On commutative elemental annihilator monoids

Abstract

In this paper we describe commutative monoids S containing a zero element in which every ideal is the annihilator of an element of S.

1 Introduction and motivation

For a nonempty subset X of a semigroup with a zero element, let ${{\mathfrak {A}}}_l(X)$, ${{\mathfrak {A}}}_r(X)$, and ${{\mathfrak {A}}}(X)$ denote the left annihilator, the right annihilator, and the annihilator of X. A semigroup S with a zero element is called a dual semigroup if ${{\mathfrak {A}}}_l({{\mathfrak {A}}}_r(L))=L$ and ${{\mathfrak {A}}}_r({{\mathfrak {A}}}_l(R))=R$ are satisfied for every left ideal L and every right ideal R of S. The notion of a dual semigroup was introduced by S. Schwarz in [6] motivated by Baer’s notion of a dual ring [1]. By Lemma 2, ${{\mathfrak {A}}}_l({{\mathfrak {A}}}_r(L))=L$ is satisfied for a left ideal L of a semigroup S containing a zero element if and only if L is the left annihilator of a nonempty subset of S. This result and its dual imply that a semigroup S with a zero element is a dual semigroup if and only if every left ideal of S is the left annihilator of a nonempty subset of S, and every right ideal of S is the right annihilator of a nonempty subset of S (Corollary 3). Motivated by Yohe’s notion of a left (resp., right) elemental annihilator ring, we introduce the notion of a left (resp., right) elemental annihilator semigroup. We say that a semigroup S with a zero element is a left elemental annihilator semigroup if every left ideal of S is the left annihilator of a one-element subset of S. A right elemental annihilator semigroup is defined analogously. In [7], C.R. Yohe proved that a commutative ring with a unit element is an elemental annihilator ring if and only if it is a direct sum of completely primary principal ideal rings. This result motivates us to investigate commutative elemental annihilator monoids. As a main result of our paper we prove that the following three conditions on a nontrivial commutative monoid S containing a zero element are equivalent: (1) S is an elemental annihilator semigroup; (2) the unique maximal ideal $M_S$ of S is a nilpotent semigroup, and the orbits of $M_S$ under the action by the unit group $S^{\times }$ of S form a cyclic nilsemigroup; (3) the factor semigroup $S/{{\mathcal {J}}}$ is a cyclic nilsemigroup with an identity adjoined, where ${\mathcal {J}}$ is Green’s equivalence on S defined by $(a,b)\in {{\mathcal {J}}}$ if and only if a and b generate the same principal ideal of S.

2 Preliminaries

Let S be a multiplicative semigroup with a zero element 0. By the left annihilator of a nonempty subset X of S we mean the subset ${{\mathfrak {A}}}_l(X)=\{s\in S:\ sX=\{ 0\}\}$. The right annihilator ${{\mathfrak {A}}}_r(X)$ of X is defined analogously. It is clear that ${{\mathfrak {A}}}_l(X)$ (resp., ${{\mathfrak {A}}}_r(X)$) is a left (resp., right) ideal of S. If S is commutative, then ${{\mathfrak {A}}}_l(X)={{\mathfrak {A}}}_r(X)$ is called the annihilator of X and denoted by ${{\mathfrak {A}}}(X)$.

The proofs of the following two lemmas are straightforward and we omit them.

Lemma 1

For an arbitrary nonempty subset X of a semigroup S containing a zero element, the following two equations hold:

• (1) ${{\mathfrak {A}}}_l({{\mathfrak {A}}}_r({{\mathfrak {A}}}_l(X)))={{\mathfrak {A}}}_l(X)$,

• (2) ${{\mathfrak {A}}}_r({{\mathfrak {A}}}_l({{\mathfrak {A}}}_r(X)))={{\mathfrak {A}}}_r(X)$.

Lemma 2

The following two conditions on a left ideal L of a semigroup S containing a zero element are equivalent:

• (1) ${{\mathfrak {A}}}_l({{\mathfrak {A}}}_r(L))=L$,

• (2) L is the left annihilator of a nonempty subset of S.

Corollary 3

A semigroup containing a zero element is a dual semigroup if and only if every left ideal of S is the left annihilator of a nonempty subset of S, and every right ideal of S is the right annihilator of a nonempty subset of S.

Proof

By Lemma 2 and its dual, the assertion of the corollary is obvious. $\square$

Lemma 4

For an arbitrary element x of a semigroup S containing a zero element, ${{\mathfrak {A}}}_l(x)={{\mathfrak {A}}}_l(x\cup xS)$ and ${{\mathfrak {A}}}_r(x)={{\mathfrak {A}}}_r(x\cup Sx)$.

Proof

Let x be an arbitrary element of a semigroup S containing a zero element 0. If $t\in {{\mathfrak {A}}}_l(x)$, then $tx=0$, hence $txS=\{ 0\}$. Thus $t\in {{\mathfrak {A}}}_l(x\cup xS)$ from which it follows that ${{\mathfrak {A}}}_l(x)\subseteq {{\mathfrak {A}}}_l(x\cup xS)$. It is obvious that ${{\mathfrak {A}}}_l(x\cup xS)\subseteq {{\mathfrak {A}}}_l(x)$. Hence ${{\mathfrak {A}}}_l(x)={{\mathfrak {A}}}_l(x\cup xS)$. The equation ${{\mathfrak {A}}}_r(x)={{\mathfrak {A}}}_r(x\cup Sx)$ can be proved in a similar way. $\square$

We say that an equivalence relation $\sigma$ on a semigroup S is a left congruence on S if $(a,b)\in \sigma$ implies $(sa, sb)\in \sigma$ for every $a, b, s\in S$. The notion of the right congruence on S is defined analogously.

Let S be a semigroup containing a zero element. Let $\alpha _{{{\mathfrak {A}}}_l}$ denote the equivalence relation on S defined as follows: $(a, b)\in \alpha _{{{\mathfrak {A}}}_l}$ if and only if ${{\mathfrak {A}}}_l(a)={{\mathfrak {A}}}_l(b)$. The equivalence relation $\alpha _{{{\mathfrak {A}}}_r}$ on S is defined analogously.

Proposition 5

Let S be a semigroup with a zero element. Then $\alpha _{{{\mathfrak {A}}}_l}$ is a left congruence on S, and $\alpha _{{{\mathfrak {A}}}_r}$ is a right congruence on S.

Proof

Assume $(a, b)\in \alpha _{{{\mathfrak {A}}}_l}$ for elements $a, b\in S$. Let $s\in S$ and $t\in {{\mathfrak {A}}}_l(sa)$ be arbitrary elements. Then $ts\in {{\mathfrak {A}}}_l(a)={{\mathfrak {A}}}_l(b)$, and hence $t\in {{\mathfrak {A}}}_l(sb)$. Thus ${{\mathfrak {A}}}_l(sa)\subseteq {{\mathfrak {A}}}_l(sb)$. Similarly, ${{\mathfrak {A}}}_l(sb)\subseteq {{\mathfrak {A}}}_l(sa)$, and consequently ${{\mathfrak {A}}}_l(sa)={{\mathfrak {A}}}_l(sb)$, that is, $(sa, sb)\in \alpha _{{{\mathfrak {A}}}_l}$. Hence $\alpha _{{{\mathfrak {A}}}_l}$ is a left congruence on S. The proof that $\alpha _{{{\mathfrak {A}}}_r}$ is a right congruence on S is similar. $\square$

Two elements of a semigroup S are said to be ${\mathcal {L}}$-equivalent if they generate the same principal left ideal of S. The ${\mathcal {R}}$-equivalence is defined dually. ${\mathcal {L}}$ is a right congruence and ${\mathcal {R}}$ is a left congruence on an arbitrary semigroup.

Proposition 6

If S is a semigroup containing a zero element, then ${{\mathcal {R}}}\subseteq \alpha _{{{\mathfrak {A}}}_l}$ and ${{\mathcal {L}}}\subseteq \alpha _{{{\mathfrak {A}}}_r}$.

Proof

Let S be a semigroup containing a zero element. If $(a, b)\in {{\mathcal {R}}}$ for elements $a, b\in S$, then $a\cup aS=b\cup bS$. Using Lemma 4, we get ${{\mathfrak {A}}}_l(a)={{\mathfrak {A}}}_l(a\cup aS)={{\mathfrak {A}}}_l(b\cup bS)={{\mathfrak {A}}}_l(b)$, that is, $(a, b)\in \alpha _{{{\mathfrak {A}}}_l}$. Thus ${{\mathcal {R}}}\subseteq \alpha _{{{\mathfrak {A}}}_l}$. The proof of ${{\mathcal {L}}}\subseteq \alpha _{{{\mathfrak {A}}}_r}$ is similar. $\square$

Definition 1

A semigroup S with a zero element is called a left elemental annihilator semigroup if, for every left ideal L of S, there exists an element $x\in S$ such that $L={{\mathfrak {A}}}_l(x)$. The right elemental annihilator semigroup is defined analogously.

For notions and notations not defined but used in this paper, we refer to the books [2] and [4].

3 Left elemental annihilator semigroups

Proposition 7

If S is a left elemental annihilator semigroup and L is a left ideal of S, then we have ${{\mathfrak {A}}}_l({{\mathfrak {A}}}_r(L))=L$.

Proof

By Lemma 2, it is obvious. $\square$

By Proposition 7 and its dual, we have the following corollary.

Corollary 8

If a semigroup is a left elemental annihilator semigroup and a right elemental annihilator semigroup, then it is a dual semigroup.

Lemma 9

If a semigroup S is a left elemental annihilator semigroup and a right elemental annihilator semigroup, then $x\in Sx$ and $x\in xS$ for every $x\in S$.

Proof

See Corollary 8 and [6, Lemma 1.6]. $\square$

The next proposition is about the relationship between the right congruence $\alpha _{{{\mathfrak {A}}}_r}$ and Green’s right congruence ${\mathcal {L}}$ on a left elemental annihilator semigroup.

Proposition 10

If S is a left elemental annihilator semigroup, then $\alpha _{{{\mathfrak {A}}}_r}={{\mathcal {L}}}$.

Proof

Let S be a left elemental annihilator semigroup. Assume $(a, b)\in \alpha _{{{\mathfrak {A}}}_r}$ for elements $a, b\in S$. Then, by Lemma 4, ${{\mathfrak {A}}}_r(a\cup Sa)={{\mathfrak {A}}}_r(a)={{\mathfrak {A}}}_r(b)={{\mathfrak {A}}}_r(b\cup Sb)$, and hence ${{\mathfrak {A}}}_l({{\mathfrak {A}}}_r(a\cup Sa))={{\mathfrak {A}}}_l({{\mathfrak {A}}}_r(b\cup Sb))$. By Proposition 7, we get $a\cup Sa=b\cup Sb$. Thus $(a, b)\in {{\mathcal {L}}}$, and consequently $\alpha _{{{\mathfrak {A}}}_r}\subseteq {{\mathcal {L}}}$. By Proposition 6, ${{\mathcal {L}}}\subseteq \alpha _{{{\mathfrak {A}}}_r}$ is satisfied in an arbitrary semigroup with a zero element. Thus we have $\alpha _{{{\mathfrak {A}}}_r}={{\mathcal {L}}}$. $\square$

For a congruence $\alpha$ on a semigroup S, let $[a]_{\alpha }$ denote the $\alpha$-class of S containing the element a of S.

Proposition 11

If $\alpha$ is a congruence on a left elemental annihilator semigroup S such that $[0]_{\alpha }=\{ 0\}$, then the factor semigroup $S/\alpha$ is a left elemental annihilator semigroup.

Proof

Let S be a left elemental annihilator semigroup. Assume that $\alpha$ is a congruence on S such that $[0]_{\alpha }=\{ 0\}$. Let L be a left ideal of the factor semigroup $S/\alpha$. Then $L'=\{ s\in S:\ [s]_{\alpha }\in L\}$ is a left ideal of S. Since S is a left elemental annihilator semigroup, there is an element $b\in S$ such that ${{\mathfrak {A}}}_l(b)=L'$. We show that ${{\mathfrak {A}}}_l([b]_{\alpha })=L$ from which it already follows that $S/\alpha$ is a left elemental annihilator semigroup. If $[x]_{\alpha }\in L$, then $x\in L'$, and hence $[x]_{\alpha }[b]_{\alpha }=[xb]_{\alpha }=[0]_{\alpha }$. Thus $[x]_{\alpha }\in {{\mathfrak {A}}}_l([b]_{\alpha })$, and consequently $L\subseteq {{\mathfrak {A}}}_l([b]_{\alpha })$. To prove the converse inclusion, assume $[y]_{\alpha }\in {{\mathfrak {A}}}_l([b]_{\alpha })$. Then $[yb]_{\alpha }=[y]_{\alpha }[b]_{\alpha }=[0]_{\alpha }$, and hence $yb=0$ by hypothesis for $[0]_{\alpha }$. Thus $y\in {{\mathfrak {A}}}_l(b)=L'$ from which we get $[y]_{\alpha }\in L$. Hence ${{\mathfrak {A}}}_l([b]_{\alpha })\subseteq L$. Consequently ${{\mathfrak {A}}}_l([b]_{\alpha })=L$. $\square$

Proposition 12

Let $\alpha$ be a congruence on a semigroup S containing a zero element 0 such that $[0]_{\alpha }=\{ 0\}$ and the factor semigroup $S/\alpha$ is a left elemental annihilator semigroup. Then, for every left ideal L of S which is a union of $\alpha$-classes of S, there is an element $b\in S$ such that ${{\mathfrak {A}}}_l(b)=L$.

Proof

Let L be a left ideal of S which is a union of $\alpha$-classes of S. Let $L'=\{[s]_{\alpha }:\ s\in L\}$. Then $L'$ is a left ideal of $S/\alpha$. Since $S/\alpha$ is a left elemental annihilator semigroup, there is an element $[b]_{\alpha }\in S/\alpha$ such that ${{\mathfrak {A}}}_l([b]_{\alpha })=L'$. We show that ${{\mathfrak {A}}}_l(b)=L$. Let $x\in {{\mathfrak {A}}}_l(b)$. Then $xb=0$, and hence $[x]_{\alpha }[b]_{\alpha }=[xb]_{\alpha }=[0]_{\alpha }$ which means that $[x]_{\alpha }\in {{\mathfrak {A}}}_l([b]_{\alpha })=L'$. Thus $x\in L$, and hence ${{\mathfrak {A}}}_l(b)\subseteq L$. To show the converse inclusion, assume that $y\in L$ be an arbitrary element. Then $[y]_{\alpha }\in L'={{\mathfrak {A}}}_l([b]_{\alpha })$ from which it follows that $[yb]_{\alpha }=[y]_{\alpha }[b]_{\alpha }=[0]_{\alpha }$. Thus $yb=0$ by hypothesis for $[0]_{\alpha }$, that is, $y\in {{\mathfrak {A}}}_l(b)$, and hence $L\subseteq {{\mathfrak {A}}}_l(b)$. Consequently $L={{\mathfrak {A}}}_l(b)$ . $\square$

Theorem 13

Let S be a semigroup containing a zero element, and let $\alpha$ be a congruence on S such that every left ideal of S is the union of $\alpha$-classes of S. Then S is a left elemental annihilator semigroup if and only if the factor semigroup $S/\alpha$ is a left elemental annihilator semigroup.

Proof

Since $[0]_{\alpha }=\{ 0\}$ by hypothesis, the assertion of the theorem is an immediate consequence of Proposition 11 and Proposition 12. $\square$

By a monoid we mean a semigroup containing an identity element. Let S be a semigroup and let 1 be a symbol not representing any element of S. Extend the operation on S to $S\cup 1$ such that $11=1$ and $x1=1x=x$ for every $x\in S$. Then $S\cup 1$ is a monoid in which 1 is the identity element. We say that this semigroup is obtained from the semigroup S by the adjunction of an identity element 1 to S. If S is a semigroup, then $S^1$ denotes the following monoid: $S^1=S$ if $|S|\ge 2$ and S has an identity element; $S^1=S\cup 1$ otherwise. Recall that if S is a one-element semigroup, then $S^1$ is a two-element monoid.

Proposition 14

Let S be a semigroup with a zero element 0 such that $S\ne S^1$. Then the semigroup $S^1$ is a left elemental annihilator semigroup if and only if every nonzero left ideal of S is the left annihilator of a nonzero element of S.

Proof

Assume that $S^1$ is a left elemental annihilator semigroup. Let $L\ne \{ 0\}$ be a left ideal of S. Then L is a left ideal of $S^1$, and hence there is an element $x\in S^1$ such that ${{\mathfrak {A}}}_l(x)=L$ in $S^1$. Since ${{\mathfrak {A}}}_l(1)=\{ 0\}$ in $S^1$, we have $x\in S$. Since ${{\mathfrak {A}}}_l(0)=S^1$ in $S^1$, we have $x\ne 0$ and so ${{\mathfrak {A}}}_l(x)=L$ is also satisfied in S. Thus every nonzero left ideal of S is the left annihilator of a nonzero element of S.

Conversely, assume that every nonzero left ideal of S is the left annihilator of a nonzero element of S. It is clear that, in $S^1$, ${{\mathfrak {A}}}_l(1)=\{ 0\}$ and ${{\mathfrak {A}}}_l(0)=S^1$. Let L be a left ideal of $S^1$ with $L\ne \{ 0\}$ and $L\ne S^1$. Then L is a nonzero left ideal of S and hence there is a nonzero element $x\in S$ such that ${{\mathfrak {A}}}_l(x)=L$ in S. It is obvious that ${{\mathfrak {A}}}_l(x)=L$ is also satisfied in $S^1$, because $x\ne 0$. Consequently $S^1$ is a left elemental annihilator semigroup. $\square$

4 Commutative elemental annihilator semigroups

A commutative semigroup S with a zero element is called an elemental annihilator semigroup if every ideal of S is the annihilator of an element of S.

For an element a of a semigroup S, let J(a) denote the principal ideal of S generated by a. It is known that $J(a)=a\cup aS\cup Sa\cup SaS$. If S is commutative, then $J(a)=a\cup aS$. From Lemma 9 it follows that if S is a commutative elemental annihilator semigroup then $J(a)=aS$.

A semigroup S is called a principal ideal semigroup if every ideal I of S is principal, that is, $I=J(a)$ for some $a\in S$.

Theorem 15

Every commutative elemental annihilator semigroup is a principal ideal semigroup.

Proof

Let A be an arbitrary ideal of a commutative elemental annihilator semigroup S. Since ${{\mathfrak {A}}}(A)$ is an ideal of S, there is an element $x\in S$ such that ${{\mathfrak {A}}}(A)={{\mathfrak {A}}}(x)$. By Lemmas 4 and 9, ${{\mathfrak {A}}}(x)={{\mathfrak {A}}}(xS)$. Using Proposition 7, we get $A={{\mathfrak {A}}}({{\mathfrak {A}}}(A))={{\mathfrak {A}}}({{\mathfrak {A}}}(x))={{\mathfrak {A}}}({{\mathfrak {A}}}(xS))=xS$, because A and xS are ideals of S. Thus A is a principal ideal. Consequently S is a principal ideal semigroup. $\square$

Proposition 16

Let S be a commutative elemental annihilator semigroup. Then, for arbitrary elements a and b of S, ${{\mathfrak {A}}}(b)=aS$ if and only if ${{\mathfrak {A}}}(a)=bS$.

Proof

Assume ${{\mathfrak {A}}}(b)=aS$ for elements $a, b\in S$. By Proposition 7, Lemmas 4, and 9, ${{\mathfrak {A}}}(a)={{\mathfrak {A}}}(aS)={{\mathfrak {A}}}({{\mathfrak {A}}}(b))={{\mathfrak {A}}}({{\mathfrak {A}}}(bS))=bS$. This proves the proposition. $\square$

If S is a commutative semigroup, then Green’s equivalences ${\mathcal {L}}$, ${\mathcal {R}}$ and ${\mathcal {J}}$ are congruences on S, and ${{\mathcal {L}}}={{\mathcal {R}}}={{\mathcal {J}}}$ .

Proposition 17

If S is a commutative elemental annihilator semigroup, then ${{\mathcal {J}}}=\alpha _{{{\mathfrak {A}}}}$, where $\alpha _{{{\mathfrak {A}}}}$ is the congruence on S defined by $(a, b)\in \alpha _{{{\mathfrak {A}}}}$ if and only if ${{\mathfrak {A}}}(a)={{\mathfrak {A}}}(b)$.

Proof

It is an immediate consequence of Proposition 10. $\square$

5 Commutative elemental annihilator monoids

By Corollary 8, every commutative elemental annihilator monoid is a dual semigroup. Results related to dual monoids can be found in Chapter 7 of [6], however these results cannot be used in our investigation. For example, it is assumed in [6, Theorem 7.2] that the examined dual monoid S contains a nilpotent radical, and it is proved (using further conditions) that the nilpotent radical is the unique maximal ideal of S such that the complement of the nilpotent radical in S is a subgroup of S. In our study, we start from the fact that every nontrivial commutative monoid S containing a zero element is a disjoint union $S=M_S\cup S^{\times }$, where $S^{\times }$ is the unit group of S and $M_S$ is the unique maximal ideal of S. We prove in Theorem 18 that if S is a nontrivial commutative elemental annihilator monoid, then the maximal ideal $M_S$ is nilpotent. Using this result, we give a characterization of nontrivial commutative elemental annihilator monoids in Theorem 23.

An element a of a semigroup S with a zero element 0 is said to be nilpotent if there is a positive integer n such that $a^n=0$. A semigroup containing a zero element is called a nilsemigroup if all its elements are nilpotent. We say that a semigroup S containing a zero element 0 is a nilpotent semigroup if there is a positive integer n such that $S^n=\{ 0\}$.

Theorem 18

If S is a nontrivial commutative elemental annihilator monoid, then the unique maximal ideal $M_S$ of S is a nilpotent semigroup.

Proof

Let S be a nontrivial commutative elemental annihilator monoid. Then S is a principal ideal semigroup by Theorem 15, and hence the ideals of S form a chain with respect to inclusion by [5, 1.1. Theorem]. Let 0 denote the zero element of S, and let a be an arbitrary element of $M_S$. We show that a is nilpotent. Let $g\in S^{\times }$ be an arbitrary element. Then obviously $J(g) = S$. Since $J(a)\subseteq M_S$, we have $(a,g)\notin {{\mathcal {J}}}$. By Proposition 17, ${{\mathcal {J}}}=\alpha _{{{\mathfrak {A}}}}$. Thus $(a,g)\notin \alpha _{{{\mathfrak {A}}}}$. Since ${{\mathfrak {A}}}(g)=\{ 0\}$, we have ${{\mathfrak {A}}}(a)\ne \{ 0\}$. Then, by [5, 1.5. Theorem], a is a nilpotent element. Consequently $M_S$ is a nilsemigroup. As every ideal of S is a principal ideal, there exists an element $b\in M_S$ such that $M_S=J(b)=bS$. Since b is nilpotent, there is a positive integer k such that $b^k=0$. Then, for arbitrary $x_1, \dots , x_k\in M_S$, $x_1\cdots x_k\in b^kS=\{ 0\}$, and hence $(M_S)^k=\{ 0\}$. Consequently $M_S$ is a nilpotent semigroup. $\square$

A semigroup S is called a cyclic semigroup if S is generated by a single element of S. A semigroup S with a zero element is called a cyclic nilsemigroup if S is generated by a single nilpotent element. A semigroup S is called a cyclic nilsemigroup with an identity adjoined if S is the result of adjoining an identity to a cyclic nilsemigroup, i.e., S has an identity 1 and $S\setminus \{1\}$ is a cyclic nilsemigroup.

Proposition 19

Let N be a commutative nilsemigroup. Then every nonzero ideal of N is the annihilator of a nonzero element of N if and only if N is a cyclic nilsemigroup.

Proof

The assertion of the proposition is trivial in that case when N contains one element. Thus we can suppose that N is nontrivial.

Let N be a nontrivial cyclic nilsemigroup. We show that every nonzero ideal of N is the annihilator of an element of N. Let $N=\{ b, b^2, \dots , b^{k-1}, 0\}$, where $k\ge 2$ is the least integer with the property $b^k=0$. The ideals of N are $\{ 0\}$ and $N^t=J(b^t)=\{b^t, \dots , b^{k-1}, 0\}$ ($t=1, \dots , k-1$). Since $N^t={{\mathfrak {A}}}(b^{k-t})$ ($t=1, \dots , k-1$), every nonzero ideal of N is the annihilator of a nonzero element of N.

To prove the converse assertion, assume that N is a nontrivial commutative nilsemigroup having the property that every nonzero ideal of N is the annihilator of a nonzero element of N. Since $N\ne N^1$, Proposition 14 implies that $N^1$ is a commutative elemental annihilator monoid. Then, by Theorem 15, $N^1$ is a principal ideal semigroup. If I is an ideal of N, then I is an ideal of $N^1$, and hence there is an element $x\in N$ such that $I=xN^1=x\cup xN$. Thus I is a principal ideal of N. Hence N is a principal ideal nilsemigroup. Let b be an element of N such that $N=b\cup bN$. Since $N\ne \{ 0\}$, we have $b\ne 0$. Let k be the least positive integer with the property $b^k=0$. Then $k\ge 2$. We show that $N=\{ b, b^2, \dots , b^{k-1}, 0\}$. Let $t\in \{1, 2, \dots , k-1\}$ be an arbitrary integer. Then $b^t\ne 0$ and $N^t\ne \{ 0\}$. Since $N^t=(b\cup bN)^t\subseteq b^t\cup b^tN\subseteq N^t$, we have

$$\begin{aligned} N^t=b^t\cup b^tN. \end{aligned}$$

If $b^t=b^tx$ for some $x\in N$, then $b^t=b^tx^n$ for every positive integer n, from which it follows that $b^t=0$. This is a contradiction. Thus

$$\begin{aligned} b^t\notin b^tN. \end{aligned}$$

Since $N^{t+1}=N^tN=(b^t\cup b^tN)N=b^tN\cup b^tN^2\subseteq b^tN\subseteq N^{t+1}$, we have

$$\begin{aligned} b^tN=N^{t+1}. \end{aligned}$$

Thus

$$\begin{aligned}&N=b\cup bN=b\cup N^2=b\cup (b^2\cup b^2N)=\\&\quad =\{ b, b^2\}\cup b^2N=\{b, b^2\}\cup N^3=\cdots =\{b, b^2, \dots , b^{k-1}, 0\}. \end{aligned}$$

$\square$

Proposition 20

For a commutative nilsemigroup N, the monoid $N^1$ is an elemental annihilator semigroup if and only if N is a cyclic nilsemigroup.

Proof

If N is a commutative nilsemigroup, then $N\ne N^1$. Thus, by Proposition 14, $N^1$ is an elemental annihilator semigroup if and only if every nonzero ideal of N is the annihilator of a nonzero element of N. By Proposition 19, every nonzero ideal of N is the annihilator of a nonzero element of N if and only if N is a cyclic nilsemigroup. $\square$

It is obvious that if S is a nontrivial commutative monoid with a zero element, then the binary relation $\alpha _{orb}$ on S defined by $(a, b)\in \alpha _{orb}$ if and only if $aS^{\times }=bS^{\times }$ is a congruence on S. The $\alpha _{orb}$-classes of S are precisely the orbits of S under the action by the unit group $S^{\times }$. The subgroup $S^{\times }$ is a single orbit, and the remaining orbits form a subsemigroup (in $S/\alpha _{orb}$) denoted by $M_S/\alpha _{orb}$. Thus $S/\alpha _{orb}=(M_S/\alpha _{orb})^1$.

Lemma 21

If S is a nontrivial commutative monoid with a zero element, then every ideal of S is a union of $\alpha _{orb}$-classes of S.

Proof

Let S be a nontrivial commutative monoid with a zero element. Let K be an arbitrary ideal of S. Assume $a\in K$ and $(a, b)\in \alpha _{orb}$ for elements $a, b\in S$. Since the identity element of $S^{\times }$ is the identity element of S, we have $b\in bS^{\times }=aS^{\times }\subseteq K$. Thus K is a union of $\alpha _{orb}$-classes of S. $\square$

Proposition 22

A nontrivial commutative monoid S with a zero element is an elemental annihilator semigroup if and only if the factor semigroup $S/\alpha _{orb}$ is an elemental annihilator semigroup.

Proof

By Theorem 13 and Lemma 21, it is obvious. $\square$

In the next theorem, we characterize nontrivial commutative elemental annihilator monoids.

Theorem 23

The following three conditions on a nontrivial commutative monoid S containing a zero element are equivalent.

• (1) S is an elemental annihilator semigroup.

• (2) The unique maximal ideal $M_S$ of S is a nilpotent semigroup, and the orbits of $M_S$ under the action by the unit group $S^{\times }$ of S form a cyclic nilsemigroup.

• (3) The factor semigroup $S/{{\mathcal {J}}}$ is a cyclic nilsemigroup with an identity adjoined.

Proof

(1) implies (2): Assume that S is an elemental annihilator semigroup. By Proposition 22, $S/\alpha _{orb}$ is an elemental annihilator monoid. By Theorem 18, the unique maximal ideal $M_S$ of S is a nilpotent semigroup. Then the semigroup $M_S/\alpha _{orb}$ is a commutative nilsemigroup. Since $S/\alpha _{orb}=(M_S/\alpha _{orb})^1$, $M_S/\alpha _{orb}$ is a cyclic nilsemigroup by Proposition 20.

(2) implies (3): Assume that the unique maximal ideal $M_S$ of S is a nilpotent semigroup and the orbits of $M_S$ under the action by the unit group $S^{\times }$ of S form a cyclic nilsemigroup. Then S is an elementary semigroup [3], and hence $\alpha _{orb}={{\mathcal {H}}}$ in S by [3, Proposition 5.1]. Since S is commutative, ${{\mathcal {H}}}={{\mathcal {J}}}$. Thus $S/{{\mathcal {J}}}=S/\alpha _{orb}=(M_S/\alpha _{orb})^1$ is a cyclic nilsemigroup with an identity adjoined.

(3) implies (1): Assume that the factor semigroup $S/{{\mathcal {J}}}$ is a cyclic nilsemigroup with an identity adjoined. Then, by Proposition 20, $S/{{\mathcal {J}}}$ is an elemental annihilator semigroup. As every ideal of S is the union of ${\mathcal {J}}$-classes of S, Theorem 13 implies that S is an elemental annihilator semigroup. $\square$

Next, we give a construction that can be used to obtain a nontrivial commutative elemental annihilator monoid whose factor semigroup modulo $\alpha _{orb}$ is isomorphic to a given cyclic nilsemigroup with an identity adjoined.

If I is an ideal of a semigroup S, then the relation $\varrho _I$ on S defined by $(a, b)\in \varrho _I$ if and only if $a=b$ or $a, b\in I$ is a congruence on S which is called the Rees congruence on S determined by I. The equivalence classes of S mod $\varrho _I$ are I itself and every one-element set $\{ a\}$ with $a\in S\setminus I$. The factor semigroup $S/\varrho _I$ is called the Rees factor semigroup of S modulo I. We shall write S/I instead of $S/\varrho _I$. We may describe S/I as the result of collapsing I into a single (zero) element, while the elements of S outside of I retain their identity.

Let G be a group and H be a semigroup with a zero element 0. Then the direct product $G\times \{ 0\}$ is an ideal of the direct product $G\times H$. Let $G\triangle H$ denote the Rees factor semigroup $(G\times H)/(G\times \{ 0\})$.

Theorem 24

Let G be a commutative group and $N^1$ be a cyclic nilsemigroup with an identity adjoined. Then $G\triangle N^1$ is a nontrivial commutative elemental annihilator monoid such that the factor semigroup $(G\triangle N^1)/\alpha _{orb}$ is isomorphic to $N^1$.

Proof

Let J be an ideal of the direct product $G\times N^1$. Let I be the set of all elements $x\in N^1$ with the property that $(g,x)\in J$ for some $g\in G$. Then $J\subseteq G\times I$. Let e denote the identity element of G. If $x\in I$, that is, $(g,x)\in J$ for some $g\in G$, then $(g,xy)=(g,x)(e,y)\in J$ for all $y\in N^1$, which implies that I is an ideal of $N^1$. It is clear that $G\times I$ is an ideal of $G\times N^1$. We show that $J=G\times I$. By the above inclusion, it is sufficient to show that $G\times I\subseteq J$. Let $(g,x)\in G\times I$ be an arbitrary element. Since $x\in I$, there is an element $h\in G$ such that $(h, x)\in J$. Since G is a group, there is an element $\xi \in G$ such that $g=h\xi$. Thus $(g,x)=(h\xi , x)=(h,x)(\xi ,1)\in J$, where 1 is the identity element of $N^1$. Consequently $G\times I\subseteq J$, and hence $J=G\times I$. Thus, for every ideal J of $G\times N^1$, there is an ideal I of $N^1$ such that $J=G\times I$. It is clear that $G\triangle N^1=(G\times N^1)/(G\times \{0\})$ is a commutative monoid containing a zero element. Let $0_{\triangle }$ denote the zero of $G\triangle N^1$. Recall that $G\triangle N^1$ can be considered as the result of collapsing $G\times \{ 0\}$ into the element $0_{\triangle }$, while the elements of $G\times N^1$ outside of $G\times \{ 0\}$ retain their identity. We show that $G\triangle N^1$ is an elemental annihilator semigroup. Let J be an arbitrary ideal of $G\triangle N^1$. Since ${{\mathfrak {A}}}(0_{\triangle })=G\triangle N^1$, we can suppose that $J\ne G\triangle N^1$. By the above, there is an ideal I of $N^1$ with $I\ne N^1$ such that $J=((G\times I)\setminus (G\times \{ 0\})\cup \{ 0_{\triangle }\}$. Since $N^1$ is an elemental annihilator semigroup, there is an element $x\in N^1$ such that ${{\mathfrak {A}}}(x)=I$ in $N^1$. Because of $I\ne N^1$, we get $x\ne 0$. We show that ${{\mathfrak {A}}}((g,x))=J$ for an arbitrary $g\in G$. Since $x\ne 0$, we have $(g,x)\ne 0_{\triangle }$. Let (a, h) be an arbitrary nonzero element of $G\triangle N^1$. If $(a,h)\in J$, then $0\ne h\in I$, and hence $(g,x)(a,h)=(ga, xh)=0_{\triangle }$, because $xh=0$ by $h\in {{\mathfrak {A}}}(x)$. If $(a,h)\notin J$, then $h\notin I$. Thus $(g,x)(a,h)=(ga,xh)\ne 0_{\triangle }$, because $xh\ne 0$ by $h\notin {{\mathfrak {A}}}(x)$. Consequently ${{\mathfrak {A}}}((g,x))=J$. Thus $G\triangle N^1$ is a commutative elemental annihilator monoid. The unit group of $G\triangle N^1$ is $G\times \{ 1\}$. The orbits of $G\triangle N^1$ under the action by its unit group are $\{ 0_{\triangle }\}$ and the subsets $G\times \{ h\}$ where h is an arbitray nonzero element of $N^1$. Let $\phi$ denote the mapping of $N^1$ onto $(G\triangle N^1)/\alpha _{orb}$ defined in the following way: $\varphi (h)=G\times \{h\}$ if $h\in N^1\setminus \{0\}$; $\varphi (h)=0_{\triangle }$ if $h=0$. It is clear that $\phi$ is bijective. Let $x, y\in N^1$ be arbitrary elements. If one of them is 0, then one of $\phi (x)$ and $\phi (y)$ is $0_{\triangle }$, and hence $\phi (xy)=\phi (0)=0_{\triangle }=\phi (x)\phi (y)$. If $xy=0$ and $0\notin \{x, y\}$, then $\phi (xy)=\phi (0)=0_{\triangle }=(G\times \{ x\})(G\times \{y\})=\phi (x)\phi (y)$. If $xy\ne 0$, then $\phi (xy)=G\times \{xy\}=(G\times \{ x\})(G\times \{ y\})=\phi (x)\phi (y)$. Consequently $\phi$ is a homomorphism. Hence $\phi$ is an isomorphism of $N^1$ onto $(G\triangle N^1)/\alpha _{orb}$. $\square$

Example 1

Let S be a semigroup, where $S=\{1, a, b, c, 0\}$, and the operation on S is defined by the following Cayley table:

$$\begin{aligned} \begin{array}{l|lllll} &{} 1 &{} a &{} b &{} c &{} 0 \\ \hline 1 &{} 1 &{} a &{} b &{} c &{} 0 \\ a &{} a &{} 1 &{} c &{} b &{} 0 \\ b &{} b &{} c &{} 0 &{} 0 &{} 0 \\ c &{} c &{} b &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ \end{array} \end{aligned}$$

It is a matter of checking to see that S is isomorphic to the semigroup $G\triangle N^1$, where G is a two-element group and N is a two-element cyclic nilsemigroup. Then, by Theorem 24, S is a commutative elemental annihilator monoid. The unit group of S is $\{ 1, a\}$, and the $\alpha _{orb}$-classes of S are $\{ 0\}$, $\{ b, c\}$ and $\{1, a\}$. Thus $S/\alpha _{orb}\cong N^1$, as Theorem 24 states.

Appendix

If S is a commutative elemental annihilator semigroup, then every principal ideal of S is the annihilator of an element of S. The next example shows that the converse is not true, in general. We show an example of a commutative monoid S with a zero element in which every principal ideal is the annihilator of an element of S, but S is not an elemental annihilator semigroup.

Example 2

Let $A=\{ 1, a_1, a_2, \dots \}$ and $B=\{ 0, b_{-1}, b_{-2}, \dots \}$ be sets such that $A\cap B=\emptyset$. We define an operation on $S=A\cup B$ as follows:

$$\begin{aligned}&1s=s1=s\quad \hbox {and}\quad 0s=s0=0\quad \hbox {for every }s\in S,\\&a_ia_j=a_{i+j}\quad \hbox {and}\quad b_{-i}b_{-j}=0 \quad \hbox {for every positive integers } i \hbox { and } j,\\&a_ib_{-j}=b_{-j}a_i=\left\{ \begin{array}{ll} 0, &{} \text {if }\, i-j\ge 0, \\ b_{i-j}, &{} \text {if}\, i-j<0. \end{array} \right. \end{aligned}$$

It is a matter of checking to see that this operation is associative. We present a case as an example. If $i-j+k\ge 0$, then

$$\begin{aligned} a_ib_{-j}=\left\{ \begin{array}{ll} 0, &{} \text {if }\, i-j\ge 0, \\ b_{i-j}, &{} \text {if }\,i-j<0\end{array} \right. \quad \hbox {and}\quad b_{-j}a_k=\left\{ \begin{array}{ll} 0, &{} \text {if }\,-j+k\ge 0, \\ b_{-j+k}, &{} \text {if }\,-j+k<0.\end{array} \right. \end{aligned}$$

Thus

$$\begin{aligned} (a_ib_{-j})a_k=0=a_i(b_{-j}a_k). \end{aligned}$$

If $i-j+k<0$, then $i-j<0$ and $-j+k<0$ from which we get

$$\begin{aligned} (a_ib_{-j})a_k=b_{i-j}a_k=b_{i-j+k}=a_i(b_{-j+k})=a_i(b_{-j}a_k). \end{aligned}$$

The operation is also commutative. 1 is the identity element and 0 is the zero element of S. Thus S is a commutative monoid with a zero element. It is easy to see that

$$\begin{aligned} 1S=S={{\mathfrak {A}}}(0),\quad 0S=\{ 0\}={{\mathfrak {A}}}(1) \end{aligned}$$

and, for every positive integer i,

$$\begin{aligned}&a_iS=\{a_i, a_{i+1}, \dots , ;0; b_{-1}, b_{-2}, \dots \}={{\mathfrak {A}}}(b_{-i}),\\&b_{-i}S=\{b_{-i}, b_{-(i-1)}, \dots b_{-1}, 0\}={{\mathfrak {A}}}(a_i). \end{aligned}$$

Thus every principal ideal of S is the annihilator of an element of S. The ideal B of S is not the annihilator of an element of S. Hence S is not an elemental annihilator semigroup.