On Singular and non Singular Rings

Nazar H. Shuker College of Computer Sciences and Mathematics University of Mosul nazarh_2013@yahoo.com Husam Q. Mohammad College of Computer Sciences and Mathematics University of Mosul husam_alsabawi@yahoo.com Shaimaa H. Ahmad College of Computer Sciences and Mathematics University of Mosul Shaymashatim@uomosul.edu.iq Received on:4/10/2005 Accepted on: 20/11/2005 ABSTRACT In this work, we study singular and non singular rings and we give some new basic properties of such rings and its relation with other rings. Finally, we consider rings for which R/Y(R) is regular Key word: singular ring, singular ideal, regular ring تاقلحلا لوح  ةدرفنملا ريغو ةدرفنملا  ركش نودمح رازن .د.أ  دمحم مساق ماسح .د.م.أ & دمحم مساق ماسح .د.م.أ &  تايضايرلاو تابساحلا مولع ةيلك  لصوملا ةعماج  .ةمظتنملا تاقلحلا , ةدرفنملا تايلاثملا , ةذاشلا تاقلحلا : ةيحاتفملا تاملكلا  صخلملا  يف  اذه  ثحبلا  انسرد  تاقلحلا  ةدرفنملا  تاقلحلاو  ريغ  ةدرفنملا  انيطعأو  ةدايدج ااصاوخ  لذاكو ااهل  انسرد  ضعب  تاقلاعلا  نيب  هذه  تااقلحلا  تااقلحو  راخأ . ارايخأو  اناسرد  تااقلحلا  ااهيق نواكو ياتلا R/Y(R) ةقلح  ةمظتنم .


1.Introduction:
Throughout this paper, R denotes an associative ring with identity and all modules are unitary right R-modules. Recall that: (1) A ring R is called right duo-ring if every right ideal is a two-sided ideal in R [5] . (2)A ring R is called strongly right bounded (briefly SRB), if every non-zero right ideal contains a non-zero two sided ideal of R [1]. (3) A right R-module M is called a general right principally injective (briefly GP-injective) if for any R a   0 , there exists a positive integer n such that 0  n a and any right R-homomorphism of a n R into M extends to one of R into M [5].(4) A ring R is called Von Neumann regular ring if for any aR there exists bR such that a=aba. (5) r(x), l(x) , J(R) will stand respectively for the right annihilator of x, left annihilator of x and the Jacobson radical of R. (6) A ring R is said to be reduced if R contains no non zero nilpotent elements.

Singular Ideal (Basic Properties):
Recall that the right singular ideal of R is Y(R)={xR| r(x) is essential in R } . Ring R such that Y(R)=R is called singular, and ring R such that Y(R)=0 is called non-singular.
We start this section with the following results: Lemma 2.1: Let I be a reduced ideal in R . Then aR r(a)=0 for any element a in I.

Theorem 2.2:
Let R be a ring. Then Y(R) I=0 for any reduced ideal I in R. Proof: Let I be a reduced ideal and Y(R) I 0 . Let 0 a Y(R) I implies that r(a) is essential and a I, since I is a reduced ideal then by Lemma (2.1) we get aRr(a)=0. But r(a)is essential therefore a=0 ,a contradiction .Thus Y(R) I=0. Following [4] a ring R is called zero commutative if for every a,b in R with a.b=0 implies that b.a=0.

Theorem2.3:
Let R be zero-commutative ring. Then Y(R) I=0 if and only if I is a reduced ideal in R . Proof: Suppose that Y(R) I=0, let 0 a I and a 2 =0 implies that ar(a) .
We have to prove that r(a) is an essential right ideal of R, if not, then there exists a non-zero right ideal K in R such that r(a) K=0. Then Ka K l(a),since R zero-commutative then l(a)r(a), so KaKr(a)=0 implies that Ka=0 , so K l(a) r(a) a contradiction .Therefore r(a) must be essential .
Thus aY(R), but a I, so aY(R) I=0. Thus a=0 a contradiction, therefore I must be reduced a ideal in R.

Theorem 2.4:
If R is SRB ring, then NY(R), where N is the set of all nilpotent elements of R. Proof: Let a 2 =0 implies that a r(a) . If r(a) is not essential there exists a non-zero right ideal K such that r(a) K is essential right ideal in R . Since R is SRB ,there exists a non-zero two-sided ideal I of R such that I K. Now, aIaRI r(a) K=0 ,hence I r(a) K=0.This is a contradiction . Thus r(a) must be an essential right ideal of R, implies that aY(R), therefore, NY(R) . Corollary 2.5: Let R be a non-singular SRB ring. Then R is a reduced ring.
Theorem 2.6: Every right duo-nil ring is singular. Proof: Let 0 a R such that a n =0, now we have to prove that r(a) is essential. If not, there exists a non-zero right ideal I such that I r(a)=0 ,since a n =0 implies that a n-1  r(a) and so Ia n-1  I l(a) .But l(a) r(a) and Ir(a)=0 therefore Ia n-1 =0 .Similarly we get Ia=0 so I l(a) r(a) therefore I=0 contradiction so r(a) is essential .Thus R is singular . A ring R is said to be a right strongly prime if every non-zero ideal I of R contains a finite subset F such that r(F)= 0. An ideal P of a ring R is called strongly prime if the ring R/P is strongly prime [3].
Consequently, r(F) is an essential right ideal of R, a contradiction. Therefore Y(R)=0 .
Let R be a non-nilpotent finitely generated ring. For every natural number n, the ring R n is finitely generated as well. Hence applying Zorn's lemma one can find an ideal M of R maximal with respect to R n  M for n=1, 2…….. The ring R/M is strongly prime. Indeed, if I/M is a non-zero ideal of R/M, then the maximality of M implies that there exists a natural number n, such that R n  I. Let F be a finite set generating the ring R n . Then rR/M(F+M)= rR/M(R n +M/M)=J/M for an ideal J of R. If MJ , then the maximality of M implies that R m J for a natural number m. However, in this case R n+m R n JM, a contradiction. Now from Proposition (2.7) it follows that the ring R/M is nonsingular. Taking in particular R a finitely generated non-nilpotent nil ring one gets that R/M is a non singular nil ring . This shows that Theorem (2.6) does not hold for non-commutative nil ring.
Nam [5] proved the following : Lemma 2.8: If aCent(R) with a=ara for some r R, then there exists bCent(R) such that a=aba [where Cent(R) is the Center of R] . Now we give the following main result. Theorem 2.9: Let R be a non-singular ring with every simple singular right R-module is GP-injective. Then Cent(R) is a reduced Von Neumann regular ring. Proof: First we have to prove that Cent(R) is reduced. Let a 0 Cent(R) and a 2 =0 implies that a r(a) .If r(a) is essential then aY(R)=0 implies that a=0 .We are done . If r(a) is not essential , there exists a right ideal I in R such that r(a) I=0 and I 0.Then Ia I r(a) [ aCent(R) ] but Ir(a)=0 implies that Ia=0 and we get I l(a)=r(a) so I=0 contradiction .Therefore a=0 ,so Cent(R) is a reduced ring . Now we shall show that aR+r(a)=R for any a Cent(R) . If not ,there exists a maximal right ideal M of R such that aR+r(a) M, observe that M is an essential right ideal of R. If not ,then M is a direct summamd of R.So we can write M=r(e) for some 0e=e 2  R. Since a M and aCent(R), ae=ea=0. Thus er(a)M=r(e), whence e=0 . It is a contradiction. Therefore M must be an essential right ideal of R . Therefore R/M is GP-injective .So there exists a positive integer n such that any Rhomomorphism of a n R into R/M extends to one of R into R/M.Let f:a n R→R/M be defined by f(a n r)=r+M .Since aCent(R)and Cent(R) is reduced ,then f is a well-defined R-homomorphism . Now R/M is GPinjective so there exists cR such that 1+M=f(a n )=ca n +M .Hence 1-ca n M and so 1 M. which is a contradiction; therefore, aR+r(a)=R for any aCent(R) and we have a=ara for some r R. Applying Lemma (2.8) Cent(R) is Von Neumann regular ring .

Ring for which R/Y(R) Regular
In this section, we consider rings for which R/Y(R) is regular and we characterize such rings in terms of any ideal I and Y(R).
We shall begin this section with the following result.
Theorem 3.1: If R/Y(R) is regular ring, then for any reduced ideal IR is an idempotent. Proof: Let I be a reduced ideal in R and R/Y(R) is regular , and let 0aI, then there exists rR such that a-araY(R) ,but a-araI . So a-araIY(R) applying Theorem (2.2) we get a-ara=0 implies that a=ara. Therefore a I 2  I=I 2 .
Next we give the following results

Theorem3.2:
Let R be a ring. Then R/Y(R) is regular if and only if for any right ideal I in R , I=I 2 +(Y(R)I) . Proof: Let R/Y(R) be a regular ring and a I, then there exists r R such that a-araY(R), but a-ara I so a-araY(R) I . Now take k=a-ara, then a=ara+k ,but ara I 2 , therefore aI 2 +(Y(R)I) implies that I 2 +(Y(R) I) I and clearly I 2 +(Y(R)I) I .Therefore I=I 2 +(Y(R) I) .
Conversely , let aR, take I=aR and aR=(aR) 2 +(Y(R)aR), then there exists rR and xY(R)aR such that a=ara+x implies that a-ara=xY(R), so a+Y(R)=ara+Y(R) .Hence R/Y(R) is regular