Biordered sets of lattices and homogeneous basis
aa r X i v : . [ m a t h . R A ] J un BIORDERED SETS OF LATTICES AND HOMOGENEOUSBASIS
P. G. ROMEO AND AKHILA. R Abstract.
In this paper, we discuss the properties of the biordered set E P ( L ) obtained from the complemented modular lattice L , defines anoperation ⊕ using the sandwich elements of the biordered set. Further,we describe the biordered subset E P ( L ) of E P ( L ) satisfying certain condi-tions, so that the complemented modular lattice admits a homogeneousbasis. Finally, analogous to von Neumann’s coordinatization theoremwe describe the coordinatization theorem for complemented modularlattice using the biordered set of idempotents E P ( L ) . Introduction
The concept of regular rings was introduced by John von Neumann in apaper "On Regular Rings" in 1936. He used regular rings as an algebraictool for studying certain lattice of projections on algebra of operators ona Hilbert space. A lattice L is said to be coordinatized by a regular ring R , if L is isomorphic to the principal left[right] ideals of the regular ring.von Neumann proved that every complemented modular lattice with ordergreater than or equal to 4 is coordinalizable (see [8]).The multiplicative reduct of a regular ring is a regular semigroup andthus it is obvious that the study of regular semigroups play a singificantrole in the study of regular rings. In order to study the structure of regularsemigroup, in his celebrated 1973, paper K.S.S. Nambooripad introducedthe concept of a biordered sets to study the structure of idempotents of aregular semigroup (cf.[9]). He defined a biordered set as an order structureto represent the set of idempotents of a semigroup and identified a partialbinary operation on the set of all idempotents E ( S ) of the semigroup S aris-ing from the underlined binary operation in S and defined two quasiorders ω r and ω l on the set E ( S ) satisfying certain axioms (see the definition be-low) which is abstracted as a biordered set.In [2] we extend the biordered set approch from regular semigroups toregular rings by explicitly desceribing the structure of the multiplicativeidempotents E R of a regular ring R with two quasi orders ω l and ω r as a Mathematics Subject Classification.
Key words and phrases.
Biordered Sets, Sandwich sets, Homogeneous Basis, Comple-mented Modular Lattices. AND AKHILA. R bounded and complemented biorderd set . In [1], we describe the biorderideals generated by elements in E R as { ω l ( e ) : e ∈ E R } and { ω r ( e ) : e ∈ E R } and are the complemented modular lattice Ω L and Ω R respectively. Furtherwe also discuss certain interesting propertes of the complemented modularlattice such as perspectivity and independence of the lattce and obtained thenecessary and suffient condition at which the complementd modular latticeis of order n .The converse problem of describing the structure which coordinatise agiven complemented modular lattice is attempted by Pastjin in the caseof a strongly regular baer semigroup (see cf.[10]). He defined the normalmappings on a complemented modular lattice L using complementary pairsand it is shown that these normal mappings form a semigroup P ( L ) and theset of idempotents E P ( L ) of P ( L ) is the biordered set of the complementedmodular lattice and L can be coordinatized by P ( L ).Here we extend Pastjin’s approach to obtain a regular ring coordinatizingthe complemented modular lattice L . For, we note that the biordered set E P ( L ) obtained from the complemented modular lattice L is a biorderedset with least and greatest element, each element having an inverse and forelements satisfying v i ≤ n j in L , an operation in E P ( L ) can be defined usingthe sandwich sets and we observe some properties of this operation suchas cancellation, inverse. It is also seen that on some conditions in E P ( L ) ,the complemented modular lattice L admits a homogeneous basis and thislattice L corresponds to the lattice Ω L the lattice of all ω l -ideals of a ring R (cf.[1]). 2. Preliminaries
Let S be a regular semigroup, and E ( S ) the set of idempotents of S .The relation ω l [ ω r ] on E ( S ) is defined by e ω l f [ e ω r f ] if and only if ef = e [ f e = e ] is a quasiorder on E ( S ); the sets, ω r ( e ) = { f : ef = f } , ω l ( e ) = { f : f e = f } are principal ideals and the relations R = ω r ∩ ( ω r ) − , L = ω l ∩ ( ω l ) − ,and ω ( e ) = ω r ( e ) ∩ ω l ( e ) for all e ∈ S are equivalences and partial orderrespectively. Definition 1.
Let E be a partial algebra. Then E is a biordered set if thefollowing axioms and their duals hold:(1) ω r and ω l are quasi orders on E and D E = ( ω r ∪ ω l ) ∪ ( ω r ∪ ω l ) − (2) f ∈ ω r ( e ) ⇒ f R f eωe (3) gω l f and f, g ∈ ω r ( e ) ⇒ geω l f e. (4) gω r f ω r e ⇒ gf = ( ge ) f IORDERED SETS OF LATTICES AND HOMOGENEOUS BASIS 3 (5) gω l f and f, g ∈ ω r ( e ) ⇒ ( f g ) e = ( f e )( ge ) . Let M ( e, f ) denote the quasi ordered set ( ω l ( e ) ∩ ω r ( f ) , < ) where < is defined by g < h ⇔ eg ω r eh, and gf ω l hf. Then the set S ( e, f ) = { h ∈ M ( e, f ) : g < h for all g ∈ M ( e, f ) } is called the sandwich set of e and f .(6) f, g ∈ ω r ( e ) ⇒ S ( f, g ) e = S ( f e, ge )We shall often write E = h E, ω l , ω r i to mean that E is a biordered setwith quasi-orders ω l , ω r . The relation ω defined is a partial order and ω ∩ ( ω ) − ⊂ ω r ∩ ( ω l ) − = 1 E . The biordered set E is said to be regular if S ( e, f ) = ∅ for all e, f ∈ E .Regular semigroups which determine the same biordered set will be calledbiorder isomorphic. Definition 2.
Let e and f be idempotents in a semigroup S . By an E -sequence from e to f , we mean a finite sequence e = e, e , e , ..., e n − , e n = f of idempotents such that e i − ( L∪R ) e i for i = 1 , , ..., n ; n is called the lengthof the E -sequence. If there exists an E -sequence from e to f , d ( e, f ) is the length of theshortest E -sequence from e to f ; and d ( e, e ) = 1. If there is no E -sequencefrom e to f , we define d ( e, f ) = 0. Definition 3. (cf. [5] ) If P is a partially ordered set and Φ : P → P is anisotone(order preserving) mapping, the Φ will be called normal if(1) im Φ is a principal ideal of P and(2) whenever x Φ = y , then there exists some z ≤ x such that Φ mapsthe principal ideal P ( z ) isomorphically onto the principal ideal P ( y ) . Definition 4. (cf. [5] ) The partially ordered set P will be called regular iffor every e ∈ P , P ( e ) = im Φ for some normal mapping Φ : P → P with Φ = Φ . If P is a partially ordered set, then it is easy to see that the set S ( P )[ S ∗ ( P )]of normal mappings of P into itself, considered as left [right] operators forma regular semigroup. Definition 5.
A lattice is a partially ordered set in which each pair ofelements has the least upper bound and the greatest lower bound. If a and b are elements of a lattice, we denote their greatest lower bound (meet) andleast upper bound (join) by a ∧ b and a ∨ b respectively. It is easy to see that a ∨ b and a ∧ b are unique. Definition 6.
A lattice is called modular (or a Dedekind lattice) if ( a ∨ b ) ∧ c = a ∨ ( b ∧ c ) for all a ≤ c. P. G. ROMEO AND AKHILA. R A lattice is bounded if it has both a maximum element and a minimumelement, we use the symbols 0 and 1 to denote the minimum element andmaximum element of a lattice. A bounded lattice L is said to be comple-mented if for each element a of L , there exists at least one element b suchthat a ∨ b = 1 and a ∧ b = 0. The element b is referred to as a complement of a . It is quite possible for an element of a complemented lattice to have manycomplements. An element x is called a complement of a in b if a ∨ x = b and a ∧ x = 0 . Definition 7.
The elements x , x , · · · x n of a lattice are called independentif ( x ∨ · · · ∨ x i − ∨ x i +1 ∨ · · · ∨ x n ) ∧ x i = 0 for every i . Definition 8.
Two elements a and b in a lattice L are said to be perspective(in symbols a ∼ b ) if there exists x in L such that a ∨ x = b ∨ x, a ∧ x = b ∧ x = 0 . Such an element x is called an axis of perspective. Definition 9.
Let L be a complemented, modular lattice with zero andunit . By a basis of L is meant a system ( a i : i = 1 , , . . . n ) of n elementsof L such that ( a i ; i = 1 , , . . . n ) are independent , a ∨ a ∨ . . . ∨ a n = 1 A basis is homogeneous if its elements are pairwise perspective. a i ∼ a j ( i, j = 1 , , . . . n ) The number of n elements in a basis is called the order of the basis. Definition 10.
A complemented modular lattice L is said to have order m in case it has a homogeneous basis of order m . Biordered set of the complemented modular lattice and itsHomogeneous basis
Let L be a complemented modular lattice and ( n ; v ) be any pair of com-plementary elements of L Let ( n ; v ) : L → L, be the map defined by x → v ∧ ( n ∨ x ) for all x in L and ( n ; v ) ′ : L → L defined by x → n ∨ ( v ∧ x ) for all x in L are idempotent order preserving normal mappings of L onto [0 , v ] of ( L, ∧ )acts on L as right operators denoted by P ( L ) the subsemigroup of S ∗ ( L )which is generated by these idempotent normal mappings ( n ; v ) , n, v ∈ L andthe mappings ( n ; v ) ′ which are order preserving idempotent normal mappingof L onto the principal ideal [1 , n ] of ( L, ∨ ) into itself. Letting ( n ; v ) ′ acton L as left operators, denote by P ( L ) ′ the subsemigroup of S ( L ) which is IORDERED SETS OF LATTICES AND HOMOGENEOUS BASIS 5 generated by these idempotent normal mappings ( n ; v ) ′ , n, v ∈ L .Let E P ( L ) = { ( n ; v ) : n, v ∈ L n ∨ v = 1 , n ∧ v = 0 } and E P ( L ) ′ = (cid:8) ( n ; v ) ′ : n, v ∈ L n ∨ v = 1 , n ∧ v = 0 (cid:9) we refer to the elements ( n ; v )[( n ; v ) ′ ] as idempotent generators of P ( L )[ P ( L )] ′ . Theorem 1 (cf.[10], Theorem 1) . Let L be a complemented modularlattice. Then(1) P ( L ) is a regular subsemigroup of S ∗ ( L ) and E P ( L ) = { ( n ; v ) : n, v ∈ L n ∨ v = 1 , n ∧ v = 0 } . (2) In ( E P ( L ) , ω l , ω r ) we have ( n ; v ) ω l ( n ; v ) ⇐⇒ v ≤ v in L and then ( n ; v )( n ; v ) = ( n ∨ ( v ∧ n ); v ); we have ( n ; v ) ω r ( n ; v ) ⇐⇒ n ≤ n in L and then ( n ; v )( n ; v ) = ( n ; v ∧ ( n ∨ v )) (3) Let ( n ; v ) and ( n ; v ) be any idempotent of P ( L ) . Let n be anycomplement of v ∨ n in [ n , ; let v be any complement of v ∧ n in [0 , v ] ; the n and v are complementary in L and ( n ; v ) is an elementin the sandwich set S (( n ; v ) , ( n ; v )) . Conversely, any element inthe sandwich set S (( n ; v ) , ( n ; v )) can be obtained in this way. The above theorem provides a biordered set E P ( L ) from a complementedmodular lattice L . The zero of P ( L ) is (1; 0) and the identity is (0; 1).Obviously (1; 0) and (0; 1) are in E P ( L ) and for any ( n ; v ) in biordered set E P ( L ) , ( n ; v ) ω (0; 1) and (1; 0) ω ( n ; v ). Lemma 1.
Let ( n ; v ) , ( n ; v ) ∈ E P ( L ) , then(1) S (( n ; v ) , ( n ; v )) = S (( n ; v ) , ( n ; v )) = { (1; 0) } if and only if v ≤ n and v ≤ n .(2) For v ≤ n and v ≤ n , ( v ∨ v ; n ∧ n ) is the unique element in S (( v ; n ) , ( v ; n )) and S (( v ; n ) , ( v ; n )) .Proof. (1) ( n ; v ) , ( n ; v ) ∈ E P ( L ) , v ≤ n and ( n ; v ) ∈ S (( n ; v ) , ( n ; v )).Then by definition of sandwich set n is a complement of n in [ n ,
1] and v isa complement of v in [0 , v ]. Since, ( n ; v ) ω l ( n ; v ) and ( n ; v ) ω r ( n ; v ),it follows that n ≤ n and n ∨ n = 1 implies n = 1 and v ≤ v and v ∧ v = 0 implies v = 0. Thus S (( n ; v ) , ( n ; v )) = { (1; 0) } . Similarly, P. G. ROMEO AND AKHILA. R S (( n ; v ) , ( n ; v )) = { (1; 0) } when v ≤ n . The converse follows immedi-ately.(2) For v ≤ n and v ≤ n , by definition,( v ; n )( v ; n ) = ( v ∨ ( n ∧ v ); n ∧ ( v ∨ n )) = ( v ∨ v ; n ∧ n )and ( v ; n )( v ; n ) = ( v ∨ ( n ∧ v ); n ∧ ( v ∨ n )) = ( v ∨ v ; n ∧ n )thus ( v ; n )( v ; n ) = ( v ; n )( v ; n ) and is easy to see that ( v ∨ v ) is acomplement of v ∨ n in [ v ,
1] and ( n ∧ n ) is a complement of v ∧ n in[0 , n ]. Thus ( v ∨ v ; n ∧ n ) ∈ S (( v ; n ) , ( v ; n )). Similarly, ( v ∨ v ) isa complement of v ∨ n in [ v ,
1] and ( n ∧ n ) is a complement of v ∧ n in [0 , n ]. Therefore, ( v ∨ v ; n ∧ n ) ∈ S (( v ; n ) , ( v ; n )).It remains to prove the uniqueness of this element, suppose there existsanother element ( a ; a ′ ) ∈ S (( v ; n )( v ; n )) ∩ S (( v ; n )( v ; n )). Then v ≤ a, a ′ ≤ n , v ≤ a, a ′ ≤ n and from the definition of sandwich set it can beseen that a ∨ n = 1 , a ′ ∧ v = 0 , a ∨ n = 1 , a ′ ∧ v = 0 and a ∧ n ≤ v , n ≤ a ′ ∨ v , a ∧ n ≤ v , n ≤ a ′ ∨ v . Thus a = v ∨ v and a ′ = n ∧ n . and( a ; a ′ ) = ( v ; n )( v ; n ) is unique. (cid:3) Thus for each ( n ; v ) ∈ E P ( L ) there exists an element ( v ; n ) ∈ E P ( L ) suchthat ( n ; v )( v ; n ) = ( v ; n )( n ; v ) = (1; 0) and call this element ( v ; n ) the inverseof ( n ; v ). Also • ( n ; v ) ω l ( n ; v ) ⇐⇒ ( v ; n ) ω r ( v ; n ). • v ≤ n ⇐⇒ S (( n ; v )( n ; v )) = (1; 0)from here onwards we consider the biordered subset of E P ( L ) satisfying v i ≤ n j for all ( n i ; v i ), ( n j ; v j ) and i = j . Clearly in this biordered subset( v i ; n i )( v j ; n j ) = ( v j ; n j )( v i ; n i ) = ( v i ∨ v j ; n j ∧ n i ). Define( n i ; v i ) ⊕ ( n j ; v j ) = ( n i ∧ n j ; v i ∨ v j ). Lemma 2.
Consider the biordered subset of E P ( L ) with v i ≤ n j for i = j and let ( p ; q ) denotes ( n i ; v i ) ⊕ ( n j ; v j ) . Then we have the following:(1) ( n i ; v i ) , ( n j ; v j ) ∈ ω (( p ; q )) (2) If ( r ; s ) ∈ E P ( L ) with ( n i ; v i ) , ( n j ; v j ) ∈ ω l (( r ; s )) then ( p ; q ) ∈ ω l (( r ; s )) .(3) If ( r ; s ) ∈ E P ( L ) with ( n i ; v i ) , ( n j ; v j ) ∈ ω r (( r ; s )) , then ( p ; q ) ∈ ω r (( r ; s )) .Proof. (1) Note that ( p ; q ) ∈ S (( n i ; v i )( n j ; v j )) ∩ S (( n j ; v j ) , ( n i ; v i )). There-fore, ( q ; p ) ω l ( v i ; n i ) , ( q ; p ) ω r ( v j ; n j ) , ( q ; p ) ω l ( v j ; n j ) and ( q ; p ) ω r ( v i ; n i ).Thus p ≤ n i , p ≤ n j , v j ≤ q, v i ≤ q and ( n i ; v i ) ω ( p ; q ) and ( n j ; v j ) ω ( p ; q ). IORDERED SETS OF LATTICES AND HOMOGENEOUS BASIS 7 (2) Let ( r ; s ) ∈ E P ( L ) with ( n i ; v i ) ω l ( r ; s ) and ( n j ; v j ) ω l ( r ; s ), then v i ≤ s and v j ≤ s . Then as seen above in Lemma 1.(2),( q ; p ) = ( v i ; n i )( v j ; n j ) . Thus, v j ≤ s implies ( s ; r ) ω r ( v j ; n j ), that is ( v j ; n j )( s ; r ) = ( s ; r ). Simi-larly, v i ≤ s implies ( s ; r ) ω r ( v i ; n i ), that is ( v i ; n i )( s ; r ) = ( s ; r ).Therefore, ( q ; p )( s ; r ) = ( v i ; n i )( v j ; n j )( s ; r ) = ( s ; r )that is ( s ; r ) ω r ( q ; p ) also ( p ; q ) ω l ( r ; s ). In a similar manner (3) follows. (cid:3) The next lemma shows that the opertion ⊕ is cancellative. Lemma 3.
Let ( n i ; v i ) , ( n j ; v j ) , ( n k ; v k ) ∈ E P ( L ) with v i ≤ n j , v j ≤ n i and v i ≤ n k , v k ≤ n i for i = j = k . Then ( n i ; v i ) ⊕ ( n j ; v j ) = ( n i ; v i ) ⊕ ( n k ; v k ) if and only if ( n j ; v j ) = ( n k ; v k ) .Proof. Suppose ( n j ; v j ) = ( n k ; v k ), then( n i ; v i ) ⊕ ( n j ; v j ) = ( n j ∧ n i ; v i ∨ v j ) = ( n k ∧ n i ; v i ∨ v k ) = ( n i ; v i ) ⊕ ( n k ; v k ) . Conversely suppose that ( n i ; v i ) ⊕ ( n j ; v j ) = ( n i ; v i ) ⊕ ( n k ; v k ). Then( n j ∧ n i ; v i ∨ v j ) = ( n i ∧ n j ; v j ∨ v i ) = ( n k ∧ n i ; v i ∨ v k ) = ( n i ∧ n k ; v k ∨ v i ) . Since v j ≤ n i and v i ≤ n j , ( n j ; v j )( n i ; v i ) = ( n j ; v j ) and( n j ; v j )( v k ; n k ) = ( n j ; v j )( v i ; n i )( v k ; n k )= ( n j ; v j )( v i ∨ v k ; n i ∧ n k )= ( n j ; v j )( v j ∨ v i ; n j ∧ n i )= ( n j ; v j )( v j ; n j )( v i ; n i )= (1; 0)( v i ; n i )= (1; 0)therefore, S (( n j ; v j )( v j ; n j )) = { (1; 0) } and so v j ≤ v k and( v k ; n k )( n j ; v j ) = ( v k ; n k )( v i ; n i )( n j ; v j )= ( v k ∨ v i ; n i ∧ n k )( n j ; v j )= ( v i ∨ v j ; n j ∧ n i )( n j ; v j )= ( v i ; n i )( v j ; n j )( n j ; v j )= ( v i ; n i )(1; 0)= (1; 0)Therefore, S (( v k ; n k ) , ( n j ; v j )) = { (1; 0) } and so n j ≤ n k . Interchanging( n k ; v k ) and ( n j ; v j ), n j ≤ n k , thus n j = n k and v j = v k . That is, ( n j ; v j ) =( n k ; v k ). (cid:3) Corollary 1.
Let ( n i ; v i ) , ( n j ; v j ) ∈ E P ( L ) with v j ≤ n i for i = j . Then ( n i ; v i ) ⊕ ( n j ; v j ) = (0; 1) if and only if ( n j ; v j ) = ( v i ; n i ) . P. G. ROMEO AND AKHILA. R Proof.
By Lemma.1 we have S (( n i ; v i ) , ( v i ; n i )) = S (( v i ; n i ) , ( n i ; v i )) = { (1; 0) } . Therefore S (( n i ; v i ) , ( v i ; n i )) ∩ S (( v i ; n i ) , ( n i ; v i )) = { (1; 0) } . ie., (( v i ∨ n i ; n i ∧ v i )) = (1; 0), and thus( n i ; v i ) ⊕ ( v i ; n i ) = ( n i ∧ v i ; v i ∨ n i ) = (0; 1) . Conversely, suppose ( n i ; v i ) ⊕ ( n j ; v j ) = (1; 0), since ( n i ; v i ) ⊕ ( v i ; n i ) = (1; 0)it follows that ( n j ; v j ) = ( v i ; n i ). (cid:3) Lemma 4.
Let E P ( L ) be the biordered set with v i ≤ n j , v j ≤ n i , v i ≤ n k , v k ≤ n i , v j ≤ n k , v k ≤ n j for i = j = k . Then for elements ( n i ; v i ) , ( n j ; v j ) , ( n k ; v k ) , i, j, k = 1 , , . . . , N with i = j = k in E P ( L ) , thecollection { v , v , . . . , v N } are independent elements in the lattice L .Proof. We have the set { ( n i ; v i ) : i = 1 , , . . . , N } in E P ( L ) so that the el-ements v , v , . . . , v N are in the complemented modular lattice L . Since v i ≤ n k and v j ≤ n k for i = j in E P ( L ) , we have ( v i ∨ v j ) ≤ n k for i = j = k ,then ( v i ∨ v j ) ∧ v k ≤ n k ∧ v k = 0 . Hence the collection { v i : i = 1 , . . . , N } are independent. (cid:3) Now recall from [1], that given a regular ring R whose biordered set ofidempotents E R then the principal biorder ideals ω l ( e ) [ ω r ( e )] , e ∈ E R is thecomplemented modular lattice Ω L [Ω R ]. Also we have the following theorem Theorem 2. ( [1] , Theorem 6) Let R be a regular ring with idempotents e , e , · · · , e n such that M ( e , e j ) = { } for i = j, d l ( e , e j ) ≤ and e + e + · · · + e n = 1 . Then a maximal system ω l ( e ) , · · · ω i ( e n ) in Ω L formsa homogeneous basis of rank n and the complemented modular lattice is oforder n . It is easy to observe that E P ( L ) has elements { ( n i ; v i ) : i = 1 , , . . . , N } satisfying the following properties:(1) v i ≤ n j for i = j (2) ( n ; v ) ⊕ ( n ; v ) ⊕ . . . ⊕ ( n N ; v N ) = (0; 1)(3) d l (( n i ; v i ) , ( n j ; v j )) = 3 for i = j .and this subset of E P ( L ) is again a biordered set and we denote this biorderedset as E P ( L ) . Let L be the complemented modular lattice with E P ( L ) = E P ( L ) then L is of order N .The next lemma gives a biorder isomorphism between the biordered setof idempotents in the ring R and the biordered set E P (Ω L ) . Lemma 5.
Every idempotent e in a ring R is associated with a pair ( ω l (1 − e ); ω l ( e )) of complementary biorder ideals in Ω L . The map ǫ : E R → E P (Ω L ) defined by ǫ ( e ) = ( ω l (1 − e ); ω l ( e )) is a biorder isomorphism. IORDERED SETS OF LATTICES AND HOMOGENEOUS BASIS 9
Proof.
For each e ∈ E R , ( ω l (1 − e ); ω l ( e )) is a complementary pair in thelattice Ω L and the mapping ǫ : e → ( ω l (1 − e ); ω l ( e )) for all e ∈ E R is a map of E R into E P (Ω l ) . The map ǫ is clearly injective. It followsfrom the definition of biordered set ([9], Definition 1) and the equation.1 inTheorem(1) [10] that the map ǫ preserve basic products and hence ǫ : E R → E P (Ω L ) is a biorder isomorphism. Also, it can be easily seen that this map ǫ is a regular bimorphism. (cid:3) Thus we have E P (Ω L ) and E R are biorder isomorphic. Now we show thatthere exists elements e , e , . . . , e N in E R satisfying all the conditions of E P (Ω L ) . Consider E P (Ω L ) , then there are elements (( ω l (1 − e i ); ω l ( e i ) : i =1 , , . . . , N ) such that(1) ω l (1 − e i ) ≤ ω l ( e j ) for i = j (2) ( ω l (1 − e ); ω l ( e )) ⊕ ( ω l (1 − e ); ω l ( e )) ⊕ . . . ⊕ ( ω l (1 − e N ); ω l ( e N )) =(0; 1)(3) d l (( ω l (1 − e i ); ω l ( e i )) , ( ω l (1 − e j ); ω l ( e j ))) = 3Since E P (Ω L ) and E R are biorder isomorphic, for each (( ω l (1 − e i ); ω l ( e i ) : i =1 , , . . . , N ), there exists elements e , e , . . . , e N such that(1) ω l (1 − e i ) ≤ ω l ( e j ) for i = j implies ( ω l (1 − e i ); ω l ( e i ))( ω l (1 − e j ); ω l ( e j )) = { (0; 1) } . But ω l (1 − e i ) ≤ ω l ( e j ) implies 1 − e i ω l e j thus e i ω r (1 − e j ) and so e i e j = 0 for i = j .(2) Since ( ω l (1 − e ); ω l ( e )) ⊕ . . . ⊕ ( ω l (1 − e N ); ω l ( e N )) = (0; ω l (1)) whichimplies ω l ( e ) ∨ ω l ( e ) ∨ . . . ∨ ω l ( e N ) = ω l (1). But we have from ([1],Lemma 9) e i e j = 0 for i = j implies ω l ( e ) ∨ ω l ( e ) ∨ . . . ∨ ω l ( e N ) = ω l ( e + e + . . . + e N ) = ω l (1) and hence e + e + . . . + e N = 1.(3) d l (( ω l (1 − e i ); ω l ( e i )) , ( ω l (1 − e j ); ω l ( e j ))) = 3 implies there exists ele-ments ( ω l (1 − e i ); ω l ( e i )) L ( ω l (1 − e h ); ω l ( e h )) R ( ω l (1 − e k ); ω l ( e k )) L ( ω l (1 − e j ); ω l ( e j )). Therefore, by definition of L and R , ω l ( e i ) = ω l ( e h )and ω l (1 − e h ) = ω l (1 − e k ) and ω l ( e k ) = ω l ( e j ). But ω l (1 − e h ) = ω l (1 − e k ) implies ω r ( e h ) = ω r ( e k ). Thus we get e i L e h R e k L e j andhence d l ( e i , e j ) = 3.Since E P (Ω L ) is a biorder subset of E P (Ω L ) and corresponding to each elementin E P (Ω L ) there exists elements in E R satisfying all the conditions of E P (Ω L ) as shown above, we have E P (Ω L ) and E R (where E R is the biordered setsimilar to E P ( L ) ) are biorder isomorphic.Thus it can be seen that for given a complemented modular lattice L there exists a complemnted modular lattice L admiting a biordered subset E P ( L ) of the biordered set E P ( L ) consisting of N elements and hence L admits a homogeneous basis of order N . Thus analogous to von-Neumann’scoordinatization theorem, we have the following theorem: AND AKHILA. R Theorem 3.
Let L be a complemented modular lattice admitting a biorderedsubset with at least elements, having the following properties:(1) v i ≤ n j for i = j (2) ( n ; v ) ⊕ ( n ; v ) ⊕ . . . ⊕ ( n N ; v N ) = (0; 1) (3) d l (( n i ; v i ) , ( n j ; v j )) = 3 for i = j ,then there exists a von Neumann regular ring R such that L is isomorphicto the lattice of all principal left ideals of R . References [1] Akhila R and P. G. Romeo, On the Lattice of Biorder Ideals of Regular Rings,
Asian European Journal of Mathematics , Vol 13. No.1, World Scientific PublishingCompany, (2020).[2] Akhila R, Study on Regular Rings- A Biordered Set Approach,
PhD theisis , Depart-ment of Mathematics, Cochin University of Science and Technology , Kerala, India,2018.[3] Cesar Polcino Milies and Sudarshan K. Sehgal,
An Introduction to Group Rings ,Kluwer Academic Publishers, Dordrecht/Boston/London, ISBN 1-4020-0238-6,(2002).[4] A. H. Clifford and G. B. Preston,
The Algebraic Theory of Semigroups , Volume Math. Surveys of the American. Math. Soc.7, Providence, R. I. (1964).[5] P. A. Grillet, The structure of regular semigroups. I. A representation,
SemigroupForum ., 177-183 (1974).[6] P. A. Grillet, Semigroups : An Introduction to the Structure Theory , Monographsand Textbooks in Pure and Applied Mathematics, . Marcel Dekker Inc, New York(1995).[7] J. M. Howie,
An Introduction To Semigroup Theory , Academic Press Inc. (London)(1976).[8] John von Neumann(1960): Continuous Geometry, Princeton University Press, Prince-ton, New Jersey, ISBN-13:987-069105893[9] K.S.S. Nambooripad, Structure of Regular Semigroups (
MEMOIRS , No. ), Amer-ican Mathematical Society (December 1979).[10] F.Pastjin, Biordered sets and complemented modular lattices,
Semigroup Forum
Vol. , (1980) 205-220.[11] Skornyakov L. A., Complemented Modular Lattice and Regular Rings , Oliver andBoyd(1964). Professor, Department of Mathematics, Cochin University of Science andTechnology, Kochi, Kerala, INDIA. Assistant Professor, Department of Mathematics, Amrita School of Artsand Sciences, Kochi, Amrita Vishwa Vidyapeetham, INDIA
E-mail address : romeo − parackal @ yahoo.com , akhila.ravikumar @@