Weak Singularity in Graded Rings -.:: Natural Sciences Publishing

Math. Sci. Lett. 4, No. 1, 51-53 (2015) 51 Mathematical Sciences Letters An International Journal

http://dx.doi.org/10.12785/msl/040111

Weak Singularity in Graded Rings

Gayatri Das and Helen K. Saikia∗ Department of Mathematics, Gauhati University, Guwahati, 781014 India

Received: 27 Jun. 2014, Revised: 27 Sep. 2014, Accepted: 18 Oct. 2014 Published online: 1 Jan. 2015

Abstract: In this paper we introduce the notion of graded weak singular ideals of a graded ring R. It is shown that every graded weak singular ideal of R is graded singular. A graded weakly nil ring is graded weak singular. If R is a graded weak non-singular ring then R is graded semiprime. Every graded strongly prime ring R is graded weak non-singular and the same holds for every reduced graded ring.

Keywords: Graded Ring, Graded Ideals, Graded Singular Ideals, Graded Semiprime Ring.

1 Introduction We now present the following definitions that are needed in the sequel. The notion of singularity plays a very important role in the study of algebraic structures. It was remarked by Definition 2.1: The graded singular ideal of R, denoted Miguel Ferrero and Edmund R. Puczylowski in [3] that by Z(R) is defined as studying properties of rings one can usually say more Z(R)= {x ∈ R|annR(x) ∩ H = 0 for every nonzero graded assuming that the considered rings are either singular or ideal H of R} = {x ∈ h(R)|xK = 0 for some graded nonsingular. Therefore, in the studies of rings, the essential ideal K} importance of the concept of singularity is remarkable. In Definition 2.2: A graded ring R is called graded singular [3], the authors have established some properties of provided Z(R)= R. On the other hand, R is called graded singular ideals of a ring. T. K. Dutta et al have introduced nonsingular provided Z(R)= 0. the notion of singular ideals in ternary semirings and investigated various properties of such ideals [2]. Definition 2.3: Graded weak singular ideal denoted by In this paper, we attempt to study some properties of Zw(R) is defined as graded weak singular ideals. It is shown that every graded Zw(R)= {r ∈ R|r = r1 + r2 + ...rk,ri ∈ h(R) and riK = weak singular ideal is graded singular, but the converse is 0 ∀i, for some graded essential ideal K of R} not true. It is established that a graded weak nonsingular ring is graded semi prime and hence every graded strongly Definition 2.4: R is a graded weak nonsingular ring if prime ring R is graded semi prime. Finally we prove that ZW (R)= 0 and a graded weak singular ring if ZW (R)= R. if I is a graded ideal of R and I is graded semi prime then ZW (I)= I ∩ZW (R), where ZW (R) denotes the graded weak Definition 2.5: R is called a graded weakly nil ring if singular ideal of R. This implies that the class of semiprime there is such a multiplicatively closed subset S of graded weak nonsingular rings is hereditary. nilpotent homogeneous elements from R that each homogeneous element of R is a finite sum of elements from S.

2 Preliminaries Definition 2.6: A graded ring R is said to be graded (right) strongly prime if every nonzero graded ideal I of R Throughout our discussion, we consider R to be a contains a finite subset F ⊆ h(I) such that annR(F)= 0. commutative graded ring. The basic definitions used in this paper on rings, modules and graded rings are Definition 2.7: A graded ring R is called reduced if it has available in [4,5]. no nonzero homogeneous nilpotent elements.

∗ Corresponding author e-mail: [email protected]

c 2015 NSP Natural Sciences Publishing Cor. 52 G. Das, H. K. Saikia: Weak singularity in graded rings

3 Main Results We show that ann(a j) ≤e R. 2 2 Now a = 0 implies (a1 + a2 + ... + as) = 0. In this section we present our main results. Therefore s Proposition 3.1: Every graded weak singular ideal of R is 2 ∑ a j + 2 ∑ a jai = 0 graded singular. j=1 j=i Proof: We need to prove Zw(R) ⊆ Z(R). . Let r ∈ Z (R) such that r = r + r + ... + r . Then This gives w 1 2 n s riK = 0 i,and K ≤e R 2 ( ∑ a j + 2 ∑ a jai)xk = 0 ⇒ r1K + r2K + ... + rnK = 0, K ≤e R j=1 j=i ⇒ (r1 + r2 + ... + rn)K = 0, K ≤e R . ⇒ rK = 0,K ≤e R, where r = r1 + r2 + ... + rn. Hence s Hence r ∈ Z(R). As a consequence we get 2 ∑ a xk + 2 ∑ a jaixk = 0 Zw(R) ⊆ Z(R). j Remark: The converse of the above is not true. j=1 j=i 2 For if r ∈ Z(R) such that r = r1 + r2 + ... + rk, then which implies a j xk = 0, a jaixk = 0. Thus rK = 0,K ≤e R. a jxk ∈ ann(a j), aixk ∈ ann(a j) for each But it does not imply riK = 0 where K ≤e R. j = 1,...,s; k = 1,...,t. Hence xkR ∩ ann(a j) = 0, so that For example: Let R = Z12 LZ12. Here G = Z2 , a = a1 + a2 + ... + as ∈ Zw(R). Thus we have a = 0. R0 = Z12, R1 = Z12 Cor 3.4: If R is graded weak nonsingular ring, then R is Z(R)= {x ∈ R|xK = 0,K ≤e R} graded semiprime. Zw(R)= {x = x1 + x2 + ... + xk ∈ R|xiK = 0 ∀i,K ≤e R} Proposition 3.5: Every graded strongly prime ring R is I = {(2x,y)|x,y ∈ Z12}≤e R. Then Z(R)= {(0,0),(6,0)}. graded weak non-singular. Now 6 ∈ R0, 0 ∈ R1 and 6I = 0. Proof: Let ZW (R) = 0. Thus we have (6,0) ∈/ Z(R) such that (6,0) ∈/ Zw(R). Since R is graded strongly prime, so there exists a ﬁnite Hence Z(R) ⊂ Zw(R). subset S = x,y,z,...,t ⊆ ZW (R) such that ann(S)= 0. This It is known that every nil ring is singular [1]. The implies ann(x) ∩ ann(y) ∩ ... ann(t)= 0. following result establishes the same for weaker We take x = x1 + x2 + ... + xk. conditions. Since x ∈ ZW (R), so we have xiK = 0 for all i, K ≤e R. Proposition 3.2: Every graded weakly nil ring R is graded Thus ann(xi) ≤e R, for all i. This implies ann(x) ≤e R. So weak singular. we have ZW (R) ∩ ann(S) = 0, a contradiction. So Zw(R)= 0. Hence R is graded weak nonsingular. Proof: Let x ∈ R be such that x = x + x + ... + x ,where 1 2 k Proposition 3.6: If I is a graded right ideal of R, then xi ∈ h(R),degxi > 0, for all i. And 0 = y ∈ R such that Zw(I)I ⊆ Zw(R). y = y1 + y2 + ... + ys, 0 = y j ∈ h(R),degy j > 0, for all i. Since R is graded weakly nil, we have a smallest natural Proof: Let z ∈ Zw(I). Then z ∈ I such that z = z1 + z2 + n n−1 ... + z where each z ∈ h(I). number n such that xi y jR = 0, So xi y jR = 0. Thus we t i n−1 And zt k = 0, K ≤e I. have 0 = x y j ⊆ r(xi). i Let i ∈ I such that i = i + i + ... + i . Then z ∈ Z (I)I is Also 0 = xn−1y = y xn−1 ⊆ y R. 1 2 j i w i j j i j such that Thus y jR ∩ r(xi) = 0 which implies r(xi) ≤e R, for all i. Hence x Z R giving thereby R Z R . ∈ w( ) ⊆ w( ) zi = ∑zsik, z = 1,...,t;k = 1,..., j It is obvious that Zw(R) ⊆ R. So Zw(R)= R. s,k . In [1], it is shown that if the singular ideal Z(R) of a It is clear that zi ∈ R such that zsik ∈ h(R). ring = 0, then R is a semiprime ring. It is interesting to Let 0 = H be a graded ideal of R. note that a similar results holds in the graded case. If ikH = 0, then zsikH = 0 implies H ⊆ annR(zsik). Proposition 3.3: If graded weak singular ideal ZW (R)= 0 Therefore 0 = H ⊆ annR(zsik)H ∩ annR(zsik) = 0. then R is a graded semiprime ring. This gives annR(zsik) ≤eR. Proof: We assume ZW (R)= 0. If ikH = 0, then ikH is a non-zero graded right ideal of I. Let a ∈ R such that a = a1 + a2 + ... + as, ai ∈ h(R) with Since z ∈ Zw(I),so we have annR(zs) ∩ ikH = 0, So there 2 deg a j > 0 for all i and a = 0. exists a nonzero homogeneous element h in H such that We need to show a = 0. zsikh = 0. Hence we have ann z i H 0. So ann z i R Let 0 = x ∈ R such that x = x1 +x2 +...+xt with deg x > 0 R( s k) ∩ = R( s k) ≤e giving thereby z Z R . and xk = 0. Then either ax = 0 or ax = 0. i ∈ w( ) Hence the result follows. Thus ∑a jxk = 0 or ∑a jxk = 0 which implies a jxk = 0 or a jxk = 0. Proposition 3.7: Every reduced graded ring R(with an Thus xk ∈ ann(a j) or a jxk = 0. identity) is graded weak nonsingular.

c 2015 NSP Natural Sciences Publishing Cor. Math. Sci. Lett. 4, No. 1, 51-53 (2015) / www.naturalspublishing.com/Journals.asp 53

Proof: Let r ∈ R such that r = r1 + r2 + ... + rk, r j = 0 and deg r1 < deg r2 < ... < deg rk. Let 0 = H be a nonzero homogeneous ideal of R. We consider a homogeneous element x such that x ∈ annR(r j) ∩ rH, j = 1,2,...,k. Then x = rh and 2 r jrh = 0,for each j. So (rhRr j) = 0 gives rhRr j = 0. Now xRx = rhRrh = ∑rhRr jh = 0. Thus x = 0. This implies annR(r j) ∩ rH = 0. Hence Zw(R)= 0. Proposition 3.8: Let I be a graded ideal of R such that I is graded semiprime then

Zw(I)= I ∩ Zw(R) . Proof: Let i ∈ Zw(I).Then i ∈ I such that i = i1 + i2 + ... + it , where each ik ∈ h(I). And ikann(ik)= 0, annI(ik) ≤e I. Let 0 = H be a right graded ideal of I. Then either HI = 0 or HI = 0. Suppose HI = 0. Then (IH)2 = 0. This implies IH = 0. Thus 0 = H ⊆ annR(I) ⊆ annR(i) ⊆ annR(ik). This gives H ∩ annR(ik) = 0. Hence annR(ik) ≤e R. Next we consider HI = 0. Then HI is a nonzero graded right ideal of I. So we have annI(ik)∩HI = 0. This implies annR(ik) ∩ H = 0 which proves that annR(ik) ≤e R. Hence i ∈ Zw(R). Also since i ∈ I, so i ∈ I ∩ Zw(R). Conversely, let us consider a homogeneous element i such that i ∈ I ∩ Zw(R) where i = i1 +i2 +...+it and deg i1 < degi2 < ... < deg it . Let 0 = H be a graded ideal of I. Then we have HI = 0, since I is semiprime. Now HI is a nonzero graded right ideal of R, So we have annR(ik) ∩ HI = 0. Thus annI(ik) ∩ H = 0. This implies annI(ik) ≤e I. Hence i ∈ Zw(I). Thus we have Zw(I)= I ∩ Zw(R).

References

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