Right ideal Howson semigroups

We define a semigroup S to be right ideal Howson if the intersection of any two finitely generated right ideals, or, equivalently, any two principal right ideals, is again finitely generated. We give many examples of such semigroups, including right coherent monoids, finitely aligned semigroups, and inverse semigroups. We investigate the closure of the class of right ideal Howson semigroups under algebraic constructions. For any n∈N0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n \in \mathbb {N}^0$$\end{document} we give a presentation of a right ideal Howson semigroup possessing an intersection of principal right ideals that requires exactly n generators that is, in a particular sense, universal. We give analogous presentations for commutative, and for commutative cancellative, (right) ideal Howson semigroups.

groups, free inverse semigroups have the Howson property if and only if they are free on a one-element set [17].
The aim of this article is to change tack and to consider the Howson property for semigroups regarded as semigroup acts over themselves, so that the right (left) subacts of a semigroup S are precisely its right (left) ideals. We consider ; as being a right (left) ideal with empty set of generators. Definition 1.1 A semigroup S is right (left) ideal Howson if the intersection of any two finitely generated right (left) ideals of S is finitely generated.
Note that since intersection distributes over union, a semigroup is right (left) ideal Howson if and only if the intersection of principal right (left) ideals is finitely generated; we use this fact throughout this article. We remark that a monoid is right ideal Howson if and only if it is finitely aligned [7]; for semigroups being finitely aligned is a stronger condition, as we demonstrate. From this point we will explicitly refer to and give results for right ideal Howson semigroups; clearly, the dual results hold for left ideal Howson semigroups. Certainly for a commutative semigroup, the notions of right ideal Howson and left ideal Howson coincide; similar remarks apply to related definitions.
The property of being right ideal Howson is a finiteness condition for a semigroup; that is to say any finite semigroup is right ideal Howson. In Sect. 3 we show how it is connected to other finiteness conditions that have been studied recently, such as that of being right coherent [10,11] or right Noetherian [22].
Semigroups that are right ideal Howson abound. We list some examples here, that may easily be verified by consulting any standard semigroup text such as [5,14]: groups, inverse semigroups, completely (0-)simple semigroups, free semigroups and free monoids. We present many others subsequently in this paper. The reader may note that any of the semigroups in the previous list display the extra condition that the intersection of principal right ideals is empty or principal. Monoids that satisfy this extra condition have been well-studied by Clifford, Cherubini and Petrich: the latter authors referring to this condition (for left ideals) as Clifford's condition [3]. Indeed, Clifford [4] showed that bisimple inverse monoids can be viewed as inverse hulls of right cancellative monoids satisfying Clifford's condition. This connection has been developed by a number of authors such as Lawson [19], McAlister [21] and Reilly [24]. The notion of being finitely aligned [7] is closely connected with that of being right ideal Howson, being a stronger condition (Lemma 2.1); and coincides for many semigroups, including monoids. Indeed, it is noted in [7] that finitely aligned semigroups may be called right Howson. However, the situation for semigroups is more subtle; as we show in Remark 5.8, a right ideal Howson semigroup need not be finitely aligned. We introduce the term right ideal Howson to distinguish from that of being finitely aligned, and to make clear we are talking of right ideals and not right congruences. Explicit connections between finitely aligned semigroups, higher rank graphs and constructions of C Ã -algebras are given in [7]. This article has several aims. One is to give natural, universal, examples of right ideal Howson semigroups such that the intersection of principal right ideals may require any fixed, finite number of generators. The second is to examine closure properties of the class of right ideal Howson semigroups under standard algebraic and semigroup theoretic operations. The third is to consider connections with other natural finiteness conditions. In this way we build a broader understanding of the class of right ideal Howson semigroups.
We organise this paper as follows. In Sect. 2 we recall essential terminology and fundamental results. In Sect. 3 we provide examples of right ideal Howson semigroups, with a particular focus on bands and coherent monoids. For each variety of bands, we give an explicit presentation of a right ideal Howson band belonging to the variety, and show that the lattice of varieties of bands splits into two with regard to Clifford's condition. We show that any semigroup given by a commutative presentation with finite set of relations is (right) coherent, and hence certainly (right) ideal Howson. In Sect. 4 we explore a number of closure results for the classes of right ideal Howson monoids and semigroups. We show that both the classes of right ideal Howson semigroups and right ideal Howson monoids are closed under free products. Right ideal Howson semigroups are not closed under direct products but, on the other hand, right ideal Howson monoids are closed under direct but not semidirect products. Finally, in Sect. 5, we consider a number of semigroup presentations, reflecting those given for bands in Sect. 3. We give presentations of right ideal Howson semigroups (which are also cancellative), commutative (right) ideal Howson semigroups and commutative cancellative (right) ideal Howson semigroups, all of which are universal in a given sense.

Preliminaries
We denote the natural numbers by N and put N 0 ¼ N [ f0g. For any n 2 N we define n :¼ f1; . . .; ng and n 0 :¼ n [ f0g. Throughout this paper S denotes a semigroup and S 1 is the monoid obtained from S by adjoining an identity if necessary (so that S 1 ¼ S if and only if S is a monoid). For any element a 2 S, the principal right ideal generated by a is aS 1 ¼ fag [ aS, so that aS 1 ¼ aS if and only if a 2 aS. A right ideal is finitely generated if it is the finite union of principal right ideals. Our first observation will be useful in what follows. Lemma 2.1 Let S be a semigroup such that for all a; b 2 S we have where u i 2 aS 1 \ bS 1 for all i 2 n. Then S is right ideal Howson.
Proof Let S be as given. Then, for any a; b 2 S, we may write Clearly, if a R b then aS 1 \ bS 1 ¼ aS 1 (the case where b R a is entirely dual). However, if a£ R b and b£ R a then we have by the above that aS 1 \ bS 1 ¼ aS \ bS. Therefore, for such an a; b 2 S, we have Semigroups satisfying the hypothesis of Lemma 2.1 are called finitely aligned in [7]. However, as we show in Remark 5.8, a right ideal Howson semigroup need not be finitely aligned. For semigroups that are right factorisable, that is, semigroups S such that sS ¼ sS 1 for any s 2 S, the two notions coincide. This is the situation for monoids, inverse semigroups and bands (semigroups of idempotents), for example, but not for free semigroups.
The next observation we make follows quickly from the definition of a semigroup being right ideal Howson.
Lemma 2.2 A semigroup S is right ideal Howson if and only if the monoid S 1 is right ideal Howson.
We will say a right ideal I of S is exactly n-generated for some n 2 N 0 , if there are n elements of S that generate I, but no n À 1 elements will suffice. With this in mind, we note that I ¼ ; if and only if I is exactly 0-generated. Definition 2.3 A semigroup S satisfies (Rn) for n 2 N 0 if there exists some a; b 2 S such that aS 1 \ bS 1 is exactly n-generated. The condition (Ln) is defined dually.

Semigroup presentations
For an equivalence relation h on a set A we denote the h-class of a 2 A by ½a h , or [a] if h is understood.
Let X be a non-empty set, whose elements we will refer to as letters. We denote by X þ (X Ã ) the free semigroup (free monoid) on X. We take X þ to be the set of all non-empty words over X with operation concatenation, and to obtain X Ã we adjoin the empty word, often denoted by .
The length of a word w 2 X Ã , denoted by jwj, is the number of letters in w, counting repeats. The context of w, written c(w), is the set of letters that appear in w. For a number of results in this paper, it will be useful to let aðwÞ and xðwÞ denote respectively the first and last letter of w 2 X þ .
For a subset q of X þ Â X þ , the smallest congruence relation on X þ containing q is denoted by q ] . For any w; x 2 X þ we have that ðw; xÞ 2 q ] if and only if w ¼ x or there exists a finite sequence of the form where z iÀ1 ¼ c i p i d i and z i ¼ c i q i d i with ðp i ; q i Þ 2 q [ q À1 and c i ; d i 2 X Ã for all i 2 n [14]. We say that z iÀ1 ! z i an elementary q-transition. The q ] -class ½w q ] of w 2 X þ will be denoted by ½w q (or [w] if q is understood).
Definition 2.4 A semigroup presentation hX : qi consists of non-empty set X and a subset q of X þ Â X þ , and defines the semigroup X þ =q ] . By standard convention, we may denote ðu; vÞ 2 q by the equality u ¼ v. Moreover, we may we identify the semigroup presentation with the semigroup that it defines.
Similarly, monoid presentations are given by the same notation hX : qi where q X Ã Â X Ã and define the monoid X Ã =q ] . In the case where juj ¼ jvj for every ðu; vÞ 2 q, and hence for every ðw; xÞ 2 q ] , we say that the corresponding presentation is homogeneous. In this article we will also need the notion of commutative semigroup and monoid presentations. To this end we denote by CX þ and CX Ã the free commutative semigroup and free commutative monoid on X, respectively. We view CX þ as consisting of all unordered non-empty words over X under concatenation; similarly, CX Ã consists of all unordered words over X.
Definition 2.5 A commutative semigroup presentation hX : qi consists of nonempty set X and a subset q of CX þ Â CX þ , and defines the commutative semigroup CX þ =q ] .
In Definition 2.5 we may think of q as a subset of X þ Â X þ , if convenient. Commutative monoid presentations are defined in the obvious way.

Monoid actions
Throughout this subsection we let M be a monoid. We briefly present the basic notions surrounding M-acts, in order to facilitate later discussions on coherency. Further details may be found in [18].
A set A is a (right) M-act if there exists a map A Â M ! A : ða; mÞ7 !a Á m such that for any a 2 A and s; t 2 M we have a Á 1 ¼ a and ða Á sÞ Á t ¼ a Á ðstÞ. An Msubact of an M-act A is a subset B of A closed under the action of M; a congruence on A is an equivalence relation h on A that is closed under the action of M, that is, for any a; b 2 A and s 2 M, if a h b then as h bs. If h is a congruence on A then we may form the quotient M-act A=h ¼ f½a : a 2 Ag with action ½as ¼ ½as. Clearly, M may be regarded as an M-act where the action is the multiplication in M and the Msubacts are then precisely its right ideals. To avoid confusion, if h is a congruence on the M-act M, then we refer to h as a right congruence. Of course, if M is commutative, then right congruences coincide with (monoid) congruences on M.
A congruence h on an M-act A is finitely generated if it is the smallest congruence containing a given finite set of elements of A Â A. An explicit formula for obtaining h from its generators may be found in a way analogous to that in Equation (1). Note that the identity relation is always finitely generated. The notions of being free, finitely generated and finitely presented for an M-act A are the standard ones from universal algebra, translated to the context of M-acts. Explicitly, (finitely generated) free M-acts are disjoint unions of (finitely many) copies of the M-act M; an M-act A is finitely generated if A ¼ a 1 M [ . . . [ a n M for some a i 2 A, n 2 N 0 , and finitely presented if A ffi F=h for some finitely generated free M-act F and finitely generated congruence q on F.
For any M-act, say A, and for every a 2 A we define the right congruence rðaÞ on M by rðaÞ ¼ fðu; vÞ 2 M Â M : au ¼ avg:

Coherency and noetherianity
A semigroup is weakly right Noetherian if every right ideal is finitely generated. This is a weaker condition than that of being right Noetherian, which means that every right congruence is finitely generated (a fact easily witnessed by considering a group). These are well-established notions, recently attracting new attention [22]. The following remark is worth stating explicitly.
Remark 2.6 Any weakly right noetherian semigroup is right ideal Howson.
In fact, we can weaken the hypothesis of Remark 2.6 much further.
Definition 2.7 [9,26] A monoid M is weakly right coherent if every finitely generated right ideal is finitely presented and right coherent if every finitely presented M-subact of every finitely presented M-act is finitely presented.
Any right noetherian monoid is right coherent, but weakly right noetherian monoids need not even be weakly right coherent [9]. We are interested in these notions here due to the result below, which may be regarded as analogous to that of Chase for rings [2].

.4] A monoid M is weakly right coherent if and
only if for any a; b 2 M the right congruence rðaÞ is finitely generated, and the right ideal aM \ bM is finitely generated.
A monoid M is right coherent if and only if for any finitely generated right congruence h on M and any ½a; ½b 2 M=h we have the right congruence rð½aÞ is finitely generated and the M-subact ½aM \ ½bM of M=h is finitely generated. Corollary 2.9 A weakly right coherent monoid is right ideal Howson.

Examples of right ideal Howson semigroups
We begin by summarising some previous remarks.
Remark 3.1 Any finite semigroup, (weakly) right noetherian semigroup or semigroup S such that S 1 is (weakly) right coherent is right ideal Howson.
There are many examples of right coherent monoids [6,9]. Rédei's Theorem states that the free commutative monoid CX Ã , for any finite set X, is (right) noetherian [23]. Gould used this result to show that for any set X, the free commutative monoid CX Ã is (right) coherent [9,Theorem 4.3]. Our next result is a significant extension of the latter fact. The proof makes essential use of the fact that for a commutative monoid, right congruences and congruences coincide. Proof Let q ] be a finitely generated congruence on M ¼ CX Ã =s ] . We have that q ¼ fð½a s ; ½b s Þ : ða; bÞ 2 rg for some finite subset r of CX Ã Â CX Ã . We let m ¼ s [ r so that m ] is a finitely generated congruence on CX Ã . Taking care with the generators, it follows from the second isomorphism theorem [14,Theorem 1.5.4] that there exists a monoid isomorphism For convenience we denote the q ] -class of ½w s 2 M by ½w s:q .
We are considering the monoid M acting on the right of the M-act M=q ] . To this end, note that for ½u s 2 M and ½w s:q 2 M=q ] we have Let w 2 CX Ã . Since m is finite and CX Ã is coherent, it follows that there exists a finite symmetric set of generators j for rð½w m Þ, say so that g is also finite and symmetric.
Proof That g rð½w s:q Þ follows from Proposition 3.3.
Suppose for the converse that ð½u s ; ½v s Þ 2 rð½w s:q Þ, so that ðu; vÞ 2 rð½w m Þ, again by Proposition 3.3. Since j generates rð½w m Þ there exists ' 2 N 0 and a finite sequence of the form Thus rð½w s:q Þ hgi and we therefore have equality as desired. h Let a; b 2 CX Ã and let I ¼ ½a s:q M \ ½b s:q M and J ¼ ½a m CX Ã \ ½b m CX Ã . We proceed to show that I is finitely generated. Proof Suppose that w; u; v 2 CX Ã . Then h Since m is finitely generated, and CX Ã is coherent, J is generated by a set f½ac m : c 2 Yg for some finite subset Y of X Ã . Proposition 3.5 then gives us that I is finitely generated by the set f½ac s:q : c 2 Yg and this completes the proof of Theorem 3.2. h From Corollary 2.9 and Theorem 3.2 we immediately deduce the following. In particular, any finitely presented commutative semigroup or monoid is (right) ideal Howson.
Changing tack, the following is well known. Proof For any a; b 2 S we have aS ¼ aa À1 S and bS ¼ bb À1 S. Since idempotents commute it follows that Since the free band on a finite set is finite (see [14]), any finitely generated band is finite. Hence: Lemma 3.8 Every finitely generated band is right ideal Howson.
Let X be a countable set. For some x; y 2 X þ , we say that a semigroup S satisfies an identity x ¼ y if for every choice of homomorphism h : X þ ! S we have xh ¼ yh. A semigroup variety denotes a class of semigroups containing all semigroups that satisfy a given collection of identities.
Throughout we let V be a variety of bands. We recall the variety of right regular bands RR (variety of rectangular bands RB) is determined by the identities x 2 ¼ x and xy ¼ yxy (x 2 ¼ x and x ¼ xyx) (in addition to the identity guaranteeing associativity). Proof Suppose that B 2 V. Since B is a band, we note that for any a; b 2 B we have aB \ bB abB. If B 2 RR then we have ab 2 aB and ab ¼ bab 2 bB, so that in this case aB \ bB ¼ abB is principal. On the other hand, if B 2 RB then for all a; b 2 B we have aB \ bB is either principal if a R b or empty else. h In order to prove the next therorem, we must draw upon Fennemore's result [8] concerning the defining identities of varieties of bands. This yields that if V is a variety of bands not contained in RR or RB, then V is defined by an identity V :¼ p ¼ q with the property that aðpÞ ¼ aðqÞ and cðpÞ ¼ cðqÞ.
Theorem 3.10 Let V be a variety of bands not contained in RR or RB. Then there exists a band B 1 2 V that is not right ideal Howson, and for each n 2 N 0 a band B n 2 V that satisfies (Rn).
Proof Fix some n 2 N 0 and let X n ¼ fa; b; u i ; v i : i 2 ng. If n ¼ 0 then we simply put X 0 ¼ fa; bg. Define two subsets q n and r n of X þ n Â X þ n as follows q n ¼ fðau i ; bv i Þ : i 2 ng and r n ¼ fðw; w 2 Þ : w 2 X þ n g: In this way, if n ¼ 0 then q 0 ¼ ;. Suppose V is the defining identity for V (as above) and let B n be the band with semigroup presentation hX n : q n [ r n [ Vi. From this point we will write X, q, r and B instead of X n , q n , r n and B n respectively. Moreover, for ease of notation we will let It follows immediately from Lemma 3.8 that B is right ideal Howson since it is finitely generated. Hence, it remains to show that B satisfies (Rn). We achieve this by showing that I ¼ ½aB \ ½bB is exactly n-generated.
Firstly, we suppose that there exist some s; t 2 X Ã such that ðas; btÞ 2 s ] . This implies that there exists a finite sequence of the form If z iÀ1 ! z i is an elementary ror an elementary V-transition for all i 2 ', then we immediately reach a contradiction since ðw; xÞ 2 ðr [ VÞ ] implies aðwÞ ¼ aðxÞ.
Therefore, we may assume that z iÀ1 ! z i is an elementary q-transition for some i 2 ', and, in order to avoid a similar contradiction, we may also assume that for at least one such i 2 ', we have c i ¼ . Thus, we have shown z i ¼ au j d i or bv j d i for some i 2 ' and j 2 n. It follows that the reverse inclusion is clear from the form of the presentation.
To show that I is exactly n-generated, we note that if w s ] x for some w; x 2 X Ã and u i 2 cðwÞ then fu i ; v i g \ cðxÞ 6 ¼ ;, since any elementary s-transition has this property. Thus if i; j 2 n with i 6 ¼ j, then for no t 2 X Ã do we have that au i s ] au j t. Thus ½au i B 6 ½au j B and so I requires n generators as claimed.
The existence of B 1 is proved in an entirely similar manner, starting with it is easy to see that I ¼ ½aB \ ½bB cannot be finitely generated. h We end this section with a brief discussion of another finiteness condition for a monoid M, namely R [12]. This condition arises from axiomatisability properties of classes of right M-acts and states that for any a; b 2 M the subact of the direct product right M-act M Â M given by Rða; bÞ ¼ fðu; vÞ 2 M Â M : au ¼ bvg is finitely generated. It was shown that R is independent of being weakly right noetherian [12]. The next corollary comes from Lemma 3.11 and the results of [10].

Corollary 3.12
The free inverse monoid, the free ample (restriction) monoid, and the free left ample (restriction) monoid on any set is right and left ideal Howson.

Closure results
In this section, we explore a number of closure results regarding the class of right ideal Howson semigroups. Our first result is immediate, since any free semigroup X þ is right ideal Howson.

Free products
Let K be a non-empty set and let S k be a semigroup (monoid) for every k 2 K. Let we write the product of x; y 2 X þ or X Ã as x y. Consider the subsets l and m of X þ Â X þ and X Ã Â X Ã , respectively, where where I S k denotes the identity of S k in the case of monoids. The free product of semigroups S k ; k 2 K, may be given by the quotient X þ =l ] , and the free product of monoids S k ; k 2 K, may be given by the quotient X Ã =m ] . If K ¼ n for some n 2 N, we may write S 1 Á Á Á S n for the semigroup or monoid free product, where the distinction will be clear from the context.

Proposition 4.2 The class of right ideal Howson semigroups is closed under free products of semigroups.
Proof Let S be the semigroup free product of right ideal Howson semigroups S k ; k 2 K. Let ½a; ½b 2 S where ½a ¼ ½a 1 Á Á Á a n and ½b ¼ ½b 1 Á Á Á b m and put I ¼ ½aS 1 \ ½bS 1 . We show that I is finitely generated. We may assume that n, m are least, so, for example, there does not exist an i 2 n À 1 such that a i a iþ1 2 S k for some k 2 K. Either I ¼ ;, ½aS 1 ½bS 1 , ½bS 1 ½aS 1 , or there exists some ½u; ½v 2 S such that ½a u ¼ ½b v. In the second and third cases I is principal: we consider the final case. Here we take ½u ¼ ½u 1 Á Á Á u k and ½v ¼ ½v 1 Á Á Á v ' for some least k; ' 2 N, so that ½a½u ¼ ½a 1 Á Á Á a n u 1 Á Á Á u k and ½b½v ¼ ½b 1 Let us assume n m and proceed with a case-by-case consideration.
(i) If n\m then a i ¼ b i for all i 2 n À 1 and b n ¼ a n or b n ¼ a n u 1 . In either case, I ¼ ½bS 1 is principal. Dually if m\n. (ii) If n ¼ m then again a i ¼ b i for all i 2 n À 1. Suppose that a n ; b n 2 S k ; if a n ¼ b n , a n ¼ b n w or b n ¼ a n w for some w 2 S 1 k , then clearly I is principal. If this is not the case, then a n u 1 ¼ b n v 1 . Let J ¼ a n S 1 k \ b n S 1 k . Since S k is right ideal Howson this intersection is finitely generated, say J ¼ [ i2r a n w i S 1 k : Clearly [ i2r ½a 1 . . . a n w i S 1 k I: On the other hand, a n u 1 ¼ a n w i w for some i 2 r; w 2 S k , so that ½a ½u 2 ½a 1 . . . a n w i S 1 for some i 2 r. Thus [ i2r ½a 1 . . . a n w i S 1 ¼ I so that I is finitely generated as required. h The corresponding result hold for the free product of monoids. Proof The argument runs along the same lines as that of Proposition 4.2, but with added technicalities due to the extra relations in m. Let S be the (monoid) free product of right ideal Howson monoids S k ; k 2 K. Let ½a; ½b 2 S and put I ¼ ½aS \ ½bS. We show that I is finitely generated. If [a] is the identity of S, or an element that has a right inverse, then the result is clear. Thus we may suppose ½a ¼ ½a 1 . . . a n where n 2 N; a i 2 S k i ; a i 6 ¼ I S k i for i 2 n and k i 6 ¼ k iþ1 for i 2 n À 1 . Now observe that if n 0 is greatest such that a n 0 is not right invertible, then ½a R ½a 1 . . . a n 0 . We may therefore assume from the outset that a n 2 S k n is not right invertible. Similarly, we can assume ½b ¼ ½b 1 . . . b m where m 2 N; b j 2 S s j ; b j 6 ¼ I S s j for j 2 m, s j 6 ¼ s jþ1 for j 2 m À 1 and b m 2 S s m is not right invertible.
If I ¼ ;, or if ½a ¼ ½b½v or ½b ¼ ½a½u for some ½u; ½v 2 S, then we are done. Suppose therefore that ½a½u ¼ ½b½v for some ½u; ½v 2 S. Given the fact that a n ; b m are chosen to be not right invertible, the proof proceeds as in that of Proposition 4.2. h

Direct and semidirect products
We begin with a negative result for direct products of semigroups.

Proposition 4.4 The class of right ideal Howson semigroups is not closed under direct products.
Proof To see this, consider the free monogenic semigroup S ¼ hai. Clearly S is right ideal Howson since S is (right) coherent, by Remark 3.1. Indeed, for any n; m 2 N we have that a n S 1 \ a m S 1 ¼ a k S 1 where k ¼ maxfn; mg. Let T ¼ S Â S. One may then easily verify that the intersection I ¼ ða; aÞT 1 \ ða; a 2 ÞT 1 is generated by the set K ¼ fða 2 ; a k Þ : k ! 3g of R-incomparable elements, and every generating set for I must contain K. h On the other hand we have a positive result for right factorisable semigroups.

Proposition 4.5 The class of right factorisable right ideal Howson semigroups is closed under direct products.
Proof Let S and T be right factorisable right ideal Howson semigroups. Notice that for any s 2 S and t 2 T we have sS ¼ sS 1 , tT ¼ tT 1 and also It then follows easily that S Â T is right ideal Howson. h If S is free monogenic, then S 1 Â S 1 is right ideal Howson from Proposition 4.5, but Proposition 4.4 tells us that S Â S is not right ideal Howson. Hence: We now turn our attention to semidirect products. Let S and T be semigroups such that S acts on the left of T via morphisms; we denote the resulting semidirect product by ToS. If the action of S on T is trivial, then ToS is simply the direct product T Â S. Proposition 4.4 therefore tells us that the class of right ideal Howson semigroups is not closed under semidirect products. Considering now the case for monoids S and T, where S acts as a monoid on T by monoid morphisms, the semidirect product T Â S becomes a monoid. In view of Proposition 4.5 we know that the direct product of right ideal Howson monoids is right ideal Howson. By way of contrast we have the following. Proof Let X ¼ fa; b; a i : i 2 Ng and A ¼ fa i : i 2 Ng. We consider the left action of X Ã on PðXÞ where, for all U 2 PðXÞ, we have b Á U ¼ ; ¼ a i Á U for all i 2 N and One may verify that this is a left action of X by monoid endomorphisms on the semilattice PðXÞ under union. The only case that needs thought is that of a Á ðU [ VÞ where U \ A 6 ¼ ; but V \ A ¼ ; (or the dual). In this case the second equality following since V fa; bg. We may then extend the action of X to that of the monoid X Ã . Let S ¼ PðXÞoX Ã and let We claim that I is not finitely generated. To see this, notice that for any i 2 N we have Since Z i £ R Z j for any i 6 ¼ j, it follows that I cannot be finitely generated. h

Semigroup presentations
In Corollary 3.6 we show that a commutative semigroup presentation hX : si, where s is finite, gives rise to a (right) ideal Howson semigroup. In fact, one can show that both conditions -commutativity and the fact s is finite -are strictly necessary. We illustrate this by way of the examples below.
Example 5.1 Let S be given by the commutative semigroup presentation Then S is not ideal Howson.
Proof This presentation is the commutative semigroup version of the band presentation of B 1 given in Theorem 3.10. It is easy to see from the form of the presentation that and is not finitely generated. h It is also easy to see that the (non-commutative) semigroup on the same presentation as that in Example 5.1 is not right ideal Howson; the presentation, however, is not finite. We now give an example of a finitely presented semigroup that is not right ideal Howson.
Example 5.2 Let S be given by the semigroup presentation Then S is not right ideal Howson.
Proof Let X ¼ fa; b; c; d; p; q; u; vg and define subsets q and r of X þ Â X þ by q ¼ È ðauvc; bpqdÞ É and r ¼ È ðau; uaÞ; ðub; bpÞ; ðuv; u 2 v 2 Þ É and let s ¼ q [ r. First, we note that for any t 2 X Ã we have ½au h vc ¼ ½au k vct if and only if h ¼ k and t ¼ . This is a fact witnessed by verifying, by induction on the length of a s-sequence, that ½au h vc ¼ fu t au hþ'Àt v 'þ1 c; u r bp s qd : ' ! 0; 0 t h þ '; r ! 0; s ! 0; r þ s ¼ hg: It follows that ½au h vc and ½au k vc, for any distinct h; k 2 N are incomparable under the R-order. Furthermore notice that ½au h vc 2 I ¼ ½aS 1 \ ½bS 1 for any h 2 N, since au h vc r ] u hÀ1 auvc q ] u hÀ1 bpqd r ] bp h qd: To complete the proof, we must show that if aw s ] bx for some w; x 2 X Ã then ½aw 2 ½au h vcS 1 for some h ! 1. Suppose therefore that aw s ] bx, so there exists a finite sequence of the form We claim that at least one elementary s-transition must be of the form where k ! 0; d i 2 X Ã . Suppose for contradiction that this is not the case; we show that for each i 2 n 0 we have z i ¼ u k i az 0 i for some k i ! 0; z 0 i 2 X Ã . Clearly this is true for i ¼ 0. Suppose for induction that z i ¼ u k i az 0 i as given where i 2 n 0 . Avoiding the elementary s-transition of the form above, our possibilities for z i ! z iþ1 are where z 0 i ! z 0 iþ1 under an elementary s-transition. Thus aðz n Þ 2 fa; ug, a contradiction. Thus for some i we must have z i ¼ u k auvcd i where k ! 0, and then ½ax ¼ ½u k auvcd i 2 ½au h vcS where h ¼ k þ 1 2 N. This completes the proof. h

Semigroup presentations of right and left ideal Howson semigroups
We now turn to positive results, constructing semigroup presentations that are right (left, right and left) ideal Howson, which are, by construction, universal in a specific sense. The semigroups we construct in this subsection are also all cancellative. For n; m 2 N 0 we define an alphabet X nm by X nm ¼ È a; b; u i ; v i ; p j ; q j : i 2 n; j 2 m É and a relation q nm on X þ nm by q nm ¼ È ðau i ; bv i Þ; ðp j a; q j bÞ : i 2 n; j 2 m É and, as a convention, let S q nm be the semigroup with presentation hX nm : q nm i. If n ¼ 0 then we simplify our ingredients considerably; we have X 0m ¼ È a; b; p j ; q j : j 2 m É and q 0m ¼ È ðp j a; q j bÞ : j 2 m É and similarly if m ¼ 0. If m ¼ n ¼ 0 then X ¼ fa; bg and q 00 ¼ ;. In this case, S q 00 ¼ fa; bg þ , being free, is certainly right and left ideal Howson, with intersections of right (left) ideals being empty or principal. Since n; m 2 N 0 are fixed, we simplify notation and denote X nm ; q nm and S q nm by X; q and S, respectively. To proceed we consider a specific factorisation of elements of X þ .
Let UðnÞ ¼ fu i ; v i : i 2 ng and PðmÞ ¼ fp j ; q j : j 2 mg where we regard U(n) and P(m) to be empty if n ¼ 0 and m ¼ 0 respectively. Let We call such a factorisation the q-factorisation of w with corresponding q-length equal to p. Notice that jr i j ¼ 2 or 3 for all i 2 p. Proof If w and x are as given and r i q ] s i for every i 2 p, then clearly w q ] x. For the converse, suppose that w is as given and w ! y is an elementary qtransition. then, (from the definition of q-factorisation) we must have y ¼ w 0 s 1 w 1 . . .w pÀ1 s p w p where, for all but one i 2 p, we have s i ¼ r i and for a single j 2 p we have r j ; s j 2 CðqÞ with r j q ] s j . Therefore y has the form required. The result then follows by induction on the length of a q-sequence starting from w and ending at x. h

Claim 5.4 The semigroup S is right and left ideal Howson.
Proof We show that S is right ideal Howson, the argument for left ideals being dual. Let w; x 2 X þ and put I ¼ ½wS 1 \ ½xS 1 . Suppose that I 6 ¼ ;, so that wh q ] xk for some h; k 2 X Ã . By Claim 5.3 we have q-factorisations wh ¼ w 0 r 1 w 1 . . .w pÀ1 r p w p and xk ¼ w 0 s 1 w 1 . . .w pÀ1 s p w p for some p 2 N 0 where r i q ] s i for all i 2 p. Without loss of generality we may assume jwj jxj and then consider all possible cases for the q-factorisation of h.
(i) Suppose that w ¼ w 0 r 1 w 1 . . .r iÀ1 s and h ¼ tr i . . .w pÀ1 r p w p where w i ¼ st for some i 2 p 0 and s; t 2 X Ã . Since x ¼ w 0 s 1 w 1 . . .s iÀ1 sy for some y 2 X Ã , we have that wy ¼ w 0 r 1 w 1 . . .r iÀ1 sy q ] w 0 s 1 w 1 . . .rs iÀ1 sy ¼ x and so I ¼ ½xS 1 is principal. (ii) Suppose now that w ¼ w 0 r 1 w 1 . . .w iÀ1 s and h ¼ tw i . . .w pÀ1 r p w p where r i ¼ st for some i 2 p and s; t 2 X Ã . We write x ¼ w 0 s 1 w 1 . . .w iÀ1 s i y for some y 2 X Ã and see so that I ¼ ½xS 1 is principal. (iii) Lastly, we suppose that w is as in case (ii) and x ¼ w 0 s 1 w 1 . . .w iÀ1 u where s i ¼ uv for some i 2 p and u; v 2 X Ã with jsj juj. Proof Again we only give the proof for right ideals. Let w; x 2 X þ ; we show the intersection I ¼ ½wS 1 \ ½xS 1 requires at most n generators. The only situation where I is not empty or principal is in the second situation of (iii). Here we consider all the possibilities for ðp; qÞ 2 RðqÞ; we can have jr i j ¼ js i j ¼ 2 with jsj ¼ juj ¼ 1 or jsj ¼ 1 and juj ¼ 2, or jr i j ¼ js i j ¼ 3 with jsj ¼ juj ¼ 1 or jsj ¼ 1 and juj ¼ 2, or which achieves the bound n as required. h We are now in a position to given the main result of this subsection.
Theorem 5.6 Let n; m 2 N 0 . The semigroup S q nm is cancellative, right and left ideal Howson, and satisfies (Rn) and (Lm). Further, the intersection of any two principal right (left) ideals of S requires at most n (m) generators.
Proof Cancellativity follows from the conditions given by Adjan in [1] for a semigroup given by a presentation to be embeddable into a group. The remainder of the result comes from Claims 5.3, 5.4 and 5.5. h We now show that our semigroup S q nm is in a specific sense universal.
Proposition 5.7 Let n; m 2 N 0 . Suppose U is a semigroup containing elements a; b such that aU 1 \ bU 1 and U 1 a \ U 1 b are each exactly n-and m-generated, respectively, by ac 1 ¼ bd 1 ; . . .; ac n ¼ bd n and l 1 a ¼ p 1 b; . . .; l m a ¼ p m b, respectively. Then there is a homomorphism h : S q nm ! U such that for all i 2 n and j 2 m.
Proof Let w : X þ nm ! U be given by determining its values on the elements of X nm by for all i 2 m and j 2 n. Clearly q nm ker w so that w induces a morphism h : S q nm ! U as in the statement of the proposition. h Remark 5.8 Let n ! 1. The semigroup S q 1m is not finitely right aligned. To see this, suppose that for some w k 2 X þ such that ½w k 2 ½aS 1 \ ½bS 1 for all k 2 '. We have ½au 1 ¼ ½bv 1 2 ½aS \ ½bS, so that ½au 1 ¼ ½bv 1 ¼ ½w k ½z for some k 2 ' and z 2 X þ . We must have that w k z ¼ au i or bv i . If w k ¼ a then we would not have w k 2 ½bS 1 and similarly if w k ¼ b, a contradiction.
In fact, an easier approach to obtain a semigroup presentation that satisfies (Rn) and (Lm) for some fixed n; m 2 N 0 , but producing a less tight result, runs as follows.
For n; m 2 N 0 we define an alphabet Y nm by Y nm ¼ È a; b; c; d; u i ; v i ; p j ; q j : i 2 n; j 2 m É and a relation r mn on Y þ mn by r nm ¼ È ðau i ; bv i Þ; ðp j c; q j dÞ : i 2 n; j 2 n É and, as a convention, we let T r nm be the semigroup with presentation hY nm : r nm i. If n ¼ 0 then (as before) we have Y 0m ¼ È a; b; c; d; p j ; q j : j 2 m É and r 0m ¼ È ðp j c; q j dÞ : j 2 m É and similarly when m ¼ 0. If in fact n ¼ m ¼ 0 then we simply have Y 00 ¼ fa; b; c; dg and r 00 ¼ ;. The proof for the following theorem is similar to that of Theorem 5.6 but rather simpler, since the complications of overlapping generators for r ] nm do not occur. Theorem 5.9 Let n; m 2 N 0 . The semigroup T r nm is cancellative, right and left ideal Howson, and satisfies (Rn) and (Lm). Further, the intersection of any two principal right (left) ideals of S requires at most n (m) generators.
There is also a corresponding universal type result for T r nm , analogous to that of Proposition 5.7. Namely, if U is a semigroup containing elements a; b; c and d such that aU 1 \ bU 1 and U 1 c \ U 1 d are each exactly n-and m-generated respectively, then U 1 contains a morphic image of T r nm obtained as in Proposition 5.7.

Commutative semigroup presentations of (right) ideal Howson semigroups
For any n; m 2 N 0 , the semigroup S q nm is not commutative. In this section, we provide a commutative semigroup presentation that satisfies (Rn), which now coincides with (Ln), for any fixed n 2 N. Unlike the case for S q nm we do not automatically have that the semigroups we construct are cancellative. However, we can construct natural quotients that satisfy (Rn) and are cancellative. For a fixed n 2 N 0 we define the alphabet X n by X n ¼ fa; b; u i ; v i : i 2 ng and the relations s n and t n on X þ n by s n ¼ È ðau i ; bv i Þ : i 2 n É ; t n ¼ È ðau i ; bv i Þ; ðu i v j ; u j v i Þ : i; j 2 n; i 6 ¼ j É : Similarly to the conventions in Sect. 5.1, we let S s n and S t n be the semigroups with commutative presentations hX n : s n i and hX n : t n i respectively. If n ¼ 0 then we simply put X 0 ¼ fa; bg and s 0 ¼ È ðab; baÞ; ðba; abÞ É ¼ t 0 : Since n 2 N 0 is fixed, we will write X, s, t, S s and S t instead of X n , s n , t n , S s n and S t n respectively. Also, note that we will continue to refer to words but we now mean elements of CX þ rather than X þ .
Theorem 5.10 The commutative semigroups S s n and S t n satisfy ðRðn þ 1ÞÞ.
In particular, is a partial order when restricted to any set of balanced words; certainly any s ] -class or t ] -class. Hence, let w s and w t be the unique words in ½w s and ½w t that are greatest under . We now draw upon some well-known results regarding rewriting systems to obtain w s and w t . For basic definitions surrounding rewriting systems we refer the reader to the work of Gray and Malheiro [13].
Claim 5.15 The rewriting system on ½w s for all w 2 X þ , given by the rewriting rules bv k ! au k and av i u j ! au i v j for all i; j; k 2 n with i\j, is confluent.
Proof If w ¼ ybv k and x ¼ yau k for some y 2 X Ã , then clearly w s ] x and w\x. Correspondingly, if w ¼ yav i u j and x ¼ yau i v j for some y 2 X Ã with i\j then again w s ] x and w\x. It follows that this is a noetherian rewriting system. It is routine to check that it is also locally confluent and thus confluent. h Consequently, if w 2 CX þ , then applying the rewriting rules to w yields a unique reduced word x 2 ½w s , and x is independent of the choice of w. We know that x w s and we deduce from w s x s ¼ x that x ¼ w s . We say that w s is the word in normal form in ½w s . An entirely similar argument can be made for a rewriting system consisting of rewriting rules bv k ! au k and v i u j ! u i v j (for all i; j; k 2 N such that i j) on elements of ½w t for any w 2 CX þ .
w ¼ a p 0 b q 0 u p 1 Proposition 5.20 Let S be a commutative (commutative and cancellative) semigroup such that S contains two principal right ideals aS 1 and bS 1 such that aS 1 \ bS 1 has exactly n generators ac 1 ¼ bd 1 ; . . .; ac n ¼ bd n . Then there is a homomorphism h : S s ! S (h : S t ! S) such that ½ah ¼ a; ½bh ¼ b; ½u i h ¼ c i and ½v i h ¼ d i for all i 2 n.
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