Linear Orders on General Algebras
Order (2005) 22: 41–62 DOI: 10.1007/s11083-005-9007-8
#
Springer 2005
Linear Orders on General Algebras ´ NDOR RADELECZKI* and JENO ´´ SZIGETI* SA Institute of Mathematics, University of Miskolc, Miskolc, 3515, Hungary. e-mail: [email protected], [email protected] (Received: 7 June 2004; in final form: 4 August 2005) Abstract. We answer the question, when a partial order in a partially ordered algebraic structure has a compatible linear extension. The finite extension property enables us to show, that if there is no such extension, then it is caused by a certain finite subset in the direct square of the base set. As a consequence, we prove that a partial order can be linearly extended if and only if it can be linearly extended on every finitely generated subalgebra. Using a special equivalence relation on the above direct square, we obtain a further property of linearly extendible partial orders. Imposing conditions on the lattice of compatible quasi orders, the number of linear orders can be determined. Our general approach yields new results even in the case of semi-groups and groups. Mathematics Subject Classification (1991): Primary 06F99, 06F05, Secondary 08A05, 06D15. Key Words: compatible quasi and partial orders on algebraic structures, finite extension property, linear extension, zero-distributive and zero-modular lattice.
1. Introduction One of the central problems in the theory of ordered algebras is to find necessary and sufficient conditions for the existence of a compatible linear extension R of r in a partially ordered algebraic structure (A, F, r). Such an R A A is a compatible linear order on (A, F) with r R. The well-known Szpilrajn theorem ([13]) asserts that a partial order can always be extended to a linear order, this answers the simplest case of the above problem, when F = ;. If F = {t} and t: A Y A is a unary operation, then the answer can be found in [11]: the partially ordered (mono-unary) algebra (A, {t}, r) has a compatible linear extension if and only if t is acyclic. For classical algebraic structures (when (A, F) is a group or a ring) the problem of linear extensions has been thoroughly investigated and has an abundant literature (see [4 –10]). A partially ordered abelian group (A, {+}, r) has a compatible linear extension if and only if A is torsion-free. Unfortunately, there is no such simple and sweeping result for non-commutative groups: e.g. the existence of a compatible linear order on the free group cannot be derived from the known general criteria providing compatible linear orders on arbitrary groups. We note that the compatibility of a partial (or linear) order on a ring is * Corresponding author.
42
´ NDOR RADELECZKI AND JENO ´´ SZIGETI SA
defined in a slightly different way, the multiplication does not entirely preserve the partial order relation. The implications x er y ) xz er yz and x er y ) zx er zy are required only for z’s with 0 er z. Consequently, our general results cannot be directly applied to rings. The main aim of our paper is to give a complete answer to the question of existence of a compatible linear extension of r in a partially ordered algebraic structure (A, F, r). A compatible linear order on (A, F) can be simply viewed as a compatible linear extension of the partially ordered algebra (A, F, DA), where DA A A is the trivial partial order (identity). In view of this observation, we shall derive a necessary and sufficient condition for the existence of a compatible linear order on an arbitrary algebra. It may happen that an algebra has a compatible linear order, however at the same time it has a compatible partial order which cannot be extended to a compatible linear order. Our treatment is based on the use of the so-called finite extension property of a partial (quasi) order. We shall see, that if r in a partially ordered algebraic structure (A, F, r) cannot be extended to a compatible linear order, then it is caused by a certain finite subset of A A. As a consequence, we prove the remarkable fact, that r has a compatible linear extension if and only if r has a compatible linear extension on every finitely generated sub-algebra of (A, F). The application of our general results in particular algebraic structures will provide new conditions for the existence of a compatible linear extension, even in the case of semi-groups and groups. Replacing compatible partial orders by (positive) cones, the finite extension property and the present approach to linear extensions can be adapted for rings (see [12]). Using a special equivalence relation on (A A)\DA, we obtain a further necessary condition for the existence of a compatible linear extension. If the lattice of compatible quasi orders on (A, F) is zero-distributive, then the above condition provides a complete characterization of linearly extendible compatible partial orders. The presence of a majority term on (A, F) allows us to formulate an even more transparent characterization. In the last section of our paper we investigate the number of compatible linear orders. Imposing zero-modularity or zero-distributivity on the lattice of compatible quasi orders, we obtain that this number is zero or two. These results show, that the algebraic properties of the above lattice have a strong influence on the linear orders (contained in this lattice). We know, that if (A, F) has a Malcev term, then the compatible quasi orders on (A, F) are congruences. Thus the only compatible partial order on such an algebraic structure is the trivial one (there are no compatible linear orders). Since groups are considered as semi-groups when we deal with partial orders on them and the situation in rings is even more complicated, the partial (linear) orders on classical algebraic structures are not compatible with respect to the well-known Malcev terms.
LINEAR ORDERS ON GENERAL ALGEBRAS
43
2. Partial Orders with the Finite Extension Property For a set A m ; let Quordð AÞ ¼ fq A Ajq reflexive and transitiveg denote the set of all quasi orders on A, then the relation provides a natural lattice structure on Quord(A). The infimum and supremum of the relations qi 2 Quord(A), i 2 I are as follows ^ qi ¼ \ qi and _ qi ¼ [qi ;
i2I
i2I
i2I
i2I
moreover the inverse q1 ¼ fðy; xÞjðx; yÞ 2 qg defines an involution of the complete lattice (Quord(A), $, ¦). A reflexive, antisymmetric and transitive relation r A A is a partial order, in this case we also use the notation x er y for (x, y) 2 r. If q A A is a quasi order and q r for some partial order r A A, then q is also a partial order. Let (T (A), ), idA) be the transformation monoid of all A Y A functions with e denote the submonoid of the usual composition and for an algebra (A, F), let F T (A) generated by the so-called translations. A translation of (A, F) is of the form x7! tðc1 ; . . . ; ci1 ; x; ciþ1 ; . . . ; cn Þ; where t : An Y A is an n-ary operation in F, x 2 A is a variable and the elements e are finite c1, . . . , cij1, ci+1, . . . , cn 2 A are constants. The elements of F compositions of translations and can be called the polynomial functions of e is contained in the set of all polynomial functions P ð1Þ (F) degree one. Clearly, F ð1Þ of (A, F), here P (F) denotes the set of unary functions in the clone generated by F and the constant functions. We say that a function (n-ary operation) t : An Y A preserves the binary relation r A A, if (t(a1, . . . , an), t(b1, . . . , bn)) 2 r for any choice of the pairs (a1, b1), . . . , (an, bn) 2 r. If t : An Y A preserves the reflexive relation r A A, then for x, y, c1, . . . , cij1, ci+1, . . . , cn 2 A with (x, y) 2 r, we have ðtðc1 ; . . . ; ci1 ; x; ciþ1 ; . . . ; cn Þ; tðc1 ; . . . ; ci1 ; y; ciþ1 ; . . . ; cn ÞÞ 2 r: If each operation t 2 F preserves the reflexive relation r A A, then all e preserve r. On the other hand, if the translations and all functions in F translations of (A, F), arising from an n-ary operation t : An Y A, preserve the relation r A A, then for 1 e i e n we have ðtðb1 ; . . . ; bi1 ; ai ; aiþ1 ; . . . ; an Þ; tðb1 ; . . . ; bi1 ; bi ; aiþ1 ; . . . ; an ÞÞ 2 r
44
´ NDOR RADELECZKI AND JENO ´´ SZIGETI SA
for any choice of the pairs (a1, b1), . . . , (an, bn) 2 r. If in addition r is transitive, then we obtain that ðtða1 ; . . . ; an Þ; tðb1 ; . . . ; bn ÞÞ 2 r: It follows that, if each translation of (A, F) preserves the transitive relation r A A, then all operations in F preserve r. A relation r A A is called compatible on (A, F), if each t 2 F preserves r. Thus for an algebra (A, F) and a quasi order q A A the following statements are equivalent: (1) q is compatible on (A, F). e ). (2) q is compatible on the unary algebra (A, F Let Quord(A, F) = {q A Aªq reflexive, transitive and each t 2 F preserves q} denote the set of all compatible quasi orders on the algebra (A, F), then Quord (A, F) is a complete sublattice in Quord(A), moreover q 2 QuordðA; F Þ) q1 2 QuordðA; F Þ: For a, b 2 A we shall make use of the relations n o e e pða; bÞ ¼ ð f ðaÞ; f ðbÞÞj f 2 F and f ðaÞ 6¼ f ðbÞ and
n o pða; bÞ ¼ ð f ðaÞ; f ðbÞÞj f 2 P ð1Þ ð F Þ and f ðaÞ 6¼ f ðbÞ :
Now e p (a, b) p(a, b) and e p (a, a) = p(a, a) = ;, for a m b the containment (a, e . The definition of e b) 2 e p (a, b) is a consequence of idA 2 F p(a, b) and of p(a, b) yields that pða; bÞ s1 and ða; bÞ 2 s2 ) pða; bÞ s2 ða; bÞ 2 s1 ) e e ) and s2 A A on (A, F). It is for any compatible relation s1 A A on (A, F e ) and p(a, b) is compatible on (A, F easy to see, that the reflexive relation DA ? e the reflexive relation DA ?p(a, b) is compatible on (A, F). We claim, that $A [ e pða; bÞ ¼ $A [ pða; bÞ: pða; bÞ A [ pða; bÞ and the transitive closure A [ e pða; bÞ is Indeed, A [ e e ) and also on (A, F). Since ða; bÞ 2 A [ e pða; bÞ, we obtain compatible on (A, F pða; bÞ and the claim easily follows. Let that pða; bÞ A [ e qða; bÞ ¼ $A [ e pða; bÞ ¼ $A [ pða; bÞ;
45
LINEAR ORDERS ON GENERAL ALGEBRAS
then q(a, b) 2 Quord(A, F) is the intersection of the compatible quasi orders on (A, F) containing (a, b) and q ¼ _ qða; bÞ ða;bÞ2q
holds for any compatible quasi order q 2 Quord(A, F). If r is a compatible quasi (partial) order on (A, F), then the triple (A, F, r) is called a quasi (partially) ordered algebra. Let H A A be a subset, a function : H Y A A is called an orientation of H, if ða; bÞ 2 fða; bÞ; ðb; aÞg for all (a, b) 2 H. Now (a, b) = (1 (a, b), 2 (a, b)), where 1 (a, b) and 2 (a, b) denote the first and the second coordinate of (a, b). In a partially ordered algebra (A, F, r), we say that : H Y A A is an r-extending orientation, if for any choice of pairs (a1, b1), . . . , (an, bn), (an +1, bn +1) 2 H and functions f1, f2, . . . , e with fn + 1 = f1 and (an + 1, bn + 1) = (a1, b1), the relations fn, fn + 1 2 F fi ð2 ðai ; bi ÞÞ r fi þ 1 ð1 ðai þ1 ; bi þ1 ÞÞ; 1 i n imply that fi ð1 ðai ; bi ÞÞ ¼ fi ð2 ðai ; bi ÞÞ ¼ fi þ1 ð1 ðaiþ1 ; biþ1 ÞÞ for all 1 e i e n. We note, that jf fi ð1 ðai ; bi ÞÞ; fi ð2 ðai ; bi ÞÞj1 i ngj ¼ 1 is equivalent to the last condition in the above definition. LEMMA 2.1. If (A, F, r) is a quasi ordered algebra, H A A and : H Y A A is an orientation, then the following conditions are equivalent. (1) The relation r A A is a partial order and is r-extending. (2) The supremum r ðH Þ ¼ r _ _ qðða; bÞÞ ða;bÞ2H
in Quord(A, F) is a partial order. (3) There exists a compatible partial order s A A on (A, F) such that r ? (H) s. Proof. (1) ) (2) : Since r ð H Þ is a compatible quasi order on (A, F), it is enough to prove that r ðH Þ is anti-symmetric. Suppose that r ðH Þ ¼ r [ pðða; bÞÞ [ qðða; bÞÞ ¼ r [ [ e ða;bÞ2H
ða;bÞ2H
46
´ NDOR RADELECZKI AND JENO ´´ SZIGETI SA
is not anti-symmetric. Then the properties of the partial order r, as well as the properties of the transitive closure, ensure that we have a cycle of the form ðx1 ; y1 ; . . . ; xn ; yn ; xnþ1 Þ; where {x1, y1, . . . , xn, yn} A is not a one element subset, xn + 1 = x1 and ðxi ; yi Þ 2 e pððai ; bi ÞÞ ¼ e pð1 ; ðai ; bi Þ; 2 ðai ; bi ÞÞ; yi r xiþ1 for certain pairs (ai, bi) 2 H, 1 e i e n. It follows that xi m yi and ðxi ; yi Þ ¼ ðfi ði ðai ; bi ÞÞ; fi ð2 ðai ; bi ÞÞÞ e , 1 e i e n. On rewriting yi er xi + 1, we obtain the for some functions fi 2 F inequalities fi ð2 ðai ; bi ÞÞ r fiþ1 ð1 ðaiþ1 ; biþ1 ÞÞ; 1 i n with (an+1, bn+1) = (a1, b1). The fact, that is an r-extending orientation of H, imply that fi ð1 ðai ; bi ÞÞ ¼ fi ð2 ðai ; bi ÞÞ ¼ fiþ1 ð1 ðaiþ1 ; biþ1 ÞÞ for all 1 i n i.e., that xi ¼ yi ¼ xiþ1 for all 1 i n in contradiction with x1 m y1. Hence r ð H Þ is a partial order on (A, F). (2) ) (3) : Since r ? (H) r (H), we can take s ¼ r ðH Þ. (3) ) (1) : The fact, that r is a partial order, immediately follows from r s. Consider the relations fi ð2 ðai ; bi ÞÞ r fiþ1 ð1 ðaiþ1 ; biþ1 ÞÞ; 1 i n; e , (ai, bi) 2 H for all 1 e i e n + 1 and fn + 1 = f1, (an + 1, bn + 1) = where fi 2 F (a1, b1). Since (1(ai, bi), 2(ai, bi)) = (ai, bi) 2 s, we have 1(ai, bi) es 2(ai, bi). The order preserving property of fi ensures that fi ð1 ðai ; bi ÞÞ s fi ð2 ðai ; bi ÞÞ: Using r s, we obtain the following sequence of relations: f1 ð1 ða1 ; b1 ÞÞ s f1 ð2 ða1 ; b1 ÞÞ s f2 ð1 ða2 ; b2 ÞÞ s f2 ð2 ða2 ; b2 ÞÞ s :::::: s fn1 ð2 ðan1 ; bn1 ÞÞ s fn ð1 ðan ; bn ÞÞ s fn ð2 ðan ; bn ÞÞ s f1 ð1 ða1 ; b1 ÞÞ: The antisymmetry of s yields that all of the terms in the above sequence are Ì equal, thus is r-extending.
LINEAR ORDERS ON GENERAL ALGEBRAS
47
The compatible quasi order r A A on (A, F) has the finite extension property (FEP), if for any finite subset H A A, there exists a compatible partial order s A A on (A, F) such that r s and the elements of H are comparable pairs with respect to s. In view of Lemma 2.1., it is easy to see that the FEP of r is equivalent to the following equivalent conditions: The relation r is a partial order and for any finite subset H A A, there
exists an r-extending orientation : H Y A A.
For any finite subset H A A, there exists an orientation : H Y A A
such that the supremum r ðH Þ ¼ r _ q ð ð a; b Þ Þ _ ða;bÞ2H
in Quord(A, F) is a partial order. For any finite subset H A A, there exists a compatible partial order s
A A on (A, F) such that r ? (H) s for some orientation : H Y A A. Using an orientation : H Y A A and a function " : H Y {j1, 1}, a one to one correspondence between H Y A A orientations and H Y {j1, 1} functions can be established by 1 if ða; bÞ ¼ ða; bÞ "ða; bÞ ¼ and 1 if ða; bÞ ¼ ðb; aÞ ða; bÞ if "ða; bÞ ¼ 1 : ða; bÞ ¼ ðb; aÞ if "ða; bÞ ¼ 1 Since q(b, a) = qj1 (a, b), we can derive the following equivalent of the FEP: For any finite subset H A A, there exists a function " : H Y {j1, 1},
such that the supremum "ða; bÞ ða; bÞ r" ðH Þ ¼ r _ _ q ða; bÞ2H
in Quord(A, F) is a partial order. Our last equivalent of the FEP can be easily derived from the following properties of the quasi order q: 1. The map q 7! qj1 is an involution of the complete lattice Quord(A, F). 2. q is a partial order () q $ qj1 = DA. For any sequence (a1, b1), . . . , (an, bn) 2 A A of distinct pairs, we can find
exponents "1,"2, . . . , "n 2{j1, 1} such that ½r _ q"1 ða1 ; b1 Þ _ . . . _ q"n ðan ; bn Þ ^
r1 _ q"1 ða1 ; b1 Þ _ . . . _ q"n ðan ; bn Þ ¼ $A :
48
´ NDOR RADELECZKI AND JENO ´´ SZIGETI SA
LEMMA 2.2. Let a, b 2 A and r A A be a compatible quasi order on (A, F) with the FEP. Then one (or both) of the following compatible quasi orders on (A, F) has the FEP: ra;b ¼ r _ qða; bÞ and rb;a ¼ r _ qðb; aÞ: Proof. Suppose that neither ra,b nor rb,a has the FEP. Then we have finite 0 subsets H1, H2 A A, such that for any choice of the orientations : H1 Y 00 0 00 A A and : H2 Y A A the sets ra,b ? (H1) and rb,a ? (H2) are not contained in compatible partial orders on (A, F). Consider the finite subset H1 [ H2 [ fða; bÞg A A; the FEP of r provides a compatible partial order s A A on (A, F), such that r [ ðH1 [ H2 [ fða; bÞgÞ s for some orientation : H1 [ H2 [ fða; bÞg ! A A: If (a,b) = (a,b), then (a,b) 2 s implies that q(a,b) s, whence we obtain first 0 r ? q(a,b) s and next ra,b = r ¦ q(a,b) s. Now we have ra,b ? (H1) s 0 00 with = ÊH1, a contradiction. If (a,b) = (b,a), then we get rb,a ? (H2) s 00 Ì with = ÊH2, another contradiction. LEMMA 2.3. Let (A, F) be an algebra and rw A A, w 2 W is chain (with respect to ) of compatible quasi orders on (A, F), such that all rw have the FEP. Then w2W [ rw is a compatible quasi order on (A, F) with the FEP. Proof. It is easy to see that [w2W rw is a compatible quasi order on (A, F), so we have to prove only the FEP. Suppose that r ¼w2W [ rw has no FEP, then there exists a finite set H A A such that the quasi order r ðH Þ ¼ r _ _ qðða; bÞÞ ða;bÞ2H
is not a partial order for any choice of the orientation : H Y A A. Since r ðH Þ ¼ r [
[ qðða; bÞÞ
ða;bÞ2H
is not anti-symmetric, the properties of the quasi order r, as well as the properties of the transitive closure, ensure that for any choice of the orientation : H Y A A we can find a cycle of the form x1 ; y1 ; . . . ; xnðÞ ; ynðÞ ; xnðÞþ1 ;
49
LINEAR ORDERS ON GENERAL ALGEBRAS
n
o
where x1 ; y1 ; . . . ; xnðÞ ; ynðÞ A is not a one-element subset, xnðÞþ1 ¼ x1 and xi ; yi 2 q ai ; bi ; ð*Þ yi ; xiþ1 2 r ¼ [ rw w2W for some pairs ai ; bi 2 H, 1 e i e n(). Clearly, (*) implies that yi ; xiþ1 2 rwð;iÞ for some index w(,i) 2 W. Since
rwð;iÞ is an orientation of H; 1 i nðÞ is a finite subset of the chain frw jw 2 W g, there exists an orientation s : H Y A A and an integer 1 e j e n(s), such that rw (, i ) rw (s, j) for any choice of and 1 e i e n(). Now yi ; xiþ1 2 rwð; jÞ holds for all and 1 e i e n( ), whence we deduce, that rwð; jÞ [ [ qðða; bÞÞ ¼ rwð; jÞ _ _ qðða; bÞÞ ða; bÞ2H
ða; bÞ2H
is not a partial order for any choice of the orientation : H Y A A. Thus we Ì obtained the contradiction, that rwð; jÞ does not have the FEP. THEOREM 2.4. Let (A, F, r) be a partially ordered algebra, then the following conditions on r are equivalent: (1) r has the FEP. (2) r has a compatible linear extension, i.e., there exists a compatible linear order R A A on (A, F) such that r R. Proof. (1))(2): Take R ¼ fsjr s A A; is a compatible quasi
#
Springer 2005
Linear Orders on General Algebras ´ NDOR RADELECZKI* and JENO ´´ SZIGETI* SA Institute of Mathematics, University of Miskolc, Miskolc, 3515, Hungary. e-mail: [email protected], [email protected] (Received: 7 June 2004; in final form: 4 August 2005) Abstract. We answer the question, when a partial order in a partially ordered algebraic structure has a compatible linear extension. The finite extension property enables us to show, that if there is no such extension, then it is caused by a certain finite subset in the direct square of the base set. As a consequence, we prove that a partial order can be linearly extended if and only if it can be linearly extended on every finitely generated subalgebra. Using a special equivalence relation on the above direct square, we obtain a further property of linearly extendible partial orders. Imposing conditions on the lattice of compatible quasi orders, the number of linear orders can be determined. Our general approach yields new results even in the case of semi-groups and groups. Mathematics Subject Classification (1991): Primary 06F99, 06F05, Secondary 08A05, 06D15. Key Words: compatible quasi and partial orders on algebraic structures, finite extension property, linear extension, zero-distributive and zero-modular lattice.
1. Introduction One of the central problems in the theory of ordered algebras is to find necessary and sufficient conditions for the existence of a compatible linear extension R of r in a partially ordered algebraic structure (A, F, r). Such an R A A is a compatible linear order on (A, F) with r R. The well-known Szpilrajn theorem ([13]) asserts that a partial order can always be extended to a linear order, this answers the simplest case of the above problem, when F = ;. If F = {t} and t: A Y A is a unary operation, then the answer can be found in [11]: the partially ordered (mono-unary) algebra (A, {t}, r) has a compatible linear extension if and only if t is acyclic. For classical algebraic structures (when (A, F) is a group or a ring) the problem of linear extensions has been thoroughly investigated and has an abundant literature (see [4 –10]). A partially ordered abelian group (A, {+}, r) has a compatible linear extension if and only if A is torsion-free. Unfortunately, there is no such simple and sweeping result for non-commutative groups: e.g. the existence of a compatible linear order on the free group cannot be derived from the known general criteria providing compatible linear orders on arbitrary groups. We note that the compatibility of a partial (or linear) order on a ring is * Corresponding author.
42
´ NDOR RADELECZKI AND JENO ´´ SZIGETI SA
defined in a slightly different way, the multiplication does not entirely preserve the partial order relation. The implications x er y ) xz er yz and x er y ) zx er zy are required only for z’s with 0 er z. Consequently, our general results cannot be directly applied to rings. The main aim of our paper is to give a complete answer to the question of existence of a compatible linear extension of r in a partially ordered algebraic structure (A, F, r). A compatible linear order on (A, F) can be simply viewed as a compatible linear extension of the partially ordered algebra (A, F, DA), where DA A A is the trivial partial order (identity). In view of this observation, we shall derive a necessary and sufficient condition for the existence of a compatible linear order on an arbitrary algebra. It may happen that an algebra has a compatible linear order, however at the same time it has a compatible partial order which cannot be extended to a compatible linear order. Our treatment is based on the use of the so-called finite extension property of a partial (quasi) order. We shall see, that if r in a partially ordered algebraic structure (A, F, r) cannot be extended to a compatible linear order, then it is caused by a certain finite subset of A A. As a consequence, we prove the remarkable fact, that r has a compatible linear extension if and only if r has a compatible linear extension on every finitely generated sub-algebra of (A, F). The application of our general results in particular algebraic structures will provide new conditions for the existence of a compatible linear extension, even in the case of semi-groups and groups. Replacing compatible partial orders by (positive) cones, the finite extension property and the present approach to linear extensions can be adapted for rings (see [12]). Using a special equivalence relation on (A A)\DA, we obtain a further necessary condition for the existence of a compatible linear extension. If the lattice of compatible quasi orders on (A, F) is zero-distributive, then the above condition provides a complete characterization of linearly extendible compatible partial orders. The presence of a majority term on (A, F) allows us to formulate an even more transparent characterization. In the last section of our paper we investigate the number of compatible linear orders. Imposing zero-modularity or zero-distributivity on the lattice of compatible quasi orders, we obtain that this number is zero or two. These results show, that the algebraic properties of the above lattice have a strong influence on the linear orders (contained in this lattice). We know, that if (A, F) has a Malcev term, then the compatible quasi orders on (A, F) are congruences. Thus the only compatible partial order on such an algebraic structure is the trivial one (there are no compatible linear orders). Since groups are considered as semi-groups when we deal with partial orders on them and the situation in rings is even more complicated, the partial (linear) orders on classical algebraic structures are not compatible with respect to the well-known Malcev terms.
LINEAR ORDERS ON GENERAL ALGEBRAS
43
2. Partial Orders with the Finite Extension Property For a set A m ; let Quordð AÞ ¼ fq A Ajq reflexive and transitiveg denote the set of all quasi orders on A, then the relation provides a natural lattice structure on Quord(A). The infimum and supremum of the relations qi 2 Quord(A), i 2 I are as follows ^ qi ¼ \ qi and _ qi ¼ [qi ;
i2I
i2I
i2I
i2I
moreover the inverse q1 ¼ fðy; xÞjðx; yÞ 2 qg defines an involution of the complete lattice (Quord(A), $, ¦). A reflexive, antisymmetric and transitive relation r A A is a partial order, in this case we also use the notation x er y for (x, y) 2 r. If q A A is a quasi order and q r for some partial order r A A, then q is also a partial order. Let (T (A), ), idA) be the transformation monoid of all A Y A functions with e denote the submonoid of the usual composition and for an algebra (A, F), let F T (A) generated by the so-called translations. A translation of (A, F) is of the form x7! tðc1 ; . . . ; ci1 ; x; ciþ1 ; . . . ; cn Þ; where t : An Y A is an n-ary operation in F, x 2 A is a variable and the elements e are finite c1, . . . , cij1, ci+1, . . . , cn 2 A are constants. The elements of F compositions of translations and can be called the polynomial functions of e is contained in the set of all polynomial functions P ð1Þ (F) degree one. Clearly, F ð1Þ of (A, F), here P (F) denotes the set of unary functions in the clone generated by F and the constant functions. We say that a function (n-ary operation) t : An Y A preserves the binary relation r A A, if (t(a1, . . . , an), t(b1, . . . , bn)) 2 r for any choice of the pairs (a1, b1), . . . , (an, bn) 2 r. If t : An Y A preserves the reflexive relation r A A, then for x, y, c1, . . . , cij1, ci+1, . . . , cn 2 A with (x, y) 2 r, we have ðtðc1 ; . . . ; ci1 ; x; ciþ1 ; . . . ; cn Þ; tðc1 ; . . . ; ci1 ; y; ciþ1 ; . . . ; cn ÞÞ 2 r: If each operation t 2 F preserves the reflexive relation r A A, then all e preserve r. On the other hand, if the translations and all functions in F translations of (A, F), arising from an n-ary operation t : An Y A, preserve the relation r A A, then for 1 e i e n we have ðtðb1 ; . . . ; bi1 ; ai ; aiþ1 ; . . . ; an Þ; tðb1 ; . . . ; bi1 ; bi ; aiþ1 ; . . . ; an ÞÞ 2 r
44
´ NDOR RADELECZKI AND JENO ´´ SZIGETI SA
for any choice of the pairs (a1, b1), . . . , (an, bn) 2 r. If in addition r is transitive, then we obtain that ðtða1 ; . . . ; an Þ; tðb1 ; . . . ; bn ÞÞ 2 r: It follows that, if each translation of (A, F) preserves the transitive relation r A A, then all operations in F preserve r. A relation r A A is called compatible on (A, F), if each t 2 F preserves r. Thus for an algebra (A, F) and a quasi order q A A the following statements are equivalent: (1) q is compatible on (A, F). e ). (2) q is compatible on the unary algebra (A, F Let Quord(A, F) = {q A Aªq reflexive, transitive and each t 2 F preserves q} denote the set of all compatible quasi orders on the algebra (A, F), then Quord (A, F) is a complete sublattice in Quord(A), moreover q 2 QuordðA; F Þ) q1 2 QuordðA; F Þ: For a, b 2 A we shall make use of the relations n o e e pða; bÞ ¼ ð f ðaÞ; f ðbÞÞj f 2 F and f ðaÞ 6¼ f ðbÞ and
n o pða; bÞ ¼ ð f ðaÞ; f ðbÞÞj f 2 P ð1Þ ð F Þ and f ðaÞ 6¼ f ðbÞ :
Now e p (a, b) p(a, b) and e p (a, a) = p(a, a) = ;, for a m b the containment (a, e . The definition of e b) 2 e p (a, b) is a consequence of idA 2 F p(a, b) and of p(a, b) yields that pða; bÞ s1 and ða; bÞ 2 s2 ) pða; bÞ s2 ða; bÞ 2 s1 ) e e ) and s2 A A on (A, F). It is for any compatible relation s1 A A on (A, F e ) and p(a, b) is compatible on (A, F easy to see, that the reflexive relation DA ? e the reflexive relation DA ?p(a, b) is compatible on (A, F). We claim, that $A [ e pða; bÞ ¼ $A [ pða; bÞ: pða; bÞ A [ pða; bÞ and the transitive closure A [ e pða; bÞ is Indeed, A [ e e ) and also on (A, F). Since ða; bÞ 2 A [ e pða; bÞ, we obtain compatible on (A, F pða; bÞ and the claim easily follows. Let that pða; bÞ A [ e qða; bÞ ¼ $A [ e pða; bÞ ¼ $A [ pða; bÞ;
45
LINEAR ORDERS ON GENERAL ALGEBRAS
then q(a, b) 2 Quord(A, F) is the intersection of the compatible quasi orders on (A, F) containing (a, b) and q ¼ _ qða; bÞ ða;bÞ2q
holds for any compatible quasi order q 2 Quord(A, F). If r is a compatible quasi (partial) order on (A, F), then the triple (A, F, r) is called a quasi (partially) ordered algebra. Let H A A be a subset, a function : H Y A A is called an orientation of H, if ða; bÞ 2 fða; bÞ; ðb; aÞg for all (a, b) 2 H. Now (a, b) = (1 (a, b), 2 (a, b)), where 1 (a, b) and 2 (a, b) denote the first and the second coordinate of (a, b). In a partially ordered algebra (A, F, r), we say that : H Y A A is an r-extending orientation, if for any choice of pairs (a1, b1), . . . , (an, bn), (an +1, bn +1) 2 H and functions f1, f2, . . . , e with fn + 1 = f1 and (an + 1, bn + 1) = (a1, b1), the relations fn, fn + 1 2 F fi ð2 ðai ; bi ÞÞ r fi þ 1 ð1 ðai þ1 ; bi þ1 ÞÞ; 1 i n imply that fi ð1 ðai ; bi ÞÞ ¼ fi ð2 ðai ; bi ÞÞ ¼ fi þ1 ð1 ðaiþ1 ; biþ1 ÞÞ for all 1 e i e n. We note, that jf fi ð1 ðai ; bi ÞÞ; fi ð2 ðai ; bi ÞÞj1 i ngj ¼ 1 is equivalent to the last condition in the above definition. LEMMA 2.1. If (A, F, r) is a quasi ordered algebra, H A A and : H Y A A is an orientation, then the following conditions are equivalent. (1) The relation r A A is a partial order and is r-extending. (2) The supremum r ðH Þ ¼ r _ _ qðða; bÞÞ ða;bÞ2H
in Quord(A, F) is a partial order. (3) There exists a compatible partial order s A A on (A, F) such that r ? (H) s. Proof. (1) ) (2) : Since r ð H Þ is a compatible quasi order on (A, F), it is enough to prove that r ðH Þ is anti-symmetric. Suppose that r ðH Þ ¼ r [ pðða; bÞÞ [ qðða; bÞÞ ¼ r [ [ e ða;bÞ2H
ða;bÞ2H
46
´ NDOR RADELECZKI AND JENO ´´ SZIGETI SA
is not anti-symmetric. Then the properties of the partial order r, as well as the properties of the transitive closure, ensure that we have a cycle of the form ðx1 ; y1 ; . . . ; xn ; yn ; xnþ1 Þ; where {x1, y1, . . . , xn, yn} A is not a one element subset, xn + 1 = x1 and ðxi ; yi Þ 2 e pððai ; bi ÞÞ ¼ e pð1 ; ðai ; bi Þ; 2 ðai ; bi ÞÞ; yi r xiþ1 for certain pairs (ai, bi) 2 H, 1 e i e n. It follows that xi m yi and ðxi ; yi Þ ¼ ðfi ði ðai ; bi ÞÞ; fi ð2 ðai ; bi ÞÞÞ e , 1 e i e n. On rewriting yi er xi + 1, we obtain the for some functions fi 2 F inequalities fi ð2 ðai ; bi ÞÞ r fiþ1 ð1 ðaiþ1 ; biþ1 ÞÞ; 1 i n with (an+1, bn+1) = (a1, b1). The fact, that is an r-extending orientation of H, imply that fi ð1 ðai ; bi ÞÞ ¼ fi ð2 ðai ; bi ÞÞ ¼ fiþ1 ð1 ðaiþ1 ; biþ1 ÞÞ for all 1 i n i.e., that xi ¼ yi ¼ xiþ1 for all 1 i n in contradiction with x1 m y1. Hence r ð H Þ is a partial order on (A, F). (2) ) (3) : Since r ? (H) r (H), we can take s ¼ r ðH Þ. (3) ) (1) : The fact, that r is a partial order, immediately follows from r s. Consider the relations fi ð2 ðai ; bi ÞÞ r fiþ1 ð1 ðaiþ1 ; biþ1 ÞÞ; 1 i n; e , (ai, bi) 2 H for all 1 e i e n + 1 and fn + 1 = f1, (an + 1, bn + 1) = where fi 2 F (a1, b1). Since (1(ai, bi), 2(ai, bi)) = (ai, bi) 2 s, we have 1(ai, bi) es 2(ai, bi). The order preserving property of fi ensures that fi ð1 ðai ; bi ÞÞ s fi ð2 ðai ; bi ÞÞ: Using r s, we obtain the following sequence of relations: f1 ð1 ða1 ; b1 ÞÞ s f1 ð2 ða1 ; b1 ÞÞ s f2 ð1 ða2 ; b2 ÞÞ s f2 ð2 ða2 ; b2 ÞÞ s :::::: s fn1 ð2 ðan1 ; bn1 ÞÞ s fn ð1 ðan ; bn ÞÞ s fn ð2 ðan ; bn ÞÞ s f1 ð1 ða1 ; b1 ÞÞ: The antisymmetry of s yields that all of the terms in the above sequence are Ì equal, thus is r-extending.
LINEAR ORDERS ON GENERAL ALGEBRAS
47
The compatible quasi order r A A on (A, F) has the finite extension property (FEP), if for any finite subset H A A, there exists a compatible partial order s A A on (A, F) such that r s and the elements of H are comparable pairs with respect to s. In view of Lemma 2.1., it is easy to see that the FEP of r is equivalent to the following equivalent conditions: The relation r is a partial order and for any finite subset H A A, there
exists an r-extending orientation : H Y A A.
For any finite subset H A A, there exists an orientation : H Y A A
such that the supremum r ðH Þ ¼ r _ q ð ð a; b Þ Þ _ ða;bÞ2H
in Quord(A, F) is a partial order. For any finite subset H A A, there exists a compatible partial order s
A A on (A, F) such that r ? (H) s for some orientation : H Y A A. Using an orientation : H Y A A and a function " : H Y {j1, 1}, a one to one correspondence between H Y A A orientations and H Y {j1, 1} functions can be established by 1 if ða; bÞ ¼ ða; bÞ "ða; bÞ ¼ and 1 if ða; bÞ ¼ ðb; aÞ ða; bÞ if "ða; bÞ ¼ 1 : ða; bÞ ¼ ðb; aÞ if "ða; bÞ ¼ 1 Since q(b, a) = qj1 (a, b), we can derive the following equivalent of the FEP: For any finite subset H A A, there exists a function " : H Y {j1, 1},
such that the supremum "ða; bÞ ða; bÞ r" ðH Þ ¼ r _ _ q ða; bÞ2H
in Quord(A, F) is a partial order. Our last equivalent of the FEP can be easily derived from the following properties of the quasi order q: 1. The map q 7! qj1 is an involution of the complete lattice Quord(A, F). 2. q is a partial order () q $ qj1 = DA. For any sequence (a1, b1), . . . , (an, bn) 2 A A of distinct pairs, we can find
exponents "1,"2, . . . , "n 2{j1, 1} such that ½r _ q"1 ða1 ; b1 Þ _ . . . _ q"n ðan ; bn Þ ^
r1 _ q"1 ða1 ; b1 Þ _ . . . _ q"n ðan ; bn Þ ¼ $A :
48
´ NDOR RADELECZKI AND JENO ´´ SZIGETI SA
LEMMA 2.2. Let a, b 2 A and r A A be a compatible quasi order on (A, F) with the FEP. Then one (or both) of the following compatible quasi orders on (A, F) has the FEP: ra;b ¼ r _ qða; bÞ and rb;a ¼ r _ qðb; aÞ: Proof. Suppose that neither ra,b nor rb,a has the FEP. Then we have finite 0 subsets H1, H2 A A, such that for any choice of the orientations : H1 Y 00 0 00 A A and : H2 Y A A the sets ra,b ? (H1) and rb,a ? (H2) are not contained in compatible partial orders on (A, F). Consider the finite subset H1 [ H2 [ fða; bÞg A A; the FEP of r provides a compatible partial order s A A on (A, F), such that r [ ðH1 [ H2 [ fða; bÞgÞ s for some orientation : H1 [ H2 [ fða; bÞg ! A A: If (a,b) = (a,b), then (a,b) 2 s implies that q(a,b) s, whence we obtain first 0 r ? q(a,b) s and next ra,b = r ¦ q(a,b) s. Now we have ra,b ? (H1) s 0 00 with = ÊH1, a contradiction. If (a,b) = (b,a), then we get rb,a ? (H2) s 00 Ì with = ÊH2, another contradiction. LEMMA 2.3. Let (A, F) be an algebra and rw A A, w 2 W is chain (with respect to ) of compatible quasi orders on (A, F), such that all rw have the FEP. Then w2W [ rw is a compatible quasi order on (A, F) with the FEP. Proof. It is easy to see that [w2W rw is a compatible quasi order on (A, F), so we have to prove only the FEP. Suppose that r ¼w2W [ rw has no FEP, then there exists a finite set H A A such that the quasi order r ðH Þ ¼ r _ _ qðða; bÞÞ ða;bÞ2H
is not a partial order for any choice of the orientation : H Y A A. Since r ðH Þ ¼ r [
[ qðða; bÞÞ
ða;bÞ2H
is not anti-symmetric, the properties of the quasi order r, as well as the properties of the transitive closure, ensure that for any choice of the orientation : H Y A A we can find a cycle of the form x1 ; y1 ; . . . ; xnðÞ ; ynðÞ ; xnðÞþ1 ;
49
LINEAR ORDERS ON GENERAL ALGEBRAS
n
o
where x1 ; y1 ; . . . ; xnðÞ ; ynðÞ A is not a one-element subset, xnðÞþ1 ¼ x1 and xi ; yi 2 q ai ; bi ; ð*Þ yi ; xiþ1 2 r ¼ [ rw w2W for some pairs ai ; bi 2 H, 1 e i e n(). Clearly, (*) implies that yi ; xiþ1 2 rwð;iÞ for some index w(,i) 2 W. Since
rwð;iÞ is an orientation of H; 1 i nðÞ is a finite subset of the chain frw jw 2 W g, there exists an orientation s : H Y A A and an integer 1 e j e n(s), such that rw (, i ) rw (s, j) for any choice of and 1 e i e n(). Now yi ; xiþ1 2 rwð; jÞ holds for all and 1 e i e n( ), whence we deduce, that rwð; jÞ [ [ qðða; bÞÞ ¼ rwð; jÞ _ _ qðða; bÞÞ ða; bÞ2H
ða; bÞ2H
is not a partial order for any choice of the orientation : H Y A A. Thus we Ì obtained the contradiction, that rwð; jÞ does not have the FEP. THEOREM 2.4. Let (A, F, r) be a partially ordered algebra, then the following conditions on r are equivalent: (1) r has the FEP. (2) r has a compatible linear extension, i.e., there exists a compatible linear order R A A on (A, F) such that r R. Proof. (1))(2): Take R ¼ fsjr s A A; is a compatible quasi
