ON A CLASS OF NON-FLEXIBLE ALGEBRAS BY Frank James Kosier AN ABSTRACT Submitted to the School for Advanced Graduate Studies of Michigan State University of Agriculture and Applied Science in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1960 Approved by (”Eziekfiék \& <§2§$¥m§gg== FRANK JAMES KOSIER ABSTRACT Let A be an algebra over a field F of characteristic not two such that: I. The elements of A satisfy a non-trivial identity of the form (1) a1(zx)y + 62(zy)x + a3y(zx) + ahx(zy) + a5(xz)y + a6(yz)x + a7y(xz) + a8x(yz) = 0 for fixed “1 in F. II. There is an algebra B over F such that B satisfies (1), B has an identity element, and B is non-flexible; that is, there are elements x and y in B such that x(yx) % (xy)x. These conditions are similar to those used by A. A. Albert to define almost left alternative algebras [Almost alternative algebras, Portugal. Math. vol 8 (l9h9) pp. 23-36]. Albert's paper led to the study of algebras of (TV, 8 ) type by Kleinfeld and Kokoris. Assume F to be algebraically closed. It is then shown that A is quasi-equivalent in F to an algebra A(p) where A(p) satisfies one of the following identities: (1) ny - Rny = -(Lyx - LxLy) . 2 2 _ (ii) Rx + Lx — 2LxRx 2 2 _ (iii) Rx + Lx - LxRx + RxLx (iV)(Rx + Lx)ax and Rx is called a right multiplication of A. Sim- ilarly Lx is defined as the mapping a ->xa and is called a left multiplication of A. It is then possible to rewrite identities in terms of these mappings. For example, the commutative law becomes Rx = Lx’ the flexible law becomes RxLx = LxRx’ and the associative law becomes Rny = Lny. One of the important tools in the study of associative rings is the Peirce decomposition with respect to an -6- idempotent e of the ring. This is the decomposition of the ring A into the direct sum of the modules A13 = {E: ea = 1a, ae = is} , i,J = 0,1. This may be general- ized to power-associative rings A of characteristic not two in which the equation 2x = a has a solution x in A for all a in A[3]. First we define the ring A(+) to be the same additive group as A but we define a new multiplication xoy. We set 2(xoy) = xy + yx where Juxtaposition indicates the multiplication of A. Clearly A(+) is a commutative ring and powers in A(+) agree with powers in A. It is then possible to show that any commutative power-associative ring A of characteristic not two has a decomposition into the direct sum of the submodules Ae(p) where Ae(p) = {E:2ea = pa} , p = 0,1,2. Moreover, Ae(2) and Ae(O) are orthogonal subrings of A and the following relations hold: Ae(l)Ae(1) g Ae(2) + Ae(O); Ae(u)Ae(1) + Ae(1)Ae(u.) g Ae(2-p) + Ae(l) for p = 0,2 [3; Ch. I: Theorem 2]. In the non- commutative case we can pass to the ring A(+) and since A(+) has this decomposition, A = Ae(2) + Ae(l) + Ae(O) where Ae(p) = {a:ea + as = p€7 , p = 0,1,2. We may interpret the results for A(+) in terms of the multiplication of A. Then we find that Ae(2) and Ae(O) are orthogonal submodules of A and e acts as an identity element for Ae(2) [3; Ch. I; Theorem 3]. An idempotent e is principal if there is no idempotent in Ae(O). An idempotent e is said to be primitive if e is the only idempotent in Ae(2). The degree of A is the maximum number of pairwise orthogonal idempotents that appear in any -7- AK where K is a scalar extension of the center of A. Let A be an arbitrary power-associative ring. An ideal L of A is said to be nil if every element of L is nilpotent. Then the radical of A is defined to be the maximal nil ideal of A and we say that A is semi-simple whenever A has zero radical. A is simple if'A has no proper ideals and A is semi-simple. A Jordan ring A is a commutative ring which satisfies the identity x2(ax) = (x2a)x for all x and a in A. A ring A is Jordan-admissible if the attached ring A(+) is a Jordan ring. This is equivalent to saying that A satisfies the identity (5) (Rxx + Lxx)(Rx + Lx) = (Rx + Lx)(Rxx + Lxx)° In a similar manner we define associative-admissibility for a ring A and see that this is equivalent to the ring A satisfying the identity (6) (Rx + Lx)(Ry + Ly) = (Ry + Ly)(Rx + Lx)° If we define a new product in A by [x,y] = xy - yx and denote the resulting ring by A(-), then A is said to be Lie-admissible whenever A(-) is a Lie ring. The product [x,y] = xy - yx is called the Lie product. Finally, we shall define a generalized notion of isomor- phism for algebras. Suppose A and B are algebras over a field F such that A and B are isomorphic as vector spaces. We may then consider A and B as the same vector space and we shall say that A is quasi-equivalent in F to B = A(p) if there is a p in F not equal to % such that the product xoy in B is given in terms of the product xy of A by xoy = pxy + (l-p)yx I l [3]. The multiplications Rx and Lx of A(p) are given in I terms of the multiplications Rx and Lx of A by Rx = pr ' + (l-p)Lx and Lx = (l-p.)Rx + pLx. If p = 1, this is the usual notion of isomorphism; if p = O, A and B are anti- isomorphic. This relation is actually an equivalence relation ___.iw ~ :i-efi on the class of all algebras over a field F, but the import- ance of the notion lies in the following; (i) powers in A and A(p) coincide; (ii) A and A(p) have the same ideals; (iii) A and A(p) have the same nil radical. Thus A and A(u) have the same structure. Section II. The Class C5\ By property II there is an algebra B in (IL with elements x and y such that xy # yx. Then, by a series of substitutions of the elements l,x,y in (l), we find the following relations: ul + a2 + a3 + an + as + a6 + a7 + a8 = 0 a1 + a” + a5 + a8 = a2 + u3 + a6 + d7 = 0 a1 + a2 + a3 + a6 = an + a5 + a7 + u8 = 0 cl + “2 + “h + a5 = a3 + 66 + o7 + a8 = O. Combining these we have “8 = a2, d7 = d1, d6 = -(a1 + d2 + a3), and as = -(a1 + «2 + an). Now using these relations and rewriting (l) in terms of right and left multiplications, we obtain (7) (21(12ny + LxLy) 4- (12(11ny + Lny) + a3RxLy 4- (:13ny -(a1 + a2 + ah)Lny - (a1 + 02 + G3)Lny = 0. If we interchange x and z in (l), we find (8) <11“ny + RxLy) + «2(ny + Ryx) + oz3LxI.y + :2an -(a1 + a2 + ah)Rny - (a1 + a2 + a3)Lyx = 0. Then interchanging y and z in (1), we obtain (9) “1(Lyx + ny) + 02(Lny + Rny) + Q3Ryx + ukLny .(a1 + a2 + “A)ny - (a1 + a2 + e3)Rny = o, ““T‘n’myw -10- Finally, setting y = x in (7) yields 2 2 - (10) (a1 + a2)(Rx + Lx - 2LxRx) + (a3 + “A)(RxLx - LxRx) - 0. Suppose a1 + a2 = 0. Then from (10) we must have (a3 + °h)(RxLx - LxRx) = 0, but property II implies RxLx - LxRx % 0, so that d3 + ch = 0. Suppose also that d1 = 62 = O. Substitution of these values in (8) along with the condition that not all the a1 are zero yields (11) ny - Rny = -(Lyx - LxLy)‘ In the remainder of this section we suppose A to belong to the class (JL and F to be an algebraically closed field. THEOREM 1. Suppose a1 + a2 = 0 and a1 % 3 a3. Then A is quasi-equivalent in F to an algebra A(p) satisfying identity (11). PROOF. If a1 = o the result is immediate. If a1 g o “3 and B = - 3—. f g I, then (8) can be written as 1 (12) ny - Lny - RxLy + Ryx = B(ny - Rny - LxLy + Lyx)' h v I. II O We note t at Rxoy - Rny - LxLy + Lyox _ _ _ 2 + 9(1 u)ny + u(1 MRyx + (1 n) Lyx = 0 in A(p) is equivalent to (.2ny “"Rx + (1-u)Lx)(pRy + (1-p)Ly) — (mmx + uLx)((1-p)Ry + .Ly) + p(l-p)Ryx + uZLyx + (l-u)2ny + p(1-p)ny = o in A. Simpli- fying we have 2 (13) (2» - 2p + 1)(Rxy - Rny - LXLy + Lyx) = -2p(1-p)(ny - Lny - RxLy + Ryx). -11- Now consider the equation in p, 2p2 - 2p + l = -2p(l-p)B or 2(l-B)u2 - 2(l-B)p + 1 = 0. Since B K 1 and F is algebra- ically closed, there is a p in F satisfying this equation. Suppose p = %. Then the above equation becomes 1 - B - 2(l-B) + 2 = 0. Hence, B = -l contrary to assumption. Then substitution of this value of p in (13) yields (12) and the proof is complete. Let a1 = -a3 and d1 + a2 = 0. Then (12) becomes ny - Lny - RxLy + Ryx = ny - Rny - LXLy + Lyx or (1h) ny _ yx - ny _ yx = Lny - Rny - LxLy + RxLy° If a1 = a3 and a1 + a2 = 0, we see that (12) becomes (15) ny + W + ny , yx = (Rx + Lx)(Ry + Ly) (by symmetry) = (By + Ly)(Rx + Lx)' This is exactly the condition that A be associative-admissible. THEOREM 2. If a1 + a2 # 0, then A is quasi-equivalent in F to an algebra A(p) satisfying either 2 2 _ (16) Rx +_Lx - 2LxRx or 2 2 _ (17) RI + Lx - anx + RxLx. PROOF. Since «1 + a2 fi'o, we may write (10) in the form a + a 22 _ -3“ (l8) Rx + Lx - 2LXRx - B(LxRx - RxLx) where B — 31-1-3; . v 2 I 2 I 0 Then, as before, (Rx) + (Lx) - 2LxRx = 0 in A(p) is equivalent to (pr + (l-p.)Lx)2 + (“Lx + (1-u)Rx)2 - 2(uLx + (l-MRxHuRx + (1-u)Lx) 2- x = (“A2 - “A + 1)(R: + L = O in A. Now examine the equation (hp2 — hp + l)B = -(up2 - 6p + 2) or (2p - 1)((2 + 2B)» - (2 + 5)) = o. This has a solution u #’% provided 34% -1. Thus the theorem is valid except possibly when B = -1. But then (18) becomes R: + L: = LxRx + RxLx. Linearization of (16) gives us (19) Rny + Rny + LxLy + Lny = 2LxRy + 2LyRx which is equivalent to (16) since the characteristic of F is not two. Similarly (17) is equivalent to (20) Rny + Rny + LXI.y + Lny = RxLy + Rny + Lny + Lny provided the characteristic of F is not two. We note that setting y = x in (1%) yields (17). Combining these remarks with Theorems 1 and 2 we state THEOREM 3. Let A belong to the class CD. and suppose also that F is algebraically closed. Then A is quasi- equivalent in F to an algebra A(p) where A(p) satisfies one of the identities (11), (15), (16), (17): each of which is a particular determination of (l). 2 ZLXRX) - (up - 6p + 2)(RXLX *LxRx) ar—I. -13- Section III. The Identity ny - Rny = -(Lyx - LxLy) In the following we shall be concerned with a ring A of characteristic not two such that A satisfies identity (11). Set (x,y,z) = (xy)z - x(yz). (x,y,z) is called an associator. Then (11), when applied to an element 2 of A, becomes in terms of associators (21) (x,y.2) = (Z.y.x). We now write (2) in terms of associators as (22) (x,y,z) + (X.Z.y) . (y.X.z) + (v.2.x) + (zany) + (z,y,x) = 0. Using (21) in (22) we obtain (23) (x.y.z) + (y.z,x) + (2.x.y) = 0. THEOREM #. If A is a ring of characteristic zero satisfying (21) and xx2 = x2x, then A is power-associative. PROOF. By Albert's result [1, Lemma h], we need only show that x2x2 = x3x for all x in A. Set y = x and z = x2 in (21) and (23). Then (x,x,x2) = (x2,x,x) and (x,x,x2) 2 2 3 + (x,x2,x) + (x,x,x2) = 0. Thus 2x x = x x + xx3 and 2x3x - 2xx3 = 0, so that x2x2 = x3x = xx3 and the result follows. The following example shows that xx2 = x2x is actually necessary to guarantee power-associativity. Let A be the algebra with a basis of the three elements e,u, and v over a field F of characteristic zero where multiplication is defined by uv = e, eu = u, ve = v, e2 = e, and all other products zero. Then A satisfies (11), but we observe that (u + v)2(u + v) = u and (u + v)(u + v)2 = v. We note that (21) along with (h) implies associativity while (21) and (23) together yield 2(x,x,y) = 2(y,x,x) = ~(x,y,x). Hence, for rings of characteristic not two satisfying (21) and xx2 = xzx, the notions of being assoc- iative, flexible, right alternative, and left alternative are equivalent. Hereafter, whenever we refer to a substitution or a permutation of certain elements in an identity given in terms of the right and left multiplications of the elements x,y,xy, and yx, we shall consider these multiplications as acting on the element 2. Also, we shall often make use of these identities without writing them in terms of right and left multiplications. THEOREM 5. A is both Jordan-admissible and Lie-admissible. PROOF. Permuting x,y,z in (11) we obtain (2%) Ryx - Rny - Lny + ny = O and (25) RxLy - Rny - Lny + Lny = O. Subtracting (11) from (2M) and then adding (25) to the resulting right hand member we have ny-yx - ny-yx = (By - Ly)(Rx - Lx) - (Rx - Lx)(Ry - Ly), so that A is Lie-admissible. -15- Now set y = x2 in (11). Then _ _ 2 (26) R - RxRxx + LxLxx and y — x in (2h) X(xx) + L(xx)x yields (27) R(xx)x + Lx(xx) = RxxRx + LxxLx' We also set Y = x2 in (25) to obtain (28) RxLxx - RxxLx - LxxRx + LxRxx = 0. Subtracting (27) from (26) and then adding (28) we obtain (Rx + Lx)(Rxx + Lxx) - (Rxx + Lxx)(Rx + Lx) = 0 and the theorem is proved. In the remainder of this section we suppose that A is a power-associative ring. LEMMA 1. If e is any idempotent of A, then (e,e,x) = (x,e,e) = (e,x,e) = O for all x in A. PROOF. If we set A = Ae(2) + Ae(l) + Ae(O), we then see that it is sufficient to prove the proposition for x in Ae(l). Thus, we suppose in the following that x is in Ae(l). Substitution of x,y = z = e in (21) yields (29) ex + xe = e(ex) + (xe)e and since x = ex + xe = e(ex) + (xe)e + e(xe) + (ex)e we have (30) (ex)e + e(xe) = O. I Set ex = x2 + x1 + x0 and xe = -x2 + x1 - xo where x1 is in Ae(i). Then by substitution in (31) ex = e(ex + xe) = e(ex) + e(xe) = e(ex) - (ex)e we find x1e = - %(x2 + x0), so that ex1 = %(x2 + x0) + x1. But then x1 = exl + x1e = e(exl) + (xle)e = % x2 + exl - % x2 = exl. Hence, ex1 = x1 and x2 = x0 = 0, so that by consid- eration of (29), (30), and (31), we have e(ex) - ex = xe - (xe)e = e(xe) - (ex)e = O and the result follows. THEOREM 6. If e is any idempotent of A, then every x in A may be written uniquely as x = x11 + x10 + x01 + x00 where x13 is in A13 = {}:ea = ia; ae = 35} , 1,3 = 0,1. PROOF. The theorem follows immediately from Lemma 1 just as in the associative case. Let x and y be in All' Then from (21) e(xy) + (yx)e = (ex)y + y(xe) = xy + yx. If xy 2 a11 + a10 + a01 + 800’ then yx = b11 - a10 - a01 - a00 so that a10 - a01 : O, and 2 hence, a10 = 801 = 0. Therefore A11 9 All 4- A00. If x and y are in A00, then e(xy) + (yx)e = (ex)y + y(xe) = O and, 2 as above, we find A00 ‘3 All + A00. If x is in All and y in A10, then xy + yx = (xe)y + y(ex) = x(ey) + (ye)x = xy. Thus yx = 0. But then e(xy) + (yx)e = (ex)y + y(xe) or e(xy) = xy and e(yx) + (xy)e = (ey)x + x(ye) = 0, so that (xy)e = 0. Hence AHA10 51 A10 and AlOAll = 0. Replacing y in A10 by y in A01, we find that A11A01 = O and AOlAll E 01. Then replacing x in All by x in A00, we have AOOA10 = AOlAOO = O, AIOAOO 9; A10, and AOOAOl ngl. Suppose x and y are in A10. Then yx (xe)y + y(ex) x(ey) + (ye)x = xy and e(xy) + (yx)e (ex)y + y(xe) xy. Hence e(xy) + (xy)e = xy, so that xy is in Alo + A01. But 2xy = xy + yx is in .. - 2 _ All + A00. Thus xy - yx - 0. Likewise A01 - 0. Now suppose x is in A10 and y is in A01. Then e(xy) + (yx)e = (ex)y + x(xe) = xx and e(yx) + (xy)e = (ey)x + x(ye) = xy. If we set xy = all + a10 + a0]- + a00 and yx = b11 - a10 - a01 + bOO and substitute in the above equations, we find all I 810 * bll ‘ 801 = all * alo * a01 * 8oo = bll ' 810 + a = = 0 so that 11 - a01. Hence b11 = a01 = a00 xy = all and yx = boo. Combining these results we state THEOREM 7. Suppose e is any idempotent of the ring A and A = All + A + A + A 10 00 where the A13 are defined as 01 in Theorem 6. If i and J are distinct and i,j,k,m = 0,1, 2 then Aii c; Aii + A“, Aijim = o, and if either k f 3, or 3 )1 m then also Aijjm g Akm. COROLLARY 1. L = A10A01 + A10 4- A0]. + AOIAIO is an ideal of A. PROOF. By the preceding theorem it is sufficient to show that A11(AHA31) + (AijAJiMii g L for i ,x’ 3. But if x is in A11, y in A13, z in A11, then x(yz) = (xy)z + z(yx) - (zy)x = (xy)z. Then x(yz) is in AijAji g L. Likewise (A13A31)A11 Q— L. COROLLARY 2. Let A be simple and e an idempotent of A. Then either A 10 * A01 = O °r All = AlOA01 and A00 ‘ AOlAlo' PROOF. The result is immediate from Corollary 1. We now state the main result of this section. THEOREM 8. If A is a simple power-associative ring of characteristic not two possessing an idempotent e such that A10 + A01 ;! o and satisfying identity (11), then A is associative. PROOF. Let x be in A13, y in Ars’ z in Ann“ Then x(yz) = 0 unless s = m and j = r. Similarly (xy)z = 0 unless s = m and j = r. Hence, it is sufficient to consider products of the form x(yz) where x is in AiJ’ y is in A35’ and z is in Asm' Case 1. Suppose i = j = s = m. We may represent x as x = 2::x1nxn1 where i f n since we may use Corollary 2. Then, by repeated applications of identity (11) and Theorem 7, we find (xy)z = Z ((xinxn1)y)z = Z (x1n(xniy))z = E xin((xniy)z) = Z x1n(xni(yz)) = Z (xinxnixyz) = x(yz). Case 2. Let i = 3. Then x(yz) = (xy)z + z(yx) - (zy)x and the product is associative unless i = m = 5 since other- wise the last two terms of the right hand member are zero. But, if i = m = 3, Case 1 applies so that the result holds in this instance. Case 3. Suppose J = 3. Proceed as in Case 2. Case h. Suppose s = m. Proceed as in Case 2. Thus we have reduced the proof to the case where x and z are in A13 and y is in A”, with i ;! j. Substitution of x,y, and z in (2) along with Theorem 7 yields (32) x(yz) + z(yx) = (XY)Z + (zy)x which with (11) implies x(yz) = (xy)z and (zy)x = z(yx) and the proof is complete. REMARK. If A is a semi-simple strictly power-associative algebra of characteristic not two satisfying (11) and e is a principal idempotent of A, then A10 + A01 = O. PROOF. We observe that e is also a principal idempotent (+) - of A and Ae(l) — A10 4- A01. Hence A10 + A01 4- A00 9 radical of A(+) [10, Theorem 5]. Then we claim that the ideal L defined in Corollary 1 of Theorem 7 is contained in the radical of A(+). For, if x is in A10 and y is in A01, (+). B then 2xoy = xy + yx is in the radical of A ut 2e(xoy) + 2(xoy)e = 2xy is in the radical of A(+). Thus AloA01 is contained in the radical of A(+), so that L must be a nil ideal of A. Therefore A10 + A01 = 0. Consideration of Theorem 8 tells us that if we are to find any new simple powerrassociative algebras satisfying (11), they must come from those which have no idempotent e such that A10 + A01 # O. The existence of such algebras is guaranteed by the following examples which seem to have a rather complicated structure. First, let A be the 3-dimen- sional algebra over a field of characteristic 0 with a basis e,u,v where e is the identity element, uv = -vu = e, and other products are zero. Then A is a simple power-associative algebra satisfying (11) but A is not flexible for (uv)u = -u(vu) = u. It is then possible by use of this example to construct simple power—associative algebras B of any degree satisfying (11) which are not flexible. we need only set B equal to the supplementary sum of n-copies of the algebra A described above where each pair are orthog- onal subspaces and we redefine uv = -vu = 1, where 1 is the i identity element of A, for each of the n-copies. We now give an example of a semi-simple power-associative ; algebra C which satisfies (11) but has no identity element and is not the direct sum of simple algebras. Consider C equal to the supplementary sum of the algebra A above and I I the subspace A with a basis u1,v1 where AA = A'A = 0 and ulv1 = ~v1u1 = e and all other products zero. Then, since every proper ideal of C contains A, we see that the nil radical of C is 0. One might hope for further results when A is semi-simple with an idempotent e such that A10 + A01 # 0, but examination of the algebra D = C GB Fé, where C is the algebra described Just above and F2 is the total matric algebra over F of degree 2, shows there are such algebras which do not have an identity and are not the direct sum of simple algebras. -21- 2 2 _ Section IV. The Identity Rx + Lx - 2LxRx In this section we assume A to be a ring having characteristic prime to six and having a solution to the 1 equation 2x = a for all a in A. Moreover, we shall assume that A satisfies (19) or equivalently (16). f THEOREM 9. Suppose A is a ring of characteristic prime to 30 such that A satisfies (l9) and (x2)2 = (x2x)x for all x in A. Then A is power-associative. PROOF. We define xn inductively by xn+1 = xnx. First we show that xxn = xnx for all x in A and all positive integers n. Substitution of x = y = z in (19) yields x2x + xx2 = 2x2x. Hence xx2 = x2x. Suppose xxk = xkx. Then setting x = z and y = xk in (19) yields (xkx)x + X(xxk) = 2(xxk)x so that xxk+1 = xk+1x. Thus xx“ = xnx for all x in A and all positive integers n. The result then follows from [1, Lemma #1. In the remainder of this section we suppose that A is a strictly power-associative ring. Let e be an idempotent of A. Then from (16) we find (33) (xe)e + e(ex) = 2(ex)e for all x in A. Setting y = z = e in (19) yields (3%) (ex)e + ex + e(xe) + xe = 2(xe)e + 2ex Suppose x is in Ae(l). Then x = e(ex) + e(xe) + (ex)e + (xe)e and using (33) we obtain (35) x = 3(ex)e + e(xe) for all x in Ae(l). (*) 1 Then by consideration of A we may set ex = x2 + 5 X -22- + x' + x0 and xe = -x2 + %x- x' - x0. Substitution in (33) 1 yields -x2 + - xe - x'e - % ex + ex' = 2x2 + xe + 2x'e 2 so that we have x2 - x' - x0 = ex' - 3x'e. Hence x2 = h(ex')2, x0 = H(ex')o, and x' = 3(x'e)1 - (ex')1 = ex' + x'e. Thus (ex’)1 = (x'e)1 = x'. If we now carry through the same argument with x replaced by x', we find that (ex')2 = (ex')o = O and hence, x2 = x0 = 0. We note that e(xe) = (ex)e = & x. Therefore, for any x in Ae(l), we have ex ‘ l - 2 where x' is in Ae(l) and ex' = x'e = % x'. x + x' Suppose x and y are in Ae(2). Then we set xy = a2 + a1 + a0, yx = b2 - a1 - a0. Substitution of e,x,y in (19) yields xy + yx + (yx)e + e(xy) = 2yx + 2(xy)e so that ea1 - ale = -2a1 - 2ao + 2a1e. Hence a0 = O and 2a1 = 3ale - ea1 = 2a1e + 2ea1. But then 3ea1 = ale = Q a1. For any 1, a1 in Ae(l) we have ea1 = % a1 + a' where ea' = a'e so that % a1 - ai = a a1. Then -#ai = a1 and therefore ea1 = ale. Thus we finally have Mai a subring of A. a1 = O and Ae(2) is Let x and y be in Ae(O). Then substitution of x,y,e in (19) yields (yx)e + e(xy) = 2(xy)e. we suppose xy = a2 +1) + a1 + a0 and yx -a2 - al 0. Then -a1e + eal = 2a2 + 2a1e. Hence a2 0 and ea1 = 3a1e. Then just as for Ae(2) we find a1 = 0 and Ae(O) is a subring of A. Suppose x is in Ae(2) and y is in Ae(l). Then let xy a2 + a1 + a0, yx = -a2 + b1 + b0. Then (19) becomes (xy)e + xy + e(yx) + yx = 2(yx)e + 2xy and hence ale + eb1 + bl + b = 2ble + a1 + a0. Thus a0 = b0 and ale + eb1 = 2b1e - bl or ea1 = 2b1 - 3b1e. Another substitution in (19) yields (yx)e + (ye)x + e(xy) + x(ey) = 2(xy)e + 2(ey)x. Now set ey = % y + y' where ey' = y'e = % y'. Then ble + ea1 + % a1 + xy' = a2 + 2a1e + % b1 + 2y'x. Comparing components we find a2 = (xy' - 3y'x)2 = 1+(xy')2 and (y'x)O = (xy')O = 0. Yet another substitution of x,y',e in (19) yields xy' = y'x P so that (xy')2 = (y'x)2 = 0. Thus a2 = 0. Since ea = 2bl I .. + b1 and ea1 - - 3ble a substitution of eb % b Rm“ 1 1 1 gives the relation §(b1 - a1) ai - 3bi. Then e(b1 - a1) (b1 - al)e = %(b1 - a1). If ea1 = 2b1 - 3b1e, then e(eal) 2eb1 - 3e(b1e). But then we obtain 32-(b1 - a = (2ai - ubi) 1) so that ai = bi. Finally, eb1 - ale = %(b1 - a1) + 2ai = 0. Let x be in Ae(O) and y in Ae(l). We set xy = a2 + a1 + a0 and yx = b2 + bl - a0. Substitution of x,y,e in (19) yields a2 + ale + ebl = b2 + 2ble. Then a2 = b2 and ale = 2b1e - ebl. Set ey = %'y + y'. Then, as before, y'x = xy' and substi- tution in (19) yields J2‘(xy - yx) = -2xy' so that a0 = 0. Now ale = 2b1e - eb1 implies e(ale) = 2e(ble) — e(ebl) or &(b1 - a1) = b'. We shall use these relations in our later [ at work. Suppose A has an identity element 1 = e1 + ... + et where ei are pairwise orthogonal idempotents. Then A can be decomposed as the direct sum of the modules A13 where for i = j, Aij = Aei(2) and for i ,1 3, A13: Aei(l)nAej(1) [6]. Let i,3,m,n be distinct. Then A“ 0.131135; Aim’ A130 A13 9;- A11 + A33, A13 0 Ann = 0. But Am; Aem(l), A13 gAemw) so that by our multiplication A” Amn E Aem(1) + Aem(2). Since the Aem(2) components of a13 aInn and aInn a1.j are the same, A1.1 Aum E Aem(1). Likewise A13 Amn g Aer(1) for r = 1,3,n. Thus AiJ Amn = 0. Since Amn E Aei(0) we have Aii Amn = Amn Aii = 0. Again, these results will be of use to us in later developments. A ring is said to be stable if for every idempotent e of A we have Ae(u)Ae(1) 4- Ae(1)Ae(u) E Ae(l) for p = 0,2. By our multiplication it is readily seen that a ring A satisfying (19) is stable if and only if A(+) is stable. -25- Let e be an idempotent of the ring A. Then for x in Ae(2) and w in Ae(l) we have xow = wl + wo where w1 is in Ae(l) and w0 is in Ae(O). Then the mapping w -> w1 is an endomorphism of the module Ae(l) determined by the element x of Ae(2). We denote this mapping by S(x). By Albert's result [6, Theorem 1] the mapping x -> 2S(x) of the ring Ae(I)(2) onto the special Jordan ring of endomorphisms S(x) is a homomorphism with kernel Be where Be is the set of ele- ments x of Ae(+)(2) such that xow is in Ae(O) for all w in Ae(l). This is equivalent to saying that xw + wx is in Ae(O) for all w in Ae(l). Certainly Be is an ideal of Ae(+)(2) and we shall attempt to show that Be is in fact an ideal of Ae(2). LEMMA 2. If x is in Ae(2) and w is in Ae(l), xw = a1 + a0, wx = b1 + b0, then a0 = b0 and ale = ebl. PROOF. The result follows from our earlier remarks on such products. LEMMA 3. If x is an element of Ae(2) such that xw + wx is in Ae(O) for all w in Ae(l), then xw = wx is in Ae(O). PROOF. If xw + wx is in Ae(O). then by Lemma 2 we see that a1 = -b1. From our earlier results a1 - b1 = hdl where - - l _ _ _ _ = % a1. But then d1 = 0 so that a1 = b1 = O and xw = wx = a0 is in Ae(O). LEMMA k. B is an ideal of Ae(2). e PROOF. Let y be an arbitrary element of Ae(2). We set (xy)w = 81 + 80’ w(xy) = ai + a0. (yx)w = b1 + b0. w(yx) = bi + be. By various substitutions of x,y,w in (19) we find (xy)w + (xw)y + w(xy) + y(wx) = 2(yx)w + 2(wy)x and (yx)w + (yw)x + w(xy) + x(wy) = 2(xy)w + 2(xy)x. Since xw = wx is in Ae(O) and er(l) = Ae(l)x EAe(0), the first of these equations becomes (xy)w + w(yx) = 2(yx)w and the second implies that (yx)w + w(xy) - 2(xy)w is in Ae(O). Hence, a0 = b0, a1 + bi = 2bl, and b1 + ai = 2a1. Adding we find ai + bi = a1 + b1 and then using Lemma 2 we obtain ea1 + eb1 = eai + ebi = ale + ble = aie + bie = %(a1 + bl) = %(ai + bi). Then ea1 + ebi = 2eb10r eel = 2ebl - ebi = 2eb1 - ble. If ea1 = % a1 + d1, then ab = % b1 - d1 so that % a1 + c11 = b1 - 2d1 - % bl + d1. Thus %(al - bl), = -2d1 so that e(a1 - bl) = (a1 - b1)e = %(a1 - b1). Then e(a1 + b1) + e(a1 - bl) = 32*(a1 + b1) + ‘12‘(a1 - bl) = a1. Hence d1 = O and a1 = b1 = ai = bi. We note that we have actually shown that (xy)w = (yx)w = w(xy) = w(yx). Now set 2 = e in (3). Then consideration of the preceding remark yields 2(xy)w + 2((xy)w)e + 2(yw)x + (wy)x + ((yw + wy)e) + ((yw + wy)x)e = 0. Then using the properties of x we have 2(xy)w + 2((xy)w)e is in Ae(O). Hence 2al + 2a1e = 3a1 is in Ae(O) and since the characteristic of A is not three a1 = 0. This completes the proof. We now suppose A to be a simple ring with an identity 1 such that 1 = e + e + 1 2 e3 where the ei are pairwise orthogonal idempotents. LEMMA 5. Let e be an idempotent of A such that e f l and suppose x is an element of Ae(2) with the property that xw + wx = O for all w in Ae(l). Then x = O. PROOF. By Lemmas 2 and 3 we see that (xw)1 = (wx)1= 0 = (xw)O = (wx)o. Let Ce be the set of all such x in Ae(2). We claim that C6 is an ideal of A and since e‘# 1 and e is not in Ce it will follow that C6 = 0. By the above remark CeAe(l) = Ae(l)Ce = O and since Ae(2)Ae(O) = Ae(O)Ae(2) = O the proof will be complete if we show that Ae(2)Ce + CeAe(2) 9 Ce' Let y be in Ae(2). Then as in the proof of Lemma h we find (xy)w = w(xy) = w(yx) = (yx)w. Again as in the proof of Lemma h a substitution of x,y,w,e = z in (3) yields 2(xy)w + 2((xy)w)e - 0. Thus ((xy)w)0 = O and ((xy)w)l + ((xy)w)le = O which is impossible unless ((xy)w)1 = 0. Therefore xy and yx are in C9 as was to be shown. LEMMA 6. Set g = el + e2. Then Bg = Be1 + Be2’ PROOF. From an earlier remark we may set A = All + A12 + A22 + Al3 + A23 + A33 where All + A12 + A22 = Ag(2), Al3 + A23 = Ag(l), and A33 = Ag(O). Suppose X11 + X12 + x22 is in B8 c_: Ag(2). Then x12 = LI-(xll + x12 + x22)oe1 -1+((x11 + x12 + x22)oe1)oe1 and since Bg is an ideal of Ag(2), x12 is in Bg. Then e1(x11 + x22) = x11 is in BE so that x22 is in Bg also. Suppose a13 + a23 is in Ag(l). Then 2xl2 o(a13 + a23) = 2x12 oa13 + 2x12 oa23 which is an element of A23 + A13 by our remarks on the multiplication of the A13. But x12 is in B8 so that 2x12 o(a13 + a23) must be an element of Ag(O) = A33. Hence x12 o(a13 + a23) = 0. Apply Lemma 5 to obtain x12 = O and thus Bg 9 All + A22. For a12 in A12 we find x11 oa12 is an element of A12 + A22. But x11 is in B8 which is an ideal of Ag(2) so that x11 oa12 is in A22. By the definition of Bg we have x11 oa13 is an element of Ag(O) = A33. Thus if w is an element of A12 + A13 = Ae1(1), then x11 ow is in A22 + A33 = Ae(O). Therefore x11 is in Bel and in a similar manner x22 is in B62. Hence B8 9 Be1 + Be2. If x11 is in Bel, then x11 0(A12 4- A13) 91822 4- A23 4- A33 = Ae1(0). But then xlloAg(1) = X110(A13 I A23) 3 x11°A13 9 A33 = Ag(O). Hence Bel ng d C213 . h s B + B = B . an likewise Be2 _ g T u el 92 g LEMMA 7. B = Be + B + B is an ideal of A. 1 e2 e3 PROOF. Since Bg = Be + Be is an ideal of Ag(2), we 1 2 have Ag(2)B + B Ag(2) 9B + B . If h = e + e , then e1 e1 e1 e2 1 3 Ah(2)B + B Ah(2) EB + B . Since A c_:A (o) B A e1 el 81 e3 23 e1 ' el 23 = A23Be1 = 0. Thus ABel + BelA = (All + A12 + A22)Be1 Be1(A11 + A12 + A22) + (All + A13 + A33)Be1 + 391(A11 + A13 + A33) 4' A23Bel + BelA23 E- Bel + B,22 + Be3. Interchanging subscripts we find AB + B A g B + B 4- Be and AB e2 e2 e1 e2 3 e + Be A 9 Be + Be 4- Be . Therefore B is an ideal of A. 3 1 2 3 Either B = A or B = 0 since we assumed A to be simple. B = A is impossible since B EIAll + A22 + A33. Thus we have _ - - (+) - (+) (+) (+) e 2 Ay>=ap.qy.qyanly>.gy.gy.nyam Jordan rings. We now refer the reader to the proof of Theorem 1 in Albert's paper [6] and note that his combinatorial type proof will suffice to show that A(+) is a Jordan ring. We new state the following: THEOREM 10. If A is a simple strictly power-associative ring of characteristic prime to six satisfying (19) and possessing an identity element which is sum of three pairwise orthogonal idempotents, then A is Jordan-admissible. Now suppose that e is a principal idempotent of the strictly power-associative algebra A which has characteristic -30- prime to six and satisfies (19). Then e is also a principal idempotent of A and by [10, Theorem 5] Ae(l) + Ae(O)§;radica1 of A(+). Hereafter we shall denote the radical of A(+) by RadA(+). We shall attempt to show that the ideals generated in A and A(+) by Ae(l) + Ae(O) are the same. LEMMA 8. If Ae(l) C_-'- Rad A“) and z and w are elements of Ae(l), then zw and wz are in the Rad A(+). Also (zw)2 = 2(ezow)2. PROOF. Substitution in (19) yields (ez)w + w(ze) + (ew)z + z(we) = 2(we)z + 2(ze)w or equivalently (36) zw - (ze)w + w(ze) + wz - (we)z + z(we) = 2(we)z + 2(ze)w. Another substitution yields (ze)w + w(ez) + (zw)e + e(wz) = 2(wz)e + 2(ez)w or equivalently (37) (ze)w + wz - w(ze) 4- (zw)e + e(wz) = 2(wz)e 4- 2zw - 2(ze)w. Adding (36) and (37) we obtain (38) 2wz + (zw)e + e(wz) + z(we) = zw + 2(wz)e + 3(we)z Set zw = a2 + a1 + a0, wz = b2 - a1 + b0. Equating the Ae(2) components of (38) we obtain 2b2 + a2 + b2 + (z(we))2 = 82 + 2b2 + 3((we)z)2. Thus b2 = -(z(we))2 + 3((we)z)2. Let we = % w + w'. Then b2 = - % a2 - (zw')2 + 5 b2 + 3(w'z)2 and so %(a2 - b2) = 3(w'z)2 - (zw')2. But, if we carry through the same argument with w replaced by w', we find (w'z)2 = (zw')2. Thus %(a2 - b2) = 2(w'z)2. We note that 2(ewoz)2 = ((ew)z + z(ew))2 = %(wz)2 + 32‘(zw)2 - (w'z)2 - (zw')2 so that 2(ewoz)2 = %(b2 + a2) - 2(wlz)2 = %(b2 + a2) - %(a2 - b2) = b2. But b2 = (wz)2. By consideration of the Ae(O) components of (38) we obtain 2bO + (z(we))o = a0 + 3((we)z)o. Hence 2bO - a0 = g be - % 80 + 3(wuz)0 - (zw')0 or é-(bO - a0) = 3(w'z)0 - (zw')0. As before (w'z)o (zw')0. Then 2(ewoz)0 = ((ew)z + z(ew))O = %(b0 + a0) - l - l _ _ - - 2(w'z)o - 2(b0 + ao) 2(bo a0) — aO - (zw)e. Then 1' 2(ewoz) = %(a2 + b2) +15% + b0) - FAQ - a0) - 2‘82 - b2) F = b2 + a0 which is in the Rad A(+). Hence b2 and a0 are in Rad A(*) and likewise a2 and b0 are in Rad A(+) so that zw and wz are in Red A(+). LEMMA 9. If x is in Ae(p ) and w is in Ae(l) for p = 0,2 and if either x is in Rad A(+) or Ae(l) gZRad A(+), then xw and wx are in Rad A(+). PROOF. Set we = % w + w’. Substitution in :19) yields (39) x(we) + w(xe) + (ew)x + (ex)w = 2(xe)w + 2(we)x Suppose x is in Ae(2). Then from (39) we have x(we) + wx + (ew)x + xw = 2xw + 2(we)x or xw - wx = x(we) + (ew)x - 2(we)x = x(we) + wx ~ (we)x - 2(we)x = % xw + xw' + wx - % wx - 3w'x and hence %(xw - wx) = xw' - 3w'x. But by the same argument with w replaced by w' we find xw' = w'x so that -hw'ox = %(xw - wx). Hence if x is in Rad A(+), (+) then xw - wx is in Rad A But 2xow = xw + wx is also in Rad A(+) so that xw and wx are in Rad A(+). Suppose x is in Ae(O). Then (39) becomes x(we) + (ew)x = 2(we)x and ~ _ ' _ _ ' u- r, ~" '. s' 1'. erjzme'm -32- NH“ xw + xw' + % wx - w'x = 2w'x - wx or %(xw - wx) = 3w'x - xw' = 2w'x = hw'ox. Hence if x is in Rad A(+), then so are xw and wx. The result is clear if we replace the con- dition that x is in Rad A(*) by the condition that Ae(l) is contained in Rad A(+). LEMMA 10. The ideal generated in A by Ae(l) + Ae(O) is contained in Rad A(*) if Ae(l) + Ae(O) is contained in Rad A(+). PROOF. We shall show that the ideal in A generated by Ae(l) + Ae(O) is contained in N + Ae(l) + Ae(O) where N is defined as the set of all finite linear combinations of elements of the form (xlxo)2 and (xlzl)2. By our multipli— cation we see that it is sufficient to show that NAe(2) + Ae(2)N g N. Set L = N + Ae(l) + Ae(O). Let y and 2 be in Ae(l) and x in Ae(2). Then x(yz) + y(xz) + (zy)x + (zx)y = 2(xz)y + 2(yz)x. Thus x(yz) + (zy)x - 2(yz)x is in L. Also x(zy) + z(xy) + (n) + (yx)z = 2(xy)z + 2(zy)x so that x(zy) + (zy)x - 2(zy)x is in L. Adding these two expressions we find x(yz + zy) - (yz + zy)x is in L. Now in ( 3) we substitute x,y,z,w = e obtaining 2x(zy + yz) + 2(zy + yz)x + y(xz + zx) + (xz + zx)y + z(xy + yx) + (xy + yx)z ((xy + yx)z)e + ((xy + yx)e)z + ((xz + zx)e)y + ((xz + zx)y)e + 2(xy)z + 2(xz)y + (yz)x + (yx)z + (zy)x + (zx)y + ((zy + y2)e)x + ((zy + yz)x)e. By the restrictions on x,y,z and the assumption that Ae(l) 4- Ae(O) g L we find that 2x(zy + yz) - (zy + yz)x is an element of L. From above x(yz + zy) - (yz + zy)x is in L and hence x(yoz) and (yoz)x are in L. Thus 2x(eyoz)2 is in L and by Lemma 8 x(yz)2 is in L. Now let x be in Ae(2), y in Ae(l), and z in Ae(O). Set yz = a2 + a1, zy = a2 + b1. Then substitution in (19) yields x(yz) + (zy)x = 2(yz)x or xa2 + a2x = 2a2x. Thus sax = xa2. Substitution of x,y,z,w = e in (3) yields 2x(yz + zy) +2(zy + zy)x = ((xy + yx)z)e + ((xy + yx)e)z + 2(xy)z + (yz)x + (yz)x + (yx)z + ((zy + yz)x)e + ((zy + yz)e)x. Hence 2x(yz + zy) + 2(yz + zy)x - (yz)x - ((zy + yz)x)e - ((zy + yz)e)x is an element of L so that 2x(yz + zy) - (yz)x is an element of L. Thus 2x(a2 + a2) - 82x = 3xa2 is in L. Since the characteristic of A is prime to six, xa2 = x(yz)2 is in L. Thus the ideal generated by Ae(l) + Ae(O) is contained in N + Ae(l) + Ae(O). Therefore this ideal must be equal to N + Ae(l) + Ae(O). If e is a principal idempotent of A, then e is also a principal idempotent of A(I) and as we have stated earlier Ae(l) + Ae(O) EiRad A(I). By the preceding lemma N + Ae(l) + Ae(O) is an ideal of A which is contained in Red A(+). Hence N + Ae(l) + Ae(O) is a nil ideal of A. If we now suppose A to be semi-simple, then N + Ae(l) + Ae(O) = 0 so that A = Ae(2). Therefore e is an identity for A and we state this as THEOREM 11. Every semi-simple strictly power-associative algebra of characteristic prime to six satisfying (19) has an identity element. -3h- Suppose D is an ideal of a semi-simple strictly power- associative algebra A of characteristic prime to six satis- fying (19). If D fl'O then since D is not nil D must possess an idempotent e and we may suppose e to be principal. Then D = De(2) + De(l) + De(O). Since e is an element of D we must have Ae(2) and Ae(l) contained in D. Hence De(2) = Ae(2) and De(l) = Ae(l) so that we may write D = Ae(2) + Ae(l) + De(O). Let M be the radical of D. Then since e is princi- pal, De(l) + De(O) E M. In order that M be an ideal of A we see that it is sufficient to show that Ae(O)M + MAe(O) §;M. Let x be in Ae(O) and m = m2 + m1 + m0 be an element of M. Then xm = xm1 + xmO where xm1 is in Ae(2) + Ae(l) and xmO is in De(O) since D is an ideal of A. Hence it is sufficient to show that (xml)2 is in M. Since (xml)2 = (m1!)2 we need only show (xoml)2 is in M. We note that (3) holds in A(I) which is commutative. Setting x = y and z = w = m1 in (3) 2 2 2 for A(+) we find 8x oml + 16(xoml) = ’+(x20m1)om1 + h(m§ ox)ox + 8((xom1)ox)om1 + 8((xoml)oml)ox. Thus we have ‘+(xoml)2 - 2((xoml)om1)ox = (x20ml)oml + (mi ox)ox + 2((xoml)ox)oml + 2x20mi. Remembering that m1 is in M[W‘De(l) = De(l) and De(l) 4- De(O) QM we obtain (x20ml)om1 is in M, (miox)ox is in Ae(0)n D = De(O) 9M, 2((xoml)ox)oml is in DoM EM, 2 and 2x2oml is in Ae(O)fl D = De(O) EM. Thus the right hand member of the above equation is an element of M so that w ‘1 /,.' 5 4-1- .. i ”r . . . «ev- . ' ass-.4. ~w.::‘:.' .4 m‘“ - . ~“ 4’ 2(xoml)2 - ((xoml)oml)ox is also in M. Then the De(2) component of this expression is in M. Set xoml = b2 + b1' 2 2 Then 2(b2 + 2b2 ob1 + b1) - (b2 om1)ox - (bl oml)ox is in M and the De(2)component is 2b§ + 2(bi)2-((b20m1)ox)2. Using [6, Identity 8] we obtain ((b2 om1)ox)2 = ((b2 oml)lox)2 = %(m1 ox)2 ob2. But %(m1 ox)ob2 = % b3. Therefore g'bg : l - 2(bi)2 is in M. But (bi)2 is in M since b1 is an element of De(1)C__:M. Thus since the characteristic of A is not three, b3 must be in M. Every element of M is nilpotent so that b2 is nilpotent. Now consider the ideal of D generated by M and all elements of the form b2 = (xm1)2. If we can show that y(xm1)2 is of the form (xm1)2 for all y in Ae(2) = De(2), then this ideal will be a nil ideal of D containing the maximal nil ideal M leaving b2 = (xm1)2 in M as the only remaining possibility. Substitution of x,y,m = z, and w = e in (3) yields 2y(xml + mlx) + 2(xm1 + mlx)y = 2(ym1)x + ((ym1 + mly)x)e + ((ym1 + mly)e)x + (mly)x + (mlx)y + ((mlx + xml)y)e + ((mlx + xm1)e)y. Considering the Ae(2) components and using (y(xm1))2 (y(mlx))2 = ((mlx)y)2 = ((xml)y)2 we find 8(y(xm1))2 2((le)x)2 + ((ym1 + mly)x)2 + (((yml + mly)e)x)2 + ((mly)x)2 + (y(xml))2 + 2(y(xm1))2 + 2(y(mlx))2. Simplifying we obtain 3y(xml)2 = ((3ym1 + 2m1y + (ym1 + mly)e)1x)2 -36- and we observe that (3yml + 2mly + (ym1 + mly)e)l is an element of M r)Ae(l). Thus, since the characteristic of A is not three, y(xml)2 is of the desired form and hence is nilpotent. Therefore M is a nil ideal of A which we had assumed to be semi-simple; therefore M = Ae(l) = De(O) = 0. Thus A = Ae(2) G Ae(O) where Ae(2) and Ae(O) are semi-simple t algebras with identity elements a and 1 - e respectively. Proceeding in the usual manner we may now state . » “l‘V‘J-h.ku;.l . THEOREM 12. Every semi-simple strictly power-associative algebra A over a field F of characteristic prime to six and satisfying (19) has an identity and is the direct sum of simple algebras. In the following we shall suppose A to be a strictly power-associative algebra over a field F of characteristic prime to six such that A is simple and satisfies (19). More- over, we assume that A has an identity 1 which can be written as 1 = e1 + e2 where the e1 are pairwise orthogonal idempotents of A and that A is Jordan-admissible. A is simple over its center and simple over the algebraic closure K of its center. Thus AK is a simple algebra which is Jordan-admissible and we may suppose F = K. Then we can set 1 = u + v where u is primitive idempotent of A. Since A(+)u(2) is a Jordan algebra, we may use Theorem A of Jacobson's paper [8] to write Au(2) = uK + N where N is the ideal of nilpotent elements of A(+)u(2). Suppose N is not an ideal of Au(2). Then there are elements x and y in N such that xy = u + n and yx = -u + n' where n and n' are in N. Suppose w is in Au(l). Then substitution in (19) yields (xy)w + (xw)y + y(wx) + w(yx) = 2(yx)w + 2(wx)y. Then uw + nw + (xw)y + y(wx) - wu + wn' = -2uw + 2n'w + 2(wx)y. Rearranging terms we find 3uw - wu = 2n'w - nw - wn' + 2(wx)y - y(wx) - (xw)y. Since A(+) is a Jordan algebra, it is stable [2] and by our earlier remark A is stable. Then applying Lemma 9 to each term of the right hand member of the above relation we find that the right hand member is in Bad A(+) and, consequently, the left band member is also in Rad A(+). Now we set uw = % w + w' where w'u = uw' = % w'. Since w was an arbitrary element of Au(l), 3uw' - w'u = w' .l is also in Rad A(+). But then 3uw - wu = 3 w + 3w‘ - 2 w + w' so that w is in Rad A“). Thus Au(l) Q Rad A“) and an application of Lemma 8 yields Au(l) 4- Au(l)Au(l) gRad A“). We claim Au(l) + Au(l) Au(l) is, in fact, an ideal of A. Let x be in Au(2) and y and z in Au(l). Substitution in (3) yields 2x(wz + zw) + 2(wz + zw)x + w(zx + xz) + (zx + xz)w + z(xw + wx) + (xw + wx)z = 2(xw)z + 2(xz)w + ((xz + zx)e)w + ((xz + zx)w)e + ((xw + wx)e)z + ((xw + wx)z)e + ((zw + wz)e)x + ((zw + wz)x)e + (wz)x + (wx)z + (zw)x + (zx)w. Using the stability of A we find 2x(zw + wz) - (zw + wz)x is an element of Au(l)Au(l). Two different substitutions in (19) yield the following relations: x(wz) + (zw)x - 2(wz)x is in Au(l)Au(l) and x(zw) + (wz)x - 2(zw)x is in Au(l)Au(l). Add- ing these two we have x(wz + zw) - (wz + zw)x is in Au(l)Au(l). Combining this with the above remark that . ‘3‘ fixfifiw‘hffi: "f” ' “n . :4. 2x(zw + wz) - (wz + zw)x is in Au(l)Au(l), we find x(wz + zw) is an element of Au(l)Au(l). Replacing z by us and applying Lemma 8, we see that x(zw)2 = x(uzow) is an element of Au(l)Au(l). This result along with the stability of A makes Au(l) + Au(l)Au(l) closed under multiplication by Au(2). We note that Av(2) = Au(O) and Av(O) = Au(2). In- terchanging the roles of u and v, we find that Au(l) F + Au(l)Au(l) = Av(l) + Av(l)Av(l) is closed under multipli- A J cation by Au(O). Noting the stability of A it is clear that F Au(l) + Au(l)Au(l) is closed under multiplication by Au(l). Hence, Au(l) + Au(l)Au(l) is an ideal of A which is contained in Rad A(+) and since the simple algebra A contains an idempotent, we must have Au(l) = 0. But then A = Au(2) G Au(O) which is impossible by the simplicity of A. Thus N is an ideal of Au(2). Now suppose I = e1 + ... + et where the e1 are primitive pairwise orthogonal idempotents of A. Then for an arbitrary element x of A we may write x = Z: kie1 + X: x1‘, + z: xi I where k1 are in K, x1.} are in A1.j for i f'j, and x1 are elements of Ni where N1 is the ideal of nilpotent elements of Aei(2). Setting S(x) = Z: ki defines a bilinear function on A to K and in order that 8 be an admissible trace function [7] for A and A{+) it is necessary to prove that the following conditions are satisfied by 8 : (We only show properties 1 and 2.) <1) é§(xy) = = 80:2) + S(bo) = S(yx). We next consider condition (2). We must show that S(x(yz)) = S((xyh). We restate (19) as x(yz) 4- y(xz) + y(xz) + (zy)x + (zx)y = 2(xz)y 4- 2(yz)x. Using S(xy) S(yx) we obtain S(x(y2)) + S(y(xz)) - S((zyht) S((th') = 0. Hence S(y(zx)) - 8((y2)x) = S((xzh') S(x(zy)) or S(y(zx) - (yz)x) = S((xz)y - x(zy)). Then it S(xo(y02) - (xoy)oz) = S(XU'Z) + x(zy) 4- (yz)x 4- (zy)x -(xy)z - Z(yx) - z(xy) - (yx)z) = S(dez) - (xy)z) + S(xuy) - (xz)y) + S((YZM - y(zx)) + S((zyh: - z(yx)) = 1+ 8(x(yz) - (XY)Z) (By use of 80:30 = S(yx) and 8 (y(zx) - (yz)x) = S((xzhr - x(zy))). Thus it is sufficient to show that (2) holds in A(*). Let the coefficient of ei in the decomposition of x,y,z be respec- tively 41,31,111. Then 2 S(zo(xoy)) = 2 Z: c.11Bip1 + 2 8[2: (P191 4' "jejbuijoyiji’ + S[: (<1i + ‘13)21303'13] A ' --"E'x;‘—~&W‘n tmnmiW' *Mn + S[Z (Bi 4- Bjnijoxijj + 2 8[E zst°(xij°yjt)] F + 2 S 2: zsi°(xij°yjt) + 2 8L: zito(xijost) . From the proof of (l) we see that S[ei°(x13°yij)] =% S(xijoyij) = g[eJo(xiJoyij)]. Hence if we consider the above expression with x and z interchanged, we observe I. that it will suffice to show that <¢\[ziko(xijoym)] = S(xijdzikoydkn for i,J,k distinct since 8 [zsto(xijoy3k)] = 0 unless zst is in A11! and i,J,k are distinct. We now state the Jordan identity for A”) in its linearized form. 0+0) Z (xoy)o(woz) ‘ Z ((xoy)ow)oz (symmetric in x,y,z). Set x = xij’ y = ykj’ w = 21k, and z = ei. Then we find (xijoyjk)°zik + (zikoyjk)°xij = 2((xijoy3k)°zik)°ei + (xii ozik)°yjk' Then interchanging x and z, and i and k we have (zikoyjk)°xij + (xijokaMz1k = 2((zikoyjk)°xij)°ei + (zikoxij)°yjk' Subtracting we have ((xijoyjk)°zik)°ei = ((zikoyjk)°xij)°ei' Then, using the above remarks we obtain S(((xij°yjk)°zik)°ei) = Co\(((xijo(kaozik)oe1) =32; S(xijo(yjkozik)) =% S((xijoyjk)°zik)' Therefore (2) holds. By (1) and (2) the set N8 of all x in A such that 81x3) = O for all y in A is an ideal of A. Surely Rad A(+)SED% and, since A is simple and S(el) % O, 2 Had A£+J r C. Thus A)+) is a simple Jordan algebra N3 [3; Ch. V; Theorem 8]. THEOREM 13. If A is a simple strictly power-associative algebra over field K of characteristic # 2,3 which satisfies (19) and such that A(+) is a Jordan algebra of degree t > 1, then A(+) is a simple Jordan algebra. We now reproduce Albert's argument [k] to show that under the hypotheses of the above theorem A is flexible. Set x = z in (2). Then w = (xy + yx)x - x(xy + yx) + x2y - yx2 = 0 and so wz = zw = O for all z in A. Interchanging y and z and adding we obtain ((xy)x)z + ((yx)x)z - (x(xy))z - (x(yx))z + (x2y)z - (yx2)z + ((xz)x)y + ((zx)x)y - (x(xz))y - (x(zx))y + (x2z)y - (2x2)y = 0. Then applying S; and using properties (1) and (2) repeatedly we find22 S((xy)(xz)) = 2 S((zx)(yx)). Applying (2) to the left hand member and (2) and (l) to the right hand member we find S(((xy)x)z) = S(((X(yx))z) so that S(((xy)x - x(yx))z) = 0 for all x,y,z in A. Hence (xy)x - x(yx) is in Rad A and since A is simple we are finished. THEOREM 1%. A simple strictly power-associative algebra A over a field K of characteristic 7! 2,3 which satisfies (19) is (a) a commutative Jordan algebra; (b) a quasi-associative algebra; 2 2 _ Section V. The Identities Rx + Lx - LxRx + RxLx and (Rx + Lx)(Ry + Ly) = (Ry + Ly)(Rx + Lx) We shall present in this section some examples of algebras satisfying (15) and (17). First consider the algebra A over a field F of characteristic zero with a basis e1,a, and e2 where the ei are orthogonal idempotents such that e1 + e2 = 1, ela = ae2 = 1 + % a, e2a = ae1 = -1 + % a, and all the remaining products are zero. In order to show that A is power-associative we must show xx2 = x2x and 2 2 2 (x x)x = (x ) for all x in A. Set x = a1 el + ca + a2 e2. Then x2 = mi el + e(d1 + a2)a + a: e2 and x2x = a; el + C(Gi + a1 a2 + a§)a + a; e2 = xx2. We see that (x2)2 = a: e1 + e(d1 + a2)(d§ + a§)a + a: e2 and (x2x)x (mi e1 + 0(oi + a1 a2 + a§)a + mg e2) (ul el + d a + «2 e2) 2)2. This along a: e1 + e(al + a2)(qi + a§)a + a: e2 = (x with [1, Lemma h] implies that A is power-associative. We set x = :1 e1 + a a + d2e2 and y = B1 e1 + B a+B2 e2. Then XV I (“131 ' QBl I “13 ' “29 I up2’91 I 2(“51 I “18 I “23 I “52)3 I (“232 ' “51 I “13 I “23 I up2’92 and yx = (alB1 + aBl - alB + a2B - aB2)e1 #7753117 xw'- ‘ - A ...l -hs- I 2 (“51 I u1B I “23 I “‘32)at I (“252 I GBl ' “15 I “25 ' “52’92' Hence xy - yx = 2(a1B - aBl + aB2 - 62B)e1 + 2(a15 - aBl + aB2 - GZB)e2 = 2(a1B - aBl + aB2 - a2B)l. Thus xy - yx commutes with all elements 2 of A so that z(xy - yx) = (xy - yx)z and (17) holds. Suppose L is an ideal of A and x = alel + as + a2e2 % 0 l is an element of L. Then elx + xe1 = 2a1e1 + «a is in L. : “Mr—us» , _.£.__ Thus 2a1e1 + ea - x = alel - d2e2 is in L. If either a1 or a2 is not zero then by the orthogonality of the e1 either e1 or a2 is in L. But then eia + aei = a is in L implying that ela - % a = 1 is in L. Thus L = A. Suppose a1 = a2 = 0. Then a f 0 so that a is in L and, as above, 1 is in L. Therefore A is a simple power-associative algebra satisfying (17) which is not flexible since e1(ael) = - % e1 + k a + % e2 - 1 _ 1 and (ela)e1 — 2 e1 + & a 2 e2. We may construct new examples by setting A equal to the algebra over a field F of characteristic zero with a basis e1,..., en, aiJ’ i < J = 1,...,n, where 1 = e1 + ... + en, the ei are pairwise orthogonal idempotents, eiaij = aiJeJ - l - _ _ l - l + 2 813’ eJaiJ - aiJei - l + 2 aij’ and all other products are zero. We note that the non-flexible examples given earlier which satisfy (11) also satisfy (17). This is not too surprising when we observe that any algebra A which -hé- satisfies (11) and either (15) or (17) must satisfy all three. Identities (15) and (17) are not strong enough conditions in themselves to enable us to obtain any significant results concerning algebras which satisfy either of them. It is not evident to us at this time what other conditions we might impose on these algebras. 7. 9. 10. ll. BIBLIOGRAPHY Albert, A. A., On the power-associativity of rings, Brasiliensis Mathematicae, Vol. 2 (l9h8), 1-13. , A structure theory for Jordan algebras, Ann. of Math. #8 (l9h7), 5h6-567. , Power-associative rings, Trans. Amer. Math. Soc. 6% (19h8), 552-597. , A theory of trace-admissible algebras, Proc. Nat. Acad. Sci., U.S.A., 35 (l9h9), 317-322. 1, Almost alternative algebras, Portugaliae Mathematicae, 8 (l9h9), 23-36. , A theory of power-associative commutative algebras, Trans. Amer. Math. Soc. 69 (1950), 503-527. , The structure of right alternative algebras, Ann. of Math. 59 (195%), h07-hl7. Jacobson, N., A theorem on the structure of Jordan algebras, Proc. Nat. Acad. Sci. U.S.A., M2 (1956), 1h0-1h7 e Kleinfeld, Erwin, Rings of (’3’, S ) type, Portugaliae Mathematics, 18 (1958), 107-110. Kokoris, L. A., New results on power-associative algebras, Trans. Amer. Math. Soc. 77 (195%), 363-373. , On a class of almost alternative algebras, Canad. J. of Math. 8 (1956). 250-255. -hg- 12. , 0n rings of (’3), 8 ) type, Proc. Amer. Soc. 9 (1958), 897-90N. l3. Oehmke, Robert H., 0n flexible algebras, Ann. Math. 68 (1958), 221-230. 1%. Schafer, R. D., Noncommutative Jordan algebras of characteristic 0, Proc. Amer. Math. Soc. 6 (1955), h72—h75. ,'-' .z.._ ’ .7 . .. .u‘ x '33 7- :32:- mac-m m3. ': :- . IaWK ilulllle I‘ll", l. .I‘Il! n 4' .3 I- -_.4 ' ”VII-a ._ FCS‘ CC GNU