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Серия «Математика»
2021. Т. 35. С. 87—102
Онлайн-доступ к журналу: http://mathizv.isu.ru
УДК 512.53 MSC 08A30
DOI https://doi.org/10.26516/1997-7670.2021.35.87
S-acts over a Well-ordered Monoid with Modular Congruence Lattice *
A. A. Stepanova
Far Eastern Federal University, Vladivostok, Russian Federation
Abstract. This work relates to the structural act theory. The structural theory includes the description of acts over certain classes of monoids or having certain properties, for example, satisfying some requirement for the congruence lattice. The congruences of universal algebra is the same as the kernels of homomorphisms from this algebra into other algebras. Knowledge of all congruences implies the knowledge of all the homo-morphic images of the algebra. A left 5-act over monoid S is a set A upon which S acts unitarily on the left. In this paper, we consider 5-acts over linearly ordered and over well-ordered monoids, where a linearly ordered monoid S is a linearly ordered set with a minimal element and with a binary operation max, with respect to which S is obviously a commutative monoid; a well-ordered monoid S is a well-ordered set with a binary operation max, with respect to which S is also a commutative monoid. The paper is a continuation of the work of the author in co-authorship with M.S. Kazak, which describes S-acts over linearly ordered monoids with a linearly ordered congruence lattice and S-acts over a well-ordered monoid with distributive congruence lattice. In this article, we give the description of S-acts over a well-ordered monoid such that the corresponding congruence lattice is modular.
Keywords: act over monoid, congruence lattice of algebra, modular lattice.
A significant number of works are devoted to the study of S'-acts with given conditions on their congruence lattices. In particular, in [1], unars,
1. Introduction
* Research supported by Ministry of Science and Higher Education of the Russian Federation, additional agreement from 21.04.2020 N 075-02-2020-1482-1.
that are the unary algebras with one unary operation, with linearly ordered, distributive or modular congruence lattices are described. A description of commutative unary algebras whose congruence lattices are a chain is obtained in [4]. The congruence lattices of disconnected S-acts over monoids are studied in [8]. A description of acts over certain classes of semigroups (such that semigroups of right and left zeros, rectangular bands, linearly ordered monoids) that have modular, distributive or linearly ordered congruence lattice is obtained in [2; 7; 10]. For commutative S-acts the conditions of modularity and distributivity of the congruence lattice are investigated in [3]. In this paper, we describe 5-acts over a well-ordered monoid with modular congruence lattices.
2. Preliminaries
Let us recall some definitions and facts from act theory and universal algebra (see [5;6;9]). Throughout this paper S will denote a monoid with identity 1. An algebraic system {A; s)ses of the language Ls = (s | s e S} consisting of unary operation symbols is a (left) S-act if s1(s2a) = (s1s2)a and 1a = a for all s1,s2 e S and a e A. An S-act {A; s)ses is denoted by sA. Let sA be an 5-act and sВ be a subact of sA. An equivalence relation в on sA is called a congruence on sA, if (a, b) e д implied (sa, sb) e д for a,b e A, s e S. Any subact sВ С sA defines the Rees congruence p(B) on sA, by setting (a, b) e p(B) if a,b e В or a = b.
Elements x, у of an S-act sA are called connected (denoted by x ~ y) if there exist n e ш, a0,... ,an e A, s\,... ,sn e S such that x = a0, у = an, and ai = Siai-\ or a^i = Siai. An S-act sA is called connected if we have x ~ у for any x,y e sA. It is easy to check that ~ is a congruence relation on the 5-act sA. The classes of this relation are called connected components of the 5-act sA. A coproduct of 5-acts sAi is a disjunctive union of this 5-acts. The coproduct of 5-acts sAi is denoted by Ц sAi. It
iei
is known (see [5]) that that every S-act sA can be uniquely represented as a coproduct of connected components.
For a congruence в on sA and a subact sВ С sA we define the restriction в \ В of в for sВ by в \ В = в П (В x В). Instead (a, b) e в we will write sometimes adb. The class of a e A with respect to congruence 9 is a set d(a) = (b e A | adb}. Note that the set of all congruences on sA forms a lattice according to the relation C, which is called the lattice of congruences on the S-act sA, and denoted by Con(sA).
Theorem 1. [5] Let sA be an S-act, a,b e A, 9i,02 e Con(sA). Then а (в1 Ув2) b if and only if there are the elements x0,x1,..., x2n in A such that a = xo, X2n = b, X2k #1 X2k+i and X2k+i O2 X2k+2 for all к e (0,1,...,n -1}.
A lattice (L, A, V) is called modular if (a V b) A c = a V (b A c) for all a,b,c £ L with a ^ c.
Theorem 2. [9] A lattice L is modular iff the conditions a ^ b, and a V c = b V c, a A c = b A c for some c £ L imply a = b for all a,b £ L.
A congruence d on an 5-act sA is called perforating if there are 5-acts sB, sC and elements b1,b2 £ B, c1,c2 £ C such that
sa =s b c, (61,62) £ 0, (ci, C2) £ e, (bi,ci) £ e, (b2, C2) £ e.
Theorem 3. [8] A lattice Con(sA) is modular if and only if the following conditions are true:
(1) the S-act sA contains no more than three connected components;
(2) the latices of congruences on all connected components of an S-act sA are modular;
(3) there are no perforating congruences on an S-act sA.
A lattice (L, A, V) is called distributive if (a V b) A c = (a A c) V (b A c) for all a,b,c £ L. It is clear that a distributive lattice is modular.
Let < be a total ordering on S and 1 be a minimal element in S. Then (S; ■) is a commutative monoid relative to the operation a ■ b = maxja, b} for a,b £ S, at that 1 is identity of monoid S. This monoid is called a linearly ordered monoid. If (S; <) is well-ordered set then linearly ordered monoid (S; ■) is called a well-ordered monoid.
Proposition 1. [10] Let S be a well-ordered monoid. Then the lattice on any cyclic S-act is distributive.
3. S'-acts over a well-ordered monoid with modular congruence
lattices
Lemma 1. Let S be a linearly ordered monoid, s A be an S-act, a,b £ A, Sa = Sb. Then a = b.
Proof. Let the conditions of the Lemma hold. Since Sa = Sb, we have a = sb and b = ta for some s,t £ S. Suppose that s < t. Then a = sb = sta = ta = b. □
Lemma 2. Let S be a linearly ordered monoid, s A be an S-act, a,b,c £ A, Se ç Sb ç Sa, d be a congruence on an S-act s A and adc. Then adbdc.
Proof. Since Se ç Sb ç Sa, we have b = ta and c = sb = sta for some s,t £ S. Since adc, we have b = tadtc = tsta = tsa = c. Hence bdc. Therefore, aOcOb, that is adb. □
Lemma 3. Let S be a well-ordered monoid, sA be a connected S-act. Then the following conditions are true:
(1) for any a, b £ A there is a minimal element m £ S such that Sa П Sb = Sma and ma = mb;
(2) there are ai £ A (i £ I) such that A = |J Sai and ai £ Saj for any
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different elements i,j £ I.
Proof. Let us prove (1). The existence of minimal element m £ S such that ma = mb follows from the connectivity of sA and the well-ordering of the monoid S. Let us check for equality Sa П Sb = Sma. Obviously Sma С Sa П Sb. Let с £ Sa П Sb. Then с = s0a = s\b for some so, s1 £ S. This means that с = sa = sb, where s = max{s0, s1}. In particular, s > m. Therefore, с = sa = sma £ Sma.
To prove (2), let A = {aa | a £ к}, where к is some ordinal. Then we assume I = {ß £ к | there are no 7 £ к such that Saß С Sa~(}. □
Theorem 4. Let S be a well-ordered monoid. The lattice Con(sA) is modular if and only if the following conditions are true:
(1) an S-act sA contains no more than three connected components;
(2) if a1,a2 £ A, s £ S, s = 1 and Sa1 П Sa2 = Ssa1 П Ssa2 then sa1 = sa2, or sa1 = ra1, or sa2 = ra2 for some r £ S, r < s;
(3) if a1, a2, a3 £ A and s £ S such that s = 1, ai £ Saj, sai = saj and Sai П Saj = Ssa1 for any different i,j £ {1,2,3}, then sai = rai for some r £ S, r < s, and i £ {1,2,3}.
Proof. Necessity. Let Con(sA) be a modular lattice. By Theorem 3 we have (1).
Let us prove (2). Suppose for contradiction that there are elements a1,a2 £ A and s £ S such that s = 1, Sa1 П Sa2 = Ssa1 П Ssa2, sa1 = sa2 and sai = rai for all r < s and for all i £ {1,2}. Let us define the equivalence relations 91, в2, ц on the set A as follows:
(u, v) £ 61 & u,v £ Ssa1USsa2, or u,v £ Sa1\Ssa1, or u,v £ Sa2\Ssa2, or и = v;
(u, v) £ 62 & (u, v) £ 61 or u,v £ (Sa1 \ Ssa1) U (Sa2 \ Ssa2); (u,v) £ ц & u,v £ {ta1 I t < s}, or u,v £ {ta2 | t < s}, or и = v. We show that the relation 01 is a congruence on sA. Let u,v £ Sa1 U Sa2, t £ S. If t > s then tu = tsu £ Ssa1 U Ssa2 and tv = tsv £ Ssa1 U Ssa2, that is (tu,tv) £ 61. So, we assume that t < s. If u,v £ Ssa1 U Ssa2 then tu,tv £ Ssa1 U Ssa2 and (tu, tv) £ 91. Let u,v £ Sa1 \ Ssa1 and tu £ Ssa1. Then tu = stu = su and и = la1 for some I £ S. As и £ Ssa1 then I < s. Hence tla1 = tu = su = sla1 = sa1. Because of the inequality tl < s we get the contradiction. So tu £ Ssa1. Similarly, it is shown that tv £ Ssa1. Thus, tu,tv £ (Sa1 \ Ssa1), i.e. (tu,tv) £ 91. Therefore, the relation 91 is a congruence on the 5-act sA.
Similarly, it is proved that the relation 02 is a congruence on the 5-act s A.
We show that the relation *q is a congruence on sA. Let (u,v) £ ~q, r £ S. We can assume that u = t1a1 and v = t2a1 for some t1 < s, t2 < s. Since
rt1 < s & r < s & rt2 < s
then (ru, rv) £ r/. So *q is a congruence on the S-act sA.
It is clear that d1 C 02. Now we show that d1 c d2. If a1 £ Ssa1, then Sa1 = Ssa1 and by Lemma 1 we have 1 ■ a1 = a1 = sa1, this contradicts the assumption. Therefore, a1 £ Ssa1. If a1 £ Sa2 then a1 £ Sa1 n Sa2 = Ssa1 n Ssa2 C Ssa1, contradiction. So, a1 £ Ssa1 U Ssa2 and a1 £ Sa1 \ Ssa1. Similarly, a2 £ Ssa1 U Ssa2 and a2 £ Sa2 \ Ssa2. Hence it is proved that (a1,a2) £ d2. Note that (Sa1 \ Ssa1) n (Sa2 \ Ssa2) = 0. Indeed,
(Sa1 \ Ssa1) n (Sa2 \ Ssa2) = (Sa1 n Sa2) \ (Ssa1 U Ssa2) =
= (Ssa1 n Ssa2) \ (Ssa1 U Ssa2) = 0.
So, (a1,a2) £ d1. Thus, (a1 ,a2) £ d2 \ d1.
To prove the equality d1 A ^ = 02 A ^ it is enough to check the inclusion 02 A ^ C d1. Indeed, let (u,v) £ 62 A r/. Since (u,v) £ ^ then we can assume, for example, that u = t1 a1, v = t2a1 for some t1,t2 < s. More over, since (u,v) £ 62 we can assume u,v £ (Sa1 \ Ssa1) U (Sa2 \ Ssa2) (otherwise (u, v) £ d1, as claimed). If u £ Sa2 \ Ssa2 then u £ Sa1 n Sa2 = Ssa1 n Ssa2 C Ssa2, contradiction. So u £ Sa1 \ Ssa1 and, similarly, v £ Sa1 \ Ssa1. Hence, (u,v) £ 01.
Let us prove d1 V ^ = 02 V ^ = p(Sa1 U Sa2), where p(Sa1 U Sa2) is the Rees congruence. Clearly, d1 C Q2 C p(Sa1 U Sa2) and ^ C p(Sa1 U Sa2), i.e. d1 V ^ C Q2 v ^ C p(Sa1 U Sa2). Note that a1 r]sa1d1sa2r]a2. Then (a1 ,a2) £ d1 V r/. Let t £ S. Since ta1,qa1 (if t < s) or ta1d1sa1 (if t > s) then (ta1,a1) £ d1 V r/. Analogously, (ta2,a2) £ d1 V r/. So for all u,v £ Sa1 U Sa2 we have (u, v) £ d1 V i.e. p(Sa1 U Sa2) C d1 V ^ C 62 V r/.
Thus, d1 c d2, Q1 A -q = d2 A -q and d1 V -q = d2 V -q. By Theorem 2 it contradicts the modularity of the lattice Con(sA).
To prove the condition (3) we suppose to the contrary that there are a1,a2,a3 £ A and s £ S such that s = 1, Sai n Saj = Ssa1, ai £ Saj, sai = saj for all different i, j £ {1,2,3} and sai = rai for all r < s and for all i £ {1, 2, 3}. Let us define the equivalence relations d1, 62, *q on the set A as follows:
(u,v) £ d1 & u,v £ (Sa1 U Sa2) \ Ssa1, or u,v £ Ssa1, or u,v £ Sa3 \ Ssa1, or u = v;
(u, v) £ 62 & (u, v) £ d1 or u,v £ Sa3;
(u,v) £ 'q & u,v £ (Sa2 U Sa3) \ Ssa1, or u,v £ Sa1, or u = v. We show that d1 is a congruence on sA. Let (u,v) £ d1, r £ S. The proof
is divided into three cases.
Case 1: u,v £ Ssal. Then it is obviously that ru, rv £ Ssal and (ru, rv) £ dl for any r £ S.
Case 2: u,v £ (Sal U Sa2) \ Ssal. If r > s then ru = rsu £ Ssal and rv = rsv £ Ssa\, that is ru,rv £ Ssal and (ru,rv) £ 0l. Suppose that r < s and, for example, и £ Sal \ Ssal. Then и = ta\ for some t < s, i.e. ru = rta\, where rt < s. By assumption, sal = rta\. By Lemma
1 Ssa\ С Srta\ = Sru, that is ru £ Ssa\. So ru £ (Sal U Sa2) \ Ssal. Similarly, rv £ (Sal U Sa2) \ Ssal, that is (ru, rv) £ dl.
Case 3: u,v £ Sa3 \ Ssal. This case is considered similarly to case 2. Thus, it is proved that 9l is the congruence on $ A. Similarly checked that в2, ц are the congruences on sA.
It is clear that 9l С в2. Let us show that 9l С в2. Since sal = sa3 then sa\,a3 £ Sa3, i.e. (sa\,a3) £ в2. By assumption, a3 £ Sa\, in particular, a3 £ Ssa\. So, (sa\, a3) £ Thus, (sa\,a3) £ 62 \ Q\.
To prove the equality в\ Л ц = в2 Л ц it is enough to check the inclusion 62 Л ц С в\. Indeed, let (u,v) £ 62 Л г/. Since (u,v) £ 62 then we can assume that u,v £ Sa3 (otherwise (u, v) £ , as claimed). Since (u, v) £ ц than we have either u,v £ Sa2 \ Ssa\ (then (u, v) £ в\), or u,v £ Sa3 \ Ssa\ (then (u, v) £ в\ too), or u,v £ Sa\. In the last case we have u,v £ Sa\ П Sa3 = Ssa\, i.e. (u, v) £ .
Let us prove в\Уц = в2Уц = p(Sa\USa2USa3), where p(Sa\USa2USa3) is the Rees congruence. Clearly, 0\ С 62 С p(Sa\ U Sa2 U Sa3) and ц С p(Sai U Sa2 U Sa3), i.e. У -q С 62 У -q С p(Sai U Sa2 U Sa3). Note that saldlaldla2'qa3 for all s £ S. Then saldlta2,qta3 for all s,t £ S, i.e. p(Sal U Sa2 U Sa3) С Ql у -q С в2 У -q.
Thus, Ql С 02, Ql Л -q = 62 Л -q and Ql У -q = 62 У -q. By Theorem 2 it contradicts the modularity of the lattice Con(sA). Sufficiency. Let us prove few statements first.
Lemma 4. Let the condition (2) of the theorem is true, sA be a connected S-act and a, b £ A. Then the following statements are true:
(1) if 9 £ Con(sA), adb and m £ S is the minimum element with the conditions Sa П Sb = Sma and ma = mb then sadb and adsb for all s < m;
(2) if в £ Con(sA), adb and Sa П Sb С Sc С Sa then авсвЬ;
(3) if Ql,...,Qn £ Con(s A), 0 = 0l о ... о 0n, adb and m £ S is the minimum element with the conditions Sa П Sb = Sma and ma = mb then sadb and adsb for all s < m.
Proof. Since sA is a connected 5-act then Sa П Sb = 0.
Let us prove (1). If m = 1 then a = b and (1) is satisfied. Let m > 1. If (mb, b) £ в and s < m then Smb С Ssb С Sb implies bdmb, i.e. by Lemma
2 we have adbdsb and sadsbdb. Suppose that (mb, b) £ в. Then (mb, a) £ в. We assume kl = minjr £ S | (rb,a) £ в}, k2 = minjr £ S | (rb,b) £ в}.
If k2 < ki then by the definition of ki we have b0a0k2b which contradicts the definition of k2. If ki < k2 then by the definition of k2 we have adbdkib which contradicts the definition of ki. So ki = k2 = k and 1 < k < m. Let x £ Sa n Sb. Then x = sa = sb for some s £ S .It means that sb £ Sa and by the definition of m we have k < m < s. Hence, kx = ksa = sa = x and kx = ksb = sb = x, i.e. x £ Ska n Skb. Thus, Sa n Sb = Ska n Skb. By the condition (2) of the theorem we have la = ka for some I < k, or lb = kb for some I < k, or ka = kb. In the first case, by the definition of k2 = k we have Ibdb; the condition adb implies kbdka = ladlbdb, that contradicts the definition of k2 = k. In the second case, (lb,b) = (kb,b) £ d, that contradicts the definition of k2 = k. And in the third case, since k < m, from the definition of element m we have k = m; from the definition of element k we have sadsbdbda for all s < m = k.
Let us prove (2). By Lemma 3 there exists a minimal element m £ S such that Sa n Sb = Sma and ma = mb. Since Sma C Sc C Sa then c = sa for some s < m. So by (1) we have c = sadb.
Let us prove (3). Suppose s < m. By induction on n we will prove sadb (adsb is proved similarly). If n = 1 then (3) is done by (1). Suppose n> 1, and for n — 1 (3) is done, and = 62 o... o dn. Then a9icqb for some c £ A. Since sA is a connected 5-act then Sa n Sc = 0 and Sc n Sb = 0. By Lemma 3 there exists minimal element r £ S such that Sa n Sc = Sra and ra = rc. If s < r then by (1) we have sad\c, i.e. sad\cqb and sadb. If s > r then m > r. By Lemma 3 there exists a minimal element k £ S such that Sc n Sb = Skb and kb = kc. If s < k then from the induction hypothesis we obtain scqb, i.e. sad\scr]b and sadb. Suppose that s > k. Then m > k. If r > k then the equation kb = kc implies rb = rc = ra. So, by the choice of the element m, we have r > m. Contradiction. If k > r then the equation ra = rc implies ka = kc = kb. So, by the choice of the element m we have k > m. Contradiction. □
Lemma 5. Let the conditions (2) and (3) of the theorem are true, sA be a connected S-act and c,ci,c2,c3 £ A are such that Sci n Scj = Sc for all i,j, 1 < i < j < 3. Then the following statements are true:
(1) if 6,ri £ Con(sA) and c\dc2,qc3 then c\dc or cqc3;
(2) if di,...,dn,^i,...,^m £ Con(s A), d = di o... o 0n, ^ = o... o and cidc2r]c3 then cidc or cqc3.
Proof. Let di,..., dn, <qi,...,<qm £ Con(s A), d = di o... odn, ^ = ^i o.. .o and cidc2,qc3. If Ci £ Scj for some different i,j £ {1,2,3} then Sci n Scj = Sci = Sc and by Lemma 1 we have c = Ci, that is cidc or cqc3. Suppose that Ci £ Scj for all different i, j £ {1,2,3}. By Lemma 3 for all i, j, i < j, there exists a minimal element Sij £ S such that Sci n Scj = Ssija and Sij Ci = Sij Cj. Since Sci nScj = Sc then by Lemma 1 we have c = Sij Ci where i < j. If si2 < s23 then by Lemma 4(3) we have c = si2c2r]c3>, i.e. cqc3.
Similarly, if s23 < si2 then c9ci. So, we can assume that si2 = s23 = s. Therefore, sci = sc2 = sc3 = с and Sci П Scj = Ssci for all different i,j e {1,2,3}.
Let us prove (1). By condition (3) of the Theorem there exists r < s such that sci = rci = с for some i e {1, 2, 3}. By Lemma 4 (1) we have rci9c2, ci0rc2, rc2r]c3 and c2*qrc3. If rci = с then с1вгс2вгс1 = с. Similarly, if rc3 = с then cqc3. If rc2 = с then ci0rc2 = с and с = гс2^с3.
By induction on m + n let us prove (2). If m = n = 1 then the required statement follows from (1). Let m + n > 2. Without loss of generality we can assume that n > 1. Let £ = 9i о ... о 0n-i. Then ci^x9nc2 for some x e A. Let к = min{s' e S | s'c2 = s'x}. Then by Lemma 3 Skc2 = Sc2 П Sx. In view of well-ordering of the monoid S there is one of three cases.
Case 1: s = к. In this case Sc2 П Sx = Sc. Let t = min{s' e S | s'c3 = s'x}. Then by Lemma 3 Stc3 = Sc3 П Sx. Again in view of well-ordering of the monoid S we have s ^ t or s > t. In the first case, Sc = Ssc3 5 Stc3 = Sc3 П Sx 5 Sc, so Sc = Stc3, that is by Lemma 1 we have с = tc3; thus, Sc2 П Sx = Sc3 П Sx = Sc2 П Sc3 = Sc; from the induction hypothesis applied to relation хвпс2г]с3, we obtain с1(хвпс or cqc3, i.e. с1вс or cqc3. Let s > t. Since хвпс2, Sx П Sc2 = Ssc2, sc2 = sx and t < s, then by Lemma 4 (1) we have txdnc2. Let I = min{s' e S | s'ci = s'x}. Then by Lemma 3 Slci = Sci П Sx and lci = lx. Since sci = sc2 = sx, it follows that Sx П Sci 5 Ssci and I < s. If t > I then tci = tlci = tlx = tlc3 = tc3, i.e. tci = tc3 and t > s, contradiction. Let t < I. By Lemma 4 (3) we have ci(txdnc2. Note that Sc = Sci П Sc2 5 Six П Stx = Six 5 Ssx = Ssc2 = Sc. Hence Six = Sc. By Lemma 1 we have lx = c. So Sci nStx = Six = Sc. Therefore, Sci П Stx = Sci П Sc2 = Sc for all i e {1, 2}. By the induction hypothesis, we obtain ci(c or свпс2. If свпс2 then с1(хвпс2впс, i.e. с1(хвпс and ci9c. Thus, с1вс or cqc3.
Case 2: s > k. In this case Sc = Ssc2 С Skc2. Then Sci П Skc2 = Sci П Sc2 П Skc2 = Sc П Skc2 = Sc = Ssci for all i e {1,3}. It is clear that sci = sc3 = skc2. Since s > k then by Lemma 4 (3) we have ci(kc2 and kc2r)c3, i.e. ci(kc2,qcs. By the induction hypothesis, we have cior сг/с3, i.e. с1вс or cqc3.
Case 3: s < k. Since Sx П Sc2 = Skc2, kc2 = kx and s < k, then by Lemma 4 (1) we have xOnsc2 = c. Hence с1(хвпс, i.e. ci9c. □
Let S be a well-ordered monoid and sA satisfies the conditions (1)-(3) of Theorem. If sA is a cyclic 5-act then by Proposition 1 Con(sA) is a distributive lattice, therefore, Con(sA) is a modular lattice.
Let sA = Uiei sSai be a connected S-act and II| > 1. By Lemma 3 (2) we can assume that ai £ Saj (i = j).
For в e Con(sA), we denote (9 \ Sai) U e Con(sA) by 9\ Note that (9i Л 92) = 91 Л 92 for all 9i,92 e Con(sA) and i e I.
Suppose that 91, d2, -q £ Con(s A) such that d1 C 02 ,Q1 A rf = d2 A -q, 91 V q = d2 V "q. By Theorem 2 it is enough to prove d1 = d2.
Let i £ I. Then 9\ C P2, 9\ A rf = 92 A rf.
We divide the proof into several steps.
Step I. We show that rf V 9\ = rf V 92.
Clearly, rf V 9\ C rf V 9l2. Let tai(9l2 V rf)rai and tai = rai. Then tai(92 Vrf)rai. Hence tai(91 Vrf)rai. By Lemma 1 we have Srai = Stai. Let, for example, Srai c Stai. By Theorem 1 there exist n £ u, k0,..., k2n £ I (k0 = k2n = i) and dj £ Sakj (0 < j < 2n) such that tai = do, rai = d2n, d2j91 d2j+1rqd2j+2 (0 < j < n). Since sA is a connected 5-act then Sai n Sdk = 0 (0 < k < 2n). Let Sbk = Sai n Sdk (0 < k < 2n).
By induction on n we prove (*): if d0(9l2 V rf)d2n, d2j91d2j+1rfd2j+2 for all j, 0 < j < n, and Sd2n C Sd0, then d0(9\ V •q%)d2n.
Let n = 1. Then Sd0 C Sb1, or Sb1 c Sd2, or Sd2 C Sb1 c Sd0. In the first case, we have Sd0 C Sd1; since d1,qd2 and Sd2 C Sd0 C Sd1, by Lemma 2 we have d0rfd2, i.e. d0(9\ V rf)d2. In the second case, since d091 d1 and Sd0 n Sd1 = Sb1 c Sd2 C Sd0, by Lemma 4 (2) we have d091 d2, i.e. d0(9\ V rf)d2. Let Sd2 C Sb1 c Sd0, i.e. the first and second cases are wrong. Since d1,qd2 then by Lemma 2 we have b1rqld2 and b1rfd1. Since d091d1 then by Lemma 4 (2) we have c91d0 and c91d1 for all c £ A such that Sb1 c Sc C Sd0. Since d0(9l2 V rf)d2 then by Lemma 2 we have d0(92 Viq%)b1. Theorem 1 and Lemma 2 imply the existence of element c £ A such that Sb1 c Sc C Sd0 and c^b1 or c92b1. If c^b1 then d091cqb1,qd2, i.e. d0(9\ Vrf)d2. If c92b1 then in view of 91 C 92 we have b192c92d1; since b1,qd1 then b1(92 A rj)d1; so the equality 91 A rf = 92 A rf implies b1(91 A ,q)d1, in particular, b191d1; hence d091d191b1'qd2, d091 b1,qd2 and d0(9\ Vr]%)d2. Thus, for n = 1 (*) is proved.
Let n> 1. Consider two cases.
Case 1: Sd0 C Sbj for some j > 2. Since Sd2n = Sb2n c Sd0 then we can choose j such that Sbj+1 c Sd0 C Sbj and j > 2. Since dj(dj+1, where ( £ {91,rq}, and Sdj n Sdj+1 = Sd0 n Sdj+1 = Sbj+1 c Sd0 C Sdj, then by Lemma 4 (2) we have d0(dj+1. Let j be even number. Then ( = 91 and d091 dj+1 r/dj+2. Besides that, d2k01d2k+1rfd2k+2 for all k, j + 2 < 2k < 2n. From the induction hypothesis we obtain d0(9\ V rf)d2n, as claimed. Let j be odd number. Then j > 3, ( = rf and d091d0,qdj+1. Besides that, d2k91d2k+1 rfd2k+2 for all k, j + 1 < 2k < 2n. From the induction hypothesis we obtain d0(9\ V rf)d2n, as claimed.
Case 2: Sbj+1 c Sd2n for some j < 2n — 3. Since Sd2n c Sd0 = Sb0 then we can choose j such that Sbj+1 c Sd2n C Sbj and j < 2n — 3. Since dj(dj+1, where ( £ {91,rq}, and Sdj n Sdj+1 = Sd0 n Sdj+1 = Sbj+1 c Sd2n C Sdj, then by Lemma 2 we have d2n(dj. From the reasoning given in case 1, it follows d0(9\ V rjl)d2n, as claimed.
Hence we can assume that
Sbj c Sd0 for all j > 2 and Sd2n C Sbj+1 for all j < 2n — 3.
Let n> 2. Again consider two cases.
Case 1: Sd2n C Sbj+1 c Sbj c Sd0 for some j, 2 < j < 2n — 3. Since dj(dj+1, where ( £ {01,iq}, and SdjnSdj+1 = Sd0nSdj+1 = Sbj+1 c Sbj C Sdj, then by Lemma 4 we have dj+1(bj. By Theorem 1 d0(d%2 V rf)d2n implies d0(d%2 V rf)bj, bj(92 V rf%)d2n. Let j be even number. Then j < 2n — 4, ( = O1 and djQ1bjQ1dj+1. Since j < 2n — 4, dj61bj, do(92 V rf)bj and d2k01d2k+1'qd2k+2 for all k, 0 < 2k < 2n, then from the induction hypothesis we obtain d0(d\ V rf)bj. Since j > 2, bjd1dj+1, bj(92 V rf)d2n and d2k01d2k+1rfd2k+2 for all k, 0 < 2k < 2n, then from the induction hypothesis we obtain bj(d\ V rf)d2n. Hence d0(d\ V rf)d2n. Let j be odd number. Then j > 3, ( = rf and djrfbjrfdj+1. Since j < 2n — 3, djrfbj, d0(92 V rf)bj and d2kd1d2k+1rfd2k+2 for all k, 0 < 2k < 2n, then from the induction hypothesis we obtain d0(d\ V rf)bj. Since j > 3, bjrfdj+1, bj(92 V rfl)d2n and d2kd1d2k+1rfd2k+2 for all k, 0 < 2k < 2n, then from the induction hypothesis we obtain bj(d\ V rfl)d2n. Hence d0(d\ V rf)d2n.
Case 2: Sd2n C Sbj c Sbj+1 c Sd0 for some j, 2 < j < 2n — 3. From the reasoning given above when considering case 1, it follows d0(d\ Vrf)d2n.
Thus, we can assume that for n> 2
Sb2n-1 c Sd0, Sd2n C Sb1 and Sd2n C Sb2 = Sb3 = ... = Sb2n-2 c Sd0.
Clearly, it is true for n = 2.
By Lemma 1 bi = bj = b for all i,j, 2 < i,j < 2n — 2, i.e. Sd2n C Sb c Sdo.
We show that b(d\ V rfl)d2n. If Sb C Sb2n-1 then in view of d2n-1rfd2n by Lemma 2 we obtain brid2n. If Sb2n-1 c Sb then in view of d2n-2d1d2n-1 and Sd2n-2 n Sd2n-1 = Sd0 n Sd2n-1 = Sb2n-1 c Sb C Sd2n-1, by Lemma 4 we obtain bd1d2n-1, i.e. b(d1 o rf)d2n. Since b(d22 V rf)d2n then by induction basis we have b(d\ V rf)d2n.
Hence to prove d0(d\ V rf)d2n, it is enough to prove d0(d\ V rf)b.
If Sd0 n Sd1 = Sb1 c Sb C Sd0 then in view of do01d1 by Lemma 4 (1) we have d0d1b, i.e. d0(d\ V rf)b. So we will assume that Sb C Sb1. Since d0(92 V rf)d2n, it is clear that d0(dl2 V rf%)b. The proof is divided into three cases.
Case 1: SdjnSdj+1 = Sb for some j £ {2,..., 2n—3}. Let dj-1(dj('dj+1, where (, (' are different elements {01,rq}. Since Sd0 n Sdj = Sd0 n Sdj+1 = Sdj n Sdj+1 = Sb then by Lemma 5 (2) we obtain d0d1 o ... o (b or b('dj+1, i.e. d0d1 o ... o dj('dj+1 ('b. Since j < 2n — 3, then in both cases from the induction hypothesis we obtain d0(d\ V rf)b. So, in this case * is proved.
Case 2: Sdj n Sdj+1 = Sb for all j £{2,..., 2n — 3} and Sdi n Sdk = Sb for some different elements i,k £ {2,..., 2n — 2}. Hence Sdi n Sd2n-2 = Sb
for some i £ {2,..., 2n — 3}. Let j be a minimum element such that 2 < j < 2n — 3 and Sdj n Sd2n-2 = Sb. Then Sdj+i n Sd2n-2 D Sb and Sdj n Sdj+i = Sb, that contradicts our assumption.
Case 3: Sb C Sdi n Sdk for all different elements j,k £ {2,..., 2n — 2}. By Lemma 3 (1) we have P| Sdi = Sd for some d £ A. So, Sb C Sd.
2<i<2n-2
We have three sub-cases: Sb = Sbi and Sb = Sdi n Sd, or Sb = Sbi and Sb C Sdi n Sd = Sc, or Sb C Sbi. In the thirst case we have Sdi n Sdo = Sdi n Sd = Sdo nSd, since dodidirid2 and Sdi n Sd2 = Sb C Sd C Sd2 then by Lemma 4 (2) we have d^d, and then by Lemma 5 (1) we have bdid0, i.e. d0(d\ V rf)b, or b~qd; if b~qd then by induction basis d0didiridrib implies d0(d\ V rf)b. In the second case, since d09\d\ and Sd0 n Sdi = Sb C Sc C Sdi, then by Lemma 4 (2) we have d0dic; if Sc C Sd2n-i then by Lemma 2 we have d2n-irqd2n implies cr/b, i.e. d0dicqb; by induction basis we have d0(6\ V rf)b. If Sc n Sd2n-i C Sc then by Lemma 4 (2) d2n-2did2n-i and Sd2n-2 n Sd2n-i C Sd2n-2 n SC n Sd2n-i = SC n Sd2n-i C SC C Sd2n-2 imply cdid2n-i; so d0dicdid2n-ir]d2n, hence by induction basis we have d0(6\ V •q%)d2n and by Lemma 2 we have d0(6\ V rf)b. Let Sb C Sbi. Note that by Lemma 4 (2) dirid2 implies wqdi and v^d2 for all u,v such that Sb C Su C Sbi C Sdi and Sb C Sv C Sd C Sd2, in particular, bi'qdi and d~qd2. Let Sb C Sb2n-i. Clearly, Sbi n Sb2n-i = Su where u = bi or u = b2n-i. By Lemma 2 d2n-i,qd2n and Sd2n C Sb C Su C Sd2n-i imply wqb. As noted above wqdi. Hence doQidi'qwqb, so by induction basis we have d0(d\ V r}%)b. Let Sb = Sb2n-i C Sd n Sd2n-i. By Lemma 3 (1) we have Sd n Sd2n-i = Sv for some v £ A. By Lemma 2 d2n-i^d2n implies vqb. As noted above vqd, i.e. d0didi,qd'qv'qb. By induction basis we have d0(d\ V rf )b. Let Sb2n-i C Sb or Sb2n-i = Sb = Sd n Sd2n-i. In the first case, by Lemma 4 (2) d2n-2did2n-i implies ddib. Let us prove d0(d\ V rf)b or ddib in the second case. By Lemma 4 (2) d2n-2did2n-i and Sb C Sd C Sd2n-2 imply ddid2n-i. By Lemma 5 (1) dirid9id2n-l and Sdi n Sd = Sdi n Sd2n-i = Sd n Sd2n-i = Sb imply either di'qb, i.e. dftQidi'qb and by induction basis d0(d\ V r]%)b, or bdid2n-i, i.e. bdid. It remains to prove that bdid implies d0(d\ V rf)b. Let bdid. Since d0(d%2 V rf)b then there exists u £ A such that Sb C Su C Sbi and wqb or ud2b. As noted above we have wqdi, i.e. d0didiriu and by induction basis we have d0(d\ V rf )u. If wqb then d0(d\ V ^l)u^b, i.e. d0(d\ V rf )b. Let ud2b. Since bdid then bd2d. So, u,d2bd2d. Since wqd then u(d2 Arj)d. Since d2 A= di A then udid. So, udxddxb and do(d\ V r1i)u£ib, i.e. do'(d\ V rf)b.
Therefore, we proof d0(d\ V rf)d2n. So, (*) is proved. Thus, d\ V rf = d%2 V rf and step I is finished.
Since by Proposition 1 the lattice Con(sSai) is distributive then it is modular and by Theorem 2 we have d\ = dl2.
Step II. We show that dx = d2.
We suppose to the contrary that d1 c d2. Then there are d0,d1 £ A and different k0,k1 £ I such that d0 £ Sau°, d1 £ Sau1 and (d0,d1) £ d2 \ d1. Then d0 £ Sd1 and d1 £ Sd0. By Lemma 3 (1) there exists a minimum element s £ S such that Ssd0 = Sd0 n Sd1 and sd0 = sd1 = b. Note that sd0 = rd0 and sd1 = rd1 for all r £ S, r < s. Indeed, suppose for example that sd0 = rd0 for some r £ S, r < s; by Lemma 4 (2) we have d0d2rd0d2d1; since 9X° = 02,° and 0X1 = 021 then do01rdo01d1, i.e. do01d1, contradiction.
Since do02d1 then do(02 V rj)d1. Since 01 V rf = 02 V rf then do(01 V rj)d1. By Theorem 1 d1rfd2d1 d3 ... d2n-1Vd2 n — do for some d2,... d2n-1 £ A and n > 1. Note that n> 1 (otherwise d0(02 A rj)d1, so do(01 A rj)d1 and do01d1, contradiction).
There is one of two cases.
Case 1: Sdk n Sd0 C Sb and Sdk n Sd1 C Sb for some k, 1 < k < 2n. Let i be a minimum number such that Sdi n Sd1 C Sb and 1 < i < 2n, j be a maximum number such that Sdj n Sd0 C Sb and 1 < j < 2n. Consider only the case i > 2, j < 2n — 2 as the most difficult. By Lemma 3 (1) we have Sr1d1 = Sdi-1 n Sd1 D Sb, Sr0d0 = Sdj+1 n Sd0 D Sb for some r0, r1 £ S. Note that r1 < s, r0 < s. By Lemma 4 (2) d102r1d102rodo02do. Clearly, Sdi n Sdi-1 C Sb.
Suppose that
Let Sdi n Sd0 = St0 do and di-l(di, where £ E {0\,-q}. Since Sdi-\ n Sdi Ç Sb C St0d0 C Sdi and Sdi-\ n Sdi Ç Sb C Sr\d\ ç Sdi-\, then in view of di-\(di by Lemma 4 (2) we have t0d0(r\d\. Since d002d\, Sd0nSd\ = Sb C St0d0 ç Sd0 and Sd0 n Sd\ = Sb C Sr\d\ ç Sdi, then by Lemma 4 (2) we have do02todo02ridi02dl. Since 01° = 02,° and 011 = O^1, then do0itodo and diQiridi. If ( = Oi then do0itodo0iridi0i, i.e. do0idi, contradiction. If £ = then todo(02 A ri)ridi; since Oi A = 02 A then t0do(0i A rj)ridi and do0itodo0iridi0idi, i.e. do0idi, contradiction.
Let (1) not be done. We show that
Let Sdi n Sdi C Sb. Since di(di-i, where Ç E {Oi,-q}, then in view of Sdi-inSdi C Sb ç Sdi-i and Sdi-inSdi C Sndi ç Sdi-i by Lemma 4 (2) we have b£di-i and ri^(di-i, i.e. b^ridi. Let SdinSdi = Sb. Because (1) is wrong then Sdi n Sd0 = Sb. Since di02do and Sd0 n Sdi = Sb C Sridi ç Sdi, by Lemma 4 (2) we have ridi02do. Since di(di-i, where £ E {Oi,^}, and Sdi n Sdi-i C Sridi ç Sdi-i, then by Lemma 4 (2) we have ridi^di. By Lemma 5 (1) we have do02b, or di(b. If do02b then b02do02ridi and in view oX1 = ol1 we have bOindi. If di^b then b^di^ridi.
Thus, (2) is proven. Just like (2), it is proved r0d0rib V rodo0ib.
If ridiOib or rodo0ib then ndi02b or rodo02b; hence di02b02do; in view Qk° = Qk° and oX1 = ol1 we have di9id0, contradiction. Let ri^rqbrqr0d0.
Sdi П Sdi = Sb and Sdi П Sd0 D Sb.
(1)
ridi'qb V ridiOib.
(2)
Then do02di, Sd0 n Sdi C Sr0d0 C Sd0 and Sd0 n Sdi C Sridi C Sdi by Lemma 4 (2) imply ridid2r0d0. Then ridi(d2 A ,q)r0d0, i.e. in view 62 A *q = di A *q we have didirldldlr0d0dld0, contradiction.
Case 2: Sdk n Sd0 D Sb or Sdk n Sdi D Sb for all k, 1 < k < 2n. Then we have Sdk n Sdi D Sb for all k, 1 < k < 2n, or Sdk n Sd0 D Sb for all k, 1 < k < 2n, or Sdkl n Sdo D Sb and Sdk2 n Sdi D Sb for some ki,k2, 1 < ki,k2 < 2n. In the first case by Lemma 3 (1) we have Sui = Sd2n-i n Sdi D Sb for some ui £ A. Since d2n-i'qd0 and d0d2di then by Lemma 4 (2) we have d0(^ A d2)ui and did2ui. Since 911 = d^1 then d\9\ui Since d\ A *q = 62 A *q then d0(d\ A rj)u\, in particular, d0d\u\. Therefore, d0diuididi, i.e. d0&idi, contradiction. The second case is similar to case a. Let Sdkl n Sd0 D Sb and Sdk2 n Sdi D Sb for some ki,k2, 1 < ki,k2 < 2n. Since Sdk n Sd0 D Sb or Sdk n Sdi D Sb for all k, 1 < k < 2n then Sdk n Sdi D Sb and Sdk+i n Sd0 D Sb for some k, 1 < k < 2n — 1, or Sdk n Sd0 D Sb and Sdk+i n Sdi D Sb for some k, 1 <k < 2n — 1. Let Sui = Sdk n Sdi D Sb and Su0 = Sdk+i n Sd0 D Sb for some k, 1 < k < 2n — 1 (an other sub-case is considered similarly). Then Sdk n Sdk+i C Sui and Sdk n Sdk+i C Su0. Since dk(dk+i, where £ £ {Oi,vi}, Sdk n Sdk+i C Su0 C Sdk+i and Sdk n Sdk+i C Sui C Sdk, then by Lemma 4 (2) we have u0(dk+i and ui(dk, therefore, u0(ui. Since d0d2di, Sd0 n Sdi = Sb C Su0 C Sd0 and Sd0 n Sdi = Sb C Sui C Sdi, then by Lemma 4 (2) we have u0d2d0, uid2di, hence u0d2ui, i.e. u0(d2 A rj)ui or u00iui. Since Bi A-q = d2 A-q then u00iui. Since = 62,° and B^1 = d^1 then u0didu and uiQidi. Therefore, d0diu0diuididi, i.e. d0&idi, contradiction.
Thus, the lattice Con(sA) is modular.
Step III. Let s^ not be connected 5-act. As we have proved, the lattices of congruences on connected components of sA are modular. Let us prove that the lattice Con(sA) is modular. By Theorem 3 it is enough to prove that there are no perforating congruences on sA. Suppose that sBUsC CsA, d is the perforating congruences on sA, bi, b2 £ B, ci,c2 £ C, (bi,ci) £ 6, (b2, c2) £ d, (bi,b2) £ 6, (ci, c2) £ 6. By point 1 of Theorem we can assume that sB is connected S-act.
We show that sbidbi for all i £ {1,2} and s £ S. Note that Sbi n Sa = Ssbi n Ssci = 0 for all s £ S. Suppose that there is I £ S such that (bi,lbi) £ 0. Let t = min{/ | (bi,lbi) £ 6}. Then t > 1. By point 2 of Theorem we have tbi = rbi or tci = rci for some r £ S, r < t. In the first case, we have tbi = rbiQh. Contradiction. In the second case, we have tbidtci = rcidrbi, contradiction.
Since sB is connected 5-act then nbi = r2b2 for some ri,r2 £ S. As proved above bidribi = r2b2db2, contradiction. Therefore, there are no perforating congruences on sA and by Theorem 3 the lattice Con(sA) is modular. □
4. Conclusion
In this paper, we obtain a description of 5-acts over a well-ordered monoid with modular congruence lattice. In [10], S-acts over linearly ordered monoids with linearly ordered congruence lattices and 5-acts over a well-ordered monoid with distributive congruence lattices are characterized. The questions of describing 5-acts over linearly ordered monoids with distributive congruence lattices and with modular congruence lattice remain open.
References
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Alena Stepanova, Doctor of Sciences (Physics and Mathematics), Professor, Far Eastern Federal University, 10, Ajax Bay, Russky Island, Vladivostok, 690922, Russian Federation, tel.: +7(902)5060356, e-mailstepltd@mail.ru, ORCID iD https://orcid.org/0000-0001-7484-4108.
Received 26.01.2021
S-полигоны над вполне упорядоченным моноидом с модулярной решеткой конгруэнций
А. А. Степанова
Дальневосточный федеральный университет, Владивосток, Российская Федерация
Аннотация. Исследование относится к структурной теории полигонов, подразумевающей описание полигонов над теми или иными классами моноидов или обладающих теми или иными свойствами, например удовлетворяющих какому-либо требованию, предъявляемому к решётке конгруэнций. Конгруэнции универсальной алгебры — это ядра гомоморфизмов этой алгебры в другие. Знание всех конгруэнций означает знание всех гомоморфных образов алгебры. Левый ^-полигон над моноидом S — это множество А, на котором моноид S действует слева, причем единица этого моноида действует тождественно. Рассматриваются полигоны над линейно упорядоченными и над вполне упорядоченными моноидами, где под линейно упорядоченным моноидом S понимается линейно упорядоченное множество с минимальным элементом и с бинарной операцией max, относительно которой S является, очевидно, коммутативным моноидом; под вполне упорядоченным моноидом S понимается вполне упорядоченное множество с бинарной операцией max, относительно которой S также является коммутативным моноидом. Статья является продолжением авторского исследования с М. С. Казаком, где приводится описание S-полигонов над линейно упорядоченными моноидами с линейной решеткой конгруэнций и S-полигонов над вполне упорядоченными моноидами с дистрибутивной решеткой конгруэнций. Описываются 5-полигоны над вполне упорядоченными моноидами, решетки конгруэнций которых модулярны.
Ключевые слова: полигон над моноидом, решетка конгруэнций алгебры, модулярная решетка.
Список литературы
1. Егорова Д. П. Структура конгруэнций унарной алгебры // Упорядоченные множества и решётки : межвуз. науч. сб. Саратов, 1978. Вып. 5. С. 11-44.
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Алена Андреевна Степанова, доктор физико-математических наук, профессор, Дальневосточный федеральный университет, Российская Федерация, 690922, г. Владивосток, о. Русский, кампус ДВФУ, корпуе D, каб. D646, tel.: +7(902)506 03 56, e-mail: stepltd@mail.ru, ORCID iD https://oreid.org/0000-0001-7484-4108.
Поступила в 'редакцию 26.01.2021
