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Some notes on graded weakly 1-absorbing primary ideals

  • Azzh Saad Alshehry , Rashid Abu-Dawwas EMAIL logo and Majd Al-Rashdan
Published/Copyright: November 21, 2023
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Abstract

A proper graded ideal P of a commutative graded ring R is called graded weakly 1-absorbing primary if whenever x , y , z are nonunit homogeneous elements of R with 0 x y z P , then either x y P or z is in the graded radical of P . In this article, we explore more results on graded weakly 1-absorbing primary ideals.

MSC 2010: 13A02; 13A15; 16W50

1 Introduction

In dispersion through this article, G is a group and R is a commutative ring with nonzero unity 1 unless specified differently. If R = g G R g with the property R g R h R g h for all g , h G , where R g is an additive subgroup of R for all g G , then R is aforementioned to be a graded ring (gr-R). The aspects of R g are called homogeneous of degree g . If s R , then s can be expressed uniquely as g G s g , where s g is the component of s in R g , and s g = 0 is represented by the symbol. The set of all homogeneous aspects of R is g G R g and is denoted by h ( R ) . The component R e is a subring of R and 1 R e . Let R be a gr-R and P be an ideal of R . Then, P is aforementioned to be a graded ideal (gr-I) if P = g G ( P R g ) , i.e., for p P , p g P for all g G . An ideal of a gr-R is not necessarily gr-I. For a G -gr-R R and a gr-I P of R , R P is a G -gr-R with ( R P ) g = ( R g + P ) P for all g G . For further phrasing, see [1].

A proper gr-I P of R is aforementioned to be a graded prime ideal (gr-p-I) if x y P implies either x P or y P , for all x , y h ( R ) [2]. It is clear that if P is a prime ideal of R and it is a gr-I, then P is a gr-p-I of R . Indeed, the example below demonstrates that a gr-p-I is not necessarily a prime ideal:

Example 1.1

Consider R = Z [ i ] and G = Z 2 . Then, R is gr-R by R 0 = Z and R 1 = i Z . Consider the gr-I P = p R of R , where p is a prime number with p = c 2 + d 2 , for some c , d Z . We show that P is a gr-p-I of R . Let x y P for some x , y h ( R ) .

Case 1 ̲ : Assume that x , y R 0 . In this instance, if x , y Z , where p divides x y , then either p divides x or p divides y , which implies that x P or y P .

Case 2 ̲ : Assume that x , y R 1 . In such a case, x = i a and y = i b for some a , b Z such that p divides x y = a b , and then p divides a or p divides b in Z , which suggests that p divides x = i a or p divides y = i b in R . Then, there is that x P or y P .

Case 3 ̲ : Consider that x R 0 and y R 1 . In this instance, x Z and y = i b for some b Z such that p divides x y = i x b in R , i.e., i x b = p ( α + i β ) for some α , β Z . Then, we obtain x b = p β , i.e., p divides x b in Z , and again p divides x or p divides b , which implies that p divides x or p divides y = i b in R . Thus, x P or y P .

So, P is a gr-p-I of R . On the other hand, P is not a prime ideal of R since ( c i d ) ( c + i d ) = c 2 + d 2 = p P , ( c i d ) P , and ( c + i d ) P .

Allow for P to be a gr-I of R . Then, the graded radical of P is denoted by Gr-rad ( P ) and is defined as follows:

Gr-rad ( P ) = s = g G s g R : g G , n g N s.t. s g n g P .

Recall that Gr-rad( P ) is every time a gr-I of R [2].

A proper gr-I P of R is aforementioned to be a graded primary ideal (gr-py-I) if x y P suggests either x P or y Gr-rad ( P ) , for all x , y h ( R ) [3]. In this situation, Q = Gr-rad ( P ) is a gr-p-I of R and P is allegedly graded Q -primary.

Since gr-p-I’s and gr-py-I’s are vital in commutative graded ring theory, numerous authors have looked into various generalizations of these gr-Is. Atani [4] proposed the idea of graded weakly prime ideals. A proper gr-I P of R is called a graded weakly prime ideal (gr-wp-I) whenever x , y h ( R ) and 0 x y P , then x P or y P . Atani [5] presented the impression of graded weakly primary ideals. A proper gr-I P of R is called a graded weakly primary ideal (gr-w-py-I) of R if whenever x , y h ( R ) and 0 x y P , then x P or y Gr-rad ( P ) . New generalizations of graded primary ideals and graded weakly primary ideals are, accordingly, the notions of graded 1-absorbing primary ideals and graded weakly 1-absorbing primary ideals proposed by Abu-Dawwas and Bataineh [6,7]. A proper gr-I P of R is called a graded 1-absorbing primary ideal (gr-1-ab-py-I) if whenever nonunit elements x , y , z h ( R ) and x y z P , then x y P or z Gr-rad ( P ) . A proper gr-I P of R is called a graded weakly 1-absorbing primary ideal (gr-w-1-ab-py-I) if whenever nonunit elements x , y , z h ( R ) and 0 x y z P , then x y P or z Gr-rad ( P ) . Certainly, every gr-py-I is gr-1-ab-py-I. The following example demonstrates that the converse is not true in general:

Example 1.2

[6] Consider R = K [ X , Y ] , where K is a field, and G = Z . Then, R is gr-R by R n = i + j = n , i , j 0 K X i Y j for all n Z . Recall that deg ( X ) = deg ( Y ) = 1 . Consider the gr-I P = X 2 , X Y of R . Then, Gr-rad ( P ) = X , and it is obvious that P is a gr-1-ab-py-I of R . On the contrary, P is not gr-py-I of R Example 2.11 in the study by Soheilnia and Darani [8].

It is recognizable that a gr-1-ab-py-I of R is gr-w-1-ab-py-I. However, since { 0 } is always gr-w-1-ab-py-I, a gr-w-1-ab-py-I of R is not necessarily gr-1-ab-py-I, see Example 1.3.

Example 1.3

[7] Propose R = Z 6 [ i ] and G = Z 2 . So, R is gr-R by R 0 = Z 6 and R 1 = i Z 6 . Now, P = { 0 } is a gr-w-1-ab-py-I of R . On the other hand, 2 , 3 h ( R ) such that 2.2.3 P with neither 2.2 P nor 3 Gr-rad ( P ) . Hence, P is not a gr-1-ab-py-I of R .

Definition 1.4

[6,9] Let R be a G -graded ring and P be a graded ideal of R . Assume that g G , where P g R g . Then,

  • P is supposedly a g -1-absorbing primary ideal ( g -1-ab-py-I) of R if whenever nonunit elements x , y , z R g such that x y z P , then x y P or z Gr-rad ( P ) .

  • P is presumably a g -weakly 1-absorbing primary ideal ( g -w-1-ab-py-I) of R , whenever nonunit elements x , y , z R g , where 0 x y z P , then x y P or z Gr-rad ( P ) .

  • P is repeatedly a g -prime ideal ( g -p-I) of R if whenever x , y R g , where x y P , then either x P or y P .

  • P is presumably a g -primary ideal ( g -py-I) of R , if whenever x , y R g , where x y P , then either x P or y Gr-rad ( P ) .

  • P is supposedly a g -weakly primary ideal ( g -w-py-I) of R if whenever x , y R g , where 0 x y P , then either x P or y Gr-rad ( P ) .

In this article, we explore more outcomes on graded weakly 1-absorbing primary ideals. In fact, the study by Almahdi et al. [10] inspired quite a few of the outcomes. Among a number of outcomes, we proved that if R e is a nonlocal ring and P is an e -w-1-ab-py-I of R that is not an e -w-py-I, then either P e 3 = 0 or P e 2 = s with s as an idempotent such that 1 s is a maximal ideal of R e (Theorem 2.4). In addition, we showed that if every nonzero gr-py-I of R is a gr-p-I and Gr-rad ( 0 ) is a gr-m-I of R , then either Gr-rad ( 0 ) = 0 or Gr-rad ( 0 ) is the unique nonzero proper gr-I of R (Proposition 2.8). In addition, we proved that if R is a HUN-ring and { 0 } is a gr-py-I of R , then R is a gr-loc-R with gr-m-I Gr-rad ( 0 ) (Theorem 2.12). Moreover, a nice characterization was introduced in Theorem 2.13. In addition, we showed that if R is a finitely generated gr-loc-R with gr-m-I X , R is a gr-D, and every gr-1-ab-py-I of R is a gr-w-py-I, then R is either HUN-ring or X is the unique nonzero gr-p-I of R (Theorem 2.14). Furthermore, we proved that if R is a first strongly gr-R, then every e -w-1-ab-py-I of R is an e -s-py-I if and only if Gr-rad ( 0 ) is an e -p-I of R (Proposition 2.16). Finally, we showed that if R is a reduced first strongly gr-R, then every e -w-1-ab-py-I of R is an e -1-ab-py-I if and only if R e is a domain (Proposition 2.19).

2 Results

Our results are presented in this paragraph.

Proposition 2.1

Let P be a gr-I of R such that Gr-rad ( P ) = P . If P is a g-w-1-ab-py-I of R, then P is a g-p-I of R or r 3 = 0 for all r P g .

Proof

Suppose that r P g exists, where r 3 0 . Let x , y R g in a manner that x y P . We may assume that x and y are nonunit. If x 2 y 0 , then x 2 P or y Gr-rad ( P ) . Hence, x Gr-rad ( P ) = P or y Gr-rad ( P ) = P . Similarly, if x y 2 0 , we arrive at the same result. Now, suppose that x 2 y = x y 2 = 0 . If x 2 P g 0 , then there exists s P g such that x 2 s 0 , and 0 x 2 s = x 2 ( y + s ) P . If y + s is a unit, then x P . Otherwise, x 2 P or y + s P . Thus, x P or y P . Similarly, if y 2 P g 0 , then x P or y P . Suppose that x 2 P g = y 2 P g = 0 . We have ( x 2 + r ) 2 ( y 2 + r ) = r 3 P . If x 2 + r (resp. y 2 + r ) is a unit, then y P (resp. x P ). Otherwise, ( x 2 + r ) 2 P or y 2 + r P . Thus, x P or y P . Finally, we establish that P is a g -p-I of R .□

Corollary 2.2

Let P be a gr-I of R in such a way that Gr-rad ( P ) = P . If P is a g-w-1-ab-py-I of R which is not a g-p-I, then P g Gr-rad ( { 0 } ) .

Proof

Apply Proposition 2.1.□

Lemma 2.3

Let R be a gr-R. Then, R e holds all homogeneous idempotent elements of R.

Proof

Let x h ( R ) be an idempotent. Then, x R g for some g G , and then x = x 2 = x . x R g R g R g 2 . If x = 0 , then it has been completed. Suppose that x 0 . Then, 0 x R g R g 2 , which suggests that g 2 = g , i.e., g = e . As a deduction, x R e .□

Theorem 2.4

Allow for R to be a gr-R such that R e is a nonlocal ring. If P is an e-w-1-ab-py-I of R that is not an e-w-py-I, then

  • P e 3 = 0 , or

  • P e 2 = s with s as an idempotent such that 1 s is a maximal ideal of R e .

Proof

Let us say that (2) is not met. Since P is not an e -w-py-I, there exists x , y R e in such a way that 0 x y P , x P , and y Gr-rad ( P ) . Certainly, x and y are nonunits. Suppose that v x P for all nonunit v R e . Let u be a unit in R e . If v + u is a nonunit, then ( v + u ) x P , and so u x P , a contradiction since x P . Hence, for each nonunit v R e and each unit u R e , v + u is a unit. Thus, by Lemma 1 in the study by Badawi and Celikel [11], R e is a local ring, a contradiction. Because of that, there exists a nonunit v R e such that v x P . If v x y 0 , then v x P since y Gr-rad ( P ) and P is an e -w-1-ab-py-I, a contradiction. Hence, v x y = 0 . Presume the existence of p P e such that v x p 0 . Then, 0 v x p = v x ( y + p ) P . If y + p is a unit, then v x P , a contradiction. Hence, since v x P , we obtain y + p Gr-rad ( P ) . Thus, y Gr-rad ( P ) , a contradiction. Consequently, v x P e = 0 . Consider the existence of p P e in such a way that v y p 0 . Then, 0 v y p = v ( x + p ) P . If x + p = u is a unit, then u y = x y + p y P , and so y P , a contradiction. Hence, x + p is a nonunit and v ( x + p ) P . So, v x P , a contradiction. Consequently, v y P e = 0 . Suppose that there exist p , q P , where v p q 0 . Then, 0 v p q = v ( x + p ) ( y + q ) P . As above, x + p and y + q are nonunits. Hence, v ( x + p ) P . So, v x P , a contradiction. Therefore, v P e 2 = 0 . Assume there exists p P e , where x y p 0 . Then, 0 x y p = ( v + p ) x y P . Suppose that u = v + p is a unit. Then, u p 2 = p 3 . Hence, ( p u 1 ) 3 = ( p u 1 ) 2 . Thus, s = ( p u 1 ) 2 is an idempotent. For each q , t P e , we have q t u = q t p and q p u = q p 2 . Thus, q t u 2 = t ( q p u ) = t q p 2 . Hence, q t = q t s . Then, P e 2 s P e 2 . Therefore, P e 2 = s . By assumption, 1 s is not a maximal ideal of R e . If s is a maximal ideal of R e , then P e = P e 2 = s , a contradiction since P is not an e -w-py-I. Thus, neither 1 s nor s is a maximal ideal of R e . Hence, R e R e s × R e 1 s is a product of two nonfield rings. By Theorem 13 of the study by Badawi and Yetkin [12], P is an e -py-I, a contradiction. Subsequently, v + p is a nonunit, and so ( v + p ) x P . Then, v x P , a contradiction. As a consequence, x y P e = 0 . Consider the existence of p , q P e in such a way that x p q 0 . Then, 0 x p q = x ( v + p ) ( y + q ) P . As above, v + p and y + q are nonunits. Hence, x ( v + p ) P . So, v x P , a contradiction. As a consequence, x P e 2 = 0 . Suppose that there exist p , q P e where y p q 0 . Consequently, 0 y p q = ( v + p ) ( x + q ) y P . As above, v + p and x + q are nonunits. Hence, ( v + p ) ( x + q ) P . So, v x P , a contradiction. As a consequence, y P e 2 = 0 . Let p , q , t P e in such a way that p q t 0 . Afterward, ( v + p ) ( x + q ) ( y + t ) = p q t 0 . As above, v + p , x + q , and y + t are nonunits. Then, ( v + p ) ( x + q ) P or y + t Gr-rad ( P ) . That is, v x P or y Gr-rad ( P ) , a contradiction. Hence, P e 3 = 0 .□

A gr-R R is said to be strongly graded if 1 R g R g 1 for all g G , that is equivalent to R g R h = R g h for all g , h G [1]. A gr-R R is said to be first strongly graded if 1 R g R g 1 for all g supp ( R , G ) = { g G : R g 0 } [13]. Undoubtedly, if R is strongly graded, then R is first strongly graded. The following example, however, demonstrates that the converse is not always true.

Example 2.5

Let R = M 2 ( K ) (the ring of all 2 × 2 matrices with entries from a field K ) and G = Z 4 . Then R is gr-R by

R 0 = K 0 0 K , R 2 = 0 K K 0 and R 1 = R 3 = { 0 } .

R is first strongly graded since I R 0 R 0 and I R 2 R 2 , but R is not strongly graded since R 1 R 3 = 0 R 0 .

Without any doubt, if R is strongly graded, then supp ( R , G ) = G . Besides, if R is first strongly graded, then supp ( R , G ) is a subgroup of G . Actually, R is first strongly graded on the condition that supp ( R , G ) is a subgroup of G and R g R h = R g h for all g , h supp ( R , G ) .

Theorem 2.6

Let R be a first strongly gr-R such that R e is a nonlocal reduced ring. Suppose that P is an e-w-1-ab-py-I of R. If P is not an e-w-py-I, then Gr-rad ( P e ) = P e .

Proof

If P e 3 = 0 , then P e = 0 and P g = R g P = R e R g R e P = R e R g R g R g 1 P = R g ( R e R g 1 P ) R g ( R e P ) = R g P e = 0 for all g supp ( R , G ) . Besides, for g supp ( R , G ) , R g = 0 , which implies that P g = R g P = 0 . Hence, P g = 0 for all g G , i.e., P = 0 , which is an e -w-py-I, a contradiction. So, by Theorem 2.4, P e 2 = s with s as an idempotent such that 1 s is a maximal ideal of R e . We have that R e R e s × R e 1 s with the isomorphism f ( r ) = ( r + s , r + 1 s ) . Let R 1 = R e s and K = R e 1 s . Then, without any doubt, K is a field and f ( P e 2 ) = { 0 } × K . While maintaining generality, set R e = R 1 × K and P e = I × J such that I and J are graded ideals of R 1 and K , respectively. For that reason, since P e 2 = { 0 } × K and R 1 is reduced, we conclude that P e = { 0 } × K . In addition, Gr-rad ( P e ) = Gr-rad ( { 0 } ) × K = { 0 } × K = P e since R 1 is reduced.□

A proper gr-I X of R is allegedly a graded maximal ideal (gr-m-I) of R if whenever I is a gr-I of R with X I R , then I = X or I = R . Assuredly, every gr-m-I is a gr-p-I. A gr-R R is assumed to be a graded local ring (gr-loc-R) if R has a unique gr-m-I.

Proposition 2.7

Allow for R to be a gr-loc-R with gr-m-I X. Assume that P is a gr-p-I of R such that P X . Then, PX is a gr-1-ab-py-I of R .

Proof

Take note of the fact that Gr-rad ( P X ) = P . Suppose that x y z P X for some nonunit elements x , y , z h ( R ) . If x P or y P , then without a doubt, x y P X . Assume that neither x P nor y P . Then, x y P . Since x y z P X P and x y P , we ultimately decide that z P = Gr-rad ( P X ) . Thus, P X is a gr-1-ab-py-I of R .□

Proposition 2.8

Allow for R to be a gr-R such that every nonzero gr-py-I of R is a gr-p-I. If Gr-rad ( 0 ) is a gr-m-I of R, then either Gr-rad ( 0 ) = 0 or Gr-rad ( 0 ) is the unique nonzero proper gr-I of R.

Proof

If R is a gr-D, then Gr-rad ( 0 ) = 0 . Assume that R is not a gr-D. Allow for J to be a nonzero proper gr-I of R . Then, Gr-rad ( 0 ) Gr-rad ( J ) , and then as Gr-rad ( 0 ) is a gr-m-I of R , Gr-rad ( 0 ) = Gr-rad ( J ) . So, Gr-rad ( J ) is a gr-m-I of R , which implies that J is a gr-py-I of R by Proposition 1.11 of the study by Refai and Al-Zoubi [3], and then J is a gr-p-I of R , and so J = Gr-rad ( J ) = Gr-rad ( 0 ) . As a consequence, Gr-rad ( 0 ) is the unique nonzero proper gr-I of R .□

A gr-R R is said to be a graded domain (gr-D) if R has no homogeneous zero divisors, and is said to be a graded field (gr-F) if every nonzero homogeneous element of R is unit [1]. Assuredly, if R is a domain (field) and it is graded, then R is a gr-D (gr-F). Nevertheless, Example 2.4 of the study by Abu-Dawwas [14] shows that a gr-D (gr-F) is not necessarily a domain (field). Recall from [15] and [16, Proposition 2.25], if every element of R is either nilpotent or unit, or alternatively if all of its nonunit elements are products of unit and nilpotent elements, then R is said to be a UN-ring. A straightforward UN-ring example is Z 9 Z . In fact, we present the idea of HUN-rings:

Definition 2.9

A gr-R R is presumably a HUN-ring if every homogeneous element of R is either a unit or a nilpotent.

Absolutely, if R is a UN-ring and it is graded, then R is a HUN-ring. A HUN-ring is not always a UN-ring, as the example below demonstrates:

Example 2.10

Let K be a field and u K with u 2 = 1 . Assume that R = { α + u β : α , β K } and G = Z 2 . Then R is a gr-R by R 0 = K and R 1 = u K . By Example 2.4 of the study by Abu-Dawwas [14], R is a gr-F, and then R is a HUN-ring. But R is not a UN-ring since 1 + u R is neither a unit nor a nilpotent.

For a gr-R R , the set of all homogeneous zero divisors of R , H Z ( R ) , and the set of all zero divisors of R , Z ( R ) , are not the same. Indeed, H Z ( R ) Z ( R ) , but, in Example 2.10, 1 + u Z ( R ) as ( 1 + u ) ( 1 u ) = 0 , while 1 + u H Z ( R ) as 1 + u h ( R ) . For a gr-R R , H Z ( R ) is not necessarily a gr-I of R since it is not necessarily an ideal; consider R = Z 6 [ i ] , G = Z 2 , R 0 = Z 6 , and R 1 = i Z 6 , Note that 2 , 3 H Z ( R ) with 2 + 3 = 5 H Z ( R ) . Nevertheless, if H Z ( R ) is a gr-I in some gr-R R , then H Z ( R ) should be a gr-p-I of R . To see this, let x , y h ( R ) , where x y H Z ( R ) . Then, there exists 0 z h ( R ) such that x y z = 0 . If y z 0 , then x H Z ( R ) . If y z = 0 , then y H Z ( R ) . Indeed, the following lemma exists:

Lemma 2.11

Let R be a gr-R, where H Z ( R ) is a gr-I of R. Consequently, { 0 } is a gr-py-I of R if and only if H Z ( R ) = Gr-rad ( 0 ) .

Proof

Suppose that { 0 } is a gr-py-I of R . Let x Gr-rad ( 0 ) . Then, for all g G , there exists a positive integer n g in such a way that x g n g = 0 , and then x g H Z ( R ) , for all g G , and so x H Z ( R ) as H Z ( R ) is an ideal. Hence, Gr-rad ( 0 ) Z ( R ) . Let y H Z ( R ) . Then, there exists 0 z h ( R ) , where z y = 0 , and then y Gr-rad ( 0 ) as { 0 } is a gr-py-I and z 0 . Hence, H Z ( R ) = Gr-rad ( 0 ) . Conversely, let a , b h ( R ) in such a way that a b = 0 . If a = 0 , then it is done. If a 0 , then b H Z ( R ) = Gr-rad ( 0 ) . Thus, { 0 } is a gr-py-I of R .□

Theorem 2.12

Let R be a HUN-ring. If { 0 } is a gr-py-I of R, then R is a gr-loc-R with gr-m-I Gr-rad ( 0 ) .

Proof

Since { 0 } is a gr-py-I of R , H Z ( R ) = Gr-rad ( 0 ) by Lemma 2.11, and so H Z ( R ) is a gr-I of R . Let J be a gr-I of R in such a way that H Z ( R ) J R and H Z ( R ) J . Then, there is the existence of x J , where x H Z ( R ) ; therefore, there exists g G , where x g H Z ( R ) . Note that, x g J as J is a gr-I. Since x g H Z ( R ) , x g is not a nilpotent, so x g is a unit as R is a HUN-ring, and hence J = R . Thus, H Z ( R ) is a gr-m-I of R . Allow for K to be a proper gr-I of R , and suppose that a K . Since a g K for all g G and K is a proper, a g is a nonunit for all g G , and then a g is a nilpotent for all g G , i.e., a g H Z ( R ) for all g G , then a H Z ( R ) . So, J H Z ( R ) , and hence H Z ( R ) is the only gr-m-I of R . Thus, R is a gr-loc-R with gr-m-I H Z ( R ) = Gr-rad ( 0 ) .□

In the following theorem, we give a stronger and better conclusion than Theorem 2.12. Undeniably, we investigate the notion of graded n -ideals that were appeared in the study by Al-Zoubi et al. [17]. A proper gr-I P of R is presumably a graded n -ideal (gr- n -I) of R whenever x , y h ( R ) , where x y P and x Gr-rad ( 0 ) , then y P .

Theorem 2.13

For any gr-R R, the following are interchangeable:

  • R is a HUN-ring.

  • x is a gr-n-I of R, for every x h ( R ) with x R .

  • Every proper gr-I is a gr-n-I.

  • R has a unique gr-p-I, which is Gr-rad ( 0 ) .

  • R is a gr-loc-R with gr-m-I Gr-rad ( 0 ) .

  • R ( Gr-rad ( 0 ) ) is a gr-F.

Proof

( 1 ) ( 2 ) : Let x h ( R ) with x R . Consider a , b h ( R ) in such a way that a b x and a Gr-rad ( 0 ) . So, a is a unit, and then b x , also x is a gr- n -I of R .

( 2 ) ( 3 ) : Let P be a proper gr-I of R . Presume that x , y h ( R ) , where x y P and x Gr-rad ( 0 ) . Considering x y x y and x y is a gr- n -I, y x y P . Therefore, P is a gr- n -I of R .

( 3 ) ( 4 ) : Let P be a gr-p-I of R . By equation (3) and [17, Theorem 1], P = Gr-rad ( 0 ) .

( 4 ) ( 5 ) : Since R has one gr-p-I, which is Gr-rad ( 0 ) , we conclude that R is a gr-loc-R with gr-m-I Gr-rad ( 0 ) .

( 5 ) ( 6 ) : It is obvious to see.

( 6 ) ( 1 ) : Let x h ( R ) such that x is not a nilpotent. Then, x Gr-rad ( 0 ) and x + Gr-rad ( 0 ) is a nonzero homogeneous element in R Gr-rad ( 0 ) , which implies that x + Gr-rad ( 0 ) is a unit, i.e., ( x + Gr-rad ( 0 ) ) ( y + Gr-rad ( 0 ) ) = 1 + Gr-rad ( 0 ) , for some y h ( R ) . So, x y 1 is a nilpotent, and then ( x y 1 ) + 1 = x y is a unit, which gives that x is a unit. Thus, R is a HUN-ring.□

Theorem 2.14

Let R be a finitely generated gr-loc-R with gr-m-I X. If R is a gr-D and every gr-1-ab-py-I of R is a gr-w-py-I, then either R is a HUN-ring or X is the unique nonzero gr-p-I of R.

Proof

Assume that R is not a HUN-ring. Assume that R is a gr-D. Let 0 P be gr-p-I of R that is not a gr-m-I. Then, P X is a gr-1-ab-py-I of R and Gr-rad ( P X ) = P by Proposition 2.7. Then, P X is a gr-w-py-I of R . Let 0 p P and x X P . Then, p = g G p g , where p g P as P is a gr-I, and also, there exists h G such that x h P . Note that, x h X as X is a gr-I. We have 0 p g x h P X for all g G with p g 0 , and x h P = Gr-rad ( P X ) . Thus, p g P X for all g G with p g 0 , so p P X as p g = 0 P X too. Hence, P = P X . We obtain P = 0 from the Nakayama’s lemma, a contradiction. Thus, X is the unique nonzero gr-p-I of R .□

Recall from the study by Bataineh and Abu-Dawwas [18] that a proper gr-I P of R is presumably a graded semi-primary ideal (gr-s-py-I) of R if whenever x , y h ( R ) , where x y P , consequently x Gr-rad ( P ) or y Gr-rad ( P ) , or equivalently, Gr-rad ( P ) is a gr-p-I of R [19, Proposition 4]. It has been proved in Lemma 2.7 of the study by Abu-Dawwas [14] that every gr-1-ab-py-I of R is a gr-s-py-I. We establish the concept of g -semi-primary ideals ( g -s-py-I’s), and then we present a case where every e -w-1-ab-py-I of R is an e -s-py-I.

Definition 2.15

Allow R to be a gr-R, g G and P be a gr-I of R with P g R g . Then, P is said to be a g -semi-primary ideal ( g -s-py-I) of R if Gr-rad ( P ) is a g -p-I of R .

Proposition 2.16

Assume R is a first strongly gr-R. Then, every e-w-1-ab-py-I of R is an e-s-py-I supposing that Gr-rad ( 0 ) is an e-p-I of R.

Proof

Presume that Gr-rad ( 0 ) is an e -p-I of R . Let P be an e -w-1-ab-py-I of R . Assume that x , y R e with x y Gr-rad ( P ) and x Gr-rad ( P ) . We can suppose that x is not a unit. Now, there is the existence of a positive integer n , where x n y n P . Accordingly, x n + 1 y n P and n + 1 2 . If x n + 1 y n 0 , then x n + 1 P or y n P . Thus, y Gr-rad ( P ) since x Gr-rad ( P ) . Consider x n + 1 y n = 0 . If x n + 1 P e = { 0 } Gr-rad ( 0 ) , then P e Gr-rad ( 0 ) since Gr-rad ( 0 ) is an e -p-I and x Gr-rad ( 0 ) . If g G with P g 0 , then P g = R g P e R g ( Gr-rad ( 0 ) ) Gr-rad ( 0 ) . So, P g Gr-rad ( 0 ) for all g G . Thus, P Gr-rad ( 0 ) , and then Gr-rad ( P ) = Gr-rad ( 0 ) is an e -p-I. If x n + 1 P e 0 , then there exists a P e such that x n + 1 a 0 , and so 0 x n + 1 ( a + y n ) P . If a + y n is a unit, then x n + 1 P , a contradiction. Thus, a + y n is a nonunit. Since x n + 1 P , we obtain a + y n Gr-rad ( P ) . Thus, y Gr-rad ( P ) . Consequently, Gr-rad ( P ) is an e -p-I of R . Conversely, since { 0 } is an e -w-1-ab-py-I of R , { 0 } is an e -s-py-I, and hence Gr-rad ( 0 ) is an e -p-I of R .□

Proposition 2.17

Let R be a gr-R. Then, every gr-w-1-ab-py-I of R is a gr-1-ab-py-I if and only if { 0 } is a gr-1-ab-py-I ideal of R.

Proof

Speculate that { 0 } is a gr-1-ab-py-I ideal of R . Let P be a gr-w-1-ab-py-I of R . Assume that x , y , z h ( R ) are nonunits such that x y z P and z Gr-rad ( P ) . If x y z 0 , then x y P . Now, consider x y z = 0 . Hence, x y = 0 or z Gr-rad ( 0 ) since { 0 } is a gr-1-ab-py-I. The second case, however, cannot happen since z Gr-rad ( P ) . Hence, x y = 0 P , in the desired manner. Conversely, since { 0 } is a gr-w-1-ab-py-I of R , { 0 } is a gr-1-ab-py-I of R .□

Proposition 2.18

Allow R to be a gr-R in such a way that H Z ( R ) is a gr-I of R. If { 0 } is a gr-1-ab-py-I ideal of R, then either H Z ( R ) = G r a d ( 0 ) or R is a HUN-ring with H Z ( R ) = Ann R ( x ) = { a R : a x = 0 } for some x h ( R ) .

Proof

Suppose that H Z ( R ) Gr-rad ( 0 ) . Let a H Z ( R ) ( Gr-rad ( 0 ) ) . There is the existence of 0 x h ( R ) , where a x = 0 . Suppose that R has a homogeneous nonunit regular element named s . We have s x a = 0 and a Gr-rad ( 0 ) . Then, s x = 0 , and so x = 0 , wholly unattainable. Thus, homogeneous nonunit elements of R are in H Z ( R ) . So, R is a HUN-ring. Let y H Z ( R ) . We have y x a = 0 and a Gr-rad ( 0 ) , and so y x = 0 . Thus, H Z ( R ) Ann R ( x ) . Let r Ann R ( x ) . Then, r g Ann R ( x ) for all g G as Ann R ( x ) is a gr-I by ([20], page 3, line 11), which implies that r g H Z ( R ) for all g G , and so r H Z ( R ) as H Z ( R ) is a gr-I. Thus, H Z ( R ) = Ann R ( x ) .□

Proposition 2.19

Let R be a reduced first strongly gr-R. Then, every e-w-1-ab-py-I of R is an e-1-ab-py-I on the condition that R e is a domain.

Proof

Presume that every e -w-1-ab-py-I of R is an e -1-ab-py-I. Similarly as in Lemma 2.7 of the study by Abu-Dawwas [14], one can prove that every e -1-ab-py-I of R is an e -s-py-I. So, every e -w-1-ab-py-I of R is an e -s-py-I. Hence, by Proposition 2.16, { 0 } = Gr-rad ( 0 ) is an e -p-I of R . Therefore, R e is a domain. Conversely, let P be an e -w-1-ab-py-I of R . Assume that x , y , z R e are nonunits such that x y z P . If x y z = 0 , then x = 0 or y = 0 or z = 0 as R e is a domain, and so it is done. Consider x y z 0 . Then, as P is an e -w-1-ab-py-I, either x y P or z Gr-rad ( 0 ) . Therefore, P is an e -1-ab-py-I of R .□

3 Conclusion

In this article, we looked at and explored more outcomes to graded weakly 1-absorbing primary ideals. We proved that if R e is a nonlocal ring and P is an e -w-1-ab-py-I of R that is not an e -w-py-I, then either P e 3 = 0 or P e 2 = s with s as an idempotent such that 1 s is a maximal ideal of R e (Theorem 2.4). In addition, we showed that if every nonzero gr-py-I of R is a gr-p-I and Gr-rad ( 0 ) is a gr-m-I of R , then either Gr-rad ( 0 ) = 0 or Gr-rad ( 0 ) is the unique nonzero proper gr-I of R (Proposition 2.8). In addition, we proved that if R is a HUN-ring and { 0 } is a gr-py-I of R , then R is a gr-loc-R with gr-m-I Gr-rad ( 0 ) (Theorem 2.12). Moreover, a nice characterization was introduced in Theorem 2.13. We also showed that if R is a finitely generated gr-loc-R with gr-m-I X , R is a gr-D, and every gr-1-ab-py-I of R is a gr-w-py-I, then R is either HUN-ring or X is the unique nonzero gr-p-I of R (Theorem 2.14). Furthermore, we proved that if R is the first strongly gr-R, then every e -w-1-ab-py-I of R is an e -s-py-I if and only if Gr-rad ( 0 ) is an e -p-I of R (Proposition 2.16). Finally, we showed that if R is a reduced first strongly gr-R, then every e -w-1-ab-py-I of R is an e -1-ab-py-I if and only if R e is a domain (Proposition 2.19). As a proposal for further work on the topic, we are going to introduce a deep study on the concept of graded 1-absorbing prime ideals that have been established in the study by Abu-Dawwas et al. [9].

Acknowledgments

This article was supported by Princess Nourah bint Abdulrahman University Researchers Supporting Project number (PNURSP2023R183), Princess Nourah bint Abdulrahman University, P.O. Box 84428, Riyadh 11671, Saudi Arabia. We are very grateful for the reviewer comments that helped to improve our article.

  1. Funding information: Princess Nourah bint Abdulrahman University Researchers Supporting Project number (PNURSP2023R183), Princess Nourah bint Abdulrahman University, P.O. Box 84428, Riyadh 11671, Saudi Arabia.

  2. Conflict of interest: The authors declare that there is no conflict of interest.

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Received: 2022-11-02
Accepted: 2023-08-13
Published Online: 2023-11-21

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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