Volume 2007, Article ID 84840,6pages doi:10.1155/2007/84840
Research Article
Some External Characterizations of SV-Rings and
Hereditary Rings
A. Haily and H. Rahnaoui
Received 7 November 2006; Accepted 16 April 2007
Recommended by Francois Goichot
We introduce a new class of rings which we callAI-rings. By applying this property to endomorphism rings, we give characterizations of semi-ArtinianV-rings and hereditary rings.
Copyright © 2007 A. Haily and H. Rahnaoui. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, dis-tribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
In ring theory, the notion of annihilator is an important tool for studying the structures. Many characterizations and structure theorems can be derived by using this notion. On the other hand, certain classes of rings (e.g., Baer rings and Rickart rings) are defined by considering annihilators ideals. In the present work, we introduce a class of rings which is close to the class of Rickart rings. We then investigate endomorphism rings having this property. This will enable us to obtain characterizations of certain classes of rings, namely the SV-rings and the hereditary rings.
We have divided this work into three sections. In the first we give some preliminary results and examples. In the second, we characterize SV-rings. The third section is devoted to hereditary rings.
2. Definitions and preliminary results
All rings are assumed to be associative and unitary. All modules are left unital modules. For definitions and background, we refer to [1] or [2].
Definitions 2.1. (i) LetRbe a ring. The right annihilator ofS⊂Ris the set Annr(S)=
(ii) A ring is called a leftAI-ring if every nonzero right annihilator contains a nonzero idempotent.
Examples 2.2. (i) A ringRis called left Rickart (or left PP-ring), if every left annihilator of an elements∈Ris generated by an idempotent. IfRis left Rickart, thenRis leftAI. For ifI=Annr(S) is a nonzero right annihilator, one can pick a nonzero elementainI
and consider Annl(a). SinceRis left Rickart, there exists an idempotente=1 inRsuch that Annl(a)=Re. NowS⊂Re, implying thatS(1−e)=0. That is, 0=1−e∈I.
(ii) Any prime ringRwith nonzero socle is left and rightAIsince Soc(R) is (left and right) essential, so ifI=Annr(S), thenI∩Soc(R)=0. ThusI contains a minimal right ideal which is generated by an idempotent. In particular, ifEis a normed vector space, then the algebra of bounded linear operator BL(E) is left and rightAI.
Proposition 2.3. IfRis a leftAI-ring, thenRis left nonsingular.
Proof. Suppose that the left singular ideal of R, Sing(R), is nonzero, and let 0=x∈ Sing(R). Then there exists an essential left ideal I of R such thatIx=0. But Ris left AI, thus there exists a nonzero idempotent e∈R such that Ie=0. This implies that
I⊂R(1−e), a contradiction.
Example 2.4. The converse of the preceding proposition is not true as we see by the fol-lowing construction. LetKbe any field, consider the factor ringR=K[x,y]/(xy), where K[x,y] is the polynomial ring of two commuting indeterminatesx, y, and (xy) is the ideal ofK[x,y] generated byxy.Ris nonsingular but Annx=(y) contains no nonzero idempotent.
For rings with some finiteness conditions on idempotents, the condition of being an AI-ring is equivalent to that of being Baer ring.
Theorem 2.5. LetRbe a ring having no infinite set of pairwise orthogonal idempotents. The following assertions are equivalent:
(i)Ris leftAI; (ii)Ris rightAI; (iii)Ris Baer;
(iv)Ris left and right Rickart.
For the proof, see [3, Theorem 7.55].
3. Characterization of SV-ring by theAIcondition
In this section, we will consider modules withAIendomorphism ring. This will lead us to an external characterization of the class of semi-ArtinianV-rings.
by
ux1,x2,...,xn,...=0,x1,x2,...,xn,...,
vx1,x2,...,xn,...
=x2,x3,...,xn,...
. (3.1)
(i) Leth:P→Qand f :P→Qbe defined by f(x)=(h(x), 0,..., 0,...).
Consider the endomorphismφofQ⊕P⊕Ldefined by
φ= ⎛ ⎜ ⎝
u f 0
0 0 0
0 0 IL
⎞ ⎟
⎠. (3.2)
Then, if there exists a nonzero projectionθ ofQ⊕P⊕Lsuch thatφθ=0, then there exists a nonzero projectionpofPsuch thathp=0.
(ii) Leth:P→Qand f:Q→Pbe defined by f(x1,x2,...,xn,...)=h(x1).
Consider the endomorphismφofQ⊕P⊕Ldefined by
φ=
⎛ ⎜ ⎝
v 0 0 f 0 0
0 0 IL
⎞ ⎟
⎠. (3.3)
Then, if there exists a nonzero projectionθofQ⊕P⊕Lsuch thatθφ=0, then there exists a nonzero projectionpofPsuch thatph=0.
Proof. We will only prove (i), since (ii) is dual of (i). Let
θ= ⎛ ⎜ ⎝
a1 a2 a3
b1 b2 b3
c1 c2 c3 ⎞ ⎟
⎠ (3.4)
be a nonzero projection ofQ⊕P⊕Lsuch that φθ=0. Thenci=0 for i=1, 2, 3. On
the other hand,uai+ f bi=0 fori=1, 2, 3. Composing on the left byv, using the fact that vu=IQ andv f =0, we obtainai=0 fori=1, 2, 3. The equality θ2=θ implies b2
2=b2. Nowb2=0, since otherwiseθ2=θ=0, a contradiction. So p=b2is a nonzero
idempotent of EndR(P) such that f p=0 and thenhp=0.
Recall that a ringRis said to be a (left)V-ring if every simpleR-module is injective. Or equivalently, the Jacobson radical of everyR-module is zero. A ring is said to be (left) semi-Artinian if every nonzero module has a nonzero socle.
A ring which is together aV-ring and semi-Artinian is called a semi-artinianV-ring, (SV-ring). Many authors have studied the class of SV-rings, see for example [4,5], or
[6,7], for SV-modules.
Theorem 3.2. LetRbe a ring. Then the following propositions are equivalent. (i) For everyR-moduleM, the endomorphism ring EndR(M) is leftAI.
(iii) Every nonzeroR-module contains a nonzero injective submodule. (iv)Ris a semi-artinianV-ring.
Proof. The equivalence (iii)⇔(iv) is by [6, Theorem 13].
(ii)⇒(iii). Take the injective hull ofE(M) ofM. IfNis a submodule ofM, then there exists a projectionpofE(M) such thatp(E(M))⊂N. Butp(E(M)) is a direct summand ofE(M), and it is therefore injective.
(iv)⇒(i). LetRbe semi-ArtinianV-ring,Ma nonzeroR-module, andS⊂EndR(M) such that Annr(S)=0. So, there exists a nonzerou∈Ann(S) such thatSu=0, that is,
u(M)⊂s∈SKers. SinceRis semi-Artinian, there exists a simple submoduleN⊂u(M).
On the other hand,Ris aV-ring, thusNis injective and hence a summand ofM. Now take an idempotent endomorphismpsuch thatp(M)=N, thenSp=0.
(i)⇒(ii). LetMbe anR-module andNa proper submodule ofM. InLemma 3.1, take L=N,P=M,Q=(M/N), andh=π, whereπis the canonical surjectionM→Q. Let
ψ= ⎛ ⎜ ⎜ ⎝
0 0 0
0 0 ı
0 0 0
⎞ ⎟ ⎟
⎠, (3.5)
whereı:N→M, is the canonical injection. Thenψ=0 andφψ=0. Since, EndR(Q⊕
P⊕L) is leftAI-ring, there exists a nonzero idempotentθ∈EndR(Q⊕P⊕L), such that
φθ=0. We conclude fromLemma 3.1that there exists a nonzero projectionpofMsuch
thatπ p=0. That is,p(M)⊂N.
Remark 3.3. Dually, one can prove that for a ringR, EndR(M) is rightAI for everyR
-moduleM, if and only ifRis semisimple.
In [8], Brodski and Grigorjan have given a characterization of left SV-ring which is analogous toTheorem 3.2, by using injective modules. This result can be deduced from our lemma as we will see presently.
Theorem 3.4 [8, Theorem 2]. LetRbe a ring, then the following propositions are equiva-lent.
(i) For every injective leftR-moduleMand every nonleft invertibleu∈EndR(M), there exists a nonzero idempotentpsuch thatup=0.
(ii)Ris an SV-ring.
Proof. (ii)⇒(i) is clear. For the converse, take inLemma 3.1 that L=P=M an injec-tive module,N a submodule of M,Q=E(M/N) the injective hull ofM/N,Q the di-rect product of a denumerable family of copies ofQ. Takeh:M→E(M/N) defined by h(x)=π(x). The endomorphismφinLemma 3.1is not injective. SinceQ⊕P⊕Lis an injective module, thenφis not left invertible. By hypothesis, there exists a nonzero idem-potentθ∈EndR(Q⊕P⊕L) such thatφθ=0. We conclude fromLemma 3.1that there
4. Characterization of hereditary rings
Recall that a ringRis said to be left hereditary if every submodule of a projective module is projective. Or equivalently, if every factor module of an injective module is injective.
TheAIproperty can be used to give a characterization for left hereditary ring.
Theorem 4.1. LetRbe a ring. Then the following statements are equivalent. (i)Ris left hereditary.
(ii) For every injectiveR-moduleM, for everyu∈EndR(M), Imuis a direct summand
ofM.
(iii) For every injectiveR-moduleM, EndR(M) is right Rickart.
(iv) For every injectiveR-moduleM, EndR(M) is rightAI.
(v) For every injectiveR-moduleM, and every nonright regular endomorphismuofM, there exists a nonzero idempotent endomorphismpsuch thatpu=0.
Proof. (i)⇒(ii) is clear since Imu∼=M/Keru. For (ii)⇒(iii), see [7, Theorem 39.15]. (iii)⇒
(iv) and (iv)⇒(v) are also clear. So it remains to show that (v)⇒(i). Let M be an in-jectiveR-module and N a submodule of M. We have to show that the factor module M/N is injective. Let E(M/N) be its injective hull. Consider the canonical surjection π:M→M/Nand the canonical injectionı:M/N→E(M/N). Puth=ıπand let us show thathis surjective. Otherwise, there exist an injectiveR-moduleSand a homomorphism g:E(M/N)→Ssuch thatgh=0. Now put inLemma 3.1thatL=S,P=E(M/N),Q=M, Qis the direct product of a denumerable family of copies ofM, and consider the
endo-morphismφofQ⊕P⊕Lof the lemma. Take
ψ=
⎛ ⎜ ⎜ ⎝
0 0 0
0 0 0
0 g 0
⎞ ⎟ ⎟
⎠. (4.1)
Thenψ=0 andψφ=0. SinceQ⊕P⊕Lis injective, there exists an idempotent en-domorphismθofQ⊕P⊕Lsuch thatθφ=0.Lemma 3.1implies the existence of a nonzero idempotent endomorphismpofE(M/N) such thatph=0, that is,p(M/N)=
0. This contradicts the fact thatM/N is essential inE(M/N). This contradiction shows thathis surjective. Thusıis surjective, and hence is bijective. This means thatM/N is
isomorphic toE(M/N) and is injective.
The dualization of the preceding theorem requires the existence of the dual notion of injective hull, namely the projective cover. So, we suppose that the ringRis left perfect.
Theorem 4.2. LetRbe a left perfect ring. Then the following statements are equivalent. (i)Ris left hereditary.
(ii) For every projective R-moduleM, and for every u∈EndR(M), Keru is a direct
summand ofM.
(iii) For every projectiveR-moduleM, EndR(M) is left Rickart.
(iv) For every projectiveR-moduleM, EndR(M) is leftAI.
(v) For every projectiveR-moduleM, and for every nonleft regularu∈EndR(M), there
Note that the equivalences of (i), (ii), and (iii) are true without the left perfectness assumption (see [7]).
Acknowledgment
The first author is partially supported by the Project MEC, Proyecto I+D+I MTM2004-03882, AECI, PCI 62/04/R/E (Spain).
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A. Haily: D´epartement de Math´ematiques, Facult´e des Sciences, BP 20 El Jadida, Morocco Email address:afhaily@yahoo.fr
