Cancellable elements of the lattice of monoid varieties
aa r X i v : . [ m a t h . G R ] J a n CANCELLABLE ELEMENTS OF THE LATTICEOF MONOID VARIETIES
SERGEY V. GUSEV AND EDMOND W. H. LEE
Abstract.
The set of all cancellable elements of the lattice of semigroupvarieties has recently been shown to be countably infinite. But the descriptionof all cancellable elements of the lattice
MON of monoid varieties remainsunknown. This problem is addressed in the present article. The first exampleof a monoid variety with modular but non-distributive subvariety lattice is firstexhibited. Then a necessary condition of the modularity of an element in
MON is established. These results play a crucial role in the complete description ofall cancellable elements of the lattice
MON . It turns out that there are preciselyfive such elements. Introduction and summary
The present article is concerned with the lattice
MON of all monoid varieties,where monoids are considered as semigroups with an identity element that is fixedby a 0-ary operation. For many years, results on the lattice
MON were scarce.But recently, interest in this lattice has grown significantly; in particular, the studyof its special elements was initiated in the articles [5, 6]. In the present work, wecontinue these investigations.Special elements play an important role in general lattice theory; see [3, Sec-tion III.2], for instance. We recall definitions of those types of special elements thatare relevant here. An element x of a lattice L is cancellable if ∀ y, z ∈ L : x ∨ y = x ∨ z & x ∧ y = x ∧ z −→ y = z ; modular if ∀ y, z ∈ L : y ≤ z −→ ( x ∧ z ) ∨ y = ( x ∨ y ) ∧ z. It is easy to see that every cancellable element is modular.Our main goal is to describe all cancellable elements of the lattice
MON . Toformulate our main result, we need some definitions and notation. Let X + [respec-tively, X ∗ ] denote the free semigroup [respectively, monoid] over a countably infinitealphabet X . Elements of X are called letters and elements of X ∗ are called words .Words unlike letters are written in bold. An identity is written as u ≈ v , where u , v ∈ X ∗ .Let T , SL , and MON denote the variety of trivial monoids, the variety ofsemilattice monoids, and the variety of all monoids, respectively. For any identitysystem Σ, let var Σ denote the variety of monoids given by Σ. Put C = var { x ≈ x , xy ≈ yx } and D = var { x ≈ x , x y ≈ xyx ≈ yx } . Mathematics Subject Classification.
Key words and phrases.
Monoid, variety, lattice of varieties, cancellable element of a lattice,modular element of a lattice.The first author is supported by the Ministry of Science and Higher Education of the RussianFederation (project FEUZ-2020-0016).
Then the following is our main result.
Theorem 1.1.
A monoid variety is a cancellable element of the lattice
MON ifand only if it coincides with one of the varieties T , SL , C , D or MON . Many articles were devoted to special elements of different types in the lattice
SEM of all semigroup varieties; an overview of results published before 2015 can befound in the survey [16]. It is natural to compare Theorem 1.1 with the descriptionof cancellable elements of the lattice
SEM that was found in 2019 [15]. Theorem 1.1shows that there are only five cancellable elements in the lattice
MON . In contrast,the set of all cancellable elements of the lattice
SEM is countably infinite.In general, the set of cancellable elements in a lattice need not form a sublattice.For example, the elements x and y of the lattice in Fig. 1 are cancellable but theirjoin x ∨ y is not. However, the class of all cancellable elements of SEM forms adistributive sublattice of
SEM ; see Corollary 3.14 in the extended version of thesurvey [16]. Theorem 1.1 shows that the same is true for monoid varieties too; infact, the five cancellable elements in
MON constitute a chain. ss ssss ss ss x y
Figure 1.
Now since the chain T ⊂ SL ⊂ C ⊂ D coincides with the lattice L ( D ) ofsubvarieties of D (see Fig. 2), a monoid variety V is a cancellable element of thelattice MON if and only if either V ⊆ D or V = MON . It is routinely verifiedthat the variety D can be given by the single identity x yz ≈ yxzx . Therefore itis easy to check the cancellability of proper elements of the lattice MON ; a monoidvariety is proper if it is different from
MON . Corollary 1.2.
Suppose that M is any monoid that generates a proper subvariety V of MON . Then V is a cancellable element of the lattice MON if and only if M satisfies the identity x yz ≈ yxzx . The article consists of five sections. Section 2 contains definitions, notation,certain known results and their simple corollaries. In Section 3, the first exampleof a monoid variety with modular but non-distributive subvariety lattice is given.In Section 4, a necessary condition of the modularity of an element in
MON isestablished in Proposition 4.3. Results from Sections 3 and 4 will then be used inSection 5 to prove Theorem 1.1. An extended version of this survey, which is periodically updated as new results are foundand/or new articles are published, is available at http://arxiv.org/abs/1309.0228v20.
ANCELLABLE ELEMENTS OF THE LATTICE OF MONOID VARIETIES 3 Preliminaries
Acquaintance with rudiments of universal algebra is assumed of the reader. Referto the monograph [2] for more information.Recall that a variety is periodic if it consists of periodic monoids. Equivalently,a variety is periodic if and only if it satisfies the identity x n ≈ x n + m for some n, m ≥
1. For any word w and any set X of letters, the word obtained from w bydeleting all the letters of X is denoted by w X . The content of a word w , denotedby con( w ), is the set of letters occurring in w . The partition lattice over a set X is denoted by Part( X ). Let L FIC( X ∗ ) denote the lattice of all fully invariantcongruences on the monoid X ∗ , and for any variety V of monoids, let FIC( V )denote the fully invariant congruence on X ∗ corresponding to V . It is well knownthat the mapping FIC : MON −→ L
FIC ( X ∗ ) is an anti-isomorphism of lattices;see [2, Theorem 11.9 and Corollary 14.10], for instance. For any u , v ∈ X + , we put u (cid:22) v if v = a ξ ( u ) b for some words a , b ∈ X ∗ and some endomorphism ξ of X + . Itis easily seen that the relation (cid:22) on X + is a quasi-order. For an arbitrary anti-chain A ⊆ X + under the relation (cid:22) , let L A denote the set of all monoid varieties V forwhich A is a union of FIC( V )-classes. Define the map ϕ A : L A −→ Part( A ) by therule ϕ A ( V ) = FIC( V ) | A for any V ∈ L A . Lemma 2.1 ( [4, Lemma 3]) . Let A be any anti-chain under the quasi-order (cid:22) .Suppose that for any words u , v ∈ A and any nonempty set X ⊆ con( u ) , theequalities con( u ) = con( v ) and u X = v X hold. Then (i) the set L A is a sublattice of the lattice MON ; (ii) the map ϕ A is a surjective anti-homomorpism of the lattice L A onto thelattice Part( A ) ; (iii) for any partition β ∈ Part( A ) , there exists a non-periodic monoid variety V ∈ L A such that ϕ A ( V ) = β . Recall that a band is left regular if it is a semilattice of left zero bands. It is wellknown that the class of left regular band monoids coincides with the variety
LRB = var { xy ≈ xyx } . The initial part of a word w , denoted by ini( w ), is the word obtained from w byretaining the first occurrence of each letter. The following assertion is well knownand easily verified. Lemma 2.2.
An identity u ≈ v holds in LRB if and only if ini( u ) = ini( v ) . For any n ≥
2, the variety C n = var { x n ≈ x n +1 , xy ≈ yx } is generated by the monoid h a, | a n = 0 i [1, Corollary 6.1.5]. Note that the variety C has already been introduced in Section 1. A word w is an isoterm for a variety V if the FIC( V )-class of w is singleton. The following result is easily deducedfrom [10, Lemma 3.3]. Lemma 2.3.
Let n ≥ . For any monoid variety V , the following are equivalent: a) x n is not an isoterm for V ; b) V satisfies the identity x n ≈ x n + m for some m ≥ ; c) C n +1 * V . S. V. GUSEV AND E. W. H. LEE
A monoid is completely regular if it is a union of its maximal subgroups. Avariety is completely regular if it consists of completely regular monoids. It is wellknown that a monoid variety is completely regular if and only if it satisfies theidentity x ≈ x n +1 for some n ≥ Lemma 2.4 ( [7, Lemma 2.14]) . If a monoid variety V is non-completely regularand noncommutative, then D ⊆ V . Lemma 2.5.
Let V be any monoid variety such that C ⊆ V . Suppose that V does not contain the variety E = var { x ≈ x , x y ≈ xyx, x y ≈ y x } . Then V satisfies the identity x p yx q ≈ yx r for some p, q ≥ and r ≥ .Proof. If D ⊆ V , then the result follows from [7, Lemma 4.1 and Proposition 4.2].Therefore suppose that D * V , so that by Lemma 2.4, the variety V is eithercompletely regular or commutative. But V cannot be completely regular because C ⊆ V . Hence V is commutative and satisfies the identity xyx ≈ yx . (cid:3) Monoid variety with modular but non-distributive subvarietylattice
There are many examples of monoid varieties with non-distributive subvarietylattice; see [5, 6, 13], for instance. However, all these varieties have non-modularsubvariety lattice as well. In this section, we present the first example of a monoidvariety whose subvariety lattice is modular but non-distributive. To this end, thefollowing varieties are required: the variety D generated by the monoid h a, b, | a = b = bab = 0 i = { a, b, ab, ba, aba, , } , the variety R generated by the monoid h a, b, | a = b = ba = 0 i = { a, b, a , ab, a b, , } and the variety R δ dual to R . It is proved in [9, Lemmas 2.2.8 and 2.2.9] that D = var { x ≈ x , x yzt ≈ yxzxtx,xyzxty ≈ yxzxty, xzxyty ≈ xzyxty, xtyzxy ≈ xtyzyx } , R ∨ R δ = var { x ≈ x , x yzt ≈ yxzxtx,xyzxty ≈ yxzxty, xzxyty ≈ xzyxty, xtyzxy ≈ xtyzyx } . It is easily seen that D = ( R ∨ R δ ) ∧ var { x ≈ x } . Proposition 3.1.
The lattice L ( R ∨ R δ ) of subvarieties of R ∨ R δ is given inFig. . In particular, this lattice is modular but not distributive.Proof. It is easily shown that C ⊆ R ∨ R δ . According to Lemma 2.3, any subvariety V of R ∨ R δ such that C * V satisfies the identity x ≈ x , whence V ⊆ D .Therefore, the lattice L ( R ∨ R δ ) is the disjoint union of the lattice L ( D ) and theinterval [ C , R ∨ R δ ]. It is proved in [10, Lemmas 4.4 and 4.5] that the lattice L ( D )coincides with the 5-element chain in Fig. 2. Thus it remains to describe the interval[ C , R ∨ R δ ]. It follows from [12, Proposition 4.1] that every noncommutativevariety in this interval is defined within R ∨ R δ by some of the identities xyx ≈ x y , ANCELLABLE ELEMENTS OF THE LATTICE OF MONOID VARIETIES 5 s ss ssssss ss C C DD RR ∨ R δ R δ SLT
Figure 2.
The subvariety lattice L ( R ∨ R δ ) xyx ≈ yx or x y ≈ yx . It is then routinely shown that the interval [ C , R ∨ R δ ]is as described in Fig. 2, where R = ( R ∨ R δ ) ∧ var { xyx ≈ yx } , D ∨ C = ( R ∨ R δ ) ∧ var { x y ≈ yx } , and D ∨ C = ( D ∨ C ) ∧ R = ( D ∨ C ) ∧ R δ = R ∨ R δ . The proof of thisproposition is thus complete. (cid:3) Necessary condition of the modularity of an element in
MON
Given any word w and letter x , let occ x ( w ) denote the number of occurrencesof x in w . Let λ denote the empty word. Let W = W ∪ W , where W = { y r xt r z r y r t r xz r | r , r , r , r , r , r ≥ } ,W = { y r xt r z r xy r t r xz r | r , r , r , r , r , r ≥ } . Let us fix the following two words: p = y xt z y t xz and q = y xt z xy t xz . Put K = var { p ≈ q } . Lemma 4.1.
The set W is a FIC( K ) -class.Proof. Let u ≈ v be any identity of K with u ∈ W . We need to verify that v ∈ W .By assumption, there is a deduction of the identity u ≈ v from the identity p ≈ q ,that is, a sequence w , w , . . . , w m of words such that w = u , w m = v and, foreach i = 0 , , . . . , m −
1, there are words a i , b i ∈ X ∗ and an endomorphism ξ i of X ∗ such that w i = a i ξ i ( s i ) b i and w i +1 = a i ξ i ( t i ) b i , where { s i , t i } = { p , q } . Bytrivial induction on m , it suffices to only consider the case when u = a ξ ( s ) b and v = a ξ ( t ) b for some words a , b ∈ X ∗ , an endomorphism ξ of X ∗ and words s and t such that { s , t } = { p , q } . S. V. GUSEV AND E. W. H. LEE
Since any subword of u of the form ab , where a an b are distinct letters, occursonly once in u and all letters occurring in s are multiple, the following holds:(I) For any a ∈ con( s ), either ξ ( a ) = λ or ξ ( a ) is a power of some letter.Further, since occ x ( u ) ≤ y ( s ) = occ z ( s ) = occ t ( s ) = 4, we have(II) x / ∈ con( ξ ( yzt )).We note that if ξ ( s ) = λ or ξ ( s ) is a power of some letter, then the requiredstatement is evident. So, we may assume that(III) | con( ξ ( s )) | ≥ u = y ℓ xt ℓ z ℓ x c y ℓ t ℓ xz ℓ , where c ∈ { , } and ℓ , ℓ , ℓ , ℓ , ℓ , ℓ ≥
2, and let d = ( s = p , s = q . If ξ ( x ) = λ , then ξ ( s ) = ξ ( t ), whence v = u ∈ W . So, it remains to consider thecase when ξ ( x ) = λ . Then (I) implies that ξ ( x ) is a power of some letter.Suppose that ξ ( x ) is a power of y . Then (III) implies that con( ξ ( t z x d y t ))contains one of the letters x , z and t . This is only possible when ξ ( t z x d y t ) = y p xt ℓ z ℓ x c y q for some 0 ≤ p ≤ ℓ and 0 ≤ q ≤ ℓ . But since x / ∈ { y } = con( ξ ( x ))by assumption and x / ∈ con( ξ ( yzt )) by (II), the contradiction x / ∈ con( ξ ( t z x d y t ))is deduced. Therefore, ξ ( x ) cannot be a power of y . Similarly, ξ ( x ) cannot be apower of z as well.Suppose now that ξ ( x ) is a power of t . Then (III) implies that con( ξ ( t z x d y t ))contains one of the letters x , y and z . This is only possible when ξ ( t z x d y t ) = t p z ℓ x c y ℓ t q for some 0 ≤ p ≤ ℓ and 0 ≤ q ≤ ℓ . Then by (I), either ξ ( t ) = λ or ξ ( t )is a power of t . This implies that ξ ( z x d y ) = z ℓ x c y ℓ . Taking into account that ξ ( x ) is a power of t , we apply (I) again and obtain that ξ ( z ) = z ℓ , ξ ( y ) = y ℓ and c = d = 0. This is only possible when ξ ( xt z x d ) = y r xt ℓ z ℓ x c for some0 ≤ r ≤ ℓ . But since x / ∈ { t } = con( ξ ( x )) by assumption and x / ∈ con( ξ ( yzt )) by(II), the contradiction x / ∈ con( ξ ( xt z x d )) is deduced. Therefore, ξ ( x ) cannot bea power of t .Finally, suppose that ξ ( x ) is a power of x . Then since x is not a subword of u ,we have ξ ( x ) = x .Suppose that c = 0. Then d = 0 because otherwise, occ x ( u ) < occ x ( ξ ( s )). Then ξ ( t z y t ) = t ℓ z ℓ y ℓ t ℓ . It follow from (I) that ξ ( z ) = z ℓ , ξ ( y ) = y ℓ and ξ ( t ) = t ℓ = t ℓ . Then ξ ( s ) = y ℓ xt ℓ z ℓ y ℓ t ℓ xz ℓ , a = y ℓ − ℓ and b = z ℓ − ℓ .Therefore, ξ ( t ) = y ℓ xt ℓ z ℓ xy ℓ t ℓ xz ℓ , whence v = y ℓ xt ℓ z ℓ xy ℓ t ℓ xz ℓ ∈ W ,and we are done.Suppose now that c = 1. If x ∈ con( b ), then d = 0 because otherwise, occ x ( u ) < occ x ( ξ ( s ) b ). This is only possible when a ξ ( y ) = y ℓ , xξ ( t z y t ) x = xt ℓ z ℓ x and ξ ( z ) b = y ℓ t ℓ xz ℓ . The second equality implies that ξ ( t z y t ) = t ℓ z ℓ . Clearly, ξ ( t ) = λ , whence ξ ( z y ) = t ℓ z ℓ . In view of (I), we have ξ ( z ) = t ℓ and ξ ( y ) = z ℓ . But thiscontradicts the fact that ξ ( z ) b = z ℓ t ℓ xz ℓ . Therefore, x / ∈ con( b ). Analogously,one can verify that x / ∈ con( a ). It follows that d = 1. Then a ξ ( y ) = y ℓ , xξ ( t z ) xξ ( y t ) x = xt ℓ z ℓ xy ℓ t ℓ x and ξ ( z ) b = z ℓ . ANCELLABLE ELEMENTS OF THE LATTICE OF MONOID VARIETIES 7
It follows from (I) that ξ ( z ) = z ℓ , ξ ( y ) = y ℓ and ξ ( t ) = t ℓ = t ℓ . Then ξ ( s ) = y ℓ xt ℓ z ℓ xy ℓ t ℓ xz ℓ , a = y ℓ − ℓ and b = z ℓ − ℓ . Therefore, ξ ( t ) = y ℓ xt ℓ z ℓ xy ℓ t ℓ xz ℓ , whence v = y ℓ xt ℓ z ℓ y ℓ t ℓ xz ℓ ∈ W , and we are done. (cid:3) For any n ≥
1, put B n = var { x n ≈ x n +1 } . Lemma 4.2.
Suppose that V is any proper monoid variety that is a modular ele-ment of the lattice MON . Then V is periodic.Proof. Seeking a contradiction, suppose that V is not periodic, so that V containsthe variety COM of all commutative monoids. Since V is proper and non-periodic,it satisfies some nontrivial identity u ≈ v such that every letter from con( uv )occurs n times on both sides for some n ≥
1, that is, n = occ a ( u ) = occ a ( v ) for all a ∈ con( uv ). Then by [14, Lemma 3.2], there exist two distinct letters x and y suchthat the identity obtained from u ≈ v by retaining x and y is nontrivial. Thereforewe may assume that con( u ) = con( v ) = { x, y } with n = occ x ( u ) = occ x ( v ) =occ y ( u ) = occ y ( v ).Suppose that LRB ⊆ V . In view of Lemma 2.2, we may assume without lossof generality that ini( u ) = ini( v ) = xy . Let u ′ and v ′ be words that obtainfrom u and v , respectively, by making the substitution ( x, y ) ( y, x ). Thenini( u ′ ) = ini( v ′ ) = yx . Put A = { w ∈ { x, y } + | occ x ( w ) = n + 1 , occ y ( w ) = n } . Let w , w ′ ∈ A and w (cid:22) w ′ . This means that w ′ = a ξ ( w ) b for some words a , b ∈ X ∗ and some endomorphism ξ of X + . Since the length of w equals to the length of w ′ ,we have a = b = λ . Then w ′ = ξ ( w ). But this is only possible when ξ ( x ) = x and ξ ( y ) = y because occ y ( w ) < occ x ( w ). Hence w = w ′ . So, A is an anti-chain underthe quasi-order (cid:22) . Then L A is a sublattice of MON by Lemma 2.1(i) and V ∈ L A .Clearly, u x, u ′ x, v x, v ′ x ∈ A . Evidently, ini( u x ) = ini( v x ) = xy and ini( u ′ x ) =ini( v ′ x ) = yx . In view of Lemma 2.2, the words u x and u ′ x lie in distinct FIC( V )-classes. Then, since V satisfies the nontrivial identities u x ≈ v x and u ′ x ≈ v ′ x ,the equivalence γ = ϕ ( V ) contains at least two non-singleton classes. It is verifiedin [11, Proposition 2.2] that a partition ρ ∈ Part( X ) is a modular element inPart( X ) if and only if ρ has at most one non-singleton class. This result implies that γ is not a modular element of the lattice Part( A ). Then there are α, β ∈ Part( A )such that α ⊂ β and(4.1) ( γ ∧ β ) ∨ α ⊂ ( γ ∨ α ) ∧ β. According to Lemma 2.1, we can find a non-periodic variety X ∈ L A such that ϕ ( X ) = α . Put Y = X ∧ var { w ≈ w ′ | ( w , w ′ ) ∈ β } . Clearly, Y ∈ L A and ϕ ( Y ) = β . Then( V ∧ X ) ∨ Y ⊂ ( V ∨ Y ) ∧ X because otherwise, the inclusion (4.1) does not hold. We see that V is not a modularelement of the lattice MON , which is a contradiction.Suppose now that
LRB * V . Then Lemma 2.2 allows us to assume that u starts with the letter x but v starts with the letter y . Let Z = var { x n +1 ≈ x n +2 , x n v ≈ x n +1 v } . S. V. GUSEV AND E. W. H. LEE
We note that V ∧ B n +1 ⊆ Z . Indeed, V ∧ B n +1 satisfies the identities x n v ≈ x n u ≈ x n +1 u ≈ x n +1 v and so the identity x n v ≈ x n +1 v . Clearly, the word x n v is an isoterm for both V ∨ Z and B n +1 . It follows that x n v is an isoterm for ( V ∨ Z ) ∧ B n +1 as well.However, x n v is not an isoterm for Z because Z satisfies x n v ≈ x n +1 v . Therefore,( V ∧ B n +1 ) ∨ Z = Z ⊂ ( V ∨ Z ) ∧ B n +1 . This means that V is not a modular element of the lattice MON , which again is acontradiction. (cid:3)
The following is the main result of this section.
Proposition 4.3.
Suppose that V is any proper monoid variety that is a modularelement of the lattice MON . Then V satisfies the identities x ≈ x , (4.2) x y ≈ yx . (4.3) Proof.
By Lemma 4.2, the variety V is periodic and so it satisfies the identity x n ≈ x n + m for some n, m ≥
1; we may assume n and m to be the least possible.First, suppose that n = 1, so that V is completely regular. If X is a noncom-mutative completely regular variety, then it is verified in [5, Lemma 3.1] that( X ∧ D ) ∨ C ⊂ ( X ∨ C ) ∧ D , whence X is not a modular element of the lattice MON . If X is a commutativevariety containing a nontrivial group, then it is proved in [5, Lemma 3.2] that( X ∧ B ) ∨ Q ⊂ ( X ∨ Q ) ∧ B , where Q = var { yxyzxy ≈ yxzxyxz } , whence X is again not a modular elementof the lattice MON . In view of these two facts, the variety V is commutative anddoes not contain any nontrivial group. Since V is also completely regular, it isidempotent and so is contained in SL . Obviously, SL satisfies (4.2) and (4.3).So, it remains to consider the case when n >
1. Then C n ⊆ V and C n +1 * V by Lemma 2.3. It follows from [6, Lemma 2] that E * V . Then by Lemma 2.5, V satisfies the identity x p yx q ≈ yx r for some p , q ≥ r ≥
2. Thedual arguments imply that V also satisfies the identity x p yx q ≈ x r y for some p , q ≥ r ≥
2. Since one can substitute x n for x in these identities and V satisfies x n ≈ x n + m , we may assume without loss of generality that p , p , q , q , r , r ∈ { n, n + 1 , . . . , n + m − } . Evidently, there exist ℓ and ℓ such that the identities x p yx q + ℓ ≈ yx r + ℓ and x p + ℓ yx q ≈ x r + ℓ y are equivalent modulo x n ≈ x n + m to the identities x p yx q ≈ yx r + ℓ and x p yx q ≈ x r + ℓ y, respectively. Therefore V satisfies x r + ℓ y ≈ yx r + ℓ , whence it satisfies(4.4) x k y ≈ yx k ANCELLABLE ELEMENTS OF THE LATTICE OF MONOID VARIETIES 9 for some k ≥ n . It follows that the meet V ∧ B satisfies the identities (4.2) and(4.4); it also satisfies the identity p ≈ q because p = y xt z y t xz ≈ y k xt k z k y k t k xz k (4.4) ≈ x y k z k t k (4.2) ≈ x y k z k t k (4.4) ≈ y k xt k z k xy k t k xz k (4.2) ≈ y xt z xy t xz = q . Therefore V ∧ B ⊆ K , so that ( V ∧ B ) ∨ K = K .Suppose that n > m >
1. Recall from the beginning of the section that W = { y r xt r z r y r t r xz r | r , r , r , r , r , r ≥ } . Let a ≈ b be any identity of V ∨ K with a ∈ W . If n >
2, then b ∈ W byLemmas 2.3 and 4.1. Clearly, V contains the variety A m of all Abelian groups ofexponent m . It is well known and easily verified that an identity w ≈ w ′ holdsin A m if and only if occ a ( w ) ≡ occ a ( w ′ ) (mod m ) for all a ∈ X . This fact andLemma 4.1 imply that if m >
1, then b ∈ W . We see that if n > m >
1, then b ∈ W in either case. Evidently, if B satisfies an identity c ≈ d with c ∈ W , then d ∈ W . This implies that if an identity of the form p ≈ w holds in ( V ∨ K ) ∧ B ,then w ∈ W . In particular, ( V ∨ K ) ∧ B violates p ≈ q . Therefore,( V ∧ B ) ∨ K = K ⊂ ( V ∨ K ) ∧ B . This means that V is not a modular element in MON . It follows that n = 2 and m = 1. Then V satisfies (4.2). Besides that, since (4.4) holds in the variety V , thisvariety satisfies (4.3).Proposition 4.3 is thus proved. (cid:3) Proof of Theorem 1.1
Necessity . Let V be any proper monoid variety that is a cancellable element ofthe lattice MON . Since any cancellable element is modular, Proposition 4.3 impliesthat V satisfies the identities (4.2) and (4.3). If V does not coincide with any of thevarieties T , SL , C and D , then V contains the variety D by [8, Lemma 3.3(i)].Proposition 3.1 and the fact that C * V imply that V ∨ R = V ∨ R δ and V ∧ R = V ∧ R δ = D , contradicting the assumption that V is a cancellableelement of MON . Hence V coincides with one of the varieties T , SL , C and D . Sufficiency . Obviously, T and MON are cancellable elements of
MON . Anelement x of a lattice L is costandard if ∀ y, z ∈ L : ( x ∧ z ) ∨ y = ( x ∨ y ) ∧ ( z ∨ y ) . It is easily seen that any costandard element is cancellable. It is shown in [5,Theorem 1.2] that the varieties SL and C are costandard elements of the lattice MON . Therefore, these varieties are cancellable elements of this lattice.So, it remains to establish that D is a cancellable element in MON . Let X and Y be monoid varieties such that D ∨ X = D ∨ Y and D ∧ X = D ∧ Y . If D ⊆ X , then D = D ∧ X = D ∧ Y , so that D ⊆ Y , whence X = D ∨ X = D ∨ Y = Y andwe are done. Therefore by symmetry, we may assume that D * X and D * Y .Now the subvariety lattice L ( D ) is the chain T ⊂ SL ⊂ C ⊂ D ; see Fig. 2. Itfollows that D ∧ X = D ∧ Y ∈ { T , SL , C } . If D ∧ X = D ∧ Y = T , then X and Y are varieties of groups by [7, Lemma 2.1]. Then X ∨ Y is a variety of groups tooand so SL * X ∨ Y , whence(5.1) D ∧ ( X ∨ Y ) = D ∧ X = D ∧ Y . If D ∧ X = D ∧ Y = SL , then X and Y are completely regular varieties by [7,Corollary 2.6]. Then X ∨ Y is completely regular and so C * X ∨ Y , whencethe equality (5.1) is true. Finally, if D ∧ X = D ∧ Y = C , then X and Y arecommutative by Lemma 2.4. Then X ∨ Y is commutative and so D * X ∨ Y ,whence the equality (5.1) is true again. We see that the equality (5.1) holds in anycase.Clearly,(5.2) D ∨ ( X ∨ Y ) = D ∨ X = D ∨ Y . Then X = ( D ∧ X ) ∨ X because D ∧ X ⊂ X = ( D ∧ ( X ∨ Y )) ∨ X by (5.1)= ( D ∨ X ) ∧ ( X ∨ Y ) by [6, Proposition 7]= ( D ∨ ( X ∨ Y )) ∧ ( X ∨ Y ) by (5.2)= ( X ∨ Y ) because X ∨ Y ⊂ D ∨ ( X ∨ Y ) . We see that X = X ∨ Y . By symmetry, Y = X ∨ Y , whence X = Y . Therefore, D is a cancellable element in MON . (cid:3) References [1] J. Almeida, Finite Semigroups and Universal Algebra, World Scientific, Singapore, 1994. Zbl0844.20039, MR1331143, DOI 10.1142/2481[2] S. Burris, H. P. Sankappanavar, A Course in Universal Algebra, Springer Verlag, New York,1981. Zbl 0478.08001, MR0648287[3] G. Gr¨atzer, Lattice Theory: Foundation, Springer Basel AG, 2011. Zbl 1233.06001,MR2768581, DOI 10.1007/978-3-0348-0018-1[4] S. V. Gusev,
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