Local Rings of Rings of Quotients

Local rings of rings of quotients

M. A. Gomez´ Lozano; M. Siles Molina Departamento de Algebra,´ Geometr´ıay Topolog´ıa, Universidad de Málaga, 29071 Málaga, España e-mail addresses: [email protected]; [email protected]

Abstract: The aim of this paper is to characterize those elements in a semiprime ring R for which taking local algebras 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. An intrinsic 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.

Introduction Local algebras at elements were introduced by K. Meyberg in [15]. 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. 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 [8] local rings at elements were the key tool used to give a Goldie-like theory for associative pairs. In [7] 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 [9] 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, [9, Theorem 4.11] related Fountain-Gould left orders and left quotient rings with classical left orders. A connection between Fountain-Gould left orders and classical left orders by using local rings at idempotents was established by P. N. Anh´ and L. Márkiin [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

Partially supported by the Ministerio de Educaci´ony Ciencia and Fondos Feder, jointly, trough projects MTM2004-03845 and MTM2004-06580-C02-02, and by the Junta de Andaluc´ıa,FQM-264 and FQM336.

1 2 Gomez´ and Siles

ones, and will be used here to relate theorems of Amitsur and Martindale with theorems due to Kaplansky and Posner, respectively.

In [9, 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 makes 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 [5, Corollary 1.9], where it was shown in particular that for a full 2 l ∼ l idempotent e = e in a ring R we have 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 2.1) 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 2.6): 2.6. Theorem 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 Neumann regular, hence Theorem 2.5 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 in the previous l ∼ l paragraph, that is, Qmax(Ra) = Qmax(R)a for every element a ∈ R (Corollary 2.8). From Theorem 2.6 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 2.10 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 0 0 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 Seccions 3 and 4 results similar to Theorem 2.6 for the Martindale left quotient ring and the maximal symmetric ring of a semiprime ring R. In theses cases, it is not clear which elements in R become von Neumann regular in Q.

1. Local rings at elements.

Let a be an element in an arbitrary ring R. The local ring of R at a, denoted by Ra, Local rings of rings of quotients 3

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 [15] in a Jordan context: Given an element a in the ring R, the a-homotope ring Ra (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 Ra/Ker(a). It is easy to see that the map a Ra → R /Ker(a) defined by axa 7→ x is a ring isomorphism. We will also need an extension of the notion of local ring at an element which we define as follows: Let R be a subring of a ring Q. If b ∈ Q is such that R is a subring of the b homotope Q , then bRb can be regarded as a subring of the local ring Qb of Q at b. This ring will be called the generalized local ring of R at b, and will be denoted by Rb. If b is actually in R, then the definition of generalized local ring at an element agrees, of course, with that given above.

1.1. Lemma. 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 6= x ∈ Zl(R), where Zl(R) denotes the left singular ideal of R. By the primeness of R, aRxRa 6= 0. Take 0 6= arxsa, which is in Zl(R) (because the latter is an ideal of R). We have Zl(Rarxsa) = (by [9, Proposition 2.1 (viii)]) Rarxsa = (Ra)arxsa, which must be left nonsingular because the latter is (by [9, Proposition 2.1 (vii)]), a contradiction.

Notation. For a left R-homomorphism f : RN → RM, we will write (x)f, or simply 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.

2. 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, with p 6= 0, there exists a ∈ R satisfying ap 6= 0 and aq ∈ R. A right quotient ring is deﬁned analogously. In his paper, Utumi proved that every ring without total right zero divisors (a nonzero element x ∈ R is a total right zero divisor if Rx = 0) has a l unique maximal left quotient ring. This ring, 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, with x 6= 0, there exists a ∈ R such that ax 6= 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. 4 Gomez´ and Siles

The maximal left quotient ring of a ring without total right zero divisors can be characterized as follows.

2.1. Proposition [11, Corollary in p. 99]. Let R be a ring without 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 ∈ S and L ∈ Idl(R), Ls = 0 implies s = 0.

(3) For any L ∈ Idl(R) and f ∈ HomR(RL, RR), there exists s ∈ S such that (x)f = xs for all x ∈ L.

2.2. Remark. The conditions (1) and (2) in (2.1) 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 [9, 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, with p 6= 0, by (1) there exists L ∈ Idl(R) such that Lq ⊆ R; by (2), Lp 6= 0. Then there exists an r ∈ R such that rlp 6= 0 and rlq ∈ L.

2.3. Lemma. 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, with axa 6= 0. Since L ∈ Idl(R), there exists l ∈ L such that laxa 6= 0 and lay ∈ L. By the semiprimeness of R, 0 6= aslaxa for some s ∈ R. Then asla ∈ La satisﬁes asla · axa = aslaxa 6= 0 and asla · aya = aslaya ∈ La.

2.4. Proposition. Let R be a semiprime ring and let S be a left quotient ring of R. l Take a nonzero element a ∈ R such that aba = a and bab = b for some b ∈ 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. P P P The sum is direct because sialia ∈ SaLa∩lanS(b) implies 0 = sialiaba = sialia. Now, let p and q be in S with p 6= 0. Suppose ﬁrst pua 6= 0 for some u ∈ S. By semiprimeness of S ([9, Proposition 3.1 (ii)]) there exists v ∈ S such that avpua 6= 0. Since Ra is a left quotient ring of La (by (2.3)) and Qa is a left quotient ring of Ra, see [9, Proposition 3.2 (v)], Qa is a left quotient ring of La. So, there exists ala ∈ aLa satisfying 0 6= ala · avpua = alavpua and ala · avqba = alavqba ∈ aLa. Let s ∈ S be such that salavpua 6= 0, then salavp 6= 0 and salavq = salavqba + salavq(1 − ba) ∈ SaLa ⊕ lanS(b).

Now, suppose pSa = 0. In this case p ∈ 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, we have 2 0 6= (p) ⊆ annS(SaS)(p). Hence zp 6= 0 for some z ∈ annS(SaS) ⊆ lanS(b). For this element z we have zq ∈ annS(SaS) ⊆ lanS(b). Local rings of rings of quotients 5

2.5. Remark. 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. 2.6. Theorem. 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 ([9, Proposition 3.2 (i)]) and every ring of left quotients of a semiprime ring is semiprime itself. If we suppose aqa − a 6= 0, there exists p ∈ Q such that 0 6= (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 [9, Proposition 3.2 (v)] Qa is a left quotient ring of Ra. So, by (2.2), conditions (1) and (2) of (2.1) are satisﬁed. Now, let us show (3) of (2.1) and the proof will be complete. Take aLa ∈ Idl(Ra) and f ∈ HomRa (Ra aLa, Ra Ra) and deﬁne

f : RaLa ⊕ lanR(b) −→ R P P rialia + t 7→ ri(alia)f P P 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 X X X 0 6= au ri(alia)f = auri(alia)f = auria · (alia)f = (f ∈ HomRa (Ra aLa, Ra Ra)) X X (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)), RR), with RaLa ⊕ lanR(b) ∈ Idl(R), by (2.4), we P P can apply condition (3) in (2.1) to ﬁnd q ∈ Q such that ( rialia + t)f = ( rialia + t)q P 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, and bearing in mind RaLa ⊕ lanR(b) ∈ Idl(R) and (2.1) (2) we obtain qba − q = 0. Finally, we will see that for every ala ∈ aLa, (ala)f = alaq(= ala · aqba). Take arala ∈ aRaLa. Then arala · aqba = aralaq = a(rala)f = ar(ala)f = ara · (ala)f = 6 Gomez´ and Siles

(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 [16]) 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 [6, 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 6= b ∈ R (an element b in a ring R is said to be square cancellable if b2x = b2y implies bx = by for x, y ∈ R ∪ {1} and xb2 = yb2 implies xb = yb for x, y ∈ R ∪ {1}) such that bRb is a PI ring. Now, if we put 2 2 2 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 ∼ 2.7. Corollary. 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 being Ra a PI-ring implies ([14, Theorem 1] ) that it is left nonsingular. By Lemma 1.1 the whole ring R is left nonsingular; now, apply Theorem 2.6.

l 2.8. Proposition. 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 ([9(3.1)]). Suppose that a is a PI-element. Then Ra, which is also a prime ring by [9, 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 [17, Theorem]). So, by [9, Theorem 3.2 (iii)] Ra is left nonsingular and by (1.1) R is left nonsingular too. Now, by Johnson’s Theorem [10, (13.36)], Q is a von Neumann regular ring, in particular a l l is von Neumann regular in Q. So, Mn(∆) = Qcl(Ra) = (by [10, (13.41)]) Qmax(Ra) = (by l (2.6)) (Qmax(R))a. This implies (by [9, Proposition 2.1 (v)]) a ∈ Soc(Q). 2.9. Remark. For R a unital ring, the previous result can be obtained as a consequence of [6, Corollary 6.17] as follows: If R is prime and has a nonzero PI-element, then R satisﬁes a GPI. Use [6, Corollary 6.17] to obtain that Q is primitive and has nonzero socle.

2.10. Theorem. Let R be a ring without total right zero divisors. An element a ∈ R l 0 is von Neumann regular in Qmax(R) if and only if there exist L and L , left ideals of R, such 0 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 represented as a class [f, I], where I is a dense left ideal of R and f ∈ Hom(RI,R R). 0 0 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, Local rings of rings of quotients 7

0 0 0 0 z ∈ lanR(a). Define g : L ⊕La → L by g(l +la) = l, for every l ∈ L and l ∈ L. The map g is 0 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, and let b 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 0 0 R. Define L := Ib and L := I(1 − ba). We claim that L ⊕ lanR(a) and L ⊕ La are 0 dense left ideals of R. Indeed, it is clear that L ∩ lanR(a) = 0 and La ∩ L = 0. Moreover, La⊕L0 = Iba⊕I(1−ba) = I, which is dense. Now, take r, s ∈ R, with r 6= 0. Since I is dense, there exists y ∈ I such that yr 6= 0 and ysa, ys(1 − ab), ys ∈ I. Then ys = ysab + ys(1 − ab), with ysab ∈ L and ys(1 − ab) ∈ lanR(a). This shows that L ⊕ lanR(a) ∈ Idl(R).

3. 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 [13] 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 with semiprime rings have since 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), or simply by Ie, the set of all essential ideals in a semiprime ring R. The Martindale symmetric ring of quotients of R is deﬁned as follows:

l Qs(R) := {q ∈ Qmax(R) | Iq + qI ⊆ R for some I ∈ Ie}.

l 3.1. Remark. For every semiprime ring R, Qs(R), which is a subring of Qmax(R), is a 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 a right R-homomorphism g : J → R are compatible if for every x ∈ I and y ∈ J,(x)fy = xg(y). In [10, (14.25)] there is an abstract characterization of the Martindale symmetric ring of quotients of a semiprime ring R similar to that of (2.1), but in terms of compatible R- homomorphisms. 3.2. Lemma. 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). Then za 6= 0. By semipri- 0 0 meness of R, there exist t, t ∈ R satisfying 0 6= atzat a. Apply twice aIa ∈ Ie(Ra) to 8 Gomez´ and Siles

ﬁnd aya, ay0a ∈ aIa such that 0 6= atzat0a · aya · ay0a = atzat0ayay0a ∈ atzRaIaR=0, a contradiction. Analogously we can prove that z ∈ ran(a).

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

(i) a is von Neumann regular in Qs(R). ∼ (ii) Qs(Ra) = Qs(R)a under an isomorphism which is the identity on Ra. Proof: (ii) ⇒ (i) can be proved as in Theorem 2.6.

(i) ⇒ (ii). Since R ⊆ Qs(R),Ra ⊆ (Qs(R))a. Now, we will prove the three conditions in [10, (14.25)].

(1). Take aqa ∈ (Qs(R))a, with q ∈ Qs(R). Choose I and J in Ie(R) satisfying qaI, Jaq ⊂ R. Then aqaIa, aJaqa ⊆ aRa, with aIa, aJa ∈ Ie(R), by (2.3).

(2). Let aqa be in (Qs(R))a, with q ∈ Qs(R), and suppose aIa, aJa ∈ Ie(Ra) such that aqaIa = 0 or aJaqa = 0. By (3.1), Qs(R) is a left quotient ring of R, so by [9, Proposition 3.2 (v)] (Qs(R))a is a left quotient ring of Ra. Since aJa is a dense left ideal of Ra, by (2.2) aqa = 0. Analogously we can prove that aqa = 0 if and only if aqaIa = 0.

(3). Let aIa and aJa be in Ie(Ra), and let f ∈ HomRa (Ra aIa, Ra Ra) and g ∈

HomRa (aJaRa ,RaRa ) be compatible. Deﬁne

f : RaIaR ⊕ ann(RaIaR) −→ R P P riaziasi + u 7→ ri(aziasia)f

P We see that the map f is well deﬁned: Indeed, ri(aziasia)f 6= 0 implies, by semiprime- P P P ness of R, that for some t ∈ R, 0 6= at ri(aziasia)f = atri(aziasia)f = atria · P P (aziasia)f = (f ∈ HomRa (Ra aIa, Ra Ra)) ( atria · aziasia)f = ( atriaziasia)f, and P hence riaziasi 6= 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 P P riaziasi + v 7→ 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 (3.2) (ii)) r(ayasa)fbaxataz = r[(ayasa)f · axata]z = (f and g are compatible) r[ayasa · g(axata)]y = rayasabg(axata)z = (by (3.2)(ii)) (rayasab + u)g(axata)z = (rayasab + u)g(axataz + v). The compatibility of f and g, and the fact that RaIaR ⊕ ann(RaIaR) and RaJaR ⊕ ann(RaJaR) belong to Ie(R), by (3.2) (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. This implies qba − q = 0, and analogously we obtain abq = q. Finally, it is easy to see that Local rings of rings of quotients 9

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).

4. 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 [19] and has been considered recently by Lanning in [12].

Denote by Idr(R) (or simply Idr) the family of dense right ideals of R. For a semiprime ring R, deﬁne l Qσ(R) = {q ∈ Qmax(R) | qI ⊆ R for some I ∈ Idr}, 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 [12, Proposition 2.4]). Its proof makes use of the compatibility of homomorphisms.

4.1. Remark. 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 [12, Proposition 2.1] and (2.2). 4.2. Theorem. 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 2.6.

(i) ⇒ (ii). Denote Q := Qσ(R), and let b 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 [12, Proposition 2.1] hold, which will mean that our claim is satisﬁed.

(ii) and (iii). By [9, Proposition 3.2 (v)], Qa is a right and a left quotient ring of Ra. So, by (2.2) we have conditions (ii) and (iii) of [12, Proposition 2.1]. 10 Gomez´ and Siles

(3). Take aLa ∈ Idl(Ra), aKa ∈ Idr(Ra), f ∈ HomRa (Ra aLa, Ra Ra) and g ∈

HomRa (aKaRa ,RaRa ) 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 P P pialia + u 7→ pi(alia)fb

We notice that by condition (i) in (2.4) QaLa ⊕ lanQ(b) is a dense left ideal of Q. The map P P f is 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, P for every i, and r pi(alia)f 6= 0. By the semiprimeness of R, for some s ∈ R we have 0 6= P P P asr pi(alia)f = 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. Analogously, for

g : aKaQ ⊕ ranQ(b) −→ Q P P akiapi + v 7→ bg(akia)pi we have that aKaQ ⊕ ranQ(b) ∈ Idr(Q), by (2.4) (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 [12, Proposition 2.1] applied to Q, which coincides with Qσ(Q) (by [12, The- orem 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, and aqa · aka = aqaka = (by (2)) ag(aka) = abg(aka) = g(aka). This concludes our proof. Acknowledgements. The authors would like to thank the referee for his suggestions.

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