Journal of 237, 798᎐812Ž. 2001 doi:10.1006rjabr.2000.8617, available online at http:rrwww.idealibrary.com on

The of Fractions of a Jordan Algebra

C. Martınez´ 1

Departamento de Matematicas,´ Uni¨ersidad de O¨iedo, Cr Cal¨o Sotelo srn, 33007 O¨iedo, Spain E-mail: [email protected]

Communicated by Efim Zelmano¨

Received May 15, 2000

We derive a necessary and sufficient Ore type condition for a Jordan algebra to have a ring of fractions. ᮊ 2001 Academic Press

1. INTRODUCTION

Let R be an associative ring. Let S be a subset of R which is closed under multiplication and which consists of regular elementsŽ not zero divisors.Ž. . An overring R : Q is called a right ring of fractions of R with respect to S ifŽ. 1 all elements from S are invertible in Q, Ž. 2 an arbitrary element q g Q can be represented as asy1, where a g R, s g S. OreŽ see wx8. found a necessary and sufficient condition for a right ring of fractions to exist:

The Ore Condition. For arbitrary elements a g R, s g S there exist X X X X elements a g R, s g S such that as s sa .

Jacobson et al.wx 2 proved the existence of rings of fractions of Jordan domains satisfying some Ore-type conditions. In this paper we derive a necessary and sufficient Ore-type condition for an arbitrary Jordan algebra to have a ring of fractions. Goldie’s theorems in Jordan algebrasŽ an important application of Ore localization. have been studied inwx 11, 12 . Throughout the paper we will consider over a field F, char F / 2, 3.

1 Partially supported by DGES PB97-1291-C03-01, FICYT PGI-PB99-04.

798 0021-8693r01 $35.00 Copyright ᮊ 2001 by Academic Press All rights of reproduction in any form reserved. RINGS OF FRACTIONS 799

AŽ. linear Jordan algebra is a vector space J with a binary operation Ž.x, y ª xy satisfying the following identities: Ž.J1 xy s yx, Ž.ŽJ2 xyx2 . s xyx2 Ž.. For an element s g J let RxŽ.denote the right multiplication RxŽ.: a ª ax in J. The linearization ofŽ. J2 can be written in terms of operators as

RxyzŽ.Ž .q RxRzRy Ž. Ž. Ž.q RyRzRx Ž. Ž. Ž. s RxyRzŽ.Ž.q RxzRy Ž.Ž.q RyzRx Ž.Ž. s RzRxyŽ. Ž .q RyRxz Ž. Ž .q RxRyz Ž. Ž ..

We will refer to it as the Jordan identity. For elements x, y, z in J,byÄ4x, y, z we denote their Jordan triple product, Ä4x, y, z s Ž.xy z q xyz Ž.y Ž.xz y. By UxŽ.Ž, y resp. Vx Ž.., y we will denote the operators zU Ž. x, y s Ä x, z, y4 ŽŽ.resp. zV x, y s Ä4x, y, z .. For arbitrary elements x, y g J the operator DxŽ., y s RxRy Ž.Ž.y RyRxŽ. Ž.is known to be a derivation of J Žseewx 1. . We denote UxŽ.s Ux Ž, x .. An element a of a Jordan algebra J is called regular if the operator UaŽ.is injective.

DEFINITION 1.1Ž compare to the definition inwx 2.Ž . An Ore monad or simply a monad, by short. S of J is a nonempty subset of J, consisting of X regular elements such that for arbitrary elements s, s g S X Ž.i s2, sUŽ s . g S, X Ž.ii SU Ž. s l SU Ž s . / л.

DEFINITION 1.2. Let J be a Jordan algebra. A Jordan algebra Q, J : Q is said to be a ring of fractions of J with respect to a monad S if Ž.i an arbitrary element of S is invertible in Q, Ž.ii for an arbitrary element q g Q there exists an element s g S such that q и s g J, q и s2 g J. For a Jordan algebra J let RJ²:denote its multiplication algebra, i.e., g the subalgebra of EndF Ž. J generated by all multiplications Ra Ž., a J.

DEFINITION 1.3. Let S : J be a monad of a Jordan algebra J. We say that J satisfies the Ore condition with respect to S if for an arbitrary element s g S and for an arbitrary operator W g RJ²:there exist an XXX X element s g S and an operator W g RJ²:such that WUsŽ.s Us Ž . W. 800 C. MARTINEZ´

The main theorem of this paper is

THEOREM 1.1. Let J be a Jordan algebra and let S be a monad in J. Then J has a ring of fractions with respect to S if and only if J satisfies the Ore condition with respect to S.

We construct a ring of fractions QJŽ.inside the ring of partially defined derivations of the Tits᎐Kantor᎐Koecher Lie algebra LJŽ.following the idea of Johnson, Utumi, and LambekŽ seewx 6, 8. . g s q q 2 For an element s S denote K s JUŽ. s Fs Fs , which is an inner ideal of J. Let RK²:s denote the subalgebra of RJ²:generated by all g multiplications RaŽ., a K s.

2. NECESSITY OF THE ORE CONDITION

We will start this section with some equivalent characterizations of the Ore condition.

PROPOSITION 2.1. Let J be a Jordan algebra and let S be a monad in J. The following conditions are equi¨alent:

Ž.1 J satisfies the Ore condition with respect to S. That is, for an arbitrary element s g S, an arbitrary operator W g RJ²:, there exist an XXX X element s g S and an operator W g R²: J such that W UŽ. s s Us Ž . W. X Ž.2 For arbitrary elements s g S, a g J there exist an operator W g XXX R²: J and an element s g S such that W UŽ. s s Us Ž .Ž. Ra. X Ž.3 For arbitrary elements a g J, s g S there exists an element s g

SUŽ. s which annihilates the element a modulo K s, that is, и X g Ž.i a s Ks and X g wx Ž.ii Da Ž, s .²:ŽR Ks compare with 10. . Proof. ClearlyŽ. 1 implies Ž. 2 . Ž.2 implies Ž. 1 . Let us show that the set of operators W g RJ ² :with the property that for an arbitrary element s g S there exists an element X X s g S such that UsŽ .²:Ž. Wg RJUsis a subring of RJ ²:. g Suppose that the operators W12, W RJ²:have this property and let s g be an arbitrary element of S. Then there exist elements s12, s S such g g that UsŽ11 . W RJUs ²:Ž.and Us Ž22 . W RJUs ²:Ž.. By the definition of l / л g l an Ore monad SUŽ. s12SU Ž. s . Let s 312SUŽ. s SU Ž. s . Then q g UsŽ.Ž31 W W 2 .RJUs ²:Ž.. g g There exists an element s44S such that UsŽ. W12RJUs ²:Ž.. Then g UsŽ.412 WW RJUs ²:Ž.. RINGS OF FRACTIONS 801

Since this subring contains generators RaŽ., a g J,of RJ ²:, it follows that it is equal to RJ²:. Ž.2 implies Ž. 3 . Let’s assume that J satisfies Ž. 2 . Then given arbitrary g g g elements a J, s S there exist an element s1 S and an operator X g s X W RJ²:such that Us Ž1 .Ž. Ra WUsŽ.. l / л X s g l Since SUŽ. s12SU Ž. s , let s sUsŽ.1SU Ž. s1SU Ž. s . X и s s X g X2 Hence s a sUs21Ž.Ž. Ra sWUs 2 Ž. K s. Consequently, DaŽ, s . s XXg 2 DasŽ , s .²:RKs . X2 s 22s On the other hand, s ŽŽ..sUs21sUs 12Ž.Ž. Us 1and so

X2 s 2 s 2 X g s a s12UsŽ.Ž.Ž. Us 1 Ra s 12Us Ž. WUs Ž. K s .

X2 g The element s SUŽ. s annihilates a modulo K s. Ž.3 implies Ž. 2 . Now we will assume that given arbitrary elements a g J X and s g S there exists an element s g SUŽ. s that annihilates a mod- ulo K . s X XX X We have UsŽ .Ž. Ras 2Us Ž, sa.Ž.Žy RaUs.. X X X If s s sUsŽ.then Us Ž .s UsUs Ž.Ž.Ž. Us and besides sag K . Hence XX 22s UsŽ , sa.²:Ž.g RJUs. This implies the assertion. Remark 1. We can define in a similar way the ‘‘right Ore condition.’’ Proceeding as in the proof ofŽ. 3 implies Ž. 2 above, we can prove that Ž. 3 implies that for arbitrary elements a g J, s g S there exist an element X X s g S and an operator W g RJ²:such that RaUs Ž.Ž.Ž.s UsW. Conse- quently, the ‘‘left Ore condition’’ implies ‘‘the right Ore condition.’’

LEMMA 2.1. Let S be a monad in a Jordan algebra J. Let Q be a Jordan algebra that contains J. For an arbitrary element q g Q the condition that there exists an element s g S such that qs g J, qs2 g J, is equi¨alent to the g : condition that there exists an element t S such that qK t J. : g 2 g Proof. If qK t J then qt J and qt J. g 2 g : s 2 Conversely, if s S and qs, qs J then qK t J, where t s . Indeed, qRŽ s2 .Žg J and qR s4 .Žs q y2 Rs Ž.2 RsŽ 2 .Ž.Žq 2 RsRs3.Žq Rs2 .2 . g J. Moreover, for an arbitrary element b g J we have

qRŽ. bUŽ. s2 s 2Ä4qs, bUŽ. s , s y Ä4qU Ž. s , b, s2 g J.

The lemma is proved.

PROPOSITION 2.2. If a Jordan algebra J has a ring of fractions with respect to a monad S : J, then J satisfies the Ore condition with respect to S. Proof. Let Q be a ring of fractions of J with respect to S. We need to prove that for arbitrary elements a g J, s g S there exists an element 802 C. MARTINEZ´

g g s s1 S and an operator W RJ²:such that Us Ž1 .Ž. Ra WU Ž. s ,or y1 g equivalently, UsŽ.Ž.Ž1 RaUs .²:RJ. Since RaUsŽ.Žy1 .Žq Usy1 .Ž. Ras 2Us Žy1 a, sy1. we have

y1 sy y1 q y1 y1 UsŽ.Ž.Ž111 RaUs .Us Ž.Ž Us .Ž.Ra 2Us Ž.Ž Us a, s .. Linearizing the identity UxUyŽ.Ž.s 4Vx Ž, y .2 y 2VxŽŽ.., xU y we get UxUyŽ.Ž.Ž.Ž.Ž.Ž., z s 2Vx, yV x, z q 2Vx, zV x, y y 2VxŽ., xUŽ. y, z . In particular,

y1 y1 s y1 y1 q y1 y1 UsŽ.Ž11 Us a, s . 2VsŽ.Ž.Ž.Ž., s Vs1, s a 2Vs1, s aV s1, s y y1 y1 2VsŽ.11, sUŽ. s , s a . s y1 s y1 g Denote q12s , q s a. By Lemma 2.1 there exist elements t12, t S such that qK : J, i s 1, 2. Choose an element t g SUŽ. t l SU Ž. t l iti 12 SUŽ. s . Then K : K l K . tt12 t 2 : 2 q 2 : 2 : We have VqŽ i, K t.ŽDq i, K t.ŽRqK it.²:RJ since qKitqK it : 2 : : J and DqŽ it, K .ŽDqK itt, K .²:RJ. s 2 g 2 Let s1 t K t . Then y1 y1 y1 y1 y1 g UsŽ.Ž11 Us .Ž. Ra, VsŽ.Ž.Ž.Ž., sVs11, sa, Vs, saVs1, s RJ²: y1 y1 g and it remains to show that VsŽŽ11, sUs , sa..RJ ² :. Linearizing the identity VxŽŽ..ŽŽ.., xU y s VyUx, y we get the identity VxŽŽ..ŽŽ..ŽŽ.., xU y, z s VyUx, z q VzUx, y . Hence

y1 y1 s y1 y1 q y1 y1 VsŽ.11, sUŽ s , s a .VsŽ.Us Ž.1, s a Vs Ž. ŽaU .Ž. s1, s . y1 s y1 g y1 y1 g Now, sUsŽ.1 sUtUtŽ. Ž. K t , and therefore VsŽ UsŽ.1 , sa. RJ²:. y1 s g : Similarly, ŽsaUs.Ž12 .qUtUt Ž.Ž.JU Ž. t K t , which implies y1 y1 g VsaUsŽ.Ž.Ž.1 , s RJ²:. The proposition is proved.

3. CONSTRUCTION OF THE RING OF QUOTIENTS

Throughout this section we will assume that a Jordan algebra J satisfies the Ore condition with respect to a monad S : J. RINGS OF FRACTIONS 803

Let us show that to embed J in a ring of fractions it is sufficient to embed J in a Jordan overring Q˜ in which all elements from S are invertible.

PROPOSITION 3.1. Let J : Q˜˜ and all elements from S are in¨ertible in Q. Then the subalgebra Q s ²:J, Sy1 of Q˜ generated by J and by all in¨erses sy1, s g S, is a ring of fractions of J with respect to S. U Proof. Say that an element q g Q˜ has propertyŽ. if for an arbitrary X X element s g S there exists an element s g S such that q и K : K . ssU From Lemma 2.1 it follows that elements of J have propertyŽ . . U Let us show that an arbitrary inverse ty1, t g S, has propertyŽ . . X g g : X Choose s S. There exists an element s S such that tK ssK . Then y 1 и : и : X t K sUŽt. K sst K . U It is clear that if elements a, b g Q˜ have propertyŽ. then their sum U a q b has propertyŽ . . U Suppose that an element a g Q˜ has propertyŽ. . We will show that a2 U has propertyŽ . . X Let s g S. There exists an element s g S such that a и K : K X . 1 ss1 Similarly, there exists an element s g S such that a и K : K l K X .We ss1 s g X can also assume that s K s . We have

2 2 s 2 a RKŽ.ssaRŽ. a R Ž. K : и q q и 2 aRŽ.Ž Kss R a K .aR Ž.Ž.Ž. K s R a R K saRŽ. a K s.

Furthermore, a и K : K XX, aRŽ.Ž. K R a : K и a : K . ss s s1 s 2 и 2 : X 2 и : X Hence, a K ssK and a K s2 K s. From what we proved it follows that an arbitrary element from Q s U ² J, Sy1 :Žsatisfies. . By Lemma 2.1, Q is a ring of fractions of J. The proposition is proved. ޚ s A Jordan algebra J gives rise to a -graded Lie algebra KJŽ.KJ Ž.y1 q q ᎐ ᎐ KJŽ.01KJ Ž., which is known as the Tits Kantor Koecher construc- tionŽ seewx 3, 4, 9. . Let Ä4a, b, c s Ž.ab c q abc Ž.y bac Ž.denote the so- called Jordan triple product of elements a, b, c g J. Consider two copies Jy, Jq of the vector space J. We identify an element a g J with elements ay in Jy and aq in Jq. For arbitrary elements ayg Jy, bqg Jq we ␦ yqg yq[ define a linear operator Ža , b . EndFF J End J via

yª y yq c Ä4a, b, c ␦ Ž.a , b : q ½ cqª yÄ4b, a, c . 804 C. MARTINEZ´

Jordan identities imply that for arbitrary elements a, b, c, d g J we have ␦ Ž.Ž.ayq, b , ␦ c yq, d s ␦␦Ž. Ž.ayqyq, bc, d q ␦ Ž.c y, ␦ Ž.a yq, bd q, so the linear space ␦Ž Jyq, J .Žof all operators ␦ ayq, b .; a, b g J, is a Lie algebra. Now consider the direct sum of vector spaces KJŽ.s Jy[ ␦ ŽJy, Jq .J. Define a bracketwx , on KJŽ.via w Jyy, J xws Jqq, J x s Ž.0 ; for arbi- trary elements ayyqqg J , b g J , wayq, b xwsy bqy, a x s ␦Žayq, b .; for an arbitrary element x g Jyqq J and for an operator ␦ g ␦Ž Jyq, J .wx␦, x s ␦Ž.x sywxx, ␦ ; elements from ␦Ž Jyq, J . are commutated as linear operators. A straightforward verification shows that KJŽ.is a Lie algebra. s s y q w yqx q q Denote Lss, L KKŽ. sK sK s, K sK s, a Lie subalgebra of L s KJŽ..

XX: : ª PROPOSITION 3.2. Let K ssKŽ. and so L ssL and let D : LsLbe a deri¨ation such that D < X s 0. Then D s 0. L s Before proving Proposition 3.2 we need some preliminary lemmas.

LEMMA 3.1. The centralizer of Ls in L is zero. Proof. Let us show that no nonzero element from Jy commutes with w yqx yyg w y w yqxx s s K ss, K .If a J and a , K ss, K Ž.0 then Ä4a, K ss, K Ž.0. Therefore Ä4a, s, s s a и s2 s 0 and Äa, s22, s 4 s a и s4 s 0. This implies that 4 wx 4 s lies in the annihilator AnnJ a in the sense of 10 . Since s is a regular element it follows that a s 0. q w yqx Similarly, no nonzero element from J commutes with K ss, K . Now 0 g w yqxwy 0 xws q 0 x s let a J , J lie in the centralizer of Ls.If J , a J , a Ž.0 then a0 s 0. Let us assume that 0 / wby, a0 x for some element byyg J . X By Proposition 2.1 there exists an element s g Ä4s, S, s such that wby, qy y w XXxx : K ss, K K s. y qy y qy y ww 0 xw XXxxs ww w XXxx 0 xw: 0 x s Hence, b , a , K ss, K b , K ss, K , a K s, a Ž.0, which contradicts what was proved above.

Since the centralizer of Ls is a graded algebra we conclude that it is equal toŽ. 0 . The lemma is proved.

LEMMA 3.2. For arbitrary elements a g L, s g S there exists an element X g wxX : s S such that a, LssL . Proof. Let’s assume that for given elements a, b g L there exist ele- XY ments s , s g S such that wxa, L XY: L and wxb, L : L . Choose an Z ss s s ZXY: l wxwxq ZZ: q element s such that LsssL L . Then a b, Lsa, L s wxwxwxZXY: q : b, Lssssa, L b, L L . RINGS OF FRACTIONS 805

Hence we can assume that a g Jyjw Jyq, J x j Jq. It is sufficient to assume that a g Jyj Jq. Indeed, suppose that for elements from Jyj Jq the assertion of the lemma is valid. Let ayg Jy, aqg Jq, s g S. Then X y q g w XXxw: x : there exists an element s S such that a , LssL , b , L ssL . Y y q g w YXxw: Similarly, there exists an element s S such that a , LssL , b , YXx : LssL . Then yq y q q y wxYY: q Y a , b , Lssa , b , L b , a , L s yq : XXq : a , Lsssb , L L . So let ayg Jy, s g S. We need to prove the existence of an element X yq yq y yq y g w XXxw: xww Xxx : s S such that a , K sssK , K and a , K sss, K K . X g By Proposition 2.1 there exists an element s K s that annihilates a modulo K s. Hence

y qy q yqyq qy XX X: XXX: : a , K ss, K , K s a , K sss, K , K K ss, K L s.

X3 X g X : XX Since s SUŽ s . also annihilates a modulo K ssssand K 3 Ä K , K , y XXw x : K ss4, it follows that a , L 3 Ls. The lemma is proved.

X : ª Proof of Proposition 3.2. Let LssL and let D : LsL be a deriva- X s tion such that DLŽ.s 0. Let us assume that there exists an element a g L such that DaŽ./ 0. s Y g wxYX: By Lemma 3.2 there exists an element s S such that a, LssL Y : X and we can assume without loss of generality that LssL . YYs w xw: Yx q wxY: X Then DLŽ.Ž.ss0 and DaŽ., L a, DLŽ.sDaŽ , Ls.Ž.DLs s w Y x s Ž.0 . Hence DaŽ., Ls Ž.0 , which contradicts Lemma 3.1. The propo- sition is proved. g U ª For an element s S let Dssdenote the set of all derivations d : L U s D U L. Let D sg SsD . X U X For derivations d, d g D , d ' d if and only if there exists an element X X s g S such that d, d are both defined on L and d< s d< . s L ssL Clearly, this is an equivalence relation. Consider the quotient set D s U U D r' . Abusing notation we will denote the class of a derivation d g D as d. Let’s define a structure of a vector space on D. Consider elements of D X represented by derivations d : L ª L, d : L X ª L. Let ␣, ␤ g F. There Y ss X exists an element s g S such that L Y : L l L X . Define ␣ d q ␤d : L Y X X sss s ª L, ␣ d q ␤d : a ª ␣ daŽ.q ␤daŽ.. g ޚ For an arbitrary s S the algebras Ls and L are -graded. We say that g U : sy a derivation d D has degree i if dLŽŽsk . . L kqi for k 1, 0, 1. U s 2 U Clearly, D Ýisy2Ž D .i. ޚ ޚ s 2 This -gradation induces a -gradation on D, D Ýisy2 Di. 806 C. MARTINEZ´

s s LEMMA 3.3. Dy22Ž.0 D . ª y ª q Proof. Let d : LssL be a derivation of degree 2. Then d : K J , w yqx ª q ª d : K ss, K Ž.0 and d : K s Ž.0. ␸ ª ␸ s We will define a linear mapping : K s J via a b if and only if ayd s bq. X Y g For arbitrary elements a, a , a K s we have

Xq Yy Xq Yy wxwxayy, a , ads a , a , ad Xq Yy Xq Yy Xq Yy s w ady , a , a xwq ay, ad, a xwq ay, a , adx Xq Y q sy a , ay, Ž.a␸ ,

Xq Xq since wady , a x s 0 and ads 0. XY XY Hence, Äa, a , a 4Ä␸ sy a , a, a␸4. Xq Yy Yy Xq On the other hand, way, a , a xws a , a , ayx and so

XY XY XY YX Ä a, a , a 4Ä␸ sy a , a, a␸4Äsy a , a , a␸4Äsy a␸, a , a 4 .

In particular

a5␸ s Ä a3 , a, a4 ␸ syÄ4a, a34, a␸ syŽ.Ž.a␸ Ra syÄ4a3␸, a, a s Ä4Ä4a␸, a, aa, a s Ža␸ .Ž.Ž.Ra22 Ra .

Therefore bRaŽŽ4 .Žq Ra2 .Ž Ra2 .. s 0 for b s a␸. Similarly

a7␸ s Ä a33, a, a 4 ␸ syÄ4a33␸, a , a s Ž.Ž.Ž.a␸ Ra2 Ra 4 s Ä a, a33, a 4 ␸ syÄ4a␸, a33, a syŽ.Ž.a␸ Ra6 and so bRaŽŽ6 .Žq Ra2 .Ž Ra4 .. s 0. But by the Jordan identity RaŽ 6 .Žq 2 Ra2 .3 s 3RaŽ 2 .Ž Ra4 ..So

3 bRaž/Ž.Ž.24 Ra q Ra Ž. 2s 0 s bRaŽ. Ž. 6q Ra Ž.Ž. 24 Ra 3 s bž/y2 RaŽ.224q 4Ra Ž.Ž. Ra .

This implies that bRŽ a2 .3 s 0 s bRŽ a6 .Žs bR a2 .Ž R a4 .. If a is a regular elementŽ. for example, an element from S , then bRŽ a2 .Ž U a2 .Žs 0 implies that bR a2 .Žs 0 and bR a4 .Žs bR a6 . s 0. Consequently the element a4 annihilates b Žin the sense ofwx 10. , which implies b s 0. RINGS OF FRACTIONS 807

We have proved that S␸ s Ž.0 , which implies that sdy s 0. We have yys q 2 yqw yqyx y s Ls Fs FsŽ . s , J , s . This implies that Lds Ž.0. s In the same way we can prove that Dy2 Ž.0 . The lemma is proved. Our next aim is to define a Lie bracket on D. ¨ ª PROPOSITION 3.3. For an arbitrary deri ation d : Ls L there exists an element s g SUŽ. s such that L d : L . 1 ss1 Let us show that it it sufficient to prove the proposition for homoge- ª s q neous derivations d. Indeed, let d : Ls L be a derivation, d dy10d q d , where the d ’s are homogeneous derivations. Suppose that there 1 i X exist elements s g SUŽ. s , y1 F i F 1, such that Ld: L .If s g isi is F X : y1 F iF1SUŽ. siss, then Ld L .

LEMMA 3.4. Let d : L ª L be a homogeneous deri¨ation of degree 0. s X g X : Then there exists an element s SUŽ. s such that Lss d L . Proof. Let sqd s aq, syd s by; a, b g J. There exists an element X t g SUŽ. s which annihilates both elements a, b modulo K. Let s s tUŽ. s . X We have JUŽ s .Ž.Ž.: JU t U s . Hence,

X q yy Ž.JUŽ. s d : Ž.JUŽ. t d, sqq, s q Ž.JUŽ. t , sd q, s q y q Ž.JUŽ. t , sqq, sd : qqq : q Ž.JUŽ. s Ä4a, JU Ž. t , s K s , Xq g X2 q g and sd K ss, Žsd. K . y y X : X : Similarly K ssd K . This implies Lssd L . The lemma is proved. X LEMMA 3.5. Let b g J, s g S. Then there exists an element s g SUŽ. s y X : such that Lss adŽ b . L . s w qyx Proof. Let’s consider an element of degree zero, d0 s , b .By g wx: X s Lemma 3.4 there exists t SUŽ. s such that Lt , d0 Ls. Let s tUŽ. s . w qyy q q yx : We need to prove that J , t , t , s , s , b Ls. w qyyw qyx qxwq qyyw q w qyxxx : This reduces to J , t , t , s , b , s J , t , t , s , s , b Ls. w qyy x g y But J , t , t , d0 K s by the choice of t. Then qyy q: y q: J , t , t , d0 , s K ss, K L s. w qyyx : y w q w qyxx g q As for the second summand, J , t , t K ssand s , s , b K . w qyyw q w qyxxx : Hence J , t , t , s , s , b Ls. On the other hand,

qyy q s qyy qs qyy q : J , t , t , s , d00J , t , t , d , s J , t , t , sd 0Ls 808 C. MARTINEZ´

w qyy q xwws qyyxwq w qyxxx: w yqx : since J , t , t , sd0 J , t , t , s , s , b K ss, K L s. The lemma is proved.

LEMMA 3.6. Let d be a deri¨ation of degree 1. Then there exists an X g X : element s SUŽ. s such that Lss d L . Proof. Let bys w syy, sdx. By Lemma 3.4 there exists an element g w y x : t SUŽ. s such that Lts, sd L . g w yx : By Lemma 3.5 there exists an element u SUŽ. s such that Lu, b X g l Ls. Let s SUŽ. t SU Ž u .. Then

Xq Xq Xq Xq w Jy, s , s , syy, s , d xw: Jy, s , s , syy, d, s x Xq Xq qw Jy, s , s , d, syy, s x Xq Xq q Jyyy, s , s , s , wxs , d .

But w y Xq Xq y x : J , s , s , b L2 and wxwyyyyyyXq Xq : Xq Xq x: J , s , s , s , d, s J , s , s , s , d , s Ls ,

X g X : since s SUŽ. t . Consequently LdssL . The lemma is proved. We can prove in a similar way the corresponding result for a derivation of degree y1. This finishes the proof of Proposition 3.3. X U Y U Now we can define a Lie bracket on D. Let d g D X , d g D Y . Choose X Y ss an element s g SUŽ s .Žl SU s .. By Proposition 3.3 there exists an ele- g XY: : ment t SUŽ. s such that LdtsL and Ld tL s. ª Define the derivation d : Lt L via

s X Y y Y X g xd Ž.xd d Žxd .d , x Lt .

XY Define wd r' , d r'x s dr' . It is easy to see that with the thus defined bracket D becomes a graded s q q Lie algebra, D Dy101D D . / s Since char F 2, 3 the pair of spaces P Ž.Dy11, D is a Jordan pair Žseewx 7. with respect to operations

⑀ y⑀⑀s ⑀y⑀⑀g ⑀ s " Ä4d12, d , d 3d 12, d , d 3 D⑀ , 1.

We will prove that the Jordan pair Ž Jyq, J . associated to J can be embedded into P. RINGS OF FRACTIONS 809

Define

␸ y q ª s : Ž.J , J P ŽDy11, D .via ␸ Ž.a⑀ s ad Ž. a⑀ , ⑀ s ", a g J.

By Lemma 3.1 the linear transformation ␸ is injective. Since Jordan products in Ž Jyq, J . and P are defined via Lie brackets, it follows that ␸ is a homomorphism of Jordan pairs. Let us show that for an arbitrary element s g S the pair ŽŽ␸ sy.Ž, ␸ sq.. is invertible. s s y q wxy q q q Let K˜˜˜ssssJUŽ. s , L K K˜ , K˜ssK˜ . For an arbitrary element X g XX: : s SUŽ. s we have K ssssK˜˜, L L . We will construct two derivations y ª q ª y q : L˜ssL, q : L˜ L of degrees 1, 1, respectively. Their restrictions y q X ␸ ␸ to Ls will define the inverse of ŽŽs .Ž, s ... yys qg q qys Let Kq˜ssŽ.0 . For an element ŽaU Ž s .. K we let ŽaU Ž s .. q wxyqs w y qxwg y q x a , s . Consider an element x0 Ý kii, ŽŽ..aU s K˜ ss, K˜ ; let ys ww qxxsy w yqyx xq0 Ýiik , a i, s Ý iik , s , a i. y We need to verify that q is well defined. An element from K˜s can be y q expressed in the form aUŽ. s uniquely. Hence q is well defined in K˜s . w y qxws yq However, we need to verify that Ýiik , ŽŽ..aU i s 0 implies Ýiik , s , yx s ai 0. y w qyxws yqqyxws yqqy If ÝiiŽŽ..aU s , k iÝ iia , s , s , ki 0 then Ýiia , s , s , ki, sqx s 0. Since ŽŽad sq.. 3 s 0 and the characteristic is / 3, it follows that for an yg arbitrary element b Ly1 we have

adŽ. sqyq22 ad Ž.Ž. b ad s s ad Ž.Ž.Ž. s qyq ad b ad s Ž.)

Žseewx 5. . w yqyqqx s Hence Ýiia , s , k i, s , s 0. Since s is a regular element, it implies w yqyx s that Ýiia , s , k i 0. The linear mapping qy has been well defined. Similarly, we can define a q ª mapping q : L˜s L of degree 1. Let’s prove now that qy Žresp. qq. is a derivation. 00g w yqx yg y qqqg q Let e , a K ss, K , e K s, e , a , b K s. Since we know that we0, eqyx ys 0 s weq0 yy, e xws e0, eqyyx, we only need to check: Ž.i waqq, bqx ys 0 s waqyq, q , b xwq aqqy, bqx, Ž.ii we0, eqqx ys 0 s weq0 yq, e xwq e0, eqqyx, Ž.iii we00, aqx ys weq0 y, a0 xwq e00, aqyx. 810 C. MARTINEZ´

Let aqs Ä sqyq, ␣ , s 4, bqs Ä sqyq, ␤ , s 4. Then wxwxwxwxaqqyq, b q a q, bq qys ␣ yq, s , b qq a q, ␤ yq, s sy ␣yq, s , wx␤ yqq, s , s q ␤yq, s , wx␣ yqq, s , s s ␣yqq, s , s , wx␤ yq, s y ␣ yqq, s , s , wx␤ yq, s y ␣yq, s , wx␤ yq, s , s q syw ␣yq, s , ␤ yqq, s , s xwq ␣yqq, s , s , ␤ yq, s x0 s 0 by Ž.) . This proves Ž. i . In order to proveŽ. ii , let’s assume, by linearity, that e0 s wkyq, k x, kqs Ä sqyq, c , s 4, and eqs Ä sqyq, b , s 4. Then w e0 , eq xwsy kqyq, k , e x syÄ4Ä4Ä4sqyq, c , s , k y, s qyq, b , s syÄ4sqy, Ä4c , Ä sqyq, k , s 4 , bsyq. By definition of qy, wxe0 , eqqysy Ä4c yqyqyq, Ä4s , k , sb, s sy c yqyqyq, ws , k , s x, b , s . On the other hand, ys ywx yq sy yqy qys wx yq eq0 k , c , s Ä4k , s , c and eq b , s . So we have to prove cy, wxs q, k yq, s , b yq, s s wxwxk yqy, s , c , s q, b yq, s y ky, wxwxs qyq, c , s , b yq, s . But cyqyq, ws , k , s xs c yqyq, wxs , k , s y c yqqy, s , wxs , k s wxkyqy, s , c , s qq c yq, s , wxk yq, s s w xyq, s xwq cyq, s , k yq, s xwy cyqq, s , s , k yx s 2w xyq, s xwq kqy, k x, where x s Ä4k, s, c . Hence, cy, wxs q, k yq, s , b yq, s s 2 wx yq, s , b yq, s xwxq k q, k y, b yq, s . On the other hand wxwxkyqy, s , c , s q, b yq, s y k y, k q, wxb yq, s s xy, wxs q, b yq, s y k y, k q, wxb yq, s . RINGS OF FRACTIONS 811

So we need to prove that

2wxwxwxxyq, s , b yq, s y k y, k qq, s , b ys x y, s q, b yq, s ,

which reduces to

w kyqqy, k , s , b xws xyqq, s , s , b yx.

It is sufficient to prove that wkyqq, k , s xws xyqq, s , s x. But, using Ž.) , we have w xyqq, s , s x s cy adsŽ q.Ž adky.Ž adsq. 2 s cadsy Ž q. 2 adkŽ y.Ž adsq. sykq adŽ ky.Ž ad sq. s wkyqq, k , s x. Finally, let’s proveŽ. iii . Let e0 s wkyq, k x as above and a0 s wcyq, d xw.So e00, a xwws ky, a0 x, kqxwq ky, wkq, a0 xx. UsingŽ. ii we have

wxe00, aqyqyysy kq, wxwxk , a0 y kq, aq0 yy, k

sy kqqy, k y, a0 q kqqy, a0 , kyqyy kq, a0 , ky

q aq0 yqy, k , k

s wxkyqy, kq, a00y aqyyq, wxk , k

s eq0 y, a00y aqy, e0 .

So we have proved that qy is a derivation. Finally we will prove that the pairs ŽŽ␸ sy.Ž, ␸ sq..and Žqyq, q . are mutually inverse. s w yq␸ x s g Denote l q , Ž.s . For an arbitrary element a bUŽ. s K˜s we have

wxaqqyqqyqqyqq, l s a , q , ␸ Ž.s s s , b , ␸ Ž.s s ws , b , s xs a .

X Furthermore, for arbitrary elements x g J, s g SUŽ. s

Xq Xq Xq Xq Xq Xq Xq Xq wxwxwxwxxyyyy, s , s , l s x , l, s , s q 2 x , s , s s x , s , s ,

Xq Xq which implies ww xy, lx q xy, s , s xws 0. Hence xy, lx syxy. Similarly we conclude that w xq, lx s xq. s q Now let d1 be a derivation of degree 1 defined on LttŽ.L y1 Ž.Lt 0 q g g Ž.Lt 1, t S. For an arbitrary element bitiŽ.L we have wxws xwxy s q wxwxy bi , d1 , l bi , d1 , l bi , l, d1 Ž.i 1 bi , d1 ibi , d1 s wx bi , d1 . ww xy xs wxs Hence Lt , d11, l d Ž.0 . By Lemma 3.1, d11, l d . 812 C. MARTINEZ´

wxsy y Similarly dy1 , l dy1 for a derivation of degree 1. wwq ␸ y xx sy g Arguing as above one can show that dii, q , Žs . id for diDi, i sy1, 0, 1. Since char F / 2 it implies that the pairs ŽŽ␸ sy.Ž, ␸ sq..and Žqyq, q . are mutually inverse. Now we can finish the proof of Theorem 1.1. Without loss of generality we will assume that the algebra J is unital. If not consider the unital hull Jˆˆs J q F1of J with the same monad S : J. It is easy to see that Jˆstill satisfies the Ore condition with respect to S. Let us show that ŽŽ␸ 1,y.Ž␸ 1q.. is an identity of the Jordan pair s wx P Ž.ŽDy11, D see 7. . Clearly, e s ŽŽ␸ 1,y.Ž␸ 1q.. is an idempotent of P. ⑀ ⑀ If s g S then we showed above that D⑀ s Ä␸Žs .Ž, Dy⑀ , ␸ s .4, ⑀ s "1, which implies that the whole P lies in the 1-Peirce component of e. Hence e is an identity of P. и Let J˜ be the Jordan algebra on Dy1 with the multiplication xy1 yy1 s ␸ q ª ª ␸ y Ä xy1 , Ž.1,yy14. Then the mapping J J˜, a Ž.a is an embed- ding and for an arbitrary element s g S its image ␸Žsy. has the inverse qyg J˜.

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