A Class Op Simple Gamma Delta Rings
A CLASS OP SIMPLE GAMMA DELTA RINGS
DISSERTATION Presented in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the Graduate School of The Ohio State University
By CARL CHRISTOPHER MANERI, B. S
-Yc The Ohio State University 195>9 Approved by: Srvoiw>. kJ^uJUXct ______Adviser Department of Mathematics ii I wish to express my appreciation to Professor Erwin Kleinfeld for his encouragement and guidance in the research for this dissertation. TABLE OP CONTENTS Introduction ...... 1 Section 1: Preliminary Discussion and Definitions .... 6 Section 2: General Identities ...... 10 Section 3: Identities Involving anIdempotent ...... 18 Section ij.: Pierce Decomposition ...... 2l± Section 5>i Further Results ...... 33 Section 6: Appendix ...... 39 Bibliography ...... k 2 Autobiography ...... I4I4- iil INTRODUCTION The study of non-associative algebras began when Cayley introduced the Cayley numbers In 18J4.5 rs) • Their generalization to the Cayley-DIckson algebras gives a class of simple alternative rings. It was found by E. Kleinfeld that except for characteristic 3, these algebras are the only simple alternative rings which are not associative. For characteristic 3 he showed that the Cayley-Dickson algebras are the only simple alternative rings which are not either associative or nil £7*8,93 • The study of simple rings was initiated in 1908 when Wedderburn classified all simple finite dimensional associative algebras as either trivial algebras with all products zero or total matric algebras over a division ring Ll6]J . Semi-simple associative algebras of finite dimension are the direct sura of a finite number of simple algebras. From the above examples one can see the importance of looking at simple rings in the study of any class of rings. Simple rings are one of the fundamental building blocks of general rings. A. A. Albert introduced the study of right alternative rings in [2] . He proved that a simple finite dimensional right alternative algebra of characteristic not 2 is alternative [3} • Skornyakov proved that a right alternative division ring of characteristic not 2 is alternative[l3Q . This was later generalized by Kleinfeld when he proved that a right alternative ring of characteristic not 2, with the property that (x,y,z) = 0 implies (x,y,z) = 0, is alternative [ 6"\ . R. L. San Soucio found an example of a right alternative division ring of characteristic 2 which is not alternative tiiQ • San Soucie gives the necessary and sufficient conditions for a right alternative division ring of characteristic 2 to be alternative r u ] - In 19ij-9 Albert introduced a new class of algebras which he called almost alternative algebras (their definition as well as others to be discussed will be given in Section 1). Within the class of almost alternative algebras are what Albert called the ( Y, % ) algebras. Albert proved in Cl] that practically all of the almost alternative algebras are equivalent to either right alternative algebras or to (Yj £ ) algebras ( a more precise form of this statement appears in Section 1). This equivalence preserves ideals, nilpotence of elements and therefore preserves nil ideals. Thus one can get information about the general almost alternative algebras from a study of ( Yj ^ ) algebras. Some of the results on ( YjS ) algebras and rings will now be discussed. Kokoris showed that a simple ( £ ) algebra with £ ^ 0,1 and characteristic not 2, 3 or 5> is either associative or possesses an absolutely primitive unity element r i a l . Kleinfeld showed that an idempotent of a simple ( £) ring R, with i f 0,1, which is of characteristic not 2 or 3 and which is not associative is the unity element of R. This appears as the Appendix of [ 13 *) . Kokoris in proved that an idempotent e in a simple (1,1) ring R of characteristic not 2 is either the unity of R or R is associative. In a paper on (1,1) algebras i ih , Kleinfeld showed that simple (1,1) rings which are not associative and have characteristic not 2 or 3 possess no proper left ideals C ul . He also showed in this paper that a (1,1) ring of characteristic not 2 or 3 which has no proper right ideals is associative. His main result in this paper is that simple (1,1) algebras of characteristic not 2, 3 t or 5 are associative. In this same paper Kleinfeld showed that a ( V> £ ) ring R, with £ ^ 0,1, possesses no proper left or right ideals if R is simple and of characteristic not 2 or 3 and is not associative. The principal result in this dissertation is that an idempotent e of a simple (-1,1) ring R is the unity element of R if R is simple, not associative and of characteristic not 2 or 3. This eliminates one class of simple (-1,1) rings. One might conjecture that all simple (-1,1) rings (with the properly chosen characteristic) are associative. We shall also show that, if R is a simple not associative (-1,1) ring of characteristic not 2 or 3, then R possesses no proper left ideals. The attempt to do the same for right ideals was unsuccessful. The results of Albert on right alternative algebras tell us that a simple (-1,1) algebra of characteristic not 2 or 3 is associative (this is discussed in Section 1). These results can be summarized as follows: 1. Any simple ( Vj % ) ring which possesses an idempotent is either associative or that idempotent is the unity of the ring. 2. Any simple ( £ ) algebra with $ = 0,1 is associative. 3- Any simple (Vj &) ring cannot contain proper left ideals and proper right ideals at the same time, unless it is associative. If & ^ 0,1 it can contain neither, unless it is associative. Section 1 gives background material and definitions. In Section 2 we derive some identities in an arbitrary (-1,1) ring. Section 3 brings in the Idempotent and simplicity. Section ij. on the Pierce decomposition contains the main result. Section 5 is the result on left Ideals mentioned above. Section 6 contains an example of a (-1,1) algebra which possesses an Idempotent not the unity. 5 The entire dissertation is self contained except for the proof of identity (28). This requires two lemmas from Kleinfeld1s paper on right alternative rings 16^ . * 1. PRELIMINARY DISCUSSION AND DEFINITIONS Since we are dealing with, non-associative rings, we define the associator of x,y,z, where x,y,z are elements of our ring, to be (x,y,z) = (xy)z - x(yz). We also define the commutator to be (x,y) = xy - yx. The alternative laws are (x,x,y) « 0 and (y,x,x) = 0. The first is the left alternative law while the second is the right alternative law. An alternative ring is one which satisfies both of these laws. The flexible law is (x,y,x) = 0. A ring or algebra is defined to be of type ( Yj £ ) if the following identities hold: (1) A(x,y, z) sa (x,y, z) + (y,z,x) + (z,x,y) = 0, (2) (z,x,y) + Y(x,z,y) + S(y,z,x) = 0, with y 2 - S2 + £ = 1. Albert defined an almost left alternative algebra to be a finite dimensional algebra A over a field F of characteristic not 2 with the following properties: I. For fixed T-,T\ * F 6 7 (3) z(xy) = o((zx)y + Q (zy)x + V ( xz)y + &(yz)x + 6 y( zx) +t^x(zy) + 0-y(xz) +Tx(yz), for any x,y,z in A. II. (x,x,x) = 0 for x ^ A . III. There exists an algebra B over F such that B is not commutative, possesses a unity element and satisfies identity (3). An algebra A over F is called almost right alternative if I, II and III are valid in A but with identity (3) replaced by an identity of the same form except that z(xy) is replaced by (xy)z. An algebra A over F is called almost alternative if it is both almost left alternative and almost right alternative. For algebras over a given field F, of characteristic not 2, an equivalence relation is defined in the following way. Let A be an algebra over F. Let B be the algebra over F consisting of the same vector space as A but with a different multiplication. In other words A and B are identical as vector spaces. Let x,y£A. Then xy will denote the product in A and x*y will denote the product in B. A and B are called quasiequivalent if the identity x*y = ^xy + (l- *X)yx holds for arbitrary x , y 6 A and with a fixed > 6 F where X t i. Albert shows in E 1} that if A is an almost alternative algebra which is not flexible with vl + P / 3/Kt then there is a finite extension K of P such that the algebra ak , where AK is the algebra over K obtained by enlarging the base field from P to K, is quasiequivalent with ^ ^ K to an algebra B where B is a left alternative algebra or is a ( & ) algebra. We can note very easily that ideals are preserved by quasiequivalence as well as nilpotence of an element. Hence nil ideals are preserved. In a (-1,1) ring (2) reduces to (z,x,y) - (x,z,y) + (y,z,x) = 0. If this is subtracted from (1) we get (i+) B(x,y,z) = (x,y,z) + (x,z,y) = 0. If we set y = x in (z,x,y) - (x,z,y) + (y,z,x) = 0, we get (5) (Z,x,x) = 0. Thus in a (-1,1) ring the right alternative law is valid, and all theorems about right alternative rings hold for (-1,1) rings. Albert has proved that a semi-simple right alternative algebra of characteristic not 2 is alternative . Prom this we can conclude that a semi-simple (-1,1) algebra A of characteristic not 2 Is alternative. Prom , since A is alternative, we have the identities (x,y,z) = (y,z,x) = (z,x,y). If A is also of characteristic not 3, A(x,y,z) = 3(x,y,z) - 0 Implies that A is associative. A simple (-1,1) algebra which is not nil is semi- simple, hence associative if the characteristic is not 2 or 3- In an algebra, the characteristic is always the same as that of the field. For rings the situation is different. Elements In a ring may be of different addi tive order. However when we say that a ring R has characteristic not n we mean that there are no elements of R which have additive order n. In a simple ring all non-zero elements have the same additive order (see Appendix). Thus In a simple ring there Is a uniquely defined characteristic. Kleinfeld*s result that a right alternative ring of characteristic not 2 which has no nilpotent elements Is alternative will now be applied to (-1,1) rings. We have that a (-1,1) ring of characteristic not 2 or 3 which has no nilpotent elements is alternative hence associative by the above remarks. 2. GENERAL IDENTITIES We shall assume R is a (-1,1) ring, that Is R satisfies identities (1), (2^) and (5). Let x,y,z,w be arbitrary elements in R. In an arbitrary ring we have the identity, (xy,z) - x(y,z) - (x,z)y - (x,y,z) + (x,z,y) - (z,x,y) = 0. If we subtract B(x,y,z) = 0 from this we get (6) C(x,y,z) - (x y ,z) - x(y,z) - (x,z)y - 2(x,y,z) - (z,x,y) = 0. In an arbitrary ring we have (7) F(w,x,y,z) = (wx,y,z) - (w,xy,z) + (w,x,yz) - w(x,y,z) - (w,x,y)z « 0. Since R satisfies (1), the following identity, 0 = F(x,y,z,w) - F(y,z,w,x) + F(z,w,x,y) - F(w,x,y,z) = (xy,z,w) - (x,yz,w) + (x,y,zw) - x(y,z,w) - (x,y,z)w - (yz,w,x) + (y,zw,x) - (y,z,wx) + y(z,w,x) + (y,z,w)x + (zw,x,y) - (z,wx,y) + (z,w ,xy) - z(w,x,y) - (z,w,x)y - (wx,y,z) + (w,xy,z) - (w,x,yz) + w(x,y,z) + (w,x,y)z = A(xy,z,w) + A(zw,x,y) - A(yz,w,x) - A(wx,y,z) + (w,(x,y,z)) - (x,(y,z,w)) + (y,(z,w,x)) - (z,(w,x,y)), reduces to 10 11 (8) G(w,x,y,z) = (w,(x,y,z)) - (x,(y,z,w)) + (y,(z,w,x)) - (z,(w,x,y)) = 0. The following three identities in R are a conse quence of the right alternative law if R is assumed to have characteristic not 2, and are proved in £6} . However for the benefit of the reader they are proved here. (9) H(x,y,z) = (x,y,yz) - (x,y,z)y = 0. (10) J(x,w,y,z) = (x,w,yz) + (x,y,wz) - (x,w,z)y - (x,y,z)w = 0. (11) K(x,y,z) = (x,y2,z) - (x,y,yz + zy) = 0. To prove (9) we compute 0 = F(x,y,y,z) - F(x,z,y,y) + F(x,y,z,y) = (xy,y,z) - (x,y2,z) + (x,y,yz) - x(y,y,z) - (x,y,y)z - (xz,y,y) + (x,zy,y) - (x,z,y2) + x(z,y,y) + (x,z,y)y + (xy,z,y) - (x,yz,y) + (x,y,zy) - x(y,z,y) - (x,y,z)y = B(xy,y,z) - B(x,y2,z) - xB(y,y,z) + B(x,zy,y) + (x,z,y)y - (x,y,z)y + (x,y,yz) - (x,yz,y) = (x,z,y)y - (x,y,z)y + (x,y,yz) - (x,yz,y) - B(x,z,y)y + B(x,yz,y) = 2(x,y,yz) - 2(x,y,z)y. Thus (9) results if the ring has characteristic not 2. Note that B(x,y,z) = 0 results from the right alternative law (x,y,y) = 0 by replacing y with y+z. Identity (10) results from (9) by replacing y with y+w. 12 Now 0 = F(x,z,y,y) = (xz,y,y) - (x,zy,y) + (x,z,y2 ) - x(z,y,y) - (x,z,y)y = (x,z,y2) - (x,zy,y) - (x,z,y)y + B(x,z,y)y + H(x,y,z) - B(x,z,y2) + B(x,zy,y) = (x,y,yz) + (x,y,zy) - (x,y2,z) - - K(x,y,z). In the following we shall assume R is a (-1,1) ring (I.e. satisfies (1) and (Ij.) ) and has characteristic not 2 or 3. We compute G(x,x,x,y) + (x, B(x,y,x)) = (x,(x,x,y)) - (x,(x,y,x)) + (x,(x,y,x)) + (x,(x,x,y)) = 2(x,(x,x,y))=0. Thus (12) (x,(x,x,y)) = 0 , (x,(x,y,x)) = 0. The second Identity in (12) follows from (x,B(x,y,x)) = 0. From G(x,y,x,y) = (x,(y,x,y)) - (y,(x,y,x)) + (x,(y,x,y) - (y»(x,y,x)) = 2(x,(y,x,y)) - 2(y,(x,y,x)) = 0, we get (x,(y,x,y)) - (y,(x,y,x)) = 0. Thus (x,B(y,x,y)) - (y,B(x,y,x)) - (x,(y,x,y)) + (y,(x,y,x)) = (x,(y,y,x)) - (y/(x,x,y)) = 0. On the other hand G(x,y,y,x) = (x,(y,y,x)) + (y,(x,x,y)) = 0. Combining these last two we get 2(x,(y,y,x)) = 0 and this gives (13) (x,(y,y,x)) = 0. If we replace x with x+z in (13) w© g©t (111) L(x,y,x) = (x, (y,y, z)) + (z,(y,y,x)) = 0. If we replace y in (13) with y+z we get (15) M(x,y,z) = (x,(y,z,x)) + (x,(z,y,x)) = 0. 13 We use (12) in the following, and the fact that (x,x,x) = 0 which follows from the right alternative law. F(x,x,x,y) = (x2 ,x,y) - (x,x2 ,y) + (x,x,xy)- x(x,x,y) = (x2 ,x,y) - (x,x2 ,y) + (x,x,xy) - (x,x,y)x - (x,(x,x,y)) = (x2,x,y) - (x,x2,y) + H(x,x,y) = (x2 ,x,y) - (x,x2,y) = 0. Thus (x2,x,y) = (x,x ,y). Hence M(y,x2,x) = (y,(x2,x,y)) + (y,(x,x2,y)) = 2(y,(x2,x,y))=0. Thus (16) (y, (x2 ,x,y)) = 0. We now use B(y,x2,x) - H(y,x,x) = (y,x2,x) + (y,x,x2) - (y,x,x2) + (y,x,x)x = (y,x2,x) = 0 along with (16) to get ^(y»y>x2,x) = (y,(y,x2,x)) - (y,(x2,x,y)) + (x2,(x,y,y)) - (x,(y,y,x2)) + L(x,y,x2) = (x2,(y,y,x)) = 0. (17) (x2 ,(y,y,x)) = 0. The identity (xy,z) + (yz,x) + (zx,y) = (x,y,z) + (y,z,x) + (z,x,y) holds in any ring; hence in a (-1,1) ring we have (18) D(x,y,z) = (xy,z) + (yz,x) + (zx,y) = 0, as a consequence of (1). p Using (13) and (17)» which say that x and x commute with (y,y,x), we have D(x,x,(y,y,x)) = (x2,(y,y,x)) + (x(y,y,x),x) + ((y,y,x)x,x) = 2((y,y,x)x,x) = 0, and thus ((y,y,x)x,x) = 0. Now we observe o = -((y,y,x)x,x) + (B(y,y,x)x,x) = ((y,x,y)x,x) = ((y,x,y)x,x)'+ (H(y,x,y),x) = ((y,x,xy),x). Thus (19) (x,(y,x,xy)) = 0. Replacing y by y+w in (19) we get (x,(y,x,xy)) + (x,(y,x,xw)) + (x,(w,x,xy)) + (x,(w,x,xw)) = (x,(y,x,xw)) + (x,(w,x,xy)) = 0. (20) (x,(y,x,xw)) + (x,(w,x,xy)) = 0. From L(x,y,x2 ) = (x,(y,y,x2)) + (x2 ,(y,y,x)) - 0, and (17) we get (x,(y,y,x2)) = 0. Now 0 = (x,B(y,y,x2)) - (x,(y,y,x2)) = (x,(y,x2,y)) = (x,(y,x2,y)) - (x,K(y,x2,y)) = (x,(y,x,xy+ yx)) = (x,(y,x,yx)) using (19). Now we have (x,B(y,x,yx)) - (x,(y,x,yx)) = (x,(y,yx,x)) = 0. We also have from this 0 = -M(x,y,yx) + (x,(y,yx,x))= -(x,(y,yx,x)) - (x,(yx,y,x)) + (x,(y,yx,x)) = (x,(yx,x,y)). (21) (x,(yx,x,y)) = 0. Now (21), (x,(y,yx,x)) = 0 and (x,A(y,yx,x)) = 0 give (22) (x,(x,y,yx)) = 0. Replacing y with y+w in (21) we get (23) (x,(wx,x,y)) + (x,(yx,x,w)) = 0. A similar substitution in (22) gives (2i|) (x, (x,w,yx)) + (x,(x,y,wx)) = 0. We now use F(w,x,x,y) = (wx,x,y) - (w,x2 ,y) + (w,x,xy)~ w(x,x,y) = 0 to calculate w(x,x,y). w(x,x,y) = (wx,x,y) - (w,x2,y) + (w,x,xy) + K(w,x,y) = (wx,x,y) - (w,x,xy) - (w,x,yx) + (w,x,x.y) = (wx,x,y) - (w,x,yx). Hence (x,w(x,x,y)) = (x,(wx,x,y)) - (x ,(w ,x,yx)) + (x,B(w,x,yx)) = (x,(wx,x,y)) + + (x,(w,yx,x)) - M(x,w,yx) = (x,(wx,x,y)) - (x ,(yx,w ,x)) + (x,B(yx,w,x)) = (x,(wx,x,y)) + (x ,(yx jX,w)). Thus from (23) we have (2$) (x,w(x,x,y)) = 0. The next step is to use F{x,y,x,w) + F(x,y,w,x) = (xy,x,w) - (x,yx,w) + {x,y,xw) - x(y,x,w) - (x,y,x)w + (xy,w,x) - (x,yw,x) + (x,y,wx) - x(y,w,x) - (x,y,w)x - 0 to compute (x,y,x)w. This gives (x,y,x)w = (xy,x,w) - (x,yx,w) + (x,y,xw) » x(y,x,w) + (xy,w,x) - (x,yw,x) + (x,y,wx) - x(y,w,x) - {x,y,w)x = B(xy,x,w) - xB(y,x,w) - (x,yx,w) + (x,y,xw) - (x,yw,x) + (x,y,wx) - (x,y,w)x - - (x,yx,w) + (x,y,xw) - (x,yw,x) + (x,y,wx) - (x,y,w)x + B(x,yx,w) = (x,w,yx) + (x,y,wx) - (x,yw,x) + (x,y,xw) - (x,y,w)x. We note that x commutes with (x,w,yx) + (x,y,wx) from (2l|). We also note that x commutes with - (x,yw,x) by (12). We must reduce (x,y,xw) - (x,y,w)x further. (x,y,xw) - (x,y,w)x = (x,y,xw) - (x,y,w)x + A(x,y,w)x = (x,y,xw) + (y,w,x)x + (w,x,y)x - B(y,w,x)x = (x,y,xw) - (y,x,w)x + (w,x,y)x - H(y,x,w) + H(w,x,y) = (x,y,xw) - (y,x,xw) + (w,x ,xy) + B(y,x,xw) = (x,y,xw) + (y,xw,x) + (w,x,xy) - A(x,y,xw) = (w,x,xy) - (xw,x,y). Now (x,(w,x,xy) - (xw,x,y)) = (x,(w,x,xy) - (xw,x,y)) + (x,B(xw,x,y)) = (x,{w,x,xy)) + (x,(xw,y,x)) - M(x,xw,y) = (x ,(w ,x ,xy)) - (x,(y,xw,x)) + (x,B(y,xw,x)) = (x,(w,x,xy)) + (x,(y,x,xw)) and this is zero by (20). Thus (x,(x,y,x)w) = 0 and from B(x,y,x) = 0 we get (26) (x,(x,x,y)w) = 0. Now 0 = D((x,x,y),w,x) - D{w, (x,x,y),x) = ((x,x,y)w,x) + (wx,(x,x,y)) + (x(x,x,y),w) - (w(x,x,y),x) - (xw,(x,x,y)) - ((x,x,y)x,w) = ((w,x),(x,x,y)) using (25), (26) and (12). From this we get L(y,x,(w,x)) - ((w,x),(x,x,y)) = (y,(x,x,(w,x))) + ((w,x),(x,x,y)) - ((w,x),(x,x,y)) = (y,(x,x,(w,x))) = 0, or (27) (y, (x,x, (x,w))) =5 0. We shall now use some results from C6 3 . We define the class S(a,b) in R, for fixed a and b in R, in the following way. S(a,b) is the class of elements x in R which have the property that (x,a,b) = x(b,a). From Lemma of we have (x,x,y) , (x,x,y)x £ S(x,y). Now we have from (12) that (x,x,y)x = x(x,x,y). Thus 17 x(x,x,y) £ S(x,y). Now Lemma 3 of 163 says that since ^ (x,x,y) and x(x,x,y) both belong to S(x,y), we have (28) (x,x,y)(x,x,y) = 0. Let U be the class of elements u in R such that (u,R) = 0. Let u6U. Then C(x,x,u) =* (x2,u) - x(x,u) - (x,u)x - 2(x,x,u) - (u,x,x) = - 2(x,x,u) = 0. Thus since we have characteristic not 2, (x,x,u) - 0 and from B(x,x,u) = 0 we have (x,u,x) = 0. (29) (x,x,u) = (x,u,x) = 0 for u^ U . Replacing x by x+y in each identity of (29) gives (30) (x,y,u) - - (y,x,u) and (x,u,y) = - (y,u,x) for u€U. 3. IDENTITIES INVOLVING AN IDEMPOTENT Let e be an idempotent of R, that is = e / 0. Then 0 = (x,K(e,e,y)) ■ (x,(e,e,y)) - (x,(e,e,ey)) - (x,(e,e,ye)) = (x,(e,e,y)) - 2(x,(e,e,ey)) using (27). We also use (12), i.e. the fact that (e,(e,e,y)) = 0, to observe that G(e,e,(e,e,y)) = (e,(e,e,y)) - o(e,(e,e,y)) - (e,(e,e,y))e - 2(e,e,(e,e,y)) = - 2(e,e,(e,e,y)) = 0. Thus (e,e,(e,e,y)) = 0. Therefore we can replace y by ey in (x,(e,e,y)) - 2(x,(e,e,ey)) = 0 to get (x,(e,e,ey)) - 2(x,(e,e,ey)) = - (x,(e,e,ey)) - 0, and this gives (31) (x,(e,e,y)) = 0. From (28) we have (e,e,y)^ = 0, and we replace y by y+w to get (e,e,y)(e,e,w) + (e,e,w)(e,e,y) = 2(e,e,y)(e,e,w) = 0 as a result of (31). Hence (32) (e,e,y)(e,e,w) = 0. Let us now substitute u for (e,e,y) and use F(e,a,x,u) = (e,x,u) - (e,ex,u) + (e,e,xu) - e(e,x,u) - (e,e,x)u = 0. We note from (32) that (e,e,x)u = 0. We also have (e,e,xu) = 0 from C(e,e,xu) = 0 and the fact that (e,xu) = 0 which follows from (2£) (characteristic not 2 is used also). Thus we have (33) (e,x,u) = (e,ex,u) + e(e,x,u). 18 19 In order to use (33) we need three observations. The first is that (e,(e,x,y)) = 0 . This follows from G(e,e,x,y) = (e,(e,x,y)) - (e,(x,y,e)) + (x,(y,e,e)) - (y#(e»®*x)) “ (e,(e,x,y)) - (e,(x,y,e)) = 0 since (y,(e,e,x)) = 0 by (31). Now since (e,(e,x,y)) = (e,(x,y,e)), 0 = (e,A(e,x,y)) = (e,(e,x,y)) + (e,(x,y,e)) + (e,(y,e,x)) = 2(e,(x,y,e)) + (e,(y,e,x)) - (e,B(y,e,x)) = 2(e,(x,y,e)) - (e,(y,x,e)) + M(e,y,x) = 2(e,(x,y,e)) - (e,(y,x,e)) + (e,(y,x,e)) + (e,(x,y,e)) = 3(e,(x,y,e)). Hence (e,(x,y,e)) = 0 and thus (e,(e,x,y)) = 0. This reduces (33) to (3U) (e,x,u) = (e,ex,u) + (e,x,u)e. The second observation is that (e,x,u)e = (e,ex,u). This follows first from the fact that u fcU and thus A(e,x,u) = (e,x,u) + (x,u,e) + (u,e,x) = (e,x,u) - (e,u,x) + (u,e,x) = (e,x,u) - (e,u,x) + (u,e,x) + B(e,u,x) = 2(e,x,u) + (u,e,x) = 0. Similarly A(e,ex,u) = 2(e,ex,u) + (u,e,ex) = 0. Now A(e,x,u)e = 2(e,x,u)e + (u,e,x)e + H(u,e,x) = 2(e,x,u)a + (u,e,ex) = 0. Subtracting this from A(e,ex,u) = 2(e,ex,u) + (u,e,ex) = 0 gives 2(e,x,u)e - 2(e,ex,u) = 0. Thus for characteristic not 2 we have (e,x,u)e = (e,ex,u) and (3U) becomes (35) (e,x,u) = 2(e,ex,u). 20 The third observation is that any associator with two elements from U in it is zero. The identities in (30) give A(x,ui,U2) = (xjU^Ug) + (u-^u^x) + (u2,x,u1) = (x,u1#u2) - (x,^,^) - (x,U2,ux) + 2B(x,u2,u1) - 3(x,u1,u2) = 0. Thus (x,u1#u2) = 0. Now since (e,e,x)^U and u £ U we have (e,(e,e,x),u)=0 and (e,e(ex),u) =* (e,ex,u). We can now replace x in (35) by ex to get (e,ex,u) = 2(e,ex,u) and thus (e,ex,u) = 0 . Prom (35) we then get (36) (e,x,u) = 0 . Since A(e,x,u) = 2(e,x,u) + (u,e,x) = 0 we have (u,e,x) = 0 and thus A(e,x,u) = (e,x,u) + (u,e,x) + (x,u,e) = 0 gives (x,u,e) = 0 . (37) (u,e,x) = (x,u,e) = 0 . Since ue = (e,e,y)e + H(e,e,y) = (e,e,ey)6 U by (31) we have C(ue,x) ~ (ue,x) - u(e,x) - (u,x)e - 2(u,e,x) - (x,u,e) = 0 = u(e,x) = 0 by (37). Thus using (31) again (38) (e,e,y)(e,x) = (e,x)(e,e,y) « 0 . We now use J(e,w,e,(e,x)) - (e,w,e(e,x)) + (e,e,w(e,x)) - (e,w,(e,x))e - (e,e,(e,x))w = 0 to compute (e,e,(e,x))w - (e,e,w(e,x)). This gives (e,e,(e,x))w - (e,e,w(e,x)) =» (e,w,e(e,x)) - (e,w,(e,x))e - A(e,w,e(e,x)) + A(e,w,(e,x))e = - (w,e(e,x),e) - (e(e,x),e,w) + (w,(e,x),e)e 21 + ((e,x),e,w)e + B(w,e(e,x),e) - B(w,(e,x),e)e = (w,e,e(e,x)) - (w,e,(e,x))e - (e(e,x),e,w) + ((e,x),e,w)e = H(w,e,(e,x)) - (e(e,x),e,w) + ((e,x),e,w)e = ((e,x),e,w)e - (e(e,x),e,w). Continuing with this calculation we have ((e,x),e,w)e - (e(e,x),e,w) + H((e,x),e,w) = ((e,x),e,ew) - (e(e,x),e,w) = ((e,x),e,ew) - (e(e,x),e,w) - (C(e,e,x),e,w) = ((e,x),e,ew) - (e(e,x), e,w) - ((e,x),e,w) + (e(e,x),e,w) + ((e,x)e,e,w) + 2((e,e,x),e,w) = ((e,x)e,e,w) - ((e,x),e,w) + ((e,x),e,ew) because of (37) which says ((e,e,x),e,w) = 0. Now 0 = P((e,x),e,e,w) = ((e,x)e,e,w) - ((e,x),e,w) + ((e,x),e,ew) - (e,x)(e,e,w) = ((e,x)e,e,w) - ((e,x),e,w) + ((e,x),e,ew) because of (38). Thus (39) ( e,e,w(e,x)) = (e,e,(e,x))w. LEMMA. 1. Let R be a (-1,1) ring of characteristic not 3. Then if R is commutative it is also associative. PROOF. For a commutative (-1,1) ring C(y,x,x) = - (x,y,x) = 0. Thus (x,y,x) = 0. Replacing x by x+z, this gives (x,y,z) = - (z,y,x). From B(x,y,z) = 0 we have (x,y,z) = - (x,z,y). Thus 0 = A(x,y,z) = (x,y,z) + (y,z,x) + (z,x,y) = (x,y,z) - (x,z,y) - (z,y,x) = (x,y,z) + (x,y,z) + (x,y,z) = 3(x,y,z). Hence (x,y,z) = 0 and R is associative. 22 LEMMA 2. Let R be a simple not associative ring of type (-1,1) of characteristic not 2 or 3* Let S = ^sfeR|s6U and syfcU for any y6Rj . Then 3 = 0. PROOF. Let stS and x,y£ R. Then using the fact that s and sxfcU and the identities in (30) we compute the following: (sx)y = (s,x,y) + s(xy) and (sx)y = y(sx) = y(xs) = - (y,x,s) + (yx)s = (x,y,s) + (yx)s. Combining these, we have 3((sx)y) = (s,x,y) + 2(x,y,s) + s(xy) + 2((yx)s) = (s,x,y) + 2(x,y,s) + s(xy + 2yx). Now since A(s,x,y) = (s,x,y) + 2(x,y,s) = 0, 3((sx)y) = s(xy+2yx)£U. Now the class of elements T = ^3x| x€ rJ forms an ideal in any ring. Thus in R, T is an ideal and T ^ 0 since R is of characteristic not 3. Thus T = R. Since 3(sx)y = (sx)(3y), we have the fact that an arbitrary multiple of sx is in U. Thus sx = xs fcS, and S is an ideal of R. By Lemma 1, S ^ R since S is commutative. Hence S = 0. Now since (e,e,(e,x))£ U by (31) and since (e,e,(e,x))w = (e,e,w(e,x)) by (39), we have (e,e,(e,x) )w*.U. Thus (e,e,(e,x))€ S. Now if we take the hypothesis of Lemma 2 for our ring R, we have UO) (e,e,(e,x)) = 0. 23 Now K(e,e,x) = (e,e,x) - (e,e,ex+xe) = (e,e,x) - 2( e,e,ex) = 0 because of (2+0) . We can replace x by ex in this last equation, observing again that (e,e,(e,e,x)) = 0, to get (e,e,ex) - 2(e,e,ex) = - (e,e,ex) =s o. Thus (e,e,x) = 0 and because of B(e,e,x) = 0 we have (l+l) ( e, e,x) = (e,x,e) = 0. 4. PIERCE DECOMPOSITION We assume R is a (-1,1) ring which is simple, not associative, of characteristic not 2 or 3, We also assume R has an idempotent e. Then Identity (ij.1) and the right alternative law permit us to write x = exe + (ex-exe) + (xe-exe) + (x-ex-xe+exe) where x is an arbitrary element of R. This is just the Pierce decomposition with: R-^ = ^ x 6R | ex = xe = x J R10 = £ x6 R \ ex = x , xe = oj R0i = £ x € R I ex = 0 , xe = x] Rqq = (xfeR \ ex = xe = 0 J , since exe£R-j^> ex - exefiR^, xe - exe 6 Rq-j_ and x - ex - xe + exe 6 Rq q . We can compute the multiplication table for these subspaces as follows: w o o R01 R10 R11 Roo R00 R01 0 0 R01 0 Roo* Roo R01 H o R10 R11 Rll* 0 0 R11 0 R10 R11 2k 25> The asterisks denote the two cases where this multi plication differs from that of an associative ring. We will now verify these 16 entries. 1. x 6 Rqo * 76 R00 It is obvious directly that (x,e,y) = 0 and that (y,e,x) = 0. Prom B(x,e,y) = 0 we see that (x,y,e) = 0. Thus from A(y,e,x) = 0 we get (e,x,y) = 0. The two important identities are (x,y,e) = 0 and (e,x,y) = 0. These give (xy)e = x(ye) = 0 e(xy) = (ex)y = 0. Hence xy6 RQ0, and R00R00 — R00* 2. x 6 Rqq , y 6 RQ1 We compute directly that (x,e,y) = 0 . We also have K(y,e,x) = (y,e,x) - (y,e,ex+xe) = (y,e,x) =0. Now B(x,e,y) = 0 and B(y,e,x) = 0 give (x,y,e) - 0 and (y*x,e) = 0. Finally A{x,y,e) = 0 gives (e,x,y) = 0. These computations give e(xy) = (ex)y « 0 (xy)e = x(ye) = xy Hence xy£ R01, and Rq 0R0i ^ R01* Now (y,e,x) = 0 gives (ye)x = y(ex) and yx = 0. Hence yx = 0 and Rq ^Rqq ~ 0* 26 3. x * r q0 * y t R io' We see directly that (y,e,x) = 0. Thus A(e,x,y) - B(x,y,e) = (e,x,y) + (x,y,e) + (y,e,x) - (x,y,e) - (x,e,y) = (e,x,y) - (x,e,y) = 0, and (ex)y - e(xy) - (xe)y + x(ey) = - e(xy) + xy = 0 and e(xy} = xy. Prom (y,e,x) = 0 and B(y,e,x) = 0 we have (y,x,e) = 0, so that A(e,y,x) » (e,y,x) + (x,e,y) = 0. This gives (ey)x - e(yx) + (xe)y - x(ey) = yx - e(yx) - xy = 0. Multiplying this last by e on the left we get e(xy) = 0. Hence xy = 0 and Rqqr io = Now since xy = 0 we have yx - e(yx) = 0 and e(yx) = yx. Since (y,x,e) = 0, (yx)e = 0. Hence yx6R10 and R ^ R ^ C R ^ . it. x £ Rqq * K(x,e,y) - (x,e,y) - (x,e,ey+ye) = (x,e,y) - 2(x,e,y) = - (x,e,y) = 0. Thus (x,e,y) = (xe)y - x(ey) = - xy = 0, and xy = 0. Hence R ^ R ^ = 0. K(y,e,x) = (y,e,x) - (y,e,ex+xe) = (y,e,x) = 0. Thus (ye)x = y(ex) and yx = 0. Hence Ru Hqq = 0. 5. x ^ R qi , y ^ R 0i. H(x,e,y) - (x,e,ey) - (x,e,y)e = - (x,e,y)e = 0. It was observed just before identity (3^) that 27 (e,(x,y',e)) = 0. Using B(x,y,e) = 0 we have (e,(x,e,y)) = 0. Thus e(x,e,y) = (x,e,y)e = 0 and (x,e,y)aoo. But (x,e,y) = (xe)y - x(ey) = xy£RQ0. Thus H01H01^ R00* 6. x < B q1 , y6R10. Clearly (x,e,y) = 0 and (y,e,x) = 0 from direct computation. B(x,e,y) = 0 and B(y,e,x) = 0 give (x,y,e) = 0 and (y,x,e) » 0 respectively. Then A(x,y,e)=0 and A(y,x,e) = 0 give (e,x,y) = 0 and (e,y,x) = 0. From these we have e(xy) = (ex)y = 0 and (xy)e = x(ye) = 0 and hence R01R1 0 - R00* We also have e(yx) = (ey)x = yx and (yx)e = y(xe) - yx. Hence R]_o^oi^ R11 • 7. xfeR01 , y£ R11. We compute directly that (x,e,y) = 0. B(x,y,e) = 0 gives (x,y,e) = 0. Thus (xy)e = x(ye) = xy. Now H(y,e,x) = (y,e,ex) - (y,e,x)e = - (y,e,x)e = 0. Since A(e,x,y) = (e,x,y) + (y,e,x) = 0 we have (e,x,y)e = 0. Since (e,(e,x,y)) = 0, this gives e(e,x,y) = 0. Now e(e,x,y) = e((ex)y - e(xy)) = e(-e(xy)) » - e(xy) = 0. Thus e(xy) = 0. Hence R01 * 28 Since e(xy) = 0, we have (e,x,y) = (ex)y - e(xy) = 0. Combining this with A(e,x,y) = (e,x,y) + (y,e,x) = 0 gives (y,e,x) = 0 and (ye)x = y(ex) or yx = 0. Hence R n Roi " 8. X feH10 , y£R1(). H{x,e,y) = (x,e,ey) - (x,e,y)e = (x,e,y) - (x,ety)e = 0. Thus (x,e,y)e = (x,e,y). Since (e, (x, e,y)) = 0, e(x,e,y) = (x,e,y) and (x,ej)6 Hn , (x,e,y) = (xe)y - x(ey) = - *y<=Ri:L‘ Thus xy^Rlx and R10Ri0€.Rii* 9. xiE-^Q , y Hence yx6R10 and RXXRX0 ~ R10* 10. x RX1 , y ^ R xl (x,e,y) = 0 and (y,e,x) = 0 by direct calculation. B(x,e,y) = 0 gives (x,y,e) = 0 and thus A(x,y,e) = 0 gives (e,x,y) = 0, Hence 29 e(xy) = (ex)y = xy and (xy)e = x(ye) = xy and Eh H ^ C r ^ , Prom the multiplication table we observe that for any x,y of R (i+2) (e,x,y) = 0. This follows from the fact that R = R11 + R10 + R01 + Rq q » tilat is x = X11 + x10 + X01 + x00 and y = J n + y10 + y0i + y00 wher9 x ij> Hence (e,x,y) breaks up into 16 associators all of which are zero. Two typical examples are (e’x01’y01) = (n,x,y) = (x,n,y) = (x,y,n) = 0 for arbitrary x and y in R. LEMMA i|. Let R be a (-1,1) ring which is simple, not associative and of characteristic not 2 or 3. bet n be an element of R which satisfies the identity (n,x,y) = 0. Then n is in the center of R. PROOF. M(x,n,y) = (x,(n,y,x)) + (x,(y,n,x)) - (x,{y,n,x)) = 0, since (n,y,x) = 0. Thus from (x,A{n,x,y)) = 0 we get (x,(x,y,n)) = 0. Combining (x,(y,n,x)) = 0 and (x,(x,y,n)) = 0 with G(x,y,n,x) = (x,(y,n,x)) - (y,(n,x,x)) + (n,(x,x,y)) - (x,(x,y,n)) = 0 gives (n,(x,x,y)) = 0. 31 Replacing x by x+z in this last equation gives {n, (x,z,y)) + (n,(z,x,y)) = 0 . Because of the above identity, we have (n,A(x,y,z)) = (n,A(x,y,z)) + (n,(x,y,z)) + (n,(y,x,z)) - (n,(x,z,y)) - (n,(z,x,y)) + (n,B(x,y,z)) = (n,(x,y,z)) + (n,(y,z,x)) + (n,(z,x,y)) + (n,(x,y,z)) + (n,(y,x,z)) - (n,(x,z,y)) - (n,(z,x,y)) + (n,(x,y,z)) + + (n,(x,z,y)) = 3(n,(x,y,z)) + (n,B(y,z,x)) = 3(n,(x,y,z)) = 0 . Hence (n,(x,y,z)) = 0 . Prom this we conclude that for any x in R, (n,x) = 0, because of Lemma 3- Now for any x,y£R, since (n,x,y) = 0 and n ^ U we have A(n,x,y) = (x,y,n) + (y,n,x) = (x,y,n) - (x,n,y) + B(x,y,n) = 2(x,y,n) = 0. Thus (x,y,n) = 0 and since A(n,x,y) = 0, (y,n,x) = 0. This proves the lemma. Prom identity (I+2) and Lemma I4., we have for any x £R, (i+3) (e,x) = 0. THEOREM. Let R be a simple, not associative, (-1,1) ring of characteristic not 2 or 3 with an idempotent e. Then e is the unity element of R. PROOF. Since (e,x) = 0 for any x frR, RQ1 = R1Q = 0. This means then that Rqq is an ideal of R because R11R00 = R00R11 = 0 and RooRo o £ Roo- How slnce R00 i3 an ideal of R and e is not in Rq q , Rqq = 0* Thus R = but e is the unity element of Rq j . s. FURTHER RESULTS In the study of simple (-1,1) rings If all such rings are associative (with the properly chosen characteristic), then the following lemma represents a step in that direction. If not it eliminates a class of the ones which are not associative. LEMMA. 4. A simple not associative (-1,1) ring of characteristic not 2 or 3 has no proper left ideals. PROOF. Let T be.a left ideal of R, a simple .(-1,1) ring of characteristic not 2 or 3> such that T ^ R. We wish to show that T - 0. Let x,y be arbitrary in R and t£T. Then (x,y,t) - (xy)t - x(yt)£T. Since B(x,y,t) = 0, (x,t,y)£T. Since x and y are arbitrary we have (y,t,x)£T. Thus A(x,y,t) = 0 gives (t,x,y)£T. Let T 1 be the class T 1 = £tfe\T|tx€.T for any.x,fR^ . Now let t'^T' Then t»x and xtf are in T. If we can show that (t'x)y and (xt1)y both belong to T, we will have proved that T' is an ideal of R. This is because we would then have t1 x€ T! and xt1 £ T1 . Now (t’x)y = (t',-X,y) + t'(xy). Since (t!,x,y)£T and t'(xy)6 T we have (t’x)y^T. (xt')y = (x,t',y) + x(t»y). Since (x,t»,y)€T, and x(t’y)£T, because t’y6T and T is a left ideal, we have (xt’)y^T. This proves that T 1 is an ideal of R. Since T' T f R, we have T 1 =0. 33 3b Now let t,t'fi T and x,y^R. F(t,t*,x,y) = (tt',x,y) - (t,t'x,y) + (t,t',xy) - t(t»,x,y) - (t,t»,x)y = 0 . The first four terms of this identity are elements of T, hence (t,t*,x)y is also in T. Thus (t,t*,x)6 T* and (t,t*,x) = 0. Now B(t,tf,x) = 0 gives (t,x,t') =* 0. Since t and t' are arbitrary in T, we have (t',x,t) = 0, and A(t*,x,t) = 0 gives (x,t,t ') = 0 . (t,t' ,x) = (t'>x,t) = (x,t,t' ) = 0 . Now take t £ T and x, y£R. G(t,x,y,t) = (t,(x,y,t)) - (x,(y,t,t)) + (y,(t,t,x)) - (t,(t,x,y)) = (t,(x,y,t>) - (t,(t,x,y)) = 0 because of (ijif-)* Thus {t,A(x,y,t)) = 2(t,(x,y,t)) + (t,(y,t,x)) = 2(t,(x,y,t)) + (t,(y ,t,x)) - (t,B(y,t,x)) + M(t,x,y) =* 2{t,(x,y,t)) + (t, (y,t ,x)) - (t,(y,t,x)) - (t,(y,x,t)) + (t,(x,y,t)) + (t,(y,x,t)) = 3(t,(x,y,t)) = 0. Since R has characteristic not 3» (U5) (t,(x,y,t)) = 0. Replacing t by t + t 1 in (i+5>), where t,t l€ T, we get (t,(x,y,t»)) + (t’,(x,y,t)) = 0. G(t,x,y,t') = (t,(x,y,t»)) - ( x,(y,t1,t)) + (y,(t',t,x)) - (t*,(t,x,y)) = (t,(x,y,t1)) - (tf,(t,x,y)) = 0 by {I4.I4.). Combining this and (t,(x,y,tf)) + (t',(x,y,t)) = 0 we have (t»,(t,x,y)) + (t',(x,y,t)) = 0. This and (t 1,A(t,x,y)) = 0 give (t 1,(y,t,x)) = 0. Since y and x are arbitrary, (t1 ,(x:,t,y)) = 0, and (t 1 ,B(x,t,y)) = 0 gives (t *,(x,y,t))=0. Then (t 1,A(x,y,t)) = 0 gives (t»,(t,x,y)) = 0. 35 (1*6 ) (t1,(t,x,y)) = (t1 ,(x,y,t)) = (t',(y,t,x)) = 0 . Now F(w,x,x,t) = (wx,x,t) - (w,x^,t) + (w,x,xt) - w(x,x,t) = 0 along with (1*6 ) gives (t1 ,w(x,x,t)) = 0 since xt£T. (t»,wB(x,x,t)) = 0 then gives (1*7 ) (t • ,w(x,t,x)) = 0 . we now observe F(w,x,t,x) + B(w,xt,x) - H(w,x,t) = (wx,t,x) - (w,xt,x) + (w,x,tx) - w(x,t,x) - (w,x,t)x + (w,xt,x) + (w,x,xt) - (w,x,xt) + (w,x,t)x = (wx,t,x) + (w,x,tx) - w(x,t,x) - 0 . Thus from (1*6) and (i*7)» we get (1*8 ) (t' ,(w,x,tx)) = 0 . Replacing x by x + y in (1*8) gives (t',(w,x,ty)) + (t',(w,y,tx)) = 0 . Subtracting (t>,B(w,y,tx)) from this gives (1*9 ) (t1 ,(w,x,ty)) - (t1 ,(w,tx,y)) = 0 . Now since xt,wt£T, using (1*6) we have 0 = (t',F(w,x,t,y) + F(w,t,x,y)) ® (t»,(wx,t,y) - (w,xt,y) + (w,x,ty) ~ w(x,t,y) - (w,x,t)y + (wt,x,y) - (w,tx,y) + (w,t,xy) - (w,t,x)y - w(t,x,y)) = (t1,(w,x,ty)) - (t»,(w,tx,y)) - (tf,(w,x,t)y) - (t»,(w,t,x)y) - (t1,w(t,x,y)) - (t1,w(x,t,y)) = - (t«,B(w,x,t)y) - (t',w(t,x,y)) - (t1 ,w(x,t,y)) because of (1*9). Thus we have (50 ) (t«,w(x,t,y)) = - (t1,w(t,x,y)). 36 Replacing x by x + y in (47) gives (5D (t<,w(x,t,y)) = - (t' ,w(y,t,x)). (t«,wB(x,t,y)) = 0 gives (52) (t«,w(x,t,y)) = - (t',w(x,y,t)). Adding (50), (5D and (52) gives 3(t*,w(x,t,y)) * - (t1,wA(x,y,t)) = 0. Thus (t1,w(x,t,y)) = 0, and from (52) we get (53) (t1>w(x,y,t)) = 0 . Now F(w,x,y,fr)=(wx,y,t) - (w,xy,t) + (w,x,yt) - w(x,y,t) - (w,x,y)t = 0. Since t' commutes with the first three terms of this by (46) and with the fourth by (53)» we have (54) (t',(w,x,y)t) = 0. Since any element of R is the sum of associators, because of Lemma 3> we have for any x in R (55) (t',xt) - 0 . Now using H(w,x,t) = (w,x,xt) - (w,x,t)x = 0 we have (56) (t»,(w,x,t)x) = 0 by (I46) since xt£T. Replacing x by (x + w) in (56) gives (57) (t',(w,w,t)x) + (t',(w,x,t)w) = 0. We have (t»,(w,x,t)w) = (t1,(w,x,t)w) - (t*,A(w,x,t)w) = - (t1,(x,t,w)w) - (t',(t,w,x)w) + (t',B(x,t,w)w) = (t1,(x,w,t)w) - (tf,(t,w,x)w) + (t1,H(x,w,t)) - (t1,H(t,w,x)) = (t*,(x,w,wt)) - (t',(t,w,wx)) = 0 37 by (1+6) sines wt£T. Hence from ($7) we get (58) (t »,(w,w,t)x) = 0 = (t',(w,x,t)w) Now F(w,w,t,x) = (w2,t,x) - (w,wt,x) + (w,w,tx) - w(w,t,x) - (w,w,t)x = 0. Thus from (58), (1+6) and (55), since (w,t,x)<»T, we have (59) (t1,(w,w,tx)) = 0. If we replace w by w + x in (59), we get {t 1,(w,x,tx)) + (t',(x,w,tx)) = 0. From (L+8) we have (t ' , (w,x, tx)) = 0, hence (t 1,(x,w,tx)) = 0. This last combines with (t 1,B(x,w,tx)) = 0 to give (t ',(x,tx,w)) = 0. Of the identities just proved the two important ones are (t »,(w,x,tx)) = (t *,(x,tx,w)) = 0. These combine with (t 1,A(tx,w,x)) = 0 to give 0 = (t1,(tx,w,x)) = (tr,(tx,w,x)) - (t 1,B(tx,w,x)) = - (t 1,(tx,x,w)). Thus (60) (t *,(tx,x,w)) = 0. Now (60) and (1+6) combined with F(t,x,x,w) = (tx,x,w) - (t,x2,w) + (t,x,xw) - t(x,x,w) = 0 give (61) (t1,t(x,x,w)) = 0. If we replace x with x + y in (61) we get (t •,t(x,y,w)) + (t»,t(y,x,w)) = 0. Since x,y, and w are arbitrary, we can write (t*,t(x,w,y)) + (t1,t(w,x,y)) = 0. Hence we form the sum 38 0 = (t',t(x,y,w)) + (t',t(y,x,w)) - (t*,t(x,w,y)) - (t *,t(w,x,y)) + (t1,tA(x,y,w)) + (t 1,tB(x,w,y)) = 3(t1,t(x,y,w)) + (t ',tB(y,x,w)) = 3(t*,t(x,y,w)). Thus (62) (t*,t(x,y,w)) = 0. Because of Lemma 3 again we have for any x in R (63) (t ',tx) = 0. Now D(t,t ',x) = (tt',x) + (t'x,t) + (xt,t>) = (tt',x) = 0 because of (55) and (63). (6J+) (tt' ,x) = 0 . The class T2 thus is contained in U. This makes T2 a two sided ideal since x(tt') ** (xt)t'€T2 and (tt')x = x(tt') = (xt)t'6-T2 . Since T 2 ^ R we have T2 = 0. Next we consider T + TR. An element of this class is the sum of a finite number of elements of the form t 1 and tx, where t,t'£T and x if R. Now for y £R (t»+tx)y = t1y+(tx)y = t'y + (t,x,y) + t(xy) = (t,x,y) + Ct(xy) + t'y] Thus T k R and this is a contradiction. Thus T + TR = 0 and T = 0. 6. APPENDIX LEMMA. Let R be a simple ring. (No assumptions are made about multiplication except for the distributive laws). Then either all non zero elements of R are of infinite additive order or they are all of the same finite order. PROOF. Let x ^ 0 be an element of R such that x has finite additive order n and no non zero element of R has additive order less than n. Consider now the ideal I, generated by x. Since x / 0 and R is simple, I = R. We shall now show that all non zero elements of I have additive order n. We denote by Ry the mapping of R into R given by t •» tRy = ty for arbitrary t in R and fixed y in R. We denote by Ly the mapping t -- » tLy = yt for arbitrary t in R and fixed y in R. Now let S^, S2, denote arbitrary mappings from either the class of all Ly or the class of all Ry. We see that a typical element of I Is the finite sum of elements of the form xSlS2 Sk’ and an Integral multiple of x. However n £ x S1S2 ••• S^"] = (nx)S1S2 ••• Sk = 0 and n times any Integral multiple of x Is zero. Thus any 39 ^0 element of I has additive order at most n, but since n is minimal all non zero elements of I have additive order n. EXAMPLE. Let F be a field and let us define an algebra Since (e,e,u) = eu - e(eu) = v - ev = v, Q. is not alternative or associative. Thus we see that even for a finite dimensional (-1,1) algebra it is necessary to assume simplicity to prove the main theorem of this dissertation. In fact the critical identity, (e,e,x) = 0, fails. We shall now show that satisfies identity (1). Let p,q,r be basis elements of (X>. Then (p,q,r) = 0 if at least two of'the elements p,q,r are not the element e. Thus when we consider A(p,q,r) we need only show that A(p,q,r) = 0 for p = q = e, and r = u,v,w or z. Now since is right alternative, A(e,e,r) = (e,e,r) + (e,r,e) = B(e,e,r) = 0. We thus have that A(p,q,r) = 0 if p,q,r are basis elements of . Let a-j_ - e, a2 = u, = v, a ^ = = Let p = , q = 2 ^ P i^i > r ~ 5 ^ ^i ai where V ^ F . Then A(p,q,r) = (p,q,r) + (q,r,p) + (r,p,q) = 2 ©C. p y A(a, ,a.,a, ) = 0. Hence i,j,k 1 J K 1 J is a (-1,1) algebra. BIBLIOGRAPHY 1. Albert, A. A., Almost alternative algebras, Portugaliae Math., vol. 8 (1949) pp. 23-26. 2. , On right alternative algebras, Ann, of Math. vol. ^0 (1949) PP. 318-328. 3 . , The structure of right alternative algebras, Ann. of Math. vol. 59 (1954) PP* l|08-i|17. 4. Bruck, R. H. and Kleinfeld, E., The structure of alternative division rings, Proc. Amer. Math. Soc. vol. 2 (1951) PP. 878-8 9 0 . 5. Cayley, A., On Jacobi's elliptic functions, in reply to the Rev. Brice Bronwin and on quaternions, Phil. Mag., London, vol. 26 (1845) pp. 210-213. 6 . Kleinfeld, B., Right alternative rings, Proc. Amer. Math. Soc. vol. 4 (1953) pp. 939-914-4• 7. , On simple alternative rings, Am. J, of Math. vol. 75 (1953) PP. 98-104. 8 * Simple alternative rings, Ann. of Math. vol. 58 (1953) pp. 544-547. 9. . Alternative nil rings, Ann. of Math, vol. 66 (1957) pp. 395-399. 10. . Rings of ( yfi £ ) type, Portugaliae Math. vol. 18 (1959) pp. 107-110. 11. , Simple algebras of type (1,1) are associative, ("To appear) . 12. Kokoris, L. A., On a class of almost alternative algebras,. Canad. J. of Math. vol. 8 (1956) pp. 250-255. 13. , On rings of ( £ ) type, Proc. Amer. Math. Soc. vol. 9 (1958) PP. 897-904- 14. San Soucie, R. L., Right alternative division rings of characteristic 2, Proc. Amer. Math. Soc. vol. 6 (1955) PP. 291-296. 15. Skornyakov, L. A., Right alternative fields, Izvestia Akad. Nauk SSSR Ser. Mat. vol. 15 (1951) pp. 177-184. 42 i+3 16. Wedderburn, J. H. M., On hypercomplex numbers, Proc. London Math, Soc., Ser. 2, vol. 6 (1908) pp. 77-H 7 • AUTOBIOGRAPHY I, Carl Christopher Maneri, was born in Cleveland, Ohio, January 25, 1933* I received my high school diploma from Parma Schaaf High School in Parma, Ohio, in 1950. I received a Bachelor of Science degree in physics from Case Institute of Technology in 1954* While there I studied enough mathematics to enroll in graduate school at Ohio State University in mathematics in 195i|. While working toward my doctorate there, I was employed as an instructor of mathematics. For two quarters in 1956 I was a research fellow in projective geometry, and In the summer of 1959 I was a National Science fellow.