{\perp}=\{0\}$. ∎

Now we are ready to define Vasconcelos monoids.

Definition 4.4**.**

A Vasconcelos monoid is a Bézout monoid $S$ that satisfies the following conditions:

1. (1)

$S$ is semiprime, 2. (2)

$S$ contains a zero divisor $a\not=0$ such that $P=a^{\perp}$ is an $m$-prime filter of $S$ which consists of idempotents, and 3. (3)

for the submonoid $C=S_{1}\cup S_{a}$ of $S$ (where $S_{1}S_{a}\subseteq S_{a}$), $C\cup\{0\}$ is isomorphic to $S/P=S/a^{\perp}$ via the map $x\mapsto x/{\equiv_{P}}$ ($x\in C\cup\{0\}$), hence $C\cup\{0\}$ is a Bézout monoid, and $C$ is isomorphic to the positive cone of the lattice ordered abelian group of quotients of $S_{a}$.

We will need several consequences of this definition, which we state and prove in the next proposition.

Proposition 4.5**.**

Let $S$ be a Vasconcelos monoid, let $a\in S$ be as in Definition 4.4 above, and let $P=a^{\perp}$. Furthermore, let $B$ denote the Boolean algebra of all idempotent elements of $S$.

1. (1)

$B\setminus P$* is the set of orthogonal complements of the elements of $P$.* 2. (2)

$ce=e$* holds for all $c\in S_{1}$ and $e\in P$; hence, $P\subseteq cS$ for all $c\in S_{1}$.* 3. (3)

$de=0$* and $df=d$ hold for all $d\in S_{a}\cup\{0\}$, $e\in P$, and $f\in B\setminus P$.* 4. (4)

$S$* is the disjoint union of its subsets*

[TABLE]

Moreover, every element of $S_{1}(B\setminus P)$ can be written uniquely as a product $cf$ with $c\in S_{1}$, $f\in B\setminus P$, and every element of $(S_{a}\cup\{0\})\wedge P$ can be written uniquely as a meet $d\wedge e$ with $d\in S_{a}\cup\{0\}$, $e\in P$. 5. (5)

If the elements of $S$ are written in the form described in (4), then the monoid operation of $S$ takes the following form: for all $c_{1},c_{2}\in S_{1}$, $d_{1},d_{2}\in S_{a}$, $e_{1},e_{2}\in P$, and $f_{1},f_{2}\in B\setminus P$ where $f_{i}$ is the orthogonal complement of $e_{i}$ for $i=1,2$,

[TABLE]

Proof.

To prove (1), recall that $P\subseteq B$ is an $m$-prime filter of $S$. Therefore, if $e,f$ is any pair of orthogonal idempotents in $B$, then $ef=0\in P$ and $e\wedge f=1\notin P$ imply that exactly one of $e,f$ belongs to $P$, which proves statement (1).

For (2), let $c\in S_{1}$ and $e\in P$. Then $ce\in P$, since $P$ is a filter of $S$, and $ce$ is idempotent, since $P\subseteq B$. But $c\in S_{1}$ implies that $(ce)^{\perp}=e^{\perp}$, therefore it must be that $ce=e$. Now the inclusion $P\subseteq cS$ is clear for all $c\in S_{1}$.

Statement (3) is trivial if $d=0$, so let us assume that $d\in S_{a}$. In view of statement (1), we may assume without loss of generality that $f\in B\setminus P$ is the orthogonal complement of $e\in P$. Since $d\in S_{a}$, we have $d^{\perp}=a^{\perp}=P$, so $de=0$. Hence, $d=d1=d(e\wedge f)=de\wedge df=0\wedge df=df$.

To prove the first statement in (4), let $s\in S$. By item (3) in Definition 4.4, there exists $c\in C\cup\{0\}=S_{1}\cup S_{a}\cup\{0\}$ such that $s/{\equiv}_{P}=c/{\equiv}_{P}$, that is, $s\wedge e^{\prime}=c\wedge e^{\prime}$ for some $e^{\prime}\in P$. Let $f^{\prime}$ denote the orthogonal complement of $e^{\prime}$. Then $e^{\prime}f^{\prime}=0$ and $e^{\prime}\wedge f^{\prime}=1$, hence $sf^{\prime}=sf^{\prime}\wedge e^{\prime}f^{\prime}=(s\wedge e^{\prime})f^{\prime}=(c\wedge e^{\prime})f^{\prime}=cf^{\prime}\wedge e^{\prime}f^{\prime}=cf^{\prime}$. Notice also that $se^{\prime}\in P$, so $se^{\prime}\in B$ (i.e., $se^{\prime}$ is idempotent). First, let $c\in S_{1}$. Then, by statement (2), $ce^{\prime}=e^{\prime}$. Since $se^{\prime}\in B$, we get that $f=f^{\prime}\wedge se^{\prime}\in B$. Clearly, $f\notin P$, because $f\leq f^{\prime}$ and $f^{\prime}\notin P$. Furthermore,

[TABLE]

where the third equality follows from the equalities $cf^{\prime}=sf^{\prime}$ and $ce^{\prime}=e^{\prime}$ proved earlier. This show that in the case $c\in S_{1}$, we have that $s\in S_{1}(B\setminus P)$. Now let $c\in S_{a}\cup\{0\}$. If $c=0$, that is, if $s\wedge e^{\prime}=0\wedge e^{\prime}=e^{\prime}$ for some $e^{\prime}\in P$, then $s\geq e^{\prime}$ and $s\in P$. Hence $s\in\{0\}\wedge P$. It remains to consider the case $c\in S_{a}$. Since $P$ is an $m$-prime filter of $S$ and $s,c,f^{\prime}\notin P$, we get that $sf^{\prime}=cf^{\prime}\notin P$. Next we want to argue that $cf^{\prime}\in S_{a}$, that is, $(cf^{\prime})^{\perp}=a^{\perp}=P$. The inclusion $\supseteq$ is clear, because $c\in S_{a}$. For $\subseteq$, let $x\in(cf^{\prime})^{\perp}$, that is, $cf^{\prime}x=0$. Then $cf^{\prime}x\in P$ and $cf^{\prime}\notin P$ imply that $x\in P$, as required. Thus, using again the equality $sf^{\prime}=cf^{\prime}$ we get that

[TABLE]

with $cf^{\prime}\in S_{a}$ and $se^{\prime}\in P$, which proves that $s\in S_{a}\wedge P$ in this case.

Our argument above shows that the set $S_{1}(B\setminus P)$ is the union of the equivalence classes $c/{\equiv}_{P}\in S/P$ with $c\in S_{1}$, while the set $(S_{a}\cup\{0\})\wedge P$ is the union of the equivalence classes $c/{\equiv}_{P}\in S/P$ with $c\in S_{a}\cup\{0\}$. Therefore condition (3) in Definition 4.4 ensures that the sets $S_{1}(B\setminus P)$ and $S_{1}(B\setminus P)$ are disjoint. This completes the proof of the first statement in (4).

Now we prove the uniqueness statement in (4). Let $c_{1},c_{2}\in S_{1}$ and $f_{1},f_{2}\in B\setminus P$ be such that $c_{1}f_{1}=c_{2}f_{2}$. Then, by Proposition 4.3(4), $c_{1}f_{1}=c_{2}f_{2}$ is an element of $f_{i}S_{1}=S_{f_{i}}$ for $i=1,2$. Therefore $S_{f_{1}}$ and $S_{f_{2}}$ are not disjoint, so by Proposition 4.3(3) it must be that $f_{1}=f_{2}$. To show that $c_{1}=c_{2}$, consider the orthogonal complement $e$ of $f_{1}=f_{2}\in B\setminus P$, and recall from statements (1) and (2) that $e\in P$ and $c_{1}e=e=c_{2}e$. Thus, $c_{i}=c_{i}1=c_{i}(e\wedge f_{i})=c_{i}e\wedge c_{i}f_{i}=e\wedge c_{i}f_{i}$ for $i=1,2$, whence $c_{1}=e\wedge c_{1}f_{1}=e\wedge c_{2}f_{2}=c_{2}$. This proves the uniqueness statement for the subset $S_{1}(B\setminus P)$ of $S$. Now let $d_{1},d_{2}\in S_{a}\cup\{0\}$ and $e_{1},e_{2}\in P$ be such that $d_{1}\wedge e_{1}=d_{2}\wedge e_{2}$. Then all four products $d_{1}e_{1},d_{1}e_{2},d_{2}e_{1},d_{2}e_{2}$ are [math] by statement (3). It follows that $d_{1}^{2}=d_{1}d_{2}$, because

[TABLE]

Since $S_{a}$ is a cancellative submonoid of $S$ (by Proposition 4.3(2)), this forces $d_{1}=d_{2}$. A similar calculation shows that $e_{1}^{2}=e_{1}e_{2}$ holds as well:

[TABLE]

By switching the roles of $d_{1}\wedge e_{1}$ and $d_{2}\wedge e_{2}$ we also get that $e_{2}^{2}=e_{1}e_{2}$. Hence $e_{1}=e_{1}^{2}=e_{1}e_{2}=e_{2}^{2}=e_{2}$, which proves the uniqueness statement for the subset $(S_{a}\cup\{0\})\wedge P$ of $S$. The proof of statement (4) is now complete.

For the proof of statement (5) let $c_{1},c_{2},d_{1},d_{2},e_{1},e_{2},f_{1},f_{2}$ satisfy the assumptions. For the elements $c_{1},c_{2},d_{1},d_{2}$, Proposition 4.3(2)(i) implies that $c_{1}c_{2}\in S_{1}$ and $d_{1}d_{2}\in S_{a}\cup\{0\}$. The fact that $c_{1}d_{2}\in S_{a}\cup\{0\}$ is clear if $d_{2}=0$, and follows from $(c_{1}d_{2})^{\perp}=d_{2}^{\perp}=a^{\perp}$, since $c_{1}\in S_{1}$ and $d_{2}\in S_{a}$. Statement (1) and the fact that $P$ is a filter clearly imply that $e_{1}\vee e_{2}\in P$, $e=f_{1}e_{2}\in P$, and the idempotent $f_{1}\vee f_{2}$, which is the orthogonal complement of $e_{1}\wedge e_{2}\in P$ belongs to $B\setminus P$. Furthermore, it is straightforward to verify that $e=f_{1}e_{2}$ is the relative complement of $e_{1}e_{2}$ in $e_{2}$; indeed, $e_{1}e_{2}\wedge e=e_{1}e_{2}\wedge f_{1}e_{2}=(e_{1}\wedge f_{1})e_{2}=1e_{2}=e_{2}$ and $e_{1}e_{2}\vee e=e_{1}e_{2}\vee f_{1}e_{2}=(e_{1}\vee f_{1})e_{2}=0e_{2}=0$.

Now we verify the three equalities in (5). For the first one, clearly, $(c_{1}f_{1})(c_{2}f_{2})=(c_{1}c_{2})(f_{1}f_{2})$ where $f_{1}f_{2}=f_{1}\vee f_{2}$ by Proposition 4.2(2). For the second one, we have $(d_{1}\wedge e_{1})(d_{2}\wedge e_{2})=d_{1}d_{2}\wedge d_{1}e_{2}\wedge d_{2}e_{1}\wedge e_{1}e_{2}$ where $d_{1}e_{2}=0=d_{2}e_{1}$ by statement (3). Hence, $(d_{1}\wedge e_{1})(d_{2}\wedge e_{2})=d_{1}d_{2}\wedge e_{1}e_{2}$ where $e_{1}e_{2}=e_{1}\vee e_{2}$ as before. Finally, for the third equality, $(c_{1}f_{1})(d_{2}\wedge e_{2})=c_{1}f_{1}d_{2}\wedge c_{1}f_{1}e_{2}$ where $d_{2}f_{1}=d_{2}$ and $c_{1}e_{2}=e_{2}$ hold by statements (3) and (2), respectively. Therefore, $(c_{1}f_{1})(d_{2}\wedge e_{2})=c_{1}d_{2}\wedge f_{1}e_{2}=c_{1}d_{2}\wedge e$, completing the proof. ∎

The main result of this section is the following representation theorem.

Theorem 4.6**.**

For every Vasconcelos monoid $S$ there exists a Bézout ring $R$ such that the divisibility theory of $R$ is isomorphic to $S$.

Proof.

Let $S$ be a Vasconcelos monoid, and let $B$ denote the Boolean algebra of all idempotent elements of $S$. By Definition 4.4, $S$ is semiprime, and $S$ contains a zero divisor $a\not=0$ such that conditions (2)–(3) of the definition hold for $P=a^{\perp}$ and the submonoid $C=S_{1}\cup S_{a}$ of $S$. Recall that $C\cup\{0\}$ is Bézout monoid isomorphic to $S/P$, and $C$ is a cancellative monoid isomorphic to the positive cone of the lattice ordered abelian group of quotients of $S_{a}$.

For the construction of the Bézout ring $R$ with divisibility theory $S$ we will use parts of the construction developed for the main theorem of [3], which we will recall now.

Let $L$ be an arbitrary field, and let $H$ be the localization of the semigroup algebra $LC$ at the set

[TABLE]

which is a Bézout domain with divisibility theory isomorphic to $C\cup 0$. Since $S$ is a Vasconcelos monoid, condition (3) in Definition 4.4 implies that the ideal $Q$ of $H$ generated by all elements $s\in S_{a}$ is a prime ideal, and hence $H/Q$ is an integral domain. Let $K$ denote the field of fractions of $H/Q$, and note that $K$ is isomorphic to the field of fractions of the semigroup algebra $LS_{1}$.

The second piece of the construction we need from [3] is the ring $C_{K}(\mathcal{X}(S))$ of all continuous functions from the spectrum $\mathcal{X}(S)$ of the Boolean algebra $B$ of all idempotents of $S$ endowed with the Zariski topology, into the field $K$ endowed with the discrete topology. For each $e\in B$ let $D_{e}$ denote the set of all maximal ideals of $B$ not containing $e$. The sets $D_{e}$ ($e\in B$) form a basis for the Zariski topology on $\mathcal{X}(S)$, which consists of clopen sets. If $U$ is a clopen subset of $\mathcal{X}(S)$, then its characteristic function ${\chi}_{U}$ (which is $1$ on $U$ and [math] on $\mathcal{X}(S)\setminus U$) is trivially an element of $C_{K}(\mathcal{X}(S))$. If $U=D_{e}$ for some $e\in B$, we will denote ${\chi}_{U}$ simply by ${\chi}_{e}$. Since $\mathcal{X}(S)$, endowed with the Zariski topology, is compact, a function $\varphi:\mathcal{X}(S)\to K$ is continuous if and only if there exist finitely many pairwise orthogonal idempotents $e_{i}\in B$ with $\bigwedge_{i=0}^{n-1}e_{i}=1$ such that $\varphi$ is equal to a linear combination of the form $\sum_{i=0}^{n-1}a_{i}{\chi}_{e_{i}}$ with all $a_{i}\in K$. We may, and will, assume without loss of generality that in all such linear combinations $a_{0}=0$, $a_{1},\dots,a_{n-1}\not=0$, and $e_{1},\dots,e_{n-1}\not=0$; that is, $D_{e_{0}}$ is the subset $\varphi^{-1}(0)$ of $\mathcal{X}(S)$ (which may be the empty set), while $\bigcup_{i=1}^{n-1}D_{e_{i}}=\varphi^{-1}(K\setminus\{0\})$ is the support of $\varphi$, where $D_{e_{1}},\dots,D_{e_{n-1}}\not=\emptyset$. It is proved in [3] that $C_{K}(\mathcal{X}(S))$ is a Bézout ring with divisibility theory isomorphic to $B$.

Now we are in a position to construct our Bézout ring $R$ with divisibility theory $S$. Let $I$ be the ideal of $C_{K}(\mathcal{X}(S))$ generated by all elements ${\chi}_{e}$ with $e\in P$. Hence, $I$ consists of all linear combinations $\sum_{i=0}^{n-1}a_{i}\chi_{e_{i}}$ where $e_{0},\dots,e_{n-1}$ are pairwise orthogonal idempotents such that $\bigwedge_{i=0}^{n-1}e_{i}=1$, $e_{1},\dots,e_{n-1}$ are nonzero idempotents in $P$, and for the coefficients $a_{0},a_{1},\dots,a_{n-1}$ we have that $a_{0}=0$ and $0\not=a_{i}\in K$ for all $i>0$ ($i\leq n-1$). Clearly, $I$ is a regular ring without $1$. It is also easy to see that the idempotent elements of $I$ are exactly the elements $\chi_{e}$ with $e\in P$. Furthermore, the principal ideals of $I$ are the ideals $\chi_{e}I=\chi_{e}C_{K}(\mathcal{X}(S))$ with $e\in P$, so they form a semigroup order-isomorphic to $P$.

Next we want to define an action of $H$ on $I$. Since $K$ is the field of fractions of $H/Q$, we see that for each $h\in H$ the element $\widehat{h}=h+Q\,(\in H/Q)$ is in $K$. Hence the constant function $\widehat{h}\chi_{1}$, where $\chi_{1}=\chi_{\mathcal{X}(S)}$ is the constant function with value $1$, is in the ring $C_{K}(\mathcal{X}(S))$. Clearly, multiplication by $\widehat{h}\chi_{1}$ is an $I$-module endomorphism of $I$, which is the zero endomorphism if $h\in Q$, and is an automorphism if $h\in H\setminus Q$, because in that case $\widehat{h}\in K\setminus\{0\}$. It follows, in particular, that multiplication by $\widehat{h}\chi_{1}$ ($h\in H$) maps every ideal of $I$ into itself. For each element $h\in H$ we define the action of $h$ on $I$ by

[TABLE]

Our ring $R$ is now defined, as in [12, Example 3.2], on the product set $H\times I$ with componentwise addition and with multiplication

[TABLE]

where $h_{1}i_{2}$ and $h_{2}i_{1}$ are obtained by the action of $H$ on $I$ defined above. Clearly, $H\to R$, $h\mapsto(h,0)$ is an embedding of the (unital) ring $H$ in $R$. It is easy to see that $R$ is a commutative unital ring with unity $1=(1,0)$. We also have an embedding of the (non-unital) ring $I$ in $R$, namely $I\to R$, $i\mapsto(0,i)$. If there is no danger of confusion, we may write $h$ in place of $(h,0)$ and $i$ in place of $(0,i)$.

First we determine the idempotents in $R$. Assume that $r=(h,g)\in R$ is idempotent. Then the equality $(h^{2},2hg+g^{2})=r^{2}=r=(h,g)$ implies that $h^{2}=h$, so $h\in\{0,1\}$ (since $H$ is an integral domain). Hence, either $h=0$ and $g^{2}=g$, or $h=1$ and $g^{2}=(-g)^{2}=-g$. Equivalently, either $r=(0,g)$ is an idempotent in $I$, or $r=(1,g)=(1,0)-(0,-g)=1-(0,-g)$ where $(0,-g)$ is an idempotent in $I$.

Next we determine the principal ideals of $R$. Let $r=(h,g)$ be an arbitrary element of $R$, so $h\in H$ and $g\in I$. Since the divisibility theory of the Bézout domain $H$ is isomorphic to $C\cup\{0\}$, there exist $c\in C\cup\{0\}$ and a unit $u\in H$ such that $uh=c$. Therefore $rR=urR=(c,ug)R$, so we may assume from now on that $r=(c,g)$ with $c\in C\cup\{0\}=S_{1}\cup S_{a}\cup\{0\}$ and $g\in I$. By the description of the principal ideals of $I$ discussed earlier, there exists $e\in P$ such that $gI=\chi_{e}I=\chi_{e}R$. Consequently, there exist $x,y\in I$ with $xg={\chi}_{e}$ and $g=y{\chi}_{e}$. If $c\in S_{a}\cup\{0\}\,(\subseteq Q)$, then $ci=0$ for all $i\in I$, so both of the elements $rx=(c,g)(0,x)=(0,gx)=gx=\chi_{e}$ and $\chi_{e}y=g=(0,g)$ are in the ideal $rR$. Therefore the element $c=(c,0)=(c,g)-(0,g)$, too, belongs to $rR$, which implies that $rR=cR\oplus gR$. Similarly, by switching the roles of $g$ and $\chi_{e}$ (along with those of $x$ and $y$), we get that $(c,\chi_{e})R=cR\oplus\chi_{e}R$. But $gR=gI=\chi_{e}I=\chi_{e}R$, therefore we conclude that

[TABLE]

Now let $c\in S_{1}$. By the definition of $I$, we have $g=\sum_{j=0}^{n-1}a_{j}{\chi}_{e_{j}}$ where $e_{0},\dots,e_{n-1}$ are pairwise orthogonal idempotents with $\bigwedge_{j=0}^{n-1}e_{j}=1$, $0\not=e_{1},\dots,e_{n-1}\in P$, and the coefficients $a_{0},\dots,a_{n-1}$ satisfy the following conditions: $a_{0}=0$ and $0\neq a_{j}\in K$ for every $j>0$ ($j\leq n-1$). Let $e=\bigwedge_{j=1}^{n-1}e_{j}$. Clearly, $e\in P$. Furthermore, if $g=0$, then $n=1$, $e_{0}=1$, $a_{0}=0$, and $e=0$, while if $g\not=0$, then we must have $n>1$ and $e\not=0$. Our aim is to prove that $rR=c(1,-\chi_{e^{\prime}})R$ for some $e^{\prime}\in P$. This is clearly true if $g=0$, because then $r=c=c(1,0)=c(1,-\chi_{0})$. Therefore we will assume from now on that $g\not=0$, and hence $n>1$. Notice first that for every subscript $\ell>0$ ($\ell\leq n-1$),

[TABLE]

Let $b_{\ell}=a_{\ell}+c$ for each $\ell>0$ ($\ell\leq n-1$), and let

[TABLE]

Then $e_{=},e_{\neq}\in P$, $e_{=}e_{\neq}=0$, and $e_{=}\wedge e_{\neq}=e$. Hence,

[TABLE]

Let $i=\sum_{j=1}^{n-1}b_{j}\chi_{e_{j}}=\sum_{j,\,b_{j}\not=0}b_{j}\chi_{e_{j}}$. For every $j$ with $b_{j}\not=0$ we have that $\chi_{e_{j}}I=b_{j}\chi_{e_{j}}I$, therefore $\chi_{e_{\neq}}R=\chi_{e_{\neq}}I=iI=iR$. By (4.2), we also have that $i\in rR$, and hence

[TABLE]

Thus, both principal ideals $\chi_{e_{\neq}}R=iR$ and $c(1,-\chi_{e})R$ are contained in $rR$. Their product is $\{0\}$, because $\chi_{e_{\neq}}(1,-\chi_{e})=(0,\chi_{e_{\neq}})(1,-\chi_{e})=(0,\chi_{e_{\neq}}-\chi_{e}\chi_{e_{\neq}})=(0,0)=0$, and their sum $c(1,-\chi_{e})R+\chi_{e_{\neq}}R=c(1,-\chi_{e})R+iR$ is $rR$, because $c(1,-\chi_{e})+i=r$. This proves that $rR=c(1,-\chi_{e})R\oplus\chi_{e_{\neq}}R$. Finally, we claim that

[TABLE]

For the proof of the second equality, the inclusion $\subseteq$ follows by observing that $c(1,-\chi_{e})=c(1,-\chi_{e_{=}})(1,-\chi_{e})$ and $\chi_{e_{\neq}}=c(1,-\chi_{e_{=}})(c^{-1}\chi_{e_{\neq}})$, while the reverse inclusion $\supseteq$ is a consequence of the equality $c(1,-\chi_{e_{=}})=c(1,-\chi_{e})+c\chi_{e_{\neq}}$.

To summarize our discussion in the last two paragraphs, we have proved that the set of principal ideals of $R$ is

[TABLE]

Next we will show that $R$ is a Bézout ring, that is, every finitely generated ideal of $R$ is principal. It suffices to prove that for any two principal ideals $J_{1},J_{2}$ of $R$, the ideal $J=J_{1}+J_{2}$ is principal. We will use the description of $\mathcal{I}_{\textrm{princ}}(R)$ in (4.3), and will distinguish three cases according to which of the two subsets of $\mathcal{I}_{\textrm{princ}}(R)$ indicated in (4.3) $J_{1}$ and $J_{2}$ belong to.

Assume first that both ideals $J_{i}$ ($i=1,2$) have the form $J_{i}=c_{i}(1,-\chi_{e_{i}})R$ where $c_{i}\in S_{1}$ and $e_{i}\in P$. Then $c_{1}(1,-\chi_{e_{1}})\chi_{e_{2}}=c_{1}(\chi_{e_{2}}-\chi_{e_{1}e_{2}})\in J_{1}\cap I$, so since the action of $c_{1}\in S_{1}$ on $I$ is an $I$-module automorphism which maps every ideal of $I$ into itself, we get that $\chi_{e_{2}}-\chi_{e_{1}e_{2}}\in J_{1}$. Hence, $c_{2}(1,-\chi_{e_{1}e_{2}})=c_{2}(\chi_{e_{2}}-\chi_{e_{1}e_{2}})+c_{2}(1,-\chi_{e_{2}})\in J_{1}+J_{2}=J$. By switching the roles of $J_{1}$ and $J_{2}$, we get similarly that $c_{1}(1,-\chi_{e_{1}e_{2}})\in J$. We want to show that for the element $c=c_{1}\wedge c_{2}$, which belongs to $S_{1}$ by Proposition 4.3(2), we also have $c(1,-\chi_{e_{1}e_{2}})\in J$. If $c_{1}=c_{2}$, then $c=c_{1}=c_{2}$, so our claim $c(1,-\chi_{e_{1}e_{2}})\in J$ trivially follows from $c_{i}(1,-\chi_{e_{1}e_{2}})\in J$ ($i=1,2$). So, assume from now on that $c_{1}\neq c_{2}$. Using the hypernormality of $S$ (see Definition 4.1(4)) we get that there exist elements $w_{1},w_{2}\in S$ such that $c_{1}=cw_{1},c_{2}=cw_{2}$, and $w_{1}\wedge w_{2}=1$. The equalities $c_{1}=cw_{1}$ and $c_{2}=cw_{2}$ imply that $c,w_{1},w_{2}$ are all in $S_{1}$. Since $S_{1}$ is cancellative (see Proposition 4.3(2)) and $c_{1}\neq c_{2}$, we get that $w_{1}\neq w_{2}$. Hence, by the definition of $H$, the element $h=w_{1}+w_{2}\in H$ has an inverse in $H$. Therefore, the following calculation shows that $c(1,-\chi_{e_{1}e_{2}})\in J$:

[TABLE]

In fact, we also have that $J=c(1,-\chi_{e_{1}e_{2}})R$, because the equality $c_{i}(1,-\chi_{e_{i}})=w_{i}c(1,-\chi_{e_{1}e_{2}})(1,-\chi_{e_{i}})$ implies that $J_{i}\subseteq c(1,-\chi_{e_{1}e_{2}})R$ for $i=1,2$. This finishes the proof that $J$ is a principal ideal in this case.

Next, let us assume that $J_{1}=c_{1}(1,-\chi_{e_{1}})R$ and $J_{2}=(d_{2},\chi_{e_{2}})R$ where $c_{1}\in S_{1}$, $d_{2}\in S_{a}\cup\{0\}$, and $e_{1},e_{2}\in P$. Recall from (4.1) that $J_{2}=d_{2}R\oplus\chi_{e_{2}}R$, and hence $d_{2}=(d_{2},0)\in J_{2}$ and $\chi_{e_{2}}\in J_{2}$. Let $e$ be the relative complement of $e_{1}e_{2}$ in $e_{1}$; that is, let $e\in B$ be such that $e_{1}e_{2}\wedge e=e_{1}$ and $e_{1}e_{2}\vee e=0$. As before, let $c=c_{1}\wedge d_{2}$, and notice that $c\in S_{1}$, because by Proposition 4.3(1), $c^{\perp}=c_{1}^{\perp}\cap d_{2}^{\perp}=\{0\}\cap a^{\perp}=\{0\}$. Furthermore, let us use the hypernormality of the Bézout monoid $C\cup\{0\}$ to obtain elements $w_{1},w_{2}\in C\cup\{0\}$ such that $c_{1}=cw_{1}$, $d_{2}=cw_{2}$, and $w_{1}\wedge w_{2}=1$. We note that if $d_{2}\in S_{a}$, then $w_{2}\in S_{a}$, because the equality $d_{2}=cw_{2}$ with $d_{2}\in S_{a}$, $c\in S_{1}$ implies that $w_{2}\notin S_{1}\cup\{0\}$, hence it must be that $w_{2}\in S_{a}$. If, in turn, $d_{2}=0$, then the equality $0=d_{2}=cw_{2}$ with $c\in S_{1}$ forces $w_{2}=0$. Now let $h=w_{1}+w_{2}\in H$. If $d_{2}=0$, and hence $w_{2}=0$, then $w_{1}\wedge w_{2}=1$ implies that $w_{1}=1$, so $h=1$. If, in turn, $d_{2}\in S_{a}$, then $w_{1}\in S_{1}$, $w_{2}\in S_{a}$ imply that $w_{1},w_{2}$ are distinct elements of $C$. Therefore the definition of $H$ yields that in both cases $h$ has an inverse in $H$. Therefore, the following calculation shows that $c(1,-\chi_{e_{1}})\in J$:

[TABLE]

where the third equality holds, because $d_{2}\in S_{a}$ acts on $I$ by the zero map, and hence $d_{2}\chi_{e_{1}}=0$. Thus,

[TABLE]

which implies that $c(1,-\chi_{e})R\subseteq J$. To prove that $J=c(1,-\chi_{e})R$ we need to establish that $J_{1},J_{2}\subseteq c(1,-\chi_{e})R$. For $J_{1}$, this follows from the equality $c_{1}(1,-\chi_{e_{1}})=c(1,-\chi_{e})d_{1}(1,-\chi_{e_{1}})$. For $J_{2}=d_{2}R\oplus\chi_{e_{2}}R$, the desired conclusion follows from the equalities $\chi_{e_{2}}=c(1,-\chi_{e})(c^{-1}\chi_{e_{2}})$ ($c^{-1}\in K$) and d_{2}=w_{2}c=(w_{2}c,0)=\bigl{(}w_{2}c,w_{2}c(-\chi_{e})\bigr{)}=w_{2}c(1,-\chi_{e}), where the third $=$ is valid, since $w_{2}\in S_{a}$ acts on $I$ by the zero map, forcing $w_{2}\chi_{e}=0$. This proves that $J$ is a principal ideal in this case as well.

Finally, assume that both ideals $J_{i}$ ($i=1,2$) have the form $J_{i}=(d_{i},\chi_{e_{i}})R=d_{i}R\oplus\chi_{e_{1}}R$ where $d_{1},d_{2}\in S_{a}\cup\{0\}$ and $e_{1},e_{2}\in P$. Then it is easy to see that

[TABLE]

where $d_{1}\wedge d_{2}\in S_{a}\cup\{0\}$, so $J$ is a principal ideal in this final case as well. Thus, the proof that $R$ is a Bézout ring is now complete.

Finally, we prove that the divisibility theory of $R$, i.e., the Bézout monoid $\mathcal{I}_{\textrm{princ}}(R)$ with ideal multiplication, is isomorphic to $S$. We will use the description of $\mathcal{I}_{\textrm{princ}}(R)$ in (4.3) and the description of $S$ in Proposition 4.5(4)–(5). Statements (4) and (1) of Proposition 4.5, combined with (4.3), imply that the following assignment yields a well-defined and surjective function $S\to\mathcal{I}_{\textrm{princ}}(R)$:

[TABLE]

To see that $\Phi$ is also injective, we need to argue that the principal ideals shown in (4.3) with different generators are different ideals. This follows from the following observations:

• •

If $c\in S_{1}$ and $e\in P$, then $c(1,-\chi_{e})R+I=cH+I$, and the annihilator ideal of $c(1,-\chi_{e})$ in $R$ is $\chi_{e}R$.

• •

If $d\in S_{a}\cup\{0\}$ and $e\in P$, then $(d,\chi_{e})R=dR\oplus\chi_{e}R$ (see (4.1)), therefore $(d,\chi_{e})R\cap H=dH$ and $(d,\chi_{e})R\cap I=\chi_{e}H$.

It remains to show that $\Phi$ is a monoid isomorphism, that is, $\Phi(s_{1}s_{2})=\Phi(s_{1})\Phi(s_{2})$ holds for all $s_{1},s_{2}\in S$. Checking this is fairly straightforward, by distinguishing three cases according to which of the two disjoint subsets $S_{1}(B\setminus P)$ and $(S_{a}\cup\{0\})\wedge P$ of $S$ (see Proposition 4.5(4)) the elements $s_{1},s_{2}$ belong to. In Case 1 we will assume that $s_{1},s_{2}\in S_{1}(B\setminus P)$, that is, both elements $s_{i}$ ($i=1,2$) have the form $s_{i}=c_{i}f_{i}$ where $c_{i}\in S_{1}$, $f_{i}\in B\setminus P$, so for the orthogonal complement $e_{i}$ of $f_{i}$ we have $e_{i}\in P$; in Case 2, $s_{1},s_{2}\in(S_{a}\cup\{0\})\wedge P$, that is, both elements $s_{i}$ ($i=1,2$) have the form $s_{i}=d_{i}\wedge e_{i}$ where $d_{i}\in S_{a}\cup\{0\}$ and $e_{i}\in P$; finally, in Case 3, $s_{1}\in S_{1}(B\setminus P)$ and $s_{2}\in(S_{a}\cup\{0\})\wedge P$, that is, $s_{1}=c_{1}f_{1}$ and $s_{2}=d_{2}\wedge e_{2}$ where $c_{1}\in S_{1}$, $d_{2}\in S_{a}\cup\{0\}$, $f_{1}\in B\setminus P$, and $e_{2}\in P$, so for the orthogonal complement $e_{1}$ of $f_{1}$ we have $e_{1}\in P$. The following calculations prove the equality $\Phi(s_{1}s_{2})=\Phi(s_{1})\Phi(s_{2})$ in all three cases. In each case the first and second to last equalities follow from the definition of $\Phi$, while the last equality follows from Proposition 4.5(5). So, in Case 1, we have

[TABLE]

In Case 2, we have $d_{1}\chi_{e_{2}}=0=d_{2}\chi_{e_{1}}$, because $d_{1},d_{2}\in S_{a}\cup\{0\}$ act on $I$ by the zero map. Therefore,

[TABLE]

Finally, in Case 3, we have $d_{2}\chi_{e_{1}}=0$ for the same reason as in the preceding case. Hence, if $e$ denotes the relative complement of $e_{1}e_{2}$ in $e_{2}$, then

[TABLE]

where the first and fourth $=$’s on line 3 are based on (4.1) (and also on the fact that $d_{2},c_{1}d_{2}\in S_{a}\cup\{0\}$), and the third equality on line 3 holds, because $c_{1}\in S_{1}$ acts on $I$ by an $I$-module automorphism which maps every ideal of $I$ into itself.

This completes the proof of Theorem 4.6. ∎