CHAPTER 6 WEAKLY π-REGULAR NI...

CHAPTER 6

WEAKLY π-REGULAR NI

NEAR-RINGS

6.1 Introduction

In this Chapter, we characterize 2-primal near-rings by using its minimal 0-

prime ideals and NI near-rings by using its minimal strongly 0-prime ideals.

Kim and Kwak [32] asked one question that, " Is a ring R 2-primal if OP ⊆ P

for each P ∈ mSpec(R)?". In Section 6.2, we prove that if OP has the IFP for

each P ∈ mSpec(N ), then OP ⊆ P for each P ∈ mSpec(N ) if and only if N

is a 2-primal near-ring.

In [29], Hong et al. observed that the condition "R/N ∗(R) is right weakly

π-regular" in Proposition 18 cannot be replaced by the condition "R is right

weakly π-regular " in the case of rings. In Section 6.3, we show that in the case

of NI near-ring which satisfies (CZ2) the condition "N/N ∗(N ) is left weakly

π-regular" can be replaced by the condition "N is left weakly π-regular".

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6.2. CHARACTERIZATION OF 2-PRIMAL NEAR-RINGS

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1

6.2 Characterization of 2-primal near-rings

In this section, we give some characterization of 2-primal near-ring by using its

0-prime ideals.

Proposition 6.2.1. For each P ∈ Spec(N ), O(P ) and N (P ) are ideals of N .

Proof. Let P be a 0-prime ideal of N and let a1, a2 ∈ O(P ). Then

a1N <b1> = 0 for some b1 ∈ N \ P and a2N <b2> = 0 for some b2 ∈ N \ P .

Since b1, b2 ∈ N \ P and N \ P is an m−system, there exist br ∈< b1 >

and br ∈< b2 > such that br br ∈ N \ P . Let b3 = br br . For any n ∈ N 2 1 2 1 2

and x ∈ <b3>, (a1 − a2)nx = 0 implies a1 − a2 ∈ O(P ). Let x ∈ O(P ).

Then xN = 0. Thus for n, nr, n1 ∈ N and br ∈ , we have

(n(nr + x) − nnr)n1br = 0 implies n(nr + x) − nnr ∈ O(P ) and (xn)n1br = 0

implies xn ∈ O(P ). Thus O(P ) is an ideal of N . Similarly, N (P ) is an ideal

of N .

Example 6.2.2.

D D Let N =

0 D

where D is a division ring.

Then N is a near-ring under addition and multiplication of matrices. We also

observe the following:

(i)

D D 0 D P =

0 0

are 0-prime ideals of N ;

and Q = 0 D


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(ii)

0 D I =

0 0

is a two sided ideal of N , but not a 0-prime ideal of N since P Q ⊆ I but

P ¢ I and Q ¢ I ;

(iii)

D 0 is a left ideal but not a right ideal of N;

0 0

(iv)

0 0 is a right ideal but not a left ideal of N.

0 D

(v)

D 0 is neither a left nor a right ideal of N;

0 D

(vi)

P0(N ) =

0 D = N (N ).

0 0

Therefore N is a 2-primal near-ring.

Also O (P ) = (0); O(P ) = P0(N );

D 0 OP =

0 0

OP = OP ; N (P ) = N (P ) = NP = N P = P .

O (Q) = O(Q) = OQ = OQ = Q and N (Q) = N (Q) = NQ =

N Q = P0(N ).

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The following results might be helpful for the criterion for a certain class of

rings to be 2-primal.

Theorem 6.2.3. For a near-ring N , the following statements are equivalent:

(i) N is 2-primal;

(ii) P0(N ) is a completely semiprime ideal of N ;

(iii) N (P ) is a completely semiprime ideal of N for each P ∈mSpec(N );

(iv) N P = N (P ) = N (P ) for each P ∈ mSpec(N );

(v) N (P ) = NP for each P ∈ mSpec(N );

(vi) NP ⊆ P for some P ∈mSpec(N );

(vii) NP /P0 (N ) ⊆ P /P0(N ) for each P ∈mSpec(N ).

Proof. (i) ⇒ (ii) : Since P0(N ) = N (N ), for any x in N, x2 ∈ P0(N ) im-

plies x2 is nilpotent and hence x ∈ N (N ) = P0(N ). Therefore, P0(N ) is a

completely semiprime ideal of N .

(ii) ⇒ (iii) : Let P be a minimal 0-prime ideal of N . Let x ∈ N be such

that x2 ∈ N (P ). Then x2N ⊆ P0(N ) for some b ∈ N \ P . Since

P0(N ) is a completely semiprime ideal of N , it has the IFP. So xN xN ⊆

P0(N ) which implies xN ⊆ P0(N ). Thus x ∈ N (P ) and hence N (P ) is

completely semiprime.

(iii) ⇒ (i) : Let a ∈ N (N ). Then an = 0 for some positive integer n.

If a ∈/ P0(N ), then there exists a minimal 0-prime ideal P of N such that

a ∈/ P . Since N (P ) is a completely semiprime ideal, an = 0 ∈ N (P ) implies

a ∈ N (P ) ⊆ P , a contradiction. Hence a ∈ P0(N ).

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(ii) ⇒ (iv) : Let P be a minimal 0-prime ideal of N and let a ∈ N P . Then

an ∈ NP for some positive integer n. Thus anb ∈ P0(N ) for some b ∈ N \ P .

Since P0(N ) is completely semiprime ideal of N, it has the IFP. By Theorem

1.2.18, ab ∈ P0(N ). Therefore aN ⊆ P0(N ) for some b ∈ N \ P and so

a ∈ N (P ). Thus N P ⊆ N (P ). But N (P ) ⊆ NP ⊆ N P and N (P ) ⊆ N P .

Therefore, N P = N (P ) = N (P ) for each P ∈ mSpec(N ).

(iv) ⇒ (v) ⇒ (vi) : These are obvious.

(vi) ⇒ (vii) : Let P be a minimal 0-prime ideal of N . Let N = N/P0(N )

and P = P /P0(N ). Let a = a + P0(N ) ∈ NP for some a ∈ N . Then there

exists b ∈ N \ P such that ab ∈ P0(N ) = 0. Thus ab ∈ P0(N ) and so

a ∈ NP ⊆ P . Therefore a ∈ P and hence NP ⊆ P .

(vii) ⇒ (i) : Suppose that N = N/P0(N ) is not reduced. Then there exists

a ∈ N such that a2 = 0 and a = 0. Thus a ∈/ P0(N ) and hence a ∈/ P for some

P ∈ mSpec(N ). Then a ∈/ P and so a ∈ N \ P . But since a2 = 0, we obtain

a ∈ NP ⊆ P , which is a contradiction. Therefore P0(N ) = N (N ) and hence

N is 2-primal.

Corollary 6.2.4. For a near-ring N , assume that N is 2-primal. If P = N (P )

for each P ∈Spec(N ), then P is completely prime ideal of N .

Proof. Suppose that N is a 2-primal near-ring. Let xy ∈ P = N (P ). Then

there exists b ∈ N \ P such that (xy)N ⊆ P0(N ). Since P0(N ) has the

IFP, we have (xN y)N ⊆ P0(N ) ⊆ P and so xN y ⊆ P since b ∈/ P.

Hence x ∈ P or y ∈ P since P is a 0-prime ideal of N. Therefore, P is a

completely prime ideal of N .

Proposition 6.2.5. For a near-ring N , we have the following:

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(i) N (N ) ⊆ T

P ∈Spec(N )

(ii) P0 (N ) ⊆ T

P ∈Spec(N )

O (P ) ⊆ T

Q∈mSpec(N )

N (P ) = T

Q∈mSpec(N )

O (Q) ;

N (Q) .

Proof. (i) Let a ∈ N (N ). Then an = 0 for some positive integer n. Let P

be any 0-prime ideal and let b ∈ N \ P . Since an = 0, anN = 0. Thus

an ∈ O(P ) and hence a ∈ O (P ). Therefore a ∈ T

P ∈Spec(N )

O (P ) . The other

inclusion is obvious.

(ii) Let a ∈ P0 (N ). Let P be any 0-prime ideal of N . Then

aN ⊆ P0(N ) for any b ∈ N \ P which implies that a ∈ N (P ) and so

a ∈ T

P ∈Spec(N )

N (P ). Therefore,

P0 (N ) ⊆ \

P ∈Spec(N )

N (P ) .

But T

P ∈Spec(N )

N (P ) ⊆ T

Q∈mSpec(N )

N (Q) always.

Since N (Q) ⊆ Q for each Q ∈ mSpec(N ) , T

Q∈mSpec(N )

N (Q) ⊆ P0 (N ) .

Our next result indicates that our characterization of minimal 0-prime ideal

P in terms of N (P ) holds.

Theorem 6.2.6. For a near-ring N , assume that N is 2-primal. Then for each

P ∈ Spec(N ), the following statements are equivalent:

(i) P ∈mSpec(N );

(ii) N (P ) = P.

Proof. (i) ⇒ (ii) : Let P be a minimal 0-prime ideal of N and let a ∈ P . Sup-

pose a ∈/ N (P ). Let S = {a, a2, a3, · · · }. If 0 ∈ S, then ak = 0 for some posi-

tive integer k and hence a ∈ N (N ) = P0(N ), which implies that a ∈ N (P ) by

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i

12 y 123...n−2 y

123 n 1 ∈ (y y

3

Proposition 6.2.5, a contradiction. So 0 ∈/ S. Thus S is a multiplicative system

that does not contain 0. Let L = N \ P , i.e., L is an m−system. Let T be the

set of all non-zero elements of N of the form at0 x1at1 x2 · · · atn−1 xnatn , where

xi ∈ L and the tr s are positive integers with t0 and tn allowed to be zero. Clearly,

L ⊆ T . Let M = T ∪ S. We show that M is an m−system. Let x, y ∈ M. If

x, y ∈ S , then xy ∈ S ⊆ M and we are done. Let x ∈ S and y ∈ T , say x = as

and y = at0 y1at1 y2at2 · · · ynatn . If xy = 0, then xy ∈ T . Suppose xy = 0. Since

y1, y2 ∈ L, there exist yr ∈ <y1> and yr ∈ <y2> such that yr yr ∈ L. Since 1 2 1 2

1y2, y3 ∈ L, there exist y12 ∈ <y1y2> ⊆ (<y1><y2>) and y3 ∈< y3 > such

yr r

r r r r

that yr r ∈ L. Continuing this process, we get yr

r n−1 , yn ∈ L. Then

there exist yr ··· − r 123...n−2

r n−1 ) ⊆ (· · · (((y1) (y2)) (y3)) · · · (yn−1)) and

n ∈< yn > such that w = y123···n−1yn ∈ L. Since xy = 0, xy ∈ P0(N ). Thus

yr r r

asat0 y1at1 y2 · · · ynatn ∈ P0(N ). Since P0(N ) = N (N ), P0(N ) is completely

semiprime ideal of N and hence y1y2 · · · ynas+t0 +t1 +···+tn ∈ P0(N ). Choose

m = s + t0 + t1 + · · · + tn. Then y1y2 · · · ynam ∈ P0(N ). Since P0(N ) has the

IFP, <y1><y2> · · · <yn><am> ⊆ P0(N ). Continuing this process, we obtain

m

< · · · <<< y1 >< y2 >>< y3 >> · · · < yn−1 >>< yn >< a >⊆ P0 (N )

and so yr yr am ∈ P0(N ). Hence wam ∈ P0(N ), where w = yr yr . 123···n−1 n 123···n−1 n

Since P0(N ) is a completely semiprime ideal, (aw)m∈ P0(N ) and hence

aw ∈ P0(N ). Thus a ∈ NP = N (P ), which is a contradiction. Therefore,

if x ∈ S, y ∈ T , then xy = 0 and so xy ∈ T .

Similarly, one can show that if x, y ∈ T then xy = 0 and xy ∈ T . This

shows that M is an m−system that is disjoint from (0). Hence, by Proposition

1.2.15 there is a 0-prime ideal Q that is disjoint from M such that a ∈/ Q and

Q ⊆ P . Since P is a minimal 0-prime ideal, P = Q. Therefore a ∈/ P , which


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is a contradiction. Consequently, a ∈ N (P ).

(ii) ⇒ (i) : If Q ⊆ P for Q ∈ m Spec(N ), then

N (P ) ⊆ N (Q) ⊆ Q ⊆ P = N (P )

. Therefore, P ∈ m Spec(N ).

Theorem 6.2.7. For a near-ring N , the following statements are equivalent:

(i) N is 2-primal;

(ii) OP ⊆ P for each P ∈ mSpec(N );

(iii) N (N ) = T

P ∈mSpec(N )

OP = P0(N ).

Proof. (i) ⇒ (ii) : Note that OP ⊆ N P for each P ∈ mSpec(N ). By The-

orem 6.2.3, we have N P = N (P ) ⊆ P and therefore, OP ⊆ P for each

P ∈mSpec(N ).

(ii) ⇒ (iii) : Since OP ⊆ P for each P ∈ mSpec(N ), T

P ∈mSpec(N )

OP ⊆

P0(N ). Let a ∈ N (N ). Then am = 0 ∈ O(P ) for some integer m and any

P ∈mSpec(N ). Hence a ∈ T

P ∈mSpec(N )

OP . Thus

N (N ) ⊆ \

P ∈mSpec(N )

.

(iii) ⇒ (i) : It is obvious.

OP ⊆ P0(N ) ⊆ N (N )

Proposition 6.2.8. Assume that O(P ) is a 0-prime ideal of near-ring N for each

P ∈ mSpec(N ). Then O(P ) has the IFP for each P ∈ mSpec(N ) if and only

if N is a 2-primal near-ring.

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Proof. Assume that N is a 2-primal near-ring. Let P be a minimal 0-prime ideal

of N such that O(P ) is a 0-prime ideal of N . Let xy ∈ O(P ) for x, y ∈ N . This

implies that xyN <z> = 0 for z ∈ N \ P . Then xyN <z> ⊆ P . Since z ∈/ P

and P is 0-prime, xy ∈ P . Therefore, O(P ) ⊆ P . Since P is a minimal 0-

prime ideal, O(P ) = P . Since N is 2-primal and P ∈ mSpec(N ), N (P ) = P

by Theorem 6.2.6. Therefore P is completely prime by Corollary 6.2.4 . Since

P = O(P ), O(P ) is completely prime. In particular, O(P ) has the IFP.

Conversely, suppose that O(P ) has the IFP for each P ∈ mSpec(N ). Let

x ∈ N (N ). This implies that xn = 0 for some positive integer n. So that

xn ∈ O(P ). If x ∈/ P0(N ), then there exists a minimal 0-prime ideal P of N

such that x ∈/ P . Since P is a 0-prime ideal, there exist r1, r2, · · · , rn−1∈N

such that xr1x · · · xrn−1x ∈/ P . But since O(P ) has the IFP, xr1x · · · xrn−1x ∈

O(P ). Since O(P ) ⊆ P , xr1x · · · xrn−1x ∈ P , a contradiction. Thus x ∈

P0(N ). Therefore N (N ) ⊆ P0(N ). Always P0(N ) ⊆ N (N ). Hence N (N ) =

P0(N ).

Proposition 6.2.9. If O(P ) has the IFP for each P ∈ mSpec(N ), then for every

P ∈ mSpec(N ), O(P ) is a 0-prime ideal if and only if O(P ) is a completely

prime ideal of N .

Proof. Suppose that O(P ) is a 0-prime ideal for every P ∈ mSpec(N ). Let

xy ∈ O(P ) for x, y ∈ N . If x ∈ O(P ), we have done. Suppose x ∈/ O(P ).

Since xy ∈ O(P ) and O(P ) has the IFP, xN y ⊆ O(P ). This implies that

xN yN <z> = 0 for z ∈ N \ P . This implies that xN yN <z> ⊆ P. Since P

is 0-prime, xN y ⊆ P and therefore x ∈ P or y ∈ P . By Proposition 6.2.8,

P = O(P ). Since x ∈/ O(P ), x ∈/ P . Therefore y ∈ P = O(P ). Hence O(P )

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is completely prime. The Converse is obvious.

Proposition 6.2.10. Let N be a near-ring with unity. Let O(P ) be a 0-prime

ideal of N for each P ∈ mSpec(N ). Then the following are equivalent:

(i) N is a 2-primal near-ring;

(ii) O(P ) has the IFP for each P ∈ mSpec(N );

(iii) O(P ) is a completely semiprime ideal for each P ∈ mSpec(N );

(iv) O(P ) is a symmetric ideal for each P ∈ mSpec(N );

(v) xy ∈ O(P ) implies yN x ⊆ O(P ) for x, y ∈ N and for each

P ∈ mSpec(N ).

Proof. (i) ⇒ (ii) : It follows from Proposition 6.2.8.

(ii) ⇒ (iii) : By Proposition 6.2.9, O(P ) is a completely prime ideal and

hence O(P ) is completely semiprime.

(iii) ⇒ (iv) : Suppose that O(P ) is a completely semiprime ideal for each

P ∈ mSpec(N ). Therefore it has the IFP. Let a, b, c ∈ N be such that

abc ∈ O(P ). We shall prove that acb ∈ O(P ). Since abc ∈ O(P ), there exists

s ∈ N \ P such that abcN <s> = 0. So that abcN <s> ⊆ O(P ). Since

O(P ) has the IFP , acbcN <s> ⊆ O(P ). Suppose that cN <s> ¢ O(P ).

If acb ∈/ O(P ), since O(P ) is 0-prime there exists some n ∈ N such that

acbncN <s> ¢ O(P ). It contradicts the IFP of O(P ). Therefore acb ∈ O(P ).

Suppose that cN <s> ⊆ O(P ). Since O(P ) has the IFP, cbN <s> ⊆ O(P ).

Since O(P ) is 0-prime and s ∈/ P = O(P ), cb ∈ O(P ). Therefore acb ∈ O(P ).

Hence O(P ) is a symmetric ideal in N .


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(iv) ⇒ (v) : Suppose that xy ∈ O(P ) for P ∈ mSpec(N ). Since O(P )

is symmetric and N has unity, yx ∈ O(P ). Since O(P ) has the IFP, yN x ⊆

O(P ).

(v) ⇒ (i) : Let x ∈ N (N ). Then xr = 0 for some r. So that xr ∈ O(P )

for P ∈ mSpec(N ). Suppose that x ∈/ P0(N ). Since P0(N ) = T

P , P ∈Spec(N )

x ∈/ P . Since P is a 0-prime ideal, there exist n1, n2, · · · , nr−1 ∈ N such

that xn1x · · · xnr−1x ∈/ P . Since xy ∈ O(P ), by hypothesis yN x ⊆ O(P ).

Therefore xn1x · · · xnr−1x ∈ O(P ) ⊆ P , a contradiction. Thus x ∈ P0(N ).

Hence N (N ) ⊆ P0(N ). Always P0(N ) ⊆ N (N ) and consequently N is a

2-primal near-ring.

Theorem 6.2.11. Let O(P ) be a 0-prime ideal for each P ∈ mSpec(N ). Then

the following are equivalent;

(i) N is a 2-primal near-ring;

(ii) O(P ) has the IFP;

(iii) Every minimal 0-prime ideal of N is a completely prime ideal of N .

Proof. (i) ⇒ (ii) : It follows from Proposition 6.2.8.

(ii) ⇒ (iii) : Let P be a minimal 0-prime ideal of N . Let a, b ∈ N be such

that ab ∈ P . If b ∈ P , we have done. Suppose that b ∈/ P . Since O(P ) = P ,

ab ∈ O(P ). Since O(P ) has the IFP, aN b ⊆ O(P ) = P . Since P is 0−prime

and b ∈/ P , a ∈ P . Hence, P is a completely prime ideal.

(iii) ⇒ (i) : Let x ∈ N (N ). Then xr = 0 for some r. So that xr ∈ P,

where P ∈ mSpec(N ). Since every minimal 0-prime ideal is completely prime,

x ∈ P for every P ∈ mSpec(N ).

Since P0(N ) = T

P ∈mSpec(N )

P , x ∈ P0(N ). Thus N (N ) ⊆ P0(N ).


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Theorem 6.2.12. Let O(P ) be a 0-prime ideal of N for every P ∈ mSpec(N ).

Then N is a 2-primal near-ring if and only if P = O(P ) for every minimal

0-prime ideal P of N .

Proof. Suppose that N is a 2-primal near-ring. Then O(P ) is a completely

prime ideal of N by Proposition 6.2.8. Let a ∈ O(P ). Then am ∈ O(P ). Since

O(P ) is completely prime, a ∈ O(P ). Therefore O(P ) ⊆ O(P ). Clearly,

O(P ) ⊆ O(P ). Thus O(P ) = O(P ). Since O(P ) is a 0-prime ideal of N ,

P = O(P ). Hence P = O(P ).

Conversely, assume that P = O(P ) for every minimal 0-prime ideal P of

N . Let x ∈ N (N ). This implies that xn = 0 for some n. So xn ∈ P for

every P ∈ mSpec(N ). Since P = O(P ) = O(P ), xn ∈ O(P ). Since O(P )

is completely prime, x ∈ O(P ) = O(P ) = P . This implies that x ∈ P0(N ).

Thus N (N ) ⊆ P0(N ) and consequently N is a 2-primal near-ring.

In [32], Kim and Kwak asked one question that "Is a ring R 2-primal if

OP ⊆ P for each P ∈ mSpec(R)?". Here we prove the following theorem for

near-rings.

Theorem 6.2.13. If OP has the IFP for each P ∈ mSpec(N ), then OP ⊆ P

for each P ∈ mSpec(N ) if and only if N is a 2-primal near-ring.

Proof. Let x ∈ N (N ). Then xn = 0 for some n. So that xn ∈ O(P ) ⊆

OP . Suppose x ∈/ P0(N ). Since P0(N ) = T

P ∈mSpec(N )

P , there exists

P ∈ mSpec(N ) such that x ∈/ P . Since P is a 0-prime ideal, there exist

r1, r2, · · · , rn−1 ∈ N such that xr1x · · · xrn−1x ∈/ P . But since OP has the

IFP, xr1x · · · xrn−1x ∈ OP . Again since OP ⊆ P , xr1x · · · xrn−1x ∈ P , a

contradiction.Thus x ∈ P0(N ). Hence N (N ) ⊆ P0(N ).

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6.3. NI NEAR-RINGS WHICH ARE WEAKLY π-REGULAR

Conversely, assume that N is a 2-primal near-ring. By Theorem 6.2.7, OP ⊆

P for each P ∈ mSpec(N ). Since OP ⊆ OP , OP ⊆ P for each P ∈

mSpec(N ).

6.3 NI near-rings which are weakly π-regular

In this section, we show that if N is an NI near-ring which satisfies the condition

(CZ2), then (i) every strongly 0-prime ideal is maximal if and only if N is left

weakly π-regular. (ii) P ∈ mSSpec(N ) if and only if P = O (P ).

For a near-ring N , let (m)SSpec(N ) be the set of all (minimal) strongly

0-prime ideals of N . For P ∈ SSpec(N ), we have

O(P ) = {a ∈ N | aN = 0 for some b ∈ N \ P } ,

O (P ) = {a ∈ N | am ∈ O(P ) for some positive integer m} ,

OP = {a ∈ N | ab = 0 for some b ∈ N \ P } ,

OP = {a ∈ N | am ∈ OP for some positive integer m} ,

N (P ) = {a ∈ N | aN ⊆ N ∗(N ) for some b ∈ N \ P } .

Example 6.3.1. Consider the near-ring (N, +, .) defined on the Klein’s four

group (N, +) with N = {0, a, b, c} where . is defined as follows (as per scheme

2,p.408 [35]).

· 0 a b c

0 0 0 0 0

a 0 0 a a

b 0 a b b

c 0 a c c

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Clearly {0, a} is a strongly 0-prime ideal, since the ideals are {0}, {0, a}, and

{0, a, b, c}. Let P = {0, a}. Then O {P } = OP = {0} and O(P ) = OP =

N {P } = N (P ) = NP = N P = P .

Also we observe that N (N ) = {0, a} = N ∗(N ). Therefore N is a NI near-ring.

Theorem 6.3.2. For a near-ring N the following are equivalent:

(i) N is an NI near-ring;

(ii) Every minimal strongly 0-prime ideal of N is completely prime.

Proof. (i) ⇒ (ii) : Let P be a minimal strongly 0-prime ideal of N . Let a, b ∈ N

be such that ab ∈ P and b ∈/ P . We will show that a ∈ P .

Case (i): Suppose that (ab)k = 0 for some k. Then (ab)k ∈ N ∗(N ). Since

N ∗(N ) is completely semiprime by Theorem 1.2.17, ak bk ∈ N ∗(N ). Since b ∈/

P , there exist z1, z2, z3, · · · , zk−1 ∈ N such that bz1bz2b · · · zk−1b ∈/ P and since

N ∗(N ) has the IFP, ak N (bz1bz2 · · · zk −1b) ∈ N ∗(N ). Again by completely

semiprimeness of N ∗(N ), we have aN bz1bz2 · · · zk−1b ∈ N ∗(N ). Hence

a ∈ N (P ) ⊆ P .

Case (ii): Suppose that (ab)k = 0 for all k > 0. Let S = {(ab)r| r ≥ 1},

L = N \ P and T = {n ∈ N | n = 0, n = (ab)t0 x1(ab)t1 x2(ab)t2 x3 · · · (ab)tr ,

where ti ≥ 1, i = 1, 2, · · · , r − 1, ti ≥ 0, i = 0, n and xi ∈ L for all

i}. Clearly, S = {0} and L ⊆ T . Let M = S ∪ T . We shall prove

that M is an m-system in N \ {0}. Let x, y ∈ M . If x, y ∈ S, then

xy ∈ S ⊆ M . If x ∈ S and y ∈ T , then let x = (ab)q for some q > 0 and

y = (ab)t0 x1(ab)t1 x2 · · · xr(ab)tr . Suppose that xy = 0. Since x1, x2 ∈ L, there

exist xr ∈< x1 > and xr ∈< x2 > such that xr xr ∈ L. Since xr xr , x3 ∈ L, 1 2 1 2 1 2

there exist xr ∈< xr xr >⊆<< x1 >< x2 >> and xr ∈< x3 > such that 12 1 2 3

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123 r 1 ∈< x x

r 123···r−1 x

123···r−1 r

w = x r

r

12x3 ∈ L. Continuing this process, we get x123···r−2xr−1, xr ∈ L. Then there

xr r r r

exist xr ··· − r 123···r−2

r r−1 >⊆< · · · <<< x1 >< x2 >>< x3 >> · · · <

xr−1 >> and xr ∈< xr > such that w = xr r

∈ L. Since xy = 0, xy ∈

N ∗(N ). Thus (ab)q (ab)t0 x1(ab)t1 x2 · · · xr(ab)tr ∈ N ∗(N ). Since N ∗(N ) is

completely semiprime ideal of N , x1x2 · · · xr(ab)q+t0 +···+tr ∈ N ∗(N ). Choose

m = q + t0 + · · · + tr. Then x1x2 · · · xr(ab)m ∈ N ∗(N ). Since N ∗(N ) has

the IFP, < x1 >< x2 > · · · < xr >< (ab)m >⊆ N ∗(N ). This implies that

m

< · · · <<< x1 >< x2 >>< x3 >> · · · < xr−1 >>< xr >< (ab) >⊆

N ∗(N ) and so xr xr (ab)m ∈ N ∗(N ). Hence w(ab)m ∈ N ∗(N ), where

r 123···r−1 x

r . Since N ∗(N ) is completely semiprime, ((ab)w)m ∈ N ∗(N )

and hence by case(i) a ∈ P . Suppose that xy = 0. Then, clearly, from the

definition of T , xy ∈ T ⊆ M . Similarly, we can show that if x, y ∈ T then

xy ∈ T ⊆ M . Thus we have M is an m−system in N \ {0}. By Proposition

1.2.15, there is a 0-prime ideal Q that is disjoint from M such that ab ∈/ Q and

Q ⊆ P . Now we claim that Q is strongly 0-prime. Suppose I /Q is a nonzero

nil ideal of N/Q. Since Q ⊆ I , I ∩ M = ∅. If (ab)m ∈ I for some positive

integer m, then (ab)m + Q is a nilpotent element in N/Q. Thus (ab)mk ∈ Q

for some positive integer k, which is a contradiction. So we choose x ∈ I ∩ T .

Then x ∈ T implies 0 = xt ∈ T for any positive integer t. Since x + Q is nilpo-

tent in N/Q, xs ∈ Q for some positive integer s which is again a contradiction.

Therefore, Q is a strongly 0-prime ideal of N such that ab ∈/ Q and Q ⊆ P .

Since P ∈ mSSpec(N ), Q = P . Therefore ab ∈/

and consequently a ∈ P .

P which is a contradiction

(ii) ⇒ (i) : Let xn = 0 for some positive integer n. Then xn ∈ P for all

minimal strongly 0-prime ideal P of N . Since every minimal strongly 0-prime

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ideal of N is completely prime, x ∈ P . Thus x ∈ T

P ∈mSSpec(N )

P = N ∗(N ) by

Lemma 1.2.16. Therefore, N is an NI near-ring.

The following definitions are critical to our characterization of minimal

strongly 0-prime ideals.

Definition 6.3.3. Let x, y ∈ N and n a positive integer. We say N satisfies

1) (CZ1) condition if whenever (xy)m = 0 then xmym = 0, for some positive

integer m.

2) (CZ2) condition if whenever (xy)m = 0 then <x>mN <y>m = 0, for

some positive integer m.

Hong et al. [29] observed that the condition "R/N ∗(R) is right weakly

π-regular" in Proposition 18 cannot be replaced by the condition "R is right

weakly π-regular " in the case of rings. But the following theorem shows that

in the case of NI near-ring which satisfies (CZ2) the condition "N/N ∗(N ) is

left weakly π-regular" can be replaced by the condition "N is left weakly π-

regular".

Theorem 6.3.4. Let N be an NI near-ring with unity satisfying (CZ2). Then

the following are equivalent:

(i) N is left weakly π-regular;

(ii) N/N ∗(N ) is left weakly π-regular;

(iii) Every strongly 0-prime ideal of N is maximal.

Proof. (i) ⇒ (ii) is clear.

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12 b 123···n−2 b n

123 n 1 ∈< b b

n 123···n−1 b

3

n

(ii) ⇒ (iii) : Let P ∈ SSpec(N ). Then there exists a minimal strongly 0-

prime ideal I ⊆ P which is completely prime by Theorem 1.2.18. Let

N = N/I . Then N is an integral left weakly π-regular near-ring. Let x be a non

zero element in N . There exists a positive integer k such that xk ∈< xk > xk .

Then xk = yxk , where y ∈< xk >. Therefore xk − yxk = 0 which implies that

(1 − y)xk = 0. Hence N/I is a simple near-ring. Thus I is a maximal ideal and

so is P .

(iii) ⇒ (i) : Suppose that N is not left weakly π-regular. Then there

exists an element a ∈ N such that a is not left weakly π-regular. So we

have ak ∈/< ak > ak for every positive integer k. Hence ak = 0 for all

k > 0 and a ∈/< a > a. Then < a > is contained in a maximal ideal

which is also a strongly 0-prime ideal. Let T be the union of all strongly

0-prime ideals which contain a. Let S = N \ T . Since every strongly 0-

prime ideal is maximal, every strongly 0-prime ideal is minimal. Since every

minimal strongly 0-prime ideal is completely prime, by Theorem 6.3.2, S is

multiplicatively closed. Let F = {at0 b1at1 · · · bnatn = 0| bi ∈ S and ti ∈

{0} ∪ N, where N is the set of all positive integers}. Let L = {a, a2, · · · } and

let M = L ∪ F . Clearly, S ⊆ F ⊆ M . We shall claim that M is an m-system in

N \ {0}. Let x, y ∈ M . Assume x ∈ L and y ∈ F . Suppose that xy = 0. Take

x = ar and y = at0 b1at1 b2 · · · bnatn . Since xy = 0, arat0 b1a

t1 b2 · · · bnatn = 0.

Since b1, b2 ∈ S, there exist br ∈< b1 > and br ∈< b2 > such that br br ∈ S. 1 2 1 2

Since br , b3 ∈ S, there exist br ∈ ⊆<< b1 >< b2 >> and br ∈< b3 >

2 12 1 2 3

such that br r ∈ S. Continuing this process, we get br

r n−1 , b

r ∈ S.

Then there exist br ··· − r 123···n−2

r n−1 >⊆< · · · <<< b1 >< b2 >><

b3 >> · · · < bn−1 >> and br ∈< bn > such that br r

∈ S.

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Since xy = 0 and N ∗(N ) has the IFP, ar+t0 +t1 +···+tn b1b2 · · · bn ∈ N ∗(N ).

Choose m = r + t0 + t1 + · · · + tn. Then amb1b2 · · · bn ∈ N ∗(N ). Since

N ∗(N ) has the IFP, < am >< b1 >< b2 > · · · < bn >⊆ N ∗(N ) and so

ambr r ∗ m ∗ r r

123···n−1bn ∈ N (N ). Hence a w ∈ N (N ), where w = b123···n−1bn. Since

N ∗(N ) is completely semiprime, (wa)m ∈ N ∗(N ) and hence (wa)k = 0 for

some k. Since N satisfies (CZ2), < wk > N < ak >= 0 for some k. By the

definition of S and T , a strongly 0-prime ideal cannot contain both ak and wk .

Hence

< ak > + < wk > = N.

So

< ak > ak + < wk > ak = N ak .

Since < wk > N < ak >= 0, < wk > ak = 0. Therefore ak ∈< ak > ak .

This shows that a is left weakly π-regular, a contradiction. Hence 0 = xy ∈ M .

Similarly, we can prove that if x, y ∈ F, then 0 = xy ∈ M . Thus M is an

m−system in N \ {0}. By Proposition 1.2.15, there is a 0-prime ideal Q that

is disjoint from M such that a ∈/ Q. As in the proof of Theorem 6.3.2, we

obtain Q is a strongly 0-prime ideal of N . Since a ∈/ Q, Q+ < a >= N .

Hence 1 = b + c for some b ∈ Q and c ∈< a >. This gives b ∈/ T . So

that b ∈ S ⊆ F ⊂ M which implies that Q ∩ M = ∅, a contradiction and

consequently N is left weakly π-regular.

Corollary 6.3.5. Let N be a 2-primal near-ring with unity satisfying condition

(CZ2). Then the following are equivalent:

(i) N is left weakly π- regular;

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(ii) N/N ∗(N ) is left weakly π-regular;

(iii) N/P0(N ) is left weakly π-regular;

(iv) Every 0-prime ideal of N is maximal;

(v) Every strongly 0-prime ideal of N is maximal.

Proof. (i) ⇒ (ii) and (ii) ⇒ (v) follows from Theorem 6.3.4.

(i) ⇒ (iii) is clear.

(iii) ⇒ (iv) : Let P be a 0-prime ideal of N . Then, by Corollary 6.2.4 and

Theorem 6.2.6, there exists a minimal 0-prime ideal X ⊆ P which is com-

pletely prime. Let N = N/X . Then N is an integral left weakly π−regular

near-ring. Let a be a non-zero element in N . There exists a positive integer

k such that ak ∈< ak > ak . Then ak = yak where y ∈< ak >. Therefore

ak − yak = 0 which implies that (1 − yk )ak = 0. Hence N/X is a simple

near-ring. Thus X is a maximal ideal and so is P .

(iv) ⇒ (v) : Since every strongly 0-prime ideal is 0-prime and every 0-prime

ideal is maximal, every strongly 0-prime ideal is maximal.

We now characterize the minimal strongly 0-prime ideal P of N in terms of

OP and O (P ).

Theorem 6.3.6. Let N be an NI near-ring and P a strongly 0-prime ideal of N .

1. If N satisfies (CZ1), then P is a minimal strongly 0-prime ideal of N if

and only if P = OP .

2. If N satisfies (CZ2), then P is a minimal strongly 0-prime ideal of N if

and only if P = O (P ).

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Proof. 1) : Let P be a minimal strongly 0-prime ideal of N . Then by Theorem

6.3.2, P is completely prime and so S = N \ P is an m−system. For a ∈ P , if

we suppose that ak = 0 for some k > 0, then there is nothing to prove. Assume

that ak = 0 for all k > 0. Construct M as in the proof of Theorem 6.3.4. Let

x, y ∈ M . Then either xy = 0 or xy = 0. By the similar method to that of

Theorem 6.3.4, (aw)k = 0 for some k, w ∈ N \ P or Q ∩ M = ∅ for some

strongly 0-prime ideal Q. Suppose the latter is true. Then Q ⊆ P . Since P

is minimal strongly 0-prime, Q = P . So that a ∈ Q and hence Q ∩ M = ∅,

a contradiction. Thus (aw)k = 0 for some k > 0. Since N satisfies (CZ1),

aq wq = 0 for some q > 0. Hence aq ∈ OP , because wq ∈ S and consequently

a ∈ OP . Hence P ⊆ OP .

On the other hand, let x ∈ OP . Then there exist a positive integer n and

s ∈ N \ P such that xns = 0 and so xns ∈ N ∗(N ). Since N is NI , N ∗(N ) is

completely semiprime and therefore we obtain xs ∈ N ∗(N ). Since N ∗(N ) has

the IFP, xN s ⊆ N ∗(N ) ⊆ P . Since P is strongly 0-prime, x ∈ P . Therefore

OP ⊆ P . Thus OP = P .

Conversely, assume that OP = P . We have to show that P is a minimal

strongly 0-prime ideal of N . Suppose that there is a strongly 0-prime ideal Q of

N such that Q ⊆ P . Then P = OP ⊆ OQ ⊆ Q. So that P = Q. Therefore, P

is a minimal strongly 0-prime ideal of N.

2) Let P be a minimal strongly 0-prime ideal of N and let a ∈ P . By

a similar method used in part(1), we obtain (aw)k = 0 for some k > 0 and

w ∈ N \ P . Since N satisfies (CZ2), < aq > N < wq >= 0 for some q > 0.

Since P is minimal strongly 0-prime, P is completely prime by Theorem 6.3.2.

Hence wq ∈ N \ P and so that aq ∈ O(P ) and consequently a ∈ O (P ). For

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the reverse inclusion, let x ∈ O (P ). Then xnN <s> = 0 for some n > 0 and

s ∈ N \P . Hence xnN <s> ⊆ N ∗(N ). Since N ∗(N ) is completely semiprime,

xN <s> ⊆ N ∗(N ) ⊆ P . Since P is strongly 0-prime, x ∈ P . Therefore

O (P ) ⊆ P . Thus P = O (P ). The converse is similar to the converse of

part(1).

Corollary 6.3.7. Let N be a near-ring with unity which satisfies the condition

(CZ1). Then N is NI if and only if P = OP for every minimal strongly 0-prime

ideal P of N .

Proof. Suppose that P = OP for every minimal strongly 0-prime ideal P of N .

Then

N ∗(N ) = \

P ∈mSSpec(N )

P = \

OP

P ∈mSSpec(N )

. Let xn = 0 for some n > 0. Then xn ∈ OP for all P ∈ mSSpec(N ). This

implies that x ∈ N ∗(N ). Thus N is NI. The converse follows from Theorem

6.3.6.

CHAPTER 6 WEAKLY π-REGULAR NI...

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