The Ring of Fractions of a Jordan Algebra

Transcription

1 Journal of Algebra 237, doi:1.16jabr , available online at on The Ring of Fraction of a Jordan Algebra C. Martınez 1 Departamento de Matematica, Unieridad de Oiedo, C Calo Sotelo n, 337 Oiedo, Spain chelo@pinon.ccu.uniovi.e Communicated by Efim Zelmano Received May 15, 2 We derive a neceary and ufficient Ore type condition for a Jordan algebra to have a ring of fraction. 21 Academic Pre 1. INTRODUCTION Let R be an aociative ring. Let S be a ubet of R which i cloed under multiplication and which conit of regular element Žnot zero divior.. An overring R Q i called a Ž right. ring of fraction of R with repect to S if Ž. 1 all element from S are invertible in Q, Ž. 2 an arbitrary 1 element q Q can be repreented a a, where a R, S. Ore Žee. 8 found a neceary and ufficient condition for a right ring of fraction to exit: The Ore Condition. For arbitrary element a R, S there exit element a R, S uch that a a. Jacobon et al. 2 proved the exitence of ring of fraction of Jordan domain atifying ome Ore-type condition. In thi paper we derive a neceary and ufficient Ore-type condition for an arbitrary Jordan algebra to have a ring of fraction. Goldie theorem in Jordan algebra Žan important application of Ore localization. have been tudied in 11, 12. Throughout the paper we will conider algebra over a field F, char F 2, 3. 1 Partially upported by DGES PB C3-1, FICYT PGI-PB $35. Copyright 21 by Academic Pre All right of reproduction in any form reerved. 798

2 RINGS OF FRACTIONS 799 A Ž linear. Jordan algebra i a vector pace J with a binary operation Ž x, y. xy atifying the following identitie: Ž J1. xy yx, Ž J2. Ž 2 x y. x x 2 Ž yx.. For an element J let RŽ x. denote the right multiplication RŽ x.: a ax in J. The linearization of Ž J2. can be written in term of operator a RŽ Ž xy. z. RŽ x. RŽ z. RŽ y. RŽ y. RŽ z. RŽ x. RŽ xy. RŽ z. RŽ xz. RŽ y. RŽ yz. RŽ x. RŽ z. RŽ xy. RŽ y. RŽ xz. RŽ x. RŽ yz.. We will refer to it a the Jordan identity. For element x, y, z in J, by x, y, z we denote their Jordan triple product, x, y, z Ž xy. z xž yz. Ž xz. y. By UŽ x, y. Žrep. VŽ x, y.. we will denote the operator zuž x, y. x, z, y Žrep. zvž x, y. x, y, z.. For arbitrary element x, y J the operator DŽ x, y. RŽ x. RŽ y. RŽ y. RŽ x. i known to be a derivation of J Žee. 1. We denote UŽ x. UŽ x, x.. An element a of a Jordan algebra J i called regular if the operator Ua i injective. DEFINITION 1.1 Žcompare to the definition in. 2. An Ore monad Žor imply a monad, by hort. S of J i a nonempty ubet of J, coniting of regular element uch that for arbitrary element, S Ž. i 2, UŽ. S, Ž ii. SU SUŽ.. DEFINITION 1.2. Let J be a Jordan algebra. A Jordan algebra Q, J Q i aid to be a ring of fraction of J with repect to a monad S if Ž. i an arbitrary element of S i invertible in Q, Ž ii. for an arbitrary element q Q there exit an element S uch that q J, q 2 J. For a Jordan algebra J let R² J: denote it multiplication algebra, i.e., the ubalgebra of End Ž J. generated by all multiplication Ra, a J. F DEFINITION 1.3. Let S J be a monad of a Jordan algebra J. We ay that J atifie the Ore condition with repect to S if for an arbitrary element S and for an arbitrary operator W R² J: there exit an element S and an operator W R² J: uch that WU U Ž. W.

3 8 C. MARTINEZ The main theorem of thi paper i THEOREM 1.1. Let J be a Jordan algebra and let S be a monad in J. Then J ha a ring of fraction with repect to S if and only if J atifie the Ore condition with repect to S. We contruct a ring of fraction QŽ J. inide the ring of partially defined derivation of the TitKantorKoecher Lie algebra LJ following the idea of Johnon, Utumi, and Lambek Žee 6, For an element S denote K JU F F, which i an inner ideal of J. Let R² K : denote the ubalgebra of R² J: generated by all multiplication Ra, a K. 2. NECESSITY OF THE ORE CONDITION We will tart thi ection with ome equivalent characterization of the Ore condition. PROPOSITION 2.1. Let J be a Jordan algebra and let S be a monad in J. The following condition are equialent: Ž. 1 J atifie the Ore condition with repect to S. That i, for an arbitrary element S, an arbitrary operator W R² J :, there exit an element S and an operator W R² J : uch that W U U Ž. W. Ž. 2 For arbitrary element S, a J there exit an operator W R² J : and an element S uch that W U U Ž. Ra. Ž. 3 For arbitrary element a J, S there exit an element SU which annihilate the element a modulo K, that i, Ž. i a K and Ž ii. Da, Ž. R² K : Žcompare with 1.. Proof. Clearly Ž. 1 implie Ž. 2. Ž. 2 implie Ž. 1. Let u how that the et of operator W R² J: with the property that for an arbitrary element S there exit an element S uch that U Ž. W R² J: U i a ubring of R² J :. Suppoe that the operator W, W R² J: 1 2 have thi property and let be an arbitrary element of S. Then there exit element 1, 2 S uch that U W R² J: U and U W R² J: UŽ By the definition of an Ore monad SUŽ. SUŽ.. Let SUŽ. SUŽ Then U Ž W W. R² J: UŽ There exit an element S uch that U W R² J: UŽ Then U WW R² J: UŽ.. 1 2

4 RINGS OF FRACTIONS 81 Since thi ubring contain generator Ra, a J, of R² J :, it follow that it i equal to R² J :. Ž. 2 implie Ž. 3. Let aume that J atifie Ž. 2. Then given arbitrary element a J, S there exit an element 1 S and an operator W R² J: uch that U Ra WU. 1 Since SUŽ. SU, let U SUŽ. SUŽ Hence a U Ra WU K. Conequently, Da, Ž Da, Ž. R² K :. 2 On the other hand, ŽU. 2 U 2 U and o a 2 U U R a 2 U W 1 Ž 2. Ž 1. 1 Ž 2. U K. 2 The element SU annihilate a modulo K. Ž. 3 implie Ž. 2. Now we will aume that given arbitrary element a J and S there exit an element SU that annihilate a modulo K. We have U Ž. Ra 2UŽ, a. RaU Ž.. Ž If U then U. UU U 2 2 and beide ak. Hence U, Ž a. R² J: UŽ.. Thi implie the aertion. Remark 1. We can define in a imilar way the right Ore condition. Proceeding a in the proof of Ž. 3 implie Ž. 2 above, we can prove that Ž. 3 implie that for arbitrary element a J, S there exit an element S and an operator W R² J: uch that RaU Ž. UW. Conequently, the left Ore condition implie the right Ore condition. LEMMA 2.1. Let S be a monad in a Jordan algebra J. Let Q be a Jordan algebra that contain J. For an arbitrary element q Q the condition that there exit an element S uch that q J, q 2 J, i equialent to the condition that there exit an element t S uch that qk t J. Proof. If qk t J then qt J and qt 2 J. Converely, if S and q, q 2 J then qk t J, where t 2. Indeed, qrž 2. J and qrž. qž2 R 2 R Ž 2. 2 RR Ž 3. R Ž J. Moreover, for an arbitrary element b J we have qr buž 2. 2 q, buž., quž., b, 2 J. The lemma i proved. PROPOSITION 2.2. If a Jordan algebra J ha a ring of fraction with repect to a monad S J, then J atifie the Ore condition with repect to S. Proof. Let Q be a ring of fraction of J with repect to S. We need to prove that for arbitrary element a J, S there exit an element

5 82 C. MARTINEZ S and an operator W R² J: uch that U Ra WUŽ. 1 1, or equivalently, U RaU Ž 1. R² J : 1. Since RaU Ž 1. U Ž 1. Ra 2UŽ 1 a, 1. we have UŽ 1. RŽ a. UŽ 1. UŽ 1. UŽ 1. RŽ a. 2UŽ 1. UŽ 1 a, Ž. Linearizing the identity U x U y V x, y 2V x, xu y we get In particular, UŽ x. UŽ y, z. 2VŽ x, y. VŽ x, z. 2VŽ x, z. VŽ x, y. 2VŽ x, xuž y, z... UŽ 1. UŽ 1 a, 1. 2VŽ 1, 1. VŽ 1, 1 a. 2VŽ 1, 1 av. Ž, VŽ 1, U 1 Ž 1, 1 a... Denote q1 1, q2 1 a. By Lemma 2.1 there exit element t 1, t2 S uch that qk J, i 1, 2. Chooe an element t SUŽ t. SUŽ t. i t 1 2 i SUŽ.. Then K t K t K t. 1 2 Ž 2. Ž 2. Ž 2. ² : 2 We have V q i, K t Dq, i K t RqK i t R J ince qk i t qk i t J and Dq, Ž K 2. DqK, Ž K. R² J : i t i t t. Let 1 t 2 K t 2. Then Ž 1. Ž 1. Ž 1. Ž 1. Ž U U R a, V, V, a, V, a V, R² J: and it remain to how that VŽ, UŽ 1, 1 a.. R² J : 1 1. Linearizing the identity VŽx, xuž y.. VŽyUŽ x., y. we get the identity VŽx, xuž y, z.. VŽyUŽ x., z. VŽzUŽ x., y.. Hence Ž 1 1. Ž 1. Ž 1. V, UŽ 1, 1 a. V 1 UŽ., 1 a V Ž 1 au. Ž., Ž. Ž. Ž 1 1 Now, U U t U t K, and therefore V U, a. 1 t 1 R² J :. Similarly, Ž 1 a. UŽ. qutut JUŽ t. K, which implie The propoition i proved. 1 2 t Ž V Ž a. UŽ., R² J :. 3. CONSTRUCTION OF THE RING OF QUOTIENTS Throughout thi ection we will aume that a Jordan algebra J atifie the Ore condition with repect to a monad S J.

6 RINGS OF FRACTIONS 83 Let u how that to embed J in a ring of fraction it i ufficient to embed J in a Jordan overring Q in which all element from S are invertible. PROPOSITION 3.1. Let J Q and all element from S are inertible in Q. 1 Then the ubalgebra Q ² J, S : of Q generated by J and by all inere 1, S, i a ring of fraction of J with repect to S. Proof. Say that an element q Q ha property Ž. if for an arbitrary element S there exit an element S uch that q K K. From Lemma 2.1 it follow that element of J have property Ž.. 1 Let u how that an arbitrary invere t, t S, ha property Ž.. Chooe S. There exit an element S uch that tk K. Then t 1 K UŽt. K t K. It i clear that if element a, b Q have property Ž. then their um a b ha property Ž.. 2 Suppoe that an element a Q ha property. We will how that a ha property Ž.. Let S. There exit an element 1 S uch that a K K. 1 Similarly, there exit an element S uch that a K K K. We 1 can alo aume that K. We have Ž. Ž. a 2 R K 2 arž a. R K 2 arž K. RŽ a K. arž K. RŽ a. RŽ K. arž a K 2.. Furthermore, a K K, arž K. RŽ a. K a K. 1 Hence, a 2 K 2 K and a 2 K 2 K. From what we proved it follow that an arbitrary element from Q ² 1 J, S : atifie Ž.. By Lemma 2.1, Q i a ring of fraction of J. The propoition i proved. A Jordan algebra J give rie to a -graded Lie algebra KŽ J. KŽ J. 1 KŽ J. KŽ J. 1, which i known a the TitKantorKoecher contruction Žee 3,, 9.. Let a, b, c Ž ab. c abc bac denote the ocalled Jordan triple product of element a, b, c J. Conider two copie J, J of the vector pace J. We identify an element a J with element a in J and a in J. For arbitrary element a J, b J we Ž. define a linear operator a, b End J End J via ½ c a, b, c Ž a, b. : c b, a, c. F F

7 8 C. MARTINEZ Jordan identitie imply that for arbitrary element a, b, c, d J we have Ž. Ž. a, b, c, d a, b c, d c, a, b d, Ž. Ž. o the linear pace J, J of all operator a, b ; a, b J, i a Lie algebra. Now conider the direct um of vector pace KŽ J. J Ž J, J. J. Define a bracket, on K J via J, J J, J Ž. ; for arbi- trary element a J, b J, a, b b, a Ža, b.; for an arbitrary element x J J and for an operator Ž J, J., x Ž x. x, ; element from Ž J, J. are commutated a linear operator. A traightforward verification how that KŽ J. i a Lie algebra. Denote L, L K K K K, K K, a Lie ubalgebra of L KŽ J.. PROPOSITION 3.2. Let K K Ž and o L L. and let D : L Lbe a deriation uch that D L. Then D. Before proving Propoition 3.2 we need ome preliminary lemma. LEMMA 3.1. The centralizer of L in L i zero. Proof. Let u how that no nonzero element from J commute with K, K. If a J and a, K, K Ž. then a, K, K Ž Therefore a,, a and a,, a. Thi implie that lie in the annihilator Ann J a in the ene of 1. Since i a regular element it follow that a. Similarly, no nonzero element from J commute with K, K. Now let a J, J lie in the centralizer of L.If J, a J, a Ž. then a. Let u aume that b, a for ome element b J. By Propoition 2.1 there exit an element, S, uch that b, K, K K. Hence, b, a, K, K b, K, K, a K, a Ž., which contradict what wa proved above. Since the centralizer of L i a graded algebra we conclude that it i equal to Ž.. The lemma i proved. LEMMA 3.2. For arbitrary element a L, S there exit an element S uch that a, L L. Proof. Let aume that for given element a, b L there exit ele- ment, S uch that a, L L and b, L L. Chooe an element uch that L L L. Then a b, L a, L b, L a, L b, L L.

8 RINGS OF FRACTIONS 85 Hence we can aume that a J J, J J. It i ufficient to aume that a J J. Indeed, uppoe that for element from J J the aertion of the lemma i valid. Let a J, a J, S. Then there exit an element S uch that a, L L, b, L L. Similarly, there exit an element S uch that a, L L, b, L L. Then a, b, L a, b, L b, a, L a, L b, L L. So let a J, S. We need to prove the exitence of an element S uch that a, K K, K and a, K, K K. By Propoition 2.1 there exit an element K that annihilate a modulo K. Hence a, K, K, K a, K, K, K K, K L. 3 Since SUŽ. alo annihilate a modulo K and K 3 K, K, K, it follow that a, L 3 L. The lemma i proved. Proof of Propoition 3.2. Let L L and let D : L L be a deriva- tion uch that DŽ L.. Let u aume that there exit an element a L uch that Da. By Lemma 3.2 there exit an element S uch that a, L L and we can aume without lo of generality that L L. Then DŽ L. Ž. and Da, L a, DŽ L. DŽa, L. DŽ L. Ž.. Hence Da, L Ž., which contradict Lemma 3.1. The propo- ition i proved. For an element S let D denote the et of all derivation d : L L. Let D D S. For derivation d, d D, d d if and only if there exit an element S uch that d, d are both defined on L and d d. L L Clearly, thi i an equivalence relation. Conider the quotient et D D. Abuing notation we will denote the cla of a derivation d D a d. Let define a tructure of a vector pace on D. Conider element of D repreented by derivation d : L L, d : L L. Let, F. There exit an element S uch that L L L. Define d d : L L, d d : a da d a. For an arbitrary S the algebra L and L are -graded. We ay that a derivation d D ha degree i if džž L.. L for k 1,, 1. 2 Clearly, D Ý Ž D. i2 i. Thi -gradation induce a -gradation on D, D Ý 2 i2 D. i k ki

9 86 C. MARTINEZ Ž. LEMMA 3.3. D D. 2 2 Proof. Let d : L L be a derivation of degree 2. Then d : K J, Ž. d : K, K and d : K Ž.. We will define a linear mapping : K J via a b if and only if a d b. For arbitrary element a, a, a K we have a, a, a d a, a, a d a d, a, a a, a d, a a, a, a d a, a, a, ince a d, a and a d. Hence, a, a, a a, a, a. On the other hand, a, a, a a, a, a and o a, a, a a, a, a a, a, a a, a, a. In particular a a, a, a a, a, a Ž a. RŽ a a, a, a a, a, aa, a Ž a. RŽ a. RŽ a.. Ž Ž. Ž 2. Ž 2.. Therefore bra Ra Ra for b a. Similarly a a, a, a a, a, a Ž a. RŽ a. RŽ a a, a, a a, a, a Ž a. RŽ a. and o bra Ž Ž 6. Ra Ž 2. Ra Ž... But by the Jordan identity Ra Ž 6. 2 Ra Ž RŽa 2. Ra Ž..So ž / b RŽ a. RŽ a. RŽ a. b RŽ a. RŽ a. RŽ a ž / b 2 R a R a R a. Thi implie that brža 2. 3 brža 6. brža 2. RŽa.. If a i a regular element Ž for example, an element from S., then brža 2. UŽa 2. implie that brža 2. and brža. brža 6.. Conequently the element a annihilate b Žin the ene of 1., which implie b.

10 RINGS OF FRACTIONS 87 Ž. We have proved that S, which implie that d. We have Ž 2. L F F, J,. Thi implie that L d Ž.. In the ame way we can prove that D Ž.. The lemma i proved. Our next aim i to define a Lie bracket on D. PROPOSITION 3.3. For an arbitrary deriation d : L L there exit an element SU uch that L d L. 1 1 Let u how that it it ufficient to prove the propoition for homogeneou derivation d. Indeed, let d : L L be a derivation, d d1 d d 1, where the d i are homogeneou derivation. Suppoe that there exit element SUŽ. i, 1 i 1, uch that L di L. If i SUŽ., then L d L. 1 i1 i LEMMA 3.. Let d : L L be a homogeneou deriation of degree. Then there exit an element SU uch that L d L. Proof. Let d a, d b ; a, b J. There exit an element t SU which annihilate both element a, b modulo K. Let tuž.. We have JUŽ. JUŽ t. UŽ.. Hence, JUŽ. d JUŽ t. d,, JUŽ t., d, 2 JUŽ t.,, d JU a, JUŽ t., K, and d K Ž 2., d K. Similarly K d K. Thi implie L d L. The lemma i proved. LEMMA 3.5. Let b J, S. Then there exit an element SU uch that L Ž ad b. L. Proof. Let conider an element of degree zero, d, b. By Lemma 3. there exit t SU uch that L, d L. Let tuž. t. We need to prove that J, t, t,,, b L. Thi reduce to J, t, t,, b, J, t, t,,, b L. But J, t, t, d K by the choice of t. Then J, t, t, d, K, K L. A for the econd ummand, J, t, t K and,, b K. Hence J, t, t,,, b L. On the other hand, J, t, t,, d J, t, t, d, J, t, t, d L

11 88 C. MARTINEZ ince J, t, t, d J, t, t,,, b K, K L. The lemma i proved. LEMMA 3.6. Let d be a deriation of degree 1. Then there exit an element SU uch that L d L. Proof. Let b, d. By Lemma 3. there exit an element t SU uch that L, d t L. By Lemma 3.5 there exit an element u SU uch that L, b u L. Let SUŽ. t SUŽ u.. Then But J,,,,, d J,,,, d, J,,, d,, J,,,,, d. J,,, b L2 and J,,,, d, J,,,, d, L, ince SUŽ. t. Conequently L d L. The lemma i proved. We can prove in a imilar way the correponding reult for a derivation of degree 1. Thi finihe the proof of Propoition 3.3. Now we can define a Lie bracket on D. Let d D, d D. Chooe an element SUŽ. SUŽ.. By Propoition 3.3 there exit an ele- ment t SU uch that Ld t L and Ld t L. Define the derivation d : Lt L via xd Ž xd. d Ž xd. d, x L t. Define d, d d. It i eay to ee that with the thu defined bracket D become a graded Lie algebra, D D1 D D 1. Since char F 2, 3 the pair of pace P Ž D, D. 1 1 i a Jordan pair Žee. 7 with repect to operation d 1, d 2, d3 d 1, d 2, d3 D, 1. We will prove that the Jordan pair ŽJ, J. aociated to J can be embedded into P.

12 RINGS OF FRACTIONS 89 Define : Ž J, J. P Ž D 1, D1. via Ž a. adž a.,, a J. By Lemma 3.1 the linear tranformation i injective. Since Jordan product in Ž J, J. and P are defined via Lie bracket, it follow that i a homomorphim of Jordan pair. Let u how that for an arbitrary element S the pair ŽŽ., Ž.. i invertible. Let K JUŽ., L K K, K K. For an arbitrary element SU we have K K, L L. We will contruct two derivation q : L L, q : L L of degree 1, 1, repectively. Their retriction to L Ž Ž. Ž will define the invere of,... Ž. Ž. Let K q. For an element au K we let au q a,. Conider an element x Ýk, ŽaU. K, K i i ; let x q Ý k, a, Ý k,, a i i i i i i. We need to verify that q i well defined. An element from K can be expreed in the form au uniquely. Hence q i well defined in K. Ž. However, we need to verify that Ýi k i, au i implie Ýi k i,, a i. Ž. If Ýi au i, ki Ýi a i,,, ki then Ýi a i,,, k i,. Since ŽadŽ.. 3 and the characteritic i 3, it follow that for an arbitrary element b L we have ad ad b ad ad ad b ad Žee. 5. Hence Ý a,, k,, i i i. Since i a regular element, it implie that Ý a,, k i i i. The linear mapping q ha been well defined. Similarly, we can define a mapping q : L L of degree 1. Let prove now that q Žrep. q. i a derivation. Let e, a K, K, e K, e, a, b K. Since we know that e, e q eq, e e, e q, we only need to check: Ž. i a, b q a, q, b a, b q, Ž ii. e, e q eq, e e, e q, Ž iii. e, a q eq, a e, aq.

13 81 C. MARTINEZ Let a,,, b,,. Then a q, b a, b q,, b a,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, by Ž.. Thi prove Ž i.. In order to prove ii, let aume, by linearity, that e k, k, k, c,, and e, b,. Then e, e k, k, e, c,, k,, b, By definition of q,, c,, k,, b. e, e q c,, k, b, c,, k,, b,. On the other hand, e q k, c, k,, c and e q b,. So we have to prove c,, k,, b, k,, c,, b, k,, c,, b,. But c,, k, c,, k, c,,, k k,, c, c,, k, x, c,, k, c,,, k 2 x, k, k, where x k,, c. Hence, c,, k,, b, 2x,, b, k, k, b,. On the other hand k,, c,, b, k, k, b, x,, b, k, k, b,.

14 RINGS OF FRACTIONS 811 So we need to prove that 2x,, b, k, k,, b x,, b,, which reduce to k, k,, b x,,, b. It i ufficient to prove that k, k, x,,. But, uing, we have x,, c ad Ž. adk Ž. ad Ž. 2 Ž. 2 Ž. Ž. Ž. Ž. c ad ad k ad k ad k ad k, k,. Finally, let prove Ž iii.. Let e k, k a above and a c, d.so e, a k, a, k k, k, a. Uing Ž ii. we have e, a q k q, k, a k, a q, k k q, k, a k q, a, k k q, a, k aq, k, k k, k q, a aq, k, k eq, a aq, e. So we have proved that q i a derivation. Finally we will prove that the pair ŽŽ., Ž.. and Žq, q. are mutually invere. Denote l q, Ž.. For an arbitrary element a bu K we have a, l a, q,, b,, b, a. Furthermore, for arbitrary element x J, SU x,,, l x, l,, 2 x,, x,,, which implie x, l x,,. Hence x, l x. Similarly we conclude that x, l x. Now let d be a derivation of degree 1 defined on L Ž L. Ž L. Ž L., t S. For an arbitrary element b Ž L. we have 1 t t 1 t t 1 i t i b, d, l b, d, l b, l, d Ž i 1. b, d ib, d i 1 i 1 i 1 i 1 i 1 b, d. i 1 Ž. Hence L, d, l d. By Lemma 3.1, d, l d. t

15 812 C. MARTINEZ Similarly d, l 1 d1 for a derivation of degree 1. Arguing a above one can how that d, q, Ž. i idi for di D i, i 1,, 1. Since char F 2 it implie that the pair ŽŽ., Ž.. and Žq, q. are mutually invere. Now we can finih the proof of Theorem 1.1. Without lo of generality we will aume that the algebra J i unital. If not conider the unital hull Jˆ J F1 of J with the ame monad S J. ˆ It i eay to ee that Jˆtill atifie the Ore condition with repect to S. Let u how that ŽŽ1.,Ž1.. i an identity of the Jordan pair P Ž D, D. Žee Clearly, e Ž Ž 1.,Ž1.. i an idempotent of P. If S then we howed above that D Ž., D, Ž., 1, which implie that the whole P lie in the 1-Peirce component of e. Hence e i an identity of P. Let J be the Jordan algebra on D1 with the multiplication x1 y1 Ž. Ž x, 1, y. Then the mapping J J, a a. 1 1 i an embedding and for an arbitrary element S it image Ž. ha the invere q J. REFERENCES 1. N. Jacobon, Structure and Repreentation of Jordan Algebra, Amer. Math. Soc. Colloq. Publ., Amer. Math. Soc., Providence, N. Jacobon, K. McCrimmon, and M. Parvathi, Localization of Jordan algebra, Comm. Algebra 6 Ž 1978., I. L. Kantor, Claification of irreducible tranitively differential group, Soiet Math. Dokl. 5 Ž 1965., M. Koecher, Imbedding of Jordan algebra into Lie algebra, I, II, Amer. J. Math. 89 Ž 1967., ; 9 Ž 1968., A. I. Kotrikin, On Burnide problem, Iz. Akad. Nauk. SSSR Ser. Mat. 21 Ž 1959., J. Lambek, Lecture on Ring and Module, Blaidel, Waltham, MA, O. Loo, Jordan Pair, Lecture Note in Math., Vol. 6, Springer-Verlag, Berlin Heidelberg, L. H. Rowen, Ring Theory, Academic Pre, San Diego, J. Tit, Une clae d algebre ` de Lie en relation avec de algebre ` de Jordan, Indag. Math. 2 Ž 1962., E. I. Zelmanov, Jordan algebra with finitene condition, Algebra and Logic 17 Ž 1978., E. I. Zelmanov, Goldie theorem for Jordan algebra, Siberian Math. J. 28 Ž 1987., E. I. Zelmanov, Goldie theorem for Jordan algebra, II, Siberian Math. J. 29 Ž 1988., K. A. Zhevlakov, A. M. Slinko, I. P. Shetakhov, and A. I. Shirhov, Ring That Are Nearly Aociative, Academic Pre, New York, 1982.