Cancellable elements of the lattice of semigroup varieties
We completely determine all commutative semigroup varieties that are cancellable elements of the lattice SEM of all semigroup varieties. In particular, we verify that a commutative semigroup variety is a cancellable element of the lattice SEM if and only if it is a modular element of this lattice.
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| Zitieren: | Cancellable elements of the lattice of semigroup varieties / S.V. Gusev, D.V. Skokov, B.M. Vernikov // Algebra and Discrete Mathematics. — 2018. — Vol. 26, № 1. — С. 34–46. — Бібліогр.: 13 назв. — англ. |
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| citation_txt | Cancellable elements of the lattice of semigroup varieties / S.V. Gusev, D.V. Skokov, B.M. Vernikov // Algebra and Discrete Mathematics. — 2018. — Vol. 26, № 1. — С. 34–46. — Бібліогр.: 13 назв. — англ. |
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| description | We completely determine all commutative semigroup varieties that are cancellable elements of the lattice SEM of all semigroup varieties. In particular, we verify that a commutative semigroup variety is a cancellable element of the lattice SEM if and only if it is a modular element of this lattice.
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 26 (2018). Number 1, pp. 34–46
c© Journal “Algebra and Discrete Mathematics”
Cancellable elements of the lattice
of semigroup varieties∗
Sergey V. Gusev, Dmitry V. Skokov
and Boris M. Vernikov
Communicated by Yu. V. Zhuchok
Abstract. We completely determine all commutative semi-
group varieties that are cancellable elements of the lattice SEM of
all semigroup varieties. In particular, we verify that a commutative
semigroup variety is a cancellable element of the lattice SEM if
and only if it is a modular element of this lattice.
1. Introduction and summary
The collection of all semigroup varieties forms a lattice with respect
to class-theoretical inclusion. This lattice is denoted by SEM. The lat-
tice SEM has been intensively studied since the beginning of 1960s. A
systematic overview of the material accumulated here is given in the
survey [8]. There are a number of article devoted to an examination of
special elements of different types in the lattice SEM (see [8, Section 14]
or the recent survey [11] devoted specially to this subject). The present
article continues these investigations.
In the lattice theory, special elements of many different types are
considered. We recall definitions of three types of elements that appear
∗The work is partially supported by Russian Foundation for Basic Research (grant
17-01-00551) and by the Ministry of Education and Science of the Russian Federation
(project 1.6018.2017).
2010 MSC: Primary 20M07; Secondary 08B15.
Key words and phrases: semigroup, variety, cancellable element of a lattice,
modular element of a lattice.
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S. V. Gusev, D. V. Skokov, B. M. Vernikov 35
below. An element x of a lattice 〈L;∨,∧〉 is called neutral if
(∀y, z ∈ L) (x ∨ y) ∧ (y ∨ z) ∧ (z ∨ x) = (x ∧ y) ∨ (y ∧ z) ∨ (z ∧ x).
It is well known that an element x is neutral if and only if, for all y, z ∈
L, the sublattice of L generated by x, y and z is distributive (see [1,
Theorem 254]). Neutral elements play an important role in the general
lattice theory (see [1, Section III.2], for instance). An element x ∈ L is
called
modular if (∀y, z ∈ L) y 6 z −→ (x ∨ y) ∧ z = (x ∧ z) ∨ y,
cancellable if (∀y, z ∈ L) x ∨ y = x ∨ z & x ∧ y = x ∧ z −→ y = z.
It is easy to see that any cancellable element is a modular one. A valuable
information about modular and cancellable elements in abstract lattices
can be found in [5], for instance.
Modular elements of the lattice SEM were examined in [3, 6, 9]. In
particular, commutative semigroup varieties that are modular elements of
SEM are completely determined in [9, Theorem 3.1]. Here we describe
commutative semigroup varieties that are cancellable elements of SEM.
In particular, we verify that, for commutative varieties, the properties of
being modular and cancellable elements are equivalent.
To formulate the main result of the article, we need some notation.
We denote by F the free semigroup over a countably infinite alphabet. As
usual, elements of F are called words. Words unlike variables are written
in bold. Two parts of an identity we connect by the symbol ≈, while the
symbol = denotes the equality relation on F . Note that a semigroup S
satisfies the identity system wx ≈ xw ≈ w where the variable x does not
occur in the word w if and only if S contains a zero element 0 and all
values of w in S are equal to 0. We adopt the usual convention of writing
w ≈ 0 as a short form of such a system and referring to the expression
w ≈ 0 as to a single identity. We denote by T the trivial semigroup variety
and by SL the variety of all semilattices.
The main result of the article is the following
Theorem 1.1. For a commutative semigroup variety V, the following
are equivalent:
a) V is a cancellable element of the lattice SEM;
b) V is a modular element of the lattice SEM;
c) V = M ∨N where M is one of the varieties T or SL, while N is a
variety satisfying the identities x2y ≈ 0 and xy ≈ yx.
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36 Cancellable elements of the lattice of varieties
It can be verified by fairly easy calculations that any proper subvariety
of the variety W given by the identities x2y ≈ 0 and xy ≈ yx is given
within W either by the identity x2 ≈ 0 or by the identity x1x2 · · ·xn ≈ 0 for
some natural n or by both these identities. Thus, in actual fact, Theorem 1.1
gives an exhaustive list of the varieties we consider.
The article consists of three sections. Section 2 contains an auxiliary
facts, while Section 3 is devoted to verification of Theorem 1.1.
2. Preliminaries
2.1. Preliminaries on lattices
We start with several observations dealing with cancellable or modular
elements in abstract lattices.
Lemma 2.1. Let L be a lattice with 0 and a an atom and neutral element
of L. An element x ∈ L is cancellable if and only if the element x ∨ a is
cancellable.
Proof. Necessity. Let x be a cancellable element and y, z ∈ L. We need
to verify that
y ∧ (x ∨ a) = z ∧ (x ∨ a) & y ∨ (x ∨ a) = z ∨ (x ∨ a) −→ y = z.
If a 6 x then this implication is evident because x ∨ a = x and x is
cancellable. Let now a 66 x. Throughout all the proof we will use the fact
that the element a is neutral without explicit references. We can assume
without loss of generality that either a 6 y and a 6 z or a � y and a � z
or a 6 y but a � z.
If a 6 y and a 6 z then
(y ∧ x) ∨ a = (y ∧ x) ∨ (y ∧ a) = y ∧ (x ∨ a)
= z ∧ (x ∨ a) = (z ∧ x) ∨ (z ∧ a) = (z ∧ x) ∨ a
and
(y ∧ x) ∧ a = y ∧ (x ∧ a) = y ∧ 0 = 0
= z ∧ 0 = z ∧ (a ∧ x) = (z ∧ x) ∧ a.
Thus, (y ∧ x) ∨ a = (z ∧ x) ∨ a and (y ∧ x) ∧ a = (z ∧ x) ∧ a. The element
a is cancellable because it is neutral. Therefore, y ∧ x = z ∧ x. Further,
y ∨ x = (y ∨ a) ∨ x = y ∨ (x ∨ a)
= z ∨ (x ∨ a) = (z ∨ a) ∨ x = z ∨ x.
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S. V. Gusev, D. V. Skokov, B. M. Vernikov 37
Thus, y ∧ x = z ∧ x and y ∨ x = z ∨ x. Since x is cancellable, we have
y = z.
If a � y and a � z then
y ∧ x = (y ∧ x) ∨ 0 = (y ∧ x) ∨ (y ∧ a) = y ∧ (x ∨ a)
= z ∧ (x ∨ a) = (z ∧ x) ∨ (z ∧ a) = (z ∧ x) ∨ 0 = z ∧ x.
Thus, y ∧ x = z ∧ x. Further,
(y ∨ x) ∧ a = (y ∧ a) ∨ (x ∧ a) = 0 ∨ 0
= (z ∧ a) ∨ (x ∧ a) = (z ∨ x) ∧ a.
Thus, (y ∨ x) ∧ a = (z ∨ x) ∧ a. By the hypothesis,
(y ∨ x) ∨ a = y ∨ (x ∨ a) = z ∨ (x ∨ a) = (z ∨ x) ∨ a.
Since a is neutral and every neutral element is cancellable, we have y∨x =
z ∨ x. Taking into account that the element x is cancellable, we have that
y = z.
Finally, if a 6 y but a � z then
z ∧ x = (z ∧ x) ∨ 0 = (z ∧ x) ∨ (z ∧ a) = z ∧ (x ∨ a)
= y ∧ (x ∨ a) = (y ∧ x) ∨ (y ∧ a) = (y ∧ x) ∨ a.
Thus, z ∧ x = (y ∧ x) ∨ a. Then a 6 z ∧ x 6 z, a contradiction.
Sufficiency. Let x ∨ a be a cancellable element and y, z are elements
of L with y ∧ x = z ∧ x and y ∨ x = z ∨ x. We have to verify that y = z.
If a 6 x then the desirable conclusion is evident because x ∨ a = x and
the element x ∨ a is cancellable. Let now a 66 x. We note that
y ∨ (x ∨ a) = (y ∨ x) ∨ a = (z ∨ x) ∨ a = z ∨ (x ∨ a),
i.e., y ∨ (x ∨ a) = z ∨ (x ∨ a). Since the element x ∨ a is cancellable, it
remains to check that y∧ (x∨a) = z∧ (x∨a). As in the proof of necessity,
we can assume without loss of generality that either a 6 y and a 6 z or
a � y and a � z or a 6 y but a � z.
If a 6 y and a 6 z then
y ∧ (x ∨ a) = (y ∨ a) ∧ (x ∨ a) = (y ∧ x) ∨ a
= (z ∧ x) ∨ a = (z ∨ a) ∧ (x ∨ a) = z ∧ (x ∨ a),
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38 Cancellable elements of the lattice of varieties
i.e., y ∧ (x ∨ a) = z ∧ (x ∨ a). If a � y and a � z then
y ∧ (x ∨ a) = (y ∧ x) ∨ (y ∧ a) = (y ∧ x) ∨ 0 = y ∧ x
= z ∧ x = (z ∧ x) ∨ 0 = (z ∧ x) ∨ (z ∧ a) = z ∧ (x ∨ a),
i.e., y ∧ (x ∨ a) = z ∧ (x ∨ a) again. Finally, if a 6 y but a � z then
a = a ∨ (x ∧ a) = (y ∧ a) ∨ (x ∧ a) = (y ∨ x) ∧ a
= (z ∨ x) ∧ a = (z ∧ a) ∨ (x ∧ a) = 0 ∨ 0 = 0,
a contradiction.
Lemma 2.2. Let L be a lattice with 0, a an atom and neutral element of
L and x ∈ L. If, for any y, z ∈ L, the equalities x ∨ (y ∨ a) = x ∨ (z ∨ a)
and x∧ (y∨a) = x∧ (z∨a) imply that y∨a = z∨a then x is a cancellable
element.
Proof. Let y, z ∈ L, x ∨ y = x ∨ z and x ∧ y = x ∧ z. We need to verify
that y = z. It is evident that
x ∨ (y ∨ a) = (x ∨ y) ∨ a = (x ∨ z) ∨ a = x ∨ (z ∨ a).
Since the element a is neutral, we have
x ∧ (y ∨ a) = (x ∧ y) ∨ (x ∧ a) = (x ∧ z) ∨ (x ∧ a) = x ∧ (z ∨ a).
In view of the hypothesis, we have that y ∨ a = z ∨ a. We can assume
without loss of generality that either y, z � a or y, z > a or y > a but
z � a. If y, z � a then we apply the fact that a is neutral and have
y = (y ∨ a) ∧ y = (z ∨ a) ∧ y = (z ∧ y) ∨ (a ∧ y) = (z ∧ y) ∨ 0
= (z ∧ y) ∨ (z ∧ a) = z ∧ (y ∨ a) = z ∧ (z ∨ a) = z,
i.e., y = z. If y, z > a then y = y ∨ a = z ∨ a = z. Finally, let y > a
and z � a. If x > a then x ∧ y > a and x ∧ z � a. Then x ∧ y 6= x ∧ z,
contradicting the choice of y and z. Let now x � a. Then x ∧ a = 0 and
z ∧ a = 0. Since a is neutral, we have that
(x ∨ z) ∧ a = (x ∧ a) ∨ (z ∧ a) = 0 ∨ 0 = 0,
whence x ∨ z � a. On the other hand, x ∨ y > a. Therefore, x ∨ y 6= x ∨ z
that contradicts the choice of y and z again.
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S. V. Gusev, D. V. Skokov, B. M. Vernikov 39
Lemma 2.3. Let x be a modular but not cancellable element of a lattice
L and let y and z be different elements of L such that x ∨ y = x ∨ z
and x ∧ y = x ∧ z. Then there is an element x′ ∈ L such that x′ 6 x,
x′ ∨ y = x′ ∨ z, x′ ∧ y = x′ ∧ z and y ∨ z = x′ ∨ y.
Proof. Put x′ = x ∧ (y ∨ z). Clearly, x′ 6 x. Note that
x′ ∧ y = x ∧ (y ∨ z) ∧ y = x ∧ y = x ∧ z = x ∧ (y ∨ z) ∧ z = x′ ∧ z.
It remains to verify that x′∨y = x′∨z = y∨z. Clearly, x′ 6 y∨z, whence
y ∨ x′ 6 y ∨ z. Then x ∧ (y ∨ z) = x′ 6 y ∨ x′ 6 y ∨ z, and therefore,
(y ∨ z) ∧ x = (y ∨ x′) ∧ x. (1)
Further, the equality x ∨ y = x ∨ z implies that z 6 y ∨ x. Since x′ 6 x,
we have that
(y ∨ z) ∨ x = (y ∨ x) ∨ z = y ∨ x = y ∨ (x′ ∨ x) = (y ∨ x′) ∨ x.
Thus,
(y ∨ z) ∨ x = (y ∨ x′) ∨ x. (2)
Combining these observations, we have that
y ∨ z = (x ∨ (y ∨ z)) ∧ (y ∨ z)
= (x ∨ (y ∨ x′)) ∧ (y ∨ z) by (2)
= (x ∧ (y ∨ z)) ∨ (y ∨ x′) because x is modular
and y ∨ x′ 6 y ∨ z
= (x ∧ (y ∨ x′)) ∨ (y ∨ x′) by (1)
= y ∨ x′.
Thus, we prove that y ∨ z = y ∨ x′. Similar arguments allow us to show
that y ∨ z = z ∨ x′. Therefore, y ∨ x′ = y ∨ z = z ∨ x′.
2.2. Preliminaries on semigroup varieties
Now we return to semigroup varieties. Let X be a semigroup variety. If
nilpotency index of any nil-semigroup in X is not exceeded some natural
number n and n is the least number with such a property then n is called
a degree of the variety X and is denoted by deg(X); otherwise we put
deg(X) = ∞. For a given word w, we denote by ℓ(w) the length of w,
and by con(w) the content of w, i.e., the set of all variables occurring
in w. The equivalence of the claims a) and c) of the following lemma is
verified in [10, Proposition 2.11], the implication c) −→ b) is evident, and
the implication b) −→ a) follows from [4, Lemma 1].
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40 Cancellable elements of the lattice of varieties
Lemma 2.4. For a semigroup variety V, the following are equivalent:
a) deg(V) 6 n;
b) V satisfies an identity of the form x1x2 · · ·xn ≈ v for some word v
with ℓ(v) > n;
c) V satisfies an identity of the form
x1x2 · · ·xn ≈ x1x2 · · ·xi−1(xi · · ·xj)
ℓxj+1 · · ·xn (3)
for some ℓ > 1 and 1 6 i 6 j 6 n.
The following claim is evident.
Lemma 2.5. If X and Y are semigroup varieties then deg(X ∧Y) =
min{deg(X), deg(Y)}.
Lemma 2.6 ([10, Lemma 2.13]). If X is a semigroup variety and Y is a
nil-variety of semigroups then deg(X ∨Y) = max{deg(X), deg(Y)}.
We need the following two well known and easily verified technical
remarks about identities of nilsemigroups.
Lemma 2.7. Let V be a nil-variety of semigroups.
(i) If the variety V satisfies an identity u ≈ v with con(u) 6= con(v)
then V satisfies also the identity u ≈ 0.
(ii) If the variety V satisfies an identity of the form u ≈ vuw where at
least one the words v and w is non-empty then V satisfies also the
identity u ≈ 0.
The first statement of the following lemma is generally known (see [8,
Section 1], for instance). The second claim also is well known and is verified
explicitly in [13, Proposition 2.4] (see also [8, Section 14]).
Lemma 2.8. The variety SL is
(i) an atom of the lattice SEM;
(ii) a neutral element of SEM.
3. Proof of the main result
In this section we prove Theorem 1.1. The implication a) −→ b) is
evident, while the equivalence of the claims b) and c) is checked in [9,
Theorem 3.1]. It remains to prove the implication c) −→ a). Lemmas 2.1
and 2.8 allow us to assume that V = N. Suppose that N is non-cancellable
element of SEM. Hence there are semigroup varieties Y and Z with
N ∨Y = N ∨ Z, N ∧Y = N ∧ Z and Y 6= Z.
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S. V. Gusev, D. V. Skokov, B. M. Vernikov 41
Lemma 3.1. deg(Y) = deg(Z).
Proof. Put deg(Y) = r, deg(Z) = s and deg(N) = t (here r, s, t ∈
N ∪ {∞}). Suppose that r 6= s. We can assume without any loss that
r < s. Then Lemmas 2.5 and 2.6 imply that
if t > s then deg(N ∧Y) = r < s = deg(N ∧ Z);
if r < t < s then deg(N ∧Y) = r < t = deg(N ∧ Z);
if t 6 r then deg(N ∨Y) = r < s = deg(N ∨ Z).
The first and the second cases contradict the equality N ∧Y = N ∧ Z,
while the third case is impossible because N ∨Y = N ∨ Z.
Since the claims b) and c) of Theorem 1.1 are equivalent, N is a
modular element of SEM. In view of Lemma 2.3, there is a variety N
′
such that
N
′ ⊆ N, N′ ∨Y = N
′ ∨ Z = Y ∨ Z and N
′ ∧Y = N
′ ∧ Z.
Being a subvariety of N, the variety N
′ satisfies the identities x2y ≈ 0
and xy ≈ yx.
Since Y 6= Z, we can assume without loss of generality that there
is an identity u ≈ v that holds in Y but is false in Z. If this identity
is satisfied by the variety N
′ then it holds in N
′ ∨Y = N
′ ∨ Z, and
therefore, in Z. Thus, u ≈ v is wrong in N
′. A word w is called linear
if any variable occurs in w at most once. Recall that N
′ satisfies the
identities x2y ≈ 0 and xy ≈ yx. Therefore, any non-linear word except
x2 equals to 0 in N
′. Thus, we may assume without loss of generality
that either u = x2 or u = x1x2 · · ·xk for some k. Lemmas 2.2 and 2.8
allow us to assume that Y,Z ⊇ SL. This implies that con(u) = con(v).
Combining the observations given above, we have that u ≈ v is either an
identity of the form x2 ≈ xm for some m 6= 2 or an identity of the form
x1x2 · · ·xk ≈ v where con(v) = {x1, x2, . . . , xk}.
Case 1: u ≈ v is an identity of the form x2 ≈ xm for some m 6= 2.
Suppose at first that m = 1. This means that Y is a variety of bands. Then
Z ∧N
′ = Y ∧N
′ = T. If N′ = T then Y = Y ∨N
′ = Z ∨N
′ = Z, and
we are done. Otherwise, N′ contains the variety ZM of all semigroups with
zero multiplication. Since Z ∧N
′ = T, we have that Z + ZM, whence the
variety Z is completely regular. If Z contains a non-trivial group variety G
then G ⊆ Z ∨N
′ = Y ∨N
′. But all groups in Y ∨N
′ are trivial because
this variety satisfies the identity x3 ≈ x4. Thus, Z is a completely regular
variety without non-trivial groups, i.e., a band variety. We see that the
identity u ≈ v holds in Z, a contradiction.
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42 Cancellable elements of the lattice of varieties
Let now m > 2. If N′ satisfies the identity x2 ≈ 0 then the identity
x2 ≈ xm holds in the variety N
′ ∨Y = N
′ ∨ Z, and therefore, in Z. But
this contradicts the choice of the identity u ≈ v. Thus we can assume that
the identity x2 ≈ 0 is wrong in N′. Recall that a word w is called an isoterm
for a variety V if V does not satisfy any non-trivial identity of the form
w ≈ w
′. Lemma 2.7 implies that the word x2 is an isoterm for the variety
N
′. Further, Lemma 2.7(ii) implies that the variety N
′ ∧ Z = N
′ ∧Y
satisfies the identity x2 ≈ 0. Therefore, there is a deduction of this identity
from identities of the varieties N
′ and Z. In particular, one of these
varieties satisfies a non-trivial identity of the form x2 ≈ w. Since x2 is an
isoterm for N
′, this identity holds in Z. Since Z ⊇ SL, this identity has
the form x2 ≈ xk for some k > 2. Let m be the least number with the
property that x2 ≈ xm holds in Y but does not hold in Z, while k the
least number with the property that x2 ≈ xk holds in Z.
Suppose that k < m. Then m = k + j for some natural j. It is clear
that the identity x2+j ≈ xk+j = xm holds in N
′. Then this identity is
true also in Z ∨N
′ = Y ∨N
′. Hence x2+j ≈ xm ≈ x2 holds in Y. Since
2 + j < m, this contradicts the choice of m.
Finally, let m < k. Then k = m + j for some natural j. Clearly,
the identity x2+j ≈ xm+j = xk holds in N
′. Therefore, this identity
holds in Y ∨N
′ = Z ∨N
′. This means that Z satisfies the identities
x2+j ≈ xk ≈ x2. But 2+ j < m+ j = k and we have a contradiction with
the choice of k.
Case 2: u ≈ v is an identity of the form x1x2 · · ·xk ≈ v where
con(v) = {x1, x2, . . . , xk}. Clearly, ℓ(v) > k. If ℓ(v) = k then the identity
u ≈ v has the form
x1x2 · · ·xk ≈ x1πx2π · · ·xkπ
where π is a non-trivial permutation on the set {1, 2, . . . , k}. This identity
holds in N
′ because N
′ is commutative. But this is false. Therefore,
ℓ(v) > k. Put deg(Y) = n. Then deg(Z) = deg(Y) = n by Lemma 3.1.
Lemma 2.4 implies that n 6 k. Recall that Y ∨ Z = N
′ ∨Y = N
′ ∨ Z.
Clearly, deg(Y ∨ Z) > n. Suppose at first that deg(Y ∨ Z) = n. Then
deg(N′) 6 deg(N′ ∨Y) = deg(Y ∨ Z) = n.
Being a nil-variety, N′ satisfies the identity x1x2 · · ·xn ≈ 0 in this case.
Since ℓ(v) > k > n, the identity x1x2 · · ·xk ≈ v holds in N
′ as well. This
contradicts the choice of the identity u ≈ v.
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S. V. Gusev, D. V. Skokov, B. M. Vernikov 43
Let now deg(Y ∨ Z) > n. Since deg(Y) = n, Lemma 2.4 implies that
Y satisfies an identity of the form (3) for some ℓ > 1 and 1 6 i 6 j 6 n.
The same lemma implies that this identity is false in Y ∨ Z because
deg(Y ∨ Z) = n otherwise. Therefore, (3) is wrong in Z. Analogously,
there are r > 1 and 1 6 i′ 6 j′ 6 n such that the identity
x1x2 · · ·xn ≈ x1x2 · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xn (4)
holds in Z but does not hold in Y. We will assume without any loss that
i 6 i′.
Suppose at first that j < j′. Then we substitute (xi′ · · ·xj′)
r−1xj′+1
into xj′+1 in (3) whenever j′ < n or multiply (3) by (xi′ · · ·xj′)
r−1 on the
right whenever j′ = n. We obtain the identity
x1x2 · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xn ≈ x1x2 · · ·xi−1·
· (xi · · ·xj)
ℓxj+1 · · ·xj′(xi′ · · ·xj′)
r−1xj′+1 · · ·xn.
(5)
Clearly, the identity (5) holds in the variety N
′. Then it satisfies in Z
as well because N
′ ∨Y = N
′ ∨ Z. Substitute xi−1(xi · · ·xj)
ℓ−1 into xi−1
in (4) whenever i > 1 or multiply (4) by (xi · · ·xj)
ℓ−1 on the left whenever
i = 1. As a result, we obtain the identity
x1x2 · · ·xi−1(xi · · ·xj)
ℓxj+1 · · ·xn ≈ x1x2 · · ·xi−1·
· (xi · · ·xj)
ℓ−1xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xn.
(6)
This identity holds in Z too. Note that the right parts of the identities (5)
and (6) coincide. Indeed,
x1x2 · · ·xi−1(xi · · ·xj)
ℓxj+1 · · ·xj′(xi′ · · ·xj′)
r−1xj′+1 · · ·xn
= x1x2 · · ·xi−1(xi · · ·xj)
ℓ−1xi · · ·xjxj+1 · · ·xj′(xi′ · · ·xj′)
r−1xj′+1 · · ·xn
= x1x2 · · ·xi−1(xi · · ·xj)
ℓ−1xi · · ·xi′−1xi′ · · ·xj′(xi′ · · ·xj′)
r−1xj′+1 · · ·xn
= x1x2 · · ·xi−1(xi · · ·xj)
ℓ−1xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xn.
Since the variety Z satisfies the identities (4), (5) and (6), this variety
satisfies also the identity (3). We have a contradiction.
It remains to consider the case when j′ 6 j. Suppose at first that
i = i′ and j = j′. Substitute (xi′ . . . xj′)
rxj′+1 into xj′+1 in (3) whenever
j′ < n or multiply (3) by (xi′ · · ·xj′)
r on the right whenever j′ = n. Then
we obtain the identity
x1x2 · · ·xi−1(xi · · ·xj)
rxj+1 · · ·xn ≈ x1x2 · · ·xi−1(xi · · ·xj)
r+ℓ·
·xj+1 · · ·xn.
(7)
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44 Cancellable elements of the lattice of varieties
Clearly, this identity holds in N
′. The equality Y ∨N
′ = Z ∨N
′ implies
that it holds in Z too. Similar arguments show that Z satisfies the identity
x1x2 · · ·xi−1(xi · · ·xj)
ℓxj+1 · · ·xn ≈ x1x2 · · ·xi−1(xi · · ·xj)
r+ℓ·
·xj+1 · · ·xn.
(8)
Combining the identities (4), (7) and (8), we have that Z satisfies the
identity (3), contradicting with the choice of this identity.
Thus, either i < i′ or j′ < j. Suppose without loss of generality that
i < i′. Substitute xi′−1(xi′ · · ·xj′)
r−1 into xi′−1 in (3). We obtain the
identity
x1x2 · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xn ≈ x1x2 · · ·xi−1·
· (xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)
ℓxj+1 · · ·xn.
(9)
Clearly, the identity (9) holds in the variety N
′. Besides that, it holds in
Z because N
′ ∨Y = N
′ ∨ Z. For an arbitrary word w, we suppose w
0 to
be the empty word. Let t > 0 and s > 0. Now we multiply the identity (4)
by the word
(xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)
s
on the left whenever i = 1 or substitute the word
xi−1(xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)
s
into xi−1 in (4) whenever i > 1. Besides that, we multiply (4) by the
word (xi · · ·xj)
t−1 on the right whenever j = n or substitute the word
(xi · · ·xj)
t−1xj+1 into xj+1 in (4) whenever j < n. Then we obtain the
identity
x1x2 · · ·xi−1(xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)
s(xi · · ·xj)
t·
·xj+1 · · ·xn
≈ x1x2 · · ·xi−1(xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)
s+1(xi · · ·xj)
t−1·
·xj+1 · · ·xn.
(10)
“adm-n3” — 2018/10/20 — 9:02 — page 45 — #51
S. V. Gusev, D. V. Skokov, B. M. Vernikov 45
For convenience, we write below w
ε
≈ w
′ in the case when the identity
w ≈ w
′ follows from the identity ε. The variety Z satisfies the identities
x1x2 · · ·xn
(4)
≈ x1x2 · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xn
(9)
≈ x1x2 · · ·xi−1(xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)
ℓ·
·xj+1 · · ·xn
(10)
≈ x1x2 · · ·xi−1(xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)
ℓ−1·
· (xi · · ·xj)xj+1 · · ·xn
(10)
≈ x1x2 · · ·xi−1(xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)
ℓ−2·
· (xi · · ·xj)
2xj+1 · · ·xn
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(10)
≈ x1x2 · · ·xi−1(xi · · ·xi′−1(xi′ · · ·xj′)
rxj′+1 · · ·xj)·
· (xi · · ·xj)
ℓ−1xj+1 · · ·xn
(10)
≈ x1x2 · · ·xi−1(xi · · ·xj)
ℓxj+1 · · ·xn.
Here we use the identity (10) for the first time with s = ℓ− 1 and t = 1,
for the second time with s = ℓ − 2 and t = 2, . . . , for the penultimate
time with s = 1 and t = ℓ − 1, finally, for the last time with s = 0 and
t = ℓ. We prove that the identity (3) holds in Z, a contradiction. This
completes the proof of Theorem 1.1.
At the conclusion of the article, we formulate some open questions.
Question 3.2. Does there exist a semigroup variety that is a modular
but not a cancellable element of the lattice SEM?
A semigroup variety is called 0-reduced if it may be given by identities
of the form w ≈ 0 only. It is known that any 0-reduced semigroup variety
is a modular element of the lattice SEM. This fact was noted for the first
time in [12, Corollary 3] and rediscovered (in different terminology) in [3,
Proposition 1.1]. In actual fact, it readily follows from [2, Proposition 2.2].
Question 3.3. Is any 0-reduced semigroup variety a cancellable element
of the lattice SEM?
Evidently, the negative answer to Question 3.3 immediately implies the
negative answer to Question 3.2. An affirmative answer to Question 3.3
would also have an interesting corollary. To formulate it, we recall that
an element x of a lattice L is called lower-modular if
(∀ y, z ∈ L) x 6 y −→ x ∨ (y ∧ z) = y ∧ (x ∨ z).
“adm-n3” — 2018/10/20 — 9:02 — page 46 — #52
46 Cancellable elements of the lattice of varieties
Lower-modular elements of the lattice SEM are completely determined
in [7]. This result easily implies that if an answer to Question 3.3 is
affirmative then every lower-modular element of SEM is cancellable.
References
[1] G.Grätzer, Lattice Theory: Foundation, Birkhäuser, Springer Basel AG, 2011.
[2] J.Ježek, The lattice of equational theories. Part I: modular elements, Czechosl.
Math. J., 31, 1981, pp.127–152.
[3] J.Ježek, R.N.McKenzie, Definability in the lattice of equational theories of semi-
groups, Semigroup Forum, 46, 1993, pp.199–245.
[4] M.V.Sapir, E.V.Sukhanov, On varieties of periodic semigroups, Izvestiya VUZ.
Matematika, No.4, 1981, pp.48–55 [Russian; Engl. translation: Soviet Math. (Iz.
VUZ), 25, No.4, 1981, pp.53–63].
[5] B.Šešelja, A.Tepavčević, Weak Congruences in Universal Algebra, Institute of
Mathematics, Novi Sad, 2001 (Novi Sad, Symbol).
[6] V.Yu.Shaprynskǐı, Modular and lower-modular elements of lattices of semigroup
varieties, Semigroup Forum, 85, 2012, pp.97–110.
[7] V.Yu.Shaprynskǐı, B.M.Vernikov, Lower-modular elements of the lattice of semi-
group varieties. III, Acta Sci. Math. (Szeged), 76, 2010, pp.371–382.
[8] L.N.Shevrin, B.M.Vernikov, M.V.Volkov, Lattices of semigroup varieties, Izvestiya
VUZ. Matematika, No.3, 2009, pp.3–36 [Russian; Engl. translation: Russian Math.
(Iz. VUZ), 53, No.3, 2009, pp.1–28].
[9] B.M.Vernikov, On modular elements of the lattice of semigroup varieties, Comment.
Math. Univ. Carol., 48, 2007, pp.595–606.
[10] B.M.Vernikov, Upper-modular elements of the lattice of semigroup varieties, Algebra
Universalis, 58, 2008, pp.405–428.
[11] B.M.Vernikov, Special elements in lattices of semigroup varieties, Acta Sci. Math.
(Szeged), 81, 2015, pp.79–109.
[12] B.M.Vernikov, M.V.Volkov, Lattices of nilpotent semigroup varieties, in L.N.Shevrin
(ed.), Algebraic Systems and their Varieties, Sverdlovsk, Ural State University,
1988, pp.53–65 [Russian].
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eral Algebra, 16, 2005, pp.275–288.
Contact information
S. V. Gusev,
D. V. Skokov,
B. M. Vernikov
Ural Federal University, Institute of Natural
Sciences and Mathematics, Lenina 51, 620000
Ekaterinburg, Russia
E-Mail(s): sergey.gusb@gmail.com,
dmitry.skokov@gmail.com,
bvernikov@gmail.com
Web-page(s): http://kadm.imkn.urfu.ru/
pages.php?id=vernikov
Received by the editors: 30.03.2017.
|
| id | nasplib_isofts_kiev_ua-123456789-188372 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-12-07T16:34:06Z |
| publishDate | 2018 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Gusev, S.V. Skokov, D.V. Vernikov, B.M. 2023-02-26T12:17:33Z 2023-02-26T12:17:33Z 2018 Cancellable elements of the lattice of semigroup varieties / S.V. Gusev, D.V. Skokov, B.M. Vernikov // Algebra and Discrete Mathematics. — 2018. — Vol. 26, № 1. — С. 34–46. — Бібліогр.: 13 назв. — англ. 1726-3255 2010 MSC: Primary 20M07; Secondary 08B15 https://nasplib.isofts.kiev.ua/handle/123456789/188372 We completely determine all commutative semigroup varieties that are cancellable elements of the lattice SEM of all semigroup varieties. In particular, we verify that a commutative semigroup variety is a cancellable element of the lattice SEM if and only if it is a modular element of this lattice. The work is partially supported by Russian Foundation for Basic Research (grant 17-01-00551) and by the Ministry of Education and Science of the Russian Federation (project 1.6018.2017). en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics Cancellable elements of the lattice of semigroup varieties Article published earlier |
| spellingShingle | Cancellable elements of the lattice of semigroup varieties Gusev, S.V. Skokov, D.V. Vernikov, B.M. |
| title | Cancellable elements of the lattice of semigroup varieties |
| title_full | Cancellable elements of the lattice of semigroup varieties |
| title_fullStr | Cancellable elements of the lattice of semigroup varieties |
| title_full_unstemmed | Cancellable elements of the lattice of semigroup varieties |
| title_short | Cancellable elements of the lattice of semigroup varieties |
| title_sort | cancellable elements of the lattice of semigroup varieties |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/188372 |
| work_keys_str_mv | AT gusevsv cancellableelementsofthelatticeofsemigroupvarieties AT skokovdv cancellableelementsofthelatticeofsemigroupvarieties AT vernikovbm cancellableelementsofthelatticeofsemigroupvarieties |