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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# Lie n superderivations and generalized Lie n superderivations of superalgebras

He Yuan
• Department of Mathematics, Jilin Normal University, Siping, 136000, China
• School of Mathematics and Statistics, Northeast Normal University, Changchun, 130024, China
• Other articles by this author:
/ Liangyun Chen
Published Online: 2018-03-13 | DOI: https://doi.org/10.1515/math-2018-0018

## Abstract

In the paper, we study Lie n superderivations and generalized Lie n superderivations of superalgebras, using the theory of functional identities in superalgebras. We prove that if A = A0A1 is a prime superalgebra with deg(A1) ≥ 2n + 5, n ≥ 2, then any Lie n superderivation of A is the sum of a superderivation and a linear mapping, and any generalized Lie n superderivation of A is the sum of a generalized superderivation and a linear mapping.

MSC 2010: 17A70; 16W10; 16W55

## 1 Introduction

Let A be an associative algebra. A linear mapping d : AA is called a derivation if d(xy) = d(x)y + xd(y) for all x, yA. A Lie derivation δ of A is a linear mapping from A into itself satisfying δ[x, y] = [δ(x), y] + [x,δ(y)] for all x, yA. A Lie triple derivation is a linear mapping ψ : AA which satisfies ψ[[x, y], z] = [[ψ(x), y], z] + [[x, ψ(y)], z]+[[x, y], ψ(z)] for all x, y, zA. Obviously, each derivation is a Lie derivation and each Lie derivation is a Lie triple derivation. Brešar [1] described the structure of Lie derivations and Lie triple derivations on prime rings and obtained that each Lie derivation or Lie triple derivation of a prime ring is the sum of a derivation and an additive mapping. Wang [2] studied the structure of Lie superderivations of superalgebras in 2016. Lie n-derivations were introduced by Abdullaev [3], where the form of Lie n-derivations of a certain von Neumann algebra was described. In 2012, Benkovič and Eremita [4] gave the form of Lie n-derivations on triangular rings, which has been generalized to generalized matrix algebras in [5].

The concept of a generalized derivation was introduced by Brešar [6] and generalized by Hvala [7], who has proved in [8] that each generalized Lie derivation of a prime ring is the sum of a generalized derivation and a central mapping which vanishes on all commutators.

A functional identity can be described as an identical relation involving elements in a ring together with functions. The goal when studying a functional identity is to describe the form of these functions or to determine the structure of the ring admitting the functional identity in question. The theory of functional identities in rings originated from the results of commuting mappings [9]. The name “functional identity” was introduced by Brešar in [10]. The crucial tool in the theory of functional identities in rings is the d-free set, which was developed by Beidar and Chebotar in [11, 12]. Making use of the theory of functional identities in rings, Herstein’s conjectures on Lie mappings in rings have been settled [13,14,15]. After this, Wang [16] established the theory of functional identities in superalgebras and gave the definition of d-superfree sets. As an application, Wang [17] described Lie superhomomorphisms from the set of skew elements of a superalgebra with superinvolution into a unital superalgebra. The knowledge of functional identities and d-superfree sets of superalgebras refer to [16], [18] and [19].

In the paper, our purpose is to study Lie n superderivations and generalized Lie n superderivations of superalgebras, using the theory of functional identities in superalgebras. Section 2 presents some preliminaries. In the third section we discuss the structure of Lie n superderivations. In Section 4 the results of generalized Lie n superderivations are stated and proved.

## 2 Preliminaries

Throughout the paper, by an algebra we shall mean an algebra over a fixed unital commutative ring Φ. We assume without further mentioning that $\begin{array}{}\frac{1}{2}\in \mathit{\Phi }.\end{array}$

An associative algebra A over Φ is said to be an associative superalgebra if there exist two Φ-submodules A0 and A1 of A such that A = A0A1 and AiAjAi+j, i, jZ2. We call A0 the even and A1 the odd part of A. The elements of Ai are homogeneous of degree i and we write |Ai| = i for all AiAi. For a superalgebra A, we define σ:AA by (A0 + A1)σ = A0A1, then σ is an automorphism of A such that σ2 = 1. On the other hand, for an algebra A, if there exists an automorphism σ of A such that σ2 = 1, then A becomes a superalgebra A = A0A1, where Ai = {xA|xσ = (–1)i x, i = 0, 1. A superalgebra A is called a prime superalgebra if and only if aAb = 0 implies a = 0 or b = 0, where at least one of the elements a and b is homogeneous.

On a superalgebra A, define for any x, yA0A1 the Lie superproduct

$[x,y]s=xy−(−1)|x||y|yx.$

Thus

$[a,b]s=[a0,b0]s+[a1,b0]s+[a0,b1]s+[a1,b1]s,$

where a = a0 + A1, b = b0 + b1.

In [20] Montaner obtained that a prime superalgebra A is not necessarily a prime algebra but a semiprime algebra. Hence one can define the maximal right ring of quotients Qmr of A, and the useful properties of Qmr can be found in [21]. By [21, proposition 2.5.3] σ can be uniquely extended to Qmr. Therefore, Qmr is also a superalgebra. Moreover, we can get that Qmr is a prime superalgebra.

On the other hand, we will introduce some important concepts of the theory of functional identities in superalgebras.

Let Q = Q0Q1 be a unital superalgebra with grading automorphism σ and center C = C0C1 satisfying [C, Q] = 0. Fix an element ωQ as follows : If either σ = 1 or σ is outer, we set ω = 0. Otherwise, we denote ω as an invertible element in Q such that σ(x) = ωxω–1 for all xQ. It is easy to check that ωQ0, ω2C0, ωx0 = x0ω for all x0Q0, and ωx1 = –x1ω for all x1Q1. We shall call the ω the grading element of Q. If A = b + cω, b, cC, we set ā = bcω.

Let mN*, 𝓤1,𝓤2, …, 𝓤m are subsets of Q such that either 𝓤iQ0 or 𝓤iQ1 for every 1 ≤ im. Set εi = ± 1, where either εi = 1 if 𝓤iQ0 or εi = – 1 if 𝓤iQ1. 𝓢1, 𝓢2,…, 𝓢m are nonempty sets, 𝓘, 𝓙 ⊆ {1, 2, …, m} and δl : 𝓢l → 𝓤l, l ∈ 𝓘 ∪ 𝓙, are surjective maps. Set $\begin{array}{}\stackrel{^}{\mathcal{S}}=\prod _{k=1}^{m}{\mathcal{S}}_{k},\stackrel{^}{\mathcal{U}}=\prod _{k=1}^{m}{\mathcal{U}}_{k}\end{array}$ and Δ = {δl|l ∈ 𝓘 ∪ 𝓙}.

We shall consider functional identities on Ŝ of the following form

$∑i∈IEii(X¯m)δi(Xi)+∑j∈Jδj(Xj)Fjj(X¯m)=0;$(1)

$∑i∈IEii(X¯m)δi(Xi)+∑j∈Jδj(Xj)Fjj(X¯m)∈C+Cω,$(2)

for all mŜ, where Ei : ∏ki𝓢kQ and Fj : ∏kj𝓢kQ.

Suppose that ω = 0 or each 𝓤iQ0. There exist maps

$pij:∏k≠i,jSk→Q,i∈I,j∈J,i≠j,$

$λl:∏k≠lSk→C+Cω,l∈I∪J,$

such that

$Eii(X¯m)=∑j∈Jj≠iδj(Xj)pijij(X¯m)+λii(X¯m);Fjj(X¯m)=−∑i∈Ii≠jpijij(X¯m)δi(Xi)−λjj(X¯m),$(3)

for all mŜ, where λl = 0 if l ∉ 𝓘 ∩ 𝓙.

Otherwise, there exist maps

$pij:∏k≠i,jSk→Q,i∈I,j∈J,i≠j,λl,μl:∏k≠lSk→Cl∈I∪J,$

such that

$Eii(X¯m)=∑j∈Jj≠iδj(Xj)pijij(X¯m)+λii(X¯m)+μii(X¯m)ω;Fjj(C¯m)=−∑i∈Ii≠jpijij(X¯m)δi(Xi)−λjj(X¯m)−ϵjμjj(X¯m)ω,$(4)

for all mŜ, where λl = 0 = μl if l ∉ 𝓘 ∩ 𝓙. We shall refer to (3) and (4) as a standard solution of (1) and (2).

#### Definition 2.1

([16, Definition 3.1]). Let dN*. A triple (𝓢̂;δ; 𝓤̂) is called d-superfree if the following conditions are satisfied:

1. For all m ∈ 𝓝* and 𝓘, 𝓙 ⊆ {1, 2, …, m} with max{|𝓘|, |𝓙|} < d + 1, we have that (1) implies (3) and (4).

2. For all m ∈ 𝓝* and 𝓘, 𝓙 ⊆ {1, 2, …, m} with max{|𝓘|,|𝓙|} < d, we have that (2) implies (3) and (4).

If each 𝓢k = 𝓤k and each δl = id𝓤l, then the 𝓤̂ is said to be d-superfree provided that (𝓢̂;δ; 𝓤̂) is so. Let R = R0R1 be a graded Φ-submodule of Q. For every 1 ≤ im, either 𝓤i = R0 or 𝓤i = R1. Then R is said to be d-superfree provided that each 𝓤̂ is d-superfree. And, we can get the following result.

#### Lemma 2.2

([16, Theorem 4.16]). Let A = A0A1 be a prime superalgebra. If deg(A1) ≥ 2d + 1, then A is d-superfree.

Let {x1, x2, …, xm} be a finite set of variables and k be a nonnegative integer such that km. We denote by $\begin{array}{}{\mathcal{M}}_{m}^{k}\end{array}$ the set of all multilinear monomials of degree k in the variables {x1, x2, …, xm}. It is understood that $\begin{array}{}{\mathcal{M}}_{1}^{0}\end{array}$ = {1}. We write $\begin{array}{}{\mathcal{M}}_{m}=\bigcup _{k=0}^{m}{\mathcal{M}}_{m}^{k}.\end{array}$ For a given monomial M = xi1xiu ∈ 𝓜m where umk. We denote by $\begin{array}{}{\mathcal{M}}_{m}^{k}\left(M\right)\end{array}$ the set of all multilinear monomials of degree k in the variables {x1, …, xm}\{xi1, …, xiu} and write $\begin{array}{}{\mathcal{M}}_{m}\left(M\right)=\bigcup _{k=0}^{m-u}{\mathcal{M}}_{m}^{k}\left(M\right).\end{array}$

For every 1 ≤ tm, let 𝓢t and 𝓡t be two sets and let δt : 𝓢t → 𝓡t be a surjective mapping. We set

$M(s¯m)=δi1(si1)δi2(si2)⋯δiu(siu),$

where sit ∈ 𝓢it.

We set 𝓢̂ = $\begin{array}{}\stackrel{^}{\mathcal{S}}=\prod _{i=1}^{m}{\mathcal{S}}_{i}\end{array}$ and $\begin{array}{}\stackrel{^}{\mathcal{S}}\left(M\right)=\prod _{t=1}^{m-u}{\mathcal{S}}_{{j}_{t}}.\end{array}$ For any given F : 𝓢̂(M) → Q we introduce a mapping FM : 𝓢̂ → Q by the rule

$FM(s¯m)=FM(s1,⋯,sm)=F(sj1,⋯,sjm−u),$

for any m ∈ 𝓢̂.

Let $\begin{array}{}M\in {\mathcal{M}}_{m}^{k}\end{array}$ and let λM : 𝓢̂(M) → C + Cω. A mapping 𝓢̂ → Q defined by the rule $\begin{array}{}{\overline{s}}_{m}\to {\lambda }_{M}^{M}\left({\overline{s}}_{m}\right)M\left({\overline{s}}_{m}\right)\end{array}$ for any m ∈ 𝓢̂ is called a superquasi-monomial and is denoted by λM M. A sum ∑L ∈ 𝓜m λL L of different superquasi-monomials will be called a superquasi-polynomial.

An element xA0A1 is said to be algebraic over C of degree ≤ n if there exist C0, C1, …, cnC, not all zero and such that $\begin{array}{}\sum _{i=0}^{n}{c}_{i}{x}^{n-i}=0.\end{array}$ The element x is said to be algebraic over C of degree n if it is algebraic over C of degree ≤ n and is not algebraic over C of degree ≤ n – 1. By deg(x) we shall mean the degree of x over C (if x is algebraic over C) or ∞ (if x is not algebraic over C). Given a nonempty subset SA0A1, we set

$deg(S)=sup{deg(x)|x∈S}.$

Let A be a superalgebra. For i ∈ {0, 1}, a superderivation of degree i is actually a Φ-linear mapping di : AA which satisfies di(Aj) ⊆ Ai + j, jZ2, and di(ab) = di(a)b + (–1)i|a|adi(b) for all A, bA0A1. If d = d0 + d1, then d is called a superderivation.

Let A be a superalgebra. For i ∈ {0, 1}, a Φ-linear mapping gi : AA is called a generalized superderivation of degree i if gi(Aj) ⊆ Ai + j, jZ2, and gi(xy) = gi(x)y + (–1)i|x|xdi(y) for all x, yA0A1, where di is a superderivation of degree i. If g = g_{0}+g1, then g is called a generalized superderivation.

The following identity will be used frequently,

$[aibj,ck]s=[ai,bjck]s+(−1)ij+ik[bj,ckai]sai,bj,ck∈A0∪A1,$(5)

where i, j, k ∈ {0, 1}.

## 3 Lie n superderivations of superalgebras

In the section, we describe the structure of Lie n superderivations on a superalgebra.

#### Definition 3.1

Let A be a superalgebra. For m ∈ {0, 1}, a Lie superderivation of degree m is actually a Φ-linear mapping αm : AA which satisfies αm(Aj) ⊆ Am+j, jZ2, and αm([x, y]s) = [αm(x), y]s + (–1)m|x|[x, αm(y)]s for all x, yA0A1. If α = α0 + α1, then α is called a Lie superderivation on A.

Obviously, each superderivation is a Lie superderivation on A.

#### Definition 3.2

Let A be a superalgebra. For m ∈ {0, 1}, a Φ-linear mapping βm : AA is called a Lie triple superderivation of degree m if βm(Aj) ⊆ Am + j, jZ2, and

$βm([[x,y]s,z]s)=[[βm(x),y]s,z]s+(−1)m|x|[[x,βm(y)]s,z]s+(−1)m(|x|+|y|)[[x,y]s,βm(z)]s,$

for all x, y, zA0A1. If β = β0 + β1, then β is called a Lie triple superderivation on A.

Let us define the following sequence of polynomials: p1(x) = x and

$pn(x1,x2,…,xn)=[pn−1(x1,x2,…,xn−1),xn]sn⩾2.$

Thus, p2 (x1, x2) = [x1, x2]s, p3 (x1, x2, x3) = [[x1, x2]s, x3]s, etc.

#### Definition 3.3

Let n ⩾ 2 be an integer. Let A be a superalgebra. For m ∈ {0, 1}, a Φ-linear mapping γm : AA is called a Lie n superderivation of degree m if γm(Aj) ⊆ Am + j, jZ2, and

$γm(pn(x1,x2,…,xn))=pn(γm(x1),x2,…,xn)+∑i=2n(−1)m(|x1|+|x2|+…+|xi−1|)pn(x1,x2,…,xi−1,γm(xi),xi+1,…,xn),$

for all x1, x2, …, xnA0A1. If γ = γ0 + γ1, then γ is called a Lie n superderivation on A.

#### Theorem 3.4

Let Q = Q0Q1 be a unital superalgebra with center C = C0C1. Let A = A0A1 be a superalgebra and a subalgebra of Q. Suppose that γ : AQ is a Lie n superderivation, n ≥ 2. If A is an (n + 2)-superfree subset of Q, then γ = d + h, where d : AQ is a superderivation and h : AC + C ω is a linear mapping.

#### Proof

By the definition of Lie n superderivations, we assume that γm is a Lie n superderivation of degree m, m ∈ {0, 1}. According to (5), we have

$[…[[aibj,ck]s,x3]s,…,xn]s=[…[[ai,bjck]s,x3]s,…,xn]s+(−1)ij+ik[…[[bj,ckai]s,x3]s,…,xn]s,$(6)

for all Ai, bj, ck, x3, …, xnA0A1. Applying γm to (6), we have

$0=[…[[γm(aibj),ck]s,x3]s,…,xn]s+(−1)m(i+j)[…[[aibj,γm(ck)]s,x3]s,…,xn]s+(−1)m(i+j+k)[…[[aibj,ck]s,γm(x3)]s,…,xn]s+…+(−1)m(i+j+k+…+|xn−1|)[…[[aibj,ck]s,x3]s,…,γm(xn)]s−[…[[γm(ai),bjck]s,x3]s,…,xn]s−(−1)mi[…[[ai,γm(bjck)]s,x3]s,…,xn]s−(−1)m(i+j+k)[…[[ai,bjck]s,γm(x3)]s,…,xn]s−…−(−1)m(i+j+k+…+|xn−1|)[…[[ai,bjck]s,x3]s,…,γm(xn)]s−(−1)ij+ik([…[[γm(bj),ckai]s,x3]s,…,xn]s+(−1)mj[…[[bj,γm(ckai)]s,x3]s,…,xn]s+(−1)m(i+j+k)[…[[bj,ckai]s,γm(x3)]s,…,xn]s+…+(−1)m(i+j+k+…+|xn−1|)[…[[bj,ckai]s,x3]s,…,γm(xn)]s).$

It follows from (5) that

$0=[…[[γm(aibj),ck]s+(−1)m(i+j)[ai,bjγm(ck)]s+(−1)mi+mj+ij+mi+ik[bj,γm(ck)ai]s−[γm(ai)bj,ck]s+(−1)mj+ij+mk+ik[bj,ckγm(ai)]s−(−1)mi[ai,γm(bjck)]s−(−1)ij+ik[γm(bj)ck,ai]s+(−1)ij+ik+mk+jk+mi+ji[ck,aiγm(bj)]s−(−1)ij+ik+mj[bj,γm(ckai)]s,x3]s,…,xn]s,$

for all Ai, bj, ck, x3, …, xnA0A1. Since [Ai, bj]s = –(–1)ij[bj, Ai]s, it follows that

$0=[…[[γm(aibj),ck]s−[γm(ai)bj,ck]s−(−1)mi[aiγm(bj),ck]s−(−1)mi([ai,γm(bjck)]s−[ai,γm(bj)ck]s−(−1)mj[ai,bjγm(ck)]s)−(−1)ij+ik+mj([bj,γm(ckai)]s−[bj,γm(ck)ai]s−(−1)mk[bj,ckγm(ai)]s),x3]s,…,xn]s.$(7)

Define 𝓑 : A × AQ by

$B(x,y)=γm(xy)−γm(x)y−(−1)m|x|xγm(y),$

for all x, yA0A1. It follows from (7) that

$[…[[B(ai,bj),ck]s−(−1)mi[ai,B(bj,ck)]s−(−1)ij+ik+mj[bj,B(ck,ai)]s,x3]s,…,xn]s=0,$(8)

for all Ai, bj, ck, x3, …,xnA0A1. Since A is an (n + 2)-superfree subset of Q, n ≥ 2, [16, Theorem 3.8] implies

$B(x0,y0)=λ1x0y0+λ1′y0x0+μ1(x0)y0+μ1′(y0)x0+ν1(x0,y0);B(x0,y1)=λ2x0y1+λ2′y1x0+μ2(x0)y1+μ2′(y1)x0+ν2(x0,y1);B(x1,y0)=λ3x1y0+λ3′y0x1+μ3(x1)y0+μ3′(y0)x1+ν3(x1,y0);B(x1,y1)=λ4x1y1+λ4′y1x1+μ4(x1)y1+μ4′(y1)x1+ν4(x1,y1),$(9)

where λk, $\begin{array}{}{\lambda }_{k}^{\prime }\in {C}_{m}+{C}_{m}\omega ,{\mu }_{k},{\mu }_{k}^{\prime }:{A}_{i}\to {C}_{m+i}+{C}_{m+i}\omega ,{\nu }_{k}:\end{array}$ Ai × AjC + Cω, k ∈ {1, 2, 3, 4}, i, j ∈ {0, 1}.

We shall now compute γm(xyz) in two different ways. On the one hand,

$γm(xyz)=B(xy,z)+γm(xy)z+(−1)m(|x|+|y|)xyγm(z)=B(xy,z)+(B(x,y)+γm(x)y+(−1)m|x|xγm(y))z+(−1)m(|x|+|y|)xyγm(z).$

On the other hand,

$γm(xyz)=B(x,yz)+γm(x)yz+(−1)m|x|xγm(yz)=B(x,yz)+γm(x)yz+(−1)m|x|x(B(y,z)+γm(y)z+(−1)m|y|yγm(z)).$

Comparing the above expressions, we get

$B(xy,z)−B(x,yz)+B(x,y)z−(−1)m|x|xB(y,z)=0,$(10)

for all x, y, zA0A1.

When |x| = |y| = |z| = 0, it follows from (10) that

$λ1xyz+λ1′zxy+μ1(xy)z+μ1′(z)xy+ν1(xy,z)−λ1xyz−λ1′yzx−μ1(x)yz−μ1′(yz)x−ν1(x,yz)+λ1xyz+λ1′yxz+μ1(x)yz+μ1′(y)xz+ν1(x,y)z−λ1xyz−λ1′xzy−μ1(y)xz−μ1′(z)xy−ν1(y,z)x=0.$

An easy computation shows that:

• The coefficient of zxy is $\begin{array}{}{\lambda }_{1}^{\prime };\end{array}$

• The coefficient of xz is $\begin{array}{}{\mu }_{1}^{\prime }\left(y\right)-{\mu }_{1}\left(y\right);\end{array}$

• The coefficient of z is μ1(xy) + ν1(x, y).

By [16, Theorem 3.7], we have

$λ1′=0;μ1(y0)=μ1′(y0);μ1(x0y0)=−ν1(x0,y0).$

When |x| = |z| = 0 and |y| = 1, it follows from (10) that

$λ3xyz+λ3′zxy+μ3(xy)z+μ3′(z)xy+ν3(xy,z)−λ2xyz−λ2′yzx−μ2(x)yz−μ2′(yz)x−ν2(x,yz)+λ2xyz+λ2′yxz+μ2(x)yz+μ2′(y)xz+ν2(x,y)z−λ3xyz−λ3′xzy−μ3(y)xz−μ3′(z)xy−ν3(y,z)x=0.$

An easy computation shows that:

• The coefficient of zxy is $\begin{array}{}{\lambda }_{3}^{\prime };\end{array}$

• The coefficient of yzx is $\begin{array}{}-{\lambda }_{2}^{\prime };\end{array}$

• The coefficient of xz is $\begin{array}{}{\mu }_{2}^{\prime }\left(y\right)-{\mu }_{3}\left(y\right);\end{array}$

• The coefficient of x is $\begin{array}{}-{\mu }_{2}^{\prime }\left(yz\right)-{\nu }_{3}\left(y,z\right);\end{array}$

• The coefficient of z is μ3(xy) + ν2(x, y)

By [16, Theorem 3.7], we have

$λ2′=λ3′=0;μ3(y1)=μ2′(y1);μ2′(y1z0)=−ν3(y1,z0);μ3(x0y1)=−ν2(x0,y1).$

When |x| = 0 and |y| = |z| = 1, it follows from (10) that

$λ4xyz+λ4′zxy+μ4(xy)z+μ4′(z)xy+ν4(xy,z)−λ1xyz−λ1′yzx−μ1(x)yz−μ1′(yz)x−ν1(x,yz)+λ2xyz+λ2′yxz+μ2(x)yz+μ2′(y)xz+ν2(x,y)z−λ4xyz−λ4′xzy−μ4(y)xz−μ4′(z)xy−ν4(y,z)x=0.$

An easy computation shows that:

• The coefficient of xyz is λ2 – λ1;

• The coefficient of zxy is $\begin{array}{}{\lambda }_{4}^{\prime };\end{array}$

• The coefficient of yz is μ2(x) – μ1(x);

• The coefficient of xz is $\begin{array}{}{\mu }_{2}^{\prime }\left(y\right)\end{array}$μ4(y);

• The coefficient of x is – $\begin{array}{}{\mu }_{1}^{\prime }\left(yz\right)\end{array}$ν4(y, z).

By [16, Theorem 3.7], we have

$λ1=λ2;λ4′=0;μ1(x0)=μ2(x0);μ2′(y1)=μ4(y1);μ1′(y1z1)=−ν4(y1,z1).$

By the definition of 𝓑, we have

$B([…[[x1,x2]s,x3]s,…,xn]s,ω1)−B(ω1,[…[[x1,x2]s,x3]s,…,xn]s)=γm([[…[[x1,x2]s,x3]s,…,xn]s,ω1]s)−γm([…[[x1,x2]s,x3]s,…,xn]s)ω1−[…[[x1,x2]s,x3]s,…,xn]sγm(ω1)+γm(ω1)[…[[x1,x2]s,x3]s,…,xn]s+ω1γm([…[[x1,x2]s,x3]s,…,xn]s)=[[…[γm([x1,x2]s),x3]s,…,xn]s,ω1]s+[[…[[x1,x2]s,γm(x3)]s,…,xn]s,ω1]s+…+[[…[[x1,x2]s,x3]s,…,γm(xn)]s,ω1]s−B([…[[x1,x2]s,x3]s,…,xn−1]s,xn)ω1+B(xn,[…[[x1,x2]s,x3]s,…,xn−1]s)ω1+ω1B([…[[x1,x2]s,x3]s,…,xn−1]s,xn)−ω1B(xn,[…[[x1,x2]s,x3]s,…,xn−1]s)−γm([…[[x1,x2]s,x3]s,…,xn−1]s)xnω1−[…[[x1,x2]s,x3]s,…,xn−1]sγm(xn)ω1+γm(xn)[…[[x1,x2]s,x3]s,…,xn−1]sω1+xnγm([…[[x1,x2]s,x3]s,…,xn−1]s)ω1+ω1γm([…[[x1,x2]s,x3]s,…,xn−1]s)xn+ω1[…[[x1,x2]s,x3]s,…,xn−1]sγm(xn)−ω1γm(xn)[…[[x1,x2]s,x3]s,…,xn−1]s−ω1xnγm([…[[x1,x2]s,x3]s,…,xn−1]s)=[[…[γm([x1,x2]s),x3]s,…,xn]s,ω1]s−B([…[[x1,x2]s,x3]s,…,xn−1]s,xn)ω1+B(xn,[…[[x1,x2]s,x3]s,…,xn−1]s)ω1+ω1B([…[[x1,x2]s,x3]s,…,xn−1]s,xn)−ω1B(xn,[…[[x1,x2]s,x3]s,…,xn−1]s)−B([…[[x1,x2]s,x3]s,…,xn−2]s,xn−1)xnω1+B(xn−1,[…[[x1,x2]s,x3]s,…,xn−2]s)xnω1−…−B([x1,x2]s,x3)x4…xnω1+B(x3,[x1,x2]s)x4…xnω1−γm([x1,x2]s)x3…xnω1+…+ω1γm([x1,x2]s)x3…xn,$

for all x1, x2, …, xn, ω1A0. Since the coefficient of x1x2xnω1 is (n−1)λ1, [16, Theorem 3.7] yields λ1 = 0.

On the other hand, we have

$B([…[[x1,x2]s,x3]s,…,xn]s,ω1)−B(ω1,[…[[x1,x2]s,x3]s,…,xn]s)=γm([[…[[x1,x2]s,x3]s,…,xn]s,ω1]s)−γm([…[[x1,x2]s,x3]s,…,xn]s)ω1−[…[[x1,x2]s,x3]s,…,xn]sγm(ω1)+γm(ω1)[…[[x1,x2]s,x3]s,…,xn]s+ω1γm([…[[x1,x2]s,x3]s,…,xn]s)=[[…[γm([x1,x2]s),x3]s,…,xn]s,ω1]s+[[…[[x1,x2]s,γm(x3)]s,…,xn]s,ω1]s+…+[[…[[x1,x2]s,x3]s,…,γm(xn)]s,ω1]s+[[…[[x1,x2]s,x3]s,…,xn]s,γm(ω1)]s−[…[[γm(x1),x2]s,x3]s,…,xn]sω1−(−1)m[…[[x1,γm(x2)]s,x3]s,…,xn]sω1−[…[[x1,x2]s,γm(x3)]s,…,xn]sω1−…−[…[[x1,x2]s,x3]s,…,γm(xn)]sω1−[…[[x1,x2]s,x3]s,…,xn]sγm(ω1)+γm(ω1)[…[[x1,x2]s,x3]s,…,xn]s+ω1[…[[γm(x1),x2]s,x3]s,…,xn]s+(−1)mω1[…[[x1,γm(x2)]s,x3]s,…,xn]s+ω1[…[[x1,x2]s,γm(x3)]s,…,xn]s+…+ω1[…[[x1,x2]s,x3]s,…,γm(xn)]s=[[…[B(x1,x2)+B(x2,x1),x3]s,…,xn]s,ω1]s,$

for all x1, x2A1, x3, …, xn, ω1A0. Since the coefficient of x1 x2xnω1 is λ1 – λ4, [16, Theorem 3.7] implies λ4 = λ1 = 0.

When |y| = 0 and |x| = |z| = 1, it follows from (10) that

$μ4(xy)z+μ4′(z)xy+ν4(xy,z)−μ4(x)yz−μ4′(yz)x−ν4(x,yz)+λ3xyz+μ3(x)yz+μ3′(y)xz+ν3(x,y)z−(−1)mμ2(y)¯xz−(−1)mμ2′(z)¯xy−(−1)mν2(y,z)¯x=0.$

Since the coefficient of xyz is λ3 and the coefficient of xz is $\begin{array}{}{\mu }_{3}^{\prime }\left(y\right)\end{array}$ – (–1)m μ2(y), [16, Theorem 3.7] yields

$λ3=0andμ3′(y0)=(−1)mμ2(y0)¯.$

When |x| = |y| = |z| = 1, it follows from (10) that

$μ2(xy)z+μ2′(z)xy+ν2(xy,z)−μ3(x)yz−μ3′(yz)x−ν3(x,yz)+μ4(x)yz+μ4′(y)xz+ν4(x,y)z−(−1)mμ4(y)¯xz−(−1)mμ4′(z)¯xy−(−1)mν4(y,z)¯x=0.$

Since the coefficient of xz is $\begin{array}{}{\mu }_{4}^{\prime }\left(y\right)\end{array}$ – (–1)mμ4(y), [16, Theorem 3.7] implies $\begin{array}{}{\mu }_{4}^{\prime }\left({y}_{1}\right)\end{array}$ = (–1)mμ4(y1).

According to (9), we have

$B(x0,y0)=μ1(x0)y0+μ1(y0)x0−μ1(x0y0);B(x0,y1)=μ1(x0)y1+μ4(y1)x0−μ4(x0y1);B(x1,y0)=μ4(x1)y0+(−1)mx1μ1(y0)−μ4(x1y0);B(x1,y1)=μ4(x1)y1+(−1)mx1μ4(y1)−μ1(x1y1).$(11)

Set

$μm(x)=μ1(x)x∈A0,μ4(x)x∈A1.$

It follows from (11) that:

• γm(x0y0) + μm(x0y0) = γm(x0)y0 + μm(x0)y0 + x0γm(y0) + x0μm(y0);

• γm(x0y1) + μm(x0y1) = γm(x0)y1 + μm(x0)y1 + x0γm(y1) + x0μm(y1);

• γm(x1y0) + μm(x1y0) = γm(x1)y0 + μm(x1)y0 + (–1)m(x1γm(y0) + x1μm(y0));

• γm(x1y1) + μm(x1y1) = γm(x1)y1 + μm(x1)y1 + (–1)m(x1γm(y1) + x1μm(y1)).

Let γm + μm = dm and h = –μ0μ1, then γ = d + h, where d = d0 + d1 is a superderivation and h : AC + is a linear mapping.  □

By Lemma 2.2 and the above result, we have

#### Corollary 3.5

Let A = A0A1 be a prime superalgebra with maximal right ring of quotients Q and extended centroid C. Suppose that γ : AQ is a Lie n superderivation, n ≥ 2. If deg (A1) ≥ 2n + 5, then γ = d + h, where d : AQ is a superderivation and h : AC + Cω is a linear mapping.

In particular, we get the following results, which will be used in the next section.

#### Theorem 3.6

Let Q = Q0Q1 be a unital superalgebra with center C = C0C1. Let A = A0A1 be a superalgebra and a subalgebra of Q. Suppose that α : AQ is a Lie superderivation. If A is a 4-superfree subset of Q, then α = d + h, where d : AQ is a superderivation and h : AC + C ω is a linear mapping.

#### Corollary 3.7

Let A = A0A1 be a prime superalgebra with maximal right ring of quotients Q and extended centroid C. Suppose that α : AQ is a Lie superderivation. If deg (A1) ≥ 9, then α = d + h, where d : AQ is a superderivation and h : AC + Cω is a linear mapping.

#### Theorem 3.8

Let Q = Q0Q1 be a unital superalgebra with center C = C0C1. Let A = A0A1 be a superalgebra and a subalgebra of Q. Suppose that β : AQ is a Lie triple superderivation. If A is a 5-superfree subset of Q, then β = d + h, where d : AQ is a superderivation and h : AC + Cω is a linear mapping.

#### Corollary 3.9

Let A = A0A1 be a prime superalgebra with maximal right ring of quotients Q and extended centroid C. Suppose that β : AQ is a Lie triple superderivation. If deg (A1) ≥ 11, then β = d + h, where d : AQ is a superderivation and h : AC + Cω is a linear mapping.

## 4 Generalized Lie n superderivations of superalgebras

In the section, we describe the structure of generalized Lie n superderivations on a superalgebra.

#### Definition 4.1

Let A be a superalgebra. For m ∈ {0, 1}, a Φ-linear mapping ηm : AA is called a generalized Lie superderivation of degree m if ηm(Aj)⊆ Am+j, jZ2, and

$ηm([x,y]s)=ηm(x)y−(−1)|x||y|ηm(y)x+(−1)m|x|xαm(y)−(−1)m|y|+|x||y|yαm(x),$

for all x, yA0A1, where αm is a Lie superderivation of degree m on A. If η = η0 + η1, then η is called a generalized Lie superderivation on A.

#### Definition 4.2

Let A be a superalgebra. For m ∈ {0, 1}, a Φ-linear mapping θm : AA is called a generalized Lie triple superderivation of degree m if θm(Aj) ⊆ Am+j, jZ2, and

$θm([[x,y]s,z]s)=θm(x)yz−(−1)|x||y|θm(y)xz−(−1)|x||z|+|y||z|θm(z)xy+(−1)|x||y|+|x||z|+|y||z|θm(z)yx+(−1)m|x|xβm(y)z−(−1)m|y|+|x||y|yβm(x)z−(−1)m|z|+|x||z|+|y||z|zβm(x)y+(−1)m|z|+|x||y|+|y||z|+|x||z|zβm(y)x+(−1)m|x|+m|y|xyβm(z)−(−1)m|x|+m|y|+|x||y|yxβm(z)−(−1)m|x|+m|z|+|x||z|+|y||z|zxβm(y)+(−1)m|y|+m|z|+|x||y|+|x||z|+|y||z|zyβm(x),$

for all x, y, zA0A1, where βm is a Lie triple superderivation of degree m on A. If θ = θ0+θ1, then θ is called a generalized Lie triple superderivation on A.

According to the definition of Lie n superderivations, we give the definition of generalized Lie n superderivations.

#### Definition 4.3

Let n ≥ 2 be an integer. Let A be a superalgebra. For m ∈ {0, 1}, a Φ-linear mapping ϑm : AA is called a generalized Lie n superderivation of degree m if ϑm(Aj) ⊆ Am+j, jZ2, and

$ϑm(pn(x1,x2,…,xn))=∑r=12n−1∑l=1nArτlρr(x1,x2,…,xn),$

for all x1, x2, …, xnA0A1, where

$A1=1,A2i−2+1=−(−1)|xi|(|x1|+…+|xi−1|),i∈{2,3,…,n},A2i−2+j=A2i−2+1Aj,2⩽j⩽2i−2,ρ1(xk1,xk2,…,xkn)=(xk1,xk2,…,xkn),ρ2i−2+1(xk1,xk2,…,xkn)=(xki,xk1,…,xki−1,xki+1,…,xkn),i∈{2,3,…,n},ρ2i−2+j(xk1,xk2,…,xkn)=ρ2i−2+1ρj(xk1,xk2,…,xkn),2⩽j⩽2i−2,τ1(xt1,xt2,…,xtn)=ϑm(xt1)xt2…xtn,τ2(xt1,xt2,…,xtn)=(−1)m|xt1|xt1γm(xt2)…xtn,τ3(xt1,xt2,…,xtn)=(−1)m(|xt1|+|xt2|)xt1xt2γm(xt3)…xtn,⋯τn(xt1,xt2,…,xtn)=(−1)m(|xt1|+…+|xtn−1|)xt1xt2…$

γm is a Lie n superderivation of degree m on A.

That is, Ar = 1 or Ar = −1, and ρr is a permutable. If ϑ = ϑ0 + ϑ1, then ϑ is called a generalized Lie n superderivation on A.

The expression of generalized Lie n superderivations is too complicated, so we will study the structure of generalized Lie triple superderivations firstly. In the same manner we can get the structure of generalized Lie n superderivations.

#### Theorem 4.4

Let Q = Q0Q1 be a unital superalgebra with center C = C0C1. Let A = A0A1 be a superalgebra and a subalgebra of Q. Suppose that θ : AQ is a generalized Lie triple superderivation. If A is a 5-superfree subset of Q, then θ = g + l, where g : AQ is a generalized superderivation and l : AC + C ω is a linear mapping.

#### Proof

By the definition of generalized Lie triple superderivations, we assume that θm is a generalized Lie triple superderivation of degree m and βm is a Lie triple superderivation of degree m, m ∈ {0, 1}. According to (5) we have

$[[aibj,ck]s,dl]s=[[ai,bjck]s,dl]s+(−1)ij+ik[[bj,ckai]s,dl]s,$(12)

for all ai, bj, ck, dlA0A1. Applying θm to (12), we get

$0=θm(aibj)ckdl−(−1)ik+jkθm(ck)aibjdl−(−1)il+jl+klθm(dl)aibjck+(−1)ik+jk+il+jl+klθm(dl)ckaibj+(−1)mi+mjaibjβm(ck)dl−(−1)mk+ik+jkckβm(aibj)dl−(−1)ml+il+jl+kldlβm(aibj)ck+(−1)ml+ik+jk+il+jl+kldlβm(ck)aibj+(−1)mi+mj+mkaibjckβm(dl)−(−1)mi+mj+mk+ik+jkckaibjβm(dl)−(−1)mi+mj+ml+il+jl+kldlaibjβm(ck)+(−1)mk+ml+ik+jk+il+jl+kldlckβm(aibj)−θm(ai)bjckdl+(−1)ij+ikθm(bjck)aidl+(−1)il+jl+klθm(dl)aibjck−(−1)ij+ik+il+jl+klθm(dl)bjckai−(−1)miaiβm(bjck)dl+(−1)mj+mk+ij+ikbjckβm(ai)dl+(−1)ml+il+jl+kldlβm(ai)bjck−(−1)ml+ij+ik+il+jl+kldlβm(bjck)ai−(−1)mi+mj+mkaibjckβm(dl)+(−1)mi+mj+mk+ij+ikbjckaiβm(dl)+(−1)mi+ml+il+jl+kldlaiβm(bjck)−(−1)mj+mk+ml+ij+ik+il+jl+kldlbjckβm(ai)−(−1)ij+ik(θm(bj)ckaidl−(−1)jk+jiθm(ckai)bjdl−(−1)il+jl+klθm(dl)bjckai+(−1)jk+ji+il+jl+klθm(dl)ckaibj−(−1)mk+mi+jk+jickaiβm(bj)dl+(−1)mjbjβm(ckai)dl−(−1)ml+il+jl+kldlβm(bj)ckai+(−1)ml+jk+ji+il+jl+kldlβm(ckai)bj+(−1)mi+mj+mkbjckaiβm(dl)−(−1)mi+mj+mk+jk+jickaibjβm(dl)−(−1)mj+ml+il+jl+kldlbjβm(ckai)+(−1)mi+mk+ml+jk+ji+il+jl+kldlckaiβm(bj)),$

for all ai, bj, ck, dlA0A1. An easy computation shows that

$0=θm(aibj)ckdl−θm(ai)bjckdl−(−1)ik+jkθm(ck)aibjdl+(−1)ij+ik+jk+jiθm(ckai)bjdl+(−1)ij+ikθm(bjck)aidl−(−1)ij+ikθm(bj)ckaidl+(−1)mi+mjaibjβm(ck)dl−(−1)miaiβm(bjck)dl+(−1)mj+mk+ij+ikbjckβm(ai)dl−(−1)mj+ij+ikbjβm(ckai)dl−(−1)mk+ik+jkckβm(aibj)dl+(−1)mi+mk+ik+jkckaiβm(bj)dl−(−1)ml+il+jl+kldlβm(aibj)ck+(−1)ml+il+jl+kldlβm(ai)bjck−(−1)mi+mj+il+jl+ml+kldlaibjβm(ck)+(−1)mi+ml+il+jl+kldlaiβm(bjck)+(−1)ml+ik+jk+il+jl+kldlβm(ck)aibj−(−1)ml+ik+jk+il+jl+kldlβm(ckai)bj+(−1)mk+ml+ik+jk+il+jl+kldlckβm(aibj)−(−1)mi+mk+ml+ik+jk+kl+jl+ildlckaiβm(bj)−(−1)ml+ij+ik+il+jl+kldlβm(bjck)ai+(−1)ml+ij+ik+jl+kl+ikdlβm(bj)ckai−(−1)mj+mk+ml+ij+ik+jl+kl+ildlbjckβm(ai)+(−1)mj+ml+jl+kl+il+ij+ikdlbjβm(ckai),$(13)

for all ai, bj, ck, dlA0A1. By Theorem 3.8, we have βm(x) = dm(x) + hm(x), where dm is a superderivation of degree m and hm : AC + is a linear mapping. Define 𝓑 : A × AQ by

$B(x,y)=θm(xy)−θm(x)y−(−1)m|x|xdm(y),$

for all x, yA0A1. We can rewrite (13) as

$B(ai,bj)ckdl+(−1)ij+ikB(bj,ck)aidl+(−1)ik+jkB(ck,ai)bjdl+∑λLLL=0,$

where $\begin{array}{}\sum {\lambda }_{L}^{L}L\end{array}$ is a superquasi-polynomial. By A is a 5-superfree subset of Q, [16, Theorem 3.8] implies

$B(x0,y0)=λ1x0y0+λ1′y0x0+μ1(x0)y0+μ1′(y0)x0+ν1(x0,y0);B(x0,y1)=λ2x0y1+λ2′y1x0+μ2(x0)y1+μ2′(y1)x0+ν2(x0,y1);B(x1,y0)=λ3x1y0+λ3′y0x1+μ3(x1)y0+μ3′(y0)x1+ν3(x1,y0);B(x1,y1)=λ4x1y1+λ4′y1x1+μ4(x1)y1+μ4′(y1)x1+ν4(x1,y1),$(14)

where λk, $\begin{array}{}{\lambda }_{k}^{\prime }\end{array}$Cm + Cmω, μk, $\begin{array}{}{\mu }_{k}^{\prime }\end{array}$ : aiCm+i + Cm+i ω, νk : Ai × AjC + , k ∈ {1, 2, 3, 4}, i, j ∈ {0, 1}.

By computing θm(xyz) in two different ways, we have

$B(xy,z)+B(x,y)z−B(x,yz)=0,$(15)

for all x, y, zA0A1.

By substituting (14) into (15), we get

$0=λ1xyz+λ1′zxy+μ1(xy)z+μ1′(z)xy+ν1(xy,z)+λ1xyz+λ1′yxz+μ1(x)yz+μ1′(y)xz+ν1(x,y)z−λ1xyz−λ1′yzx−μ1(x)yz−μ1′(yz)x−ν1(x,yz),$

for all x, y, zA0;

$0=λ3xyz+λ3′zxy+μ3(xy)z+μ3′(z)xy+ν3(xy,z)+λ2xyz+λ2′yxz+μ2(x)yz+μ2′(y)xz+ν2(x,y)z−λ2xyz−λ2′yzx−μ2(x)yz−μ2′(yz)x−ν2(x,yz),$

for all x, zA0, yA1;

$0=λ4xyz+λ4′zxy+μ4(xy)z+μ4′(z)xy+ν4(xy,z)+λ2xyz+λ2′yxz+μ2(x)yz+μ2′(y)xz+ν2(x,y)z−λ1xyz−λ1′yzx−μ1(x)yz−μ1′(yz)x−ν1(x,yz),$

for all xA0, y, zA1;

$0=λ4xyz+λ4′zxy+μ4(xy)z+μ4′(z)xy+ν4(xy,z)+λ3xyz+λ3′yxz+μ3(x)yz+μ3′(y)xz+ν3(x,y)z−λ4xyz−λ4′yzx−μ4(x)yz−μ4′(yz)x−ν4(x,yz),$

for all yA0, x, zA1;

$0=λ1xyz+λ1′zxy+μ1(xy)z+μ1′(z)xy+ν1(xy,z)+λ4xyz+λ4′yxz+μ4(x)yz+μ4′(y)xz+ν4(x,y)z−λ4xyz−λ4′yzx−μ4(x)yz−μ4′(yz)x−ν4(x,yz),$

for all zA0, x, yA1. By [16, Theorem 3.7], we have

$λ1=λ1′=μ1′=0,μ1(x0y0)=−ν1(x0,y0);λ3=λ3′=λ2′=μ3′=μ2′=0,μ3(x0y1)=−ν2(x0,y1);λ4=−λ2,λ4′=μ4′=0,μ1(x0)=μ2(x0);μ3(x1)=μ4(x1),μ4(x1y0)=−ν3(x1,y0);μ1(x1y1)=−ν4(x1,y1).$

By the definition of 𝓑, we have

$B([[x0,w0],y0],z1)−B(z1,[[x0,w0],y0])=θm([[[x0,w0],y0],z1])−θm([[x0,w0],y0])z1−[[x0,w0],y0]dm(z1)+θm(z1)[[x0,w0],y0]+(−1)mz1dm([[x0,w0],y0])=θm([x0,w0])y0z1−θm(y0)[x0,w0]z1−θm(z1)[x0,w0]y0+θm(z1)y0[x0,w0]+[x0,w0]βm(y0)z1−y0βm([x0,w0])z1−(−1)mz1βm([x0,w0])y0+(−1)mz1βm(y0)[x0,w0]+[x0,w0]y0βm(z1)−y0[x0,w0]βm(z1)−(−1)mz1[x0,w0]βm(y0)+(−1)mz1y0βm([x0,w0])−B([x0,w0],y0)z1+B(y0,[x0,w0])z1−θm([x0,w0])y0z1−[x0,w0]dm(y0)z1+θm(y0)[x0,w0]z1+y0dm([x0,w0])z1−[[x0,w0],y0]dm(z1)+θm(z1)[[x0,w0],y0]+(−1)mz1dm([[x0,w0],y0]),$

for all x0, y0, w0A0, z1A1. Since the coefficient of x0w0y0z1 is λ2, it follows from [16, Theorem 3.7] that λ2 = 0 and λ4 = 0. Therefore,

$B(x0,y0)=μ1(x0)y0−μ1(x0y0);B(x0,y1)=μ1(x0)y1−μ4(x0y1);B(x1,y0)=μ4(x1)y0−μ4(x1y0);B(x1,y1)=μ4(x1)y1−μ1(x1y1).$(16)

Set

$μm(x)=μ1(x)x∈A0,μ4(x)x∈A1.$

It follows from (16) that θm(xy) + μm(xy) = θm(x)y + (−1)m|x|xdm(y) + μm(x)y, for all x, yA0A1.

Let θm + μm = gm and l = −μ0μ1, then θ = g + l, where g = g0 + g1 is a generalized superderivation and l : AC + is a linear mapping.

By Lemma 2.2 and the above result, we have

#### Corollary 4.5

Let A = A0A1 be a prime superalgebra with maximal right ring of quotients Q and extended centroid C. Suppose that β : AQ is a generalized Lie triple superderivation. If deg (A1) ≥ 11, then β = g + l, where g : AQ is a generalized superderivation and l : AC + Cω is a linear mapping.

According to the proof of Theorem 3.4 and Theorem 4.4, we have

#### Theorem 4.6

Let Q = Q0Q1 be a unital superalgebra with center C = C0C1. Let A = A0A1 be a superalgebra and a subalgebra of Q. Suppose that ϑ : AQ is a generalized Lie n superderivation, n ≥ 2. If A is an (n + 2)-superfree subset of Q, then ϑ = g + l, where g : AQ is a generalized superderivation and l : AC + Cω is a linear mapping.

By Lemma 2.2 and the above result, we have

#### Corollary 4.7

Let A = A0A1 be a prime superalgebra with maximal right ring of quotients Q and extended centroid C. Suppose that ϑ : AQ is a generalized Lie n superderivation, n ≥ 2. If deg (A1) ≥ 2n + 5, then ϑ = g + l, where g : AQ is a generalized superderivation and l : AC + Cω is a linear mapping.

In particular, we have

#### Theorem 4.8

Let Q = Q0Q1 be a unital superalgebra with center C = C0C1. Let A = A0A1 be a superalgebra and a subalgebra of Q. Suppose that η : AQ is a generalized Lie superderivation. If A is a 4-superfree subset of Q, then η = g + l, where g : AQ is a generalized superderivation and l : AC + Cω is a linear mapping.

#### Corollary 4.9

Let A = A0A1 be a prime superalgebra with maximal right ring of quotients Q and extended centroid C. Suppose that η : AQ is a generalized Lie superderivation. If deg (A1) ≥ 9, then η = g + l, where g : AQ is a generalized superderivation and l : AC + Cω is a linear mapping.

## Acknowledgement

Supported by NNSF of China (Nos. 11771069 and 11471090) and the project of Jilin Science and Technology Department (Nos. 20170520068JH and 20170101048JC).

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Accepted: 2017-12-22

Published Online: 2018-03-13

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 196–209, ISSN (Online) 2391-5455,

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