# Properties of [phi]-Primal Graded Ideals.

1. Introduction

In  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 [[phi].sub.J]-P/J-primal graded ideals and in Section 5 we study the properties of [[phi].sub.S]-primal graded ideals of [R.sub.S].

Let G 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 = [[direct sum].sub.g[member of]G][R.sub.g] such that if a,b [member of] G, then [R.sub.a][R.sub.b] =[subset or equal to] [R.sub.ab]. Let h(R) = [U.sub.g[member of]G] [R.sub.g]; then h(R) is the set of homogeneous elements in R and 1 [member of] [R.sub.e], 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 [not equal to] R.

We define a proper graded ideal P of R to be prime if ab [member of] P implies that a [member of] P or b [member of] P, where a,b [member of] h(R). Let I be a proper graded ideal of R, an element a [member of] h(R) is called prime to I if ar [member of] I, where r [member of] h(P);then r [member of] I. If I is a proper graded ideal of R and [rho](I) is the set of homogeneous elements of R that are not prime to I, then we define I to be primal if the set P = [[direct sum].sub.g[member of]G] ([rho](I) [intersection] [R.sub.g]) 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 [member of] R : rJ [subset or equal to] I} is a graded ideal of R which is denoted by (I: J). Let f : I(R) [right arrow] I(R) [union] {0} be any function and let I [member of] I(R); we say that I is a [phi]-prime if whenever x, y [member of] h(R), with x, y [member of] I - [phi](I), then x [member of] I or y [member of] I. Since I - [phi](I) = I - [[phi](I) [intersection] I], there is no loss of generality to assume that [phi](I) [subset] I for every proper graded ideal I of R.

Let f : I(R) [right arrow] I(R) [union] {0} be any function. In this paper we always assume that, for any I, J [member of] I(R), [phi](J) [subset or equal to] [phi](I) if J [subset or equal to] I.

Given two functions [[psi].sub.1], [[psi].sub.2]: I(R) [right arrow] I(R) [union] {0}, we define [[psi].sub.1] [less than or equal to] [[psi].sub.2] if [[psi].sub.1](I) [subset or equal to] [[psi].sub.2](I) for each I [member of] I(R).

Let [phi]: I(R) [right arrow] I(R) [union] {0} be any function; then an element a [member of] h(R) is [phi]-prime to I, if whenever ra [member of] I - [phi](I), where r [member of] h(R), then r [member of] I. That is, a [member of] h(R) is [phi]-prime to I if

h((I:a)) - h(([phi](I):a)) [subset or equal to] h(I). (1)

An element a = [[summation].sub.g[member of]G] [a.sub.g] [member of] R is [phi]-prime to I if at least one component [a.sub.g] of a is [phi]-prime to I. Therefore, a = [[summation].sub.g[member of]G][a.sub.g] [member of] R is not [phi]-prime to I if each component [a.sub.g] of a is not [phi]-prime to I. Denote by [v.sub.[phi]] (I) the set of all elements in R that are not [phi]-prime to I. Note that if a [member of] h(R) is not [phi]-prime to I, then a [member of] [v.sub.[phi]](I).

We define I to be 0-primal if the set P = [v.sub.[phi]](I) + [phi](I) (if [phi] [not equal to] [[phi].sub.0]) or P = [v.sub.[phi]](I) (if [phi] = [[phi].sub.0]) forms a graded ideal of R. In this case we say that I is a [phi]-P-primal graded ideal of R, and P is the adjoint graded ideal of I.

In the next example we give some famous functions [phi] : I(R) [right arrow] I(R) [union] {0} and their corresponding 0-primal graded ideals.

Example 1.

[[phi].sub.0]: [[phi].sub.0](I) = 0, [for all]I [member of] I(R) primal graded ideal

[[phi].sub.0]: [[phi].sub.0](I) = {0}, [for all]I [member of] I(R) weakly primal graded ideal

[[phi].sub.2]: [[phi].sub.2](I) = [I.sup.2], [for all]I [member of] I(R) almost primal graded ideal

[[phi].sub.n]: [[phi].sub.n](I) = [I.sup.n], [for all]I [member of] I(R) n-almost primal graded ideal

[[phi].sub.[omega]]: [[phi].sub.[omega]](I) = [[intersection].sup.[infinity].sub.n=1] [I.sup.n], [for all]I [member of] I(R) [omega]-primal graded ideal

Observe that [[phi].sub.0] [less than or equal to] [[phi].sub.0] [less than or equal to] [[phi].sub.[omega]] [less than or equal to] ... [less than or equal to] [[phi].sub.n+1] [less than or equal to] [[phi].sub.n] < ... < [[phi].sub.2].

In this paper we study various properties of [phi]-primal graded ideals. Some of these properties for the nongraded case have been studied by Atani and Darani in [2, 3].

2. [phi]-Primal Graded Ideals

The next example shows that [v.sub.[phi]](I) need not to be a graded ideal of R (see this example also in ).

Example 2. Let R = [Z.sub.24] + u[Z.sub.24], where [u.sup.2] = 0, be a commutative [Z.sub.2]-graded ring and assume that [phi] = [[phi].sub.0]. Let I = 8[Z.sub.24] + m[Z.sub.24]. Then I is a graded ideal of R. Since 0 [not equal to] [bar.2] x [bar.4] [member of] I with [bar.2], [bar.4] [not member of] I, then we get that [bar.2] and [bar.4] are not [phi]-prime to I, and hence 2 and 4 are in [v.sub.[phi]](I). Easy computations imply that [bar.2] + [bar.4] = [bar.6] is [phi]-prime to I. Thus, [bar.6] [??] [v.sub.[phi]](I) which implies that [v.sub.[phi]](I) is not a graded ideal of R.

Next we give two examples of [phi]-primal Z-graded ideals of a given Z-graded ring R (where [phi] = [phi]0).

Example 3. Let R = Z[x], G = Z, and [absolute value of (x)] = 1. Then Z[x] = Z [direct sum] Zx [direct sum] Z[x.sup.2] [direct sum] ..., where Z[[x].sub.0] = Z, Z[[x].sub.1] = Zx, Z[[x].sub.2] = Z[x.sup.2], and Z[[x].sub.n] = {0} for n < 0, is a Z-graded ring. Let I = 4Z[x] be a graded ideal in R, and let [phi] = [[phi].sub.0]. Then (I) = 2Z[x] is a graded ideal of R; hence I is [phi]-primal graded ideal of R.

Example 4. Let R = [Z.sub.4][x] and [absolute value of (x)] = 1. Then R is Z-graded ring, where Z4[[x].sub.n] = [[direct sum].sup.[infinity].sub.n=0][Z.sub.4][x.sup.n] and [Z.sub.4][[x].sub.n] = {0} for n < 0. Let I = {0}. Then, for [phi] = [[phi].sub.0], by easy computations [v.sub.[phi]](I) = M[x] where M = {0,2}. Hence I = {0} is 0-primal graded ideal of R.

Lemma 5. Let I be a proper graded ideal of R, and let [phi] : I(R) [right arrow] I(R) [union] {0} be any function. Then h(I) - h([phi](I)) [subset or equal to] [v.sub.[phi]](I).

Proof. Let a [member of] h(I) - h([phi](I)). Since I is proper, 1 [not member of] I. Then a = a. 1 [member of] h(I) - h([phi](I)) implies that a is not [phi]-prime to I. Thus a [member of] [v.sub.[phi]](I).

We recall from Lemma 5 that a = [[summation].sub.g[member of]G] [a.sub.g] [member of] [v.sub.[phi]](I) if each component [a.sub.g] of a is in [v.sub.[phi]](I). Thus, by Lemma 5, each element a [member of] I can be written in the form c+d, where c [member of] [v.sub.[phi]](I) and d [member of] [phi](I). Therefore, I [subset or equal to] [v.sub.[phi]](I) + [phi](I).

Theorem 6. Let I be a proper graded ideal of R, and let [phi]: I(R) [right arrow] I(R) [union] {0} be any function. If the set [v.sub.[phi]](I) is an ideal of R, then it is [phi]-prime graded ideal of R.

Proof. Clearly [v.sub.[phi]](I) is a graded ideal of R. Now, let a,b [member of] h(R) with ab [member of] [v.sub.[phi]](I) - [phi]([v.sub.[phi]](I)). Then ab is not [phi]-prime to I. So there exists r [member of] h(R) - h(I) with abr [member of] I - [phi](I). If a [not member of] [v.sub.[phi]](I), then a is [phi]-prime to I. So abr [member of] I - [phi](I) implies that br [member of] I. Hence br [member of] I - [phi](7), since if br [member of] [phi](I), then abr [member of] [phi](I), a contradiction. Thus, b [member of] [v.sub.[phi]](I), since r [member of] h(R) - h(I).

Corollary 7. Let I be a proper graded ideal of R, and let [phi]: I(R) [right arrow] I(R) [union] {0} be any function. If the set [v.sub.[phi]](I) is an ideal of R, then

[mathematical expression not reproducible] (2)

is [phi]-prime graded ideal of R.

Proof. let a,b [member of] h(R) with ab [member of] P - [phi](P). Then ab [not member of] [phi](I), since [phi](I) [subset or equal to] [phi](P). Therefore, ab [member of] [v.sub.[phi]](I) or ab = [c.sub.g] + [d.sub.g], where [c.sub.g] [member of] ([v.sub.[phi]](I))g and [d.sub.g] [member of] [([phi](I)).sub.g].

If ab [member of] [v.sub.[phi]] (I), then, by Theorem 6, a [member of] [v.sub.[phi]](I) [subset or equal to] p or b [member of] [v.sub.[phi]] (I) [subset or equal to] p. So, P is [phi]-prime graded ideal of R.

If ab = [c.sub.g] + [d.sub.g], where [c.sub.g] [member of] ([v.sub.[phi]](I))g and [d.sub.g] [member of] [([phi](I)).sub.g], then there exists r [member of] h(R) - h(I) with [c.sub.g]r [epsilon] I - [phi](I). So, abr = ([c.sub.g] + [d.sub.g])r [member of] I - [phi](I), since [d.sub.g]r [member of] 0(7). Therefore, ab [member of] [v.sub.[phi]] (I) and by Theorem 6, a [member of] v[[gamma].sub.[phi]](I) [subset or equal to] P or b [member of] [v.sub.[phi]](I) [subset or equal to] P.

Definition 8. Let I be a proper graded ideal of R, and let [phi] : I(R) [right arrow] I(R) [union] {0} be any function. Then I is called f-Pprimal graded ideal of R, where P = [v.sub.[phi]](I) + [phi](I), if the set [v.sub.[phi]](I) forms an ideal of R. By the above corollary, P is always [phi]-prime graded ideal of R. In this case P is called the adjoint of 7.

Proposition 9. Let I, P be proper graded ideals of R. Then the following statements are equivalent.

(1) I is [phi]-primal graded ideal of R with the adjoint graded ideal P.

(2) For x [member of] h(R) with x [not member of] h(P) - h([phi](I)) we have h((I : x)) = h(I) [union] h(([phi](I) : x)). If x [member of] h(P) - h([phi](I)), then h(d:x))[??]h(I)[union]h(([phi](I):x)).

Proof. (1) [??] (2) If x [member of] h(P) - h([phi](I)), then x [member of] [v.sub.[phi]](I). So there exists r [member of] h(R) - h(I) with rx [member of] I - [phi](I). Thus r [member of] h((I: x)) and r [not member of] h(I) [union] h(([phi](I) : x)). Since it is easy to see that h((I : x)) [contain or equal to] h(I) [union] h(([phi](I) : x)), we have that h((I : x)) [??] h(I) [union] h(([phi](D:x)).

Now let x [not member of] h(P) - h([phi](I)), where x [member of] h(R). Then x [not member of] [v.sub.[phi]] (I) and hence x is [phi]-prime to I. Let r [member of] h((I : x)). If rx [not member of] [phi](I), then r [member of] h(I). If rx [member of] [phi](I), then r [member of] h(([phi](I) : x)). Hence

h((I:x)) [subset or equal to] h(I) [union] h(([phi](I):x)) [subset or equal to] h((I:x)). (3)

(2) [??] (1) From part (2) we have h(P)-h([phi](I)) = [v.sub.[phi]] (7). Thus I is [phi]-primal graded ideal of R.

We say that P is [phi]-primal graded ideal of R if P itself is the adjoint of P. The next result shows that every [phi]-prime graded ideal of R is 0-primal.

Theorem 10. Every f-prime graded ideal of R is f-primal.

Proof. Let P be [phi]-prime graded ideal of R; we show that P is [phi]-P-primal graded ideal of R. Thus we must prove that

[mathematical expression not reproducible]. (4)

If P = [phi](P), then it is easy to check that [v.sub.[phi]](P) = 0; hence P is 0-P-primal graded ideal of R. Therefore, we may assume that P = f(P). We show that h(P) - h(f(P)) = [v.sub.[phi]] (P). Let a [member of] h(P) - h(f(P)). Then a.1 [member of] P - f(P) with 1 t P, so a [member of] [v.sub.[phi]] (P). On the other hand let a [not member of] h(P) - h(f(P)). If a [member of] h(f(P)), then ra [member of] f(P) for all r [member of] h(R), so a is [pi]-prime to P and hence a [not member of] [phi](P). If at h(f(P)), then a [not member of] P, so for any [r.sub.[alpha]] [member of] [R.sub.[alpha]] with [r.sub.[sigma]]a [member of] P-[phi](P) we have that [r.sub.[alpha]] [member of] [P.sub.[alpha]], since P is [phi]-prime. Thus a is [phi]-prime to P and hence a [not member of] (P). Therefore, h(P) - h(f(P)) = [v.sub.[phi]](P) which implies that P is [phi]-P-primal graded ideal of R.

Now we give an example of [phi]-P-primal graded ideal I of R such that I itself is not [phi]-prime.

Example 11. Let [phi] = [[phi].sub.0] and let R = [Z.sub.8] + u[Z.sub.8] where [u.sup.2] = 0. Then R is a commutative [Z.sub.2]-graded ring with unity. If I = 4[Z.sub.8] + m[Z.sub.8], then I is not a [phi]-prime graded ideal of R, since [bar.2] x [bar.2] [not equal to] 0 [member of] I, but [bar.2] [not member of] I. Let P = 2[Z.sub.8] + u[Z.sub.8]; we show that I is 0-P-primal graded ideal of R. It is enough to show that v(I) = h(P) - {0}. Let 0 = [bar.a] [member of] h(P), if [bar.a] [member of] 2[Z.sub.8]; then [bar.a] = 2k [member of] [Z.sub.8]. If k is an odd number, then 0 [not equal to] [bar.2d] [member of] I, but [bar.2] [not equal to] I, and if k is an even number 0 [not equal to] [bar.1a] [member of] I with [bar.1] [not member of] I; hence [bar.a] [member of] v(I). If [bar.a] [member of] u[Z.sub.8], then [bar.a] [member of] I [subset or equal to] v(I). On the other hand, if [bar.a] [member of] h(R)-h(P), then a is an odd number in Z8. If 0 [not equal to] [bar.a m] [member of] I for some [bar.m] [member of] [Z.sub.8], then 4 divides am and so 4 divides m since (4, a) = 1; hence [bar.m] [epsilon] I. Thus I is [phi]-P-primal graded ideal of R.

Let [phi] : I(R) [right arrow] I(R) [union] {0} be any function. We assume that, for any I, J [member of] I(R), [phi](I) [subset or equal to] [phi](I) if J [subset or equal to] I. Now we prove one of the main results in this section.

Theorem 12. Suppose that [[psi].sub.1] [less than or equal to] [f.sub.2], where and [[psi].sub.2] are maps from I(R) into I(R) U {0}, and let I be a [[psi].sub.2]-P-primal graded ideal of R, with [I.sub.e][I.sub.[alpha]] [not equal to] [[psi].sub.2][(I).sub.[alpha]] for all [alpha] [member of] G, where e is the identity element in G. If P is a prime graded ideal of R, then I is [[psi].sub.1]-P-primal.

Proof. Since I is [[psi].sub.2]-P-primal graded ideal of R, then

[mathematical expression not reproducible]. (5)

To show that I is [[psi].sub.2]-P-primal graded ideal of R we must prove that

[mathematical expression not reproducible]. (6)

If [[psi].sub.2] = [[phi].sub.0], then [[psi].sub.1] = [[psi].sub.2] and hence we have that [mathematical expression not reproducible] which implies that I is [[psi].sub.1]-P-primal graded ideal of R. Now we may assume that [[psi].sub.2] [not equal to] [[phi].sub.0]. Let [mathematical expression not reproducible]. Then there exists r [member of] h(R) - h(I) with rs [member of] I - [[psi].sub.2](I) [subset or equal to] I - [[psi].sub.1](I), so [mathematical expression not reproducible] which implies that

[mathematical expression not reproducible]. (7)

Now, let a [member of] h([[psi].sub.2](I)). If a [not equal to] [[psi].sub.1](I), then a [member of] I - [[psi].sub.1](I). So 1.a [member of] I - [[psi].sub.1](I) with a [not member of] I. Hence [mathematical expression not reproducible]. Therefore,

[mathematical expression not reproducible]. (8)

From (7) and (8) we have that

[mathematical expression not reproducible] (9)

Since [[psi].sub.1](I) [subset or equal to] [[psi].sub.2](I) [subset or equal to] P, by (9)

[mathematical expression not reproducible]. (10)

Let [mathematical expression not reproducible]. Then there exists [r.sub.[beta]] [member of] [R.sub.[beta]] - [I.sub.[beta]] with a[r.sub.[beta]] [member of] I - [[psi].sub.1](I). if a[r.sub.[beta]] [member of] I - [[psi].sup.2] (I), then [mathematical expression not reproducible]. So we may assume that a[r.sub.[beta]] [not member of] I - [[psi].sub.2](I); hence a[r.sub.[beta]] [member of] [[psi].sub.2](I). First suppose that a[I.sub.[beta]] [not subset or equal to] [([[psi].sub.2](I)).sub.[alpha][beta]], say a[s.sub.[beta]] [member of] [I.sub.[alpha][beta]] - [([[psi].sub.2](I)).sub.[alpha][beta]] with [s.sub.[beta]] [member of] [I.sub.[beta]]. Then a([r.sub.[beta]] + [S.sub.[beta]]) = a[r.sub.[beta]] + a[s.sub.[beta]] [not equal to] [[psi].sub.2](I) with [r.sub.[beta]] + [s.sub.[beta]] [member of] [R.sub.[beta]] - [I.sub.[beta]]. Hence [mathematical expression not reproducible]. Therefore, we may assume that a[I.sub.[beta]] [subset or equal to] ([[psi].sub.2](I))[alpha][beta].

Now suppose that [r.sub.[beta]][I.sub.e] [not subset or equal to] [([[psi].sub.2](I)).sub.[beta]]. Then there exists c [member of] [I.sub.e] with [r.sub.[beta]]c [member of] [I.sub.[beta]] - [([[psi].sub.2](I)).sub.[beta]]. Since [a.sup.2] [member of] [R.sub.e], then ([a.sup.2] + c)[r.sub.[beta]] [member of] [I.sub.[beta]] - [([[psi].sub.2](I)).sub.[beta]] with [r.sub.[beta]] [not member of] [I.sub.[beta]]. So [a.sup.2] + c [member of] [P.sub.e] but c [member of] [I.sub.e] [subset or equal to] [P.sub.e]; therefore, [a.sup.2] [member of] P and hence a [member of] P, since P is a prime graded ideal. So we may assume that [r.sub.[beta]][I.sub.e] [subset or equal to] [([[psi].sub.2](I)).sub.[beta]]. Since ([I.sub.e][I.sub.[beta]]) [not equal to] [([[psi].sub.2](I)).sub.[beta]] there exists [bar.a] [member of] [I.sub.e] and [bar.b] [member of] [I.sub.[beta]] with [bar.ab] [not member of] [([[psi].sub.2](I)).sub.[beta]]. Thus, ([a.sup.2] + [bar.a])([r.sub.[beta]] + [bar.b]) = [a.sup.2][r.sub.[beta]] + [a.sup.2][bar.b] + [bar.a][r.sub.[beta]] + [bar.ab] [not member of] [[psi].sub.2](I), so ([a.sup.2] + [bar.a])([r.sub.[beta]] + [bar.b]) [member of] I - [[psi].sup.2](I) with [r.sub.[beta]] + [bar.b] [member of] [R.sub.[beta]] - [I.sub.[beta]] which implies that [mathematical expression not reproducible]. Hence [a.sup.2] [member of] [P.sub.e] [subset or equal to] P, and so a [member of] P, since P is a prime graded ideal of R. Therefore, [mathematical expression not reproducible] so

[mathematical expression not reproducible] (11)

and hence I is [[psi].sub.1]-P-primal graded ideal of R.

3. Conditions on [phi]-Primal Graded Ideals

In this section we introduce some conditions under which 0primal graded ideals are primal.

Let [phi]: I(R) [right arrow] I(R) [union] {0} be any function. We have to recall that if I is 0-P-primal graded ideal of R, then

[mathematical expression not reproducible] (12)

is [phi]-prime graded ideal of R.

Definition 13. Let r be a homogeneous element in R. Then [absolute value of (r)] = [alpha] if r [member of] [R.sub.[alpha]] for some [alpha] [member of] G.

In the next theorem we provide some conditions under which [phi]-primal graded ideal is primal.

Theorem 14. Let R be a commutative graded ring with unity and let [phi] : I(R) [right arrow] I(R) [union] {0} be any function. Suppose that I is [phi]-P-primal graded ideal of R with [I.sub.[gamma]][I.sub.[delta]] [not subset or equal to] [phi](I) for each [gamma], [delta] [member of] G. If P is a prime graded ideal of R, then I is P-primal.

Proof. Assume that a is a homogeneous element in P. Then a [member of] [phi](I) or a [member of] [([v.sub.[phi]](I)).sub.[alpha]] for some [alpha] [member of] G or a = [b.sub.[beta]] + [c.sub.[beta]] where [b.sub.[beta]] [member of] [([v.sub.[phi]](I)).sub.[beta]] and [c.sub.[beta]] [member of] [phi](I) for some [beta] [member of] G. If the first two cases hold, then a is not prime to I, since it is not [[phi].sub.prime] to I. In the last case, let d be a homogeneous element in R such that d [not member of] I with [b.sub.[beta]]d [member of] I - [phi](I). Then ad = [b.sub.[beta]]d + [c.sub.[beta]]d [member of] I - [phi](1), because ad [member of] [phi](I) implies that [b.sub.[beta]]d [member of] [phi](I), since [c.sub.[beta]]d [member of] [phi](I) which is a contradiction. Thus a is not [phi]-prime to I and hence a is not prime to I. Now assume that b [member of] h(R) is not prime to I, so rb [member of] I for some homogeneous element r [member of] R - I. If rb [not member of] [phi](I), then b is not [phi]-prime to I, so b [member of] P. Thus assume that rb [member of] [phi](I). Suppose that [absolute value of (r)] = [alpha]. First suppose that b[I.sub.[alpha]] [??] [phi](I). Then, there exists r' [member of] [I.sub.[alpha]] such that br' [not member of] [phi](I). So b(r + r') [member of] I - [phi](I), where r + r is a homogeneous element in R - I, implies that b is not 0-prime to I; that is, b [member of] P. Therefore, we may assume that b[I.sub.[alpha]] [subset or equal to] [phi](I). Let [absolute value of (b)] = p. If r[I.sub.[beta]] [subset or equal to] [phi](I), then rc [member of] [phi](I) for some c [member of] [I.sub.[beta]]. In this case r(b + c) [member of] I - [phi](I) with r [member of] R - I; that is, b + c [member of] P, and hence b [member of] P, since c [member of] I [subset or equal to] P. So we may assume that rip C [phi](I). Since [I.sub.[alpha]][I.sub.[beta]] [not subset or equal to] [phi](I), there are b' [member of] [I.sub.[alpha]] and a' [member of] [I.sub.[beta]] with b'a' [not member of] [phi](I). Now, (b + a)(r + b') el- [phi](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' [member of] I [subset or equal to] P, so that b [member of] 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.

Let R 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 x S have the form P x S or R x 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 x S. We leave the easy proof for this proposition to the reader. For the trivial case (i.e., [R.sub.g] = [S.sub.g] = {0} for all g [not equal to] e) the next proposition is proved in [5, Lemma 13].

Proposition 15. Let R and S be commutative G-graded rings. If I is P-primal graded ideal of R and I is Q-primal graded ideal of S, then I x S (resp., R x I) is P x S- (resp., R x Q-) primal graded ideal of R x S.

Next, we generalize [6, Theorem 16] to the graded case. Then we use this generalization to prove Theorem 18.

Theorem 16. Let [R.sub.1] and [R.sub.2] be commutative G-graded rings with unities and let [[psi].sub.i] : I(R) [right arrow] I(R) [union] {0} be functions. Let [phi] = [[psi].sub.1] x [[psi].sub.2]. Then [phi]-primes of [R.sub.1] x [R.sub.2] have exactly one of the following three types:

(1) [I.sub.1] x [I.sub.2] where It is a proper graded ideal of [R.sub.i] with [[psi].sub.i]([I.sub.i]) = [I.sub.i]

(2) [I.sub.1] x [R.sub.2] where [I.sub.1] is [[psi].sub.1]-prime of [R.sub.1] which must be prime if [[psi].sub.2]([R.sub.2]) [not equal to] [R.sub.2]

(3) [R.sub.1] x [I.sub.2] where [I.sub.2] is [[psi].sub.2]-prime of [R.sub.2] which must be prime if [[psi].sub.1]([R.sub.1]) [not equal to] [R.sub.1]

Proof. We first note that a graded ideal of [R.sub.1] x [R.sub.2] having one of these three types is [phi]-prime. Case (1) is clear since [phi](I) = I. If [I.sub.1] is a prime graded ideal of [R.sub.1], certainly [I.sub.1] x [R.sub.2] is prime and hence [phi]-prime. So suppose that [I.sub.1] is [[psi].sub.1]-prime and [[psi].sub.2]([R.sub.2]) = [R.sub.2]. Let [mathematical expression not reproducible]. Then [mathematical expression not reproducible]. The proof for case (3) is similar.

Next, suppose that [I.sub.1] x [I.sub.2] is [phi]-prime. Let a, b be homogeneous elements in [R.sub.1] such that ab [member of] [I.sub.1] - [[psi].sub.1] ([I.sub.1]). Then (a, 0)(b, 0) = (ab, 0) [member of] [I.sub.1] x [I.sub.2] - [phi]([I.sub.1] x [I.sub.2]), so (a, 0) [member of] [I.sub.1] x [I.sub.2] or (b, 0) [member of] [I.sub.1] x [I.sub.2]; that is, a [member of] [I.sub.1] or b [member of] [I.sub.1]. So [I.sub.1] is [[psi].sub.1]-prime. Likewise, [I.sub.2] is [[psi].sub.2]-prime. Suppose that [I.sub.1] x [I.sub.2] [not member of] [[psi].sub.1] ([I.sub.1]) x [[psi].sub.2] ([I.sub.2]), say, [I.sub.1] [not equal to] [[psi].sub.1] ([I.sub.1]). Let p be a homogeneous element in [R.sub.1] such that p [member of] [I.sub.1] - [[psi].sub.1] ([I.sub.1]) and let q be a homogeneous element in [I.sub.2]. Then (p, 1)(1, q) = (p, q) [member of] [I.sub.1] x [I.sub.2] -[phi]([I.sub.1] x [I.sub.2]). So (1, q) [member of] [I.sub.1] x [I.sub.2] or (p,1) [member of] [I.sub.1] x [I.sub.2]. Hence [I.sub.1] = [R.sub.1] or [I.sub.2] = [R.sub.2]. Suppose

that [I.sub.2] = [R.sub.2]. So [I.sub.1] x[R.sub.2] is [phi]-prime where [I.sub.1] is [[psi].sub.1]-prime. It remains to show that if [[psi].sub.2]([R.sub.2]) = [R.sub.2], then [I.sub.1] is actually prime graded ideal of [R.sub.1]. Let a, b be homogeneous elements in [R.sub.1] such that ab [member of] [I.sub.1]. Now 1 [member of] [R.sub.2] is a homogeneous element in [R.sub.2] not in [[psi].sub.2]([R.sub.2]). Then (a, 1)(b, 1) = (ab, 1) [member of] [I.sub.1] x [R.sub.2] - [phi]([I.sub.1] x [R.sub.2]), so (a,1) [member of] [I.sub.1] x [R.sub.2] or (b,1) [member of] [I.sub.1] x [R.sub.2]; that is, a [member of] [I.sub.1] or b [member of] [I.sub.1].

Let [R.sub.1], [R.sub.2] be commutative G-graded rings with unities and let R = [R.sub.1] x [R.sub.2]. Let [phi] : I(R) [right arrow] I(R) [union] {0} be a function. In the next theorem we also provide some conditions under which [phi]-primal graded ideal of R is primal, but first we start with the following remark.

Remark 17. Let I be a proper graded ideal of a commutative graded ring R and let [phi] : I(R) [right arrow] I(R) [union] {0} be a function. If a homogeneous element a [member of] R is not [phi]-prime to I, then there is a homogeneous element r in R-I such that ar [member of] I-[phi](I) [subset or equal to] I so a is not prime to I.

Theorem 18. Let [R.sub.1], [R.sub.2] be commutative G-graded rings with unities and let R = [R.sub.1] x [R.sub.2]. Let : I(R) [right arrow] I(R) [union] {0} be functions with [[psi].sub.i] ([R.sub.i]) [not equal to] [R.sub.i] for i = 1, 2. Let [phi] = [[psi].sub.1] x [[psi].sub.2]. Assume that P is a graded ideal of R with [phi](P) = P. If I is a [phi]-P-primal graded ideal of R, then either I = [phi](I) or I is P-primal.

Proof. Suppose [phi](I) [not equal to] I. By Corollary 7, P is [phi]-prime graded ideal of R. Therefore, by Theorem 16, P has one of the following three cases.

Case 1. P = [P.sub.1] x [P.sub.2], where [P.sub.i] is a proper graded ideal of [R.sub.i] with [[psi].sub.i]([P.sub.i]) = [P.sub.i] for i = 1, 2. In this case [phi](P) = P, a contradiction.

Case 2. P = [P.sub.1] x [R.sub.2] where [P.sub.1] is [[psi].sub.1]-prime graded ideal of [R.sub.1]. Since [[psi].sub.2]([R.sub.2]) = [R.sub.2], by Theorem 16(2), [P.sub.1] is a prime graded ideal of [R.sub.1] and so P is a prime graded ideal of R. We show that [I.sub.2] = [R.sub.2]. Since I [not equal to] [phi](I), there exists a homogeneous element (a, b) in 1 - [phi](I). So (a, 1)(1, b) = (a, b) [member of] 1 - [phi](1). If (a, 1) [not member of] I, then (1, b) is not [phi]-prime to 7; hence (1,b) [member of] P = [P.sub.1] x [R.sub.2], so 1 [member of] [P.sub.1], a contradiction. Thus (a, 1) [member of] I = [I.sub.1] x [I.sub.2]; that is, 1 [member of] [I.sub.2] and [I.sub.2] = [R.sub.2]. Now we prove that [I.sub.1] is [P.sub.1]-primal graded ideal of [R.sub.1]. Let [a.sub.1] be a homogeneous element in [P.sub.1]. Then ([a.sub.1], 0) [member of] [P.sub.1] x [R.sub.2] = P. If ([a.sub.1], 0) [member of] [phi](I) = [[psi].sub.1] ([I.sub.1]) x [[psi].sub.2]([R.sub.2]), then [a.sub.1] [member of] [[psi].sub.1] ([I.sub.1]) [subset or equal to] [I.sub.1] so [a.sub.1] is not prime to [I.sub.1]. Therefore, we may assume that ([a.sub.1], 0) [member of] [v.sub.[phi]](I). In this case there exists a homogeneous element ([r.sub.1], [r.sub.2]) [member of] R - I such that ([a.sub.1], 0)([r.sub.1], [r.sub.2]) [member of] I - [phi](I) so [a.sub.1][r.sub.1] [member of] [I.sub.1] - [[psi].sub.1]([I.sub.1]) with [r.sub.1] [member of] [R.sub.1] - [I.sub.1], since R - I = ([R.sub.1] - [I.sub.1]) x [R.sub.2], implies that [a.sub.1] is not [[psi].sub.1]-prime to [I.sub.1]; hence by Remark 17, [a.sub.1] is not prime to [I.sub.1]. Conversely, let [b.sub.1] be a homogeneous element in [R.sub.1] such that [b.sub.1] is not prime to [I.sub.1]. Then there exists a homogeneous element [c.sub.1] in [R.sub.1] - [I.sub.1] with [b.sub.1][c.sub.1] [member of] [I.sub.1]. Since [[psi].sub.2]([R.sub.2]) [not equal to] [R.sub.2], ([b.sub.1], 1)([c.sub.1], 1) = ([b.sub.1][c.sub.1], 1) [member of] [I.sub.1] x [[psi].sub.2] - ([I.sub.1] x [[psi].sub.2] ([R.sub.2])) [subset or equal to] I - [phi](I) with ([c.sub.1], 1) [member of] R - I. Hence ([b.sub.1], 1) is not [phi]-prime to I which implies that ([b.sub.1], 1) [member of] P = [P.sub.1] x [R.sub.2] and so [b.sub.1] [member of] [P.sub.1]. We have already shown that the set of homogeneous elements in [P.sub.1] consists exactly of the homogeneous elements of [R.sub.1] that are not prime to [I.sub.1]. Hence [I.sub.1] is [P.sub.1]-primal graded ideal of [R.sub.1] so by Proposition 15, I is P-primal graded ideal of R.

Case 3. P = [R.sub.1] x [P.sub.2], where [P.sub.2] is [[psi].sub.2]-primal graded ideal of [R.sub.2]. The proof of Case 3 is similar to that of Case 2.

4. [[phi].sub.J]-P/J-Primal Graded Ideals

Let R be a commutative graded ring with unity and let J be a proper graded ideal of R. Let 0 : I(R) [right arrow] I(R) [union] {0} be any function. As a generalization of , we define : I(R/J) [right arrow] I(R/J) [union] {0} by [[phi].sub.J] (I/J) = ([phi](I) + J)/J for every graded ideal I [member of] I(R) with J [subset or equal to] I (and [[phi].sub.J](I/J) = 0 if [phi] = [[phi].sub.0]).

We leave the trivial proof of the next lemma to the reader.

Lemma 19. Let R be a commutative graded ring with unity and let J be a proper graded ideal of R. Let [phi]: I(R) [right arrow] I(R) [union] {0} be any function. If P is [phi]-prime graded ideal of R containing J, then P/J is [[phi].sub.J]-prime graded ideal of R/J.

Lemma 20. Let R be a commutative graded ring with unity and let J be a proper graded ideal of R, and let 0 : I(R) [right arrow] I(R) [union] {0} be any function. Let P be a graded ideal of R containing J. If P/J is [phi]-prime graded ideal of R/J with J [subset or equal to] [phi](P), then P is [phi]-prime graded ideal of R.

Proof. Let a, b be homogeneous elements in R with ab [member of] (P) - [phi](P). Then ab [member of] J + P and ab [not member of] J + [phi](P) = [phi](P). Thus, ab [member of] (J + P)-(J + [phi](P)), so [bar.ab] [member of] (J + P)/J - (J + [phi](P))/J which implies that [bar.ab] [member of] P/J - [[phi].sub.J](P/J); that is, [bar.a] [member of] P/J or [bar.b] [member of] P/J so a [member of] P or b [member of] P. Therefore, P is [phi]-prime graded ideal of R.

In the next result and under the condition that J [subset or equal to] [phi](I) we prove that I is [phi]-primal graded ideal of R if and only if J/J is [[phi].sub.J]-primal graded ideal of R/J.

Theorem 21. Let R be a commutative graded ring with unity and let [phi] : I(R) [right arrow] I(R) [union] {0} be any function. Let I be a proper graded ideal of R, and let J be a graded ideal of R with J [subset or equal to] [phi](I). Then I is [phi]-P-primal graded ideal of R if and only if J/J is [[phi].sub.J]-P/J-primal graded ideal of R/J.

Proof. Suppose that I is 0-P-primal graded ideal of R with J [subset or equal to] I. Then, by Corollary 7, P is [phi]-prime graded ideal of R containing 7, so, by Lemma 19, P/J is [[phi].sub.J]-prime graded ideal of R/J. We show that J/J is [[phi].sub.J]-P/J-primal graded ideal of R/J. That is, we must prove that

[mathematical expression not reproducible] (13)

Let [bar.a] [member of] h(P/J); then a [member of] h(P) is not [phi]-prime to I. So there exists r [member of] h(R)-h(J) with ra [member of] I-[phi](J). If a [less than or equal to] J + [phi](I) = [phi](I), then a [member of] [[phi].sub.J](I/J). So we may assume that a [not member of] J + [phi](I) = [phi](I). Therefore, [bar.r] [bar.a] [member of] I/J - (J + [phi](I))/J = I/J - [[phi].sub.J](I/J) and because [bar.r] [member of] I/J we get that [mathematical expression not reproducible].

Now, assume that b is a homogeneous element in R/J such that [mathematical expression not reproducible]. Then there exists a homogeneous element [bar.r] in R/J-I/J such that [bar.r][bar.b] [member of] I/J - [[phi].sub.J](I/J), so rb [member of] I - [phi](I) with r [not member of] I. Thus, b is not [phi]-prime to I which implies that b [member of] P, and hence [bar.b] [member of] P/J. Therefore,

[mathematical expression not reproducible] (14)

and so I/J is [[phi].sub.J]-P/J-primal graded ideal of R/J.

Conversely, suppose that I/J is [[phi].sub.J]-P/J-primal graded ideal of R/J with the adjoint graded ideal P/J. We show that I is 0-P-primal graded ideal of R. Now, by Corollary 7, P/J is [[phi].sub.J]-prime graded ideal of R/J with J [subset or equal to] P, so, by Lemma 20, P is [phi]-prime graded ideal of R. To finish the proof we need to show that

[mathematical expression not reproducible]. (15)

Clearly, [phi](I) [subset or equal to] I [subset or equal to] P. Let a [member of] h([v.sub.[phi]](I)); then there exists a homogeneous element r [member of] R - I with [bar.r][bar.a] [member of] I - [phi](I). Since ra [not equal to] [phi](I) = J + [phi](7) we get that [bar.r][bar.a] [member of] I/J - (7 + [phi](I))/J = I/J - [[phi].sub.J](I/J) and [bar.r] [not member of] I/J. So, [bar.a] [member of] P/J and hence a [member of] P.

Now, let a [member of] h(P). Suppose that [bar.a] [member of] I/J. Then a [member of] I. If a [member of] [phi](7), then we are done. Assume that a [not member of] [phi](I). Then a [member of] I - [phi](I) and, so, a is not [phi]-prime to I; hence a [member of] [v.sub.[phi]](I). Therefore, we may assume that [bar.a] [not member of] I/J, so there exists [bar.r] [member of] h(R/J)-h(I/J) with [bar.r][bar.a] [member of] I/J - [[phi].sub.J](I/J), and so [bar.r][bar.a] [not member of] (J + [phi](I)); that is, ra [not member of] J + [phi](I) = [phi](I). Therefore, ra [member of] I - [phi](I) with r [not member of] I; that is, a [member of] [v.sub.[phi]] (I).

By Theorem 21, we get the following result.

Corollary 22. Let R be a commutative graded ring with unity, and let [phi] : I(R) [right arrow] I(R) [union] {0} be any function. let J be a graded ideal of R. Then there is one-to-one correspondence between [phi]-[P.sup.I]-primal graded ideals I of R containing J with J [subset or equal to] [phi](I) and [[phi].sub.J]-[P.sup.1]/J-primal graded ideals of R/J.

5. [[phi].sub.S]-Primal Graded Ideals of [R.sub.S]

Let R 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 [R.sub.S] the ring of fractions [S.sup.-1]R. We define a grading on [R.sub.S] by setting

[mathematical expression not reproducible]. (16)

It is easy to see that [R.sub.S] is G-graded ring (for short graded ring). Also, for G-graded ideal I of R, [I.sub.S] = [S.sup.-1] I is G-graded ideal of [R.sub.S].

Consider the canonical homomorphism [rho] : R [right arrow] [R.sub.S] which is defined by r [??] r/1 for all r [member of] h(R). Then [rho] is a homogenous graded homomorphism of degree 0.

Now let [phi] : I(R) [right arrow] I(R) [union] {0} be any function, we define [[phi].sub.S]: I([R.sub.S]) [right arrow] I([R.sub.S]) [union] {0} by [[phi].sub.S](J) = [([phi]([[rho].sup.- 1](J))).sub.S] for every J [member of] I([R.sub.S]). Note that [[phi].sub.S](J) [subset or equal to] J, since, for J [member of] 1([R.sub.S]), we have that [phi]([[rho].sup.-1](J)) [subset or equal to] [[rho].sup.-1](J) implies [[phi].sub.s](J) [subset or equal to] [([[rho].sup.-1](J)).sub.S] [subset or equal to] J.

Example 23. Let R = [Z.sub.6] + u[Z.sub.6] with [u.sup.2] = 0. Let S = {1,2,4}, [phi] = [[phi].sub.2]. Then S is a multiplicatively closed subset of h(R). If P = {0}, then one can easily check that P is [[phi].sub.2]-P-primal graded ideal of R. Moreover, [P.sub.S] = [([phi](P)).sub.S] = {0}; hence [[phi].sub.S]([P.sub.S]) = [P.sub.S], since [[rho].sup.-1]([P.sub.S]) = {0, 3,3u}, where p : R [right arrow] [R.sub.S] is the canonical homomorphism. Therefore, [P.sub.S] is [[phi].sub.S]-[P.sub.S]-primal graded ideal in [R.sub.S].

We start by proving the following properties about [phi]-prime graded ideals of R, where [rho] : R [right arrow] [R.sub.S] is the canonical homomorphism and S is a multiplicatively closed subset of h(R).

Lemma 24. Let [phi] : I(R) [right arrow] I(R) [union] {0} be any function, and let I be [phi]-prime graded ideal of R with I [intersection] S = 0; then [I.sub.S] is [[phi].sub.S]-prime graded ideal of [R.sub.S].

Proof. Let x/s, y/t be homogeneous elements in [R.sub.S] with (x/s)(y/t) [member of] [I.sub.S] - [[phi].sub.S](h); then, for some u [member of] S, xyu [member of] I-[phi](I), so x [member of] I or yu [member of] I, and thus x/s [member of] [I.sub.S] or y/t [member of] [I.sub.S]; hence [I.sub.S] is [[phi].sub.S]-prime graded ideal of [R.sub.S].

Theorem 25. Let [phi]: I(R) [right arrow] I(R) [union] {0} be any function, and let P be [phi]-prime graded ideal of R with h(P) [intersection] S = 0. If [[rho].sup.-1]([([phi](P)).sub.S]) [subset or equal to] P, then [[rho].sup.-1]([P.sub.S]) = P.

Proof. It is easy to see that P [subset or equal to] [[rho].sup.-1] ([P.sub.S]).

Conversely, let x be a homogeneous element in [[rho].sup.-1]([P.sub.S]), then for some s [member of] S, xs [member of] P. If xs [member of] [phi](P), then xs [member of] P - [phi](P) and s [member of] P so x [member of] P. Therefore, we may assume that xs [member of] [phi](P), so x is a homogeneous element in [[rho].sup.-1](([phi](P))s). Thus,

[[rho].sup.-1] ([P.sub.S]) [subset or equal to] P [union] [[rho].sup.-1] ([([phi](P)).sub.S]), (17)

and since [[rho].sup.-1] ([([phi](P)).sub.S]) [subset or equal to] p, we have that [[rho].sup.-1]([P.sub.S]) = P.

Lemma 26. Let [phi] : I(R) [right arrow] I(R) [union] {0} be any function, and let I be [phi]-P-primal graded ideal of R with h(P) [intersection] S = 0. If a/s [member of] h([I.sub.S]) - h([([phi](I)).sub.S]), then a [member of] I. Moreover, if h([[rho].sup.-1]([([phi](I)).sub.S])) [subset or equal to] h(I), then I = [[rho].sup.-1]([I.sub.S]).

Proof. Let a/s [member of] h([I.sub.S]) - h([([phi](I)).sub.S]), so a/s = b/t for some b [member of] h(I) and t [member of] S. In this case uta = usb [member of] h(I) for some u [member of] S. If uta [member of] [phi](I), then a/s = uta/uts [member of] [([phi](I)).sub.S] a contradiction. So we have that uta [member of] I - [phi](I). If a [not member of] h(I), then ut is not [phi]-prime to I; so ut [member of] h(P) [intersection] S which contradicts the hypothesis. Therefore, a [member of] I.

For the last part, it is clear that h(I) [subset or equal to] h([[rho].sup.-1]([I.sub.S])). Now let a be a homogeneous element in [[rho].sup.-1](Is). Then as [member of] h(I) for some s [member of] S. If as [not member of] [phi](I) and a [not member of] I, then s is not [phi]-prime to I, so s [member of] P [intersection] S a contradiction. So a must be in I. If as [member of] [phi](I), then a/1 = (as)/s [member of] [([phi](I)).sub.S], and so a [member of] [[rho].sup.-1]([([phi](I)).sub.S]). Therefore, h([[rho].sup.-1]([I.sub.S])) = h(I) [union] h([[rho].sup.-1] (([phi](I))g)) = h(I), since h([[rho].sup.-1]([([phi](I)).sub.S])) [subset or equal to] h(I). Hence I = [[rho].sup.-1]([I.sub.S]).

Lemma 27. Let [phi] : I(R) [right arrow] I(R) [union] {0} be any function, and let I be [phi]-P-primal graded ideal of R with h(P) [intersection] S = 0. Then h([[rho].sup.-1]([I.sub.S])) - h([[rho].sup.-1]([[phi].sub.s]([I.sub.S]))) [subset or equal to] h(I) - h([phi](I)).

Proof. Let a be a homogeneous element in [[rho].sup.-1]([I.sub.S]) such that a [not member of] h([[rho].sup.-1]([[phi].sub.S]([I.sub.S]))); then a/1 [member of] h([I.sub.S]) - h([[phi].sub.S]([I.sub.S])) [subset or equal to] h([I.sub.S]) h([([phi](I)).sub.S]) and, by Lemma 26, a [member of] I. If a [member of] [phi](I), then a/1 [member of] [([phi](I)).sub.S] [subset or equal to] [[phi].sub.S]([I.sub.S]) implies that a [member of] h([[rho].sup.-1]([[phi].sub.S]([I.sub.S]))) a contradiction. Therefore, a [member of] h(I) - h([phi](I)).

Let R be a commutative graded ring with unity and M be R-graded module. An element a [member of] h(R) is called a zerodivisor on M if am = 0 for some m [member of] h(M). We denote by [Z.sub.R](M) the set all zero-divisors of h(R) on M.

Corollary 28. Let [phi] : I(R) [right arrow] I(R) [union] {0} be any function, and let I be [phi]-P-primal graded ideal of R with h(P) [intersection] S = 0, S [intersection] [Z.sub.R](R/[phi](I)) = 0. If [[rho].sup.-1]([([phi](I)).sub.S]) [subset or equal to] I, then [mathematical expression not reproducible].

Proof. By Lemma 26, if [[rho].sup.-1]([([phi](I)).sub.S]) [subset or equal to] I, then [p.sup.-1]([I.sub.S]) = I. Let x/s be a homogeneous element in [([v.sub.[phi]](I)).sub.S] - [[phi].sub.S]([I.sub.S]); then x/s = y/t, where y [member of] [v.sub.[phi]](I). If y [member of] I, then [mathematical expression not reproducible]. Therefore we may assume that y [not member of] I. If [mathematical expression not reproducible]. Therefore we may assume that y/1 [not member of] [I.sub.S]. So, y/1 is a homogeneous element in [([v.sub.[phi]](I)).sub.S] - [I.sub.S]. So uy [member of] [v.sub.[phi]](I) for some u [member of] S and uy [not member of] I. So there exists r [member of] h(R) - h(I) such that ruy [member of] I - [phi](I). If ry [not member of] I, then u [member of] [v.sub.[phi]](I) [subset or equal to] p a contradiction. Therefore, ry [member of] I - [phi](I). So (r/1)(y/1) [member of] [I.sub.S]. If (r/1)(y/1) [member of] [[phi].sub.S]([I.sub.S]), then there exists t [member of] S with try [member of] [phi]([[rho].sup.-1]([I.sub.S])) = [phi](I), so t [member of] S [intersection] [Z.sub.R](R/[phi](I)) a contradiction. Thus (r/1)(y/1) [member of] [I.sub.S] - [[phi].sub.S]([I.sub.S]) and r/1 [not member of] [I.sub.S]. So, [mathematical expression not reproducible]. Hence [mathematical expression not reproducible].

We recall that if I is a graded ideal in R, then I [subset or equal to] [[rho].sup.-1]([I.sub.S]); therefore, we may assume that [([phi](I)).sub.S] [subset or equal to] [[phi].sub.S]([I.sub.S]).

Under the condition that [[rho].sup.-1] ([([phi](I)).sub.S]) [subset or equal to] I for all proper graded ideals I of R, we have the following propositions.

Proposition 29. Let S be a multiplicatively closed subset of h(R) with 1 [member of] S, let [phi] : I(R) [right arrow] I(R) [union] {0} be any function, and let I be [phi]-P-primal graded ideal of R with h(P) [intersection] S = 0, S [intersection] [Z.sub.R](R/[phi](I)) = 0. Then [I.sub.S] is [[phi].sub.S]-[P.sub.S]-primal graded ideal of [R.sub.S].

Proof. By Lemma 24, [P.sub.S] is [[phi].sub.S]-prime graded ideal of [R.sub.S].

To show that [I.sub.S] is [[phi].sub.S]-[P.sub.S]-primal graded ideal of [R.sub.S], we must prove that

[mathematical expression not reproducible]. (18)

Clearly, [[phi].sub.S]([I.sub.S]) [subset or equal to] [P.sub.S]; let a/s be a homogenous element in [mathematical expression not reproducible]. Then there exists r/u [member of] h([R.sub.S]) - h([I.sub.S]) with (r/u) x (a/s) [member of] [I.sub.S] - [[phi].sub.S]([I.sub.S]) [subset or equal to] [I.sub.S] - [([phi](I)).sub.S], so ra [not member of] [phi](I) and, by Lemma 26, ra [member of] I. So, ra [member of] I - [phi](I), and r [not member of] h(I). Thus a [member of] [v.sub.[phi]](I) [subset or equal to] P and hence a/s [member of] [P.sub.S].

Conversely, let a/s [member of] h(Ps) such that a/s i fg(Ig). Then a e [[rho].sup.-1](Ps) = P. If a/s [member of] [I.sub.S], then (1/1)(a/s) [member of] [I.sub.S] - [[phi].sub.S]([I.sub.S]), (1/1) [not member of] [I.sub.S], so a/s is not [[phi].sub.S]-prime to [I.sub.S]; thus a/s [member of] ([I.sub.S]). Therefore, we may assume that a/s [not equal to] [I.sub.S]; that is, ta [not member of] I for every t [member of] S. So, a [not member of] I. Therefore, a [member of] h(P) - h(I) [subset or equal to] h([v.sub.[phi]](I)). Thus, a/s [member of] h([([v.sub.[phi]](I)).sub.S]) - h([[phi].sub.S]([I.sub.S])). Since, by Corollary 28, [mathematical expression not reproducible] we have that [mathematical expression not reproducible].

Proposition 30. Let [phi] : I(R) [right arrow] I(R) [union] {0} be any function, and let J be [[phi].sub.S]-Q-primal graded ideal of [R.sub.S]. Then [[rho].sup.-1] (Q) is [phi]-prime graded ideal of R and [[rho].sup.-1](J) is f-primal graded ideal of R with the adjoint graded ideal [[rho].sup.-1] (Q) with [[rho].sup.-1] (Q) [intersection] S = 0, S [intersection] [Z.sub.R] R/[phi]([[rho].sup.-1](J))) = 0. Moreover, J = ([[rho].sup.-1] (J))s.

Proof. To show that [[rho].sup.-1] (Q) is 0-prime graded ideal of R, it is enough to prove that [[rho].sup.-1] (J) is 0-primal graded ideal of R with

the adjoint graded ideal [[rho].sup.-1](Q). Then, by using Corollary 7, [[rho].sup.-1](Q) will be 0-prime graded ideal of R.

Now, to prove that [[rho].sup.-1] (J) is [phi]-primal graded ideal of R we must show that

[mathematical expression not reproducible]. (19)

But [phi]([[rho].sup.-1] (J)) [subset or equal to] [[rho].sup.-1] (J) [subset or equal to] [[rho].sup.-1] (Q). Let a be a homogenous element in [v.sub.[phi]]([[rho].sup.-1](J)) with a [not member of] [phi]([[rho].sup.-1](J)); then a/1 [member of] ([v.sub.[phi]]([[rho].sup.-1](J)))s - [[phi].sub.S](J), since S [intersection] [Z.sub.R](R/[phi](I)) = 0 and, by Corollary 28, [mathematical expression not reproducible]. Thus, [mathematical expression not reproducible] and hence a [member of] [[rho].sup.-1](Q).

Conversely, let a be a homogeneous element in [[rho].sup.-1](Q). Then a/1 in Q. We may assume that a [not member of] [phi]([[rho].sup.-1](J)), since S [intersection] [Z.sub.R](R/[phi]([p.sup.-1](J))) = 0, a/1 [not member of] [[phi].sub.S](J). If a/1 [member of] J, then (a/1) [member of] J - [[phi].sub.S](J) and since [phi]([[rho].sup.-1](D) [subset or equal to] [[rho].sup.- 1]([[phi].sub.S](J)), a [member of] [[rho].sup.-1](J) - [[rho].sup.-1]([[phi].sub.S](J)) [subset or equal to] [[rho].sup.-1](J) - [phi]([[rho].sup.-1](J)), but 1 [not member of] [[rho].sup.-1](J), so a [member of] [v.sub.[phi]]([[rho].sup.-1](J)). If a/1 [not member of] J, then a/1 [member of] Q - J and so a/1 [member of] [v.sub.[phi]s] (J). Let x/s be a homogeneous element in [R.sub.S] - J, with (a/1)(x/s) [member of] J - [[phi].sub.S](J); then ax [member of] [[rho].sup.-1](J)- [[rho].sup.- 1]([[phi].sub.S](J)) [subset or equal to] [[rho].sup.-1](J) - [phi]([[rho].sup.-1](J)), since ax/1 [member of] J and ax/1 [not member of] [[phi].sub.S](J), for if ax/1 [member of] [[phi].sub.S](J), then ax/s [less than or equal to] [[phi].sub.S](J), a contradiction. Thus we have that ax [less than or equal to] [[rho].sup.-1](J) - f([[rho].sup.-1](J)) and x [not equal to] [[rho].sup.- 1](J), since x/s [not member of] J. Therefore, a [member of] [v.sub.[phi]]([[rho].sup.-1] (J)) and so [[rho].sup.-1](J) is [phi]- primal graded ideal of R with the adjoint graded ideal [[rho].sup.-1] (Q).

Finally, we show that J = ([[rho].sup.-1](J))g. Clearly, J [subset or equal to] [([[rho].sup.-1](J)).sub.S]. Conversely, let x/s be a homogeneous element in ([[rho].sup.-1](J))g. Then xt [member of] [[rho].sup.-1](J) for some t [member of] S. Thus, xt/1 [member of] [rho]([[rho].sup.-1](J)) = J, and hence (xt/1)(1/st) = x/s [member of] J. Therefore, J = [([[rho].sup.-1](J)).sub.S].

Under the condition that [[rho].sup.-1](([phi](I))g) [subset or equal to] I for all proper graded ideals I of R and by using Propositions 29 and 30 we have the following result.

Corollary 31. Let R be a commutative graded ring with unity. Let S be a multiplicatively closed subset of h(R), and let [phi]: I(R) [right arrow] I(R) [union] {0} be any function. Then there is one-to-one correspondence between [phi]-[P.sup.I]-primal graded ideals I of R and [[phi].sub.S]-[P.sup.I.sub.S]-primalgraded ideals [I.sub.S] of [R.sub.S], where [P.sup.I] is [phi]-prime graded ideal of R with [P.sup.I] [intersection] S = 0, S [intersection] [Z.sub.R](R/[phi](I)) = 0.

I would like to remark that "[phi]-theory" introduced here has nothing in common with "[phi]-theories" considered previously by Badawi and Lucas in .

https://doi.org/ 10.1155/2017/3817479

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.

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 A. Y. Darani, "Generalizations of primal ideals in commutative rings," Matematichki Vesnik, vol. 64, no. 1, pp. 25-31, 2012.

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Ameer Jaber

Department of Mathematics, The Hashemite University, Zarqa 13H5, Jordan

Correspondence should be addressed to Ameer Jaber; ameerj@hu.edu.jo

Received 28 February 2017; Revised 28 April 2017; Accepted 11 May 2017; Published 4 June 2017

Academic Editor: Naihuan Jing
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Title Annotation: Printer friendly Cite/link Email Feedback Research Article Jaber, Ameer Journal of Mathematics Report Jan 1, 2017 10534 Some Hermite-Hadamard-Fejer Type Integral Inequalities for Differentiable [eta]-Convex Functions with Applications. Some Properties of Serre Subcategories in the Graded Local Cohomology Modules. Mathematical research Rings (Algebra) Rings (Mathematics)

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