# Properties ofϕ-Primal Graded Ideals

Authors: Ameer Jaber
Journal: Journal of Mathematics
 (2017)

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## Abstract

LetR be a commutative graded ring with unity 1≠0. A proper graded ideal of R is a graded ideal I of R such that I≠R. Let ϕ:I(R)→I(R)∪{∅} be any function, where I(R) denotes the set of all proper graded ideals of R. A homogeneous element a∈R is ϕ-prime to I if ra∈I-ϕ(I) where r is a homogeneous element in R; then r∈I. An element a=∑g∈Gag∈R is ϕ-prime to I if at least one component ag of a is ϕ-prime to I. Therefore, a=∑g∈Gag∈R is not ϕ-prime to I if each component ag of a is not ϕ-prime to I. We denote by νϕ(I) the set of all elements in R that are not ϕ-prime to I. We define I to be ϕ-primal if the set P=νϕ(I)+ϕ(I) (if ϕ≠ϕ∅) or P=νϕ(I) (if ϕ=ϕ∅) forms a graded ideal of R. In the work by Jaber, 2016, the author studied the generalization of primal superideals over a commutative super-ring R with unity. In this paper we generalize the work by Jaber, 2016, to the graded case and we study more properties about this generalization.


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## Body

## 1. Introduction

In [1] the author studied the generalization of primal superideals over a commutative super-ring R with unity. In this paper we generalize this work to the graded case and we study more properties about this generalization. For example in Section 4 we study the properties of ϕJ-P/J-primal graded ideals and in Section 5 we study the properties of ϕS-primal graded ideals of RS.LetG be an abelian group and let R be any commutative ring with unity; then R is called a G-graded ring (for short graded ring), if R=⨁g∈GRg, such that if a,b∈G, then RaRb⊆Rab. Let h(R)=⋃g∈GRg; then h(R) is the set of homogeneous elements in R and 1∈Re, where e is the identity element in G. By a proper graded ideal of R we mean a graded ideal I of R such that I≠R.We define a proper graded idealP of R to beprime if ab∈P implies that a∈P or b∈P, where a,b∈h(R). Let I be a proper graded ideal of R, an element a∈h(R) is calledprime toI if ar∈I, where r∈h(R); then r∈I. If I is a proper graded ideal of R and ρ(I) is the set of homogeneous elements of R that are not prime to I, then we define I to beprimal if the set P=⨁g∈G(ρ(I)∩Rg) forms a graded ideal of R. In this case we say that I is a P-primal graded ideal of R. Moreover, if I is a P-primal graded ideal of R, then it is easy to check that P is a prime graded ideal of R.Throughout,R will be a commutative graded ring with unity. We will denote the set of all proper graded ideals of R by I(R). If I and J are in I(R), then the set {r∈R:rJ⊆I} is a graded ideal of R which is denoted by (I:J). Let ϕ:I(R)→I(R)∪{∅} be any function and let I∈I(R); we say that I is a ϕ-prime if whenever x,y∈h(R), with xy∈I-ϕ(I), then x∈I or y∈I. Since I-ϕ(I)=I-[ϕ(I)∩I], there is no loss of generality to assume that ϕ(I)⊆I for every proper graded ideal I of R.Letϕ:I(R)→I(R)∪{∅} be any function. In this paper we always assume that, for any I,J∈I(R), ϕ(J)⊆ϕ(I) if J⊆I.Given two functionsψ1,ψ2:I(R)→I(R)∪{∅}, we define ψ1≤ψ2 if ψ1(I)⊆ψ2(I) for each I∈I(R).Letϕ:I(R)→I(R)∪{∅} be any function; then an element a∈h(R) is ϕ-prime to I, if whenever ra∈I-ϕ(I), where r∈h(R), then r∈I. That is, a∈h(R) is ϕ-prime to I if (1)hI:a-hϕI:a⊆hI.An element a=∑g∈Gag∈R is ϕ-prime to I if at least one component ag of a is ϕ-prime to I. Therefore, a=∑g∈Gag∈R is not ϕ-prime to I if each component ag of a is not ϕ-prime to I. Denote by νϕ(I) the set of all elements in R that are not ϕ-prime to I. Note that if a∈h(R) is not ϕ-prime to I, then a∈νϕ(I).We defineI to be ϕ-primal if the set P=νϕ(I)+ϕ(I) (if ϕ≠ϕ∅) or P=νϕ(I) (if ϕ=ϕ∅) forms a graded ideal of R. In this case we say that I is a ϕ-P-primal graded ideal of R, and P is the adjoint graded ideal of I.In the next example we give some famous functionsϕ:I(R)→I(R)∪{∅} and their corresponding ϕ-primal graded ideals.Example 1.
ϕ
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ω-primal graded idealObserve that ϕ∅≤ϕ0≤ϕω≤⋯≤ϕn+1≤ϕn≤⋯≤ϕ2.In this paper we study various properties ofϕ-primal graded ideals. Some of these properties for the nongraded case have been studied by Atani and Darani in [2, 3].


## 2.ϕ-Primal Graded Ideals

The next example shows thatνϕ(I) need not to be a graded ideal of R (see this example also in [4]).Example 2.
LetR=Z24+uZ24, where u2=0, be a commutative Z2-graded ring and assume that ϕ=ϕ0. Let I=8Z24+uZ24. Then I is a graded ideal of R. Since 0≠2¯·4¯∈I with 2¯,4¯∉I, then we get that 2¯ and 4¯ are not ϕ-prime to I, and hence 2¯ and 4¯ are in νϕ(I). Easy computations imply that 2¯+4¯=6¯ is ϕ-prime to I. Thus, 6¯∉νϕ(I) which implies that νϕ(I) is not a graded ideal of R.Next we give two examples ofϕ-primal Z-graded ideals of a given Z-graded ring R (where ϕ=ϕ∅).Example 3.
LetR=Z[x], G=Z, and |x|=1. Then Z[x]=Z⊕Zx⊕Zx2⊕⋯, where Z[x]0=Z, Z[x]1=Zx, Z[x]2=Zx2, and Z[x]n={0} for n<0, is a Z-graded ring. Let I=4Z[x] be a graded ideal in R, and let ϕ=ϕ∅. Then νϕ(I)=2Z[x] is a graded ideal of R; hence I is ϕ-primal graded ideal of R.Example 4.
LetR=Z4[x] and |x|=1. Then R is Z-graded ring, where Z4[x]n=⨁n=0∞Z4xn and Z4[x]n={0} for n<0. Let I={0}. Then, for ϕ=ϕ∅, by easy computations νϕ(I)=M[x] where M={0,2}. Hence I={0} is ϕ-primal graded ideal of R.Lemma 5.
LetI be a proper graded ideal of R, and let ϕ:I(R)→I(R)∪{∅} be any function. Then h(I)-h(ϕ(I))⊆νϕ(I).Proof.
Leta∈h(I)-h(ϕ(I)). Since I is proper, 1∉I. Then a=a.1∈h(I)-h(ϕ(I)) implies that a is not ϕ-prime to I. Thus a∈νϕ(I).We recall from Lemma5 that a=∑g∈Gag∈νϕ(I) if each component ag of a is in νϕ(I). Thus, by Lemma 5, each element a∈I can be written in the form c+d, where c∈νϕ(I) and d∈ϕ(I). Therefore, I⊆νϕ(I)+ϕ(I).Theorem 6.
LetI be a proper graded ideal of R, and let ϕ:I(R)→I(R)∪{∅} be any function. If the set νϕ(I) is an ideal of R, then it is ϕ-prime graded ideal of R.Proof.
Clearlyνϕ(I) is a graded ideal of R. Now, let a,b∈h(R) with ab∈νϕ(I)-ϕ(νϕ(I)). Then ab is not ϕ-prime to I. So there exists r∈h(R)-h(I) with abr∈I-ϕ(I). If a∉νϕ(I), then a is ϕ-prime to I. So abr∈I-ϕ(I) implies that br∈I. Hence br∈I-ϕ(I), since if br∈ϕ(I), then abr∈ϕ(I), a contradiction. Thus, b∈νϕ(I), since r∈h(R)-h(I).Corollary 7.
LetI be a proper graded ideal of R, and let ϕ:I(R)→I(R)∪{∅} be any function. If the set νϕ(I) is an ideal of R, then(2)P=νϕI+ϕIifϕ≠ϕ∅νϕIifϕ=ϕ∅is ϕ-prime graded ideal of R.Proof.
leta,b∈h(R) with ab∈P-ϕ(P). Then ab∉ϕ(I), since ϕ(I)⊆ϕ(P). Therefore, ab∈νϕ(I) or ab=cg+dg, where cg∈(νϕ(I))g and dg∈(ϕ(I))g.
Ifab∈νϕ(I), then, by Theorem 6, a∈νϕ(I)⊆P or b∈νϕ(I)⊆P. So, P is ϕ-prime graded ideal of R.
Ifab=cg+dg, where cg∈(νϕ(I))g and dg∈(ϕ(I))g, then there exists r∈h(R)-h(I) with cgr∈I-ϕ(I). So, abr=(cg+dg)r∈I-ϕ(I), since dgr∈ϕ(I). Therefore, ab∈νϕ(I) and by Theorem 6, a∈νϕ(I)⊆P or b∈νϕ(I)⊆P.Definition 8.
LetI be a proper graded ideal of R, and let ϕ:I(R)→I(R)∪{∅} be any function. Then I is called ϕ-P-primal graded ideal of R, where P=νϕ(I)+ϕ(I), if the set νϕ(I) forms an ideal of R. By the above corollary, P is always ϕ-prime graded ideal of R. In this case P is called the adjoint of I.Proposition 9.
LetI, P be proper graded ideals of R. Then the following statements are equivalent.(1)
I is ϕ-primal graded ideal of R with the adjoint graded ideal P.(2)
Forx∈h(R) with x∉h(P)-h(ϕ(I)) we have h((I:x))=h(I)∪h((ϕ(I):x)). If x∈h(P)-h(ϕ(I)), then h((I:x))⫌h(I)∪h((ϕ(I):x)).Proof.
(
1
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⇒
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2
) If x∈h(P)-h(ϕ(I)), then x∈νϕ(I). So there exists r∈h(R)-h(I) with rx∈I-ϕ(I). Thus r∈h((I:x)) and r∉h(I)∪h((ϕ(I):x)). Since it is easy to see that h((I:x))⊇h(I)∪h((ϕ(I):x)), we have that h((I:x))⫌h(I)∪h((ϕ(I):x)).
Now letx∉h(P)-h(ϕ(I)), where x∈h(R). Then x∉νϕ(I) and hence x is ϕ-prime to I. Let r∈h((I:x)). If rx∉ϕ(I), then r∈h(I). If rx∈ϕ(I), then r∈h((ϕ(I):x)). Hence(3)hI:x⊆hI∪hϕI:x⊆hI:x.(2)⇒(1) From part (2) we have h(P)-h(ϕ(I))=νϕ(I). Thus I is ϕ-primal graded ideal of R.We say thatP is ϕ-primal graded ideal of R if P itself is the adjoint of P. The next result shows that every ϕ-prime graded ideal of R is ϕ-primal.Theorem 10.
Everyϕ-prime graded ideal of R is ϕ-primal.Proof.
LetP be ϕ-prime graded ideal of R; we show that P is ϕ-P-primal graded ideal of R. Thus we must prove that(4)P=νϕP+ϕPifϕ≠ϕ∅νϕPifϕ=ϕ∅.If P=ϕ(P), then it is easy to check that νϕ(P)=∅; hence P is ϕ-P-primal graded ideal of R. Therefore, we may assume that P≠ϕ(P). We show that h(P)-h(ϕ(P))=νϕ(P). Let a∈h(P)-h(ϕ(P)). Then a.1∈P-ϕ(P) with 1∉P, so a∈νϕ(P). On the other hand let a∉h(P)-h(ϕ(P)). If a∈h(ϕ(P)), then ra∈ϕ(P) for all r∈h(R), so a is ϕ-prime to P and hence a∉νϕ(P). If a∉h(ϕ(P)), then a∉P, so for any rα∈Rα with rαa∈P-ϕ(P) we have that rα∈Pα, since P is ϕ-prime. Thus a is ϕ-prime to P and hence a∉νϕ(P). Therefore, h(P)-h(ϕ(P))=νϕ(P) which implies that P is ϕ-P-primal graded ideal of R.Now we give an example ofϕ-P-primal graded ideal I of R such that I itself is not ϕ-prime.Example 11.
Letϕ=ϕ0 and let R=Z8+uZ8 where u2=0. Then R is a commutative Z2-graded ring with unity. If I=4Z8+uZ8, then I is not a ϕ-prime graded ideal of R, since 2¯·2¯≠0∈I, but 2¯∉I. Let P=2Z8+uZ8; we show that I is ϕ-P-primal graded ideal of R. It is enough to show that ν(I)=h(P)-{0}. Let 0≠a¯∈h(P), if a¯∈2Z8; then a¯=2k∈Z8. If k is an odd number, then 0≠2¯a¯∈I, but 2¯∉I, and if k is an even number 0≠1¯a¯∈I with 1¯∉I; hence a¯∈ν(I). If a¯∈uZ8, then a¯∈I⊆ν(I). On the other hand, if a¯∈h(R)-h(P), then a¯ is an odd number in Z8. If 0≠a¯m¯∈I for some m¯∈Z8, then 4 divides am and so 4 divides m since (4,a)=1; hence m¯∈I. Thus I is ϕ-P-primal graded ideal of R.Letϕ:I(R)→I(R)∪{∅} be any function. We assume that, for any I,J∈I(R), ϕ(J)⊆ϕ(I) if J⊆I. Now we prove one of the main results in this section.Theorem 12.
Suppose thatψ1≤ψ2, where ψ1 and ψ2 are maps from I(R) into I(R)∪{∅}, and let I be a ψ2-P-primal graded ideal of R, with IeIα≠ψ2(I)α for all α∈G, where e is the identity element in G. If P is a prime graded ideal of R, then I is ψ1-P-primal.Proof.
SinceI is ψ2-P-primal graded ideal of R, then(5)P=νψ2I+ψ2Iifψ2≠ϕ∅νψ2Iifψ2=ϕ∅.To show that I is ψ1-P-primal graded ideal of R we must prove that(6)P=νψ1I+ψ1Iifψ2≠ϕ∅νψ1Iifψ1=ϕ∅.If ψ2=ϕ∅, then ψ1=ψ2 and hence we have that P=νψ1(I) which implies that I is ψ1-P-primal graded ideal of R. Now we may assume that ψ2≠ϕ∅. Let a∈h(νψ2(I)). Then there exists r∈h(R)-h(I) with rs∈I-ψ2(I)⊆I-ψ1(I), so a∈h(νψ1(I)) which implies that(7)νψ2I⊆νψ1I.Now, let a∈h(ψ2(I)). If a∉ψ1(I), then a∈I-ψ1(I). So 1.a∈I-ψ1(I) with a∉I. Hence a∈h(νψ1(I)). Therefore,(8)ψ2I⊆νψ1I+ψ1Iifψ2≠ϕ∅νψ1Iifψ1=ϕ∅.From (7) and (8) we have that(9)P⊆νψ1I+ψ1Iifψ2≠ϕ∅νψ1Iifψ1=ϕ∅.Since ψ1(I)⊆ψ2(I)⊆P, by (9)(10)P=νψ1I+ψ1Iifψ2≠ϕ∅νψ1Iifψ1=ϕ∅ifνψ1I⊆P.Let a∈(νψ1(I))α. Then there exists rβ∈Rβ-Iβ with arβ∈I-ψ1(I). If arβ∈I-ψ2(I), then a∈νψ2(I)⊆P. So we may assume that arβ∉I-ψ2(I); hence arβ∈ψ2(I). First suppose that aIβ⊈(ψ2(I))αβ, say asβ∈Iαβ-(ψ2(I))αβ with sβ∈Iβ. Then a(rβ+sβ)=arβ+asβ∉ψ2(I) with rβ+sβ∈Rβ-Iβ. Hence a∈νψ2(I)⊆P. Therefore, we may assume that aIβ⊆(ψ2(I))αβ.
Now suppose thatrβIe⊈(ψ2(I))β. Then there exists c∈Ie with rβc∈Iβ-(ψ2(I))β. Since a2∈Re, then (a2+c)rβ∈Iβ-(ψ2(I))β with rβ∉Iβ. So a2+c∈Pe but c∈Ie⊆Pe; therefore, a2∈P and hence a∈P, since P is a prime graded ideal. So we may assume that rβIe⊆(ψ2(I))β. Since (IeIβ)≠(ψ2(I))β there exists a¯∈Ie and b¯∈Iβ with a¯b¯∉(ψ2(I))β. Thus, (a2+a¯)(rβ+b¯)=a2rβ+a2b¯+a¯rβ+a¯b¯∉ψ2(I), so (a2+a¯)(rβ+b¯)∈I-ψ2(I) with rβ+b¯∈Rβ-Iβ which implies that a2+a¯∈(νψ2(I))e⊆Pe. Hence a2∈Pe⊆P, and so a∈P, since P is a prime graded ideal of R. Therefore, νψ1(I)⊆P, so(11)P=νψ1I+ψ1Iifψ2≠ϕ∅νψ1Iifψ1=ϕ∅and hence I is ψ1-P-primal graded ideal of R.


## 3. Conditions onϕ-Primal Graded Ideals

In this section we introduce some conditions under whichϕ-primal graded ideals are primal.Letϕ:I(R)→I(R)∪{∅} be any function. We have to recall that if I is ϕ-P-primal graded ideal of R, then(12)P=νϕI+ϕIifϕ≠ϕ∅νϕIifϕ=ϕ∅is ϕ-prime graded ideal of R.Definition 13.
Letr be a homogeneous element in R. Then |r|=α if r∈Rα for some α∈G.In the next theorem we provide some conditions under whichϕ-primal graded ideal is primal.Theorem 14.
LetR be a commutative graded ring with unity and let ϕ:I(R)→I(R)∪{∅} be any function. Suppose that I is ϕ-P-primal graded ideal of R with IγIδ⊈ϕ(I) for each γ,δ∈G. If P is a prime graded ideal of R, then I is P-primal.Proof.
Assume thata is a homogeneous element in P. Then a∈ϕ(I) or a∈(νϕ(I))α for some α∈G or a=bβ+cβ where bβ∈(νϕ(I))β and cβ∈ϕ(I) for some β∈G. If the first two cases hold, then a is not prime to I, since it is not ϕ-prime to I. In the last case, let d be a homogeneous element in R such that d∉I with bβd∈I-ϕ(I). Then ad=bβd+cβd∈I-ϕ(I), because ad∈ϕ(I) implies that bβd∈ϕ(I), since cβd∈ϕ(I) which is a contradiction. Thus a is not ϕ-prime to I and hence a is not prime to I. Now assume that b∈h(R) is not prime to I, so rb∈I for some homogeneous element r∈R-I. If rb∉ϕ(I), then b is not ϕ-prime to I, so b∈P. Thus assume that rb∈ϕ(I). Suppose that |r|=α. First suppose that bIα⊈ϕ(I). Then, there exists r′∈Iα such that br′∉ϕ(I). So b(r+r′)∈I-ϕ(I), where r+r′ is a homogeneous element in R-I, implies that b is not ϕ-prime to I; that is, b∈P. Therefore, we may assume that bIα⊆ϕ(I). Let |b|=β. If rIβ⊈ϕ(I), then rc∉ϕ(I) for some c∈Iβ. In this case r(b+c)∈I-ϕ(I) with r∈R-I; that is, b+c∈P, and hence b∈P, since c∈I⊆P. So we may assume that rIβ⊆ϕ(I). Since IαIβ⊈ϕ(I), there are b′∈Iα and a′∈Iβ with b′a′∉ϕ(I). Now, (b+a′)(r+b′)∈I-ϕ(I), where r+b′ is a homogeneous element in R-I, implies that b+a′ is a homogeneous element in P. On the other hand a′∈I⊆P, so that b∈P. We have already shown that P is exactly the set of all elements of R that are not prime to I. Hence I is P-primal.LetR and S be commutative G-graded rings, where G is an abelian group. It is easy to prove that the prime graded ideals of R×S have the form P×S or R×Q where P is a prime graded ideal of R and Q is a prime graded ideal of S. Also we have the following proposition about primal graded ideals of R×S. We leave the easy proof for this proposition to the reader. For the trivial case (i.e., Rg=Sg=0 for all g≠e) the next proposition is proved in [5, Lemma 13].Proposition 15.
LetR and S be commutative G-graded rings. If I is P-primal graded ideal of R and J is Q-primal graded ideal of S, then I×S (resp., R×J) is P×S- (resp., R×Q-) primal graded ideal of R×S.Next, we generalize [6, Theorem 16] to the graded case. Then we use this generalization to prove Theorem 18.Theorem 16.
LetR1 and R2 be commutative G-graded rings with unities and let ψi:I(R)→I(R)∪{∅} be functions. Let ϕ=ψ1×ψ2. Then ϕ-primes of R1×R2 have exactly one of the following three types:(1)
I1×I2 where Ii is a proper graded ideal of Ri with ψi(Ii)=Ii(2)
I1×R2 where I1 is ψ1-prime of R1 which must be prime if ψ2(R2)≠R2(3)
R1×I2 where I2 is ψ2-prime of R2 which must be prime if ψ1(R1)≠R1Proof.
We first note that a graded ideal ofR1×R2 having one of these three types is ϕ-prime. Case (1) is clear since ϕ(I)=I. If I1 is a prime graded ideal of R1, certainly I1×R2 is prime and hence ϕ-prime. So suppose that I1 is ψ1-prime and ψ2(R2)=R2. Let a1, b1 be homogeneous elements in R1 and let a2, b2 be homogeneous elements in R2 with (a1b1,a2b2)=(a1,a2)(b1,b2)∈I1×R2-ψ1(I1)×ψ2(R2)=I1×R2-ψ1(I1)×R2=(I1-ψ1(I1))×R2. Then a1b1∈I1-ψ1(I1)⇒a1∈I1 or b1∈I1, so (a1,a2)∈I1×R2 or (b1,b2)∈I1×R2. So, I1×R2 is ϕ-prime. The proof for case (3) is similar.
Next, suppose thatI1×I2 is ϕ-prime. Let a, b be homogeneous elements in R1 such that ab∈I1-ψ1(I1). Then (a,0)(b,0)=(ab,0)∈I1×I2-ϕ(I1×I2), so (a,0)∈I1×I2 or (b,0)∈I1×I2; that is, a∈I1 or b∈I1. So I1 is ψ1-prime. Likewise, I2 is ψ2-prime. Suppose that I1×I2≠ψ1(I1)×ψ2(I2), say, I1≠ψ1(I1). Let p be a homogeneous element in R1 such that p∈I1-ψ1(I1) and let q be a homogeneous element in I2. Then (p,1)(1,q)=(p,q)∈I1×I2-ϕ(I1×I2). So (1,q)∈I1×I2 or (p,1)∈I1×I2. Hence I1=R1 or I2=R2. Suppose that I2=R2. So I1×R2 is ϕ-prime where I1 is ψ1-prime. It remains to show that if ψ2(R2)≠R2, then I1 is actually prime graded ideal of R1. Let a, b be homogeneous elements in R1 such that ab∈I1. Now 1∈R2 is a homogeneous element in R2 not in ψ2(R2). Then (a,1)(b,1)=(ab,1)∈I1×R2-ϕ(I1×R2), so (a,1)∈I1×R2 or (b,1)∈I1×R2; that is, a∈I1 or b∈I1.LetR1, R2 be commutative G-graded rings with unities and let R=R1×R2. Let ϕ:I(R)→I(R)∪{∅} be a function. In the next theorem we also provide some conditions under which ϕ-primal graded ideal of R is primal, but first we start with the following remark.Remark 17.
LetI be a proper graded ideal of a commutative graded ring R and let ϕ:I(R)→I(R)∪{∅} be a function. If a homogeneous element a∈R is not ϕ-prime to I, then there is a homogeneous element r in R-I such that ar∈I-ϕ(I)⊆I so a is not prime to I.Theorem 18.
LetR1, R2 be commutative G-graded rings with unities and let R=R1×R2. Let ψi:I(R)→I(R)∪{∅} be functions with ψi(Ri)≠Ri for i=1,2. Let ϕ=ψ1×ψ2. Assume that P is a graded ideal of R with ϕ(P)≠P. If I is a ϕ-P-primal graded ideal of R, then either I=ϕ(I) or I is P-primal.Proof.
Supposeϕ(I)≠I. By Corollary 7, P is ϕ-prime graded ideal of R. Therefore, by Theorem 16, P has one of the following three cases.
Case
1. P=P1×P2, where Pi is a proper graded ideal of Ri with ψi(Pi)=Pi for i=1,2. In this case ϕ(P)=P, a contradiction.
Case
2. P=P1×R2 where P1 is ψ1-prime graded ideal of R1. Since ψ2(R2)≠R2, by Theorem 16(2), P1 is a prime graded ideal of R1 and so P is a prime graded ideal of R. We show that I2=R2. Since I≠ϕ(I), there exists a homogeneous element (a,b) in I-ϕ(I). So (a,1)(1,b)=(a,b)∈I-ϕ(I). If (a,1)∉I, then (1,b) is not ϕ-prime to I; hence (1,b)∈P=P1×R2, so 1∈P1, a contradiction. Thus (a,1)∈I=I1×I2; that is, 1∈I2 and I2=R2. Now we prove that I1 is P1-primal graded ideal of R1. Let a1 be a homogeneous element in P1. Then (a1,0)∈P1×R2=P. If (a1,0)∈ϕ(I)=ψ1(I1)×ψ2(R2), then a1∈ψ1(I1)⊆I1 so a1 is not prime to I1. Therefore, we may assume that (a1,0)∈νϕ(I). In this case there exists a homogeneous element (r1,r2)∈R-I such that (a1,0)(r1,r2)∈I-ϕ(I) so a1r1∈I1-ψ1(I1) with r1∈R1-I1, since R-I=(R1-I1)×R2, implies that a1 is not ψ1-prime to I1; hence by Remark 17, a1 is not prime to I1. Conversely, let b1 be a homogeneous element in R1 such that b1 is not prime to I1. Then there exists a homogeneous element c1 in R1-I1 with b1c1∈I1. Since ψ2(R2)≠R2, (b1,1)(c1,1)=(b1c1,1)∈I1×R2-(I1×ψ2(R2))⊆I-ϕ(I) with (c1,1)∈R-I. Hence (b1,1) is not ϕ-prime to I which implies that (b1,1)∈P=P1×R2 and so b1∈P1. We have already shown that the set of homogeneous elements in P1 consists exactly of the homogeneous elements of R1 that are not prime to I1. Hence I1 is P1-primal graded ideal of R1 so by Proposition 15, I is P-primal graded ideal of R.
Case
3. P=R1×P2, where P2 is ψ2-primal graded ideal of R2. The proof of Case3 is similar to that of Case2.


## 4.ϕJ-P/J-Primal Graded Ideals

LetR be a commutative graded ring with unity and let J be a proper graded ideal of R. Let ϕ:I(R)→I(R)∪{∅} be any function. As a generalization of [6], we define ϕJ:I(R/J)→I(R/J)∪{∅} by ϕJ(I/J)=(ϕ(I)+J)/J for every graded ideal I∈I(R) with J⊆I (and ϕJ(I/J)=∅ if ϕ=ϕ∅).We leave the trivial proof of the next lemma to the reader.Lemma 19.
LetR be a commutative graded ring with unity and let J be a proper graded ideal of R. Let ϕ:I(R)→I(R)∪{∅} be any function. If P is ϕ-prime graded ideal of R containing J, then P/J is ϕJ-prime graded ideal of R/J.Lemma 20.
LetR be a commutative graded ring with unity and let J be a proper graded ideal of R, and let ϕ:I(R)→I(R)∪{∅} be any function. Let P be a graded ideal of R containing J. If P/J is ϕJ-prime graded ideal of R/J with J⊆ϕ(P), then P is ϕ-prime graded ideal of R.Proof.
Leta, b be homogeneous elements in R with ab∈P-ϕ(P). Then ab∈J+P and ab∉J+ϕ(P)=ϕ(P). Thus, ab∈(J+P)-(J+ϕ(P)), so a¯b¯∈(J+P)/J-(J+ϕ(P))/J which implies that a¯b¯∈P/J-ϕJ(P/J); that is, a¯∈P/J or b¯∈P/J so a∈P or b∈P. Therefore, P is ϕ-prime graded ideal of R.In the next result and under the condition thatJ⊆ϕ(I) we prove that I is ϕ-primal graded ideal of R if and only if I/J is ϕJ-primal graded ideal of R/J.Theorem 21.
LetR be a commutative graded ring with unity and let ϕ:I(R)→I(R)∪{∅} be any function. Let I be a proper graded ideal of R, and let J be a graded ideal of R with J⊆ϕ(I). Then I is ϕ-P-primal graded ideal of R if and only if I/J is ϕJ-P/J-primal graded ideal of R/J.Proof.
Suppose thatI is ϕ-P-primal graded ideal of R with J⊆I. Then, by Corollary 7, P is ϕ-prime graded ideal of R containing J, so, by Lemma 19, P/J is ϕJ-prime graded ideal of R/J. We show that I/J is ϕJ-P/J-primal graded ideal of R/J. That is, we must prove that(13)PJ=νϕJIJ+ϕJIJifϕJ≠ϕ∅νϕJIJifϕJ=ϕ∅.Let a¯∈h(P/J); then a∈h(P) is not ϕ-prime to I. So there exists r∈h(R)-h(I) with ra∈I-ϕ(I). If a∈J+ϕ(I)=ϕ(I), then a¯∈ϕJ(I/J). So we may assume that a∉J+ϕ(I)=ϕ(I). Therefore, r¯a¯∈I/J-(J+ϕ(I))/J=I/J-ϕJ(I/J) and because r¯∉I/J we get that a¯∈νϕJ(I/J).
Now, assume thatb¯ is a homogeneous element in R/J such that b¯∈νϕJ(I/J). Then there exists a homogeneous element r¯ in R/J-I/J such that r¯b¯∈I/J-ϕJ(I/J), so rb∈I-ϕ(I) with r∉I. Thus, b is not ϕ-prime to I which implies that b∈P, and hence b¯∈P/J. Therefore,(14)PJ=νϕJIJ+ϕJIJifϕJ≠ϕ∅νϕJIJifϕJ=ϕ∅and so I/J is ϕJ-P/J-primal graded ideal of R/J.
Conversely, suppose thatI/J is ϕJ-P/J-primal graded ideal of R/J with the adjoint graded ideal P/J. We show that I is ϕ-P-primal graded ideal of R. Now, by Corollary 7, P/J is ϕJ-prime graded ideal of R/J with J⊆P, so, by Lemma 20, P is ϕ-prime graded ideal of R. To finish the proof we need to show that(15)P=νϕI+ϕIifϕ≠ϕ∅νϕIifϕ=ϕ∅.Clearly, ϕ(I)⊆I⊆P. Let a∈h(νϕ(I)); then there exists a homogeneous element r∈R-I with ra∈I-ϕ(I). Since ra∉ϕ(I)=J+ϕ(I) we get that r¯a¯∈I/J-(J+ϕ(I))/J=I/J-ϕJ(I/J) and r¯∉I/J. So, a¯∈P/J and hence a∈P.
Now, leta∈h(P). Suppose that a¯∈I/J. Then a∈I. If a∈ϕ(I), then we are done. Assume that a∉ϕ(I). Then a∈I-ϕ(I) and, so, a is not ϕ-prime to I; hence a∈νϕ(I). Therefore, we may assume that a¯∉I/J, so there exists r¯∈h(R/J)-h(I/J) with r¯a¯∈I/J-ϕJ(I/J), and so r¯a¯∉(J+ϕ(I))/J; that is, ra∉J+ϕ(I)=ϕ(I). Therefore, ra∈I-ϕ(I) with r∉I; that is, a∈νϕ(I).By Theorem21, we get the following result.Corollary 22.
LetR be a commutative graded ring with unity, and let ϕ:I(R)→I(R)∪{∅} be any function. let J be a graded ideal of R. Then there is one-to-one correspondence between ϕ-PI-primal graded ideals I of R containing J with J⊆ϕ(I) and ϕJ-PI/J-primal graded ideals of R/J.


## 5.ϕS-Primal Graded Ideals of RS

LetR be a commutative G-graded ring (for short graded ring) with unity, let S be a multiplicatively closed subset of h(R), and denote by RS the ring of fractions S-1R. We define a grading on RS by setting(16)RSg=as∈RS:a,sare homogeneous withas-1=g.It is easy to see that RS is G-graded ring (for short graded ring). Also, for G-graded ideal I of R, IS=S-1I is G-graded ideal of RS.Consider the canonical homomorphismρ:R→RS which is defined by r↦r/1 for all r∈h(R). Then ρ is a homogenous graded homomorphism of degree 0.Now letϕ:I(R)→I(R)∪{∅} be any function, we define ϕS:I(RS)→I(RS)∪{∅} by ϕS(J)=(ϕ(ρ-1(J)))S for every J∈I(RS). Note that ϕS(J)⊆J, since, for J∈I(RS), we have that ϕ(ρ-1(J))⊆ρ-1(J) implies ϕS(J)⊆(ρ-1(J))S⊆J.Example 23.
LetR=Z6+uZ6 with u2=0. Let S={1,2,4}, ϕ=ϕ2. Then S is a multiplicatively closed subset of h(R). If P={0}, then one can easily check that P is ϕ2-P-primal graded ideal of R. Moreover, PS=(ϕ(P))S={0}; hence ϕS(PS)=PS, since ρ-1(PS)={0,3,3u}, where ρ:R→RS is the canonical homomorphism. Therefore, PS is ϕS-PS-primal graded ideal in RS.We start by proving the following properties aboutϕ-prime graded ideals of R, where ρ:R→RS is the canonical homomorphism and S is a multiplicatively closed subset of h(R).Lemma 24.
Letϕ:I(R)→I(R)∪{∅} be any function, and let I be ϕ-prime graded ideal of R with I∩S=∅; then IS is ϕS-prime graded ideal of RS.Proof.
Letx/s, y/t be homogeneous elements in RS with (x/s)(y/t)∈IS-ϕS(IS); then, for some u∈S, xyu∈I-ϕ(I), so x∈I or yu∈I, and thus x/s∈IS or y/t∈IS; hence IS is ϕS-prime graded ideal of RS.Theorem 25.
Letϕ:I(R)→I(R)∪{∅} be any function, and let P be ϕ-prime graded ideal of R with h(P)∩S=∅. If ρ-1((ϕ(P))S)⊆P, then ρ-1(PS)=P.Proof.
It is easy to see thatP⊆ρ-1(PS).
Conversely, letx be a homogeneous element in ρ-1(PS), then for some s∈S, xs∈P. If xs∉ϕ(P), then xs∈P-ϕ(P) and s∉P so x∈P. Therefore, we may assume that xs∈ϕ(P), so x is a homogeneous element in ρ-1((ϕ(P))S). Thus,(17)ρ-1PS⊆P∪ρ-1ϕPS,and since ρ-1((ϕ(P))S)⊆P, we have that ρ-1(PS)=P.Lemma 26.
Letϕ:I(R)→I(R)∪{∅} be any function, and let I be ϕ-P-primal graded ideal of R with h(P)∩S=∅. If a/s∈h(IS)-h((ϕ(I))S), then a∈I. Moreover, if h(ρ-1((ϕ(I))S))⊆h(I), then I=ρ-1(IS).Proof.
Leta/s∈h(IS)-h((ϕ(I))S), so a/s=b/t for some b∈h(I) and t∈S. In this case uta=usb∈h(I) for some u∈S. If uta∈ϕ(I), then a/s=uta/uts∈(ϕ(I))S a contradiction. So we have that uta∈I-ϕ(I). If a∉h(I), then ut is not ϕ-prime to I; so ut∈h(P)∩S which contradicts the hypothesis. Therefore, a∈I.
For the last part, it is clear thath(I)⊆h(ρ-1(IS)). Now let a be a homogeneous element in ρ-1(IS). Then as∈h(I) for some s∈S. If as∉ϕ(I) and a∉I, then s is not ϕ-prime to I, so s∈P∩S a contradiction. So a must be in I. If as∈ϕ(I), then a/1=(as)/s∈(ϕ(I))S, and so a∈ρ-1((ϕ(I))S). Therefore, h(ρ-1(IS))=h(I)∪h(ρ-1((ϕ(I))S))=h(I), since h(ρ-1((ϕ(I))S))⊆h(I). Hence I=ρ-1(IS).Lemma 27.
Letϕ:I(R)→I(R)∪{∅} be any function, and let I be ϕ-P-primal graded ideal of R with h(P)∩S=∅. Then h(ρ-1(IS))-h(ρ-1(ϕS(IS)))⊆h(I)-h(ϕ(I)).Proof.
Leta be a homogeneous element in ρ-1(IS) such that a∉h(ρ-1(ϕS(IS))); then a/1∈h(IS)-h(ϕS(IS))⊆h(IS)-h((ϕ(I))S) and, by Lemma 26, a∈I. If a∈ϕ(I), then a/1∈(ϕ(I))S⊆ϕS(IS) implies that a∈h(ρ-1(ϕS(IS))) a contradiction. Therefore, a∈h(I)-h(ϕ(I)).LetR be a commutative graded ring with unity and M be R-graded module. An element a∈h(R) is called a zero-divisor on M if am=0 for some m∈h(M). We denote by ZR(M) the set all zero-divisors of h(R) on M.Corollary 28.
Letϕ:I(R)→I(R)∪{∅} be any function, and let I be ϕ-P-primal graded ideal of R with h(P)∩S=∅, S∩ZR(R/ϕ(I))=∅. If ρ-1((ϕ(I))S)⊆I, then h((νϕ(I))S)-h(ϕS(IS))⊆h(νϕS(IS)).Proof.
By Lemma26, if ρ-1((ϕ(I))S)⊆I, then ρ-1(IS)=I. Let x/s be a homogeneous element in (νϕ(I))S-ϕS(IS); then x/s=y/t, where y∈νϕ(I). If y∈I, then y/t=x/s∈IS-ϕS(IS)⊆νϕS(IS). Therefore we may assume that y∉I. If y/1∈IS, then y/t=x/s∈IS-ϕS(IS)⊆νϕS(IS). Therefore we may assume that y/1∉IS. So, y/1 is a homogeneous element in (νϕ(I))S-IS. So uy∈νϕ(I) for some u∈S and uy∉I. So there exists r∈h(R)-h(I) such that ruy∈I-ϕ(I). If ry∉I, then u∈νϕ(I)⊆P a contradiction. Therefore, ry∈I-ϕ(I). So (r/1)(y/1)∈IS. If (r/1)(y/1)∈ϕS(IS), then there exists t∈S with try∈ϕ(ρ-1(IS))=ϕ(I), so t∈S∩ZR(R/ϕ(I)) a contradiction. Thus (r/1)(y/1)∈IS-ϕS(IS) and r/1∉IS. So, y/1∈h(νϕS(IS)). Hence x/s=y/t∈h(νϕS(IS)).We recall that ifI is a graded ideal in R, then I⊆ρ-1(IS); therefore, we may assume that (ϕ(I))S⊆ϕS(IS).Under the condition thatρ-1((ϕ(I))S)⊆I for all proper graded ideals I of R, we have the following propositions.Proposition 29.
LetS be a multiplicatively closed subset of h(R) with 1∈S, let ϕ:I(R)→I(R)∪{∅} be any function, and let I be ϕ-P-primal graded ideal of R with h(P)∩S=∅, S∩ZR(R/ϕ(I))=∅. Then IS is ϕS-PS-primal graded ideal of RS.Proof.
By Lemma24, PS is ϕS-prime graded ideal of RS.
To show thatIS is ϕS-PS-primal graded ideal of RS, we must prove that(18)PS=νϕSIS+ϕSISifϕS≠ϕ∅νϕSISifϕS=ϕ∅.Clearly, ϕS(IS)⊆PS; let a/s be a homogenous element in νϕS(IS). Then there exists r/u∈h(RS)-h(IS) with (r/u)·(a/s)∈IS-ϕS(IS)⊆IS-(ϕ(I))S, so ra∉ϕ(I) and, by Lemma 26, ra∈I. So, ra∈I-ϕ(I), and r∉h(I). Thus a∈νϕ(I)⊆P and hence a/s∈PS.
Conversely, leta/s∈h(PS) such that a/s∉ϕS(IS). Then a∈ρ-1(PS)=P. If a/s∈IS, then (1/1)(a/s)∈IS-ϕS(IS), (1/1)∉IS, so a/s is not ϕS-prime to IS; thus a/s∈νϕS(IS). Therefore, we may assume that a/s∉IS; that is, ta∉I for every t∈S. So, a∉I. Therefore, a∈h(P)-h(I)⊆h(νϕ(I)). Thus, a/s∈h((νϕ(I))S)-h(ϕS(IS)). Since, by Corollary 28, h((νϕ(I))S)-h(ϕS(IS))⊆h(νϕS(IS)) we have that a/s∈νϕS(IS).Proposition 30.
Letϕ:I(R)→I(R)∪{∅} be any function, and let J be ϕS-Q-primal graded ideal of RS. Then ρ-1(Q) is ϕ-prime graded ideal of R and ρ-1(J) is ϕ-primal graded ideal of R with the adjoint graded ideal ρ-1(Q) with ρ-1(Q)∩S=∅, S∩ZR(R/ϕ(ρ-1(J)))=∅. Moreover, J=(ρ-1(J))S.Proof.
To show thatρ-1(Q) is ϕ-prime graded ideal of R, it is enough to prove that ρ-1(J) is ϕ-primal graded ideal of R with the adjoint graded ideal ρ-1(Q). Then, by using Corollary 7, ρ-1(Q) will be ϕ-prime graded ideal of R.
Now, to prove thatρ-1(J) is ϕ-primal graded ideal of R we must show that(19)ρ-1Q=νϕρ-1J+ϕρ-1Jifϕ≠ϕ∅νϕρ-1Jifϕ=ϕ∅.But ϕ(ρ-1(J))⊆ρ-1(J)⊆ρ-1(Q). Let a be a homogenous element in νϕ(ρ-1(J)) with a∉ϕ(ρ-1(J)); then a/1∈(νϕ(ρ-1(J)))S-ϕS(J), since S∩ZR(R/ϕ(I))=∅ and, by Corollary 28, (νϕ(ρ-1(J)))S-ϕS(J)⊆νϕS(J). Thus, a/1∈νϕS(J)⊆Q and hence a∈ρ-1(Q).
Conversely, leta be a homogeneous element in ρ-1(Q). Then a/1 in Q. We may assume that a∉ϕ(ρ-1(J)), since S∩ZR(R/ϕ(ρ-1(J)))=∅, a/1∉ϕS(J). If a/1∈J, then (a/1)∈J-ϕS(J) and since ϕ(ρ-1(J))⊆ρ-1(ϕS(J)), a∈ρ-1(J)-ρ-1(ϕS(J))⊆ρ-1(J)-ϕ(ρ-1(J)), but 1∉ρ-1(J), so a∈νϕ(ρ-1(J)). If a/1∉J, then a/1∈Q-J and so a/1∈νϕS(J). Let x/s be a homogeneous element in RS-J, with (a/1)(x/s)∈J-ϕS(J); then ax∈ρ-1(J)-ρ-1(ϕS(J))⊆ρ-1(J)-ϕ(ρ-1(J)), since ax/1∈J and ax/1∉ϕS(J), for if ax/1∈ϕS(J), then ax/s∈ϕS(J), a contradiction. Thus we have that ax∈ρ-1(J)-ϕ(ρ-1(J)) and x∉ρ-1(J), since x/s∉J. Therefore, a∈νϕ(ρ-1(J)) and so ρ-1(J) is ϕ-primal graded ideal of R with the adjoint graded ideal ρ-1(Q).
Finally, we show thatJ=(ρ-1(J))S. Clearly, J⊆(ρ-1(J))S. Conversely, let x/s be a homogeneous element in (ρ-1(J))S. Then xt∈ρ-1(J) for some t∈S. Thus, xt/1∈ρ(ρ-1(J))=J, and hence (xt/1)(1/st)=x/s∈J. Therefore, J=(ρ-1(J))S.Under the condition thatρ-1((ϕ(I))S)⊆I for all proper graded ideals I of R and by using Propositions 29 and 30 we have the following result.Corollary 31.
LetR be a commutative graded ring with unity. Let S be a multiplicatively closed subset of h(R), and let ϕ:I(R)→I(R)∪{∅} be any function. Then there is one-to-one correspondence between ϕ-PI-primal graded ideals I of R and ϕS-PSI-primal graded ideals IS of RS, where PI is ϕ-prime graded ideal of R with PI∩S=∅, S∩ZR(R/ϕ(I))=∅.I would like to remark that “ϕ-theory” introduced here has nothing in common with “ϕ-theories” considered previously by Badawi and Lucas in [7].



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