Local Rings of Rings of Quotients

Algebr Represent Theor DOI 10.1007/s10468-008-9087-4

M. A. Gómez Lozano · M. Siles Molina

Received: 4 April 2006 / Accepted: 28 April 2007 © Springer Science + Business Media B.V. 2008

Abstract The aim of this paper is to characterize those elements in a semiprime ring R for which taking local rings at elements and rings of quotients are commuting operations. If Q denotes the maximal ring of left quotients of R, then this happens precisely for those elements if R which are von Neumann regular in Q.Anintrinsic characterization of such elements is given. We derive as a consequence that the maximal left quotient ring of a prime ring with a nonzero PI-element is primitive and has nonzero socle. If we change Q to the Martindale symmetric ring of quotients, or to the maximal symmetric ring of quotients of R, we obtain similar results: an element a in R is von Neumann regular if and only if the ring of quotients of the local ring of R at a is isomorphic to the local ring of Q at a.

Keywords Semiprime ring · Local ring at an element · Ring of quotients · PI-element

Mathematics Subject Classiﬁcations (2000) 16E60 · 16E50 · 16R20.

Presented by Susan Montgomery. Partially supported by the Ministerio de Educación y Ciencia and Fondos Feder, jointly, trough projects MTM2004-03845, MTM2007-61978 and MTM2004-06580-C02-02, MTM2007-60333, by the Junta de Andalucía, FQM-264, FQM336 and FQM02467 and by the Plan de Investigación del Principado de Asturias FICYT-IB05-017. M. A. Gómez Lozano (B) · M. Siles Molina Departamento de Álgebra, Geometría y Topología, Universidad de Málaga, 29071 Málaga, Spain e-mail: [email protected] M. Siles Molina e-mail: [email protected] M.A. Gómez Lozano, M. Siles Molina

1 Introduction

Local algebras at elements were introduced by Meyberg in a nonassociative context [14]. These algebras have played a fundamental role in the structure theory of Jordan systems, and they have also proved to be a very useful tool in the setting of associative systems (here, introduced in [7]). Local rings at elements make it possible to relate the structure of pairs and triple systems to that of rings, so that the machinery of rings can be applied to these contexts. In Fernández López et al. [7] local rings at elements were the key tool used to give a Goldie-like theory for associative pairs. In Compañy Cabezos et al. [6] they were a useful device for the study of non-unital exchange rings, which allowed the authors of that paper to show that the exchange property is Morita invariant as well as to extend the exchange property to associative pairs. In [8] the authors presented Fountain-Gould left orders in a different way. Here, the use of local rings at elements made it possible to relate several types of rings of quotients. For instance, Gómez Lozano and Siles Molina [8, Theorem 4.11] related Fountain-Gould left orders and left quotients rings with classical left orders. A con- nection between Fountain-Gould left orders and classical left orders by using local rings at idempotents was established by Ánh and Márki [2]. We want to emphasize that “local” properties in a ring R can be characterized by global properties of some of its local rings. For example, the ﬁnite left local uniform dimension of a ring is equivalent to the ﬁnite left uniform dimension of its local rings. Another example are the semiprime rings with GPI, which are precisely the semiprime rings with at least one PI local ring. This result makes it possible to relate Fountain-Gould left orders with classical ones, and will be used here to relate theorems of Amitsur and Martindale with theorems due to Kaplansky and Posner, respectively. In [8, Proposition 3.2(v)] we showed that the local ring Qx at any nonzero element x ∈ R of a left quotient ring Q of a semiprime ring R is again a left quotient ring of Rx, hence it has sense to compare Qx to the maximal ring of left quotients of Rx. Having in mind this result, the question that must be considered is: for which elements x in R taking local rings at elements and maximal rings of left quotients are commuting operations. A result in this line was given in [4, Corollary 1.9], where it was shown in particular that for a full idempotent e2 = e in a ring R we have l ( ) =∼ l ( ) = Qmax eRe eQmax R e (recall that an idempotent e in a ring R is full if ReR R). The characterization of the maximal left quotient ring of a ring R without total right zero divisors (Proposition 1) and the relation between the dense left ideals of a semiprime ring R and those of its local rings at elements allow us to obtain one of the main results in this paper (Theorem 1):

Theorem 1 Let R be a semiprime ring. For a nonzero element a ∈ R the following conditions are equivalent:

l ( ) (i) a is von Neumann regular in Qmax R . l ( ) =∼ l ( ) (ii) Qmax Ra Qmax R a under an isomorphism which is the identity on Ra.

The natural question that arises at this point is to characterize those elements of a semiprime ring R that become von Neumann regular in the maximal ring of left quotients Q of R. By Johnson’s Theorem, R is left nonsingular if and only if Q is von Local rings of rings of quotients

Neumann regular, hence Theorem 1 for left nonsingular rings implies that taking rings of left quotients and local rings at elements are commuting operations. In particular, a prime ring R containing a PI-element (that is, an element a such that Ra is a PI-ring), is left nonsingular, therefore we may apply the conclusion l ( ) =∼ l ( ) ∈ in the previous paragraph, that is, Qmax Ra Qmax R a for every element a R (Corollary 1). From Theorem 1 we obtain, as a corollary of Posner’s Theorem, a Martindale like l ( ) Theorem, namely, if R is a prime ring with PI-elements, then Qmax R is primitive and has nonzero socle. At the end of this section, Theorem 2 provides with a necessary and sufﬁcient condition for an element a in a ring R to be von Neumannn regular in the maximal ring of left quotients of R: the existence of left ideals L and L (of R) such that L ⊕ lanR(a) and L ⊕ La are dense left ideals of R. With the same techniques we obtain in Sections 3 and 4 results similar to Theorem 1 for the Martindale left quotient ring and the maximal symmetric ring of a semiprime ring R. In these cases, it is not clear which elements in R become von Neumann regular in Q.

2 Local Rings at Elements

Let a be an element in an arbitrary ring R.Thelocal ring of Rata, denoted by Ra, is deﬁned as the ring obtained from the abelian group (aRa, +) by considering the product given by

axa · aya = axaya.

For example, if e is an idempotent in a ring R, then the local ring of R at e, Re, coincides with the corner eRe (the (1,1)-Pierce component of R relative to this idempotent). This notion turns out to be equivalent to the one established by K. Meyberg in [14] in a Jordan context: Given an element a in the ring R,thea-homotope ring a R (the abelian group associated to R with product xa·y = xay) has as an ideal the set Ker(a) ={x ∈ R | axa = 0}. Meyberg defined the local ring of R at a as a a R /Ker(a). It is easy to see that the map Ra → R /Ker(a) defined by axa → x is a ring isomorphism. We will also need an extension of the notion of local ring at an element change by: Let R be a subring of a ring Q.Ifb ∈ Q is such that R is a subring of the homotope b Q ,thenbRb can be regarded as a subring of the local ring Qb of Q at b.Thisring will be called the generalized local ring of Ratb, and will be denoted by Rb .Ifb is actually in R, then the definition of generalized local ring at an element agrees, of course, with that given above.

Lemma 1 Let R be a prime ring and suppose that for some a ∈ R the ring Ra is left nonsingular. Then R is left nonsingular too.

Proof Suppose 0 = x ∈ Zl(R),whereZl(R) denotes the left singular ideal of R.By the primeness of R, aRxRa = 0.Take0 = arxsa, which is in Zl(R) (because the latter is an ideal of R). We have Zl(Rarxsa) = (by [8, Proposition 2.1 (viii)]) Rarxsa = M.A. Gómez Lozano, M. Siles Molina

(Ra)arxsa, which must be left nonsingular because the latter is (by [8, Proposition 2.1 (vii)]), a contradiction.

Notation For a left R-homomorphism f : R N → R M, we will write (x) f ,orsimply xf, to denote the action of f on an arbitrary element x ∈ N. We will denote the action of right R-homomorphisms in a symmetrical way.

3 The Maximal Left Quotient Ring

The notion of left quotient ring for a ring without total right zero divisors was introduced by Utumi in 1956 (see [18]). An overring Q of a ring R is said to be a left quotient ring of R if given p, q ∈ Q,withp = 0, there exists a ∈ R satisfying ap = 0 and aq ∈ R. A right quotient ring is deﬁned analogously. In his paper, Utumi proved that every ring without nonzero total right zero divisors (an element x ∈ R is a total right zero divisor if Rx = 0) has a unique maximal left quotient ring.Thisring, l ( ) denoted by Qmax R , is called the maximal left quotient ring of R. A left ideal L of a ring R is said to be dense if for every x, y ∈ R,withx = 0, there exists a ∈ R such that ax = 0 and ay ∈ L. It is not difﬁcult to show that this is equivalent to saying that R is a left quotient ring of L. Denote by Idl (R) (or simply by Idl ) the family of dense left ideals of R. The maximal left quotient ring of a ring without nonzero total right zero divisors can be characterized as follows.

Proposition 1 ([10], Corollary in p. 99) Let R be a ring without nonzero total right l ( ) zero divisors, and let S be a ring containing R. Then S is isomorphic to Qmax R , under an isomorphism which is the identity on R, if and only if S has the following properties:

(1) For any s ∈ S there exists L ∈ Idl (R) such that Ls ⊆ R. (2) For s ∈ SandL∈ Idl (R),Ls= 0 implies s = 0. (3) For any L ∈ Idl (R) and f ∈ HomR(R L, R R), there exists s ∈ S such that (x) f = xs for all x ∈ L.

Remark 1 The conditions (1) and (2) in Proposition 1 are equivalent to the following one: (4) S is a left quotient ring of R. Indeed, if S is a left quotient ring of R, (1) follows by [8, Proposition 3.3(i)]; moreover, if L is a dense left ideal of R, R is a left quotient ring of L and so S is a left quotient ring of L (see [18, (1.5)]), which implies (2). The converse is immediate: given p, q ∈ S,withp = 0, by (1) there exists L ∈ Idl (R) such that Lq ⊆ R;by(2), Lp = 0. Then there exists an r ∈ R such that rlp = 0 and rlq ∈ L.

Lemma 2 Let R be a semiprime ring and let a be a nonzero element of R. Then for every L ∈ Idl (R) the generalized local ring of L at a, La, is an element of Idl (Ra).

Proof Take axa, aya ∈ Ra,withaxa = 0.SinceL ∈ Idl (R), there exists l ∈ L such that laxa = 0 and lay ∈ L. By the semiprimeness of R, 0 = aslaxa for some s ∈ R. Then asla ∈ La satisﬁes asla · axa = aslaxa = 0 and asla · aya = aslaya ∈ La. Local rings of rings of quotients

Proposition 2 Let R be a semiprime ring and let S be a left quotient ring of R. Take ∈ = = ∈ = l ( ) a nonzero element a R such that aba aandbab bforsomeb Q Qmax R . Then for every aLa ∈ Idl (Ra) we have SaLa ⊕ lanS(b) ∈ Idl (S).

Proof Notice that we can consider R ⊂ S ⊂ Q. ∈ ∩ ( ) = = The sum is direct because sialia SaLa lanS b implies 0 sialiaba sialia.Now,letp and q be in S with p = 0. Suppose ﬁrst pua = 0 for some u ∈ S. By semiprimeness of S ([8, Proposition 3.1 (ii)]) there exists v ∈ S such that avpua = 0.SinceRa is a left quotient ring of La (by Lemma 2) and Qa is a left quotient ring of Ra,see[8, Proposition 3.2 (v)], Qa is a left quotient ring of La. So, there exists ala ∈ aLa satisfying 0 = ala · avpua = alavpua and ala · avqba = alavqba ∈ aLa.Lets ∈ S be such that salavpua = 0,then salavp = 0 and salavq = salavqba + salavq(1 − ba) ∈ SaLa ⊕ lanS(b). Now, suppose pSa = 0.Inthiscasep ∈ lanS(SaS) = annS(SaS), which is an ideal of the semiprime ring S, and therefore, if (p) denotes the ideal of S generated by p, 2 we have 0 = (p) ⊆ annS(SaS)(p).Hencezp = 0 for some z ∈ annS(SaS) ⊆ lanS(b). For this element z we have zq ∈ annS(SaS) ⊆ lanS(b).

Remark 2 There are important classes of semiprime rings which contain elements l ( ) that become von Neumann regular in Qmax R . For example, if R is left nonsingular, then Q is von Neumann regular, so every element in R satisﬁes condition (i) in the following Theorem. Consequently, for R left nonsingular, taking rings of left quotients and local rings at elements are commuting operations.

Theorem 1 Let R be a semiprime ring. For a nonzero element a ∈ R the following conditions are equivalent:

l ( ) (i) a is von Neumann regular in Qmax R . l ( ) =∼ l ( ) (ii) Qmax Ra Qmax R a under an isomorphism which is the identity on Ra.

= l ( ) Proof Denote Q Qmax R . (ii) ⇒ (i). Since the maximal ring of left quotients of a ring is unital, we have that Qa has a unity, say aqa. We want to prove that aqa = a. The semiprimeness of R implies that Qa is semiprime too, because Ra is semiprime ([8, Proposition 3.2 (i)]) and every ring of left quotients of a semiprime ring is semiprime itself. If we suppose aqa − a = 0, there exists p ∈ Q such that 0 = (aqa − a)p(aqa − a) = aqapaqa − aqapa − apaqa + apa = (aqa is the unity of Qa)= apa − apa − apa + apa = 0, a contradiction. (i) ⇒ (ii). Let b be in Q such that a = aba and b = bab .By[8, Proposition 3.2 (v)] Qa is a left quotient ring of Ra. So, by Remark 1, conditions (1) and (2) of Proposition 1 are satisﬁed. Now, let us show (3) in Proposi- tion 1 and the proof will be complete. Take aLa ∈ Idl (Ra) and f ∈ ( , ) HomRa Ra aLa Ra Ra and deﬁne

f : RaLa ⊕ lanR(b) −→ R rialia + t → ri(alia) f M.A. Gómez Lozano, M. Siles Molina = + ∈ ⊕ ( ). The map f is well deﬁned: suppose 0 rialia t RaLa lanR b Then ri(alia) f must be zero. Otherwise, by semiprimeness of R there exists u ∈ R such that 0 = au ri(alia) f = auri(alia) f = · ( ) = ( ∈ ( , )) auria alia f f HomRa Ra aLa Ra Ra au rialia f = rialia = 0 (0) f = 0,

a contradiction. Moreover, f is a homomorphism of left R-modules: let rala + t and s be in RaLa ⊕ lanR(b) and R, respectively. Then (s(rala + t)) f = (srala + st) f = sr(ala) f = s(r(ala) f ) = s((rala + t) f). Since f ∈ HomR(R(RaLa ⊕ lanR(b)), R R),withRaLa ⊕ lanR(b) ∈ I ( ) dl R , by Proposition 2, we can apply condition (3) in Proposi- ∈ + = + tion 1 to ﬁnd q Q such that rialia t f rialia t q for every rialia + t ∈ RaLa ⊕ lanR(b). Now we will prove q = qba: For every rala + t ∈ RaLa ⊕ lanR(b), (rala + t)qba = (rala + t) fba = r(ala) fba = ((ala) f ∈ aRa)r(ala) f = (rala + t) f = (rala + t)q. This implies (RaLa ⊕ lanR(b)) (qba − q) = 0,andbearinginmindRaLa ⊕ lanR(b) ∈ Idl (R) and Proposition 1 (2) we obtain qba − q = 0. Finally, we will see that for every ala ∈ aLa,(ala) f = alaq(= ala · aqba).Takearala ∈ aRaLa.Thenarala · aqba = aralaq = a(rala) f = ( ) = · ( ) = ( ∈ ( , )) ( · ) ar ala f ara ala f f HomRa Ra aLa Ra Ra ara ala f and since aRaLa ∈ Idl (Ra) we have ala · aqba = (ala) f for every ala ∈ aLa.

We will say that an element a in a semiprime ring R is a PI-element (notion introduced by Montaner in [15]) if Ra is a PI-ring, i.e., a ring which satisfies a polynomial identity. For instance, if R is a unital prime ring and satisfies a generalized polynomial identity (GPI), then by Martindale’s Theorem (see for example [5, Theorem 6.1.6]) there is a nonzero idempotent e ∈ R such that eRe satisfies a polynomial identity. This means that e is a PI-element. Another example of a PI- element is the following: suppose R to be a prime ring satisfying a GPI; by [3, Theorem] there exists a square cancellable element 0 = b ∈ R (an element b in a ring R is said to be square cancellable if b 2x = b 2 y implies bx = byfor x, y ∈ R ∪{1} and xb 2 = yb 2 implies xb = yb for x, y ∈ R ∪{1}) such that bRb is a PI ring. Now, 2 2 2 if we put a := b then the map ϕ : bRb → Ra given by ϕ(bxb) = b xb is a ring isomorphism. So, Ra satisfies a polynomial identity, i.e., a is a PI-element.

l ( ) =∼ Corollary 1 Let R be a prime ring with a nonzero PI-element. Then Qmax Ra l ( ) ∈ Qmax R a for every a R.

Proof Left nonsingularity of Ra follows because Ra a PI-ring implies ([13, Theorem 1]) that it is left nonsingular. By Lemma 1 the whole ring R is left nonsingular; now, apply Theorem 1. Local rings of rings of quotients

:= l ( ) Proposition 3 If R is a prime ring with a nonzero PI-element, then Q Qmax R is a primitive ring with nonzero socle.

Proof The primeness of Q follows since it is a left quotient ring of the prime ring R ([8, (3.1)]). Suppose that a is a PI-element. Then Ra, which is also a prime ring by [8, Proposition 2.1 (ii)], is a classical left (and right) order in Mn() for some division ring and some natural number n,where is ﬁnite dimensional over its center (see [16, Theorem]). So, by [8, Theorem 3.2 (iii)] Ra is left nonsingular and by Lemma 1 R is left nonsingular too. Now, by Johnson’s Theorem [9, (13.36)], Q is a von Neumann regular ring, in particular a is von Neumann regular in Q. So, M () = l ( ) = l ( ) = ( l ( )) n Qcl Ra (by [9, (13.41)]) Qmax Ra (by Theorem 1) Qmax R a.This implies (by [8, Proposition 3.1 (v)]) a ∈ Soc(Q).

Remark 3 For R a unital ring, the previous result can be obtained as a consequence of [5, Corollary 6.17] as follows: If R is prime and has a nonzero PI-element, then R satisﬁes a GPI. Use [5, Corollary 6.17] to obtain that Q is primitive and has nonzero socle.

Theorem 2 Let R be a ring without total right zero divisors. An element a ∈ Risvon l ( ) Neumann regular in Qmax R if and only if there exist L and L , left ideals of R, such that L ⊕ lanR(a) and L ⊕ La are dense left ideals of R.

l ( ) Proof Denote Qmax R by Q. Recall that a standard element of Q can be repre- sented as a class [ f, I], where I is a dense left ideal of R and f ∈ Hom(R I,R R). Suppose first that there exist L and L , left ideals of R such that L ⊕ lanR(a), L ⊕ La ∈ Idl (R). Identify a with the element [ f, L ⊕ lanR(a)]∈Q,where f (l + z) = la, for every l ∈ L, z ∈ lanR(a). Define g : L ⊕ La → L by g(l + la) = l, for every l ∈ L and l ∈ L.Themapg is well-defined because L ∩ La = 0 and L ∩ lanR(a) = 0, and it is an R-module homomorphism. Moreover, for b ∈ L, c ∈ lanR(a),we have ( fgf)(b + c) = ( fg)(ba) = f (b) = ba = f (b + c). This shows that x is von Neumann regular in Q. Conversely, assume that a is von Neumann regular in Q,andletb in Q be such that a = aba and b = bab . It is known that I ={x ∈ R | xb ∈ R} is a dense left ideal of R. Define L := Ib and L := I(1 − ba). We claim that L ⊕ lanR(a) and L ⊕ La are dense left ideals of R. Indeed, it is clear that L ∩ lanR(a) = 0 and La ∩ L = 0. Moreover, La ⊕ L = Iba ⊕ I(1 − ba) = I, which is dense. Now, take r, s ∈ R,with r = 0.SinceI is dense, there exists y ∈ I such that yr = 0 and ysa, ys(1 − ab), ys ∈ I. Then ys = ysab + ys(1 − ab),withysab ∈ L and ys(1 − ab) ∈ lanR(a). This shows that L ⊕ lanR(a) ∈ Idl (R).

4 The Martindale Symmetric Ring of Quotients

The kind of rings of quotients we will deal with in this section was introduced in 1969 by W. S. Martindale [12] for prime rings, and by S. A. Amitsur [1] in 1972 for semiprime rings. Martindale’s original theory was designed for applications to rings satisfying a polynomial identity. These rings of quotients associated to semiprime M.A. Gómez Lozano, M. Siles Molina rings have been proved to be useful not only for the theory of rings with polynomial identities, but also for the Galois theory of noncommutative rings. Denote by Ie(R),orsimplybyIe, the set of all essential ideals in a semiprime ring R.TheMartindale symmetric ring of quotients of R is deﬁned as follows: ( ) := ∈ l ( ) | + ⊆ ∈ I . Qs R q Qmax R Iq qI R for some I e

( ) l ( ) Remark 4 For every semiprime ring R, Qs R , which is a subring of Qmax R ,isa left and a right quotient ring of R.

Given I ∈ Ie and J ∈ Ie, we will say that a left R-homomorphism f : I → R and arightR-homomorphism g : J → R are compatible if for every x ∈ I and y ∈ J, (x) fy = xg(y). In [9, (14.25)] there is an abstract characterization of the Martindale symmetric ring of quotients of a semiprime ring R similar to that of Proposition 1, but in terms of compatible R-homomorphisms.

Lemma 3 Let R be a semiprime ring, and let a be a nonzero element in R. Then, for every aIa ∈ Ie(Ra), we have:

(i) RaIaR ⊕ ann(RaIaR) ∈ Ie(R). (ii) ann(RaIaR) ⊆ lanR(a) ∩ ranR(a).

Proof

(i) In general, for any ideal J, J ⊕ annR(J) is an essential ideal of R. (ii) Take z ∈ ann(RaIaR) and suppose z ∈/ lan(a).Thenza = 0. By semiprimeness of R, there exist t, t ∈ R satisfying 0 = atzat a. Apply twice aIa ∈ Ie(Ra) to ﬁnd aya, aya ∈ aIa such that 0 = atzata · aya · aya = atzatayaya ∈ atzRaIaR=0, a contradiction. Analogously we can prove that z ∈ ran(a).

Theorem 3 Let R be a semiprime ring. For a nonzero element a ∈ R the following conditions are equivalent: (i) a is von Neumann regular in Q (R). ∼ s (ii) Qs(Ra) = Qs(R)a under an isomorphism which is the identity on Ra.

Proof (ii) ⇒ (i) can be proved as in Theorem 1. (i) ⇒ (ii) Since R ⊆ Qs(R), Ra ⊆ (Qs(R))a. Now, we will prove the three conditions in [9, (14.25)].

(1) Take aqa ∈ (Qs(R))a,withq ∈ Qs(R). Choose I and J in Ie(R) satisfying qaI, Jaq ⊂ R.ThenaqaIa, aJaqa ⊆ aRa,withaIa, aJa ∈ Ie(R), by Lemma 2. (2) Let aqa be in (Qs(R))a,withq ∈ Qs(R), and suppose aIa, aJa ∈ Ie(Ra) such that aqaIa = 0 or aJaqa = 0. By Remark 4, Qs(R) is a left quotient ring of R,soby[8, Proposition 3.2 (v)] (Qs(R))a is a left quotient ring of Ra.SinceaJa is a dense left ideal of Ra,by Local rings of rings of quotients

Remark 1 aqa = 0. Analogously we can prove that aqa = 0 if and only if aqaIa = 0. I ( ) ∈ ( , ) (3) Let aIa and aJa be in e Ra ,andlet f HomRa Ra aIa Ra Ra and ∈ ( , ) g HomRa aJaRa Ra Ra be compatible. Deﬁne

f : RaIaR ⊕ ann(RaIaR) −→ R riaziasi + u → ri(aziasia) f We see that the map f is well deﬁned: Indeed, ri(aziasia) f = 0 implies, by semiprimeness of R, that for some t ∈ R, 0 = ( ) = ( ) = · ( ) = ( ∈ at ri aziasia f atri aziasia f atria aziasia f f ( , ) ( · ) = ( ) HomRaRa aIa Ra Ra ) atria aziasia f atriaziasia f ,and hence riaziasi = 0. Moreover, f is easily seen to be a homomorphism of left R-modules. Analogously we have that the map

g : RaJaR ⊕ ann(RaJaR) −→ R riaziasi + v → g(ariazia)si

is well deﬁned and it is a homomorphism of right R-modules. Besides, f and g are compatible: Choose b ∈ Qs(R) satisfying a = aba and b = bab . For every rayas + u ∈ RaIaR ⊕ ann(RaIaR) and every xataz+v ∈ RaJaR ⊕ ann(RaJaR)(rayas+u) f(xataz+v) = r(ayasa) f (xataz + v) = r(ayasa) fba(xataz + v) = (by Lemma 3 (ii)) r(ayasa) fbaxataz = r[(ayasa) f · axata]z = ( f and g are compatible) r[ayasa · g(axata)]y = rayasabg(axata)z = (by Lemma 3(ii)) (rayasab + u)g(axata)z = (rayasab + u)g(axataz + v).Thecom- patibility of f and g, and the fact that RaIaR ⊕ ann(RaIaR) and RaJaR ⊕ ann(RaJaR) belong to Ie(R), by Lemma 3 (i), imply that there exists q ∈ Qs(R) such that (rayas + u) f = (rayas + u)q and g(xataz + v) = q(xataz + v). We claim that q = abqba: (rayas+ u)(qba − q) = (rayasa) fba− (rayasa) f = 0 for every rayas + u ∈ RaIaR ⊕ ann(RaIaR), which is a dense left ideal of R.Thisim- plies qba − q = 0, and analogously we obtain abq = q. Finally, it is easy to see that for every arayasa ∈ aRaIaRa and every atazaua ∈ aRaJaRa, (arayasa) f = arayasa · q and g(atazaua) = q · atazaua, and hence (aya) f = aya · q and g(aza) = q · aza for every aya ∈ aIa and aza ∈ aJa, because aRaIaRa and aRaJaRa are dense left and right ideals of aIa and aJa, respectively, and two left (right) Ra- homomorphisms which coincide on a dense left (right) ideal of aIa (aJa) coincide on aIa (aJa).

5 The Maximal Symmetric Ring of Quotients

Let R be a semiprime ring. The maximal symmetric ring of quotients of R, denoted by Qσ (R), is a symmetric version of the maximal ring of quotients of R.The construction of this ring of quotients was ﬁrst given by Schelter in [17] and has been considered recently by Lanning in [11]. M.A. Gómez Lozano, M. Siles Molina

Denote by Idr(R) (or simply Idr) the family of dense right ideals of R. For a semiprime ring R, deﬁne ( ) = ∈ l ( ) | ⊆ ∈ I , Qσ R q Qmax R qI R for some I dr l ( ) which is a subring of Qmax R , called the maximal symmetric ring of quotients of R. In a semiprime ring R, given L ∈ Idl and K ∈ Idr, we will say that a left R- homomorphism f : L → R and a right R-homomorphism g : K → R are compatible if, for every x ∈ L and y ∈ K, (x) fy = xg(y). l ( ) ( ) As in the cases of Qmax R and Qs R , the maximal symmetric ring of quotients has an abstract characterization (see [11, Proposition 2.4]). Its proof makes use of the compatibility of homomorphisms.

Remark 5 Notice that Qσ (R) is a left and a right quotient ring of R. The ﬁrst assertion follows from the deﬁnition and the second one is a consequence of [11, Proposition 2.1] and Remark 1.

Theorem 4 Let R be a semiprime ring. For a nonzero element a ∈ R the following conditions are equivalent:

(i) a is von Neumann regular in Qσ (R). ∼ (ii) Qσ (Ra) = Qσ (R)a under an isomorphism which is the identity on Ra.

Proof (ii) ⇒ (i) can be proved as in Theorem 1. (i) ⇒ (ii) Denote Q := Qσ (R),andletb be in Q satisfying aba=a and bab =b First of all we observe that Ra ⊆ Qa. In what follows, we will prove that the conditions (ii), (iii) and (iv) in [11, Proposition 2.1] hold, which will mean that our claim is satisﬁed. (ii) and (iii). By [8, Proposition 3.2 (v)], Qa is a right and a left quotient ring of Ra. So, by Remark 1 we have conditions (ii) and (iii) of [11, Proposition 2.1]. ∈ I ( ) ∈ I ( ) ∈ ( , ) (iv). Take aLa dl Ra , aKa dr Ra , f HomRa Ra aLa Ra Ra ∈ ( , ) and g HomRa aKaRa Ra Ra such that f and g are compatible. We will prove that there exists aqa ∈ Qa such that (axa) f = axa · aqa and g(aya) = aqa · aya for every axa ∈ aLa and aya ∈ aKa. Consider

f : QaLa ⊕ lanQ(b) −→ Q pialia + u → pi(alia) fb

We notice that by condition (i) in Proposition 2 QaLa ⊕ lanQ(b) is a dense left ideal of Q.Themap fis well deﬁned: Suppose 0 = pialia + u ∈ QaLa ⊕ lanQ(b).Then pi(alia) fb must be zero. Otherwise, since Q is a left quotient ring of R there is an r ∈ R ∈ ( ) = such that rpi R, for every i,andr pi alia f 0. By the semiprime- ∈ = ( ) = · ness of R,forsomes R we have 0 asr pi aliaf asrpia (alia) f = ( f is a left Ra-homomorphism) as(rpi)a · alia f = (0) f = 0, a contradiction. Moreover, it is not difﬁcult to see that f is a homomorphism of left Q-modules. Local rings of rings of quotients

Analogously, for

g : aKaQ ⊕ ranQ(b) −→ Q akiapi + v → bg(akia)pi

we have that aKaQ ⊕ ranQ(b) ∈ Idr(Q), by Proposition 2 (ii); g is well deﬁned and it is a homomorphism of right Q-modules. Moreover, f and g are compatible: Take pala + u ∈ QaLa ⊕ lanQ(b) and akat+ v ∈ aKaQ ⊕ ranQ(b).Then(pala + u) f(akat + v) =[p(ala) fb](akat + v) = (v ∈ ranQ(b))p(ala) fbakat = p[(ala) f · aka]t = ( f and g are compatible)p[ala · g(aka)]t = p[alabg(aka)]t = (u ∈ lanQ(b))(pala + u) bg(aka)t = (pala + u) g(akat + v). By condition (iii) of [11, Proposition 2.1] applied to Q, which coincides with Qσ (Q) (by [11, Theorem 2.5]), there exists q ∈ Q such that for every pala + u ∈ QaLa ⊕ lanQ(b),and every akat + v ∈ aKaQ ⊕ ranQ(b), (1) (pala + u) f = (pala + u)q = (pala + u)qab,and (2) g(akat + v) = q(akat + v) = baq(akat + v). Finally, take ala ∈ aLa and aka ∈ aKa. We have ala · aqa = alaqa = (by (1)) (ala) fa= (ala) fba= (ala) f ,andaqa · aka = aqaka = (by (2)) ag(aka) = abg(aka) = g(aka). This concludes our proof.

Acknowledgement The authors would like to thank the referee for his suggestions.

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