# On Regular Elements in an Incline

Authors: A. R. Meenakshi; S. Anbalagan
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

Inclines are additively idempotent semirings in which products are less than (or) equal to either factor. Necessary and sufficient conditions for an element in an incline to be regular are obtained. It is proved that every regular incline is a distributive lattice. The existence of the Moore-Penrose inverse of an element in an incline with involution is discussed. Characterizations of the set of all generalized inverses are presented as a generalization and development of regular elements in a∗-regular ring.


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## Body

## 1. Introduction

The notion of inclines and their applications are described comprehensively in Cao et al. [1]. Recently, Kim and Roush have surveyed and outlined algebraic properties of inclines and of matrices over inclines [2]. Multiplicative semigroups unlike matrices over a field are not regular; that is, it is not always possible to solve the regularity equation axa=a. If there exists x, x is called a g-inverse of a and the element a is said to be regular. This concept of regularity of elements in a ring goes back to Neumann [3]. If every element in a ring is regular, then it is called a regular ring. Regular rings are important in many branches of mathematics, especially in matrix theory, since the regularity condition is a linear condition that solves linear equations and takes the place of canonical decomposition.In [4], Hartwig has studied on existence and construction of various g-inverses associated with an element in a *-regular ring, that is, regular ring with an anti-automorphism and developed a technique for computing g-inverses mainly by using star cancellation law. In semirings one of the most important aspects of structure is a collection of equivalence relations called Green’s relations and the corresponding equivalence classes. In [2], it is stated that an element is regular if and only if the equivalence 𝔇 class contains an idempotent.In this paper, we exhibit that Green’s equivalence relations on a pair of elements in an incline reduce to the equality of elements. This leads to the characterization of regular element in an incline that is, an element in an incline is regular if and only if it is idempotent and structure of set of allg-inverses of an element in an incline with involution. In Section 2, we present the basic definitions, notations, and required results on inclines. In Section 3, some characterization of regular elements in an incline are obtained as a generalization of regular elements in a *-regular ring studied by Hartwig and as a development of results available in a Fuzzy algebra. The invariance of the product ba-cfor elements a,b,c in a regular incline and a g-inverse a- of a is discussed. For elements in a regular incline it is proved that equality of right ideals coincides with equality of left ideals. In Section 4, equivalent conditions for the existence of the Moore-Penrose inverse of an element in an incline with involution-T are determined.Green’s equivalence relation reduces to equality of elements. We conclude that the proofs are purely based on incline property without using star cancellation law as in the work of Hartwig [4].


## 2. Preliminaries

In this section, we give some definitions and notations.Definition 2.1.
An incline is a nonempty setR with binary operations addition and multiplication denoted as +, · defined on R·R→Rsuch that for all x, y, z∈R(2.1)x+y=y+x,x+(y+z)=(x+y)+z,x(y+z)=xy+xz,(y+z)x=yx+zx,x(yz)=(xy)z,x+x=x,x+xy=x,y+xy=y.Definition 2.2.
An inclineR is said to be commutative if xy=yx for all x,y∈R.Definition 2.3.
(R,≤) is an incline with order relation “≤” defined on R such that for x,y∈R, x≤y if and only if x+y=y. If x≤y, then y is said to dominate x.Property 2.4.
Forx,y in an incline R, x+y≥xand x+y≥y.
Forx+y=(x+x)+y=x+(x+y), and x+y=x+(y+y)=(x+y)+y
Thusx+y≥x and x+y≥y.Property 2.5.
Forx, y in an incline R, xy≤x and xy≤y.
Throughout letR denote an incline with order relation ≤. For an element a∈R, aR={ax/x∈R} is the right ideal of a and Ra={xa/x∈R} is the left ideal of a.Definition 2.6 (Green’s relation [5]).
For any two elementsa, b in a semigroup S.(i)
aℒb if there exist x,y∈S such that xa=b and yb=a.(ii)
aℛb if there exist x,y∈S such that ax=b and by=a.(iii)
a𝒥b if there exist w,x,y,z such that wax=b, ybz=a.(iv)
aℋb if aℛb and aℒb.(v)
a𝔇b if there exists c∈S such that aℛc and cℒb.


## 3. Regular Elements in an Incline

In this section, equivalent conditions for regularity of an element in an incline are obtained and it is proved that a regular commutative incline is a distributive lattice. The equality of right (left) ideals of a pair of elements in a regular incline reduces to the equality of elements. This leads to the invariance of the productba-c for all choice a- of a and a,b,c in a regular incline. Characterization of the set of all g-inverses of an element in terms of a particular g-inverse is determined.Just for sake of completeness we will introduceg-inverses of an element in an incline.Definition 3.1.
a∈R is said to be regular if there exists an element x∈R such that axa=a. Then x is called a generalized inverse, in short g-inverse or 1-inverse of a and is denoted as a-. Let a{1} denotes the set of all 1-inverses of a.Definition 3.2.
An elementa∈R is called antiregular, if there exists an element x∈R such that xax=x. Then x is called the 2-inverse of a. a{2} denotes the set of all 2-inverses of a.Definition 3.3.
Fora∈R if there exists x∈R such that axa=a, xax=x, and ax=xa, then x is called the Group inverse of a. The Group inverse of a is a commuting 1-2 inverse of a.
An inclineR is said to be regular if every element of R is regular.Example 3.4.
The Fuzzy algebraℱ with support [0,1] under the max. min. operation is an incline [2]. Each element in ℱ is regular as well as idempotent [6, page 212]. Thus ℱ is a regular incline.Example 3.5.
LetD={a,b,c} and R=(𝒫(D),∪,∩), where (𝒫(D)) the power set of D is an incline. Here for each element x∈𝒫(D), x2=x∩x=x. Hence x is idempotent and x is regular (refer Proposition 3.7). Thus R is a regular incline.Lemma 3.6.
Leta∈R be regular. Then a=ax=xa for all x∈a{1}.Proof.
Ifa is regular, then by Property 2.5(3.1)a=axa≤ax≤a.
Thereforeax=a.
Similarly, froma≤xa≤a, it follows that a=xa. Thus, a=xa=ax for all x∈a{1}.Proposition 3.7.
Fora∈R, ais regular if and only if a is idempotent.Proof.
Leta∈R be regular. Then by Lemma 3.6,  a=ax=xa for all x∈a{1}. a=axa=(ax)a=a·a=a2. Thus a is idempotent.
Converse is trivial.Example 3.8.
Let us consider the example72=([0,1],sup(x,y),xy) of an incline given in [2]. Here xy is usual multiplication of real numbers. Hence for each x∈72, x2≤x and x, is not idempotent. Therefore by Proposition 3.7, 72 is not a regular incline.Proposition 3.9.
Ifa is regular, then a is the smallest g-inverse of a, that is, a≤x for all x∈a{1}.Proof.
Leta be regular, then by Proposition 3.7, a∈a{1}. By Lemma 3.6  a=ax for all x∈a{1}. Hence by Property 2.5  a≤x. Thus a is the smallest g-inverse of a.It is well known that [7] every distributive lattice is an incline, but an incline need not be a distributive lattice. Now we shall show that regular commutative incline is a distributive lattice in the following.Proposition 3.10 (see [1]).
A commutative incline is a distributive lattice as (semiring) if and only ifx2=x for all x∈X.Lemma 3.11 (see [7]).
DL is a distributive lattice. (DL is the set of all idempotent elements in an incline L.)Proposition 3.12.
LetR be a commutative incline, R is regular ⇔R is a distributive lattice.Proof.
LetR is commutative incline.
R is regular: ⇔ every element in R is idempotent (by Proposition 3.7),
⇔DR=R, where DR is the set of all idempotent elements of R,
⇒R=DR is distributive lattice (by [7, Lemma  2.1]).
Conversely, ifR is a distributive lattice then by Proposition 111 in [1] every element of R is idempotent, again by Proposition 3.7R is a regular incline.Next we shall see some characterization of regular elements in an incline.Theorem 3.13.
Fora∈R, the following are equivalent: (i)
a is regular,(ii)
a is idempotent,(iii)
a{1,2}={a},(iv)
group inverse ofa exists and coincides with a,(v)
a=va2 for some v∈a{1},(vi)
a=a2u for some u∈a{1}.In either case v,u,vau are all g-inverses of a and vau is invariant for all choice of u,v∈a{1}. vau is the smallest g-inverse of a.Proof.
(i)⇔(ii) This is precisely Proposition 3.7.
To prove the theorem it is enough to prove the following implications:
(ii)⇒(iii)⇒(iv)⇒(i); (i)⇒(v)⇒(ii) and (i)⇒(vi)⇒(ii).
(ii)⇒(iii) If a is idempotent, then a∈a{1}. For any x∈a{1,2} we have x=xax and by
Lemma3.6 we get x=(xa)x=ax=a. Therefore a{1,2}={a}.
Thus (iii) holds.
(iii)⇒(iv) If a{1,2}={a} then a is the only commuting 1-2 inverse of a.
Therefore by Definition3.3 the Group inverse of a exists and coincides with a.
(iv)⇒(i) This is trivial.
(i)⇒(v) Let a be regular, then by Lemma 3.6, for some v∈a{1},(3.2)a=(av)a=(va)a=va2.
Thus (v) holds.
(v)⇒(ii) Let a=va2 for some v∈a{1}. By Property 2.5,(3.3)a=va2≤va≤a⇒a=va=va2a=va2=(va)a=a2.
Thereforea is idempotent
Thus (ii) holds.
(i)⇒(vi)⇒(ii) can be proved along the same lines and hence omitted.
Now ifa=va2 holds then we can show that v∈a{1}(3.4)a=va2≤va≤a.
Therefore a=va2=va(3.5)a2=a·a=va2=a,
and ava=a2=a
Thusv∈a{1}.
In a similar manner we can showu∈a{1}.
Now consider,x=vau, where u,v∈a{1}. It can be verified that(3.6)x=vau∈a{1,2}={a}.
Hence,a=vau for all v,u∈a{1}.
Thusvau is invariant for all choice of g-inverse of a. By Proposition 3.9, a=vau is the smallest g-inverse of a.Remark 3.14.
Ifa is regular, then (i) aR=a2R and (ii) Ra=Ra2 automatically holds. The converse holds for an incline with unit.Remark 3.15.
Let us illustrate the relation between various inverses associated with an element in an incline in the following.
LetR={0,a,b,c,d,1} be a lattice ordered by the following Hasse graph. Define ·: R×R→R by x·y=d for all x,y∈{1,b,c,d} and 0 otherwise. Then (R,∨, ·) is an incline which is not a distributive lattice.In this inclineR, the only two elements 0, d are regular which satisfies the Theorem 3.13.(1)
d·x·d=d for each x∈{b,c,d,1}.Henced{1}={b,c,d,1} and 0{1}=R.(2)
Sinced∈R, x·d·x=x for x=0, and x=d.Henced is antiregular
(3.7)d{2}={0,d}.(3)
d{1}≠d{2} and d{1,2}=d{1}∩d{2}=d.Theorem 3.16.
LetR be a regular incline. For a,b,c∈R the following hold:(i)
(3.8)b=ya⇔b=ba⇔Rb⊆Ra,c=ax⇔c=ac⇔cR⊆aR.(ii)
x is a 1-inverse of a(3.9)⇔(ax)2=ax,axR=aR,⇔(xa)2=xa,Rxa=Ra.(iii)
x is a 2-inverse of a(3.10)⇔(xa)2=xa,xaR=xR,⇔(ax)2=ax,Rax=Rx.(iv)
qaR⊆qapR and Ra⊆Rqa implies pq=a-.(v)
Ifc=ax and b=ya then ba-c is invariant under all choice of 1-inverse of a.(vi)
(3.11)pwq=a-⇔{qapwR=qaRandw=(qap)-,Rqa=Ra.Proof.
Whenever two symmetric results are involved we shall prove the first leaving the second.(i)
Letb=ya, since a is regular. 										(3.12)ba=ba-a=yaa-a=ya=b.
Thus b=ya⇒ba=b.
Letba=b then for z∈Rb(3.13)z=xbfor somex∈R=(xb)a∈Ra.
Thus b=ba⇒Rb⊆Ra.
Sinceb is regular, by Lemma 3.6,  b=xb∈Rb, since Rb⊆Ra, b=xb=ya. Thus Rb⊆Ra⇒b=ya.
Hence (i) holds.
(ii)
Letx∈a{1} then by Lemma 3.6 and Proposition 3.7 we have
(3.14)(ax)2=ax=(xa)=a.
Hence, axR=aR and Rxa=Ra.
Conversely, letaxR=aR and (ax)2=ax.
Then,aR⊆axR⇒axa=a (by (i))(3.15)⇒x∈a{1}.(iii)
Interchangex and a in (ii) then (iii) holds.(iv)
LetqaR⊆qapR and Ra⊆Rqa(3.16)qaR⊆qapR⇒qapqa=qa(by(i))
Ra⊆Rqa⇒aqa=a, that is, a is regular with q∈a{1}.
Now,(3.17)qapqa=qa,(aqa)pqa=aqa,apqa=a.
Therefore pq∈a{1}.
Thus (iv) holds.
(v)
Letc=ax and b=ya for some x,y∈R(3.18)ba-c=y(aa-a)x=yax.
Which is independent of a- and ba-c is invariant for all choice of a- of a.(vi)
Letpwq=a-⇒a=a(pwq)a (By Definition 3.1).
From the statement (ii), we have
Ra=Rpwqa⊆Rqa⊆Ra(3.19)⇒Ra=Rqa.
Thereforeq∈a{1} (by (ii)).
Now,(3.20)apwqa=a⇒qapwaqa=qa,qaR⊆qapwR⊆qaR,qaR=qapwR,qapwqa=qa,(aqa)pwqa=aqa,a(pwq)a=a,⇒pwq∈a{1}.
Thus (vi) holds.Corollary 3.17.
For a, b in a regular incline one has the following:(3.21)Ra=Rb⇔aR=bR⇔a=b.Proof.
SinceRa=Rb,Ra⊆Rb,andRb⊆Ra.
By Theorem3.16(i) we have
(3.22)Ra⊆Rb⇒a=ab⇒a≤b(byProperty2.5)Rb⊆Ra⇒b=ba⇒b≤a(byProperty2.5).
Thereforea=b. In a similar manner we can show aR=bR⇒a=b.
On the other handa=b automatically implies Ra=Rb and aR=bR.Remark 3.18.
We note that Corollary3.17 fails for regular matrices over an incline.
Let us considerB=(1110) and A=(1011)=PB, where P=(0110).
SinceP2=I2,PA=B. Here A and B are regular.
By Proposition2.4 in [8, page 297], R(A)⊆R(B) and R(B)⊆R(A).
HenceR(A)=R(B) but A≠B.It is well known that [9, page 26] if a- is a particular g-inverse of a in a ring with unit, then the general solution of the equation axa=a is given by a-+h-a-ahaa-, where h is arbitrary. Here we shall generalize this for incline.Theorem 3.19.
Leta∈R and a- be any particular 1-inverse of a then ag-{1}={a-+h/hisarbitraryelementinR} is the set of all g-inverses of a dominating a-. Furthermore, a{1}=∪ag-{1}, union over all g-inverses of a.Proof.
Let𝒜 denote the set {a-+h/hisarbitraryelementinR}. Suppose that x is arbitrary element of ag-{1} then x≥a- which implies x+k≥a-+k for k∈R and by Property 2.4 we have a-+k≥a-.
Thereforex+k≥a-+k≥a-
Pre- and postmultiplication bya we get(3.23)axa+aka≥a(a-+k)a≥aa-a,a+aka≥a(a-+k)a≥a
(By Definition 3.1).
By Property2.5  aka≤a, hence a+aka=a.(3.24)a≥a(a-+k)a≥a,a(a-+k)a=a.
Therefore (a-+k)∈a{1}.
Thus for eachx∈ag-{1} there exists an element in 𝒜. Hence ag-{1}⊆𝒜.
On the other hand for anyy∈𝒜,y=a-+h≥a- by Property 2.4.
From Definition3.1 and Property 2.5, we get(3.25)aya=a(a-+h)a=a+aha=a.
Hence y∈ag-{1}, which implies 𝒜⊆ag-{1}. Therefore 𝒜=ag-{1}(3.26)a{1}=setofallg-inversesofa=⋃ag-{1},unionoverallg-inversesofa.


## 4. Projection on an Incline with Involution-T

In this section, the existence of the Moore-Penrose inverse of an element in an incline with involution-T is discussed as a generalization of that for elements is a *-regular ring and for elements in a Fuzzy algebra studied by Hartwig [4], Kim and Roush [8] and Meenakshi [6], respectively. Characterization of the set of all {1,3}, {1,4} inverses and a formula for Moore-Penrose are obtained analogous to those of the result established for fuzzy matrices in [6, 8].An involution-T of an incline R is an involutary anti-automorphism, that is, (aT)T=a,(a+b)T=aT+bT,(ab)T=bTaT,aT=0 if and only if a=0 for all a,b∈R.Definition 4.1.
An elementa∈R is said to be a projection if aT=a=a2, that is a is symmetric and idempotent.Definition 4.2.
Fora in an incline R with involution-T, we say that x∈R is a 3-inverse of a if (ax)T=ax, and we say that y∈R is a 4-inverse of a if (ya)T=ya.Definition 4.3.
An elementx∈R is said to be Moore-Penrose inverse of a, if x satisfies the following: (i) axa=a, (ii) xax=x, (iii) (ax)T=ax, and (iv) (xa)T=xa, denoted as a†.
In [2] it is stated that for an element a in an incline with involution-T, a† exists if and only if aℋaaTa. Here we derive equivalent condition for the existence of a† in terms of the weaker relation aℒaaTa.
First we shall show that Green’s equivalence relation on an inclineR reduces to equality of elements in R.Lemma 4.4.
Fora,b∈R the following hold:
(i)aℒb⇒a=b,  (ii)aℛb⇒a=b.
Converse holds for elements in a regular incline or incline with unit.Proof.
(i) If aℒb then by Definition 2.6 there exist x,y∈R such that xa=b and yb=a. By Property 2.5 we have xa=b⇒b≤a and yb=a⇒a≤b.
Therefore,aℒb⇒a=b.
(ii) This can be proved in a similar manner and hence omitted.
The converse holds fora regular incline. For, if a,b are regular, then by Lemma 3.6  a=b=yb=by and b=a=xa=ax for some x,y∈a{1}. Hence a=b⇒aℒb and aℛb. aℒb and aℛb trivially hold for incline with unit.Theorem 4.5.
Let R  be an incline with involution-T. Fora∈R the following are equivalent: (i)
a is a projection,(ii)
a has 1-3 inverse,(iii)
a has 1-4 inverse,(iv)
a† exists and equals a,(v)
aTax = aT has a solution in R,(vi)
xaaT = aT has a solution in R,(vii)
a is regular and aT∈a{1},(viii)
aℒaaTa,(ix)
aℛaaTa,Proof.
(i)⇔(ii)
Leta be a projection, by Definition 4.1  a is symmetric idempotent. a is regular follows from Proposition 3.7. Thus a has 1-inverse x (say) and by Lemma 3.6  a=ax=xa. Since a is symmetric, a=aT. Therefore x is a 1-3 inverse of a. Thus a has 1–3 inverses. Coverersly if a has 1–3 inverses, then again by Lemma 3.6 there exists x∈R, such that a=ax=xa and ax=(ax)T. Hence a is symmetric idempotent. Thus (i) holds.(i)⇔(iii)
This can be proved along the same lines as that of (i)⇔(iii), hence omitted.(i)⇔(iv)
This equivalence can be proved directly by verifying thata satisfy the four equations in Definition 4.3.(ii)⇔(v)
Leta has 1–3 inverses, x (say) then(4.1)aTax=aT(ax)=aT(ax)T=aTxTaT=(axa)T=aT,aTax=aT,
(by Definition 4.2).
Conversely, ifaTax=aT, then aT≤aTa≤aT⇒aT=aTa and therefore a=aTa and a is symmetric. Hence the given condition aTax=aT reduces to a x=aT
Now,axa=(ax)a=aTa=a and ax=aT=a=(ax)T
x∈a{1,3}. Thus a has 1–3 inverses.
(iii)⇔(vi)
This can be proved in the same manner and hence omitted.(vii)⇔(i)
a is regular and aT∈a{1}⇔
a is regular and a=aaTa⇔
a is idempotent and a=aTa=aaT (by Proposition 3.7 and Lemma 3.6)⇔
a is symmetric and idempotent⇔
a is a projection.
(vii)⇔(viii)⇔(ix) follow from Lemma 4.4.Remark 4.6.
It is well known that [4] for an element a in a *-regular ring if a† exists then a†=a(1,4)aa(1,3). We observe that for an element a in an incline with involution-T if a is regular, then by Lemma 3.6 it follows that a{1,3}=a{1,4}. If a† exists it is unique and given by a†=a(1,3)aa(1,3).Remark 4.7.
Let us consider the inclineR in Remark 3.15 under the identity involution-T on R. Here each element in R is symmetric and the 3-inverse of the element d is R and 4-inverse also the same.
Hence0,d are the only projections in R(4.2)d†=d=d{1,2}≠d{1,3}=d{1,4}=d{1}={b,c,d,1},b·d·c=d=d†∀b,c∈d{1,4}.Theorem 4.8.
Let R  be an incline with involution-T. For any elementa∈R and x∈a{1,3} given, then ag{1,3}={x+h/hisarbitraryelementinR} is the set of all {1,3} inverses of a  dominating x.Proof.
This can be proved along the same lines as that of Theorem3.19 and hence omitted.


## 5. Conclusion

The main results in the present paper are the generalization of the available results shown in the reference for elements in a *-regular ring [4] and for elements in a Fuzzy algebra [8]. We have proved the results by using Property 2.5 without using star cancellation law.In [2] it is stated that an element is regular if and only if 𝔇 class contains an idempotent. By Lemma 4.4 the 𝔇 class {b/b𝔇a}={a} and by Proposition 3.1  a is regular if and only if a is idempotent.Figure 1



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