CUBO, A Mathematical Journal Vol. 24, no. 02, pp. 291–305, August 2022 DOI: 10.56754/0719-0646.2402.0291 Graded weakly 1-absorbing prime ideals Ünsal Tekir1, B Suat Koç2 Rashid Abu-Dawwas3 Eda Yıldız4 1Department of Mathematics, Marmara University, Istanbul, Turkey. utekir@marmara.edu.tr B 2Department of Mathematics, Istanbul Medeniyet University, Istanbul, Turkey. suat.koc@medeniyet.edu.tr 3Department of Mathematics, Yarmouk University, Jordan. rrashid@yu.edu.jo 4Department of Mathematics, Yildiz Technical University, Istanbul, Turkey. edyildiz@yildiz.edu.tr ABSTRACT In this paper, we introduce and study graded weakly 1- absorbing prime ideals in graded commutative rings. Let G be a group and R be a G-graded commutative ring with a nonzero identity 1 6= 0. A proper graded ideal P of R is called a graded weakly 1-absorbing prime ideal if for each nonunits x, y, z ∈ h(R) with 0 6= xyz ∈ P , then either xy ∈ P or z ∈ P . We give many properties and characterizations of graded weakly 1-absorbing prime ideals. Moreover, we in- vestigate weakly 1-absorbing prime ideals under homomor- phism, in factor ring, in rings of fractions, in idealization. RESUMEN En este art́ıculo, introducimos y estudiamos ideales primos débilmente 1-absorbentes en anillos conmutativos gradados. Sea G un grupo y R un anillo conmutativo G-gradado con identidad no cero 1 6= 0. Un ideal gradado propio P de R se llama ideal primo gradado débilmente 1-absorbente si para cualquiera x, y, z ∈ h(R) no-unidades con 0 6= xyz ∈ P , entonces o bien xy ∈ P o z ∈ P . Entregamos muchas propiedades y caracterizaciones de ideales primos grada- dos débilmente 1-absorbentes. Más aún, investigamos ide- ales primos débilmente 1-absorbentes bajo homomorfismo, en anillos cociente, en anillos de fracciones, en idealización. Keywords and Phrases: graded ideal, 1-absorbing prime ideal, weakly 1-absorbing prime ideal, graded weakly 1-absorbing prime ideal. 2020 AMS Mathematics Subject Classification: 13A02, 13A15, 16W50. Accepted: 6 June, 2022 Received: 24 November, 2021 c©2022 Ü. Tekir et al. This open access article is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. http://cubo.ufro.cl/ http://dx.doi.org/10.56754/0719-0646.2402.0291 mailto:utekir@marmara.edu.tr https://orcid.org/0000-0003-0739-1449 https://orcid.org/0000-0003-1622-786X https://orcid.org/0000-0001-8998-7590 https://orcid.org/0000-0002-6469-6698 mailto:utekir@marmara.edu.tr mailto:suat.koc@medeniyet.edu.tr mailto:rrashid@yu.edu.jo mailto:edyildiz@yildiz.edu.tr 292 Ü. Tekir, S. Koç, R. Abu-Dawwas & E. Yıldız CUBO 24, 2 (2022) 1 Introduction Throughout the paper, we focus only on graded commutative rings with a nonzero identity. R will always denote such a ring and G denotes a group with identity e. u(R), N(R) and reg(R) denote the set of all unit elements, all nilpotent elements and all regular elements of R, respectively. Over the years, several types of ideals have been developed such as prime, maximal, primary, etc. The concept of prime ideals and its generalizations have a significant place in commutative algebra since they are used in understanding the structure of rings [6, 19, 11, 4]. R is said to be G-graded if R = ⊕ g∈G Rg with RgRh ⊆ Rgh for all g, h ∈ G where Rg is an additive subgroup of R for all g ∈ G. Sometimes we denote the G-graded ring R by G(R). The elements of Rg are called homogeneous of degree g. If x ∈ R, then x can be written as ∑ g∈G xg, where xg is the component of x in Rg. Also, we set h(R) = ⋃ g∈G Rg. The support of G(R) is defined as supp(G(R)) = {g ∈ G : Rg 6= {0}}. Moreover, as shown for example in [13] that Re is a subring of R and 1 ∈ Re. Let P be an ideal of a graded ring R. Then P is said to be graded ideal if P = ⊕ g∈G (P ∩ Rg), i.e., for x ∈ P , x = ∑ g∈G xg where xg ∈ P for all g ∈ G. It is known that an ideal of a graded ring need not be graded. Let R be a G-graded ring and P be a graded ideal of R. Then R/P is G-graded by (R/P) g = (Rg +P)/P for all g ∈ G. If R and S are G-graded rings, then R×S is a G-graded ring by (R × S) g = Rg ×Sg for all g ∈ G. Lemma 1.1 ([9, Lemma 2.1]). Let R be a G-graded ring. (1) If P and Q are graded ideals of R, then P + Q, PQ and P ⋂ Q are graded ideals of R. (2) If x ∈ h(R), then Rx = (x) is a graded ideal of R. Let P be a proper graded ideal of R. Then the graded radical of P is denoted by Grad(P) and it is defined as follows: Grad(P) =    x = ∑ g∈G xg ∈ R : for all g ∈ G, there exists ng ∈ N such that x ng g ∈ P    . Note that Grad(P) is always a graded ideal of R (see [15]). In [15], Refai et al. defined and studied graded prime ideals. A proper graded ideal P of a graded ring R is called graded prime ideal if whenever xy ∈ P for some x, y ∈ h(R) then either x ∈ P or y ∈ P . Clearly, if P is a prime ideal of R and it is a graded ideal of R, then P is a graded prime ideal of R. On the other hand, the importance of graded prime ideals comes from the fact that graded prime ideals are not necessarily prime ideals, as we see in the next example. Example 1.2. Consider R = Z[i] and G = Z2. Then R is G-graded by R0 = Z and R1 = iZ. CUBO 24, 2 (2022) Graded weakly 1-absorbing prime ideals 293 Consider the graded ideal I = 17R of R. We show that I is a graded prime ideal of R. Let xy ∈ I for some x, y ∈ h(R). Case (1): Assume that x, y ∈ R0. In this case, x, y ∈ Z such that 17 divides xy, and then either 17 divides x or 17 divides y as 17 is a prime number, which implies that x ∈ I or y ∈ I. Case (2): Assume that x, y ∈ R1. In this case, x = ia and y = ib for some a, b ∈ Z such that 17 divides xy = −ab, and then 17 divides a or 17 divides b in Z, which implies that 17 divides x = ia or 17 divides y = ib in R. Then we have that x ∈ I or y ∈ I. Case (3): Assume that x ∈ R0 and y ∈ R1. In this case, x ∈ Z and y = ib for some b ∈ Z such that 17 divides xy = ixb in R, that is ixb = 17(α + iβ) for some α, β ∈ Z. Then we obtain xb = 17β, that is 17 divides xb in Z, and again 17 divides x or 17 divides b, which implies that 17 divides x or 17 divides y = ib in R. Thus, x ∈ I or y ∈ I. One can similarly show that x ∈ I or y ∈ I in other cases. So, I is a graded prime ideal of R. On the other hand, I is not a prime ideal of R since (4 − i)(4 + i) ∈ I, (4 − i) /∈ I and (4 + i) /∈ I. Several generalizations of graded prime ideals attracted attention by many authors. In [14], Refai and Al-Zoubi introduced graded primary ideals which is a generalization of graded prime ideals. A proper graded ideal P of a graded ring R is called graded primary ideal if xy ∈ P for some x, y ∈ h(R) implies that either x ∈ P or y ∈ Grad(P). They also studied graded primary G- decomposition related to graded primary ideals. Atani defined a generalization of graded prime ideals as graded weakly prime ideals in [5]. A proper graded ideal P of a graded ring R is said to be graded weakly prime ideal if whenever x, y ∈ h(R) such that 0 6= xy ∈ P then either x ∈ P or y ∈ P . They gave some characterizations of graded weakly prime ideals and their homogeneous components. In [12], Naghani and Moghimi introduced 2-absorbing version of graded prime ideals and graded weakly prime ideals. A proper graded ideal P of a graded ring R is called graded 2- absorbing (graded weakly 2-absorbing) if whenever x, y, z ∈ h(R) such that xyz ∈ P (0 6= xyz ∈ P) then xy ∈ P or yz ∈ P or xz ∈ P . They investigated some properties of this new class of graded ideals. Yassine et al. studied 1-absorbing prime ideals which is a generalization of prime ideals in [19]. A proper ideal P of R is said to be 1-absorbing prime ideal if for some nonunit elements x, y, z ∈ R such that xyz ∈ P implies that either xy ∈ P or z ∈ P . Authors determined 1-absorbing prime ideals in some special rings such as principal ideal domains, valuation domains and Dedekind domains. Currently, Koç et al. defined weakly 1-absorbing prime ideals which is a generalization of 1-absorbing prime ideals in [11]. A proper ideal P of R is called weakly 1-absorbing prime ideal if 0 6= xyz ∈ P for some nonunits x, y, z ∈ R implies xy ∈ P or z ∈ P . They gave certain properties of this new concept and characterized rings that every proper ideal is weakly 1-absorbing ideal. More recently, in [1], Dawwas et al. defined graded version of 1-absorbing prime ideals which is a 294 Ü. Tekir, S. Koç, R. Abu-Dawwas & E. Yıldız CUBO 24, 2 (2022) generalization of both graded prime ideals and 1-absorbing prime ideals. A proper graded ideal P of a graded ring R is called graded 1-absorbing prime ideal if whenever for some nonunits x, y, z in h(R) such that xyz ∈ P then either xy ∈ P or z ∈ P . Moreover, many studies have been made by researchers related to graded versions of known structures [3, 7, 8, 10, 16]. In this paper, we define graded weakly 1-absorbing prime ideal which is a generalization of graded 1-absorbing prime ideals. A proper graded ideal P of a graded ring R is said to be graded weakly 1-absorbing prime ideal if whenever for some nonunits x, y, z in h(R) such that 0 6= xyz ∈ P then either xy ∈ P or z ∈ P . Every graded 1-absorbing prime ideal is a graded weakly 1- absorbing prime ideals but the converse is not true in general (see, Example 3.2). In addition to many properties of this new class of graded ideals, we also investigate behavior of graded weakly 1- absorbing ideals under homomorphism, in factor ring, in rings of fractions, in idealization (see, Theorem 3.15, Proposition 3.14, Theorem 3.16, Theorem 3.18 and Theorem 3.23). 2 Motivation Graded prime ideals play an essential role in graded commutative ring theory. Indeed, graded prime ideals are interesting because they correspond to irreducible varieties and schemes in the graded case and because of their connection to factorization. Also, graded prime ideals are important because they have applications to combinatorics and they have structural significance in graded ring theory. Thus, this concept has been generalized and studied in several directions. The significance of some of these generalizations is same as the graded prime ideals. In a feeling of animate being, they determine how far an ideal is from being graded prime. Several generalizations of graded prime ideals attracted attention by many authors. For instance, graded weakly prime ideals, graded primary ideals, graded almost prime ideals, graded 2-absorbing ideals, graded 2-absorbing primary ideals and graded 1-absorbing prime ideals. In continuation of these generalizations, we present the concept of graded weakly 1-absorbing prime ideals, as a new generalization to graded prime ideals, in order to benefit from this new concept in more applications, and to make the study of graded prime ideals more flexible. 3 Graded weakly 1-absorbing prime ideals Definition 3.1. Let R be a G-graded ring and P be a proper graded ideal of R. Then, P is called graded weakly 1-absorbing prime ideal of R if whenever 0 6= xyz ∈ P for some nonunit elements x, y, z in h(R) then xy ∈ P or z ∈ P. Example 3.2. Every graded 1-absorbing prime ideal is a graded weakly 1-absorbing prime ideal. The converse may not be true. Let R = Z21 and consider the trivial grading on R. P = (0̄) is graded CUBO 24, 2 (2022) Graded weakly 1-absorbing prime ideals 295 weakly 1-absorbing prime ideal. But it is not graded 1-absorbing prime ideal since 3̄.3̄.7̄ = 0̄ ∈ P, 3̄.3̄ 6∈ P and 7̄ 6∈ P. Example 3.3. Let R = Z8[i] = Z8 ⊕ iZ8. Then note that R is a Z2-graded ring and h(R) = Z8 ∪iZ8. Now, put P = (4). Since 2(1+i)(1−i) = 4 ∈ P but 2(1+i) /∈ P and (1−i) /∈ P, it follows that P is not a weakly 1-absorbing prime ideal of R. However, the set of nonunit homogeneous elements of R is {0, 2, 4, 6, 2i, 4i, 6i}. Let x, y, z ∈ h(R) be nonunit elements. Then note that xyz = 0 ∈ P, which implies that P is a graded weakly 1-absorbing prime ideal of R. N(R) denotes the set of all nilpotent elements of R. Recall that a ring R is said to be reduced if N(R) = 0. Theorem 3.4. Let R be a G-graded reduced ring and P be a graded weakly 1-absorbing prime ideal of R. Then, Grad(P) is a graded weakly prime ideal of R. Proof. Suppose that 0 6= xy ∈ Grad(P) where x, y ∈ h(R). Then there exists n ∈ N such that (xy)n ∈ P . We have 0 6= (xy)n = xkxn−kyn ∈ P for some positive integer k < n. If x or y is unit in h(R), we are done. So, assume that x and y are nonunit elements in h(R). As P is graded weakly 1-absorbing prime ideal, xn ∈ P or yn ∈ P showing that x ∈ Grad(P) or y ∈ Grad(P). Theorem 3.5. Let R be a G-graded ring and P be a graded weakly 1-absorbing prime ideal of R. Then, (P : a) is a graded weakly prime ideal of R where a is a regular nonunit element in h(R)−P. Proof. From [1, Lemma 2.4], (P : a) is a graded ideal of R. Suppose 0 6= xy ∈ (P : a) for some x, y ∈ h(R). Then 0 6= (xa)y ∈ P where xa, y ∈ h(R). If x or y is unit, there is nothing to prove. So, we can assume that x and y are nonunit elements in h(R). Since P is graded weakly 1-absorbing prime ideal of R, we get either xa ∈ P or y ∈ P . It gives x ∈ (P : a) or y ∈ (P : a), as needed. Definition 3.6. Let R be a G-graded ring and P be a graded ideal of R. Then, P is called g-weakly 1-absorbing prime ideal of R for g ∈ G with Pg 6= Rg if 0 6= xyz ∈ P for some nonunit elements x, y, z in Rg implies that xy ∈ P or z ∈ P. We say that a proper graded ideal P of a G-graded ring R is said to be a g-weakly prime for g ∈ G if Pg 6= Rg and whenever 0 6= xy ∈ P for some x, y ∈ Rg implies x ∈ P or y ∈ P . Proposition 3.7. Let R be a G-graded reduced ring and P be a gn-weakly 1-absorbing prime ideal of R for each n ∈ N. Then, Grad(P) is a g-weakly prime ideal of R. Proof. It immediately follows from Theorem 3.4. 296 Ü. Tekir, S. Koç, R. Abu-Dawwas & E. Yıldız CUBO 24, 2 (2022) Recall from [1] that a proper graded ideal P of a G-graded ring R is said to be a g-1-absorbing prime for g ∈ G if Pg 6= Rg and whenever xyz ∈ P for some nonunits x, y, z ∈ Rg implies xy ∈ P or z ∈ P . Proposition 3.8. Let R be a G-graded ring. If R has a g-weakly-1-absorbing prime ideal that is not a g-weakly prime ideal of R and (0) is a g-1-absorbing prime ideal of R, then, for each unit element u in Rg and for each nonunit element v in Rg the sum u + v is a unit element in Rg. Proof. Assume that P is a g-weakly 1-absorbing prime ideal of R that is not a g-weakly prime ideal of R. Then, there exist nonunit elements x, y ∈ Rg such that xy ∈ P but x 6∈ P and y 6∈ P . Then we have vxy ∈ P where v is a nonunit element in Rg. If vxy = 0 ∈ (0), then vx ∈ P since (0) is a g-1 absorbing prime ideal and y 6∈ P . If 0 6= vxy ∈ P , we have vx ∈ P since P is a g-weakly 1-absorbing prime ideal of R. Now we will show that u + v is a unit element in Rg where u is a unit element in Rg. Suppose to the contrary. If (u + v)xy = 0 ∈ (0), we get (u + v)x ∈ P . This implies ux ∈ P giving that x ∈ P which is a contadiction. If we assume 0 6= (u + v)xy ∈ P , then again we get a contradiction by using the fact that P is a g-weakly 1-absorbing prime ideal and so it completes the proof. Theorem 3.9. Let R be a G-graded ring and P be a proper graded ideal of R. Consider the following statements. (i) P is a graded weakly 1-absorbing prime ideal of R. (ii) If xy 6∈ P for some nonunits x, y ∈ h(R), then (P : xy) = P ∪ (0 : xy). (iii) If xy 6∈ P for some nonunits x, y ∈ h(R), then either (P : xy) = P or (P : xy) = (0 : xy). (iv) If 0 6= xyK ⊆ P for some nonunits x, y ∈ h(R) and proper graded ideal K of R, then either xy ∈ P or K ⊆ P. (v) If 0 6= xJK ⊆ P for some nonunit x ∈ h(R) and proper graded ideals J, K of R, then either xJ ⊆ P or K ⊆ P. (vi) If 0 6= IJK ⊆ P for proper graded ideals I, J, K of R, then either IJ ⊆ P or K ⊆ P. Then, (vi) ⇒ (v) ⇒ (iv) ⇒ (iii) ⇒ (ii) ⇒ (i). Proof. (vi) ⇒ (v) : Suppose that 0 6= xJK ⊆ P for some nonunit x ∈ h(R) and proper graded ideals J, K of R. Now, put I = (x). Then I is a proper graded ideal of R and 0 6= IJK ⊆ P. By (vi), we have xJ ⊆ IJ ⊆ P or K ⊆ P, which completes the proof. (v) ⇒ (iv) : Suppose that 0 6= xyK ⊆ P for some nonunits x, y ∈ h(R) and proper graded ideal K of R. Now, consider the proper graded ideal J = (y) of R and note that 0 6= xJK ⊆ P. So by (v), we get xy ∈ xJ ⊆ P or K ⊆ P. CUBO 24, 2 (2022) Graded weakly 1-absorbing prime ideals 297 (iv) ⇒ (iii) : Let x, y ∈ h(R) be nonunit elements such that xy /∈ P. It is easy to see that xy(P : xy) ⊆ P. Case 1: Assume that xy(P : xy) = 0. This gives (P : xy) ⊆ (0 : xy) ⊆ (P : xy), that is, (P : xy) = (0 : xy). Case 2: Assume that xy(P : xy) 6= 0. Then by (iv), we have (P : xy) ⊆ P which implies that (P : xy) = P. (iii) ⇒ (ii) : It is straightforward. (ii) ⇒ (i) : Let x, y, z ∈ h(R) be nonunits such that 0 6= xyz ∈ P. If xy ∈ P, then we are done. So assume that xy /∈ P. Since z ∈ (P : xy) − (0 : xy) and (P : xy) = (0 : xy) ∪ P, we have z ∈ P which completes the proof. In the following example, we show that the condition “P is a graded weakly 1-absorbing prime ideal” does not ensure that the conditions (ii)-(vi) in Theorem 3.9 hold. In fact, we will show that (i) ; (ii). Example 3.10. Let R = Z12[X], where X is an indeterminate over Z12. Then R = ⊕ n∈Z Rn is a Z-graded ring, where R0 = Z12 and Rn = Z12X n if n > 0, otherwise Rn = 0. Then note that h(R) = ⋃ n≥0 Z12X n and the set of nonunits homogeneous elements of R is nh(R) = {2k, 3k, aXn : k, a ∈ Z and n ∈ N}. Consider the graded ideal P = (X, 4) of R. Let f, g, h ∈ nh(R) such that 0 6= fgh ∈ P. If at least one of the f, g, h is of the form aXn, then we are done. So assume that f, g, h ∈ {2k, 3k : k ∈ Z}. Since 0 6= fgh ∈ P = (X, 4), we have 0 6= fgh = 4k for some k ∈ Z with gcd(k, 3) = 1. Since 4|fgh and 3 ∤ fgh, we conclude that f, g, h ∈ {2, 4, 8, 10}. This implies that fg ∈ P, that is, P is a graded weakly 1-absorbing prime ideal of R. Now, we will show that P does not satisfy the condition (ii) in Theorem 3.9. Take nonunits homogeneous elements c = 2, d = 3 of R. Then note that cd /∈ P. On the other hand, it is clear that 2 ∈ (0 : cd) − P and X ∈ P − (0 : cd). This gives z = 2 + X ∈ (P : cd) − ((0 : cd) ∪ P) . Thus, we have (P : cd) ) (0 : cd) ∪ P, i.e., P does not satisfy the condition (ii) in Theorem 3.9. Definition 3.11. Let P be a graded weakly 1-absorbing prime ideal of R and xg1, yg2, zg3 be nonunits in h(R). Then, (xg1 , yg2, zg3) is called graded 1-triple zero if xg1 yg2zg3 = 0, xg1 yg2 6∈ P and zg3 6∈ P, where g1, g2, g3 ∈ G. Theorem 3.12. Let P = ⊕ g∈G Pg be a graded weakly 1-absorbing prime ideal that is not graded 1-absorbing prime ideal and (xg1 , yg2, zg3) be a graded 1-triple zero of P, where g1, g2, g3 ∈ G. Then, (i) xg1yg2Pg3 = 0. (ii) xg1zg3 6∈ Pg1g3 and yg2zg3 6∈ Pg2g3 imply that xg1 zg3Pg2 = yg2zg3Pg1 = xg1 Pg2Pg3 = yg2Pg1 Pg3 = zg3Pg1Pg2 = 0. In particular, Pg1Pg2Pg3 = 0. 298 Ü. Tekir, S. Koç, R. Abu-Dawwas & E. Yıldız CUBO 24, 2 (2022) Proof. (i) : Let P = ⊕ g∈G Pg be a graded weakly 1-absorbing prime ideal that is not graded 1-absorbing prime ideal and (xg1 , yg2, zg3) be a graded 1-triple zero of P . Assume that xg1yg2Pg3 6= 0. Then, there exists a ∈ Pg3 = P ∩ Rg3 such that 0 6= xg1 yg2a. So, we have 0 6= xg1 yg2a = xg1yg2(zg3 + a) ∈ P . If zg3 + a is unit, then xg1yg2 ∈ P which gives a contradiction. Since P is graded weakly 1-absorbing prime ideal and xg1yg2 6∈ P , zg3 +a ∈ P . This shows zg3 ∈ P , a contradiction. (ii) : Let xg1zg3 6∈ Pg1g3 and yg2zg3 6∈ Pg2g3. Then, xg1zg3, yg2zg3 6∈ P . Now choose a ∈ Pg2. So, we have xg1(yg2 + a)zg3 = xg1azg3 ∈ P since xg1yg2zg3 = 0. If yg2 + a is unit, then we obtain xg1zg3 ∈ P , which is a contradiction. Thus, yg2 + a is not unit. If xg1azg3 6= 0, then 0 6= xg1(yg2 + a)zg3 ∈ P . Thus, xg1(yg2 + a) ∈ P or zg3 ∈ P implying that xg1 yg2 ∈ P or zg3 ∈ P , a contradiction. This shows xg1azg3 = 0 and so xg1 zg3Pg2 = 0. Similarly, yg2zg3Pg1 = 0. Now assume that xg1 Pg2Pg3 6= 0. Then there exist ag2 ∈ Pg2 , bg3 ∈ Pg3 such that xg1ag2bg3 6= 0. This gives 0 6= xg1 (yg2 + ag2)(zg3 + bg3) = xg1yg2zg3 + xg1yg2bg3 + xg1ag2zg3 + xg1ag2bg3 = xg1ag2bg3 ∈ P . If (yg2 + ag2) is unit, xg1(zg3 + bg3) ∈ P . It means that xg1 zg3 ∈ P , which is a contradiction. Hence, (yg2 + ag2) is nonunit. Similar argument shows that (zg3 + bg3) is nonunit. Since P is graded weakly 1-absorbing prime ideal, xg1(yg2 +ag2) ∈ P or zg3 +bg3 ∈ P . This proves xg1yg2 ∈ P or zg3 ∈ P which is a contradiction. So, xg1Pg2Pg3 = 0. Similarly we have yg2Pg1Pg3 = zg3Pg1Pg2 = 0. Suppose Pg1Pg2 Pg3 6= 0. Then there exist ag1 ∈ Pg1 , bg2 ∈ Pg2, cg3 ∈ Pg3 such that ag1bg2cg3 6= 0. So, we have 0 6= (ag1 + xg1 )(bg2 + yg2)(cg3 + zg3) = ag1bg2cg3 ∈ P since xg1zg3Pg2 = yg2zg3Pg1 = xg1 Pg2Pg3 = yg2Pg1Pg3 = zg3Pg1 Pg2 = 0 and xg1yg2zg3 = 0. If ag1 + xg1 is unit, (bg2 + yg2)(cg3 + zg3) ∈ P and it implies yg2zg3 ∈ P , a contradiction. So, ag1 + xg1 is not unit. Similar argument shows that bg2 + yg2, cg3 + zg3 are nonunits. Since P is graded weakly 1-absorbing prime ideal, we have either (ag1 + xg1)(bg2 + yg2) ∈ P or cg3 + zg3 ∈ P . Thus, we conclude that xg1yg2 ∈ P or zg3 ∈ P giving a contradiction. Therefore, Pg1Pg2Pg3 = 0. Let R be a G-graded ring. It is clear that for each g ∈ G, Rg is an Re-module and Pg is an Re-submodule of Rg. Theorem 3.13. Let P = ⊕ g∈G Pg be a graded 1-absorbing prime ideal of G(R) and g ∈ G. If x, y ∈ Rg are nonunits such that xy 6∈ P, then (Pg2 :Re xy) = Pe. Proof. Let z ∈ (Pg2 :Re xy), where x, y ∈ Rg are nonunits. Then, xyz ∈ Pg2 ⊆ P . If z is a unit, xy ∈ P which gives a contradiction. So, z is not unit. As P is graded 1-absorbing prime ideal and xy 6∈ P we get z ∈ P . Thus, z ∈ P ∩ Re = Pe. This shows (Pg2 :Re xy) ⊆ Pe. CUBO 24, 2 (2022) Graded weakly 1-absorbing prime ideals 299 On the other hand, suppose z ∈ Pe ⊆ P . Then, xyz ∈ P ∩ Rg2 = Pg2 proving z ∈ (Pg2 :Re xy), as desired. Proposition 3.14. Let R be a G-graded ring and J ⊆ I be proper graded ideals of R. Then the followings statements are satisfied. (i) If I is graded weakly 1-absorbing prime ideal, then I/J is graded weakly 1-absorbing prime ideal of R/J. (ii) Suppose that J consists of all nilpotent elements of R. If J is a graded weakly 1-absorbing prime ideal of R and I/J is a graded weakly 1-absorbing prime ideal of R/J, then I is a graded weakly 1-absorbing prime ideal of R. (iii) If (0) is graded 1-absorbing prime ideal of R and I is graded weakly 1-absorbing prime ideal of R, then I is graded 1-absorbing prime ideal of R. Proof. (i) : Let 0+J 6= (x+J)(y +J)(z +J) ∈ I/J for some nonunits x+J, y +J, z +J ∈ h(R/J). Then, 0 6= xyz + J ∈ I/J and so 0 6= xyz ∈ I where x, y, z are nonunits in h(R). As I is a graded weakly 1-absorbing prime ideal, either xy ∈ I or z ∈ I. Hence, xy + J ∈ I/J or z + J ∈ I/J, as desired. (ii) : Suppose 0 6= xyz ∈ I for some nonunits x, y, z ∈ h(R). Then, xyz+J = (x+J)(y+J)(z+J) ∈ I/J. If xyz ∈ J, then xy ∈ J ⊆ I or z ∈ J since J ⊆ I is graded weakly 1-absorbing prime ideal. So we can assume xyz 6∈ J. Then we have 0 + J 6= (x + J)(y + J)(z + J) ∈ I/J. As I/J is graded weakly 1-absorbing prime ideal of R/J, (x + J)(y + J) ∈ I/J or z + J ∈ I/J. It implies either xy ∈ I or z ∈ I. (iii) : Suppose that xyz ∈ I for some nonunits x, y, z ∈ h(R). If xyz 6= 0, then we are done. So, we can assume xyz = 0 ∈ (0). Then, we get either xy = 0 ∈ I or z = 0 ∈ I since (0) is graded 1-absorbing prime ideal. Therefore, we conclude that xy ∈ I or z ∈ I. Let R and S be two G-graded rings. A ring homomorphism f : R → S is said to be graded homomorphism if f(Rg) ⊆ Sg for all g ∈ G. Theorem 3.15. Let R1 and R2 be two G-graded rings and f : R1 −→ R2 be a graded homomor- phism such that f(1R1) = 1R2. The following statements are satisfied. (i) If f is injective, J is a graded weakly 1-absorbing prime ideal of R2 and f(x) is a nonunit element of R2 for all nonunit elements x ∈ h(R1), then f −1(J) is a graded weakly 1-absorbing prime ideal of R1. (ii) If f is surjective and I is a graded weakly 1-absorbing prime ideal of R1 with ker(f) ⊆ I, then f(I) is a graded weakly 1-absorbing prime ideal of R2. 300 Ü. Tekir, S. Koç, R. Abu-Dawwas & E. Yıldız CUBO 24, 2 (2022) Proof. (i) : It is clear that f−1(J) is a graded ideal of R1. Let 0 6= xyz ∈ f −1(J) for some nonunits x, y, z in h(R1). So, f(x), f(y) and f(z) are nonunits in h(R2) by the assumption. Since f is injective and xyz 6= 0, we have f(xyz) 6= 0. Then we get 0 6= f(x)f(y)f(z) = f(xyz) ∈ J. As J is a graded weakly 1-absorbing prime ideal of R2, f(x)f(y) ∈ J or f(z) ∈ J. It implies that we have either xy ∈ f−1(J) or z ∈ f−1(J). (ii) : Suppose that 0 6= abc ∈ f(I) for some nonunits a, b, c ∈ h(R2). Then, there exist nonunits x, y, z ∈ h(R1) such that f(x) = a, f(y) = b and f(z) = c. It gives that 0 6= f(x)f(y)f(z) = abc ∈ f(I). So, there exists i ∈ I such that f(xyz) = f(i). This means xyz − i ∈ ker(f) ⊆ I giving xyz ∈ I. Since I is a graded weakly 1-absorbing prime ideal and 0 6= xyz ∈ I, we conclude that xy ∈ I or z ∈ I. It shows f(x)f(y) = ab ∈ f(I) or f(z) = c ∈ f(I), as needed. Let S ⊆ h(R) be a multiplicative set and R be a G-graded ring. Then S−1R is a G-graded ring with (S−1R)g = { a s : a ∈ Rh, s ∈ S ∩ Rhg−1 } . Let I be a graded ideal of R. Then we denote the set {a ∈ R : ab ∈ I for some b ∈ R − I} by ZI(R). Theorem 3.16. Let R be a G-graded ring and S ⊆ h(R) be a multiplicatively closed subset. The following statements are satisfied. (i) If I is a graded weakly 1-absorbing prime ideal of R with I ∩ S = ∅, then S−1I is a graded weakly 1-absorbing prime ideal of S−1R. (ii) If S−1I is a graded weakly 1-absorbing prime ideal of S−1R, u(S−1R) = {x s : x ∈ u(R), s ∈ S}, S ⊆ reg(R) and S ∩ ZI(R) = ∅, then I is a graded weakly 1-absorbing prime ideal of R. Proof. (i) : Suppose that 0 6= x s y t z u ∈ S−1I for some nonunits x s , y t , z u ∈ h(S−1R). Then 0 6= a(xyz) = (ax)yz ∈ I for some a ∈ S. Here, ax, y, z are nonunits in h(R). Otherwise, we would have x s , y t , z u are units in S−1R, a contradiction. As I is a graded weakly 1-absorbing prime ideal of R, we have either axy ∈ I or z ∈ I. This implies that x s y t = axy ast ∈ S−1I or z u ∈ S−1I. Thus, S−1I is a graded weakly 1-absorbing prime ideal of S−1R. (ii) : Let 0 6= xyz ∈ I for some nonunits x, y, z ∈ h(R). Since S ⊆ reg(R), we conclude that 0 6= x 1 y 1 z 1 ∈ S−1I. Here, x 1 , y 1 , z 1 are nonunits in h(S−1R). Since S−1I is a graded weakly 1-absorbing prime ideal of S−1R, we conclude either x 1 y 1 = xy 1 ∈ S−1I or z 1 ∈ S−1I. Then there exists s ∈ S such that sxy ∈ I or sz ∈ I. We can assume that sxy ∈ I. If xy /∈ I, then we have s ∈ ZI(R)∩S which is a contradiction. Thus we have xy ∈ I. In other case, similarly, we get z ∈ I. Therefore, I is a graded weakly 1-absorbing prime ideal of R. Theorem 3.17. Let P = ⊕ g∈G Pg be a graded weakly 1-absorbing prime ideal of R and g ∈ G. Then, (Pg2 :Re xy) = Pe ∪ (0 :Re xy) where x, y ∈ Rg are nonunits such that xy 6∈ P. CUBO 24, 2 (2022) Graded weakly 1-absorbing prime ideals 301 Proof. Clearly (0 :Re xy) ⊆ (Pg2 :Re xy). Let z ∈ Pe ⊆ P . This implies that xyz ∈ P ∩ Rg2 = Pg2 and so z ∈ (Pg2 :Re xy). Hence, Pe ∪ (0 :Re xy) ⊆ (Pg2 :Re xy). Now, we will show that (Pg2 :Re xy) ⊆ (0 :Re xy) ∪ Pe. Let z ∈ (Pg2 :Re xy). Then, we have xyz ∈ Pg2 ⊆ P . If z is a unit, then we have xy ∈ P, a contradiction. Suppose that z is a nonunit of R. If xyz 6= 0, then z ∈ P ∩ Re = Pe. So assume that xyz = 0. It gives z ∈ (0 :Re xy). Thus we have z ∈ Pe ∪ (0 :Re xy). Therefore, (Pg2 :Re xy) = Pe ∪ (0 :Re xy). Let R = ⊕ g∈G Rg be a graded ring. Recall from [18] that R is said to be a graded field if every nonzero homogenous element is a unit in R. Theorem 3.18. Suppose that R1, R2 be two G-graded commutative rings that are not graded fields and R = R1 × R2. Let P be a nonzero proper graded ideal of R. The following statements are equivalent. (i) P is a graded weakly 1-absorbing prime ideal of R. (ii) P = P1 × R2 for some graded prime ideal P1 of R1 or P = R1 × P2 for some graded prime ideal P2 of R2. (iii) P is a graded prime ideal of R. (iv) P is a graded weakly prime ideal of R. (v) P is a graded 1-absorbing prime ideal of R. Proof. (i) ⇒ (ii) : Let P be a nonzero proper graded ideal of R. Then we can write P = P1×P2 for some graded ideals P1 of R1 and P2 of R2. Since P is nonzero, P1 6= 0 or P2 6= 0. Without loss of generality, we may assume that P1 6= 0. Then there exists a homogeneous element 0 6= x ∈ P1. Since P is a graded weakly 1-absorbing prime ideal and (0, 0) 6= (1, 0)(1, 0)(x, 1) ∈ P, we conclude either (1, 0) ∈ P or (x, 1) ∈ P. Then we have either P1 = R1 or P2 = R2. Assume that P1 = R1. Now we will show that P2 is a graded prime ideal of R2. Let yz ∈ P2 for some y, z ∈ h(R2). If y or z is a unit, then we have either y ∈ P2 or z ∈ P2. So assume that y, z are nonunits in h(R2). Since R1 is not a graded field, there exists a nonzero nonunit t ∈ h(R1). This implies that (0, 0) 6= (t, 1)(1, y)(1, z) = (t, yz) ∈ P. As P is a graded weakly 1-absorbing prime ideal of R, we conclude either (t, 1)(1, y) = (t, y) ∈ P or (1, z) ∈ P. Thus we get y ∈ P2 or z ∈ P2 and so P2 is a graded prime ideal of R2. In other case, one can similarly show that P = P1 × R2 and P1 is a graded prime ideal of R1. (ii) ⇒ (iii) ⇒ (iv) ⇒ (i) : It is obvious. (iii) ⇒ (v) : It is clear. (v) ⇒ (i) : It is straightforward. 302 Ü. Tekir, S. Koç, R. Abu-Dawwas & E. Yıldız CUBO 24, 2 (2022) Definition 3.19. Let R be a ring and M be an R-module. A proper submodule N of M is called a 1-absorbing R-submodule if whenever xym ∈ N where x, y ∈ R are nonunits, m ∈ M, then either xy ∈ (N :R M) or m ∈ N. Theorem 3.20. Let P = ⊕ g∈G Pg be a graded 1-absorbing prime ideal of G(R). If Pg 6= Rg, then Pg is a 1-absorbing Re-submodule of Rg. Proof. Let xyr ∈ Pg ⊆ P for some nonunits x, y ∈ Re and r ∈ Rg. As P is graded 1-absorbing prime ideal, xy ∈ P or r ∈ P . This implies that xy ∈ (Pg :Re Rg) since xyRg ⊆ PRg ⊆ P ∩Rg = Pg or r ∈ P ∩ Rg = Pg . Definition 3.21. Let P = ⊕ g∈G Pg be a graded ideal of G(R). A graded component Pg of P is called 1-absorbing prime subgroup of Rg if xyz ∈ Pg for some nonunits x, y, x ∈ h(R) implies either xy ∈ Pg or z ∈ Pg. Proposition 3.22. Let P = ⊕ g∈G Pg be a graded ideal of G(R). If Pg is a 1-absorbing prime subgroup of Rg for all g ∈ G, then P is a graded 1-absorbing prime ideal of R. Proof. Suppose xyz ∈ P for some nonunits x, y, z ∈ h(R). Then, xyz ∈ Pg for some g ∈ G. Since Pg is 1-absorbing prime subgroup of Rg, xy ∈ Pg or z ∈ Pg. This gives xy ∈ P or z ∈ P , as needed. Let M be an R-module. The idealization R⋉M = {(r, m) : r ∈ R and m ∈ M} of M is a commuta- tive ring with componentwise addition and multiplication: (x, m1)+(y, m2) = (x+y, m1 +m2) and (x, m1)(y, m2) = (xy, xm2 + ym1) for each x, y ∈ R and m1, m2 ∈ M. Let G be an Abelian group and M be a G-graded R-module. Then X = R⋉M is a G-graded ring by Xg = Rg ⋉Mg = Rg ⊕Mg for all g ∈ G. Note that, Xg is an additive subgroup of X for all g ∈ G. Also, for g, h ∈ G, XgXh = (Rg ⋉Mg)(Rh⋉Mh) = RgRh⋉(RgMh+RhMg) ⊆ Rgh⋉(Mgh+Mhg) ⊆ Rgh⋉Mgh = Xgh as G is Abelian (see [2, 17]). Theorem 3.23. Let G be an Abelian group, M be a G-graded R-module and P be an ideal of R. Then, the following statements are equivalent. (i) P ⋉ M is a graded weakly 1-absorbing prime ideal of R ⋉ M. (ii) P is a graded weakly 1-absorbing prime ideal of R and if xg1yg2zg3 = 0 such that xg1 yg2 6∈ P and zg3 /∈ P for some nonunit elements xg1, yg2, zg3 in h(R), where g1, g2, g3 ∈ G, then xg1yg2Mg3 = xg1 zg3Mg2 = yg2zg3Mg1 = 0. Proof. (i) ⇒ (ii) : By [17, Theorem 3.3], P is a graded ideal of R. Suppose that 0 6= abc ∈ P where a, b, c are nonunits in h(R). Since (0, 0) 6= (a, 0)(b, 0)(c, 0) ∈ P ⋉ M and (a, 0), (b, 0), (c, 0) CUBO 24, 2 (2022) Graded weakly 1-absorbing prime ideals 303 are nonunits in h(R ⋉ M), we get (a, 0)(b, 0) ∈ P ⋉ M or (c, 0) ∈ P ⋉ M. Thus, we con- clude that ab ∈ P or c ∈ P , as needed. Now suppose xg1 yg2zg3 = 0 such that xg1 yg2 6∈ P and zg3 /∈ P for some nonunit elements xg1 , yg2, zg3 in h(R), where g1, g2, g3 ∈ G. Let xg1yg2Mg3 6= 0. Then there exists mg3 ∈ Mg3 such that xg1 yg2mg3 6= 0. This gives (0, 0) 6= (xg1 , 0)(yg2, 0)(zg3, mg3) = (0, xg1yg2mg3) ∈ P ⋉ M for some nonunits (xg1 , 0), (yg2, 0), (zg3, mg3) ∈ h(R ⋉ M) and P ⋉ M is a graded weakly 1-absorbing prime ideal, we have (xg1 , 0)(yg2, 0) = (xg1yg2, 0) ∈ P ⋉ M or (zg3, mg3) ∈ P ⋉ M. This gives xg1yg2 ∈ P or zg3 ∈ P , a contradiction. Hence, xg1yg2Mg3 = 0. Similar argument shows that xg1zg3Mg2 = yg2zg3Mg1 = 0. (ii) ⇒ (i) : By [17, Theorem 3.3], P ⋉ M is a graded ideal of R ⋉ M. Assume that (0, 0) 6= (xg1 , mg1)(yg2 , mg2)(zg3 , mg3) = (xg1 yg2zg3, xg1yg2mg3 +xg1zg3mg2 +yg2zg3mg1) ∈ P ⋉M for some nonunits (xg1 , mg1), (yg2, mg2), (zg3, mg3) in h(R ⋉ M). Then we get xg1 yg2zg3 ∈ P for some nonunits xg1, yg2, zg3 ∈ h(R). Case 1: Assume that xg1yg2zg3 = 0. If xg1 yg2 6∈ P and zg3 6∈ P , we have xg1 yg2Mg3 = xg1zg3Mg2 = yg2zg3Mg1 = 0. This implies that xg1yg2mg3 + xg1zg3mg2 + yg2zg3mg1 = 0 and so (xg1 , mg1)(yg2, mg2)(zg3, mg3) = (0, 0) giving a contradiction. Hence, we must have xg1yg2 ∈ P or zg3 ∈ P . 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Van Oystaeyen, “About graded fields”, Nederl. Akad. Wetensch. Indag. Math., vol. 43, no. 3, pp. 273–286, 1981. [19] A. Yassine, M. J. Nikmehr and R. Nikandish, “On 1-absorbing prime ideals of commutative rings”, J. Algebra Appl., vol. 20, no. 10, Paper No. 2150175, 12 pages, 2021. Introduction Motivation Graded weakly 1-absorbing prime ideals