2. Preliminaries

In this section, we establish some basic definitions and notation. We begin by providing some set-theoretic definitions.

A relation $\leq$ on a set $P$ is said to be preorder if it is both reflexive and transitive. If a preorder is also symmetric, then it is an equivalence relation. On the other hand, if a preorder is antisymmetric then it is a partial order.

A poset is a set $P$ together with a partial order $\leq$. Given any set $X,$ a collection $P$ of subsets of $X$ forms a poset under the partial order of inclusion. In particular, the set of all right ideals of a semigroup is a poset (under $\subseteq$).

Let $(P,\, \leq _P)$ and $(Q,\, \leq _Q)$ be two posets. A map $\theta : P\to Q$ is said to be order-preserving if $x\leq _P y$ implies $x\theta \leq _Q y\theta$ for all $x,\, y\in P.$ A map $\theta : P\to Q$ is an isomorphism if both $\theta$ and $\theta ^{-1}$ are order-preserving (i.e. $x\leq _P y$ if and only if $x\theta \leq _Q y\theta$ for all $x,\, y\in P$) and $\theta$ is a bijection. We say that $P$ and $Q$ are isomorphic if there exists an isomorphism between them. Note that to show that a map $\theta : P\to Q$ is an isomorphism, it suffices to prove that $\theta$ is surjective and that $x\leq _P y$ if and only if $x\theta \leq _Q y\theta$ for all $x,\, y\in P$.

Two elements $a$ and $b$ of a poset $P$ are said to be comparable if either $a\leq b$ or $b\leq a$; otherwise, $a$ and $b$ are incomparable. A subset of $P$ in which any two elements are comparable is called a chain. An antichain of $P$ is a subset consisting of pairwise incomparable elements.

We now turn our attention to semigroups. We refer the reader to [Reference Howie13] for a more comprehensive introduction to semigroup theory. Throughout the remainder of the section, $S$ will denote a semigroup.

We denote by $S^{1}$ the monoid obtained from $S$ by adjoining an identity if necessary (if $S$ is already a monoid, then $S^{1}=S$). Similarly, we denote by $S^{0}$ the semigroup with zero obtained from $S$ by adjoining a zero if necessary.

Let $M$ be a monoid with identity $1.$ An element $a\in M$ is said to be right invertible if there exists $b\in M$ such that $ab=1.$ Left invertible elements are defined dually. An element of $M$ is called a unit if it is both right invertible and left invertible. The units of $M$ form a group, called the group of units of $M,$ which we denote by $U(M).$

We denote the set of idempotents of $S$ by $E(S).$ If $S=E(S),$ it is called a band. A semilattice is a commutative band. The multiplication in a semilattice $E$ induces the following partial order:

\[ e\geq f\iff ef=f. \]

In this way, we may view $E$ as a meet-semilattice in the order-theoretic sense. Conversely, any order-theoretic meet-semilattice may be viewed as a commutative band with meet taken as the binary operation.

If $a,\, b\in S$ are such that $a=aba$ and $b=bab,$ then $b$ is called an inverse of $a.$ The semigroup $S$ is said to be regular if every element of $S$ has an inverse. If, additionally, the inverse of each element of $S$ is unique, then $S$ is an inverse semigroup. It is well known that a semigroup is inverse if and only if it is regular and its idempotents form a semilattice [Reference Howie13, Theorem 5.1.1].

A subset $I\subseteq S$ is said to be a right ideal of $S$ if $IS\subseteq I.$ Left ideals are defined dually, and an ideal of $S$ is a subset that it is both a right ideal and a left ideal.

Given a subset $X\subseteq S,$ the right ideal generated by $X$ is the set $XS^{1}.$ A right ideal $I$ of $S$ is said to be finitely generated if it can be generated by a finite set.

Note that a right ideal of $S$ can be generated by a set as a right ideal or as a semigroup. For proper right ideals, we will always use the term ‘generate’ in the former sense. When we say that $S$ is generated by a set $X,$ we mean ‘generated as a semigroup’, unless stated otherwise, and we write $S=\langle X\rangle .$ We note that a right ideal can be finitely generated as a right ideal but not as a semigroup; e.g. any non-finitely group $G$ is certainly finitely generated as a right ideal.

A right congruence on $S$ is an equivalence relation $\rho$ on $S$ such that $(a,\, b)\in \rho$ implies $(ac,\, bc)\in \rho$ for all $a,\, b,\, c\in S$; left congruences are defined analogously. A congruence is a relation that is both a right congruence and left congruence. For a congruence $\rho$ on $S,$ we denote the congruence class of an element $a\in S$ by $[a]_{\rho }.$

Recall that a semigroup is right noetherian if every right congruence is finitely generated. (A right congruence $\rho$ on $S$ is finitely generated if there exists a finite set $X\subseteq \rho$ such that $\rho$ is the smallest right congruence on $S$ containing $X.$) Right noetherian semigroups are weakly right noetherian [Reference Miller and Ruškuc21, Lemma 2.7], but the converse certainly does not hold. Indeed, unlike the situation for rings, the lattice of right congruences on a semigroup is not, in general, isomorphic to the lattice of right ideals. For example, groups have no proper right ideals, but the lattice of right congruences on a group is isomorphic to its lattice of subgroups. Consequently, groups are trivially weakly right noetherian, but a group is right noetherian if and only if all its subgroups are finitely generated [Reference Miller and Ruškuc21, Proposition 2.14].

The most essential tools for understanding the structure of a semigroup are its Green's relations $\mathcal {L},\, \mathcal {R},\, \mathcal {H},\, \mathcal {D}$ and $\mathcal {J}.$ They are defined as follows.

Two elements $a,\, b\in S$ are $\mathcal {L}$-related if they generate the same principal left ideal, i.e. $S^{1}a=S^{1}b.$ Similarly, two elements $a,\, b\in S$ are $\mathcal {R}$-related if they generate the same principal right ideal. Green's relation $\mathcal {H}$ is defined as $\mathcal {H}=\mathcal {L}\cap \mathcal {R}.$ Two elements $a,\, b\in S$ are $\mathcal {D}$-related if there exists $c\in S$ such that $a\,\mathcal {L}\,c$ and $c\,\mathcal {R}\,b.$ Finally, if two elements $a,\, b\in S$ generate the same principal ideal (i.e. $S^{1}aS^{1}=S^{1}bS^{1}$), then they are said to be $\mathcal {J}$-related.

It is obvious from the definitions that $\mathcal {L},\, \mathcal {R},\, \mathcal {H}$ and $\mathcal {J}$ are equivalence relations on $S,$ and it turns out that $\mathcal {D}$ is also an equivalence relation. Moreover, Green's relation $\mathcal {L}$ is a right congruence on $S$ and $\mathcal {R}$ is a left congruence on $S.$

Green's relation $\mathcal {R}$ defines a preorder $\leq _{\mathcal {R}}$ on $S,$ given by

\[ a\leq_{\mathcal{R}}b\iff aS^{1}\subseteq bS^{1}. \]

The preorder $\mathcal {R}$ induces a partial order on the set of $\mathcal {R}$-classes of $S$: $R_a\leq R_b$ if and only if $a\leq _{\mathcal {R}}b.$ It is easy to see that the poset of $\mathcal {R}$-classes of $S$ is isomorphic to the poset of principal right ideals of $S,$ via the isomorphism $R_a\mapsto aS^{1}.$ Similarly, one can define preorders $\leq _{\mathcal {L}}$ and $\leq _{\mathcal {J}},$ leading to partial orders on the sets of $\mathcal {L}$-classes and $\mathcal {J}$-classes, respectively.

Note that when we need to distinguish between Green's relations on different semigroups, we will write them with the semigroup as a subscipt; i.e. $\mathcal {K}_S$ stands for $\mathcal {K},$ where $\mathcal {K}$ is any of Green's relations on $S.$

It is easy to see that the following inclusions between Green's relations hold:

\[ \mathcal{H}\subseteq\mathcal{L},\quad \mathcal{H}\subseteq\mathcal{R},\quad \mathcal{L}\subseteq\mathcal{D},\quad \mathcal{R}\subseteq\mathcal{D},\quad \mathcal{D}\subseteq\mathcal{J}. \]

It can be easily shown that every right (respectively left) ideal is a union of $\mathcal {R}$-classes (respectively $\mathcal {L}$-classes), and every ideal is a union of $\mathcal {J}$-classes. A semigroup with no proper (right) ideals is called (right) simple. A simple semigroup has a single $\mathcal {J}$-class; if it is right simple, then it has a single $\mathcal {R}$-class. If $S$ has a zero $0,$ $S^{2}\neq \{0\},$ and $\{0\}$ is the only proper ideal of $S,$ then it is called $0$-simple.

Given an ideal $I$ of $S,$ the Rees quotient of $S$ by $I,$ denoted by $S/I,$ is the set $(S{\setminus} I)\cup \{0\}$ with multiplication given by

\[ a\cdot b=\begin{cases} ab & \text{if }a, b, ab\in S {\setminus} I,\\ 0 & \text{otherwise.} \end{cases} \]

Let $J$ be a $\mathcal {J}$-class of $S.$ The principal factor of $J$ is defined as follows. If $J$ is the unique minimal ideal of $S,$ called the kernel of $S,$ its principal factor is itself. Otherwise, the principal factor of $J$ is the Rees quotient of the subsemigroup $S^{1}xS^{1},$ where $x$ is any element of $J,$ by the ideal $(S^{1}xS^{1}) {\setminus} J.$

The principal factors of $S$ are the principal factors of its $\mathcal {J}$-classes. The kernel of $S,$ if it exists, is simple; all other principal factors are either $0$-simple or null (every product of two elements equals zero).

A principal series of a semigroup $S$ is a finite chain of ideals

\[ K(S)=I_1\subset I_2\subset\cdots\subset I_n=S, \]

where $K(S)$ is the kernel of $S,$ and $I_k$ is maximal in $I_{k+1}$ for each $i\in \{1,\, \ldots ,\, n-1\}.$ The kernel $K(S)$ and the Rees quotients $I_{k+1}/I_k$ are the principal factors of $S.$

It is folklore that a semigroup has a principal series of length $n$ if and only if it has exactly $n$ $\mathcal {J}$-classes.

Closely related to the notion of one-sided ideals is that of semigroup acts. We provide some basic definitions about acts; one should consult [Reference Kilp, Knauer and Mikhalev16] for more information.

A (right) $S$-act is a non-empty set $A$ together with a map

\[ A\times S\to A, (a, s) \mapsto as \]

such that $a(st)=(as)t$ for all $a\in A$ and $s,\, t\in S.$ (If $S$ is a monoid, we also require that $a1=a$ for all $a\in A$.) For instance, $S$ itself is an $S$-act via right multiplication.

A subset $B$ of an $S$-act $A$ is a subact of $A$ if $bs\in B$ for all $b\in B$ and $s\in S.$ Note that the right ideals of $S$ are precisely the subacts of the $S$-act $S.$

Given an $S$-act $A$ and a subact $B$ of $A,$ the Rees quotient of $A$ by $B,$ denoted by $A/B,$ is the set $(A {\setminus} B)\cup \{0\}$ with action given by

\[ a\cdot s= \begin{cases} as & \text{if }as\in A {\setminus} B\\ 0 & \text{otherwise,} \end{cases} \]

and $0\cdot s=0,$ for all $a\in A {\setminus} B$ and $s\in S.$ It can be easily verified that $A/B$ is an $S$-act via the above action.

A subset $U$ of an $S$-act $A$ is a generating set for $A$ if $A=US^{1},$ and $A$ is said to be finitely generated if it has a finite generating set.

We call an $S$-act $A$ noetherian if every subact of $A$ is finitely generated; equivalently, $A$ satisfies the ascending chain condition on its subacts. In particular, the $S$-act $S$ being noetherian is equivalent to $S$ being a weakly right noetherian semigroup.

3. Equivalent formulations and elementary facts

We begin this section by presenting equivalent characterizations of weakly right noetherian semigroups in terms of the ascending chain condition and maximal condition on right ideals. The proof of this result is essentially the same as that of the analogue for rings and is omitted.

We now provide characterizations of weakly right noetherian semigroups in terms of their principal right ideals and also in terms of their $\mathcal {R}$-class structure.

Proof. $(1)\Rightarrow (2).$ By Proposition 3.1, $S$ certainly satisfies the ascending chain condition on principal right ideals. The fact that $S$ contains no infinite antichain of principal right ideal was proven in [Reference Hotzel12, Lemma 1.6], but we provide a proof for completeness.

Suppose for a contradiction that there exists an infinite antichain $\{a_iS^{1} : i\in \mathbb {N}\}$ of principal right ideals of $S.$ For each $n\in \mathbb {N},$ let $I_n=\{a_1,\, \ldots ,\, a_n\}S^{1}.$ Suppose that $I_m=I_n$ for some $m\leq n.$ Then $a_n\in a_iS^{1}$ for some $i\leq m.$ It must be the case that $i=m=n,$ for otherwise the incomparability of $a_iS^{1}$ and $a_nS^{1}$ would be contradicted. But then we have an infinite strictly ascending chain

\[ I_1\subsetneq I_2\subsetneq\cdots \]

of right ideals of $S,$ contradicting Proposition 3.1.

$(2)\Rightarrow (1).$ Suppose that $S$ is not weakly right noetherian yet the poset of principal right ideals of $S$ does satisfy the ascending chain condition. We need to construct an infinite antichain of principal right ideals of $S.$

By Proposition 3.1, there exists an infinite strictly ascending chain

\[ I_1\subsetneq I_2\subsetneq\cdots \]

of right ideals of $S.$ Choose elements $a_1\in I_1$ and $a_k\in I_k {\setminus} I_{k-1}$ for $k\geq 2.$ Then certainly $a_kS^{1}$ is not contained in any $a_jS^{1},\, j< k,$ since $a_jS^{1}\subseteq I_j$ and $a_k\in I_k {\setminus} I_{j}$.

Consider the infinite set $\{a_iS^{1} : i\in \mathbb {N}\}$ of principal right ideals of $S.$ This set contains a maximal element, say $a_{k_1}S^{1}$; that is, $a_{k_1}S^{1}$ is not contained in any $a_jS^{1},\, j\neq k_1.$ Indeed, if this were not the case, then there would exist an infinite strictly ascending chain of principal right ideals of $S,$ contradicting the assumption.

Now consider the infinite set $\{a_iS^{1} : i\geq k_1+1\}.$ Again, this set contains a maximal element, say $a_{k_2}S^{1}.$ Thus $a_{k_2}S^{1}$ is not contained in $a_jS^{1}$ for any $j>k_1,\, j\neq k_2.$ In fact, $a_{k_2}S^{1}$ is not contained in $a_jS^{1}$ for any $j\in \mathbb {N} {\setminus} \{k_2\},$ since, as observed above, $a_{k_2}S^{1}$ is not contained in any $a_jS^{1},\, j< k$.

Continuing this process ad infinitum, we obtain an infinite antichain $\{a_{k_i}S^{1} : i\in \mathbb {N}\}$ of principal right ideals of $S,$ as required.

$(2)\Leftrightarrow (3).$ This follows from the fact, established in § 2, that the poset of $\mathcal {R}$-classes of $S$ is isomorphic to the poset of principal right ideals of $S$.

The following result shows that there exist semigroups that satisfy the ascending chain condition on principal right ideals but are not weakly right noetherian.

Monoid acts play the analogous role in the theory of monoids to that of modules in the theory of rings. It is well known that a ring $R$ is right Noetherian if and only if every finitely generated right $R$-module is Noetherian (i.e. it satisfies the ascending chain condition on its submodules) [Reference Goodearl and Warfield8, Corollary 1.4]. We now present the analogue of this result for monoid acts.

Proof. $(1)\Rightarrow (2)$. Let $A$ be a right $M$-act with a finite generating set $X,$ and let $B$ be a subact of $A.$ For each $x\in X,$ we define a set

\[ I_x=\{m\in M : xm\in B\}. \]

Let $X^{\prime }$ be the set of elements in $X$ such that $I_x\neq \emptyset .$ Since $B$ is a subact of $A,$ we have that $I_x$ is a right ideal of $S$ for each $x\in X^{\prime }.$ Since $M$ is weakly right noetherian, for each $x\in X^{\prime }$ there exists a finite set $U_x\subseteq I_x$ such that $I_x=U_xM.$ We claim that $B$ is generated by the set

\[ U=\bigcup_{x\in X^{\prime}}xU_x. \]

Indeed, let $b\in B.$ Since $b\in A,$ we have that $b=xm$ for some $x\in X$ and $m\in M.$ Now $m\in I_x,$ so $m=un$ for some $u\in U_x$ and $n\in M,$ and hence $b=(xu)n\in UM.$

$(2)\Rightarrow (1)$. This follows from the fact that the right ideals of $M$ are subacts of the cyclic right $M$-act $M.$

For commutative semigroups, clearly the properties of being weakly right noetherian and being weakly left noetherian coincide. It is a well-known result, due to Rédei [Reference Rédei24], that every congruence on a finitely generated commutative semigroup is finitely generated; that is:

In the remainder of this section, we state some useful facts about weakly right noetherian semigroups. The following lemma is well known and will be used repeatedly throughout the remainder of the paper, usually without explicit mention.

Let $S$ be a semigroup. An element $s\in S$ is said to be decomposable if $s\in S^{2}.$ An element is indecomposable if it is not decomposable.

We now show that a semigroup composed of a finite union of weakly right noetherian subsemigroups is also weakly right noetherian.

4. Quotients and ideals

In this section, we consider the relationship between a semigroup and its quotients and ideals with regard to the property of being weakly right noetherian. We first show that this property is closed under quotients (or, equivalently, homomorphic images).

Proof. Let $I$ be a right ideal of $S/\rho ,$ and define a set

\[ J=\{a\in S : [a]_{\rho}\in I\}. \]

It is clear that $J$ is a right ideal of $S.$ Since $S$ is weakly right noetherian, $J$ is generated by a finite set $X.$ We claim that $\rho$ is generated by the finite set $Y=\{[x]_{\rho } : x\in X\}.$ Indeed, let $u\in I.$ Select $a\in S$ such that $[a]_{\rho }=u.$ Then $a=xs$ for some $x\in X$ and $s\in S^{1}.$ If $s=1,$ then $u=[x]_{\rho }\in Y.$ Otherwise, we have $u=[x]_{\rho }[s]_{\rho }\in Y(S/\rho ),$ as required.

If $\rho$ is a congruence contained in $\mathcal {R},$ then the converse of Lemma 4.1 holds. In fact, we have:

The following question naturally arises from Lemma 4.3.

The next result states that if a right ideal $I$ of a semigroup $S$ is weakly right noetherian and the Rees quotient of the $S$-act $S$ by $I$ (where $I$ is regarded as a subact of $S$) is noetherian, then $S$ is weakly right noetherian.

Recall that in a semigroup $S$ with a principal series $K(S)=I_1\subseteq \cdots \subseteq I_n=S,$ the kernel $K(S)$ and the Rees quotients $I_{k+1}/I_k$ are the principal factors of $S.$ Therefore, if all the principal factors are weakly right noetherian, then by successively applying Corollary 4.7, we deduce that $S$ is weakly right noetherian.

Although ideals of weakly right noetherian semigroup are not in general weakly right noetherian (see, for instance, Example 8.3), they do satisfy the ascending chain condition on principal right ideals. In fact, we prove a stronger statement:

Proof. Consider an infinite ascending chain

\[ a_1I^{1}\subseteq a_2I^{1}\subseteq\cdots \]

of principal right ideals of $I.$ Then for each $i,\, j\in \mathbb {N}$ with $i< j,$ there exists $u_{i,j}\in I^{1}$ such that $a_i=a_{j}u_{i,j}.$ Clearly, we have an infinite ascending chain

\[ a_1S^{1}\subseteq a_2S^{1}\subseteq\cdots \]

of principal right ideals of $S.$ By assumption, there exists $n\in \mathbb {N}$ such that $a_mS^{1}=a_nS^{1}$ for all $m\geq n.$ Let $m>n.$ If $u_{n,m}=1,$ then $a_n=a_m.$ Now suppose that $u_{n,m}\in I.$ There exists $s_m\in S^{1}$ such that $a_m=a_ns_m,$ and hence

\[ a_m=a_ns_m=a_mu_{n,m}s_m=a_n(s_mu_{n,m}s_m). \]

We have that $s_mu_{n,m}s_m\in I$ since $I$ is an ideal, so $a_m\in a_nI.$ Since $a_nI^{1}\subseteq a_mI^{1},$ we conclude that $a_nI^{1}=a_mI^{1}.$ Since $m$ was chosen arbitrarily, we have shown that the above ascending chain of principal right ideals of $I$ terminates at $a_nI^{1}.$

5. Subsemigroups

As mentioned previously, subsemigroups of weakly right noetherian semigroups need not be weakly right noetherian themselves. In this section, we explore various situations in which the property of being weakly right noetherian passes from a semigroup $S$ to a subsemigroup $T,$ and vice versa.

Proof. Let $I$ be a right ideal of $S.$ Then $I\cap T$ is a right ideal of $T.$ Since $T$ is weakly right noetherian, there exists a finite set $X\subseteq I\cap T$ such that $I\cap T=XT^{1}.$

Let $R_1,\, \ldots ,\, R_n$ be the $\mathcal {R}$-classes that intersect with $S {\setminus} T.$ For each $i\in \{1,\, \ldots \, n\},$ fix $r_i\in R_i.$ It is easy to see that if $I$ intersects an $\mathcal {R}$-class $R,$ then $R\subseteq I.$ We claim that $I$ is generated by the finite set

\[ Y=X\cup\{r_i : R_i\subseteq I, 1\leq i\leq n\}. \]

Indeed, let $a\in I.$ If $a\in T,$ then $a\in I\cap T=XT^{1}.$ If $a\in S {\setminus} T,$ then $a=r_is$ for some $i\in \{1,\, \ldots \, n\}$ and $s\in S^{1}.$ Hence, in either case, we have that $a\in YS^{1}.$

In [Reference Wallace27], Wallace introduced the idea of Greens’ relations taken relative to a subsemigroup. Let $S$ be a semigroup and let $T$ be a subsemigroup of $S.$ The $T$-relative Green's relation $\mathcal {R}^{T}$ on $S$ is given by

\[ a \mathcal{R}^{T} b\iff aT^{1}=bT^{1}. \]

The relation $\mathcal {L}^{T}$ is defined dually, and $\mathcal {H}^{T}=\mathcal {R}^{T}\cap \mathcal {L}^{T}.$ The relations $\mathcal {D}^{T}$ and $\mathcal {J}^{T}$ are defined in a similar way. All of these relations are equivalence relations on $S,$ and they respect $T$ in the sense that each class lies entirely in $T$ or entirely in $S {\setminus} T.$ The subsemigroup $T$ is said to have finite Green index in $S$ if there are only finitely many $\mathcal {H}^{T}$-classes in $S {\setminus} T.$ The notion of Green index for subsemigroups was introduced in [Reference Gray and Ruškuc10].

Proof. ($\Rightarrow$) Let $I$ be a right ideal of $T$. Let $I^{\prime }=IS^{1};$ that is, the right ideal of $S$ generated by $I.$ Since $S$ is weakly right noetherian, we have that $I^{\prime }$ is generated by some finite subset $X\subseteq I$ by Lemma 3.9. Let $R_1,\, \ldots ,\, R_n$ be the $\mathcal {R}^{T}$-classes in $S {\setminus} T,$ and for each $i\in \{1,\, \ldots \, n\}$ fix $r_i\in R_i.$ We claim that $I$ is generated by the finite set

\[ Y=X\cup\{xr_i\in I : x\in X, 1\leq i\leq n\}. \]

Indeed, let $a\in I.$ Then $a\in I^{\prime },$ so $a=xs$ for some $x\in X$ and $s\in S^{1}.$ If $s\in T^{1},$ then $a\in YT^{1}.$ If $s\in S {\setminus} T,$ then $s\in R_i$ for some $i\in \{1,\, \ldots ,\, n\},$ so there exist $t,\, u\in T^{1}$ such that $s=r_it$ and $r_i=su.$ Then $xr_i=(xs)u=au\in I,$ and hence $a=xs=(xr_i)t\in YT^{1},$ as required.

($\Leftarrow$) Each $\mathcal {R}^{S}$-class is a union of $\mathcal {R}^{T}$-classes by [Reference Gray and Ruškuc10, Proposition 9], so $S {\setminus} T$ is contained in a finite union of $\mathcal {R}^{S}$-classes, and hence $S$ is weakly right noetherian by Lemma 5.1.

Let $S$ be a semigroup and $T$ a subsemigroup of $S.$ It is easy to see that Green's $\mathcal {R}$-preorder on $T$ is contained in the restriction of Green's $\mathcal {R}$-preorder on $S$ to $T$; that is,

\[ \leq_{\mathcal{R}_T}\subseteq \leq_{\mathcal{R}_S}\!\cap~(T\times T). \]

We say that $T$ preserves $\mathcal {R}$ (in $S$), or is $\mathcal {R}$-preserving, if

\[ \leq_{\mathcal{R}_T} = \leq_{\mathcal{R}_S}\!\cap~(T\times T). \]

It can be easily shown that if $T$ preserves $\mathcal {R},$ then ${\mathcal {R}_T}$ is the restriction of $\mathcal {R}_S$ to $T.$ The next result states that the property of being weakly right noetherian is inherited by $\mathcal {R}$-preserving subsemigroups.

For a regular subsemigroup $T$ of a semigroup $S,$ Green's relation $\mathcal {R}_T$ is the restriction of $\mathcal {R}_S$ to $T$ (likewise, $\mathcal {L}_T$ and $\mathcal {H}_T$ are the restrictions to $T$ of $\mathcal {L}_S$ and $\mathcal {H}_S,$ respectively) [Reference Howie13, Proposition 2.4.2]. In fact, $T$ preserves $\mathcal {R}.$ Indeed, if $a,\, b\in T$ and $a\leq _{\mathcal {R}_S}b,$ then there exists $s\in S^{1}$ such that $a=bs.$ Letting $b^{\prime }$ be any inverse of $b,$ we have that $a=bb^{\prime }bs=b(b^{\prime }a)\in bT,$ so $a\leq _{\mathcal {R}_T}b,$ as required. Thus, by Proposition 5.6, we have:

We say that a semigroup $S$ has local right identities if $a\in aS$ for every $a\in S.$ It is easy to show that the class of semigroups with local right identities includes all regular semigroups, right simple semigroups and monoids. The notion of having local right identities will be crucial in § 6.1.

Proof. We prove that $I$ preserves $\mathcal {R}$ in $S.$ We just need to show that $\leq _{\mathcal {R}_S}\!\cap~(I\times I) \subseteq \leq _{\mathcal {R}_I.}$ So, let $(a,\, b)\in I\times I$ and $a\leq _{\mathcal {R}_S}b.$ Then $a\in bS^{1}.$ Since $I$ is a right ideal with local right identities, we have

\[ a\in bS^{1}\subseteq(bI)S^{1}=b(IS^{1})\subseteq bI, \]

so $a\leq _{\mathcal {R}_I}b,$ as required.

A subsemigroup $T$ of a semigroup $S$ is called right unitary (in $S$) if it satisfies the following condition: for all $a\in T$ and $b\in S,$ if $ab\in T$ then $b\in T.$

Clearly, a right unitary subsemigroup is $\mathcal {R}$-preserving, so we deduce the following corollary, first proven in [Reference Jespers and Okniński14], from Proposition 5.6.

If the complement of a subsemigroup is a left ideal, then the subsemigroup is right unitary, so we have:

The final part of this section concerns semigroups with a kernel.

Let $S$ be a semigroup with a zero $0.$ The right socle of $S,$ denoted by $\Sigma _r(S),$ is the union of $\{0\}$ and all the $0$-minimal right ideals of $S.$ It turns out that $\Sigma _r(S)$ is an ideal of $S,$ as noted in [Reference Clifford and Preston4, Section 6.3]. A similar argument to the one in the proof of Proposition 5.12 yields:

6. Constructions

In this section, we investigate the behaviour of the property of being weakly right noetherian under the following semigroup-theoretic constructions: direct products, free products, semilattices of semigroups, Rees matrix semigroups, Brandt extensions and Bruck–Reilly extensions.

6.1. Direct products

The problem of whether the property of being weakly right noetherian is preserved under direct products was previously considered in [Reference Davvaz and Nazemian6], where it was shown that the direct product of two weakly right noetherian commutative monoids is weakly right noetherian [Reference Davvaz and Nazemian6, Theorem 3.8].

The purpose of this subsection is to provide necessary and sufficient conditions for the direct product of two semigroups to be weakly right notherian.

Recall that a semigroup $S$ has local right identities if $a\in aS$ for every $a\in S.$

Theorem 6.1 Let $S$ and $T$ be two semigroups with $S$ infinite.

1. (1) Suppose $T$ is infinite. Then $S\times T$ is weakly right noetherian if and only if both $S$ and $T$ are weakly right noetherian and have local right identities.

2. (2) Suppose $T$ is finite. Then $S\times T$ is weakly right noetherian if and only if $S$ is weakly right noetherian and $T$ has local right identities.

In order to prove Theorem 6.1, we first present a couple of preliminary results.

Lemma 6.2 Let $S$ and $T$ be two semigroups with $S$ infinite. If $S\times T$ is weakly right noetherian, then $T$ has local right identities.

Proof. Let $t\in T,$ and let $I$ be the right ideal of $S\times T$ generated by the set $\{(s,\, t) : s\in S\}.$ Since $S\times T$ is weakly right noetherian, there exists a finite set $X\subseteq S$ such that $I$ is generated by the set $\{(x,\, t) : x\in X\}.$

Choose $s\in S {\setminus} X.$ Then $(s,\, t)=(x,\, t)w$ for some $x\in X$ and $w\in (S\times T)^{1}.$ Since $s\neq x,$ we conclude that $w\in S\times T,$ so $w=(u,\, v)$ for some $u\in S$ and $v\in T.$ It follows that $t=tv\in tT.$ Since $t$ was chosen arbitrarily, $T$ has local right identities.

Proposition 6.3 Let $S$ and $T$ be two semigroups with local right identities. If both $S$ and $T$ are weakly right noetherian, then $S\times T$ is weakly right noetherian.

Proof. Let $I$ be a right ideal of $S\times T.$ For each $a\in S,$ define a set

\[ I_a^{T}=\{t\in T : (a, t)\in I\}. \]

We claim that $I_a^{T}$ is a right ideal of $T.$ Indeed, let $t\in I_a^{T}$ and $u\in T.$ Since $S$ has local right identities, there exists $s\in S$ such that $as=a.$ Since $I$ is a right ideal of $S\times T,$ we have that $(a,\, tu)=(a,\, t)(s,\, u)\in I,$ so $tu\in I_a^{T}.$

Similarly, for each $b\in T$ we define a right ideal

\[ I_b^{S}=\{s\in S: (s, b)\in I\} \]

of $S$. We now make the following claim.

1. (1) There exists a finite set $X\subseteq S$ with the following property: for each $a\in S,$ there exists $x\in X$ such that $a\in xS$ and $I_a^{T}=I_x^{T}.$

2. (2) There exists a finite set $Y\subseteq T$ with the following property: for each $b\in T,$ there exists $y\in Y$ such that $b\in yT$ and $I_b^{S}=I_y^{S}.$

Proof of claim. Clearly, it is enough to prove (1). We shall just write $I_a$ for $I_a^{T}.$ Suppose there are infinitely many right ideals of the form $I_a.$ Note that for any $a,s\in S,$ we have $I_a\subseteq I_{as}.$ Write $S=J_1.$ Since $S$ is weakly right noetherian, there exists a finite set $X_1\subseteq J_1$ such that $J_1=X_1S^{1}.$ In fact, we have $J_1=X_1S,$ since $S$ having local right identities implies that $X_1\subseteq X_1S.$ By our assumption, there exists $x_1\in X_1$ such that there are infinitely many $a\in x_1S$ with $I_a\neq I_{x_1}.$ Consider the set

\[ J_2=\{a\in x_1S : I_a\neq I_{x_1}\}. \]

If $a\in J_2$ and $s\in S,$ then $I_{x_1}\subsetneq I_a\subseteq I_{as},$ so $as\in J_2,$ and hence $J_2$ is a right ideal of $S.$ Since $S$ is weakly right noetherian, there exists a finite set $X_2\subseteq J_2$ such that $J_2=X_2S,$ and there exists $x_2\in X_2$ such that there are infinitely many $a\in x_2S$ with $I_a\neq I_{x_2}.$ Continuing in this way, we obtain an infinite ascending chain

\[ I_{x_1}\subset I_{x_2}\subset\cdots \]

of right ideals of $T,$ but this contradicts the fact that $T$ is weakly right noetherian. Hence, there exists a finite set $U\subseteq S$ such that $I_a\in \{I_u : u\in U\}$ for every $a\in S.$ For each $u\in U,$ let $K_u$ be the right ideal of $S$ generated by the set

\[ H_u=\{a\in S : I_a=I_u\}. \]

Since $S$ is weakly right noetherian, there exists finite set $X_u\subseteq H_u$ such that $H_u=X_uS.$ Now set $X=\bigcup _{u\in U}X_u$. It is clear that $X$ satisfies the condition in the statement of the claim.

Returning to the proof of Proposition 6.3, we claim that $I$ is generated by the finite set $Z=I\cap (X\times Y).$ Indeed, let $(a,\, b)\in I.$ Then $a\in I_b^{S}$ and $b\in I_a^{T}.$ By the above claim, there exist $x\in X$ and $s\in S$ such that $a=xs$ and $I_a^{T}=I_x^{T},$ and there exist $y\in Y$ and $t\in T$ such that $b=yt$ and $I_b^{S}=I_y^{S}.$ We have that

\[ a\in I_b^{S}=I_y^{S}\Longrightarrow(a, y)\in I\Longrightarrow y\in I_a^{T}=I_x^{T}\Longrightarrow(x, y)\in I. \]

We conclude that

\[ (a, b)=(x, y)(s, t)\in Z(S\times T), \]

as required.

We are now ready to prove Theorem 6.1.

Proof of Theorem 6.1. Is $S\times T$ is weakly right noetherian, then both $S$ and $T,$ being homomorphic images of $S\times T,$ are weakly right noetherian by Lemma 4.1, and $T$ has local right identities by Lemma 6.2. If $T$ is infinite, then $S$ also has local right identities by Lemma 6.2.

For the case that $T$ is infinite, the converse follows immediately from Proposition 6.3. Now assume that $T$ is finite, and suppose that $S$ is weakly right noetherian and $T$ has local right identities. We have that $S^{1}$ is weakly right noetherian by Corollary 5.5, and clearly $S^{1}$ has local right identities. Therefore, by Proposition 6.3, we have that $S^{1}\times T$ is weakly right noetherian. Since $(S^{1}\times T) {\setminus} (S\times T)$ is finite, it follows from Corollary 5.5 that $S\times T$ is weakly right noetherian.

6.2. Free products

We shall define the free product of two semigroups (respectively monoids) in terms of semigroup (respectively monoid) presentations. For more information about semigroup and monoid presentations, we refer the reader to [Reference Clifford3, Section 9.1].

Given two semigroups $S$ and $T$ defined by presentations $\langle X\,|\,Q\rangle$ and $\langle Y\,|\,R\rangle ,$ respectively, the semigroup free product of $S$ and $T,$ denoted by $S\ast T,$ is the semigroup defined by the presentation $\langle X,\, Y\,|\,Q,\, R\rangle .$ If $S$ and $T$ are monoids, then the monoid free product of $S$ and $T,$ denoted by $S\ast _1\,T,$ is the monoid defined by the presentation $\langle X,\, Y\,|\,Q,\, R,\, 1_S=1_T\rangle ,$ where $1_S$ is a fixed word over $X$ representing the identity of $S$ and $1_T$ is a fixed word over $Y$ representing the identity of $T.$ In the case that $S$ and $T$ are groups, the monoid free product $S\ast _1\,T$ coincides with the group free product of $S$ and $T$; this fact is noted in [Reference Howie13, Section 8.2, p. 266].

In the following, we provide necessary and sufficient conditions for the semigroup (respectively monoid) free product of two semigroups (respectively monoids) to be weakly right noetherian.

Theorem 6.4 Let $S$ and $T$ be two semigroups. Then $S\ast T$ is weakly right noetherian if and only if both $S$ and $T$ are trivial.

Proof. We denote $S\ast T$ by $U.$

$(\Rightarrow ).$ Suppose that $T$ is non-trivial, and choose $a\in S$ and distinct elements $b,\, c\in T.$ For $i\in \mathbb {N},$ let $u_i=(ab)^{i}aca.$ Let $I$ be the right ideal of $U$ generated by the set $X=\{u_i : i\in \mathbb {N}\}.$ For any $i\in \mathbb {N},$ the element $u_i$ cannot we written as $u_jv$ for any $j\neq i$ and $v\in U,$ so $X$ is a minimal generating set for $I$ and hence $I$ is not finitely generated. Therefore, $U$ is not weakly right noetherian.

$(\Leftarrow ).$ The semigroup $U$ is defined by the presentation

\[ \langle e, f\,|\,e^{2}=e, f^{2}=f\rangle. \]

Then $U$ is the disjoint union of the following subsemigroups:

\[ \langle ef\rangle\cong\mathbb{N}, \langle fe\rangle\cong\mathbb{N}, \{(ef)^{i}e : i\geq 0\}\cong\mathbb{N}_0, \{(fe)^{i}f : i\geq 0\}\cong\mathbb{N}_0. \]

Since $\mathbb {N}$ and $\mathbb {N}_0$ are weakly right noetherian, it follows from Lemma 3.11 that $U$ is weakly right noetherian.

Before stating our next result, we first make some definitions.

Let $M$ and $N$ be two disjoint monoids. A reduced sequence over $M$ and $N$ is a sequence $(u_1,\, \ldots ,\, u_n)$ such that: $u_i\in (M\cup N) {\setminus} \{1_M,\, 1_N\}$ for each $i\in \{1,\, \ldots ,\, n\}$; $(u_i,\, u_{i+1})\in M\times N$ or $(u_i,\, u_{i+1})\in N\times M$ for each $i\in \{1,\, \ldots ,\, n-1\}.$

Every non-identity element of $M\ast _{1}N$ can be uniquely written as $u_1\cdots u_n$ for some reduced sequence $(u_1,\, \ldots ,\, u_n)$ over $M$ and $N$ [Reference Clifford and Preston4, Section 9.4]; the elements $u_i,\, 1\leq i\leq n,$ are called the free factors of $u_1\cdots u_n.$

Theorem 6.5 Let $M$ and $N$ be two monoids. Then $M\ast _1N$ is weakly right noetherian if and only if one of the following holds:

1. (1) $M$ is weakly right noetherian and $N$ is trivial, or vice versa$;$

2. (2) both $M$ and $N$ contain precisely two elements$;$

3. (3) both $M$ and $N$ are groups.

Proof. We denote $M\ast _1N$ by $U.$ If $N$ is trivial, then $M$ is isomorphic to $U,$ so we may assume that both $M$ and $N$ are non-trivial.

$(\Rightarrow ).$ Suppose for a contradiction that $|N|\geq 3$ and at least one of $M$ and $N$ is not a group. A monoid in which every element is right invertible is a group. Therefore, we can choose $a\in M {\setminus} \{1\}$ and distinct elements $b,\, c\in N {\setminus} \{1\}$ such that at least one of $a,\, b,\, c$ is not right invertible. Let $u_i=(ab)^{i}acab$ for $i\in \mathbb {N},$ and let $I$ be the right ideal of $U$ generated by $\{u_i : i\in \mathbb {N}\}$. Suppose that $I$ is finitely generated. Then it can be generated by a finite set

\[ X=\{u_i : 1\leq i\leq k\}. \]

Then $u_{k+1}=u_iv$ for some $i\in \{1,\, \ldots ,\, k\}$ and $v\in U.$ Since at least one of $a,\, b,\, c$ is not right invertible, the first $2i+2$ free factors of $u_iv$ are $a,\, b,\, \ldots ,\, a,\, b,\, a,\, c.$ But the free factor in position $2i+2$ of $u_{k+1}$ is $b,$ so we have a contradiction. Hence, $I$ is not finitely generated and $U$ is not weakly right noetherian.

$(\Leftarrow ).$ If $M$ and $N$ are both groups, then $U$ is also a group and hence weakly right noetherian.

Now suppose that both $M$ and $N$ contain precisely two elements and that $N$ is not a group. Then $N$ is isomorphic to the two-element semilattice $\{1,\, 0\},$ and $M$ is isomorphic to either $\{1,\, 0\}$ or $\mathbb {Z}_2.$

If $M\cong \{1,\, 0\},$ then $U$ is isomorphic to $V^{1}$ where $V$ is the free product of two trivial semigroups. It follows from Theorem 6.4 and Corollary 5.5 that $U$ is weakly right noetherian.

If $M\cong \mathbb {Z}_2,$ then $U$ is defined by the monoid presentation

\[ \langle a, b\,|\,a^{2}=1, b^{2}=b\rangle. \]

Let $u_1=ab,\, u_2=ba$ and $u_3=bab,$ and let $U_i=\{u_i^{n} : n\geq 0\}$ for $i=1,\, 2,\, 3.$ Also, let $U_4=\{u_1^{n}(aba) : n\geq 0\}.$ Then each $U_i$ is isomorphic to the free monogenic monoid $\mathbb {N}_0$, and $U=\bigcup _{i=1}^{4}U_i.$ Since $\mathbb {N}_0$ is weakly right noetherian, it follows from Lemma 3.11 that $U$ is weakly right noetherian.

6.3. Semilattices of semigroups

Let $Y$ be a semilattice and let $(S_{\alpha })_{\alpha \in Y}$ be a family of disjoint semigroups, indexed by $Y.$ If $S=\bigcup _{\alpha \in Y}S_{\alpha }$ is a semigroup such that $S_{\alpha }S_{\beta }\subseteq S_{\alpha \beta }$ for all $\alpha ,\, \beta \in Y,$ then $S$ is called a semilattice of semigroups, and we denote it by $S=\mathcal {S}(Y,\, S_{\alpha }).$

Now let $S=\bigcup _{\alpha \in Y}S_{\alpha },$ and suppose that for each $\alpha ,\, \beta \in Y$ with $\alpha \geq \beta$ there exists a homomorphism $\phi _{\alpha , \beta } : S_{\alpha }\to S_{\beta }.$ Furthermore, assume that:

• • for each $\alpha \in Y,$ the homomorphism $\phi _{\alpha , \alpha }$ is the identity map on $S_{\alpha }$;

• • for each $\alpha ,\, \beta ,\, \gamma \in Y$ with $\alpha \geq \beta \geq \gamma$, we have $\phi _{\alpha , \beta }\phi _{\beta , \gamma }=\phi _{\alpha , \gamma }.$

For $a\in S_{\alpha }$ and $b\in S_{\beta },$ we define

\[ ab=(a\phi_{\alpha, \alpha\beta})(b\phi_{\beta, \alpha\beta}). \]

With this multiplication, $S$ is a semilattice of semigroups. In this case, we call $S$ a strong semilattice of semigroups and denote it by $S=\mathcal {S}(Y,\, S_{\alpha },\, \phi _{\alpha , \beta }).$

In the remainder of this section, we investigate under what conditions a (strong) semilattice of semigroups is weakly right noetherian.

The following characterization of weakly right noetherian semilattices follows immediately from Theorem 3.2.

Proposition 6.6 Gould et al. [Reference Gould, Hartmann and Shaheen9, Proposition 3.1]

Let $Y$ be a (meet-)semilattice. Then $Y$ is weakly noetherian if and only if it contains no infinite strictly ascending chain or infinite antichain of elements.

Since the semilattice $Y$ is a homomorphic image of $S=\mathcal {S}(Y,\, S_{\alpha }),$ by Lemma 4.1 we have:

Lemma 6.7 Let $S=\mathcal {S}(Y,\, S_{\alpha })$ be a semilattice of semigroups. If $S$ is weakly right noetherian, then $Y$ is weakly noetherian.

For a semilattice of semigroups $\mathcal {S}(Y,\, S_{\alpha })$ to be weakly right noetherian, it is not required that all the $S_{\alpha }$ be weakly right noetherian. In order to show this, we consider the following construction, which will be used again later in the paper.

Construction 6.8 Let $S$ and $T$ be two semigroups with homomorphisms $\theta ,\, \phi : S\to T.$ Let $N_T=\{x_t : t\in T\}\cup \{0\}$ be a null semigroup disjoint from $S.$ We define a multiplication on $S\cup N_T,$ extending those on $S$ and $N_T,$ as follows:

\[ s\cdot x_t=x_{(s\theta)t},\quad x_t\cdot s=x_{t(s\phi)}. \]

With this multiplication, $S\cup N_T$ is a semigroup. We denote it by $\mathcal {U}(S,\, T; \theta ,\, \phi ).$ We simplify this expression in the case that $S=T$ by only writing $S$ once, and similarly if $\theta =\phi .$

We may view $\mathcal {U}(S,\, T; \theta ,\, \phi )$ as a semilattice of semigroups, where the structure semilattice is $\{\alpha ,\, 0\}$ and the corresponding subsemigroups are $S$ and $N_T,$ respectively. Every non-zero element of a null semigroup is indecomposable, so infinite null semigroups are not weakly right noetherian by Lemma 3.10. Therefore, the following result yields the desired counterexample.

Proposition 6.9 Let $S$ and $T$ be two semigroups with homomorphisms $\theta ,\, \phi : S\to T$ where $\phi$ is surjective, and let $U=\mathcal {U}(S,\, T; \theta ,\, \phi ).$ Then $U$ is weakly right noetherian if and only if $S$ is weakly right noetherian.

Proof. If $U$ is weakly right noetherian, then since $U {\setminus} S=N_T$ is an ideal of $U,$ we have that $S$ is weakly right noetherian by Corollary 5.10.

Conversely, suppose $S$ is weakly right noetherian, and let $I$ be a right ideal of $U.$ Now $I\cap S$ is either empty or a right ideal of $S$; in the latter case it is generated by a finite set $Y$ since $S$ is weakly right noetherian. We have that $T$ is weakly right noetherian by Lemma 4.1. In particular, $T=AT^{1}$ for some finite set $A\subseteq T.$ For each $a\in A,$ define a set

\[ I_a=\{s\in S : x_as\in I\}. \]

Let $A^{\prime }$ be the set of elements in $A$ such that $I_a\neq \emptyset .$ For each $a\in A^{\prime },$ we have that $I_a$ is a right ideal of $S,$ so it is generated by some finite set $U_a$ since $S$ is weakly right noetherian. We claim that $I$ is generated by the finite set

\[ Z=Y\cup\biggl(\bigcup_{a\in A^{\prime}}x_aU_a^{1}\biggr). \]

Let $u\in I.$ If $u\in S,$ then $u\in I\cap S=YS^{1}.$ Clearly, $0\in ZU,$ so we just need to consider the case that $u=x_t$ for some $t\in T.$ Then $t=av$ for some $a\in A$ and $v\in T^{1}.$ If $v=1,$ then $u=x_a\in Z.$ Otherwise, let $s\in S$ be such that $s\theta =v,$ so $s\in I_a=U_aS^{1}.$ It follows that $u=x_as\in (x_aU_a)S^{1},$ as required.

Remark 6.10 In general, principal factors of weakly right noetherian semigroups need not be weakly right noetherian. Indeed, let $G$ be an infinite group. Then $U=\mathcal {U}(G,\, \text {id})$ is weakly right noetherian. It has three $\mathcal {H}=\mathcal {J}$-classes: $G,$ $J=\{x_g : g\in G\}$ and $\{0\}.$ The principal factor of $J$ is isomorphic to $N_G,$ which is not weakly right noetherian.

In the case that a semigroup $S_{\beta },\, \beta \in Y,$ has local right identities, it is a necessary condition for $S=\mathcal {S}(Y,\, S_{\alpha })$ to be weakly right noetherian that $S_{\beta }$ be weakly right noetherian.

Lemma 6.11 Let $S=\mathcal {S}(Y,\, S_{\alpha })$ be a semilattice of semigroups, let $\beta \in Y,$ and suppose that $S_{\beta }$ has local right identities. If $S$ is weakly right noetherian, then so is $S_{\beta }.$

Proof. Let $Y^{\prime }=\{\alpha \in Y : \alpha \ngeq \beta \},$ and let $I=\bigcup _{\alpha \in Y^{\prime }}S_{\alpha }.$ Now, $I$ is an ideal and $T=S {\setminus} I$ is a subsemigroup of $S,$ so $T$ is weakly right noetherian by Corollary 5.10. Since $S_{\beta }$ is an ideal of $T$ with local right identities, it is weakly right noetherian by Corollary 5.8.

The following corollary follows from Lemmas 6.11 and 3.11.

Corollary 6.12 Let $S=\mathcal {S}(Y,\, S_{\alpha })$ be a semilattice of semigroups where $Y$ is finite and each $S_{\alpha }$ has local right identities. Then $S$ is weakly right noetherian if and only if each $S_{\alpha }$ is weakly right noetherian.

We now consider the situation for strong semilattices of semigroups.

Proposition 6.13 Let $S=\mathcal {S}(Y,\, S_{\alpha },\, \phi _{\alpha , \beta })$ be a strong semilattice of semigroups. If $S$ is weakly right noetherian, then $Y$ is weakly noetherian and each $S_{\alpha }$ is weakly right noetherian.

Proof. The semilattice $Y$ is weakly noetherian by Lemma 6.7. Now let $\alpha \in Y.$ We prove that $S_{\alpha }$ preserves $\mathcal {R}$ in $S,$ and hence $S_{\alpha }$ is weakly right noetherian by Proposition 5.6. We write $\mathcal {R}=\mathcal {R}_S$ and $\mathcal {R}_{\alpha }=\mathcal {R}_{S_{\alpha }}.$ We just need to show that $\leq _{\mathcal {R}}\cap~(S_{\alpha }\times S_{\alpha })\subseteq\leq _{\mathcal {R}_{\alpha }}$. So, let $a,\, b\in S_{\alpha }$ and $a\leq _{\mathcal {R}}b.$ Then $a=bs$ for some $s\in S^{1}.$ If $s=1$ then $a=b,$ so assume that $s\in S.$ Then $s\in S_{\beta }$ for some $\beta \in Y.$ Since $a=bs\in S_{\alpha }S_{\beta }\subseteq S_{\alpha \beta },$ we conclude that $\alpha \beta =\alpha .$ It follows that

\[ a=(b\phi_{\alpha, \alpha})(s\phi_{\beta, \alpha})=b(s\phi_{\beta, \alpha})\in bS_{\alpha}, \]

so $a\leq _{\mathcal {R}_{\alpha }}b,$ as required.

Proposition 6.13 and Lemma 3.11 together yield:

Corollary 6.14 Let $S=\mathcal {S}(Y,\, S_{\alpha },\, \phi _{\alpha , \beta })$ be a strong semilattice of semigroups where $Y$ is finite. Then $S$ is weakly right noetherian if and only if each $S_{\alpha }$ is weakly right noetherian.

Example 7.17 below shows that the converse of Proposition 6.13 does not hold, even in the case that each $S_{\alpha }$ is finite.

Open Problem 6.15 Find necessary and sufficient conditions for a strong semilattice of semigroups to be weakly right noetherian.

6.4. Rees matrix semigroups and Brandt extensions

Let $S$ be a semigroup, let $I$ and $J$ be two non-empty index sets, and let $P=(p_{ji})$ be a $J\times I$ matrix with entries from $S.$ The set $I\times S\times J$ becomes a semigroup under the multiplication given by

\[ (i, s, j)(k, t,l)=(i, sp_{jk}t, l), \]

and is called the Rees matrix semigroup over $S$ with respect to $P$. We denote this semigroup by $\mathcal {M}(S; I,\, J; P)$.

We now modify the Rees matrix construction as follows. Let the matrix $P$ have entries from $S^{0}.$ The set $(I\times S\times J)\cup \{0\}$ with multiplication given by

\[ (i, s, j)(k, t, l)= \begin{cases} (i, sp_{jk}t, l) & \text{if }p_{jk}\in S {\setminus} \{0\}\\ 0 & \text{if }p_{jk}=0, \end{cases} \]

and $0(i,\, s,\, j)=(i,\, s,\, j)0=0^{2}=0,$ is a semigroup. It is called the Rees matrix semigroup with zero over $S$ with respect to $P$, and is denoted by $\mathcal {M}^{0}(G; I,\, J; P)$.

A semigroup is said to be completely simple if it is simple and contains minimal left and right ideals. A semigroup with zero is said to be completely $0$-simple if it is $0$-simple and contains $0$-minimal left and right ideals. Rees [Reference Rees25] proved that a semigroup is completely $0$-simple if and only if it is isomorphic to a Rees matrix semigroup with zero $\mathcal {M}^{0}(G; I,\, J; P)$ over a group $G$ such that every row and column of $P$ contains at least one element in $G.$ Consequently, a semigroup is completely simple if and only if it is isomorphic to a Rees matrix semigroup $\mathcal {M}(G; I,\, J; P)$ over a group $G.$

We shall consider conditions under which a Rees matrix semigroup with zero $T=\mathcal {M}^{0}(S; I,\, J; P)$ is weakly right noetherian. We begin by considering what affect $T$ being weakly right noetherian has on $S$ and the index sets $I$ and $J.$

Lemma 6.16 Let $T=\mathcal {M}^{0}(S; I,\, J; P)$ be weakly right noetherian.

1. (1) The semigroup $S$ is weakly right noetherian and $I$ is finite.

2. (2) Let $U$ be the ideal of $S^{0}$ generated by the entries of $P.$ If the set $S {\setminus} U$ is non-empty, then both $S {\setminus} U$ and $J$ are finite.

Proof.

1. (1) Let $A$ be a right ideal of $S.$ Then $B=(I\times A\times J)\cup \{0\}$ is a right ideal of $T.$ Since $T$ is weakly right noetherian, there exists a finite set $U\subseteq B$ such that $B=UB^{1}.$ We may assume that $U=I_0\times X\times J_0$ for some finite sets $I_0\subseteq I,\, X\subseteq A,\, J_0\subseteq J.$ We claim that $I=I_0$ and $A=XS^{1}.$ Indeed, let $i\in I,$ $a\in A,$ and pick any $j\in J .$ Then $(i,\, a ,\, j)=(i_0,\, x,\, j_0)t$ for some $(i_0,\, x,\, j_0)\in U$ and $t\in T^{1}.$ It follows that $i=i_0\in I_0$ and $a\in xS^{1}\subseteq XS^{1},$ as required.

1. (2) Notice that the set $I\times (S {\setminus} U)\times J$ consists of indecomposable elements of $T$; it is hence finite by Lemma 3.10. In particular, both $S {\setminus} U$ and $J$ are finite.

The converse of Lemma 6.16(1) does not hold in general. In order to show this, we first present the following lemma.

Lemma 6.17 Let $T=\mathcal {M}^{0}(S; I,\, J; P),$ let $J^{\prime }$ be a subset of $J,$ let $P^{\prime }=(p_{ji})_{j\in J^{\prime },\, i\in I}$, and let $T^{\prime }=\mathcal {M}^{0}(S; I,\, J^{\prime }; P^{\prime }).$ If $T$ is weakly right noetherian, then so is $T^{\prime }.$

Proof. It is easy to see that $T^{\prime }$ is a right unitary subsemigroup of $T,$ so it is weakly right noetherian by Corollary 5.9.

Example 6.18 Let $T=\mathcal {M}^{0}(S; I,\, J; P)$ with $S$ infinite, and suppose there exists $j\in J$ such that $p_{ji}=0$ for all $i\in I.$ Then $T^{\prime }=\mathcal {M}^{0}(S; I,\, \{j\}; P^{\prime }),$ where $P^{\prime }=(p_{ji})_{i\in I},$ is not weakly right noetherian by Lemma 6.16(2) (in fact, $T^{\prime }$ is an infinite null semigroup). It follows from Lemma 6.17 that $T$ is not weakly right noetherian.

Remark 6.19 Let $T=\mathcal {M}(S; I,\, J; P)$ be a Rees matrix semigroup over an infinite semigroup $S$ with a zero $0$ adjoined, where $0\notin S.$ If there exists $j\in J$ such that $p_{ji}=0$ for all $i\in I,$ then $T^{\prime }=\mathcal {M}^{0}(S; I,\, J; P)$ is not weakly right noetherian by Example 6.18. Since $T^{\prime }$ is isomorphic to the Rees quotient of $T$ by the ideal $I\times \{0\}\times J,$ we deduce from Lemma 4.1 that $T$ is not weakly right noetherian.

The converse of Lemma 6.16(1) holds in the case that every row of the matrix contains a unit.

Proposition 6.20 Let $S=\mathcal {M}^{0}(M; I,\, J; P)$ be a Rees matrix semigroup over a monoid $M,$ and suppose that for every $j\in J$ there exists $i\in I$ such that $p_{ji}\in U(M).$ Then $S$ is weakly right noetherian if and only if $M$ is weakly right noetherian and $I$ is finite.

Proof. The direct implication follows from Lemma 6.16.

For the converse, let $A$ be a right ideal of $S.$ Note that if $(i,\, u,\, j)\in A,$ then $(i,\, u,\, l)\in A$ for all $l\in J.$ Indeed, there exists $k\in I$ such that $p_{jk}\in U(M),$ so $(i,\, u,\, l)=(i,\, u,\, j)(k,\, p_{jk}^{-1},\, l)\in A.$ Now fix $j_0\in J,$ and choose $i_0\in I$ such that $p_{j_0i_0}\in U(M).$ Let $I^{\prime }$ be the set of elements of $I$ that appear in $A.$ For each $i\in I^{\prime },$ define a set

\[ A_i=\{u\in M : (i, u, j_0)\in A\}. \]

We claim that $A_i$ is a right ideal of $S.$ Indeed, if $u\in A_i$ and $m\in M,$ then

\[ (i, um, j_0)=(i, u, j_0)(i_0, p_{j_0i_0}^{{-}1}m, j_0)\in A_i. \]

Since $M$ is weakly right noetherian, there exist finite sets $X_i\subseteq A_i,\, i\in I^{\prime },$ such that $A_i=X_iM.$ We claim that $A$ is generated by the finite set

\[ Y=\{(i, x, j_0) : i\in I^{\prime}, x\in X_i\}. \]

Indeed, if $(i,\, u,\, j)\in A$ then $(i,\, u,\, j_0)\in A,$ so $u=xm$ for some $x\in X_i$ and $m\in M.$ Hence, we have $(i,\, u,\, j)=(i,\, x,\, j_0)(i_0,\, p_{j_0i_0}^{-1}m,\, j)\in YM,$ as required.

Corollary 6.21 Let $S$ be a completely (0-)simple semigroup. Then $S$ is weakly right noetherian if and only if it has finitely many $\mathcal {R}$-classes.

Proof. The semigroup $S^{0}$ is completely $0$-simple. It follows from Proposition 6.20 and Corollary 5.5 that $S$ is weakly right noetherian if and only if it has finitely many $\mathcal {R}$-classes.

We now consider Brandt extensions. Let $S$ be a semigroup and let $I$ be a non-empty set. The set $(I\times S\times I)\cup \{0\}$ becomes a semigroup under the multiplication given by

\[ (i, s, j)(k, t,l)= \begin{cases} (i, st, l) & \text{if }j=k\\ 0 & \text{otherwise,} \end{cases} \]

and $0x=x0=0$ for all $x\in (I\times S\times I)\cup \{0\}.$ It is called the Brandt extension of $S$ by $I,$ and we denote it by $\mathcal {B}(S,\, I).$

Notice that if $S$ is a monoid, then $\mathcal {B}(S,\, I)$ is isomorphic to $\mathcal {M}^{0}(S; I,\, I; P)$ where $P$ is the $I\times I$ identity matrix. Brandt extensions of groups are precisely the completely $0$-simple inverse semigroups [Reference Petrich23, Theorem V.5.1].

Proposition 6.22 Let $S$ be a semigroup and let $I$ be a non-empty set. Then the Brandt extension $\mathcal {B}(S,\, I)$ is weakly right noetherian if and only if $S$ is weakly right noetherian and $I$ is finite.

Proof. Let $T=\mathcal {B}(S,\, I).$ We just need to prove that if $T$ is weakly right noetherian, then $I$ is finite. Indeed, if $I$ is finite, then $\mathcal {B}(S^{1},\, I) {\setminus} T$ is finite. It then follows from Corollary 5.5 and Proposition 6.20 that $T$ is weakly right noetherian if and only if $S$ is weakly right noetherian.

So, suppose that $T$ is weakly right noetherian. Then there exists a finite set $U\subseteq T$ such that $T=UT^{1}.$ Let $I_0$ be the elements of $I$ appearing in $U.$ Let $i\in I,$ and pick any $s\in S.$ Then $(i,\, s,\, i)=(i_1,\, x,\, i_2)t$ for some $(i_1,\, x,\, i_2)\in U$ and $t\in T^{1}.$ It follows that $i=i_1=i_2\in I_0,$ and hence $I=I_0$ is finite.