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On biderivations of upper triangular matrix rings
Linear Algebra and its Applications 438 (2013) 250–260
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On biderivations of upper triangular matrix rings Nader M. Ghosseiri Department of Mathematics, University of Kurdistan, P.O. Box 416, Sanandaj, Iran
ARTICLE INFO
ABSTRACT
Article history: Received 6 March 2012 Accepted 9 July 2012 Available online 13 September 2012
Let R and S be rings with identity, M be a unitary (R, S )-bimodule, ⎛ ⎞ R M ⎠ be the upper triangular matrix ring determined and T = ⎝ 0 S
Submitted by P. Sˇ emrl
by R, S and M. Let Eij be the standard matrix unit. In this paper we show that every biderivation of T is decomposed into the sum of three biderivations D, and , where D(E11 , E11 ) = 0, is an extremal biderivation and is a special kind of biderivation. Using this characterization, we determine the structure of biderivations of the ring Tn (R)(n 2) of all n × n upper triangular matrices over R, and show that in the special case when R is a noncommutative prime ring, every biderivation of Tn (R) is inner. This extends some results of Benkoviˇc (2009) [1]. © 2012 Elsevier Inc. All rights reserved.
AMS classification: 16W20 16S50 16U60 Keywords: Triangular matrix ring Derivation Biderivation Prime ring
1. Introduction Let R be a ring and Z (R) be the center of R. For each x, y in R, denote the commutator of x, y by [x, y] = xy − yx. An additive mapping d : R → R is said to be a derivation of R if d(xy) = d(x)y + xd(y) for all x, y in R. Let a be in R. The mapping Ia : R → R given by Ia (x) = [x, a] is easily seen to be a derivation. Ia is called the inner derivation induced by a. A biadditive mapping d : R × R → R is said to be a biderivation of R if it is a derivation in each argument; that is, d(xy, z )
= d(x, z)y + xd(y, z)
and
d(x, yz )
= d(x, y)z + yd(x, z)
(1)
are fulfilled for all x, y, z ∈ R. A map f : R → R is called centralizing on R if [f (x), x] ∈ Z (R) for all x ∈ R. In the special case when [f (x), x] = 0 for all x ∈ R, f is called commuting on R. The study of centralizing and commuting maps was initiated by a classical result of Posner [11] which states that if a prime ring admits a nonzero centralizing derivation, then it is commutative. A number of authors have extended Posner’s theorem in several directions. Readers interested in centralizing and commuting E-mail addresses: [email protected], [email protected]. 0024-3795/$ - see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2012.07.039
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251
mappings are referred to [2–4,10,13] and the references therein. The notion of additive commuting maps is closely related to the notion of biderivations. In fact, every commuting additive map f on R gives raise to a biderivation of R: linearizing [f (x), x] = 0, we observe that [f (x), y] = [x, f (y)] for all x, y ∈ R, so the mapping (x, y) → [f (x), y] is easily seen to be a biderivation of R that is inner in each component (induced by f (y) and f (x), respectively). A biderivation d of R is called an inner biderivation if there exists λ ∈ Z (R) such that d(x, y) = λ[x, y] for all x, y in R. Inner biderivations appear quite naturally in characterizing biderivations of rings. In [7], Brešar et al. showed that every biderivation d of a noncommutative prime ring R is of the form d(x, y) = λ[x, y] for some λ in the centroid C of R. Somewhat later, Brešar [5] extended this result to semiprime rings. Also, Benkoviˇc [1] proved that under certain conditions every biderivation of a triangular algebra decomposes into the sum of an inner biderivation and a special kind of biderivation which is called an extremal biderivation. More details on biderivations, their generalizations, and related maps can be found in [2,5–9,12,14,15]. Specially, the reader is encouraged to read the survey paper of Brešar [2] in which he has described the applications of biderivations to some other fields. We say that the mapping d : R × R → R is an extremal biderivation if d(x, y) = [x, [y, a]] for all x, y ∈ R, where a ∈ R is such that [a, [R, R]] = 0. Let R, S be rings with identity and M be a unitary (R, S )-bimodule. Let f : M × M → M be a biadditive mapping. We say that f is an (R, S )-bimodule homomorphism if it is a bimodule homomorphism in each argument; namely, (2) f rms, m = rf m, m s and f m, r m s = r f m, m s for all r , r
∈ R, s, s ∈ S, and m, m ∈ M.
⎞
⎛ R M
⎠ we denote the upper triangular matrix 0 S ring determined by R, S and M with the usual addition and multiplication of matrices. Eij stands for the standard matrix unit, aEij stands for the matrix whose ij-th position is a and zero elsewhere, I stands for the identity matrix and, as usual, Tn (R) denotes the ring of all n × n upper triangular matrices over R. Moreover, the zero elements of the rings and modules, zero subrings, and zero submodules are all denoted by 0. Recall that Eij Ers = δjr Eis , where δ is the Kronecker function. Motivated by [1], in this paper we show that without any further assumptions on R, S and M, any biderivation of the ring T decomposes into the sum of three biderivations D, and , where D(E11 , E11 ) = 0, is an extremal biderivation, and is a special kind of biderivation (Theorem 2.4). Using this characterization, we shall see that any biderivation of the ring T2 (R) is decomposed into the sum of four biderivations: an inner biderivation, an extremal biderivation, and two special kinds of biderivations (Corollary 2.5). Also, we prove that every biderivation of the ring Tn (R), n 3, is decomposed into the sum of an inner biderivation and an extremal biderivation (Corollary 2.8). Moreover, we shall see that if R is a noncommutative prime ring, then any biderivation of the ring Tn (R), n 2, is inner (Corollaries 2.5 and 2.8). These extend some results of Benkoviˇc [1]. Some (counter)examples are also presented. Let R, S and M be as above. In the sequel, by T
=⎝
2. Main results and proofs Before characterizing the biderivations of the triangular ring T, let us begin with the following lemma which will be used frequently. Lemma 2.1. Let R be a ring with identity and let d in R we have d(x, y)[u, v] d(x, 1)
: R × R → R be a biderivation. Then for all x, y, u, v
= [x, y]d(u, v);
= 0 = d(1, x) and d(x, 0) = 0 = d(0, x).
(3) (4)
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Proof. Since d is a derivation in the first argument, in view of (1), we have d(xy, uv)
= d(x, uv)y + xd(y, uv).
Also, since d is a derivation in the second argument, we then have d(xy, uv)
= (d(x, u)v + ud(x, v))y + x(d(y, u)v + u(d(y, v)) = d(x, u)vy + ud(x, v)y + xd(y, u)v + xud(y, v).
On the other hand, expanding d first on the second argument, and then on the first argument, we arrive at d(xy, uv)
= d(x, u)yv + xd(y, u)v + ud(x, v)y + uxd(y, v).
Comparing these two relations, we obtain (3). From d(x, 1)
= d(x, 1.1) = d(x, 1) + d(x, 1)
it follows that d(x, 1) verified. Now, we have
= 0. Similarly, d(1, x) = 0. The identities d(x, 0) = 0 = d(0, x) are easily
Theorem 2.2. Let T be the triangular ring given above and let d (i) There exists an element m∗ d(E11 , E11 )
: T × T → T be a biderivation.
∈ M such that
∗
= m E12 = d(E22 , E22 ),
(5)
and such that for all r , r ∈ R, s, s ∈ S we have r , r m∗ = 0 = m∗ s , s , d(rE11 , sE22 )
(6)
∗
= −rm sE12 = d(sE22 , rE11 ).
(7)
(ii) There exists a biderivation δ : R × R → R such that d rE11 , r E11 = δ r , r E11 + rr m∗ E12 for all r , r
∈ R.
(8)
(iii) There exists a biderivation γ : S × S → S such that d sE22 , s E22 = m∗ ss E12 + γ s, s E22 for all s, s
∈ S.
(9)
(iv) There exists an (R, S )-bimodule homomorphism g
: M → M such that
d(rE11 , mE12 ) = rg (m)E12 , d(sE22 , mE12 ) = −g (m)sE12 , δ r , r m = r , r g (m), mγ s, s = g (m) s, s for all r , r ∈ R, m ∈ M, and s, s ∈ S . (v) There exists an (R, S )-bimodule homomorphism h : M
∈ R, m ∈ M, and s, s ∈ S.
(11)
→ M such that
d(mE12 , rE11 ) = rh(m)E12 , d(mE12 , sE22 ) = −h(m)sE12 δ r , r m = r , r h(m), mγ s, s = h(m) s , s for all r , r
(10)
(12) (13)
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(vi) There exists an (R, S )-bimodule homomorphism f M , r , r ∈ R, and s, s ∈ S we have d mE12 , m E12 = f m, m E12 ; r , r f m , m = 0 = f m, m s , s .
253
: M × M → M such that for all m, m ∈ (14) (15)
Proof.
(i) Put d(E11 , E11 )
2 = (xij ). Applying d on both sides of (E11 , E11 ) = E11 , E11 , one observes that ∗ = 0, whence d(E11 , E11 ) = x12 E12 =: m E12 . Now, using the first equation of (4)
x11 = x22 twice, we have
d(E22 , E22 )
= −d(E11 , E22 ) = −(−d(E11 , E11 )) = d(E11 , E11 ) = m∗ E12 .
To prove the first conclusion of (5), substitute x = rE11 , y We have 0 = d rE11 , r E11 0 = d(x, y)[u, v] = [x, y]d(u, v) = r , r E11 m∗ E12 = r , r m∗ E12 ,
= r E11 , and u = v = E11 into (3).
so that r , r m∗ = 0. Substituting now x = y = E11 , u = sE22 , and v = s E22 into (3), we conclude that m∗ s, s = 0. To prove the relation d(rE11 , sE22 ) = −rm∗ sE12 , first we show that d(rE11 , E22 ) = −rm∗ E12 : set d(rE11 , E22 ) = (yij ). We have 2 (yij ) = d(rE11 , E22 ) = d rE11 , E22 = E22 (yij ) + (yij )E22 , from which one derives d(rE11 , E22 ) y12 E12
= y12 E12 . Therefore,
= d(rE11 , E22 ) = d((rE11 )E11 , E22 )) = d(rE11 , E22 )E11 + rE11 d(E11 , E22 ) = (y12 E12 )E11 − rE11 d(E11 , E11 ) = −rE11 (m∗ E12 ) = −rm∗ E12 .
Hence, d(rE11 , E22 ) = −rm∗ E12 . Now, assume that d(rE11 , sE22 ) = (zij ). Applying d on (rE11 , sE22 ) = (rE11 , E22 (sE22 )) and using the last conclusion, it follows that d(rE11 , sE22 ) = −rm∗ sE12 + z22 E22 . Thus, d(E11 , sE22 ) = E for some z ∈ S. We have now −m∗ sE12 + z22 22 22
−rm∗ sE12 + z22 E22 = d(rE11 , sE22 ) = d((rE11 )E11 , sE22 ) = (−rm∗ sE12 + z22 E22 )E11 E ) +rE11 (−m∗ sE12 + z22 22
= −rm∗ sE12 . Accordingly, z22
= 0, and d(rE11 , sE22 ) = −rm∗ sE12 .
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In a similar manner we can prove that d(sE22 , rE11 ) = −rm∗ sE12 . To prove (ii), let r , r be in R and set d(rE11 , r E11 ) = (tij ). Using (7), we have
(tij ) = d(rE11 , r E11 ) = d((rE11 )E11 , r E11 ) = (tij )E11 + rE11 d(E11 , r E11 ) = t11 E11 + (rE11 )(r m∗ E12 ) = t11 E11 + rr m∗ E12 = δ(r , r )E11 + rr m∗ E12 ,
where δ : R × R → R is a mapping that satisfies d rE11 , r E11 d is biadditive, so is δ . Let r , r1 , r2 ∈ R. We have
= δ r , r E11 + rr m∗ E12 . Since
δ(r1 r2 , r )E11 + r1 r2 rm∗ E12 = d(r1 r2 E11 , rE11 ) = d((r1 E11 )(r2 E11 ), rE11 ) = d(r1 E11 , rE11 )(r2 E11 ) +r1 E11 d(r2 E11 , rE11 ) = (δ(r1 , r )E11 + r1 rm∗ E12 )r2 E11 +r1 E11 (δ(r2 , r )E11 + r2 rm∗ E12 ) = δ(r1 , r )r2 E11 + r1 δ(r2 , r )E11 + r1 r2 rm∗ E12 = (δ(r1 , r )r2 + r1 δ(r2 , r ))E11 + r1 r2 rm∗ E12 . Consequently, δ(r1 r2 , r ) = δ(r1 , r )r2 + r1 δ(r2 , r ), whence δ is a derivation on the first component. Similarly, one shows that δ is also a derivation on the second component. Hence, δ is a biderivation of R. (iii) Let s, s be in S and put d(sE22 , s E22 ) = (uij ). By (4) and (7) we have d(sE22 , E22 ) = −d(sE22 , E11 ) = sm∗ E12 . Applying d on (sE22 , s E22 ) = (sE22 , E22 (s E22 )) yields z11 = 0, z12 = m∗ ss . Therefore, there exists a mapping γ : S × S → S such that d sE22 , s E22 = m∗ ss E12 + γ s, s E22 . The proof that γ is a biderivation of S is similar to that of δ , hence omitted. (iv) Let r ∈ R and m ∈ M. First assume that d(E11 , mE12 ) = (vij ). Applying d on (E11 , mE12 ) = 2 , mE12 ), one observes that v11 = v22 = 0. Thus, there exists a mapping g from M to itself (E11 such that d(E11 , mE12 ) = g (m)E12 . Now assume that d(rE11 , mE12 ) = (wij ). Applying d on (rE11 , mE12 ) = ((rE11 )E11 , mE12 ), in view of the preceding result, we observe that w12 = g (m), and w22 = 0. Now applying d on (rE11 , mE12 ) = (rE11 , (mE12 )E22 ) and using (7), we conclude that w11 = 0. Hence, d(rE11 , mE12 ) = rg (m)E12 , as desired. Let s ∈ S and m ∈ M. To compute d(sE22 , mE12 ), first note that, by (4) and the preceding result, d(E22 , mE12 ) = −d(E11 , mE12 ) = −g (m)E12 . Now, applying d first on (sE22 , mE12 ) = (E22 (sE22 ), mE12 ) and then on (sE22 , mE12 ) = (sE22 , E11 (mE12 )), we find that d(sE22 , mE12 )
= −g (m)sE12 .
Since d is additive on the second component, g is additive. Moreover, from g (rm)E12
= d(E11 , rmE12 ) = d(E11 , (rE11 )(mE12 )) = d(E11 , rE11 )mE12 + rE11 d(E11 , mE12 ) = (rm∗ E12 )mE12 + rE11 (g (m)E12 ) = rg (m)E12
N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
255
we infer that g is a left R-homomorphism. Likewise, applying d on both sides of (E22 , msE12 ) = (E22 , (mE12 )(sE22 )), we see that g is also a right S-homomorphism.
Next, we show that δ r , r m = r , r g (m): substituting x = rE11 , y = r E11 , u = E11 , and v = mE12 into (3), by (8) and (10) we have
δ r , r E11
+ rr m∗ E12 (mE12 ) = d(x, y)[u, v] = [x, y]d(u, v) = =
r , r E11 (g (m)E12 )
r , r g (m)E12 .
Therefore, δ r , r m = r , r g (m). Putting now x = E11 , y = mE 12 , u = sE22 and v = s E22 into the Eq. (3) and using (8), (10) again, we conclude that mγ s, s = g (m) s, s . This completes the proof of (iv). The proof of (v) is similar to that of (iv), except for the
coordinates, so it is omitted. (vi) Let m, m ∈ M be arbitrary and set d mE12 , m E12 = (xij ). Applying d on both sides of
mE12 , m E12 = E11 (mE12 ), m E12 and using (10), we obtain x22 = 0. Then applying d on mE12 , m E12 = (mE12 )E22 , m E12 and making use
obtain of (10), we x11 = 0. Therefore, there exists a mapping f : M × M → M such that d mE12 , m E12 = f m, m E12 . Biadditivity of f is inherited from d. To show that f is a left R-homomorphism on the first component, let r ∈ R, and m, m ∈ M . In view of (10) we have f (rm, m )E12
= d(rmE12 , m E12 ) = d((rE11 )(mE12 ), m E12 ) = d(rE11 , m E12 )mE12 + rE11 d(mE12 , m E12 ) = rg (m )E12 (mE12 ) + rE11 (f (m, m )E12 ) = rf (m, m )E12 .
That is, f rm, m = rf m, m . Likewise, one can show that f is also a right S-homomorphism on the first component, and an (R, S )-bimodule homomorphism on the second component as well (see Eq. (2)). It only remains to show that for all r , r ∈ R, s, s ∈ S and m, m ∈ M, the following relations hold. r , r f m , m = 0 = f m, m s , s . Substituting x = mE12 , y = r E11 , u = mE12 and v = m E12 into the Eq. (3), we get δ r , r E11 + rr m∗ E12 0 = r , r E11 f m, m E12 ,
from which it follows that r , r f m, m = 0 . Finally, substituting x sE22 and v = s E22 into (3), we obtain f m, m s, s = 0. Remark 2.3. (a) Let R, S , M , T , and the element m∗ ⎞ ⎛ 1 m∗ ⎠. A=⎝ 0 1
∈ M be as above. Define
Then, in view of (6), it is straightforward to verify that
[A, [X , Y ]] = 0
for all X , Y
∈ T.
= mE12 , y = m E12 , u =
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Therefore, the mapping
: T ×T ⎛ → T given , Y ) = [⎞X , [Y , A]] is an extremal bideriva⎞ by (X⎛
tion of T. Note that for every X
=⎝
r m
⎠ and Y
0 s
=⎝
r m 0 s
⎠ in T we have
(X , Y ) = [X , [Y , A]] = (rr m∗ − rm∗ s − r m∗ s + m∗ ss )E12 . (b) Assume that f : M × M : T × T → T by
(16)
→ M is the (R, S)-bimodule homomorphism introduced in (iv). Define
(X , Y ) = f m, m E12 .
(17)
An easy computation shows that is a biderivation of T. Now, we are ready to state the main result. Theorem 2.4. Let R, M , S and T be as above. If d : T × T → T is a biderivation, then there exists a biderivation D with D(E11 , E11 ) = 0, an extremal biderivation , and a special kind of biderivation such that d
= D + + . ⎞
⎛ Proof. Let X
= ⎝
r m
0 s (12), and (14) we have d(X , Y )
⎠,Y
⎛
= ⎝
r m 0 s
⎞ ⎠ be in T. Using biadditivity of d and conclusions (7)–(10),
= d(rE11 + mE12 + sE22 , r E11 + m E12 + s E22 ) ⎛
= ⎝
δ(r , r ) rg (m ) − g (m )s + r h(m) − h(m)s 0
γ (s, s )
⎞ ⎠
+(rr m∗ − rm∗ s − r m∗ s + m∗ ss )E12 + f (m, m )E12 . Therefore, if we define D(X , Y ) to be the first term of the latter expression, by (16) and (17) we have d(X , Y )
= D(X , Y ) + (X , Y ) + (X , Y ).
By the preceding remarks, is the extremal biderivation induced by A, and is a biderivation. Therefore, D = d − − is also a biderivation of T. Evidently, D(E11 , E11 ) = 0. Theorem 2.4 has some nice corollaries. First, assume that R = M g (1), b = h(1) and c = f (1, 1). Then, by (iv), for each r ∈ R we have ra
= S, and T = T2 (R). Set a =
= rg (1) = g (r .1) = g (1.r ) = g (1)r = ar ,
hence, a ∈ Z (R), and g (r ) = ra. Similarly, b ∈ Z (R), and h(r ) = rb. Moreover, using (2), it is easy to verify that c ∈ Z (R), and that for each r , r ∈ R we have f (r , r ) = crr = cr r . Also, by (11), δ(r , r ) = [r , r ]g (1) = a[r , r ], and, by (13), δ(r , r ) = [r , r ]h(1) = [r , r ]b = −b[r , r ]. Hence, δ(r , r ) = a[r , r ] = −b[r , r ] is an inner biderivation of R. Similarly, one shows that γ (r , r ) = a[r , r ] = −b[r , r ] = δ(r , r ). Note that by the last conclusions we have (a + b) r , r = 0 for all r , r ∈ R. (18)
N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
Now, in view of (18), it is straightforward to show that for every X
= (rij ) and Y =
257
rij in T we
have D(X , Y )
=
⎛
a r r + r r − r r a r11 , r11 11 12 12 22 11 12 ⎝
0 a r22 , r22 E +(a + b)r12 r11 − r12 12
r − r12 22
⎞ ⎠
= (aI )[X , Y ] + D (X , Y ), in which the first term is the image of (X , Y ) under the inner biderivation induced by the central diagonal matrix aI, and in the next one, D , as the difference of two biderivations, is a (special) biderivation of T2 (R). Moreover, by Remark 2.3(b) and the calculation above, we have (X , Y ) = f r12 , r12 E12 . E12 = cr12 r12 Gathering the above results together, we are led to the following corollary which is a generalization of Proposition 4.16 in [1]. Recall that a ring R is said to be prime if for every a, b ∈ R, whenever aRb = 0, then a = 0 or b = 0, and R is said to be semiprime if for every a ∈ R, aRa = 0 implies that a = 0. Corollary 2.5. Let R be a ring with identity and d be a biderivation of T2 (R). Then d is decomposed into the sum of the inner biderivation induced by the central matrix aI, and the biderivations D , and given above. In particular, if R is a noncommutative prime ring, then d is an inner biderivation. Proof. By the above argument, it remains only to prove the special case. First, we show that m∗ = 0: choose r1 , r2 ∈ R such that [r1 , r2 ] = 0, and let r ∈ R be arbitrary. From (6) and the fact that [rr1 , r2 ] = r [r1 , r2 ] + [r , r2 ]r1 , it follows that 0
= m∗ [rr1 , r2 ] = m∗ (r [r1 , r2 ] + [r , r2 ]r1 ) = m∗ r [r1 , r2 ].
Since [r1 , r2 ] = 0 and r is arbitrary, the primeness of R forces m∗ = 0, so that = 0. Similarly, in view of (15) (from which it follows that c [r , r ] = f (1, 1)[r , r ] = 0) and (18), we conclude that c = a + b = 0. Accordingly, D and are both identically zero. Consequently, d is an inner biderivation. The following examples show, respectively, that if in the particular case of the corollary above we remove the noncommutativity of R, or if R is even a noncommutative semiprime ring, the conclusion may fail. Example 2.6. Let R be an integral domain and put T
= T2 (R). Define d : T × T → T by d((rij ), (rij )) =
E . Then one observes that d is a noninner biderivation of T. r12 r12 12
Example 2.7. Let R = R1 × R2 , where R1 is a noncommutative prime ring and R2 is an integral domain. Put T = T2 (R) and define d : T × T → T by d (aij , bij ), aij , bij = 0, b12 b12 E12 . Then a straightforward computation shows that d is a noninner biderivation of T. Next, we show that if R is a ring with identity and n 3, then the biderivations of Tn (R) have more compact forms. This result generalizes partly Corollary 4.13 of Benkoviˇc [1]. Corollary 2.8. Let R be a ring with identity and assume that n 3. Then every biderivation of the ring Tn (R) is the sum of an inner biderivation and an extremal biderivation. In particular, if R is a noncommutative prime ring, then every biderivation of Tn (R) is inner.
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Proof. We have the obvious ring isomorphism ⎛ ⎞ R Rn−1 ⎠ Tn (R) ∼ =⎝ 0 Tn−1 (R) in which Rn−1 is considered as an R, Rn−1 -bimodule with the obvious scaler multiplications. So, upon
identifying these two rings, we can deal with ring in the right-hand side. Set 1 = (1, . . . , 1) ∈ Rn−1 , and assume that g (1) = (a1 , . . . , an−1 ). Since g is a right Tn−1 (R)-homomorphism on Rn−1 , we have
(a1 , . . . , an−1 ) = g (1) = g (1.(E11 + · · · + E1,n−1 )) = g (1)(E11 + · · · + E1,n−1 ) = (a1 , . . . , an−1 )(E11 + · · · + E1,n−1 ) = (a1 , . . . , a1 ). Hence, a1 = a2 = · · · = an−1 Rn−1 , for each r ∈ R we have
=: a. On the other hand, since g is also a left R-homomorphism on
(ra, . . . , ra) = r (a, . . . , a) = rg (1) = g (r .1) = g (1.(rI )) = g (1)rI = (a, . . . , a)rI = (ar , . . . , ar ). I.e., a
∈ Z (R). Moreover, for each (r1 , . . . , rn−1 ) ∈ Rn−1 we have g (r1 , . . . , rn−1 )
= g (1.(r1 E11 + · · · + rn−1 E1,n−1 )) = (a, . . . , a)(r1 E11 + · · · + rn−1 E1,n−1 ) = (ar1 , . . . , arn−1 ) = a(r1 , . . . , rn−1 ).
Now, we claim that δ is the inner biderivation induced by a: let r , r conclusion it follows that
∈ R. Then from (11) and the latter
δ(r , r )(1, . . . , 1) = [r , r ]g (1, . . . , 1) = [r , r ](a, . . . , a), so that δ(r , r ) = a[r , r ]. Setting now h(1) = (b1 , . . . , bn−1 ) and using (13), a similar approach shows that b1 = · · · = n−1 bn−1 =: b is in Z (R), for every (r1 , . . . , rn− we
1 ) ∈ R
have h(r1 , . . . , rn−1 ) = b(r1 , . . . , rn−1 ), and that, for every r , r ∈ R we have δ r , r = −b r , r . Next, we show that γ is the inner biderivation of Tn−1 (R) induced by the central matrix aI: By (11), for every S , S ∈ Tn−1 (R) we have
(1, . . . , 1)γ (S, S ) = g (1, . . . , 1)[S, S ] = (a, . . . , a)[S, S ]. Therefore, noting that g (1, 2, 1, . . . , 1)
= g ((1, . . . , 1)(E11 + 2E12 + · · · + E1,n−1 )) = g (1, . . . , 1)(E11 + 2E12 + · · · + E1,n−1 ) = (a, . . . , a)(E11 + 2E12 + · · · + E1,n−1 )) = (a, 2a, a, . . . , a),
(19)
N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
259
we find that
(1, 2, 1, . . . , 1)γ S, S = (a, 2a, a, . . . , a) S, S .
(20)
Analogous computations show that (1, 1, 2, 1, . . . , 1)γ S, S = (a, a, 2a, a, . . . , a) S, S , . . . ,
(21)
and, finally, that
(1, . . . , 1, 2)γ S, S = (a, . . . , a, 2a) S, S .
(22)
Eqs. (19)–(22) lead us to the matrix equation P γ S , S = P (aI ) S , S
(23)
= I +(1)− E11 is invertible (with P −1 = I − nj=−21 E1j − ni=−21 Ei1 +(n − 1)E11 ).
Therefore, from (23) it follows that γ (S , S ) = (aI ) S , S . We claim now that the mapping f : Rn−1 × Rn−1 → Rn−1 is identically zero, from which it follows that = 0. Suppose that f (1, 1) = (c1 , . . . , cn−1 ). By Theorem 2.2(vi) we have in Mn−1 (R), in which P
(c1 , . . . , cn−1 ) = f (1, 1(E11 + · · · + E1,n−1 )) = f (1, 1)(E11 + · · · + E1,n−1 ) = (c1 , . . . , cn−1 )(E11 + · · · + E1,n−1 ) = (c1 , . . . , c1 ), hence, c1 = · · · = cn−1 =: c, and f (1, 1) = (c , . . . , c ). By (15), for every S , S ∈ Tn−1 (R) we have (c , . . . , c )[S, S ] = f (1, 1)[S, S ] = 0. Therefore, putting S = E11 + E1,n−1 and S = E1,n−1 , we get (c , . . . , c )E1,n−1 = 0, whence c = 0. Thus, for every m, m ∈ Rn−1 , we have f (m, m )
= f (m.1, m .1) = mm f (1, 1) = mm c = 0.
We can now show that D is an inner biderivation. Let X = (rij ) and Y above conclusions, we have ⎛ ⎞ ] ar U − aU S − ar U + aUS a[r11 , r11 11 11 ⎠ D(X , Y ) = ⎝ 0 aI [S , S ]
= (rij ) be in Tn (R). Using the
= (aI )[X , Y ], , . . . , r ), and the matrices S , S in T where U = (r12 , . . . , r1n ), U = (r12 n−1 (R) are determined by 1n canceling out the first rows and columns of X , Y , respectively. Notice that in the above equations, the first identity matrix I is in Tn−1 (R), while the second one belongs to Tn (R). Consequently, for every X , Y ∈ Tn (R) we have
d(X , Y )
= D(X , Y ) + (X , Y ) + (X , Y ) = aI [X , Y ] + (X , Y ).
Finally, if R is a noncommutative prime ring, then as we saw in Corollary 2.5, m∗ also [1,9].)
= 0, so = 0. (See
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N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
Remark 2.9. In Corollary 2.8, assume that m∗ = (m∗1 , . . . , m∗n−1 ) (see (5)). We claim that m∗1 = · · · = m∗n−2 = 0. To see this, fix 1 i n − 2, and in the second equation of (6) put s = Eii + Ein and s = Ein . Then from 0
= (m∗1 , . . . , m∗n−1 )[s, s ] = (m∗1 , . . . , m∗n−1 )Ein
it follows that m∗i = 0. The following example shows, however, that the element m∗ zero.
∈ Rn−1 in Corollary 2.8 need not be
Example 2.10. Let R be a commutative ring with identity and put T = Tn (R)(n 3). Set A = I + E1n . Then it is easy to verify that for all X = (rij ), Y = (rij ) ∈ T, the relation [A, [X , Y ]] = 0 holds, so E is an − rnn that the mapping d : T × T → T defined by d(X , Y ) = [X , [Y , A]] = (r11 − rnn ) r11 1n (extremal) biderivation of T. Note that d(E11 , E11 ) = E1n , hence m∗ = (0, . . . , 0, 1) = 0. We finish by the following example which shows that in the particular case of Corollary 2.8, the noncommutativity of R can not be removed. Example 2.11. Assume assume that the ring R in the preceding example is an integral domain. Then for each a ∈ R we have aI [E11 , E11 ] = 0, while d(E11 , E11 ) = E1n = 0. Hence, d is not an inner biderivation of T. Acknowledgement The author would like to express his sincere thanks to the referees for careful reading the manuscript and useful suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
D. Benkoviˇc, Biderivations of triangular algebras, Linear Algebra Appl. 431 (2009) 1587–1602. M. Brešar, Commuting maps: a survey, Taiwanese J. Math. 8 (2004) 361–397. M. Brešar, Functional identities: a survey, Contemp. Math. 259 (2000) 93–109. M. Brešar, On a generalization of the notion of centralizing mappings, Proc. Amer. Math. Soc. 114 (1992) 641–649. M. Brešar, On certain pairs of functions of semiprime rings, Proc. Amer. Math. Soc. 120 (1994) 709–713. M. Brešar, On generalized biderivations and related maps, J. Algebra 172 (1995) 764–786. M. Brešar, W.S. Martindale 3rd, C.R. Miers, Centralizing maps in prime rings with involusion, J. Algebra 161 (1993) 342–357. W.-S. Cheung, Commuting maps of triangular algebras, J. London Math. Soc. 63 (2001) 117–127. D.R. Farkas, G. Letzter, Ring theory from symplectic geometry, J. Pure Appl. Algebra 125 (1998) 155–190. C. Lanski, Differential identities, Lie ideals, and Posner’s therorems, Pacific J. Math. 134 (1988) 275–298. E.C. Posner, Derivations in prime rings, Proc. Amer. Math. Soc. 8 (1957) 1093–1100. V.G. Skosyrskii, Strongly prime noncommutative Jordan algebras, Trudy Inst. Mat. (Novosibirsk) 16 (1989) 131–164 (in Russian). J. Vukman, Commuting and centralizing mappings in prime rings, Proc. Amer. Math. Soc. 109 (1990) 47–52. J.-H. Zhang, S. Feng, H.-X. Li, R.-H. Wu, Generalized biderivations of nest algebras, Linear Algebra Appl. 418 (2006) 225–233. Y. Zhao, D. Wang, R. Yao, Biderivations of upper triangular matrix algebras over commutative rings, Int. J. Math. Game Theory Algebra 18 (2009) 473–478.
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Linear Algebra and its Applications journal homepage: w w w . e l s e v i e r . c o m / l o c a t e / l a a
On biderivations of upper triangular matrix rings Nader M. Ghosseiri Department of Mathematics, University of Kurdistan, P.O. Box 416, Sanandaj, Iran
ARTICLE INFO
ABSTRACT
Article history: Received 6 March 2012 Accepted 9 July 2012 Available online 13 September 2012
Let R and S be rings with identity, M be a unitary (R, S )-bimodule, ⎛ ⎞ R M ⎠ be the upper triangular matrix ring determined and T = ⎝ 0 S
Submitted by P. Sˇ emrl
by R, S and M. Let Eij be the standard matrix unit. In this paper we show that every biderivation of T is decomposed into the sum of three biderivations D, and , where D(E11 , E11 ) = 0, is an extremal biderivation and is a special kind of biderivation. Using this characterization, we determine the structure of biderivations of the ring Tn (R)(n 2) of all n × n upper triangular matrices over R, and show that in the special case when R is a noncommutative prime ring, every biderivation of Tn (R) is inner. This extends some results of Benkoviˇc (2009) [1]. © 2012 Elsevier Inc. All rights reserved.
AMS classification: 16W20 16S50 16U60 Keywords: Triangular matrix ring Derivation Biderivation Prime ring
1. Introduction Let R be a ring and Z (R) be the center of R. For each x, y in R, denote the commutator of x, y by [x, y] = xy − yx. An additive mapping d : R → R is said to be a derivation of R if d(xy) = d(x)y + xd(y) for all x, y in R. Let a be in R. The mapping Ia : R → R given by Ia (x) = [x, a] is easily seen to be a derivation. Ia is called the inner derivation induced by a. A biadditive mapping d : R × R → R is said to be a biderivation of R if it is a derivation in each argument; that is, d(xy, z )
= d(x, z)y + xd(y, z)
and
d(x, yz )
= d(x, y)z + yd(x, z)
(1)
are fulfilled for all x, y, z ∈ R. A map f : R → R is called centralizing on R if [f (x), x] ∈ Z (R) for all x ∈ R. In the special case when [f (x), x] = 0 for all x ∈ R, f is called commuting on R. The study of centralizing and commuting maps was initiated by a classical result of Posner [11] which states that if a prime ring admits a nonzero centralizing derivation, then it is commutative. A number of authors have extended Posner’s theorem in several directions. Readers interested in centralizing and commuting E-mail addresses: [email protected], [email protected]. 0024-3795/$ - see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2012.07.039
N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
251
mappings are referred to [2–4,10,13] and the references therein. The notion of additive commuting maps is closely related to the notion of biderivations. In fact, every commuting additive map f on R gives raise to a biderivation of R: linearizing [f (x), x] = 0, we observe that [f (x), y] = [x, f (y)] for all x, y ∈ R, so the mapping (x, y) → [f (x), y] is easily seen to be a biderivation of R that is inner in each component (induced by f (y) and f (x), respectively). A biderivation d of R is called an inner biderivation if there exists λ ∈ Z (R) such that d(x, y) = λ[x, y] for all x, y in R. Inner biderivations appear quite naturally in characterizing biderivations of rings. In [7], Brešar et al. showed that every biderivation d of a noncommutative prime ring R is of the form d(x, y) = λ[x, y] for some λ in the centroid C of R. Somewhat later, Brešar [5] extended this result to semiprime rings. Also, Benkoviˇc [1] proved that under certain conditions every biderivation of a triangular algebra decomposes into the sum of an inner biderivation and a special kind of biderivation which is called an extremal biderivation. More details on biderivations, their generalizations, and related maps can be found in [2,5–9,12,14,15]. Specially, the reader is encouraged to read the survey paper of Brešar [2] in which he has described the applications of biderivations to some other fields. We say that the mapping d : R × R → R is an extremal biderivation if d(x, y) = [x, [y, a]] for all x, y ∈ R, where a ∈ R is such that [a, [R, R]] = 0. Let R, S be rings with identity and M be a unitary (R, S )-bimodule. Let f : M × M → M be a biadditive mapping. We say that f is an (R, S )-bimodule homomorphism if it is a bimodule homomorphism in each argument; namely, (2) f rms, m = rf m, m s and f m, r m s = r f m, m s for all r , r
∈ R, s, s ∈ S, and m, m ∈ M.
⎞
⎛ R M
⎠ we denote the upper triangular matrix 0 S ring determined by R, S and M with the usual addition and multiplication of matrices. Eij stands for the standard matrix unit, aEij stands for the matrix whose ij-th position is a and zero elsewhere, I stands for the identity matrix and, as usual, Tn (R) denotes the ring of all n × n upper triangular matrices over R. Moreover, the zero elements of the rings and modules, zero subrings, and zero submodules are all denoted by 0. Recall that Eij Ers = δjr Eis , where δ is the Kronecker function. Motivated by [1], in this paper we show that without any further assumptions on R, S and M, any biderivation of the ring T decomposes into the sum of three biderivations D, and , where D(E11 , E11 ) = 0, is an extremal biderivation, and is a special kind of biderivation (Theorem 2.4). Using this characterization, we shall see that any biderivation of the ring T2 (R) is decomposed into the sum of four biderivations: an inner biderivation, an extremal biderivation, and two special kinds of biderivations (Corollary 2.5). Also, we prove that every biderivation of the ring Tn (R), n 3, is decomposed into the sum of an inner biderivation and an extremal biderivation (Corollary 2.8). Moreover, we shall see that if R is a noncommutative prime ring, then any biderivation of the ring Tn (R), n 2, is inner (Corollaries 2.5 and 2.8). These extend some results of Benkoviˇc [1]. Some (counter)examples are also presented. Let R, S and M be as above. In the sequel, by T
=⎝
2. Main results and proofs Before characterizing the biderivations of the triangular ring T, let us begin with the following lemma which will be used frequently. Lemma 2.1. Let R be a ring with identity and let d in R we have d(x, y)[u, v] d(x, 1)
: R × R → R be a biderivation. Then for all x, y, u, v
= [x, y]d(u, v);
= 0 = d(1, x) and d(x, 0) = 0 = d(0, x).
(3) (4)
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N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
Proof. Since d is a derivation in the first argument, in view of (1), we have d(xy, uv)
= d(x, uv)y + xd(y, uv).
Also, since d is a derivation in the second argument, we then have d(xy, uv)
= (d(x, u)v + ud(x, v))y + x(d(y, u)v + u(d(y, v)) = d(x, u)vy + ud(x, v)y + xd(y, u)v + xud(y, v).
On the other hand, expanding d first on the second argument, and then on the first argument, we arrive at d(xy, uv)
= d(x, u)yv + xd(y, u)v + ud(x, v)y + uxd(y, v).
Comparing these two relations, we obtain (3). From d(x, 1)
= d(x, 1.1) = d(x, 1) + d(x, 1)
it follows that d(x, 1) verified. Now, we have
= 0. Similarly, d(1, x) = 0. The identities d(x, 0) = 0 = d(0, x) are easily
Theorem 2.2. Let T be the triangular ring given above and let d (i) There exists an element m∗ d(E11 , E11 )
: T × T → T be a biderivation.
∈ M such that
∗
= m E12 = d(E22 , E22 ),
(5)
and such that for all r , r ∈ R, s, s ∈ S we have r , r m∗ = 0 = m∗ s , s , d(rE11 , sE22 )
(6)
∗
= −rm sE12 = d(sE22 , rE11 ).
(7)
(ii) There exists a biderivation δ : R × R → R such that d rE11 , r E11 = δ r , r E11 + rr m∗ E12 for all r , r
∈ R.
(8)
(iii) There exists a biderivation γ : S × S → S such that d sE22 , s E22 = m∗ ss E12 + γ s, s E22 for all s, s
∈ S.
(9)
(iv) There exists an (R, S )-bimodule homomorphism g
: M → M such that
d(rE11 , mE12 ) = rg (m)E12 , d(sE22 , mE12 ) = −g (m)sE12 , δ r , r m = r , r g (m), mγ s, s = g (m) s, s for all r , r ∈ R, m ∈ M, and s, s ∈ S . (v) There exists an (R, S )-bimodule homomorphism h : M
∈ R, m ∈ M, and s, s ∈ S.
(11)
→ M such that
d(mE12 , rE11 ) = rh(m)E12 , d(mE12 , sE22 ) = −h(m)sE12 δ r , r m = r , r h(m), mγ s, s = h(m) s , s for all r , r
(10)
(12) (13)
N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
(vi) There exists an (R, S )-bimodule homomorphism f M , r , r ∈ R, and s, s ∈ S we have d mE12 , m E12 = f m, m E12 ; r , r f m , m = 0 = f m, m s , s .
253
: M × M → M such that for all m, m ∈ (14) (15)
Proof.
(i) Put d(E11 , E11 )
2 = (xij ). Applying d on both sides of (E11 , E11 ) = E11 , E11 , one observes that ∗ = 0, whence d(E11 , E11 ) = x12 E12 =: m E12 . Now, using the first equation of (4)
x11 = x22 twice, we have
d(E22 , E22 )
= −d(E11 , E22 ) = −(−d(E11 , E11 )) = d(E11 , E11 ) = m∗ E12 .
To prove the first conclusion of (5), substitute x = rE11 , y We have 0 = d rE11 , r E11 0 = d(x, y)[u, v] = [x, y]d(u, v) = r , r E11 m∗ E12 = r , r m∗ E12 ,
= r E11 , and u = v = E11 into (3).
so that r , r m∗ = 0. Substituting now x = y = E11 , u = sE22 , and v = s E22 into (3), we conclude that m∗ s, s = 0. To prove the relation d(rE11 , sE22 ) = −rm∗ sE12 , first we show that d(rE11 , E22 ) = −rm∗ E12 : set d(rE11 , E22 ) = (yij ). We have 2 (yij ) = d(rE11 , E22 ) = d rE11 , E22 = E22 (yij ) + (yij )E22 , from which one derives d(rE11 , E22 ) y12 E12
= y12 E12 . Therefore,
= d(rE11 , E22 ) = d((rE11 )E11 , E22 )) = d(rE11 , E22 )E11 + rE11 d(E11 , E22 ) = (y12 E12 )E11 − rE11 d(E11 , E11 ) = −rE11 (m∗ E12 ) = −rm∗ E12 .
Hence, d(rE11 , E22 ) = −rm∗ E12 . Now, assume that d(rE11 , sE22 ) = (zij ). Applying d on (rE11 , sE22 ) = (rE11 , E22 (sE22 )) and using the last conclusion, it follows that d(rE11 , sE22 ) = −rm∗ sE12 + z22 E22 . Thus, d(E11 , sE22 ) = E for some z ∈ S. We have now −m∗ sE12 + z22 22 22
−rm∗ sE12 + z22 E22 = d(rE11 , sE22 ) = d((rE11 )E11 , sE22 ) = (−rm∗ sE12 + z22 E22 )E11 E ) +rE11 (−m∗ sE12 + z22 22
= −rm∗ sE12 . Accordingly, z22
= 0, and d(rE11 , sE22 ) = −rm∗ sE12 .
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N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
In a similar manner we can prove that d(sE22 , rE11 ) = −rm∗ sE12 . To prove (ii), let r , r be in R and set d(rE11 , r E11 ) = (tij ). Using (7), we have
(tij ) = d(rE11 , r E11 ) = d((rE11 )E11 , r E11 ) = (tij )E11 + rE11 d(E11 , r E11 ) = t11 E11 + (rE11 )(r m∗ E12 ) = t11 E11 + rr m∗ E12 = δ(r , r )E11 + rr m∗ E12 ,
where δ : R × R → R is a mapping that satisfies d rE11 , r E11 d is biadditive, so is δ . Let r , r1 , r2 ∈ R. We have
= δ r , r E11 + rr m∗ E12 . Since
δ(r1 r2 , r )E11 + r1 r2 rm∗ E12 = d(r1 r2 E11 , rE11 ) = d((r1 E11 )(r2 E11 ), rE11 ) = d(r1 E11 , rE11 )(r2 E11 ) +r1 E11 d(r2 E11 , rE11 ) = (δ(r1 , r )E11 + r1 rm∗ E12 )r2 E11 +r1 E11 (δ(r2 , r )E11 + r2 rm∗ E12 ) = δ(r1 , r )r2 E11 + r1 δ(r2 , r )E11 + r1 r2 rm∗ E12 = (δ(r1 , r )r2 + r1 δ(r2 , r ))E11 + r1 r2 rm∗ E12 . Consequently, δ(r1 r2 , r ) = δ(r1 , r )r2 + r1 δ(r2 , r ), whence δ is a derivation on the first component. Similarly, one shows that δ is also a derivation on the second component. Hence, δ is a biderivation of R. (iii) Let s, s be in S and put d(sE22 , s E22 ) = (uij ). By (4) and (7) we have d(sE22 , E22 ) = −d(sE22 , E11 ) = sm∗ E12 . Applying d on (sE22 , s E22 ) = (sE22 , E22 (s E22 )) yields z11 = 0, z12 = m∗ ss . Therefore, there exists a mapping γ : S × S → S such that d sE22 , s E22 = m∗ ss E12 + γ s, s E22 . The proof that γ is a biderivation of S is similar to that of δ , hence omitted. (iv) Let r ∈ R and m ∈ M. First assume that d(E11 , mE12 ) = (vij ). Applying d on (E11 , mE12 ) = 2 , mE12 ), one observes that v11 = v22 = 0. Thus, there exists a mapping g from M to itself (E11 such that d(E11 , mE12 ) = g (m)E12 . Now assume that d(rE11 , mE12 ) = (wij ). Applying d on (rE11 , mE12 ) = ((rE11 )E11 , mE12 ), in view of the preceding result, we observe that w12 = g (m), and w22 = 0. Now applying d on (rE11 , mE12 ) = (rE11 , (mE12 )E22 ) and using (7), we conclude that w11 = 0. Hence, d(rE11 , mE12 ) = rg (m)E12 , as desired. Let s ∈ S and m ∈ M. To compute d(sE22 , mE12 ), first note that, by (4) and the preceding result, d(E22 , mE12 ) = −d(E11 , mE12 ) = −g (m)E12 . Now, applying d first on (sE22 , mE12 ) = (E22 (sE22 ), mE12 ) and then on (sE22 , mE12 ) = (sE22 , E11 (mE12 )), we find that d(sE22 , mE12 )
= −g (m)sE12 .
Since d is additive on the second component, g is additive. Moreover, from g (rm)E12
= d(E11 , rmE12 ) = d(E11 , (rE11 )(mE12 )) = d(E11 , rE11 )mE12 + rE11 d(E11 , mE12 ) = (rm∗ E12 )mE12 + rE11 (g (m)E12 ) = rg (m)E12
N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
255
we infer that g is a left R-homomorphism. Likewise, applying d on both sides of (E22 , msE12 ) = (E22 , (mE12 )(sE22 )), we see that g is also a right S-homomorphism.
Next, we show that δ r , r m = r , r g (m): substituting x = rE11 , y = r E11 , u = E11 , and v = mE12 into (3), by (8) and (10) we have
δ r , r E11
+ rr m∗ E12 (mE12 ) = d(x, y)[u, v] = [x, y]d(u, v) = =
r , r E11 (g (m)E12 )
r , r g (m)E12 .
Therefore, δ r , r m = r , r g (m). Putting now x = E11 , y = mE 12 , u = sE22 and v = s E22 into the Eq. (3) and using (8), (10) again, we conclude that mγ s, s = g (m) s, s . This completes the proof of (iv). The proof of (v) is similar to that of (iv), except for the
coordinates, so it is omitted. (vi) Let m, m ∈ M be arbitrary and set d mE12 , m E12 = (xij ). Applying d on both sides of
mE12 , m E12 = E11 (mE12 ), m E12 and using (10), we obtain x22 = 0. Then applying d on mE12 , m E12 = (mE12 )E22 , m E12 and making use
obtain of (10), we x11 = 0. Therefore, there exists a mapping f : M × M → M such that d mE12 , m E12 = f m, m E12 . Biadditivity of f is inherited from d. To show that f is a left R-homomorphism on the first component, let r ∈ R, and m, m ∈ M . In view of (10) we have f (rm, m )E12
= d(rmE12 , m E12 ) = d((rE11 )(mE12 ), m E12 ) = d(rE11 , m E12 )mE12 + rE11 d(mE12 , m E12 ) = rg (m )E12 (mE12 ) + rE11 (f (m, m )E12 ) = rf (m, m )E12 .
That is, f rm, m = rf m, m . Likewise, one can show that f is also a right S-homomorphism on the first component, and an (R, S )-bimodule homomorphism on the second component as well (see Eq. (2)). It only remains to show that for all r , r ∈ R, s, s ∈ S and m, m ∈ M, the following relations hold. r , r f m , m = 0 = f m, m s , s . Substituting x = mE12 , y = r E11 , u = mE12 and v = m E12 into the Eq. (3), we get δ r , r E11 + rr m∗ E12 0 = r , r E11 f m, m E12 ,
from which it follows that r , r f m, m = 0 . Finally, substituting x sE22 and v = s E22 into (3), we obtain f m, m s, s = 0. Remark 2.3. (a) Let R, S , M , T , and the element m∗ ⎞ ⎛ 1 m∗ ⎠. A=⎝ 0 1
∈ M be as above. Define
Then, in view of (6), it is straightforward to verify that
[A, [X , Y ]] = 0
for all X , Y
∈ T.
= mE12 , y = m E12 , u =
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Therefore, the mapping
: T ×T ⎛ → T given , Y ) = [⎞X , [Y , A]] is an extremal bideriva⎞ by (X⎛
tion of T. Note that for every X
=⎝
r m
⎠ and Y
0 s
=⎝
r m 0 s
⎠ in T we have
(X , Y ) = [X , [Y , A]] = (rr m∗ − rm∗ s − r m∗ s + m∗ ss )E12 . (b) Assume that f : M × M : T × T → T by
(16)
→ M is the (R, S)-bimodule homomorphism introduced in (iv). Define
(X , Y ) = f m, m E12 .
(17)
An easy computation shows that is a biderivation of T. Now, we are ready to state the main result. Theorem 2.4. Let R, M , S and T be as above. If d : T × T → T is a biderivation, then there exists a biderivation D with D(E11 , E11 ) = 0, an extremal biderivation , and a special kind of biderivation such that d
= D + + . ⎞
⎛ Proof. Let X
= ⎝
r m
0 s (12), and (14) we have d(X , Y )
⎠,Y
⎛
= ⎝
r m 0 s
⎞ ⎠ be in T. Using biadditivity of d and conclusions (7)–(10),
= d(rE11 + mE12 + sE22 , r E11 + m E12 + s E22 ) ⎛
= ⎝
δ(r , r ) rg (m ) − g (m )s + r h(m) − h(m)s 0
γ (s, s )
⎞ ⎠
+(rr m∗ − rm∗ s − r m∗ s + m∗ ss )E12 + f (m, m )E12 . Therefore, if we define D(X , Y ) to be the first term of the latter expression, by (16) and (17) we have d(X , Y )
= D(X , Y ) + (X , Y ) + (X , Y ).
By the preceding remarks, is the extremal biderivation induced by A, and is a biderivation. Therefore, D = d − − is also a biderivation of T. Evidently, D(E11 , E11 ) = 0. Theorem 2.4 has some nice corollaries. First, assume that R = M g (1), b = h(1) and c = f (1, 1). Then, by (iv), for each r ∈ R we have ra
= S, and T = T2 (R). Set a =
= rg (1) = g (r .1) = g (1.r ) = g (1)r = ar ,
hence, a ∈ Z (R), and g (r ) = ra. Similarly, b ∈ Z (R), and h(r ) = rb. Moreover, using (2), it is easy to verify that c ∈ Z (R), and that for each r , r ∈ R we have f (r , r ) = crr = cr r . Also, by (11), δ(r , r ) = [r , r ]g (1) = a[r , r ], and, by (13), δ(r , r ) = [r , r ]h(1) = [r , r ]b = −b[r , r ]. Hence, δ(r , r ) = a[r , r ] = −b[r , r ] is an inner biderivation of R. Similarly, one shows that γ (r , r ) = a[r , r ] = −b[r , r ] = δ(r , r ). Note that by the last conclusions we have (a + b) r , r = 0 for all r , r ∈ R. (18)
N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
Now, in view of (18), it is straightforward to show that for every X
= (rij ) and Y =
257
rij in T we
have D(X , Y )
=
⎛
a r r + r r − r r a r11 , r11 11 12 12 22 11 12 ⎝
0 a r22 , r22 E +(a + b)r12 r11 − r12 12
r − r12 22
⎞ ⎠
= (aI )[X , Y ] + D (X , Y ), in which the first term is the image of (X , Y ) under the inner biderivation induced by the central diagonal matrix aI, and in the next one, D , as the difference of two biderivations, is a (special) biderivation of T2 (R). Moreover, by Remark 2.3(b) and the calculation above, we have (X , Y ) = f r12 , r12 E12 . E12 = cr12 r12 Gathering the above results together, we are led to the following corollary which is a generalization of Proposition 4.16 in [1]. Recall that a ring R is said to be prime if for every a, b ∈ R, whenever aRb = 0, then a = 0 or b = 0, and R is said to be semiprime if for every a ∈ R, aRa = 0 implies that a = 0. Corollary 2.5. Let R be a ring with identity and d be a biderivation of T2 (R). Then d is decomposed into the sum of the inner biderivation induced by the central matrix aI, and the biderivations D , and given above. In particular, if R is a noncommutative prime ring, then d is an inner biderivation. Proof. By the above argument, it remains only to prove the special case. First, we show that m∗ = 0: choose r1 , r2 ∈ R such that [r1 , r2 ] = 0, and let r ∈ R be arbitrary. From (6) and the fact that [rr1 , r2 ] = r [r1 , r2 ] + [r , r2 ]r1 , it follows that 0
= m∗ [rr1 , r2 ] = m∗ (r [r1 , r2 ] + [r , r2 ]r1 ) = m∗ r [r1 , r2 ].
Since [r1 , r2 ] = 0 and r is arbitrary, the primeness of R forces m∗ = 0, so that = 0. Similarly, in view of (15) (from which it follows that c [r , r ] = f (1, 1)[r , r ] = 0) and (18), we conclude that c = a + b = 0. Accordingly, D and are both identically zero. Consequently, d is an inner biderivation. The following examples show, respectively, that if in the particular case of the corollary above we remove the noncommutativity of R, or if R is even a noncommutative semiprime ring, the conclusion may fail. Example 2.6. Let R be an integral domain and put T
= T2 (R). Define d : T × T → T by d((rij ), (rij )) =
E . Then one observes that d is a noninner biderivation of T. r12 r12 12
Example 2.7. Let R = R1 × R2 , where R1 is a noncommutative prime ring and R2 is an integral domain. Put T = T2 (R) and define d : T × T → T by d (aij , bij ), aij , bij = 0, b12 b12 E12 . Then a straightforward computation shows that d is a noninner biderivation of T. Next, we show that if R is a ring with identity and n 3, then the biderivations of Tn (R) have more compact forms. This result generalizes partly Corollary 4.13 of Benkoviˇc [1]. Corollary 2.8. Let R be a ring with identity and assume that n 3. Then every biderivation of the ring Tn (R) is the sum of an inner biderivation and an extremal biderivation. In particular, if R is a noncommutative prime ring, then every biderivation of Tn (R) is inner.
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Proof. We have the obvious ring isomorphism ⎛ ⎞ R Rn−1 ⎠ Tn (R) ∼ =⎝ 0 Tn−1 (R) in which Rn−1 is considered as an R, Rn−1 -bimodule with the obvious scaler multiplications. So, upon
identifying these two rings, we can deal with ring in the right-hand side. Set 1 = (1, . . . , 1) ∈ Rn−1 , and assume that g (1) = (a1 , . . . , an−1 ). Since g is a right Tn−1 (R)-homomorphism on Rn−1 , we have
(a1 , . . . , an−1 ) = g (1) = g (1.(E11 + · · · + E1,n−1 )) = g (1)(E11 + · · · + E1,n−1 ) = (a1 , . . . , an−1 )(E11 + · · · + E1,n−1 ) = (a1 , . . . , a1 ). Hence, a1 = a2 = · · · = an−1 Rn−1 , for each r ∈ R we have
=: a. On the other hand, since g is also a left R-homomorphism on
(ra, . . . , ra) = r (a, . . . , a) = rg (1) = g (r .1) = g (1.(rI )) = g (1)rI = (a, . . . , a)rI = (ar , . . . , ar ). I.e., a
∈ Z (R). Moreover, for each (r1 , . . . , rn−1 ) ∈ Rn−1 we have g (r1 , . . . , rn−1 )
= g (1.(r1 E11 + · · · + rn−1 E1,n−1 )) = (a, . . . , a)(r1 E11 + · · · + rn−1 E1,n−1 ) = (ar1 , . . . , arn−1 ) = a(r1 , . . . , rn−1 ).
Now, we claim that δ is the inner biderivation induced by a: let r , r conclusion it follows that
∈ R. Then from (11) and the latter
δ(r , r )(1, . . . , 1) = [r , r ]g (1, . . . , 1) = [r , r ](a, . . . , a), so that δ(r , r ) = a[r , r ]. Setting now h(1) = (b1 , . . . , bn−1 ) and using (13), a similar approach shows that b1 = · · · = n−1 bn−1 =: b is in Z (R), for every (r1 , . . . , rn− we
1 ) ∈ R
have h(r1 , . . . , rn−1 ) = b(r1 , . . . , rn−1 ), and that, for every r , r ∈ R we have δ r , r = −b r , r . Next, we show that γ is the inner biderivation of Tn−1 (R) induced by the central matrix aI: By (11), for every S , S ∈ Tn−1 (R) we have
(1, . . . , 1)γ (S, S ) = g (1, . . . , 1)[S, S ] = (a, . . . , a)[S, S ]. Therefore, noting that g (1, 2, 1, . . . , 1)
= g ((1, . . . , 1)(E11 + 2E12 + · · · + E1,n−1 )) = g (1, . . . , 1)(E11 + 2E12 + · · · + E1,n−1 ) = (a, . . . , a)(E11 + 2E12 + · · · + E1,n−1 )) = (a, 2a, a, . . . , a),
(19)
N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
259
we find that
(1, 2, 1, . . . , 1)γ S, S = (a, 2a, a, . . . , a) S, S .
(20)
Analogous computations show that (1, 1, 2, 1, . . . , 1)γ S, S = (a, a, 2a, a, . . . , a) S, S , . . . ,
(21)
and, finally, that
(1, . . . , 1, 2)γ S, S = (a, . . . , a, 2a) S, S .
(22)
Eqs. (19)–(22) lead us to the matrix equation P γ S , S = P (aI ) S , S
(23)
= I +(1)− E11 is invertible (with P −1 = I − nj=−21 E1j − ni=−21 Ei1 +(n − 1)E11 ).
Therefore, from (23) it follows that γ (S , S ) = (aI ) S , S . We claim now that the mapping f : Rn−1 × Rn−1 → Rn−1 is identically zero, from which it follows that = 0. Suppose that f (1, 1) = (c1 , . . . , cn−1 ). By Theorem 2.2(vi) we have in Mn−1 (R), in which P
(c1 , . . . , cn−1 ) = f (1, 1(E11 + · · · + E1,n−1 )) = f (1, 1)(E11 + · · · + E1,n−1 ) = (c1 , . . . , cn−1 )(E11 + · · · + E1,n−1 ) = (c1 , . . . , c1 ), hence, c1 = · · · = cn−1 =: c, and f (1, 1) = (c , . . . , c ). By (15), for every S , S ∈ Tn−1 (R) we have (c , . . . , c )[S, S ] = f (1, 1)[S, S ] = 0. Therefore, putting S = E11 + E1,n−1 and S = E1,n−1 , we get (c , . . . , c )E1,n−1 = 0, whence c = 0. Thus, for every m, m ∈ Rn−1 , we have f (m, m )
= f (m.1, m .1) = mm f (1, 1) = mm c = 0.
We can now show that D is an inner biderivation. Let X = (rij ) and Y above conclusions, we have ⎛ ⎞ ] ar U − aU S − ar U + aUS a[r11 , r11 11 11 ⎠ D(X , Y ) = ⎝ 0 aI [S , S ]
= (rij ) be in Tn (R). Using the
= (aI )[X , Y ], , . . . , r ), and the matrices S , S in T where U = (r12 , . . . , r1n ), U = (r12 n−1 (R) are determined by 1n canceling out the first rows and columns of X , Y , respectively. Notice that in the above equations, the first identity matrix I is in Tn−1 (R), while the second one belongs to Tn (R). Consequently, for every X , Y ∈ Tn (R) we have
d(X , Y )
= D(X , Y ) + (X , Y ) + (X , Y ) = aI [X , Y ] + (X , Y ).
Finally, if R is a noncommutative prime ring, then as we saw in Corollary 2.5, m∗ also [1,9].)
= 0, so = 0. (See
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N.M. Ghosseiri / Linear Algebra and its Applications 438 (2013) 250–260
Remark 2.9. In Corollary 2.8, assume that m∗ = (m∗1 , . . . , m∗n−1 ) (see (5)). We claim that m∗1 = · · · = m∗n−2 = 0. To see this, fix 1 i n − 2, and in the second equation of (6) put s = Eii + Ein and s = Ein . Then from 0
= (m∗1 , . . . , m∗n−1 )[s, s ] = (m∗1 , . . . , m∗n−1 )Ein
it follows that m∗i = 0. The following example shows, however, that the element m∗ zero.
∈ Rn−1 in Corollary 2.8 need not be
Example 2.10. Let R be a commutative ring with identity and put T = Tn (R)(n 3). Set A = I + E1n . Then it is easy to verify that for all X = (rij ), Y = (rij ) ∈ T, the relation [A, [X , Y ]] = 0 holds, so E is an − rnn that the mapping d : T × T → T defined by d(X , Y ) = [X , [Y , A]] = (r11 − rnn ) r11 1n (extremal) biderivation of T. Note that d(E11 , E11 ) = E1n , hence m∗ = (0, . . . , 0, 1) = 0. We finish by the following example which shows that in the particular case of Corollary 2.8, the noncommutativity of R can not be removed. Example 2.11. Assume assume that the ring R in the preceding example is an integral domain. Then for each a ∈ R we have aI [E11 , E11 ] = 0, while d(E11 , E11 ) = E1n = 0. Hence, d is not an inner biderivation of T. Acknowledgement The author would like to express his sincere thanks to the referees for careful reading the manuscript and useful suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
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