Biderivations of the Algebra of Strictly Upper Triangular Matrices over a Commutative Ring

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1 Journal of Mathematical Research & Exposition Nov., 2011, Vol. 31, No. 6, pp DOI: /j.issn: X Biderivations of the Algebra of Strictly Upper Triangular Matrices over a Commutative Ring Pei Sheng JI, Xiao Ling YANG, Jian Hui CHEN School of Mathematical Sciences, Qingdao University, Shandong , P. R. China Abstract Let N n(r) be the algebra consisting of all strictly upper triangular n n matrices over a commutative ring R with the identity. An R-bilinear map φ : N n(r) N n(r) N n(r) is called a biderivation if it is a derivation with respect to both arguments. In this paper, we define the notions of central biderivation and extremal biderivation of N n(r), and prove that any biderivation of N n(r) can be decomposed as a sum of an inner biderivation, central biderivation and extremal biderivation for n 5. Keywords biderivation; strictly upper triangular matrix; algebra. Document code A MR(2010) Subject Classification 17B30; 17B40 Chinese Library Classification O Introduction Let R be a commutative ring with the identity. Denote by N n (R), where n is a positive integer greater than 1, the R-algebra of strictly upper triangular n n matrices over R. In recent years, many significant researches have been done in automorphisms, Lie automorphisms, derivations and Lie derivations of N n (R). Cao [4] studied the R-algebra automorphisms of N n (R). Cao [5, 6] investigated the Lie automorphisms group of N n (R). Ou et al. [8] and Ji [7] determined the Lie derivations of N n (R). Wang and Li [9] studied the structure of all triple derivations of N n (R). Someone described the derivations of N n (R). Let T be an R-algebra and M a T-bimodule. Recall that an R-linear map d : T M is called a derivation if d(ab) = d(a)b + ad(b) for all a, b in T. For example, take x M, and set D x (a) = ax xa for all a in T. In the following, we denote by [x, y] the commutator (the Lie product) of the elements x, y T. A bilinear map φ : T T T is called a biderivation if it is a derivation with respect to both components, meaning that, φ(xy, z) = xφ(y, z) + φ(x, z)y and φ(x, yz) = φ(x, y)z + yφ(x, z) for all x, y, z T. If T is a noncommutative algebra, then the map φ(x, y) = λ[x, y] for all x, y T, where λ Z(T), the center of T, is a basic example of biderivation, and we call it an inner biderivation. Received April 8, 2010; Accepted May 28, 2010 Supported by the National Natural Science Foundation of China (Grant No ). * Corresponding author address: jipeish@yahoo.com (P. S. JI)

2 966 P. S. JI, X. L. YANG and J. H. CHEN Biderivations are a subject of research in various areas. Bresar et al. [3] proved that all biderivations on noncommutative prime rings are inner. Zhang et al. [10] showed that biderivations of nest algebras are usally inner. There are special cases of nest algebras with non-inner biderivations. Zhao et al. used the results in [10] to prove that every biderivation of an upper triangular matrix algebra is a sum of an inner biderivation and a special biderivation. Benkovic [1] generalized this result to triangular algebras. He proved that every biderivation of some certain triangular algebras is a sum of an inner biderivation and an extremal biderivation. The aim of this paper is to describe the biderivations of N n (R) for n 5. In the cases 2 n 4, we have already given the structure of the biderivations of N n (R). This paper is organized as follows. In Section 2, we construct certain special biderivation of N n (R) so as to build every biderivation of N n (R). In Section 3, we decompose any biderivation of N n (R) into a sum of those known ones. Now we introduce some preliminary notations and results about matrix algebras. Let e i,j be the standard matrix units for 1 i, j n. It is well known that the matrix set {e i,j : 1 i < j n} forms a basis of N n (R) and any x in N n (R) can be uniquely written as x = 1 i<j n x i,je i,j with x i,j R. Set M r = { x i,j e i,j N n (R) : x i,j R}, r = 1, 2,...,n 1, j i r and M r = 0 for r n. It is clear that each M r is an ideal of N n (R) and x 1 x 2 x r M r for any x 1, x 2,..., x r N n (R). It follows that M r M s M r+s. It is not difficult to know that the center of the R-algebra N n (R) is M = Re 1, To prove the main result of this paper, we need the following result, which might not be written down explicitly to our knowledge. Lemma 1.1 Let R be an arbitrary commutative ring with the identity and D a derivation of N n (R) for n 4. Then there is an n n diagonal matrix d, an element y N n (R) and c = (c 2,..., c ) R n 3 such that D(x) = [d, x] + [y, x] + i=2 x i,i+1c i e 1,n for all x = 1 i<j n x i,je i,j N n (R). 2. Certain standard biderivations of N n (R) In this section, we will give three types of standard derivations of N n (R), which will be used to describe arbitrary biderivations of N n (R). (A) Inner biderivations: Let d R. Then the map φ : N n (R) N n (R) N n (R), (x, y) d[x, y], is a biderivation of N n (R). We call it the inner biderivation of N n (R) induced by d. Note that the inner biderivation defined here is different from the one defined in the papers [1 3,10,11]. Since the center of N n (R) is Re 1,n, and for every x, y N n (R), we have that e 1,n [x, y] = 0. Hence the inner biderivation of N n (R) defined in [1 3,10,11] is the zero map. (B) Central biderivations: Let A be an (n 1) (n 1) matrix over R. We define an R- bilinear map φ A : N n (R) N n (R) N n (R) by φ A (y, x) = i,j=1 y i,i+1a i,j x j,j+1 e 1,n for all

3 Biderivations of the algebra of strictly upper triangular matrices over a commutative ring 967 x = 1 i<j n x i,je i,j, y = 1 i<j n y i,je i,j N n (R). Then it is easy to check that φ A is a biderivation of N n (R). We call it the central biderivation of N n (R) induced by A. (C) Extremal biderivations: Let a R. The R-bilinear map φ a : N n (R) N n (R) N n (R), which is defined by φ a (y, x) = a(y 2,3 x 1,3 e 1,n + y 2,3 x 2,3 e 2,n + y 1,3 x 2,3 e 1,n ) for all x = 1 i<j n x i,je i,j, y = 1 i<j n y i,je i,j N n (R), is a biderivation on N n (R). In fact, fix y = 1 i<j n y i,je i,j N n (R), for all x = 1 i<j n x i,je i,j, x = 1 i<j n x i,j e i,j, since (xx ) 2,3 = 0 and (xx ) 1,3 = x 1,2 x 2,3, we have that φ a(y, xx ) = ay 2,3 x 1,2 x 2,3 e 1,n. Computing directly, we get that φ a (y, x)x + xφ a (y, x ) =a(y 2,3 x 1,3 e 1,n + y 2,3 x 2,3 e 2,n + y 1,3 x 2,3 e 1,n )x + ax(y 2,3 x 1,3e 1,n + y 2,3 x 2,3e 2,n + y 1,3 x 2,3e 1,n ) =ax 1,2 y 2,3 x 2,3 e 1,n. Thus φ a (y, xx ) = φ a (y, x)x + xφ a (y, x ). This means that φ a is a biderivation with respect to the second component. Since φ a is a symmetric bilinear map, we have that φ a is a derivation with respect to the first component. Therefore φ a is a biderivation of N n (R). We call it the extremal biderivation of N n (R) induced by a R. 3. Biderivation of N n (R) In this section, we give the main result of this paper. Theorem 3.1 Let R be an arbitrary commutative ring with the identity and φ a biderivation of N n (R). If n 5, then there are elements a, d R and an (n 1) (n 1) matrix A over R such that for all x, y N n (R). φ(y, x) = d[y, x] + φ A (y, x) + φ a (y, x) In order to prove our main result, we need the following lemmas. Lemma 3.2 ([2, Corollary 2.4]) Let T be an algebra on a commutative ring R and φ : T T T a biderivation. Then for all x, y, u, v A. φ(x, y)[u, v] = [x, y]φ(u, v) In what follows, for 1 i < j n, since φ(e i,j, x) : N n (R) N n (R) is a derivation of N n (R), by Lemma 1.1, we suppose that φ(e i,j, x) = [d i,j, x] + [y i,j, x] + c i,j i x i,i+1 e 1,n for all x = 1 i<j n x i,je i,j N n (R), where d i,j = diag(d i,j 1,..., di,j n ) is a diagonal matrix over R, y i,j = 1 t<s n yi,j t,se i,j N n (R), and (c i,j 2,...,ci,j ) Rn 3. Lemma 3.3 For any (i, j) (1 i < j n), there is d i,j R such that d i,j = diag(d i,j, d i,j,...,d i,j ). i=2

4 968 P. S. JI, X. L. YANG and J. H. CHEN Hence [d i,j, x] = 0 for all x N n (R). Proof Step 1. First we will prove that if there is e t,s with s t < n 2 such that [e i,j, e t,s ] = 0, then d i,j t = d i,j s. Since [e i,j, e t,s ] = 0, by Lemma 3.2, we get that Hence φ(e i,j, e t,s )[u, v] = [e i,j, e t,s ]φ(u, v) = 0 for all u, v N n (R). φ(e i,j, e t,s ) =[d i,j, e t,s ] + [y i,j, e t,s ] + c i,j t δ t+1,s e 1,n =(d i,j t d i,j s )e t,s + [y i,j, e t,s ] + c i,j t δ t+1,s e 1,n M, where δ is the Kronecker s delta. Since s t < n 2, e t,s M. Since e t,s M s t \M s t+1 and [y i,j, e t,s ] + c i,j t δ t+1,s e 1,n M s t+1, we have d i,j t d i,j s = 0, i.e., d i,j t = d i,j s. Henceforth fix (i, j) (1 i < j n). Step 2. (i) Clearly, [e i,j, e 1,k ] = 0 for each 1 < k n 2 with k i. It follows from step 1 that d i,j 1 = d i,j k for all 1 < k n 2 with k i. (ii) Since [e i,j, e k,n ] = 0 for all 2 < k n 1 with k j, by Step 1, we have that d i,j k = di,j n for all 2 < k n 1 with k j. (iii) Since [e i,j, e i,k ] = 0 and [e i,j, e t,j ] = 0, by Step 1, we get that d i,j i i < k min{n, n 3 + i} and d i,j j = d i,j t all min{i, max{0, j n + 2}} k max{j, min{n, n 3 + i}}. Case 1 i = 1. = d i,j k for all j > k max{0, j n + 2}. Thus d i,j i = d i,j k It follows from Step 2 (i) that d 1,j 1 = = d 1,j. If j < n 2, by Step 2 (ii), we have that d 1,j = d1,j = d1,j So d 1,j 1 = d 1,j 2 = = d 1,j If j = n 2, then, by Step 2 (ii), d 1, suffices to prove that d 1, 2 = d 1, 1, we get d 1, 2 = d 1,. Thus d1, = d 1, n for all for. In this case, to complete the proof, it. Since [e 1,, e 2, ] = 0 and (n 1) 2 < n 2, by Step 1 = d 1, 2 = = d 1, If j = n 1, then, by Step 2 (ii), d 1, = d 1, Since [e 1,, e, ] = 0 and (n 1) (n 2) < n 2, by Step 1, we get d 1, = d 1,. Thus d1,1 = d 1, 2 = = d 1, When j = n, it follows from Step (ii) that d 1,n 3 = d 1,n 4 = = d 1,n Thus d 1,n 1 = d 1,n 2 = = d 1,n Case 2 i = 2. It follows from Step 2 (i) that d 2,j 1 = d 2,j 3 = = d 2,j, and from Step 2 (iii) that d2,j d 2,j 3 = = d 2,j. Thus d2,j 1 = d 2,j 2 = = d 2,j. If j n 2, by Step 2 (ii), we have that d 2,j d 2,n If j = n 1, then, by Step 2 (ii), d 1, = d2,j n = d 1, n When j = n, it follows from Step (ii) that d 2,n 3 = d 2,n. So d2,j 1 = d 2,j 2 = = d 2,j 1 = d 2, 2 =. Thus d 2, 2 = = d 2, 4 = = d 2,n Thus d2,n 1 = d 2,n 2 = =

5 Biderivations of the algebra of strictly upper triangular matrices over a commutative ring 969 Case 3 i 3. It follows from Step 2 (i) that d i,j 1 = d i,j 2 = = d i,j i 1, and from Step 2 (iii) that di,ji = d i,j i+1 = = di,j Since [e i,j, e i 1,i+1 ] = 0 and (i + 1) (i 1) < n 2, by Step 1, we get d i,j i 1 = di,j i+1. Thus di,j 1 = d i,j 2 = = d i,j This completes the proof. Henceforth, for 1 i < j n, we suppose that φ(e i,j, x) = [y i,j, x] + c i,j i x i,i+1 e 1,n for all x = 1 i<j n x i,je i,j N n (R), where y i,j = 1 t<s n yi,j t,se i,j N n (R), and (c i,j 2,...,ci,j ) R n 3. Since [e 1,n, x] = 0 for all x N n (R), we suppose that y i,j 1,n = 0 for all (i, j). In the proof of the following lemmas, we often use the following fact: Suppose c i,j R for all 1 i < j n. If 1 i<j n c i,jx i,j = 0 for all x = 1 i<j n x i,je i,j N n (R), then c i,j = 0 for all 1 i < j n. And for every z N n (R), and i=2 φ(e i,j, x)z = ([y i,j, x] + c i,j i x i,i+1 e 1,n )z = [y i,j, x]z i=2 zφ(e i,j, x) = z([y i,j, x] + c i,j i x i,i+1 e 1,n ) = z[y i,j, x]. Lemma 3.4 There exists an element d R such that (i) y 1,j = de 1,j + y 1,j 1, e 1, + y 1,j 2,n e 2,n for j < n 1; (ii) y 1,n = y 1,n 1, e 1, + y 1,n 2,n e 2,n; (iii) y 1, = y 1, 1, e 1, + y 1, 1, e 1, + y 1, 2,n e 2,n ; (iv) y i,j = de i,j + y i,j 1, e 1, + y i,j 2,n e 2,n for j n 1 and i > 2; (v) y i, = y i, 1, e 1, + y i, 1, e 1, + y 1, 2,n e 2,n + de i, for i > 2. i=2 Proof (I) For 1 < j n, it follows from e 1,2 e 1,j = 0 that 0 = φ(e 1,2 e 1,j, x) = e 1,2 φ(e 1,j, x) + φ(e 1,2, x)e 1,j = e 1,2 φ(e 1,j, x) = e 1,2 [y 1,j, x] n = ( y 1,j 2,t x t,s x 2,t yt,s)e 1,j 1,s s=4 2<t<s 2<t<s for all x = 1 i<j n x i,je i,j N n (R). Hence, for every 4 s n, 2<t<s y1,j 2,t x t,s 2<t<s x 2,ty 1,j t,s = 0 for all x = 1 i<j n x i,je i,j N n (R), which implies that y 1,j t,s = 0 for t > 2 and y 1,j 2,t = 0 for 3 t n 1. So y1,j = y1,j 1,t e 1,t + y 1,j 2,n e 2,n. (II) For j < n 1, since e 1,j e,n = 0, by (I), we have that 0 = φ(e 1,j e,n, x) = e 1,j φ(e,n, x) + φ(e 1,j, x)e,n = e 1,j [y,n, x] + [y 1,j, x]e,n n = ( y,n j,t x t,s x j,t y,n t,s )e 1,s + y 1,j 1,t x t,e 1,n s=j+2 j<t<s j<t<s for all x = 1 i<j n x i,je i,j N n (R). Therefore j<t<n y,n j,t x t,n j<t<s x j,ty,n t,n + 1,t x t, = 0 and j<t<s y,n x t,s j<t<s x j,ty,n t,s = 0 for j + 2 s < n, which y1,j j,t

6 970 P. S. JI, X. L. YANG and J. H. CHEN implies that y 1,j 1,t = y,n for all j < n 1, and y1,j 1,t = 0 for 2 t n 2 with t j. Suppose d = y,n, We have that y 1,j = de 1,j + y 1,j 1, e 1, + y 1,j 2,n e 2,n for j < n 1. (i) is proved. (III) Since e 1,n e,n = 0, by (I), we get that 0 = φ(e 1,n e,n, x) = e 1,n φ(e,n, x) + φ(e 1,n, x)e,n = φ(e 1,n, x)e,n = [y 1,n, x]e,n = y 1,n 1,t x t,e 1,n for all x = 1 i<j n x i,je i,j N n (R). Hence we have that y 1,n 1,t y 1,n = y 1,n 1, e 1, + y 1,n 2,n e 2,n. (IV) For 2 < i < n, since e 1,2 e i,j = 0, by (i), we have that = 0 for all 2 t n 2. Thus 0 = φ(e 1,2 e i,j, x) = e 1,2 φ(e i,j, x) + φ(e 1,2, x)e i,j = e 1,2 [y i,j, x] + [y 1,2, x]e i,j n = ( y i,j 2,t x t,s x 2,t yt,s)e i,j 1,s + dx 2,i e 1,j s=4 2<t<s 2<t<s for all x = 1 i<j n x i,je i,j N n (R). These imply that, for every 4 s n with s j, 2<t<s yi,j 2,t x t,s 2<t<s x 2,ty i,j t,s = 0, and that 2<t<j yi,j 2,t x t,j 2<t<j x 2,ty i,j t,j + dx 2,i = 0 for all x = 1 i<j n x i,je i,j N n (R). Therefore y i,j i,j = d and yi,j t,s = 0 unless t = 1 and (t, s) = (i, j), (2, n). So y i,j = yi,j 1,t e 1,t + y i,j 2,n e 2,n + de i,j. (V) For any (i, j) (i > 2 and j < n 1), since e i,j e,n = 0, by (IV), we have that 0 = φ(e i,j e,n, x) = e i,j φ(e,n, x) + φ(e i,j, x)e,n = e i,j [y,n, x] + [y i,j, x]e,n = dx j, e i,n + y i,j 1,t x t,e 1,n + dx j, e i,n = y i,j 1,t x t,e 1,n for all x = 1 i<j n x i,je i,j N n (R). This leads to yi,j 1,t x t, = 0, which implies that y i,j 1,t = 0 for all 2 t n 2. Thus yi,j = de i,j + y i,j 1, e 1, + y i,j 2,n e 2,n for j < n 1. (VI) For any i (i > 2), by (IV), it follows from e i,n e,n = 0 that 0 = φ(e i,n e,n, x) = e i,n φ(e,n, x) + φ(e i,n, x)e,n = φ(e i,n, x)e,n = [ y i,n 1,t e 1,t + y i,n 2,n e 2,n + de i,n, x]e,n = y i,n 1,t x t,e 1,n for all x = 1 i<j n x i,je i,j N n (R). Hence y i,n 1,t = 0 for all 2 t n 2. So y i,n = y i,n 1, e 1, + y i,n 2,n e 2,n + de i, (VII) Since e 1, e,n = 0, using (I) and (VI), we can get that 0 = φ(e 1, e,n, x) = e 1, φ(e,n, x) + φ(e 1,, x)e,n = e 1, [y,n 1, e 1, + y,n n 3 = y 1, 1,t x t, e 1,n 2,n e 2,n + de,n, x] + [ y 1, 1,t e 1,t + y 1, 2,n e 2,n, x]e,n

7 Biderivations of the algebra of strictly upper triangular matrices over a commutative ring 971 for all x = 1 i<j n x i,je i,j N n (R). Therefore n 3 x = 1 i<j n y1, x i,j e i,j N n (R), which implies that y 1, 1,t = 0 for all 2 t n 3. Thus 1,t x t, = 0 for all y 1, = y 1, 1, e 1, + y 1, 1, e 1, + y 1, 2,n e 2, (VIII) For i > 2, since e i, e,n = 0, by (VI), we have that 0 = φ(e i, e,n, x) = e i, φ(e,n, x) + φ(e i,, x)e,n n 3 = y i, 1,t x t, e 1,n for all x = 1 i<j n x i,je i,j N n (R). Therefore n 3 x = 1 i<j n yi, x i,j e i,j N n (R), which leads to that y 1, 1,t = 0 for all 2 t n 3. So 1,t x t, = 0 for all y i, = y i, 1, e 1, + y i, 1, e 1, + y 1, 2,n e 2,n + de i,. Lemma 3.5 For i = 2, we have the following equalities: (i) y 2,j = de 2,j + y 2,j 1, e 1, + y 2,j 3,n e 3,n + y 2,j 2,n e 2,n for j < n 1; (ii) y 2, = y 2, 1, e 1, + y 2, 1, e 1, + y 2, 3,n e 3,n + y 2, 2, e 2, + y 2, 2,n e 2,n ; (iii) y 2,n = y 2,n 1, e 1, + y 2,n 2,n e 2,n + y 2,n 2, e 2, + y 2,n 3,n e 3,n. Proof (I) For any j > 2, since e 1,3 e 2,j = 0, by Lemma 3.4(i), we have that 0 = φ(e 1,3 e 2,j, x) = e 1,3 φ(e 2,j, x) + φ(e 1,3, x)e 2,j = e 1,3 [y 2,j, x] + [y 1,3, x]e 2,j n = ( y 2,j 3,t x t,s s=5 3<t<s 3<t<s x 3,t y 2,j t,s )e 1,s for all x = 1 i<j n x i,je i,j N n (R). Hence, for every 5 s n, 3<t<s y2,j 3,t x t,s 3<t<s x 3,ty 2,j t,s = 0 for all x = 1 i<j n x i,je i,j N n (R). Therefore y 2,j t,s = 0 for t > 3 and y 2,j 3,t = 0 for 4 t n 1. So y2,j = y2,j 1,t e 1,t + n t=3 y2,j 2,t e 2,t + y 2,j 3,n e 3,n. (II) For 3 j < n 1, using (I) and Lemma 3.4 (iv), it follows from e 2,j e,n = 0 that 0 = φ(e 2,j e,n, x) = e 2,j φ(e,n, x) + φ(e 2,j, x)e,n = e 2,j [y,n, x] + [y 2,j, x]e,n = dx j, e 2,n + y 2,j 1,t x t,e 1,n + t=3 y 2,j 2,t x t,e 2,n y 2,j 2, x 1,2e 1,n for all x = 1 i<j n x i,je i,j N n (R). Therefore y2,j 1,t x t, y 2,j 2, x 1,2 = 0 and dx j, + t=3 y2,j 2,t x t, = 0 for all x = 1 i<j n x i,je i,j N n (R). Hence y 2,j 1,t = 0 for all 2 t

8 972 P. S. JI, X. L. YANG and J. H. CHEN n 2, d = y 2,j 2,j, and y2,j 2,t = 0 for 2 < t n 1 with t j. Thus y2,j = de 2,j + y 2,j 1, e 1, + y 2,j 3,n e 3,n + y 2,j 2,n e 2,n for j < n 1. (III) Since e 2, e,n = 0, by (I) and Lemma 3.4(iv), we have that 0 =φ(e 2, e,n, x) = e 2, φ(e,n, x) + φ(e 2,, x)e,n =e 2, [y,n 1, e 1, + y,n 2,n e 2,n + de,n, x]+ n [ n 3 = t=3 y 2, 1,t e 1,t + t=3 n 3 y 2, 2,t x t, e 2,n + y 2, 2,t e 2,t + y 2, 3,n e 3,n, x]e,n y 2, 1,t x t, e 1,n y 2, 2, x 1,2e 1,n for all x = 1 i<j n x i,je i,j N n (R). Therefore n 3 y2, 1,t x t, y 2, 2, x 1,2 = 0 and n 3 t=3 y2, 2,t x t, = 0 for all x = 1 i<j n x i,je i,j N n (R). It follows that y 2, 1,t = 0 for all 2 t n 3 and y 2, 2,t = 0 for all 3 t n 2. Thus y 2, = y 2, 1, e 1, + y 2, 1, e 1, + y 2, 3,n e 3,n + y 2, 2, e 2, + y 2, 2,n e 2, (IV) Since e 2,n e,n = 0, by (I), we have that 0 = φ(e 2,n e,n, x) = e 2,n φ(e,n, x) + φ(e 2,n, x)e,n n = φ(e 2,n, x)e,n = [ y 2,n 1,t e 1,t + y 2,n 2,t e 2,t + y 2,n 3,n e 3,n, x]e,n = y 2,n 1,t x t,e 1,n + t=3 t=3 y 2,n 2,t x t,e 2,n y 2,n 2, x 1,2e 1,n for all x = 1 i<j n x i,je i,j N n (R). Therefore y2,n 1,t x t, y 2,n 2, x 1,2 = 0 and t=3 y2,n 2,t x t, = 0 for all x = 1 i<j n x i,je i,j N n (R), which implies that y 2,n 1,t = 0 for 2 t n 2 and y 2,n 2,t = 0 for all 3 t n 1. Thus y 2,n = y 2,n 1, e 1, + y 2,n 2,n e 2,n + y 2,n 2, e 2, + y 2,n 3,n e 3,n. Lemma 3.6 (i) For all x N n (R), we have φ(e 1,n, x) = 0; (ii) For 3 < j < n, we get y 1,j = de 1,j and c 1,j t = 0 for 2 t n 2; (iii) Then y 1,3 = de 1,3, c 1,j t = 0 for 3 t n 2 and c 1,3 2 = y 2,3 3, Proof (I) By Lemma 3.4 (i) and (iv), it is easy to check that for all x N n (R). φ(e 1,n, x) =φ(e 1,3 e 3,n, x) = e 1,3 φ(e 3,n, x) + φ(e 1,3, x)e 3,n =e 1,3 [de 3,n + y 3,n 1, e 1, + y 3,n 2,n e 2,n, x]+ =0 [de 1,3 + y 1,3 1, e 1, + y 1,3 2,n e 2,n, x]e 3,n (II) Fix any i (2 < i < n 1), since φ(e 1,, x) = φ(e 1,i e i,, x) = e 1,i φ(e i,, x) +

9 Biderivations of the algebra of strictly upper triangular matrices over a commutative ring 973 φ(e 1,i, x)e i,, using Lemma 3.4 (i) and (v), we have that i.e., [y 1, 1, e 1, + y 1, 1, e 1, + y 1, 2,n e 2,n, x] + c 1, t x t,t+1 e 1,n = e 1,i [y i, 1, e 1, + y i, 1, e 1, + y 1, 2,n e 2,n + de i,, x]+ [de 1,i + y 1,i 1, e 1, + y 1,i 2,n e 2,n, x]e i,, y 1, 1, x,e 1, + y 1, 1, x,ne 1,n + y 1, 1, x,ne 1,n y 1, 2,n x 1,2 e 1,n + c 1, t x t,t+1 e 1,n = dx,n e 1,n for all x = 1 i<j n x i,je i,j N n (R). Hence y 1, 1, x, = 0 and y 1, 1, x,n + y 1, 1, x,n y 1, 2,n x 1,2 + c 1, t x t,t+1 = dx,n for all x = 1 i<j n x i,je i,j N n (R). Therefore y 1, 1, = 0, y1, 2,n = 0, y 1, 1, = d and c 1, t = 0 for 2 t n 2. So y 1, = de 1,. (III) For 4 j n 2, since φ(e 1,j, x) = φ(e 1,3 e 3,j, x) = e 1,3 φ(e 3,j, x) + φ(e 1,3, x)e 3,j, we get, using Lemma 3.4 (i, iv), that y 1,j 1, x,ne 1,n y 1,j 2,n x 1,2e 1,n + de 1,j x + c 1,j t x t,t+1 e 1,n = de 1,j x for all x = 1 i<j n x i,je i,j N n (R). Hence y 1,j 1, x,n y 1,j 2,n x 1,2 + c1,j t x t,t+1 = 0 for all x = 1 i<j n x i,je i,j N n (R), which implies that y 1,j 1, = y1,j 2,n 2 t n 2. Thus y 1,j = de 1,j. = 0 and c1,j t = 0 for (IV) Since φ(e 1,3, x) = φ(e 1,2 e 2,3, x) = e 1,2 φ(e 2,3, x)+φ(e 1,2, x)e 2,3, we see using Lemma 3.4 (i) and Lemma 3.5 (i) that y 1,3 1, x,ne 1,n y 1,3 2,n x 1,2e 1,n + de 1,3 x + c 1,3 t x t,t+1 e 1,n = de 1,3 x y 2,3 3,n x 2,3e 1,n for all x = 1 i<j n x i,je i,j N n (R). Hence y 1,3 1, x,n y 1,3 2,n x 1,2 + c1,3 t x t,t+1 = y 2,3 3,n x 2,3 for all x = 1 i<j n x i,je i,j N n (R). Therefore y 1,3 1, = y1,3 2,n = 0 for 3 t n 2. So y 1,3 = de 1,3. c 1,3 t Lemma 3.7 For 3 < j n, we get that y 2,j = de 2,j and c 2,j t = 0 for 2 t n 2. = 0, c1,3 2 = y 2,3 3,n and Proof (I) Since φ(e 2,n, x) = φ(e 2,3 e 3,n, x) = e 2,3 φ(e 3,n, x) + φ(e 2,3, x)e 3,n, by Lemma 3.4 (iv) and Lemma 3.5 (i, iii), we have that y 2,n 1, x,ne 1,n y 2,n 2,n x 1,2e 1,n y 2,n 2, x 1,2e 1, + y 2,n 2, x,ne 2,n y 2,n 3,n x 1,3e 1,n y 2,n 3,n x 2,3e 2,n + c 2,n t x t,t+1 e 1,n

10 974 P. S. JI, X. L. YANG and J. H. CHEN = dx 1,2 e 1,n for all x = 1 i<j n x i,je i,j N n (R). Therefore y 2,n 2, x 1,2 = 0, y 2,n 2, x,n y 2,n 3,n x 2,3 = 0 and y 2,n 1, x,n y 2,n 2,n x 1,2 y 2,n 3,n x 1,3+ t x t,t+1 = dx 1,2 for all x = 1 i<j n x i,je i,j N n (R), which means that y 2,n 2,n Thus y 2,n = de 2, c2,n = d, y2,n 1, = y2,n 2, = y2,n 3,n = 0, and c2,n t = 0 for 2 t n 2. (II) Since φ(e 2,, x) = φ(e 2,3 e 3,, x) = e 2,3 φ(e 3,, x) + φ(e 2,3, x)e 3,, by Lemma 3.4 (v) and Lemma 3.5 (i, ii), we have that y 2, 1, x,e 1, + y 2, 1, x,ne 1,n + y 2, 1, x,ne 1,n + y 2, 2, x,ne 2,n y 2, 2, x 1,2e 1, y 2, 2,n x 1,2 e 1,n y 2, 3,n x 1,3 e 1,n y 2, 3,n x 2,3 e 2,n + c 2, t x t,t+1 e 1,n = dx,n e 2,n dx 1,2 e 1, for all x = 1 i<j n x i,je i,j N n (R). Therefore y 2, 2, x,n y 2, 3,n x 2,3 = dx,n, y 2, 1, x, y 2, 2, x 1,2 = dx 1,2, y 2, 1, x,n + y 2, 1, x,n y 2, 2,n x 1,2 y 2, 3,n x 1,3 + c 2, t x t,t+1 e 1,n = 0 for all x = 1 i<j n x i,je i,j N n (R). This implies that y 2, 3,n = y 2, 1, = 0, y2, 2, = d, y 2, 1, = y2, 2,n = 0, and c 2, t = 0 for 2 t n 2. So y 2, = de 2,. (III) For 4 j < n 1, since φ(e 2,j, x) = φ(e 2,3 e 3,j, x) = e 2,3 φ(e 3,j, x) + φ(e 2,3, x)e 3,j, using Lemma 3.4 (iv) and Lemma 3.5 (i), we can get that y 2,j 1, x,ne 1,n y 2,j 2,n x 1,2e 1,n dx 1,2 e 1,j + de 2,j x y 2,j 3,n x 1,3e 1,n y 2,j 3,n x 2,3e 2,n + c 2,j t x t,t+1 e 1,n = de 2,j x dx 1,2 e 1,j for all x = 1 i<j n x i,je i,j N n (R). Therefore y 2,j 1, x,n y 2,j 2,n x 1,2 y 2,j 3,n x 1,3 + c 2,j t x t,t+1 = 0 for all x = 1 i<j n x i,je i,j N n (R). Hence y 2,j 1, = y2,j 2,n = y2,j 3,n 2 t n 2. So y 2,j = de 2,j. = 0, and c2,j t = 0 for Lemma 3.8 For i > 2 and j i 2, we have that y i,j = de i,j and c 2,j t = 0 for 2 t n 2. Proof (I) Since φ(e i,n, x) = φ(e i, e,n, x) = e i, φ(e,n, x) + φ(e i,, x)e,n, by Lemma 3.4 (iv) and (v), we have that [y i,n 1, e 1, + y i,n 2,n e 2,n + de i,n, x] + c i,n t x t,t+1 e 1,n

11 Biderivations of the algebra of strictly upper triangular matrices over a commutative ring 975 i.e., = e i, [de,n + y,n 1, e 1, + y,n 2,n e 2,n, x]+ [de i, + y i, 1, e 1, + y i, 2,n e 2,n + y i, 1, e 1,, x]e,n, dxe i,n + y i,n 1, x,ne 1,n y i,n 2,n x 1,2e 1,n + c i,n t x t,t+1 e 1,n = dxe i,n + y i, 1, x,e 1,n for all x = 1 i<j n x i,je i,j N n (R). Therefore y i,n 1, x,n y i,n 2,n x 1,2 + c i,n t x t,t+1 = y i, 1, x, for all x = 1 i<j n x i,je i,j N n (R). Hence y i,n 1, = yi,n 2,n = 0, yi, 1, = ci,n and c2,n t = 0 for 2 t n 3. Thus y i,n = de i, (II) Since φ(e i,, x) = φ(e i, e,, x) = e i, φ(e,, x) + φ(e i,, x)e, by Lemma 3.4 (iv) and (v), we have that dx,n e i,n dxe i, + y i, 1, x,ne 1,n y i, 2,n x 1,2e 1,n + y i, 1, x,e 1, + y i, 1, x,ne 1,n + c i, t x t,t+1 e 1,n = dx,n e i,n dxe i, for all x = 1 i<j n x i,je i,j N n (R). Therefore y i, 1, x, = 0, and y i, 1, x,n y i, 2,n x 1,2 + y i, 1, x,n + ci, t x t,t+1 = 0 for all x = 1 i<j n x i,je i,j N n (R). This implies that y i, 1, = yi, 1, = yi, 2,n = 0, and c 2, t = 0 for 2 t n 2. Thus y i, = de i,. By (I), we have that c i,n = yi, 1, = 0. (III) For any (i, j) (i > 2, i + 2 < j < n 1), since φ(e i,j, x) = φ(e i,i+1 e i+1,j, x) = e i,i+1 φ(e i+1,j, x) + φ(e i,i+1, x)e i+1,j, using Lemma 3.4 (iv), we can get that de i,j x dxe i,j + y i,j 1, x,ne 1,n y i,j 2,n x 1,2e 1,n + c i,j t x t,t+1 e 1,n = de i,j x dxe i,j for all x = 1 i<j n x i,je i,j N n (R). Therefore y i,j 1, x,n y i,j 2,n x 1,2 + ci,j t x t,t+1 = 0 for all x = 1 i<j n x i,je i,j N n (R). Hence y i,j 1, = yi,j 2,n = 0, and ci,j t = 0 for 2 t n 2. So y i,j = de i,j. Summarizing Lemmas 3.4 (i), 3.5 (i) and , we have the following result. Corollary 3.9 The following equalities hold (i) φ(e 1,n, x) = 0 for all x = 1 i<j n x i,je i,j N n (R); (ii) For 1 i n 1 with i 2, φ(e i,i+1, x) = [de i,i+1 + y i,i+1 1, e 1, + y i,i+1 2,n e 2,n, x] + x t,t+1 e 1,n for all x N n (R); ci,i+1 t (iii) For i + 2 j, except (i, j) = (1, 3), φ(e i,j, x) = [de i,j, x] for all x N n (R); (iv) φ(e 1,3, x) = [de 1,3, x] + c 1,3 2 x 2,3e 1,n for all x N n (R); (v) φ(e 2,3, x) = [de 2,3 + y 2,3 1, e 1, + y 2,3 2,n e 2,n + y 2,3 3,n e 3,n, x] + ci,i+1 t x t,t+1 e 1,n for all x N n (R), and y 2,3 3,n = c1,3 2.

12 976 P. S. JI, X. L. YANG and J. H. CHEN The Proof of Theorem 3.1 For 1 i n 1, denote y i,i+1 1, by ci,i+1, yi,i+1 2,n by c i,i+1 1, and c 1,3 2 = y 2,3 3,n by a. Then, for i 2, and for all x N n (R). Let A denote φ(e i,i+1, x) = [de i,i+1, x] + c i,i+1 t x t,t+1 e 1,n, t=1 φ(e 2,3, x) = [de 2,3, x] + [ ae 3,n, x] + c i,i+1 t x t,t+1 e 1,n, t=1 φ(e 1,3, x) = [de 1,3, x] + ax 2,3 e 1,n i,j=1 ci,i+1 j N n (R), it follows from Corollary 3.9 that φ(y, x) = y i,j φ(e i,j, x) 1 i<j n = 1 i<j n y i,j [de i,j, x] + e i,j. For y = 1 i<j n y i,je i,j, x = 1 i<j n x i,je i,j i=1 (y i,i+1 j=1 c i,i+1 j x j,j+1 e 1,n )+ y 2,3 [ ae 3,n, x] + ay 1,3 x 2,3 e 1,n =d[ y i,j e i,j, x] + (y i,i+1 c i,i+1 j x j,j+1 e 1,n )+ 1 i<j n i=1 j=1 a(y 2,3 x 1,3 e 1,n + y 2,3 x 2,3 e 2,n + y 1,3 x 2,3 e 1,n ) =d[y, x] + φ A (y, x) + φ a (y, x). We thus complete the proof of Theorem 3.1. Acknowledgments The authors would like to thank the referees for their comments. References [1] BENKOVIČ D. Biderivations of triangular algebras [J]. Linear Algebra Appl., 2009, 431(9): [2] BREŠAR B. On generalized biderivations and related maps [J]. J. Algebra, 1995, 172(3): [3] BREŠAR B, MARTINDALE W S, MIERS C R. Centralizing maps in prime rings with involution [J]. J. Algebra, 1993, 161(2): [4] CAO You an, WANG Jingtong. A note on algebra automorphisms of strictly upper triangular matrices over commutative rings [J]. Linear Algebra Appl., 2000, 311(1-3): [5] CAO You an. Automorphisms of the Lie algebra of strictly upper triangular matrices over certain commutative rings [J]. Linear Algebra Appl., 2001, 329(1-3): [6] CAO You an, TAN Zuowen. Automorphisms of the Lie algebra of strictly upper triangular matrices over a commutative ring [J]. Linear Algebra Appl., 2003, 360: [7] JI Peisheng, YUAN Huali. Lie derivations of strictly upper triangular matrices over commutative rings [J]. Acta Math. Sinica (Chin. Ser.), 2007, 50(4): (in Chinese) [8] OU Shikun, WANG Dengyin, YAO Ruiping. Derivations of the Lie algebra of strictly upper triangular matrices over a commutative ring [J]. Linear Algebra Appl., 2007, 424(2-3): [9] WANG Hengtai, LI Qingguo. Lie triple derivation of the Lie algebra of strictly upper triangular matrix over a commutative ring [J]. Linear Algebra Appl., 2009, 430(1): [10] ZHANG Jianhua, FENG Shan, LI Hongxia, et al. Generalized biderivations of nest algebras [J]. Linear Algebra Appl., 2006, 418(1):