An element a of an AG-groupoid S is called a left (right) regular element of S if there exists some x∈S such that a=a2x (a=xa2) and S is called left (right) regular if every element of S is left (right) regular.
Remark 4
Let S be an AG-groupoid with left identity. Then, the concepts of left and right regularity coincide in S.
Indeed,for every a∈Sthere exist some x,y∈Ssuch that a=xa2=a2y. As a=xa2=ex·aa=aa·xe=a2y,and a=a2y=xa2also holds in a similar way.
Let us give an example of an AG-groupoid which will be used for the converse parts of various problems in this section.
Example 2
Let us consider an AG-groupoid S={1,2,3,4,5} with left identity 4 defined in the following multiplication table.
It is easy to check that
S
is non-commutative and non-associative.
An AG-groupoid S is called left (right) duo if every left (right) ideal of S is a two-sided ideal of S and is called duo if it is both left and right duo. Similarly an AG-groupoid S is called an SI-left (SI-right) duo if every SI-left-ideal (SI-right-ideal) of S over U is an SI-two-sided-ideal of S over U, and S is called an SI-duo if it is both an SI-left and an SI-right duo.
Lemma 8
If every SI-left-ideal of an AG-groupoid S with left identity over U is anSI-interior-ideal of S over U, then S is left duo.
Proof
Let I be any left ideal of S with left identity. Now by Lemma 1, the identity soft mapping XI is an SI-left-ideal of S over U. Thus, by hypothesis, XI is anSI-interior-ideal of S over U, and by using Lemma 1 again, I is an interior ideal of S. Thus IS=I·SS=S·IS=SS·IS=SI·SS=SI·S⊆I. This shows that S is left duo. □
The converse part of Lemma 8 is not true in general. Let us consider an AG-groupoid S (from Example 2). It is easy to see that S is left duo because the only left ideal of S is {1,5} which is also a right ideal of S. Let A=S and define a soft set fA of S over U={p1,p2,p3,p4,p5,p6} as follows:
\(f_{A}(x)=\left \{ \begin {array}{c} U\ \text {if}\ x=1 \\ \{p_{1},p_{2},p_{3},p_{4}\}\ \text {if}x\ =2 \\ \{p_{2},p_{3},p_{4,},p_{5}\}\ \text {if}\ x=3 \\ \{p_{3},p_{4,},p_{5}\}\ \text {if}\ x=4 \\ \{p_{1},p_{2},p_{3},p_{4,}p_{5}\}\ \text {if}\ x=5 \end {array} \right \}.\)
Then, it is easy to see that fA is anSI -left-ideal of S over U but it is not anSI -interior-ideal of S over U because \(f_{A}(42\ast 4)\varsupsetneq f_{A}(2).\)
Corollary 2
Every interior ideal of an AG-groupoid S with left identity is a right ideal of S.
Theorem 1
Every SI-right-ideal of an AG-groupoid S with left identity is anSI-interior-ideal of S over U if and only if S is right duo.
Proof
It is simple. □
Theorem 2
Let S be a right regular AG-groupoid with left identity. Then, S is left duo if and only if every SI-left-ideal of S over U is anSI-interior-ideal of S over U.
Proof
Necessity. Let a right regular S with left identity be a left duo, and assume that fA is any SI-left-ideal of S over U. Let a,b,c∈S, then b≤xb2 for some x∈S. Since Sa is a left ideal of S, therefore by hypothesis, Sa is a two-sided ideal of S. Thus, ab·c=a(x·bb)·c=a(b·xb)·c=b(a·xb)·c=c(a·xb)·b. It follows that ab·c∈S(a·SS)·b⊆(S·aS)b=(SS·aS)b=(Sa·SS)b⊆(Sa·S)b⊆Sa·b. Thus, ab·c=ta·b for some t∈S, and therefore fA(ab·c)=fA(ta·b)⊇fA(b), implies that fA is anSI-interior-ideal of S over U.
Sufficiency. It can be followed from Lemma 8. □
By left-right dual of above Theorem, we have the following Theorem :
Theorem 3
Let S be a right regular AG-groupoid with left identity. Then,S is right duo if and only if every SI-right-ideal of S over U is anSI-interior-ideal of S over U.
Lemma 9
A non-empty subset A of a right regular AG-groupoid S with left identity is a two-sided ideal of S if and only if it is an interior ideal of S.
Proof
It is simple. □
Lemma 10
Every left ideal of an AG-groupoid S with left identity is an interior ideal of S if S is an SI-left duo.
Proof
It can be followed from Lemmas 1 and 9. □
The converse of Lemma 10 is not true in general. The only left ideal of S (from Example 2) is {1,2} which is also an interior ideal of S. Let A={2,3,4,5} and define a soft set fA of S over \( U= \mathbb {Z} \) as follows:
\(f_{A}(x)=\left \{ \begin {array}{c} 4 \mathbb {Z} \ \text {if}\ x=2 \\ 8 \mathbb {Z} \ \text {if}\ x=3 \\ 16 \mathbb {Z} \ \text {if}\ x=4 \\ 2 \mathbb {Z} \ \text {if}\ x=5 \end {array} \right \} \).
Then, it is easy to see that fA is anSI -left-ideal of S over U but it is not anSI -right-ideal of S over U because \(f_{A}(2\ast 4)\varsupsetneq f_{A}(2).\)
It is easy to see that every SI-right-ideal of S with left identity over U is anSI-left-ideal of S over U.
Remark 5
Every SI-right-ideal of an AG-groupoid S with left identity is anSI-left-ideal of S over U, but the converse is not true in general.
Theorem 4
Every right ideal of an AG-groupoid S with left identity is an interior ideal of S if and only if S is an SI-right duo.
Proof
It is straightforward. □
Theorem 5
Let S be a right regular AG-groupoid with left identity. Then,S is an SI-left duo if and only if every left ideal of S is an interior ideal of S.
Proof
The direct part can be followed from Lemma 10. The converse is simple. □
By left-right dual of above Theorem, we have the following Theorem.
Theorem 6
Let S be a right regular AG-groupoid with left identity. Then,S is an SI-right duo if and only if every right ideal of S is an interior ideal of S.
Theorem 7
Let S be an AG-groupoid with left identity and E={x∈S:x=x2}⊆S. Then the following assertions hold :
(i)Eforms a semilattice ;
(ii)Eis a singleton set,if a=ax·a,∀a,x∈S.
Proof
It is simple. □
Theorem 8
For an AG-groupoid S with left identity, the following conditions are equivalent :
(i)Sis right regular ;
(ii)For any interior idealIof S;
(a) I⊆I2,
(b)Iis semiprime.
(iii)For any SI-interior-ideal fAofSover U;
(a)\(f_{A}\overset {\sim }{\subseteq }f_{A}\circ f_{A},\)
(b) fAis soft semiprime over U.
(iv)Sis right regular and |E|=1, (a=ax·a, ∀ a,x∈E);
(v)Sis right regular and ∅≠E⊆Sis semilattice.
Proof
(i)⇒(v)⇒(iv) can be followed from Theorem 7.
(iv)⇒(iii):(a). Let fA be any SI -interior-ideal of a right regular S with left identity. Thus, for each a∈S, there exists some x∈S such that a=x·aa=a·xa=a·x(x·aa)=a·(ex)(a·xa)=a·(xa·a)(xe). Therefore,
$$\begin{array}{@{}rcl@{}} (f_{A}\circ f_{A})(a) &=&\bigcup \limits_{a=a\cdot (xa\cdot a)(xe)}\left \{ f_{A}(a)\cap f_{A}((xa\cdot a)(xe))\right \} \\ &\supseteq &f_{A}(a)\cap f_{A}((xa\cdot a)(xe))\supseteq f_{A}(a)\cap f_{A}(a)=f_{A}(a)\text{.} \end{array} $$
This shows that \(f_{A}\overset {\sim }{\subseteq }f_{A}\circ f_{A}.\)
(b). Also,
$$\begin{array}{@{}rcl@{}} a &=&x\cdot aa\leq ex\cdot aa=aa\cdot xe=(a\cdot xa^{2})(xe)=(x\cdot aa^{2})(xe)=x(ea\cdot aa)\cdot (xe) \\ &=&x(aa\cdot ae)\cdot (xe)=(aa)(x\cdot ae)\cdot (xe)=(ae\cdot x)(aa)\cdot (xe)=(ae\cdot x)a^{2}\cdot (xe). \end{array} $$
This implies that fA(a)=fA((ae·x)a2·(xe))⊇fA(a2). Hence, fA is softsemiprime.
(iii)⇒(ii):(a). Assume that I is any interior ideal of S, then by using Lemma 1, XI is an SI-interior-ideal of S over U. Let i∈I, then by using Lemma 2, we have XI(i)⊆(XI∘XI)(i)=(XI)(i)=U. Hence, I⊆I2.
(b). Let i2∈I. Then, by given assumption, we have XI(i)⊇XI(i2)=U. This implies that i∈I, and therefore I is semiprime.
(ii)⇒(i): Let a∈S with left identity. Since Sa2 is an interior ideals of S, and clearly a2∈Sa2. Thus, by using given assumption, a∈Sa2. Hence, S is right regular. □
Corollary 3
Every SI-interior-ideal of a right regular AG-groupoid S with left identity is softsemiprime over U.
Proof
Let fI be any SI-interior-ideal of a right regular S with left identity. Then, for each a∈S, there exists some x∈S such that fI(a)=fI(x·aa)=fI(a·xa)=fI(xa2·xa)⊇fI(a2). □
Corollary 4
Let I be an interior ideal of an AG-groupoid S. Then, I is semiprime if and only if XI is softsemiprime over U.
Theorem 9
Let S be an AG-groupoid with left identity. Then, S is right regular if and only if every SI-interior-ideal fA of S over U is softidempotent and softsemiprime.
Proof
Necessity : Let fA be any SI-interior-ideal of a right regular S with left identity over U. Then, clearly \(f_{A}\circ f_{A} \overset {\sim }{\subseteq }f_{A}\). Now for each a∈S, there exists some x∈S such that a=x·aa=a·xa=ea·xa=ax·ae=(ae·x)a. Thus,
$$\begin{array}{@{}rcl@{}} (f_{A}\circ f_{A})(a) &=&\bigcup \limits_{a=(ae\cdot x)a}\left \{ f_{A}(ae\cdot x)\cap f_{A}(a)\right \} \supseteq f_{A}(ae\cdot x)\cap f_{A}(a) \\ &\supseteq &f_{A}(a)\cap f_{A}(a)=(f_{A}\cap f_{A})(a)=f_{A}(a)\text{.} \end{array} $$
This shows that fA is softidempotent over U. Again a=ex·aa=aa·xe=a2·xe. Therefore, fA(a)=fA(a2·xe)⊇fA(a2). Hence, fA is softsemiprime over U.
Sufficiency : Since Sa2 is an interior ideal of S, therefore by Lemma 1, its soft characteristic function \(\phantom {\dot {i}\!}X_{Sa^{2}}\) is an SI-interior-ideal of S over U such that \(\phantom {\dot {i}\!}X_{Sa^{2}}\) is softidempotent over U. Since by given assumption, \(\phantom {\dot {i}\!}X_{Sa^{2}}\) is softsemiprime over U so by Corollary 4,Sa2 is semiprime. Since a2∈Sa2, therefore, a∈a2S. Thus, by using Lemma 2, we have \(\phantom {\dot {i}\!}X_{Sa^{2}}\circ X_{Sa^{2}}=X_{Sa^{2}},\) and \(\phantom {\dot {i}\!}X_{Sa^{2}}\circ X_{Sa^{2}}=X_{(Sa^{2}\cdot Sa^{2})}.\) Thus, we get \(\phantom {\dot {i}\!}X_{(Sa^{2}\cdot Sa^{2})}=X_{Sa^{2}}.\) This implies that \(\phantom {\dot {i}\!}X_{(Sa^{2}\cdot Sa^{2})}(a)=X_{Sa^{2}}(a)=U.\) Therefore, a∈Sa2·Sa2=a2S·Sa2=(Sa2·S)a2⊆Sa2. Hence,S is right regular. □
Lemma 11
Every SI-interior-ideal of a right regular AG-groupoid S with left identity over U is softidempotent.
Proof
Let fA be any SI-interior-ideal of a right regular S with left identity over U. Then, by using Lemma 4, \(f_{A}\circ f_{A}\overset {\sim }{\subseteq }f_{A}\). Since S right regular, therefore for every a∈S there exists some x∈S such that a=x·aa=a·xa=xa2·xa=ax·a2x=(a2x·x)a=(xx·aa)a=(aa·x2)a. Therefore,
$$\begin{array}{@{}rcl@{}} (f_{A}\circ f_{A})(a) &=&\bigcup \limits_{a=(aa\cdot x^{2})a}\left \{ f_{A}(aa\cdot x^{2})\cap f_{A}(a)\right \} \supseteq f_{A}(aa\cdot x^{2})\cap f_{A}(a) \\ &\supseteq &f_{A}(a)\cap f_{A}(a)=(f_{A}\cap f_{A})(a)\text{.} \end{array} $$
Thus, fA∘fA=fA. □
Theorem 10
Let S be an AG-groupoid with left identity and fA be any SI-interior-ideal of\(\ S\mathcal {\ }\)over U. Then,S is right regular if and only if fA=(XS∘fA)2 and fA is softsemiprime.
Proof
Necessity : Let fA be any SI-interior-ideal of a right regular S with left identity over U. Then, by using Lemmas 4 and 2, we have
$$(X_{S}\circ (X_{S}\circ f_{A}))\circ X_{S}=(X_{S}\circ f_{A})\circ X_{S}=(f_{A}\circ X_{S})\circ X_{S}=(X_{S}\circ X_{S})\circ f_{A}=X_{S}\circ f_{A}. $$
This shows that XS∘fA is anSI -interior-ideal of S over U. Now by using Lemmas 11 and 4, we have (XS∘fA)2=XS∘fA=fA. It is easy to see that fA is softsemiprime.
Sufficiency : Let fA=(XS∘fA)2 holds for any SI-interior-ideal fA of S over U. Then, by given assumption and Lemma 14, we get, \(f_{A}=(X_{S}\circ f_{A})^{2}=f_{A}^{2}.\) Thus, by using Theorem 9, S is right regular. □
Corollary 5
Let S be an AG-groupoid with left identity and fA be any SI-interior-ideal of\(\ S\mathcal {\ }\)over U. Then, S is right regular if and only if \(f_{A}=X_{S}\circ f_{A}^{2}\) and fA is soft semiprime.
Proof
From above theorem, \(f_{A}=(X_{S}\circ f_{A})^{2}=(X_{S}\circ f_{A})(X_{S}\circ f_{A})=(X_{S}\circ f_{A})\circ f_{A}=(f_{A}\circ f_{A})\circ X_{S}=(f_{A}\circ f_{A})\circ (X_{S}\circ X_{S})=(X_{S}\circ X_{S})\circ (f_{A}\circ f_{A})=X_{S}\circ f_{A}^{2}.\) □
Lemma 12
Let S be an AG-groupoid with left identity and fA be any SI-left-ideal (SI -right-ideal,SI-two-sided-ideal)of\( \ S\mathcal {\ }\)over U. Then, S is right regular if and only if fA is softidempotent.
Proof
Necessity : Let fA be anSI-left-ideal of a right regular S with left identity over U. Then, it is easy to see that \( f_{A}\circ f_{A}\overset {\sim }{\subseteq }f_{A}.\) Let a∈S, then there exists x∈S such that a=aa·x=xa·a. Thus
$$(f_{A}\circ f_{A})(a)=\bigcup \limits_{a=xa\cdot a}\{f_{A}(xa)\cap f_{A}(a)\} \supseteq f_{A}(a)\cap f_{A}(a)=f_{A}(a), $$
which implies that fA is softidempotent.
Sufficiency : Assume that fA∘fA=fA holds for all SI-left-ideal of S with a left identity over U. Since Sa is a left ideal of S, therefore by Lemma 1, it follows that XSa is anSI-left-ideal of S over U. Since a∈Sa, it follows that (XSa)(a)=U. By hypothesis and Lemma 2, we obtain (XSa)∘(XSa)=XSa and (XSa)∘(XSa)=XSa·Sa. Thus, we have (XSa·Sa)(a)=XSa(a)=U, which implies that a∈Sa·Sa. Therefore, a∈Sa·Sa=S2a2=Sa2. This shows that S is right regular. □
Theorem 11
Let S be an AG-groupoid with left identity and fA be any SI-left-ideal (SI -right-ideal,SI-two-sided-ideal) of\(\ S \mathcal {\ }\)over U. Then, S is right regular if and only if fA=(XS∘fA)∘(XS∘fA).
Proof
Necessity : Let S be a right regular S with left identity and let fA be any SI-left-ideal of S over U. It is easy to see that XS∘fA is also anSI -left-ideal of S over U. By Lemma 12, we obtain \((X_{S}\circ f_{A})\circ (X_{S}\circ f_{A})=(X_{S}\circ f_{A})\overset {\sim }{\subseteq } f_{A}.\) Let a∈S, then there exists x∈S such that a=aa·x=xa·a=(xa)(aa·x)=(xa)(xa·a). Therefore,
$$\begin{array}{@{}rcl@{}} \left((X_{S}\circ f_{A})\circ (X_{S}\circ f_{A})\right) (a) &\supseteq &(X_{S}\circ f_{A})(xa)\cap (X_{S}\circ f_{A})(xa\cdot a) \\ &\supseteq &X_{S}(x)\cap f_{A}(a)\cap X_{S}(xa)\cap f_{A}(a)=f_{A}(a), \end{array} $$
which is what we set out to prove.
Sufficiency : Suppose that fA=(XS∘fA)∘(XS∘fA) holds for all SI-left-ideal fA of S over U. Then \(f_{A}=(X_{S}\circ f_{A})\circ (X_{S}\circ f_{A})\overset {\sim }{ \subseteq }f_{A}\circ f_{A}\overset {\sim }{\subseteq }X_{S}\circ f_{A} \overset {\sim }{\subseteq }f_{A}.\) Thus, by Lemma 12, it follows that S is right regular. □
Lemma 13
Let fA be any SI-interior-ideal of a right regular AG-groupoid S with left identity\(\mathcal {\ }\)over U. Then, fA(a)=fA(a2), for all a∈S.
Proof
Let fA be any SI-interior-ideal of a right regular S with left identity over U. For a∈S, there exists some x in S such that a=ex·aa=aa·xe=(xe·a)a=(xe·a)(ex·aa)=(xe·a)(aa·xe)=aa·(xe·a)(xe)=ea2·(xe·a)(xe). Therefore fA(a)=fA(ea2·(xe·a)(xe))⊇fA(a2)=fA(aa)=fA(a(ex·aa))=fA(a(aa·xe))=fA((aa)(a·xe))=fA((xe·a)(aa))⊇fA(a). That is, fA(a)=fA(a2),∀a∈S □
The converse of Lemma 13 is not true in general. Let us consider an AG-groupoid S (from Example 2). Let A={1,2,4,5} and define a soft set fA of S over \(U=\left \{ \left [ \begin {array}{cc} 0 & 0 \\ x & x \end {array} \right ] /x\in \mathbb {Z} _{3}\right \} (\)the set of all 2×2 matrices with entries from \( \mathbb {Z} _{3})\) as follows:
\(f_{A}(x)=\left \{ \begin {array}{c} \left \{ \left [ \begin {array}{cc} 0 & 0 \\ 0 & 0 \end {array} \right ],\left [ \begin {array}{cc} 0 & 0 \\ 1 & 1 \end {array} \right ],\left [ \begin {array}{cc} 0 & 0 \\ 2 & 2 \end {array} \right ] \right \}\ \text {if}\ x=1 \\ \left \{ \left [ \begin {array}{cc} 0 & 0 \\ 1 & 1 \end {array} \right ],\left [ \begin {array}{cc} 0 & 0 \\ 2 & 2 \end {array} \right ] \right \}\ \text {if}\ x=2 \\ \left \{ \left [ \begin {array}{cc} 0 & 0 \\ 2 & 2 \end {array} \right ] \right \}\ \text {if}\ x=4 \\ \left \{ \left [ \begin {array}{cc} 0 & 0 \\ 1 & 1 \end {array} \right ],\left [ \begin {array}{cc} 0 & 0 \\ 2 & 2 \end {array} \right ] \right \}\ \text {if}\ x=\mathit {5} \end {array} \right \}.\)
It is easy to see that fA is anSI -interior-ideal of S such that fA(x)⊇fA(x2),∀x∈S but S is not right regular.
On the other hand, it is easy to see that every SI -two-sided-ideal of S over U is anSI -interior-ideal of S over U.
Remark 6
Every SI-two-sided-ideal of a right regular AG-groupoid S with left identity\(\mathcal {\ }\)over U is an SI-interior-ideal of S over U but the converse is not true in general.
Theorem 12
For an AG-groupoid S with left identity, the following conditions are equivalent :
(i)Sis right regular ;
(ii)Every interior ideal ofSis semiprime ;
(iii)Every SI-interior-ideal ofSoverUis soft semiprime ;
(iv)For every SI-interior-ideal fAofSover U, fA(a)=fA(a2), ∀ a∈S.
Proof
(i)⇒(iv) can be followed from Lemma 13.
(iv)⇒(iii) and (iii)⇒(ii) are obvious.
(ii)⇒(i): Since Sa2 is an interior ideal of S with left identity such that a2∈Sa2, therefore by given assumption, we have a∈Sa2. Thus, S is right regular. □