Jordan isomorphisms of triangular matrix algebras over a connected commutative ring

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Linear Algebra and its Applications 312 (2000) 197 201 www.elsevier.com/locate/laa Jordan isomorphisms of triangular matrix algebras over a connected commutative ring K.I. Beidar a,,m.brešar b,1, M.A. Chebotar c a Department of Mathematics, National Cheng-Kung University, Tainan, Taiwan, ROC b Department of Mathematics, University of Maribor, Maribor, Slovenia c Department of Mechanics and Mathematics, Moscow State University, Moscow, Russian Federation Received 17 December 1999; accepted 24 February 2000 Submitted by C.-K. Li Abstract Let C be a 2-torsionfree commutative ring with identity 1, and let T r (C), r 2, be the algebra of all upper triangular r r (r 2) matrices over C.ThenC contains no idempotents except 0 and 1 if and only if every Jordan isomorphism of T r (C) onto an arbitrary algebra over C is either an isomorphism or an anti-isomorphism. 2000 Published by Elsevier Science Inc. All rights reserved. Keywords: Triangular matrix algebra; Jordan isomorphism Let C be a commutative ring with identity 1 and let T and A be algebras over C. Recall that a bijective C-linear map ϕ : T A is called a Jordan isomorphism if ϕ(ab + BA) = ϕ(a)ϕ(b) + ϕ(b)ϕ(a) for all A, B T. Isomorphisms and antiisomorphisms are obvious examples of Jordan isomorphisms, and often it turns out that they are actually the only possible examples. However, not always: Let ε C be an idempotent and let A Ã be an anti-automorphism of the algebra T.ThenA εa + (1 ε)ã is a Jordan automorphism of T which is neither an automorphism nor an anti-automorphism, unless one of the ideals (1 ε)t or εt is commutative. Corresponding author. E-mail addresses: beidar@mail.ncku.edu.tw (K.I. Beidar), bresar@uni-mb.si (M. Brešar), mchebotar@tula.net (M.A. Chebotar). 1 Partially supported by a grant from the Ministry of Science of Slovenia. 0024-3795/00/$ - see front matter 2000 Published by Elsevier Science Inc. All rights reserved. PII:S0024-3795(00)00087-2

198 K.I. Beidar et al. / Linear Algebra and its Applications 312 (2000) 197 201 The study of Jordan isomorphisms of associative rings and algebras, primarily concerned with their relations to (associative) (anti)isomorphisms, goes back to Ancochea [1,2], Kaplansky [12], Hua [10], Jacobson and Rickart [11], Herstein [9] and Smiley [19]. More recently, some further progress has been made by Baxter and Martindale [3], McCrimmon [18], Brešar [6,7], Lagutina [13], Martindale [16] and Beidar et al. [4]. In all these papers, some structural properties of certain classes of rings and algebras have been used, and apparently none of them cover the problem of describing Jordan isomorphisms of T r (C), the algebra of all r r upper triangular matrices over a commutative ring C (we remark that as simple as this algebra appears to be, from the structural point of view it is incomparably more complicated than, say, the algebra of all r r matrices). Using linear algebraic techniques, Molnár and Šemrl [17] recently proved that automorphisms and anti-automorphisms are the only Jordan automorphisms of T r (F),whereF is a field with at least three elements. On the other hand, Dokovič [8] considered an analogous problem concerning Lie automorphisms on triangular matrix algebras (see also [14,15]). The main result in [8] characterizes Lie automorphisms of T r (C), wherec is a commutative ring with 1 which is connected, i.e., a ring in which the only idempotents are 0 and 1. The goal of the present paper is to obtain a Jordan analogue of this Lie type result, which also generalizes the theorem of Molnár and Šemrl. Our result is as follows. Theorem. Let C be a 2-torsionfree commutative ring with identity. The following two conditions are equivalent: (i) C is a connected ring; (ii) every Jordan isomorphism of C-algebra T r (C), r 2, onto an arbitrary C-algebra is either an isomorphism or an anti-isomorphism. The condition that C is 2-torsionfree means that 2A = 0 with A T r (C), implies A = 0. We remark that in the case when 2T r (C) = 0 the concept of a Jordan isomorphism coincides with that of a Lie isomorphism, so that for this case we can refer to [8]. Let us also mention another reason for our interest in Jordan isomorphisms of upper triangular matrix algebras. The need to know their form appears in our forthcoming paper [5] which is devoted to functional identities on these algebras. Before proving the theorem we first give some general comments on Jordan isomorphisms. Let T and A be 2-torsionfree algebras over a commutative ring C.Then every Jordan isomorphism ϕ : T A clearly satisfies ϕ(a 2 ) = ϕ(a) 2 for all A T. Further, from 2ABA = A(AB + BA) + (AB + BA)A (A 2 B + BA 2 ) we see that ϕ also satisfies ϕ(aba) = ϕ(a)ϕ(b)ϕ(a) for all A, B T. This obviously yields ϕ(abc + CBA) = ϕ(a)ϕ(b)ϕ(c) + ϕ(c)ϕ(b)ϕ(a) for all A, B, C T. Now let E T be an idempotent. Suppose that AE = EA = 0forsomeA T. In particular, EA + AE = 0 = EAE and so ϕ(e)ϕ(a) + ϕ(a)ϕ(e) = 0 = ϕ(e)ϕ(a)ϕ(e). However,ϕ(E) is an idempotent in A and so these two identities

K.I. Beidar et al. / Linear Algebra and its Applications 312 (2000) 197 201 199 readily imply ϕ(a)ϕ(e) = ϕ(e)ϕ(a) = 0. Since ϕ 1 is also a Jordan isomorphism, the same argument shows that the converse is also true. That is, given A, E T with E 2 = E, wehave EA = AE = 0 ϕ(e)ϕ(a) = ϕ(a)ϕ(e) = 0. (1) Proof of the theorem. Assume first that C is not connected, i.e., there is an idempotent ε C different from 0 and 1. From the observation at the beginning of this note it is clear that in order to construct a Jordan automorphism of T r = T r (C) which is neither an automorphism nor an anti-automorphism it is enough to find an anti-automorphism of T r. They do exist indeed. For example, the map A UA tr U 1 is an anti-automorphism (as a matter of fact, even an involution) of T r,wherea tr denotes the transpose of A and U = E 1r + E 2r 1 + +E r 12 + E r1, cf. [8,14,17] (here, E ij denotes a matrix unit). Assume now that C is connected and that ϕ is a Jordan isomorphism of T r onto an algebra A over C. We must show that ϕ is either an isomorphism or an antiisomorphism. According to (1) we have E ii A = AE ii = 0 ϕ(e ii )ϕ(a) = ϕ(a)ϕ(e ii ) = 0 (2) holds true for all A T r and i = 1,...,r. This is the key observation, as we shall see. We proceed to prove the theorem by induction on r. Soletfirstr = 2. Set e = ϕ(e 11 ), f = ϕ(e 22 ) and n = ϕ(e 12 ). Clearly, e 2 = e, f 2 = f and n 2 = 0. Further, applying (2) we see that ef = fe = 0, and from the identities E 12 = E 11 E 12 + E 12 E 11 = E 22 E 12 + E 12 E 22 and E 11 E 12 E 11 = E 22 E 12 E 22 = 0weinfern = en + ne = fn+ nf and ene = fnf = 0. Using these relations it is easy to see that ϕ is an isomorphism if and only if n = en (i.e., when ne = 0), and ϕ is an anti-isomorphism if and only if n = ne (i.e., when en = 0). Thus, it suffices to show that either en = 0oren = n.sinceϕ is a linear isomorphism, the elements e, f, n A span the C-module A. Therefore, en = λe + µf + νn for some λ, µ, ν C.However,multiplying this relation from the left- and right-hand sides by e and using ene = ef e = 0wegetλe = 0. Similarly, we see that µf = 0. Hence, en = νn which yields νn = en = e(en) = e(νn) = ν 2 n.thatis,ϕ((ν 2 ν)e 12 ) = 0andsoν 2 = ν. However, C is assumed to be connected, so it follows that ν = 0orν = 1. That is, either en = 0 or en = n. This proves the theorem for r = 2. Now let r 3. Let U be the set of all matrices A = (a ij ) 1 i j r T r such that a 11 = a 12 = =a 1r = 0. That is, U consists of matrices whose first row is zero. Similarly, let V be the set of all matrices in T r whose rth column is zero (i.e., a 1r = a 2r = =a rr = 0). Note that U and V are subalgebras of T r (moreover, U is a right and V is a left ideal of T r ), and they both are isomorphic to T r 1 (C). Clearly, A T r belongs to U if and only if AE 11 = E 11 A = 0 (actually, the condition E 11 A = 0 is sufficient, but we need both relations in order to apply (2)). Similarly, A T r belongs to V if and only if AE rr = E rr A = 0. Now, in view of (2), an element a A lies in ϕ(u) if and only if aϕ(e 11 ) = ϕ(e 11 )a = 0, and a A lies in ϕ(v) if and only if aϕ(e rr ) = ϕ(e rr )a = 0. From this we clearly

200 K.I. Beidar et al. / Linear Algebra and its Applications 312 (2000) 197 201 infer that ϕ(u) and ϕ(v) are (associative) subalgebras of A. Therefore, the restriction of ϕ to U is a Jordan isomorphism between associative algebras U and ϕ(u). Since U is isomorphic to T r 1 (C), the induction assumption can be used. Therefore, the restriction of ϕ to U is either an isomorphism or an anti-isomorphism. Similarly, the same is true for the restriction of ϕ to V. Suppose, for instance, that ϕ is an isomorphism on U and an anti-isomorphism on V. Since E 13 = E 12 E 22 E 23 + E 23 E 22 E 12, we have ϕ(e 13 ) = ϕ(e 12 )ϕ(e 22 )ϕ(e 23 ) + ϕ(e 23 )ϕ(e 22 )ϕ(e 12 ). However, since E 23 = E 23 E 33 and ϕ is an isomorphism on U, we have ϕ(e 23 ) = ϕ(e 23 )ϕ(e 33 ) and hence ϕ(e 23 )ϕ(e 22 )ϕ(e 12 ) = 0 for ϕ(e 33 )ϕ(e 22 ) = 0 by (2). Similarly, since ϕ is an anti-isomorphism on V it follows that ϕ(e 12 ) = ϕ(e 12 )ϕ(e 11 ) which yields ϕ(e 12 )ϕ(e 22 )ϕ(e 23 ) = 0. But then ϕ(e 13 ) = 0, a contradiction. Similarly, we see that ϕ cannot be an anti-isomorphism on U and an isomorphism on V. Assume now that ϕ is an isomorphism on both U and V, and let us show that in this case ϕ is an isomorphism on the whole T r. Let us first show that ϕ(e ij E 1r ) = ϕ(e ij )ϕ(e 1r ) (3) for all 1 i j r.sinceϕ is a Jordan isomorphism, (3) is equivalent to ϕ(e 1r E ij ) = ϕ(e 1r )ϕ(e ij ). (4) We have to consider different possibilities. First, if 2 j r 1, then ϕ(e ij ) = ϕ(e ij E jj ) = ϕ(e ij )ϕ(e jj ) for both E ij,e jj lie in V. But then (2) implies ϕ(e ij )ϕ(e 1r ) = 0 which proves (3) for this case. Similarly, we see that (4) is fulfilled when 2 i r 1. Thus, it remains to consider the following three possibilities: i = j = 1, i = j = r, and i = 1, j = r. The latter one is trivial since E1r 2 = 0. So, let i = j = 1. First, note that ϕ(e 1r) = ϕ(e 12 E 2r + E 2r E 12 ) = ϕ(e 12 )ϕ(e 2r ) + ϕ(e 2r )ϕ(e 12 ).Sinceϕ is an isomorphism on U this further implies that ϕ(e 1r ) = ϕ(e 12 )ϕ(e 2r )ϕ(e rr ) + ϕ(e 2r )ϕ(e 12 ). But this, together with ϕ(e rr )ϕ(e 11 ) = 0 (recall (2)) and ϕ(e 12 )ϕ(e 11 ) = ϕ(e 12 E 11 ) = 0 (namely, ϕ is an isomorphism on V) implies ϕ(e 1r )ϕ(e 11 ) = 0, proving (4). Similarly, we consider the last remaining case when i = j = r. Thus, (3) (and thereby (4)) is proved. All we still need to show is that ϕ(e 1i E jr ) = ϕ(e 1i )ϕ(e jr ) (5) for all 1 i r 1, 2 j r. Clearly, (5) is equivalent to ϕ(e jr E 1i ) = ϕ(e jr )ϕ(e 1i ). (6) Since E jr,e rr U and E 11,E 1i V, we have ϕ(e jr ) = ϕ(e jr )ϕ(e rr ) and ϕ(e 1i ) = ϕ(e 11 )ϕ(e 1i ).Butthenϕ(E jr )ϕ(e 1i ) = 0 by (2). This proves (6) (and thereby (5)). Thus, ϕ is indeed an isomorphism on T r. One completes the proof by showing in a similar manner that ϕ must be an antiisomorphism in the case when its restrictions to both U and V are anti-isomorphisms.

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