# Triple Derivations of Trivial Extensions

Authors: Fei Teng; Xiaokui Li; Ying Tian
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we describe the structure of triple derivations of trivial extensions. We then apply our results to triangular rings.


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## Body

## 1. Introduction

LetR be a ring with identity, and ZR be the center of R. Let M be a unitary R-bimodule. An additive mapping d from R into M is said to be a derivation if dab=dab+adb for all a,b∈R. The additive mapping d:R×R⟶M is called a biderivation if it is a derivation in each component; that is, dxy,z=dx,zy+xdy,z and dx,yz=dx,yz+ydx,z are fulfilled for all x,y,z∈R. Likewise, one can further develop the definition of triple derivations in an analogous manner. An additive mapping d:R×R×R⟶M is called a triple derivation if it is a derivation in each component; that is,(1)dxa,y,z=dx,y,za+xda,y,z,dx,ya,z=dx,y,za+ydx,a,z,dx,y,za=dx,y,za+zdx,y,a,for all x,y,z,a∈R. The trivial extension TR,M of R by M is the ring(2)TR,M=r,m:r∈R;m∈M,with the componentwise addition and the multiplication given by(3)r,mr′,m′=rr′,rm′+mr′,r,r′∈R;m,m′∈M.Trivial extensions have been extensively studied in algebra and analysis (see, for instance, [1–7]).LetR and S be rings with identity, M be a unitary R,S-bimodule, and T=RM0S be the triangular ring determined by R,S, and M with the usual addition and multiplication of matrices (see [8, 9] for more details of triangular rings). Then, one can easily verify that M can be made into a unitary R×S-bimodule via the scalar multiplications given by(4)r,sm=rm;mr,s=ms,r,s∈R×S,m∈M.Hence,TR×S,M is the trivial extension of R×S by M. It is clear that triangular rings are examples of trivial extensions. In [1], derivations and biderivations of trivial extensions are studied. Besides the afore-mentioned work, biderivations in different backgrounds are studied in [10–12], where further references can be found. A natural question arises: What is the representation of n-derivations on trivial extensions? Since the study of n-derivations is too complex, we study 3-derivations on trivial extensions and conjecture the representation of n-derivations.


## 2. Main Results and Proofs

We first describe triple derivations of the trivial extensionTR,M.Theorem 1.
Letd be a triple derivation of a trivial extension TR,M, TR,M. Then, there exist(i)
triple derivationsδ of R and γ:R×R×R⟶M(ii)
mappingsα1:R×M×M⟶R, α2:R×R×M⟶R, α3:R×M×R⟶R, α4:M×M×R⟶R, α5:M×R×R⟶R, α6:M×R×M⟶R(iii)
mappingsβ1:R×M×M⟶M, β2:R×R×M⟶M, β3:R×M×R⟶M, β4:M×M×R⟶M, β5:M×R×R⟶M, β6:M×R×M⟶M(iv)
a mappingf:M×M×M⟶R which is a bimodule homomorphism in each coordinate(v)
a mappingg:M×M×M⟶M
Such that(5)dr1,m1,r2,m2,r3,m3=δr1,r2,r3+α1r1,m2,m3+⋯+fm1,m2,m3,γr1,r2,r3+β1r1,m2,m3+⋯+gm1,m2,m3,holds for all r1,r2,r3∈R and m1,m2,m3∈M.Proof.
Letr1,r2,r3∈R and dr1,0,r2,0,r3,0=δr1,r2,r3,γr1,r2,r3. For any r1,r2,r3,r′∈R, we have(6)δr1r′,r2,r3,γr1r′,r2,r3=dr1r′,0,r2,0,r3,0=dr1,0r′,0,r2,0,r3,0=r1,0dr′,0,r2,0,r3,0+dr1,0,r2,0,r3,0r′,0=r1,0δr′,r2,r3,γr′,r2,r3+δr1,r2,r3,γr1,r2,r3r′,0=r1δr′,r2,r3+δr1,r2,r3r′,r1γr′,r2,r3+γr1,r2,r3r′.
Therefore,δ and γ are derivations in the first coordinate. Similarly, we get that δ and γ are derivations in the second and third coordinates. This proves (i).
To prove (ii) and (iii), letr1∈R,m2,m3∈M and set dr1,0,0,m2,0,m3=α1r1,m2,m3,β1r1,m2,m3. Let r′ be also in R. Then,(7)α1r1r′,m2,m3,β1r1r′,m2,m3=dr1r′,0,0,m2,0,m3=r1,0α1r′,m2,m3,β1r′,m2,m3+α1r1,m2,m3,β1r1,m2,m3r′,0=r1α1r′,m2,m3+α1r1,m2,m3r′,r1β1r′,m2,m3+β1r1,m2,m3r′.
It follows from the above expression thatα1,β1 are derivations in the first coordinate. Similarly, define(8)dr1,0,r2,0,0,m3=α2r1,r2,m3,β2r1,r2,m3,dr1,0,0,m2,r3,0=α3r1,m2,r3,β3r1,m2,r3,d0,m1,0,m2,r3,0=α4m1,m2,r3,β4m1,m2,r3,d0,m1,r2,0,r3,0=α5m1,r2,r3,β5m1,r2,r3,d0,m1,r2,0,0,m3=α6m1,r2,m3,β6m1,r2,m3.
We get thatαi,βi are derivations in the coordinate of the element in R, i∈1,2,3,4,5,6. Moreover, we have(9)α1r1,r′m2,m3,β1r1,r′m2,m3=dr1,0,0,r′m2,0,m3=dr1,0,r′,00,m2,0,m3=r′,0dr1,0,0,m2,0,m3+dr1,0,r′,0,0,m30,m2=r′α1r1,m2,m3,r′β1r1,m2,m3+α2r1,r′,m3m2,for all r1,r′∈R,m2,m3∈M. On the other hand, we get(10)α1r1,m2r′,m3,β1r1,m2r′,m3=dr1,0,0,m2r′,0,m3=dr1,0,0,m2r′,0,0,m3=0,m2dr1,0,r′,0,0,m3+dr1,0,0,m2,0,m3r′,0=α1r1,m2,m3r′,m2α2r1,r′,m3+β1r1,m2,m3r′,for all r1,r′∈R,m2,m3∈M. According to the above relations, we obtain that α1 is a bimodule homomorphism in the second coordinate, and(11)β1r1,r′m2,m3=r′β1r1,m2,m3+α2r1,r′,m3m2,(12)β1r1,m2r′,m3=m2α2r1,r′,m3+β1r1,m2,m3r′,for all r1,r′∈R,m2,m3∈M. Likewise, αi is a bimodule homomorphism in the coordinate of the element in M and(13)β1r1,m2,r′m3=r′β1r1,m2,m3+α3r1,m2,r′m3,(14)β1r1,m2,m3r′=m3α3r1,m2,r′+β1r1,m2,m3r′,(15)β2r1,r2,r′m3=r′β2r1,r2,m3+δr1,r2,r′m3,(16)β2r1,r2,m3r′=β2r1,r2,m3r′+m3δr1,r2,r′,(17)β3r1,r′m2,r3=r′β3r1,m2,r3+δr1,r′,r3m2,(18)β3r1,m2r′,r3=β3r1,m2,r3r′+m2δr1,r′,r3,(19)β4r′m1,m2,r3=r′β4m1,m2,r3+α3r′,m2,r3m1,(20)β4m1r′,m2,r3=β4m1,m2,r3r′+m1α3r′,m2,r3,(21)β4m1,r′m2,r3=r′β4m1,m2,r3+α5m1,r′,r3m2,(22)β4m1,m2r′,r3=β4m1,m2,r3r′+m2α5m1,r′,r3,(23)β5r′m1,r2,r3=r′β5m1,r2,r3+δr′,r2,r3m1,(24)β5m1r′,r2,r3=β5m1,r2,r3r′+m1δr′,r2,r3,(25)β6r′m1,r2,m3=r′β6m1,r2,m3+α2r′,r2,m3m1,(26)β6m1r′,r2,m3=β6m1,r2,m3r′+m1α2r′,r2,m3,(27)β6m1,r2,r′m3=r′β6m1,r2,m3+α5m1,r2,r′m3,(28)β6m1,r2,m3r′=β6m1,r2,m3r′+m3α5m1,r2,r′,for all r1,r2,r3,r′∈R,m1,m2,m3∈M.
To prove (iv) and (v), letm1,m2,m3∈M be arbitrary, and assume(29)d0,m1,0,m2,0,m3=fm1,m2,m3,gm1,m2,m3.
Since(30)fr′m1,m2,m3,gr′m1,m2,m3=d0,r′m1,0,m2,0,m3=dr′,00,m1,0,m2,0,m3=r′,0fm1,m2,m3,gm1,m2,m3+α1r′,m2,m3,β1r′,m2,m30,m1=r′fm1,m2,m3,r′gm1,m2,m3+α1r′,m2,m3m1,fm1r′,m2,m3,gm1r′,m2,m3=d0,m1r′,0,m2,0,m3=0,m1α1r′,m2,m3,β1r′,m2,m3+fm1,m2,m3,gm1,m2,m3r′,0=fm1,m2,m3r′,m1α1r′,m2,m3+gm1,m2,m3r′,for all r′∈R,m1,m2,m3∈M, we have that f:M×M×M⟶R is a bimodule homomorphism in the first coordinate, and(31)gr′m1,m2,m3=r′gm1,m2,m3+α1r′,m2,m3m1,gm1r′,m2,m3=m1α1r′,m2,m3+gm1,m2,m3r′,for all r′∈R,m1,m2,m3∈M. Similarly, we get that f:M×M×M⟶R is a bimodule homomorphism in the second and third coordinates, and(32)gm1,r′m2,m3=r′gm1,m2,m3+α6m1,r′,m3m2,gm1,m2r′,m3=gm1,m2,m3r′+m2α6m1,r′,m3,gm1,m2,r′m3=α4m1,m2,r′m3+r′gm1,m2,m3,gm1,m2,m3r′=gm1,m2,m3r′+m3α4m1,m2,r′,for all r′∈R,m1,m2,m3∈M.
In conclusion, we get(33)dr1,m1,r2,m2,r3,m3=δr1,r2,r3+α1r1,m2,m3+⋯+fm1,m2,m3,γr1,r2,r3+β1r1,m2,m3+⋯+gm1,m2,m3,for all r1,r2,r3∈R and m1,m2,m3∈M. □
According to Theorem1, we can decompose the triple derivation d on TR,M into the sum of five triple derivations.Corollary 1.
Letd and TR,M be as above. Then, the mappings D,F,G,H,K:TR,M×TR,M×TR,M⟶TR,M defined by(34)Dr1,m1,r2,m2,r3,m3=δr1,r2,r3,β2r1,r2,m3+β3r1,m2,r3+β5m1,r2,r3,Fr1,m1,r2,m2,r3,m3=fm1,m2,m3,0,Gr1,m1,r2,m2,r3,m3=α1r1,m2,m3+α4m1,m2,r3+α6m1,r2,m3,gm1,m2,m3,Hr1,m1,r2,m2,r3,m3=α2r1,r2,m3+α3r1,m2,r3+α5m1,r2,r3,β1r1,m2,m3+β4m1,m2,r3+β6m1,r2,m3,Kr1,m1,r2,m2,r3,m3=0,γr1,r2,r3,are triple derivations of TR,M, and d=D+F+G+H+K.
Next, we will use Theorem1 to study triple derivations of a triangular ring T=RM0S identified with the trivial extension TR×S,M.Theorem 2.
Letd be a triple derivation of the trivial extension TR×S,M. Then,(i)
there exist triple derivationsδ1 of R and δ2 of S such that, for every r1,r2,r3∈R and s1,s2,s3∈S, we have(35)δr1,s1,r2,s2,r3,s3=δ1r1,r2,r3,δ2s1,s2,s3,(ii)
there exists an elementm∗∈M such that, for every r1,r2,r3∈R and s1,s2,s3∈S, we have(36)γr1,s1,r2,s2,r3,s3=r1r2r3m∗+r1m∗s3s2+r2m∗s1s3+r3m∗s1s2−r1r2m∗s3−r1r3m∗s2−r2r3m∗s1−m∗s1s2s3,(iii)
the mappingsf=αi=0, i∈1,2,3,4,5,6,(iv)
for everyr1,s1,r2,s2,r3,s3∈R×S,r′,r″∈R,s′,s″∈S and m1,m2,m3∈M, we have(37)β1r1,s1,m2,m3=r1h1m2,m3−h1m2,m3s1,β4m1,m2,r3,s3=r3h4m1,m2−h4m1,m2s3,β6m1,r2,s2,m3=r2h6m1,m3−h6m1,m3s2,β2r1,s1,r2,s2,m3=r1r2h2m3−r1h2m3s2−r2h2m3s1+h2m3s1s2,β3r1,s1,m2,r3,s3=r1r3h3m2−r1h3m2s3−r3h3m2s1+h3m2s1s3,β5m1,r2,s2,r3,s3=r2r3h5m1−r2h5m1s3−r3h5m1s2+h5m1s2s3,r′,r″h1m2,m3=h1m2,m3s′,s″=0,r′,r″h4m1,m2=h4m1,m2s′,s″=0,r′,r″h6m1,m3=h6m1,m3s′,s″=0,where h1,h4,h6:M×M⟶M,h2,h3,h5:M⟶M are mappings(v)
for everyr1,r2,r3∈R,s1,s2,s3∈S,m1,m2,m3∈M, we have(38)δ1,0,r,0,0,s=δ1,01,0,r,0,0,s=1,0δ1,0,r,0,0,s+δ1,0,r,0,0,s1,0.Proof.
(i) Sinceδ is a triple derivation on R×S, we have(39)δ1,0,r,0,0,s=δ1,01,0,r,0,0,s=1,0δ1,0,r,0,0,s+δ1,0,r,0,0,s1,0,for all r∈R,s∈S. It follows that(40)δ1,0,r,0,0,s=0.
Similarly, we get that(41)δ1,0,r2,0,r3,0=δr1,0,r2,0,0,1=0.
Setδr1,0,r2,0,0,s3=t,z∈R×S. In view of (40), we have(42)t,z=δr1,01,0,r2,0,0,s3=δr1,0,r2,0,0,s31,0+r1,0δ1,0,r2,0,0,s3=t,z1,0=t,0.
This implies thatz=0. On the other hand, we have(43)t,z=δr1,0,r2,0,0,s30,1=0,s3δr1,0,r2,0,0,1+δr1,0,r2,0,0,s30,1=t,z0,1=0,z.
This yields thatt=0. Hence, δr1,0,r2,0,0,s3=0. Likewise,(44)δr1,0,0,s2,0,s3=δr1,0,0,s2,r3,0=δ0,s1,r2,0,r3,0=δ0,s1,0,s2,r3,0=δ0,s1,r2,0,0,s3=0.
Assume thatδr1,0,r2,0,r3,0=u,v. By (41), we obtain(45)δr1,0,r2,0,r3,0=δr1,01,0,r2,0,r3,0=r1,0δ1,0,r2,0,r3,0+δr1,0,r2,0,r3,01,0=u,0.
Therefore, we defineδr1,0,r2,0,r3,0=δ1r1,r2,r3,0. It follows that(46)δ1r1r′,r2,r3,0=δr1,0r′,0,r2,0,r3,0=r1,0δr′,0,r2,0,r3,0+δr1,0,r2,0,r3,0r′,0=r1δ1r′,r2,r3+δ1r1,r2,r3r′,0.
Thus,δ1 is a derivation in the first coordinate. Similarly, δ1 is a derivation in the second and third coordinates, so that δ1 is a triple derivation of R. By an analogue computation, one can show that there exists a triple derivation δ2 of S such that(47)δ0,s1,0,s2,0,s3=0,δ2s1,s2,s3,for all s1,s2,s3∈S. Hence, δr1,s1,r2,s2,r3,s3=δ1r1,r2,r3,δ2s1,s2,s3.
(ii) Assume thatγ1,0,1,0,1,0=m∗∈M, we have(48)γ1,0,1,0,0,1=−m∗,γ1,0,0,1,1,0=−m∗,γ1,0,0,1,0,1=m∗,γ0,1,1,0,1,0=−m∗,γ0,1,1,0,0,1=m∗,γ0,1,0,1,1,0=m∗,γ0,1,0,1,0,1=−m∗.
Sinceγ is a triple derivation, it follows that(49)γ1,0,r,0,1,0=γ1,0,r,01,0,1,0=r,0γ1,0,1,0,1,0+γ1,0,r,0,1,01,0=rm∗,for all r∈R. According to the above relation, we have(50)γr1,0,r2,0,1,0=γr1,01,0,r2,0,1,0=r1,0γ1,0,r2,0,1,0+γr1,0,r2,0,1,01,0=r1r2m∗,for all r1,r2∈R. This yields that(51)γr1,0,r2,0,r3,0=γr1,0,r2,0,r3,01,0=r3,0γr1,0,r2,0,1,0+γr1,0,r2,0,r3,01,0=r3r1r2m∗,for all r1,r2,r3∈R. Similarly, we get that(52)γr1,0,r2,0,r3,0=r1r2r3m∗=r1r3r2m∗=r2r1r3m∗=r2r3r1m∗=r3r2r1m∗,for all r1,r2,r3∈R. Analogue computation shows that, for every r1,r2,r3∈R and s1,s2,s3∈S, we have(53)γr1,0,r2,0,0,s3=−r1r2m∗s3=−r2r1m∗s3,γr1,0,0,s2,0,s3=r1m∗s3s2=r1m∗s2s3,γr1,0,0,s2,r3,0=−r1r3m∗s2=−r3r1m∗s2,γ0,s1,r2,0,r3,0=−r2r3m∗s1=−r3r2m∗s1,γ0,s1,r2,0,0,s3=r2m∗s1s3=r2m∗s3s1,γ0,s1,0,s2,r3,0=r3m∗s1s2=r3m∗s2s1,γ0,s1,0,s2,0,s3=−m∗s1s2s3=−m∗s1s3s2=−m∗s2s1s3=−m∗s2s3s1=−m∗s3s1s2=−m∗s3s2s1.
This proves (ii).
(iii) Sincef:M×M×M⟶R×S is a bimodule homomorphism in each coordinate, we have(54)fm1,m2,m3=f1,0m10,1,m2,m3=1,0fm1,m2,m30,1=0,0.
Sinceαi is a bimodule homomorphism in the coordinate of the element in M, we get αi=0, i∈1,2,3,4,5,6.
(iv) Recalling thatβ1:R×S×M×M⟶M is a derivation in the first coordinate, for every r1∈R,s1∈S,m2,m3∈M, we have(55)β1r1,0,m2,m3=β1r1,01,0,m2,m3=β1r1,0,m2,m31,0+r1,0β11,0,m2,m3=r1β11,0,m2,m3,(56)β10,s1,m2,m3=β10,10,s1,m2,m3=0,1β10,s1,m2,m3+β10,1,m2,m30,s1=β10,1,m2,m3s1.
Setβ11,0,m2,m3=h1m2,m3. It follows that β10,1,m2,m3=−h1m2,m3. Using (55) and (56), we have(57)β1r1,s1,m2,m3=r1h1m2,m3−h1m2,m3s1.
Similarly, there exist two mappingsh4,h6:M×M⟶M such that(58)β4m1,m2,r3,s3=r3h4m1,m2−h4m1,m2s3,β6m1,r2,s2,m3=r2h6m1,m3−h6m1,m3s2,for all r2,r3∈R,s2,s3∈S,m1,m2,m3∈M.
Setβ21,0,1,0,m3=h2m3. Since β2:R×S×R×S×M⟶M is a derivation in the first and second coordinates, it follows that(59)β2r1,0,r2,0,m3=β2r1,01,0,r2,0,m3=r1,0β21,0,r2,0,m3+β2r1,0,r2,0,m31,0=r1,0β21,0,r2,0,m3=r1,0β21,0,r2,01,0,m3=r1,0r2,0β21,0,1,0,m3+r1,0β21,0,r2,0,m31,0=r1r2β21,0,1,0,m3=r1r2h2m3,(60)β2r1,0,0,s2,m3=r1,0β21,0,0,s2,m3=r1,0β21,0,0,1,m30,s2=r1β21,0,0,1,m3s2=−r1h2m3s2β20,s1,r2,0,m3=r2,0β20,s1,1,0,m3=r2,0β20,1,1,0,m30,s1=r2β20,1,1,0,m3s1=−r2h2m3s1β20,s1,0,s2,m3=β20,1,0,s2,m30,s1=β20,1,0,1,m30,s20,s1=β20,1,0,1,m3s2s1=h2m3s2s1,for all r1,r2∈R,s1,s2∈S,m3∈M. Similarly, we have(61)β2r1,0,r2,0,m3=r2r1h2m3,β20,s1,0,s2,m3=h2m3s1s2,for all r1,r2∈R,s1,s2∈S,m3∈M. Therefore,(62)β2r1,s1,r2,s2,m3=r1r2h2m3−r1h2m3s2−r2h2m3s1+h2m3s1s2,for all r1,r2∈R,s1,s2∈S,m3∈M. Likewise, there exist two mappings h3,h5:M⟶M such that(63)β3r1,s1,m2,r3,s3=r1r3h3m2−r1h3m2s3−r3h3m2s1+h3m2s1s3,β5m1,r2,s2,r3,s3=r2r3h5m1−r2h5m1s3−r3h5m1s2+h5m1s2s3,for all r1,r2,r3∈R,s1,s2,s3∈S,m1,m2∈M. In addition, we have r1r3h3m2=r3r1h3m2,h3m2s1s3=h3m2s3s1,r2r3h5m1=r3r2h5m1,h5m1s2s3=h5m1s3s2.
By (11), we have(64)h1rm2,m3=β11,0,rm2,m3=β11,0,r,0m2,m3=r,0β11,0,m2,m3=rβ11,0,m2,m3=rh1m2,m3,for all r∈R,m2,m3∈M. Hence, hi is a left R-module on M. Similarly, according to (12)–(28), we obtain that hi is a bimodule homomorphism in each coordinate on M, i∈1,2,3,4,5,6. Since(65)r′r″h1m2,m3=h1r′m2,r″m3=r″r′h1m2,m3,we get r′,r″h1m2,m3=0 for all r′,r″∈R,m2,m3∈M. Likewise,(66)r′,r″h4m1,m2=r′,r″h6m1,m3=h1m2,m3s′,s″=h4m1,m2s′,s″=h6m1,m3s′,s″=0,for all r′,r″∈R,s′,s″∈S,m1,m2,m3∈M.
(v) Using (i), (15) and (61), we arrive at(67)r1r2r3h2m3=β2r1,0,r2,0,r3m3=β2r1,0,r2,0,r3,0m3=r3,0β2r1,0,r2,0,m3+δr1,0,r2,0,r3,0m3=r3r1r2h2m3+δ1r1,r2,r3,δ20,0,0m3=r3r1r2h2m3+δ1r1,r2,r3m3,for all r1,r2,r3∈R,m3∈M. According to (59) and (60), we have δ1r1,r2,r3m3=0. Similarly,(68)m3δ2s1,s2,s3=δ1r1,r2,r3m1=δ1r1,r2,r3m2=m1δ2s1,s2,s3=m2δ2s1,s2,s3=0,for all r1,r2,r3∈R,s1,s2,s3∈S.Corollary 2.
Letd and TR×S,M be as above. Then, the mappings D,F,G,H:TR×S,M×TR×S,M×TR×S,M⟶TR×S,M defined by(69)DX,Y,Z=δ1r1,r2,r3,δ2s1,s2,s3,r1r2h2m3−r1h2m3s2−r2h2m3s1+h2m3s1s2+r1r3h3m2−r1h3m2s3−r3h3m2s1+h3m3s1s3+r2r3h5m1−r2h5m1s3−r3h5m1s2+h5m1s2s3,FX,Y,Z=0,r1m∗s2s3−r1r3m∗s2−r2r3m∗s1+r2m∗s1s3+r3m∗s1s2−m∗s1s2s3+r1r2r3m∗−r2r1m∗s3,GX,Y,Z=0,gm1,m2,m3,HX,Y,Z=0,r1h1m2,m3−h1m2,m3s1+r3h4m1,m2−h4m1,m2s3+r2h6m1,m3−h6m1,m3s2,are triple derivations of TR×S,M, and d=D+F+G+H, for all r1,r2,r3∈R,s1,s2,s3∈S,m1,m2,m3∈M, where m∗∈M.
In conclusion, we conjecture thatn-derivations on trivial extensions can be decomposed into the sum of several n-derivations.



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*Source: 4302242-2022-11-09.xml*