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. (1)

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

  2. (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. (1)

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

  2. (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. (1)

    S is an elemental annihilator semigroup.

  2. (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. (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 (ah) 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.