On generalized Lie derivations

Abstract

In this paper, we investigate generalized Lie derivations. We give a complete characterization of when each generalized Lie derivation is a sum of a generalized inner derivation and a Lie derivation. This generalizes a result given by Benkovič. We also investigate when every generalized Lie derivation on some particular kind of unital algebras is a sum of a generalized derivation and a central map which vanishes on all commutators. Precisely, we consider both the unital algebras with nontrivial idempotents and the trivial extension algebras.

1 Introduction

Throughout the paper, ${\mathcal {R}}$ will denote a commutative ring with unity, A will be a unital ${\mathcal {R}}$-algebra with center Z(A) and M will be a unital A-bimodule. For $a\in A$ and $m\in M$, we use $a \circ m$ (resp., [a, m]) to denote the Jordan product $a m + m a$ (resp., the Lie product $a m -m a$) of a and m.

Let $d:A\longrightarrow M$ and $f:A\longrightarrow M$ be linear maps. Recall that f is said to be a generalizedd-derivation (or simply a generalized derivation) if

$$\begin{aligned} f(ab)=f(a)b+ad(b) \ \ \ (a,b\in A). \end{aligned}$$

(1.1)

Recall that a map ${\mathcal {H}}:A\rightarrow M$ is said to be an inner derivation (resp., a generalized inner derivation) if there exists $a \in A$ such that, for all $x\in A$, ${\mathcal {H}}(x)= ax-xa$ (resp., if there exist $a, b \in A$, such that, for all $x\in A$, ${\mathcal {H}}(x)= ax+xb$).

A linear map $f : A\longrightarrow M$ is said to be a generalized Lie d-derivation (or simply a generalized Lie derivation), if there exists a linear map $d: A \longrightarrow M$

$$\begin{aligned} f([a,b])=f(a)b-f(b)a+ad(b)-bd(a) \ \ \ (a,b\in A). \end{aligned}$$

(1.2)

Any Lie derivation $f : A \longrightarrow M$, a linear map which satisfies $f ([a, b]) = [f (a), b] + [a, f (b)]$ for all $a, b \in A$, is a generalized Lie derivation for $d = f$. Any generalized derivation is a generalized Lie derivation.

Let $Z(M) = \{m \in M, [m, A] = 0\}$ be the center of M. Notice that every linear map $\tau : A \longrightarrow Z(M)$ that vanishes on all commutators of A is a Lie derivation. A standard example of a Lie derivation is therefore a map of the form $f = d+\tau$, where $d : A \longrightarrow M$ is a derivation and $\tau : A \longrightarrow Z(M)$ is a linear map such that $\tau ([A, A]) = 0$.

The concept of a generalized derivation was introduced by Brešar [5] and generalized by Hvala [7] who proved in [8] that each generalized Lie derivation of a prime ring is the sum of a generalized derivation and a central map which vanishes on all commutators. Liao and Liu [9] generalized this result describing all generalized Lie derivations on a Lie ideal of a prime algebra. Recently, Benkovič [1] studied generalized Lie derivations mapping from A to M. The results established in [1] were applied to triangular algebras. Namely, he proved that under certain conditions each generalized Lie derivation of a triangular algebra A is the sum of a generalized derivation and a central map which vanishes on all commutators of A. Following Cheung’s terminology (see [6]), generalized Lie derivations which is the sum of a generalized derivation and a central map which vanishes on all commutators will be called proper generalized Lie derivations. One of the aim of this paper is to generalize Benkovič’s result to the context of trivial extension algebras. Furthermore, another kind of generalized Lie derivations is of interest. Indeed, one can show that also the sum of a generalized inner derivation and a Lie derivation is a generalized Lie derivation (see Proposition 2.1). Here, we call this kind of generalized Lie derivations strong generalized Lie derivations. In [1], Benkovič gave a sufficient condition so that a generalized Lie derivation be strong. In this paper, we give a complete characterization (see Theorem 2.5).

The paper is organized as follows:

In Sect. 2, as the first main result of this paper, we characterize when a generalized Lie derivation is strong (see Theorem 2.5).

In Sect. 3, we study when every generalized Lie derivation of unital algebras with idempotents is proper (see Theorem 3.3). This result leads to another context than the one given in [1, Theorem 3.6] and [2, Theorem 2.3].

In Sect. 4, we study when a trivial extension algebra has the generalized Lie derivation property; that is, every generalized Lie derivation is proper. Similar results were obtained in [3, Theorem 3.15] and [11, Theorem 2.3].

2 Inner derivation, Lie derivation and generalized Lie derivation

In this section we investigate the relation between inner derivations, Lie derivations and generalized Lie derivations.

Let us start with the following result which shows that a sum of a generalized inner derivation and a Lie derivation is an example of a generalized Lie derivation. This kind of a generalized Lie derivation will be called a strong generalized Lie derivation.

Proposition 2.1

Let $I:A \longrightarrow M$ be a generalized inner derivation; that is $I(x)=nx+xn'$ for all $x \in A$, where n and $n'$ are fixed elements of M. Let $\tau :A \longrightarrow M$ be a Lie derivation. Then, $I+\tau$ is a generalized Lie derivation with an associated map $d: x \mapsto [x,n'+\tau (1)]+\tau (x)$.

Proof

Let $f:=I+\tau$. By putting $m_0 = \tau (1)$, we show that, for any $x,y\in A$,

$$\begin{aligned} f[x,y]= [f(x),y]+[x,f(y)]+[x,y]m_0-xf(1)y+yf(1)x. \end{aligned}$$

We have $[f(x),y]+[x,f(y)]+[x,y]m_0-xf(1)y+yf(1)x = [nx+xn'+\tau (x),y] + [x,ny+yn'+\tau (y)] + [x,y]\tau (1) -x(n+n'+\tau (1))y + y(n+n'+\tau (1))x.$ Since $\tau$ is a Lie derivation, $\tau (1)\in Z(M)$ and so $[nx+xn'+\tau (x),y]+[x,ny+yn'+\tau (y)] + [x,y]\tau (1) - x(n+n'+\tau (1))y+y(n+n'+\tau (1))x = n[x,y]+[x,y]n'+\tau [x,y] = f [x,y].$ Therefore, f is a generalized Lie derivation with an associated map $d: x \mapsto f(x)-f(1)x+x\tau (1) =[x,n'+\tau (1)]+\tau (x)$. $\square$

In [1, page 1535], Benkovič observed, after some calculations on a generalized Lie derivation f with an associated map d, that f is proper if $d(1) \in Z(M)$. Moreover, he considered $\delta (x) = d(x)- xd(1)$ for all $x \in A$. Then, he showed that f is proper if $[A, A]d(1) = 0$.

Here we give also the following situation.

Proposition 2.2

Let $\delta : A \longrightarrow M$ and $d:A \longrightarrow M$ are two mappings satisfying $\delta ([x,y])=[\delta (x),y]+[x,\delta (y)]+[x,y]d(1)$ and $d([x,y])=[d(x),y]+[x,d(y)]+x[y,d(1)]+y[d(1),x]$.

Then, the linear maps f and g defined as follows $f:x \mapsto ax+\delta (x)$ and $gx \mapsto bx+d(x)$, for some $a,b\in A$, are a generalized Lie derivation.

Benkovič unified his remarks given before Proposition 2.2 in one result as follows:

Theorem 2.3

(Theorem 3.3, [1]) Let A be an algebra with identity and let M be a unital A-bimodule. Let $f : A \longrightarrow M$ be a generalized Lie derivation with an associated linear map d. Suppose that for each $m \in M$ satisfying

$$\begin{aligned}{}[x, y][z,m] + [y, z][x,m] + [z, x][y,m] = 0 \;\;\text {for}\;\; \text {all} \;\;x, y, z\in A. \end{aligned}$$

there exist $m_1,m_2 \in M$ such that $m = m_1 + m_2, m_1 \in Z(M)$, and $[A, A]m_2 = 0$. Then $f = \Delta + \delta$, where $\Delta : A \longrightarrow M$ is a generalized inner derivation and $\delta : A \longrightarrow M$ is a Lie derivation.

Our aim, in this section, is to generalize this result by giving equivalent conditions so that every generalized Lie derivation is proper. For this we need the following result.

Lemma 2.4

Let M is an $A-$module. The following assertions hold:

• 1. A linear map $f:A\longrightarrow M$ is a generalized derivation if and only if there exists $n \in M$ such that $f(xy)=f(x)y +xf(y)-xny$ for all $x,y\in A$.

• 2. A linear map $f:A\longrightarrow M$ is a generalized Lie derivation if and only if there exist $m,n \in M$ such that $f[x,y]= [f(x),y]+[x,f(y)]+[x,y]m-xny+ynx$ for all $x,y\in A$. Thus, f is a generalized Lie derivation with an associated map $d: x \mapsto f(x)-nx+xm$.

• 3. A linear map $f:A\longrightarrow M$ is a generalized Lie derivation if and only if there exists $m_0 \in M$ such that $f[x,y]= [f(x),y]+[x,f(y)]+[x,y]m_0-xf(1)y+yf(1)x$ for all $x,y\in A$. Thus, f is a generalized Lie derivation with an associated map $d: x \mapsto f(x)-f(1)x+xm_0$ and $d(1)=m_0$.

Proof

• 1. Suppose that f is a generalized derivation with an associated linear map d, then $d(x)=f(x)-f(1)x$ and so

  $$\begin{aligned} f(xy)=f(x)y +xd(y)=f(x)y+x(f(y)-f(1)y)=f(x)y+xf(y)-xf(1)y. \end{aligned}$$

  For the converse, put $d(x):=f(x)-nx$. Thus, f is a generalized derivation with d as an associated map.

• 2. Let f be a generalized Lie derivation with an associated map d, so we have

  $$\begin{aligned} f[x,y]=f(x)y-f(y)x+xd(y)-yd(x) \end{aligned}$$

  (2.1)

  Then, for $y=1$, we get

  $$\begin{aligned} d(x)=f(x)-f(1)x+xd(1). \end{aligned}$$

  Thus (2.1) becomes

  $$\begin{aligned} f[x,y]=[f(x),y]+[x,f(y)]+[x,y]d(1)-xf(1)y+yf(1)x. \end{aligned}$$

  For the converse, if there exist $m,n \in M$ such that

  $$\begin{aligned} f[x,y]= [f(x),y]+[x,f(y)]+[x,y]m-xny+ynx . \end{aligned}$$

  Therefore, f is a generalized Lie derivation with an associated map $d: x\mapsto f(x)-nx+xm$.

• 3. We put $y=1$ in

  $$\begin{aligned} f[x,y]= [f(x),y]+[x,f(y)]+[x,y]m-xny+ynx. \end{aligned}$$

  Then, we get $[x, f(1)-n]=0$ for each $x \in A$. It means that $f(1)-n \in Z(M)$. Thus, there exists $z \in Z(M)$ such that $n=z+f(1)$. And if we put $x=1$ and $n=z+f(1)$ in $d(x)=f(x)-nx+xm$, we get $d(1)=f(1)-z-f(1)+m=-z+m$ so $m = d(1)+z$. In addition, $z \in Z(M)$ shows that $[f(x),y]+[x,f(y)]+[x,y](z+d(1))-x(z+f(1))y +y(z+f(1))x = [f(x),y]+[x,f(y)] + [x,y]d(1)-xf(1)y+yf(1)x.$ As desired.

$\square$

Now, we are in a position to give our first main result. It is a generalization of [1, Theorem 3.3].

Theorem 2.5

Let $f:A \longrightarrow M$ be a generalized Lie derivation with an associated map d. Then, the following assertions are equivalent:

• 1. f is a strong generalized Lie derivation, that is $f=I+L$, where I is a generalized inner derivation and L is a Lie derivation.

• 2. There exists a generalized inner derivation I such that $f(1)-I(1)\in Z(M)$ and $[A,A](d(1) - f(1)+I(1)) = 0.$

• 3. There exists $n_0 \in Z(M)$ such that $[A,A](d(1) - n_0) = 0.$

Proof

$1)\Longrightarrow 2)$. Let $f =I+\tau$, where $I(x)=nx+xn'$$(n,n'\in M)$ and $\tau$ is a Lie derivation. We have $\tau ([x,y])- [\tau (x),y]-[x, \tau (y)] = f([x,y]) - I[x,y]-[f(x)-I(x),y] - [x, f(y)-I(y)] = 0$, and so, by Lemma 2.4, $[f(x),y]+[x,f(y)] + [x,y]d(1)-xf(1)y+ y f(1)x-n[x,y] - [x,y]n'- [f(x)-nx-xn',y]-[x,f(y)-ny-yn'] = 0$ and $[x,y]d(1)-x( f(1)-I(1))y+y( f(1)-I(1))x=0$. Putting $y=1$ in this equality, we get $f(1)-I(1) \in Z(M)$. Therefore, $[x,y](d(1)- f(1)+I(1))=0$.

$2)\Longrightarrow 3)$. By considering $n_0=f(1)-I(1)$, $n_0\in Z(M)$ and $[A,A](d(1)-n_0)=0$.

$3)\Longrightarrow 1)$. Suppose there exists $n_0 \in Z(M)$ such that $[A,A](d(1)-n_0)=0$. Put $I(x)=(f(1)-n_0)x$. We will show that $\tau (x):=f(x)-I(x)=f(x)-(f(1)-n_0)x$ is a Lie derivation. We have $\tau ([x,y])-[\tau (x),y]-[x,\tau (y)]= f([x,y])- (f(1)-n_0)[x,y]- [f(x)-(f(1)-n_0)x , y ] -[x,f(y)-(f(1)-n_0)y ]= [f(x),y]+[x,f(y)]+[x,y]d(1)-xf(1)y+yf(1)x-(f(1)-n_0)[x,y]- [f(x),y]+[(f(1)-n_0)x,y]-[x,f(y)]+[x,(f(1)-n_0)y]=[x,y]d(1)-xn_0 y+yn_0 x$. Therefore, since $n_0 \in Z(M)$ and $[A,A](d(1)-n_0)=0$, $\tau$ is a Lie derivation. $\square$

Remark 1

It is worth noting that if one of the equivalent assertions of Theorem 2.5 holds, then I can be of the form $I(x) = (f(1)-n_0 -n)x+xn$ for each $n \in M.$

3 Generalized Lie derivations of unital algebras with idempotents

As mentioned in the introduction several authors have been proved for some particular kind of algebras that every generalized Lie derivation is a sum of a generalized derivation and a central map which vanishes on all commutators. These kind of generalized Lie derivations will be called proper generalized Lie derivations. Also, if all generalized Lie derivations on an algebra A is proper, we say that A has the generalized Lie derivation property.

In this section, we investigate when every generalized Lie derivation on unital algebras with idempotents is proper.

Throughout of this section, we will consider the following conditions and notations:

Setup and notation. We assume that the algebra ${\mathcal {A}}$ admits a nontrivial idempotent e. Then,

$$\begin{aligned} {\mathcal {A}} = e{\mathcal {A}}e+e{\mathcal {A}}e'+e'{\mathcal {A}}e+e'{\mathcal {A}}e', \end{aligned}$$

where $e'=1-e$. Notice that $e{\mathcal {A}}e$ and $e' {\mathcal {A}} e'$ are algebras with unitary elements e and $e'$, respectively. Notice that also $e{\mathcal {A}} e'$ is an $(e{\mathcal {A}}e, e' {\mathcal {A}} e' )$-bimodule and $e' {\mathcal {A}}e$ is an $( e' {\mathcal {A}} e', e{\mathcal {A}}e)$-bimodule. We will assume that ${\mathcal {A}}$ satisfies the two following conditions:

$$\begin{aligned}&\text {for all} \;\; x \in {\mathcal {A}}, exe . e{\mathcal {A}} e' = \{0\} = e' {\mathcal {A}}e . exe \;\;\; \text {implies}\;\;\; exe = 0 \end{aligned}$$

(3.1)

$$\begin{aligned}&\text {for all} \;\; x \in {\mathcal {A}}, e{\mathcal {A}} e' . e' x e' = \{0\} = e' x e'.e' {\mathcal {A}}e \;\;\; \text {implies}\;\;\; e' x e' = 0 \end{aligned}$$

Some specific examples of unital algebras with nontrivial idempotents having the property (3.1) are triangular algebras, matrix algebras and prime (and hence in particular simple) algebras with nontrivial idempotents (see [2]).

To simplify notation we will use the following convention: $a=eae\in e{\mathcal {A}}e=A$, $m=eme'\in e{\mathcal {A}}e'=M$, $n=e'ne\in e'{\mathcal {A}}e=N$ and $b=e'be'\in e'{\mathcal {A}}e'=B$. Thus, ${\mathcal {A}}=A+M+N+B$ and each element $x\in {\mathcal {A}}$ can be represented in the form $x= eae+eme'+e'ne+e'be'=a+m+n+b$, where $a\in A$, $m\in M$, $n\in N$ and $b\in B$. We suppose also M to be faithful as a right B-module.

Recall also, from [2, Proposition 2.1], that the center of ${\mathcal {A}}$ is

$$\begin{aligned} Z ({\mathcal {A}}) = \{a + b \in e{\mathcal {A}}e + e' {\mathcal {A}} e' |\; am = mb, na = bn \;\; \text {for all} \;\; m \in e{\mathcal {A}} e', n \in e' {\mathcal {A}}e\} . \end{aligned}$$

Lemma 3.1

If there exists $x_0=a_0+m_0+n_0+b_0\in {\mathcal {A}}$ which satisfies

$$\begin{aligned}{}[x,y][z,x_0]+[y,z][x,x_0]+[z,x][y,x_0]=0 \;\;\text {for all}\;\; x,y,z \in {\mathcal {A}}, \end{aligned}$$

(3.2)

then the following assertions hold:

• 1. $b_0 \in Z(B)$.

• 2. $[A,A]m_0=[B,B]n_0=0$.

• 3. For all $a_1, a_2 \in A$, $b_1, b_2 \in B$, $m\in M$ and $n \in N$, $[a_1,a_2][m,x_0] +a_1mn_0a_2-a_2mn_0a_1=[b_1,b_2][n,x_0]+b_1nm_0b_2-b_2nm_0b_1=0$.

Proof

• 1. We show that $b_0 \in Z(B)$. Setting $x=e, y=m$ and $z=b$ in (3.2), we get $m[b, x_0] + mb[e, x_0] = 0$ for all $m \in M, b \in B.$ Then, $m[b, b_0] = 0$ for all $m \in M, b \in B.$ Since M is a faithful right B-module, we conclude that $b_0 \in Z(B)$.

• 2. We show that $[A,A]m_0=[B,B]n_0=0.$ Setting $x = a_1, y = a_2$ and $z = e'$ in (3.2), we obtain $[a_1, a_2][e' , x_0] = 0$ for all $a_1, a_2 \in A$. Consequently, $[a_1, a_2](m_0-n_0) = 0$ for all $a_1, a_2 \in A$ and hence $[A,A]m_0=0$.

  Similarly, if we take $x = b_1, y = b_2$ and $z = e'$ in (3.2), where $b_1, b_2 \in B$, we obtain $[B,B]n_0=0$.

• 3. Setting $x = a_1, y = a_2$ and $z = m$ in (3.2), we obtain $[a_1, a_2][m, x_0]+a_2m[a_1, x_0]-a_1m[a_2, x_0] = 0$, which implies that $[a_1, a_2](mn_0-n_0m+mb_0 - a_0m)+a_1mn_0a_2-a_2mn_0a_1 = 0$ for all $a_1, a_2 \in A, m \in M$.

Similarly, if we take $x = b_1, y = b_2$ and $z = n$ in (3.2), we obtain

$$\begin{aligned}{}[b_1,b_2](nm_0-m_0 n+na_0 - b_0n)+b_1nm_0b_2-b_2nm_0b_1 = 0 \end{aligned}$$

for all $n\in N.$$\square$

Lemma 3.2

If there exixts $x_0=a_0+m_0+n_0+b_0\in {\mathcal {A}}$ which satisfies

$$\begin{aligned}{}[x,y][z,x_0]+[y,z][x,x_0]+[z,x][y,x_0]=0 \;\;\;\text {for} \;\;\;\text {all}\;\;\; x,y,z \in {\mathcal {A}} \end{aligned}$$

(3.3)

and if $[e{\mathcal {A}}e,e{\mathcal {A}}e]=e{\mathcal {A}}e$ and $[e'{\mathcal {A}}e',e'{\mathcal {A}}e']=e'{\mathcal {A}}e'$, then $x_0 = a_0 + b_0 \in Z({\mathcal {A}})$.

Proof

If $[A, A]=A$ and $[B, B]=B$, then $Am_0 = 0$ and $Bn_0=0$ (by Lemma 3.1). Since $e\in A$ and $e' \in B$, $m_0=n_0=0$. Hence, by Lemma 3.1, $A(mb_0 - a_0m) = 0$ for all $m \in M$ and $B(na_0 - b_0n) = 0$ for all $n \in N.$ Since $e\in A$ and $e' \in B$$mb_0 = a_0m$ and $nb_0 = a_0n$. In this case $x_0 = a_0 + b_0 \in Z({\mathcal {A}})$ is of the desired form. $\square$

In the following second main result we give another situation where, under some conditions, every generalized Lie derivation of a unital algebra with a nontrivial idempotent is proper. Compare it with [1, Theorem 3.6] and [2, Theorem 2.3].

Theorem 3.3

Let ${\mathcal {A}}$ be a unital 2-torsion free algebra with a nontrivial idempotent e satisfying (3.1) and assume that $e{\mathcal {A}}e'$ is a faithful right $e'{\mathcal {A}}e'$-module. Let $f:{\mathcal {A}} \longrightarrow {\mathcal {A}}$ be a generalized Lie derivation with an associated map d. If the following conditions hold:

• (I) $[e{\mathcal {A}}e,e{\mathcal {A}}e]=e{\mathcal {A}}e$ and $[e'{\mathcal {A}}e',e'{\mathcal {A}}e']=e'{\mathcal {A}}e'$.

• (II) One of the following statements holds:

  • 1. $e{\mathcal {A}}e$ contains no central ideals.

  • 2. $e' {\mathcal {A}} e'$ contains no central ideals.

  • 3. $Z ({\mathcal {A}}) = \{a + b|a \in Z (e{\mathcal {A}}e) , b \in Z ( e' {\mathcal {A}} e' ) ,am_0 = m_0b\}$ for some $m_0 \in e{\mathcal {A}}e'$.

  • 4. $Z ({\mathcal {A}}) = \{a + b|a \in Z (e{\mathcal {A}}e) , b \in Z ( e' {\mathcal {A}} e' ) ,an_0 = n_0b\}$ for some $n_0 \in e'{\mathcal {A}}e$.

Then f is proper.

Proof

According to the assumption (I) and by [1, Lemma 3.2], the conditions of Lemma 3.2 are satisfied and hence Theorem 2.5 implies $f=\Delta _1+\delta$, where $\Delta _1 : {\mathcal {A}}\longrightarrow {\mathcal {A}}$ is a generalized inner derivation and $\delta : {\mathcal {A}}\longrightarrow {\mathcal {A}}$ is a Lie derivation. Using [2, Theorem 5.3] we see that $\delta =\Delta _2+\gamma$, where $\Delta _2 : {\mathcal {A}}\longrightarrow {\mathcal {A}}$ is a derivation and $\gamma : {\mathcal {A}}\longrightarrow Z({\mathcal {A}})$ is a linear map vanishing on $[{\mathcal {A}},{\mathcal {A}}]$. Denoting $\Delta =\Delta _1+\Delta _2$ we conclude that $f=\Delta +\gamma$. $\square$

Remark 2

Let ${\mathcal {A}}$ be a unital 2-torsion free algebra with a nontrivial idempotent e satisfying (3.1). If the conditions (I) and (II) of Theorem 3.3 are satisfied, then ${\mathcal {A}}$ has the generalized Lie derivation property; that is every generalized Lie derivation is proper.

4 Generalized Lie derivations on trivial extension algebras

Our aim in this section is to study generalized Lie derivations on a trivial extension algebra. We give conditions under which they are proper generalized Lie derivations.

Let us start with a general description of these kind of mapping on a trivial extension algebra.

Clearly, every linear mapping $f:A < imes M \longrightarrow A < imes M$ can be presented in the form

$$\begin{aligned} f(a,m)=(f_A(a)+h_1(m),f_M(a)+h_2(m))\ \ \ \ \ \ \ ((a,m)\in A < imes M), \end{aligned}$$

(4.1)

where the linear mappings $f_A:A\longrightarrow A$, $f_M:A\longrightarrow M$, $h_1:M\longrightarrow A$ and $h_2:M\longrightarrow M$ are given by $f_A(a)=(\pi _A\circ f)(a,0)$, $f_M(a)=(\pi _M\circ f)(a,0)$, $h_1(m)=(\pi _A\circ f)(0,m)$ and $h_2(m)=(\pi _M\circ f)(0,m)$, respectively. Here $\pi _A:A < imes M\longrightarrow A$ and $\pi _M:A < imes M\longrightarrow M$ are the natural projections given by $\pi _A(a,m)=a$ and $\pi _M(a,m)=m$, respectively.

Convention. Notice that, if f is a generalized Lie d-derivation, then $h_1=T$. For this reason, it suffices to consider, in the sequel, that f has a presentation given as in (4.1), and a linear map d on $A < imes M$ with a presentation as follows $d(a,m)=(d_A(a)+h_1(m),d_M(a)+S(m))$$((a,m)\in A < imes M)$.

The following three lemmas are obtained using standard arguments.

Lemma 4.1

A linear map $f:A < imes M \longrightarrow A < imes M$ is a generalized Lie d-derivation if and only if, for all $a,b\in A, m,n\in M$, the following conditions hold:

• 1. $f_A$ is a generalized Lie $d_A$-derivation.

• 2. $f_ M$ is a generalized Lie $d_M$-derivation.

• 3. $h_1([a,m])=[a , h_1(m)]$ for all $a\in A$ and $m \in M$.

• 4. $h_2([a,m])=f_A(a)m-h_2(m)a+aS(m)-md_A(a)$ for all $a\in A$ and $m \in M$.

• 5. $[h_1(m),n]+[m,h_1(n)]=0$ for all $m, n\in M$.

In the remainder of this section, we use some results from the paper [3]. Some of them are given with a proof for reader’s convenience.

Lemma 4.2

([3]) A linear map $f:A < imes M \longrightarrow A < imes M$ is a generalized d-derivation if and only if the following conditions hold:

• 1. $f_A$ is a generalized $d_A$-derivation.

• 2. $f_ M$ is a generalized $d_M$-derivation.

• 3. $h_1(am)=ah_1(m)$ and $h_1(ma)=h_1(m)a$ for all $a\in A$ and $m\in M$.

• 4. $h_2(am)=f_A(a)m+aS(m)$ and $h_2(ma)=h_2(m)a+md_A(a)$ for all $a\in A$ and $m\in M$.

• 5. $mh_1(n)+h_1(m)n=0$ for all $m,n \in M$.

Lemma 4.3

([3]) A linear map $f:A < imes M \longrightarrow A < imes M$ is a central map which vanishes on all commutators if and only if the following conditions hold:

• 1. $f_{ A }$ and $f_{ M }$ are central map which vanishes on all commutators.

• 2. $h_1([a,m])=[h_1(m),a]=0$ and $[f_{A}(a),m]=0$ for all $a\in A$ and $m \in M$.

• 3. $h_2([a,m])=0$ and $[h_2(m),a]=0$ for all $a\in A$ and $m\in M$.

• 4. $[h_1(m),n]=0$ for all $m, n\in M$.

Now we give the second fundamental result. It is similar to [3, Theorem 3.4].

Theorem 4.4

Every generalized Lie derivation on $A < imes M$ is proper if and only if the following conditions hold:

• 1. Every generalized Lie derivation $g:A\longrightarrow M$ is proper.

• 2. Every linear map $h:M\rightarrow A$ such that, for all $a\in A$ and $m,n\in M$, $h([a, m])=[a, h(m)]$ and $[n, h(m)]=0$, is a sum of an A-homomorphism $\delta$ and a commuting map $\beta$ which satisfy $m \delta (n)+\delta (m) n=0=\beta (m) n - n \beta (m)$ for all $m,n\in M$.

• 3. Every generalized Lie derivation f on $A < imes M$ of the form

  $$\begin{aligned} f(a,m)=(f_A(a),h_2(m)) \end{aligned}$$

  (i.e., $h_1=0$ and $f_M=0$ in the presentation (4.1)) is proper.

Proof

$\Rightarrow .$ We only need to prove (1) and (2).

• 1. Let g be a generalized Lie $d_M$-derivation from A into M. Clearly (0, g) is a generalized Lie derivation on $A < imes M$ with associated map $(0,d_M)$. Then, by hypothesis, there exists a generalized derivation $( \delta _{ A }+ {\mathcal {K}}', \delta _{ M } + {\mathcal {L}}')$ and a central map which vanishes on all commutators $(D_{ A }+ {\mathcal {K}} , D_{ M } + {\mathcal {L}})$ such that, for all $a\in A ,m\in M$,

  $$\begin{aligned} (0,g(a))=(D_{ A }(a) + {\mathcal {K}}(m)+\delta _{ A }(a) + {\mathcal {K}}'(m), D_{ M }(a) + {\mathcal {L}}(m)+\delta _{ M }(a) + {\mathcal {L}}'(m)). \end{aligned}$$

  Take $a=0$, we get ${\mathcal {L}}(m) + {\mathcal {L}}'(m)=0$. Hence $g=D_{ M }+\delta _{ M }$, we are done.

• 2. By hypotheses, (h, 0) is a generalized Lie derivation on $A < imes M$, and then, there exists a generalized derivation $( \delta _{ A }+ {\mathcal {K}}', \delta _{ M } + {\mathcal {L}}')$ and a central map which vanishes on all commutators $(D_{ A }+ {\mathcal {K}} , D_{ M } + {\mathcal {L}})$ such that, for all $a\in A ,m\in M$,

  $$\begin{aligned} ( h(m),0)=(D_{ A }(a) + {\mathcal {K}}(m)+\delta _{ A }(a) + {\mathcal {K}}'(m), D_{ M }(a) + {\mathcal {L}}(m)+\delta _{ M }(a) + {\mathcal {L}}'(m)). \end{aligned}$$

Take $m=0$, we get $D_{ A } +\delta _{ A } =0$ and $D_{ M } +\delta _{ M }=0.$ Therefore, $h={\mathcal {K}}+ {\mathcal {K}}'$. We need to show $[{\mathcal {K}}(m),a]=0$. Indeed, $[a, {\mathcal {K}}(m)]=[a,h(m)]-[a,{\mathcal {K}}'(m)]=h([a,m])-{\mathcal {K}}'([a,m])={\mathcal {K}}([a, m])=0$.

$\Leftarrow .$ Let $f:A < imes M \longrightarrow A < imes M$ be a generalized Lie d-derivation.

By hypothesis, $h_1$ is a sum of an A-homomorphism $\delta$ and a commuting map $\beta$. Also, $f_M$ is a sum of a generalized derivation $f_1$ and a central map which vanishes on all commutators $f_2$. We show that ($\beta$,0) is a central map which vanishes on all commutators. For that, we have

$$\begin{aligned} \beta ([a,m])= & {} (h_1-\delta )([a,m]) \\= & {} h_1([a,m])-\delta ([a,m]) \\= & {} [a,h_1(m)]-[a,\delta (m)]\\= & {} [a,\beta (m)] \\= & {} 0 \end{aligned}$$

On the other hand, it is clear that the linear map $(a,m)\longmapsto (f_A(a),h_2(m))$ is a generalized Lie derivation on $A < imes M$. Then, by (3), it can be written as the sum of a central map which vanishes on all commutators $\Theta$ and a generalized derivation $\Delta$. Then, $f(a,m)=((\delta (a),f_2(a))+\Delta (a,m))+((\beta (a) ,f_1(a))+ \Theta (a,m))$, where, using Lemmas 4.1, 4.2 and  4.3, $(a,m)\longmapsto (\delta (a),f_1(a))+\Delta (a,m)$ is a generalized derivation and $(a,m)\longmapsto (\beta (a) ,f_2(a))+ \Theta (a,m)$ is a central map which vanishes on all commutators. $\square$

Now, we turn to our fourth aim. We study generalized Lie derivations on $A < imes M$ when there exists a nontrivial idempotent e in A that satisfies $eme'=m$ for all $m\in M$ (where $e'=1-e$). A general study has given in Theorem 3.3. Note that, using Remark 2, A has the generalized Lie derivation property under the conditions of Theorem 3.3. In this section, we show that the trivial extension algebra requires mild conditions so that it has the generalized Lie derivation property. To this end, we need some lemmas. First recall that the existence of the above idempotent implies the following nice properties which will be used without explicit mention (see also the remark given before [11, Theorem 2.2]).

Lemma 4.5

([4], Proposition 2.5) Consider a non-trivial idempotent e of an algebra S and set $e'=1-e$. For every S-bimodule N, the following assertions are equivalent.

• 1. For every $m\in N$, $eme'=m$.

• 2. For every $m\in N$, $e'm =0=me$.

• 3. For every $m\in N$, $em =m=me'$.

• 4. For every $m\in N$ and $a\in S$, $am=eaem$ and $ma=me'ae'$.

Lemma 4.6

Suppose there exists a nontrivial idempotent e in A such that $eme'=m$ for all $m\in M,$ where $e'=1-e$. Then, for every generalized Lie derivation $f_{M}$, the following conditions holds:

• 1. $2f_{M}(e'ae)=0$.

• 2. $f_{M}(e)=-d_{M}(e')$.

Proof

• 1. Since $f_{M}$ is a generalized Lie derivation, we have

  $$\begin{aligned} f_{M}(e'ae)= & {} f_{M}([e', e'ae]) \\= & {} f_{M}(e')e'ae- f_{M}(e'ae)e'+e'd_{M}(e'ae)-e'aed_{M}(e') \\= & {} -f_{M}(e'ae). \end{aligned}$$

• 2. We have $f_{M}([a, b]) = f_{M}(a)b-f_{M}(b)a+ad_{M}(b)-bd_{M}(a)$. Then, for $a=e'$ and $b=e$, we get $0 = f_{M}(e')e-f_{M}(e)e'+e'd_{M}(e)-ed_{M}(e')$. Therefore, $f_{M}(e)=-d_{M}(e')$.

$\square$

Lemma 4.7

Assume that the A-bimodule M is 2-torsion free. Let $f_{M}:A\rightarrow M$ be a generalized Lie derivation with an associated linear map $d_M$. If there exists a nontrivial idempotent e in A such that $eme'=m$ for all $m\in M$ (where $e'=1-e$), then $f_{M}$ is a generalized derivation.

Proof

Let us consider the linear map $I_{M}$ defined by $I_{M}(a)=f_{M}(eae)+f_{M}(e'ae')$ for all $a\in A$. We show that $I_{M}$ is a generalized inner derivation. We have $0=f_{M}(eae)a-ad_{M}(e'ae')$. Then, taking $a=eae+e'$, we get $f_{M}(eae)=ad_{M}(e')$. We have $0=f_{M}(e'ae')e'-f_{M}(e')e'ae'+e'ae'd_{M}(e')-e'd_{M}(e'ae')$. Then, $f_{M}(e'ae')= f_{M}(e')a$. Hence, $I_{M}(a)=ad_{M}(e')+f_{M}(e')a$.

It remains to show that $g_{M}$ defined by $a\longmapsto f_{M}(eae')$ is a generalized derivation. We have

$$\begin{aligned} g_{M}(ab)= & {} f_{M}(eabe') \\= & {} f_{M}(eaebe')+f_{M}(eae'be') \\= & {} f_{M}([ea,ebe'])+f_{M}([ea,e'be']) \\= & {} ad_{M}(ebe')+ f_{M}(ea)e'be'+ead_{M}(e'be')\\= & {} ad_{M}(ebe') +f_{M}(eae')e'be'+f_{M}(eae)e'be'+ead_{M}(e'be')\\= & {} ad_{M}(ebe') +f_{M}(eae')e'be'+ad_{M}(e')e'be'-ead_{M}(e')b \\= & {} af_{M}(ebe') +f_{M}(eae')e'be' \\= & {} g_{M}(a)b+ag_{M}(b). \end{aligned}$$

Hence, with Lemma 4.6, the proof is complete. $\square$

Lemma 4.8

([3]) Let $h:M\rightarrow A$ be a linear map such that $h([a, m])=[a, h(m)]$ for all $a \in A, m \in M$. If there exists a nontrivial idempotent e in A such that $eme'=m$ for all $m\in M$ (where $e'=1-e$), then the map $\beta : m\longmapsto e'h(m)e$ satisfies the following assertions:

• 1. $2\beta (am)=0$ and $2\beta (ma)=0$ for all $a\in A$ and $m\in M$.

• 2. $\beta$ is an A-anti-homomorphism.

Proof

• 1. We have, for $a\in A$ and $m\in M$,

  $$\begin{aligned} \beta (ma)= & {} e'h(ma)e \\= & {} e'h([e,ma])e \\= & {} e'[e,h(ma)]e \\= & {} -e'h(ma)e \\= & {} -\beta (ma). \end{aligned}$$

  Similarly, we have

  $$\begin{aligned} \beta (am)= & {} e'h(am)e \\= & {} -e'h([e',am])e \\= & {} - e'[e',h(am)]e \\= & {} -e'h(am)e \\= & {} -\beta (am). \end{aligned}$$

• 2. We claim that $\beta$ is an A-antihomomorphism. Let $a \in A$ and $m \in M$. We have $\beta (am)=e'h(am)e=e'h([ea,m])e= -e'h(m)eae=-\beta (m)ae$. By (1), $\beta (am)=\beta (m)ae$. On the other hand, we have $\beta (m)ae'=e'h(m)eae'=e'[ h(m),eae']e'=e'h([ m,eae'])e'=0$, as desired. Similarly, we prove that $\beta (ma)=a\beta (m)$. $\square$

The following lemma shows that also the second condition of Theorem 4.4 holds true.

Lemma 4.9

([3]) Assume that A is 2-torsion free. Let $h:M\rightarrow A$ be a linear map such that $h([a, m])=[a, h(m)]$ for all $a \in A$ and $m \in M$. If there exists a nontrivial idempotent e in A such that $eme'=m$ for all $m\in M$ (where $e'=1-e$), then h is a sum of an A-homomorphism and a commuting map.

Proof

First note that $h(m)=h(em)=h([e , m])=[e,h(m)]=eh(m)-h(m)e$. Then, $eh(m)e=0$. Similarly, we get $e'h(m)e'=0$. This shows that $h=\delta + \beta$, where $\delta$ and $\beta$ are defined by $\delta (m)=eh(m)e'$ and $\beta (m)=e'h(m)e$. We claim that $\delta$ is an A-homomorphism. We have

$$\begin{aligned} \delta (am)=eh(am)e'=eh([ae, m])e'=eaeh(m)e'=ea\delta (m). \end{aligned}$$

But $e'a\delta (m)=e'aeh(m)e'=e'[e'ae, h(m)]e'=e'h([e'ae, m])e'=0$, as desired. Similarly, we can show that $\delta (ma)=\delta (m)a$. It remains to prove that $\beta$ is a commuting map. By Lemma 4.8, we have $[a,\beta (m)]=a\beta (m)-\beta (m)a=\beta (ma)-\beta (am)$. Since A is 2-torsion free and by Lemma 4.8, we get $[a,\beta (m)]=0$, as desired. $\square$

Lemma 4.10

Assume that there exists a nontrivial idempotent e in A such that $eme'=m$ for all $m\in M$ (where $e'=1-e$). Let f be a generalized Lie d-derivation of type $\Delta$. If $f_A$ is a generalized $d_{A}$-derivation, then the following assertions hold:

• 1. $h_2(a m) = f_{A}(a) m + aS(m)$ and $h_2( ma) = mf_{A}(a) +S(m)a$ for all $a \in A$ and $m\in M$.

• 2. $mf_A(e')=f_A(e)m$ for all $m\in M$.

• 3. $h_2( ma) = h_2(m)a + md_{A}(a)$ for all $a \in A$.

Consequently, the generalized Lie d-derivation f is a generalized d-derivation.

Proof

• 1. First we prove that $d_{A}(e)m=0=md_{A}(e)$.

  We have $f_A(e)=f_A(e)e+ed_A(e)$. Then, $ed_{A}(e)e=0$, hence $d_{A}(e)m=ed_{A}(e)em=0$. On the other hand, $md_{A}(e)=m(d_{A}(e)e+ed_{A}(e))=0$. Now, since f is a Lie generalized d-derivation, we have

  $$\begin{aligned} h_2(am)= & {} h_2([ae,m]) \\= & {} f_A(ae)m-h_2(m)ae+aeS(m)-md_{A}(ae) \\= & {} (f_A(a)e+ad_{A}(e))m+aS(m)-m(d_{A}(a)e+ad_{A}(e)) \\= & {} f_A(a)m+ad_A(e)m-mad_A(e)+aS(m) \\= & {} f_A(a)m+aS(m). \end{aligned}$$

  Similarly we get $h_2( ma) = mf_{A}(a) +S(m)a$ for all $a \in A$ and $m\in M$.

• 2. For every $m\in M$, $mf_A(e')+S(m)e'=h_2(me')=h_2(em)=f_A(e)m+eS(m)$. Then, $mf_A(e')=f_A(e)m$.

• 3. Since f is a generalized Lie d-derivation and by (1), we have, for all $a \in A$ and $m\in M$,

  $$\begin{aligned} h_2(ma)-h_2(m)a-md_A(a) = h_2(am)-f_{A}(a) m - aS(m) = 0. \end{aligned}$$

As desired. $\square$

Lemma 4.11

([3]) Assume that there exists a nontrivial idempotent e in A such that $eme'=m$ for all $m\in M,$ then the following assertions hold:

• 1. The center $Z(A < imes M)$ is described as follows:

  $$\begin{aligned} Z(A < imes M)= & {} \{(a,0), a\in A, eae\in Z(eAe), e'ae'\in Z(e'Ae'), eaem=me'ae', \\ neae= & {} e'ae'n, [a,x]=0, \text{ for } \text{ all } m\in eAe', n\in e'Ae, x\in M\}. \end{aligned}$$

• 2. For all $m\in M$, $[Z(A),m]=0$ if one of the following conditions holds:

  • (i) $Z(eAe)=\pi _{eAe}(Z(A < imes M))$ and $eAe'$ is faithful as a right $e'Ae'$-module.

  • (ii) $Z(e'Ae')=\pi _{e'Ae'}(Z(A < imes M))$ and $eAe'$ is faithful as a left eAe-module.

Proof

• 1. We have

  $$\begin{aligned} Z(A < imes M)= & {} \{(a,m);\; a\in Z(A), [b,m]=0=[a,y] \;\; \text{ for } \text{ all } \;\; b\in A,y\in M\} \\= & {} \pi _A(Z(A < imes M))\times \pi _M(Z(A < imes M)), \end{aligned}$$

  where $\pi _A: A < imes M \longrightarrow A$ and $\pi _M: A < imes M \longrightarrow M$ are the natural projections given by $\pi _A(a,m)=a$ and $\pi _M(a,m)=m$. On the other hand, by assumption, $eme'=m$ for all $m\in M.$ Then, $[e,m]=0$ implies $m=0$ for any $m\in M$. This leads to $\pi _M(a,m)=\{0\}$, and so

  $$\begin{aligned} Z(A < imes M)= & {} \{(a,0);\; a\in Z(A), [a,m]=0 \;\; \text{ for } \text{ all } \;\; a\in A,m\in M\}\end{aligned}$$

  (4.2)
  $$\begin{aligned}= & {} \pi _A(Z(A < imes M))\times \{0\}. \end{aligned}$$

  (4.3)

  As the algebra A enjoys the Pierce decomposition $A=eAe+eAe'+e'Ae+e'Ae'$, we get

  $$\begin{aligned}&Z(A)=\{a\in A, eae\in Z(eAe), e'ae'\in Z(e'Ae'), eaem=me'ae', neae=e'ae'n, \\&\quad \text{ for } \text{ all } \;\; m\in eAe', n\in e'Ae\}. \end{aligned}$$

  Using the last equality, we arrive at

  $$\begin{aligned} Z(A < imes M)= & {} \{(a,0),a\in A, eae\in Z(eAe), e'ae'\in Z(e'Ae'), eaem=me'ae', \\ neae= & {} e'ae'n, [a,x]=0, \;\; \text{ for } \text{ all } \;\; m\in eAe', n\in e'Ae, x\in M\}. \end{aligned}$$

• 2. (i) Let $a \in Z(A)$. Since $eae\in Z(eAe)$ and $Z(eAe)=\pi _{eAe}(Z(A < imes M))$, there exists an element $(a',0)\in Z(A < imes M)$ such that $eae=\pi _A(a',0)=ea'e$. It follows that $me'ae' = eaem = ea'em = me'a'e'$ for each $m \in eAe'$. Since $eAe'$ is a faithful right $e'Ae'$-module, we get $e'ae' = e'a'e'$, and so $a = eae+e'ae' =ea'e+e'a'e'=a'$. In particular, $(a,0)\in Z(A < imes M)$ and so $[a,x]=0$ for all $x\in M$, as claimed.

Similarly, we prove (ii). $\square$

The following result is the second main result in this section. It provides some sufficient conditions ensuring the generalized Lie derivation property for $A < imes M$. Compare it with [3, Theorem 3.15] and [11, Theorem 2.3].

Theorem 4.12

Assume that A and the A-bimodule M are 2-torsion free and there exists a nontrivial idempotent e in A such that $eme'=m$ for all $m\in M,$ then $A < imes M$ has the generalized Lie derivation property if the following two conditions are satisfied:

• 1. A has the generalized Lie derivation property.

• 2. One of the following conditions hold:

  • (i) $Z(eAe)=\pi _{eAe}(Z(A < imes M))$ and $eAe'$ is faithful as a right $e'Ae'$-module.

  • (ii) $Z(e'Ae')=\pi _{e'Ae'}(Z(A < imes M))$ and $eAe'$ is faithful as a left eAe-module.

Proof

Let f be a generalized Lie derivation on $A < imes M$ with a presentation as given in (4.1). By Lemma 4.7, $(0,f_M)$ is a generalized derivation. By Lemma 4.9, $(h_1,0)$ is a sum $(\delta ,0)+(\beta ,0)$, where $\delta$ is an A-homomorphism and $\beta$ is a commuting map. By hypothesis, A has the generalized Lie derivation property. Then, $(f_A,h_2)=(\Delta ,h_2)+(\tau _A,0)$. By Lemma 4.10, $(\Delta ,h_2)$ is a generalized derivation. Using Lemma 4.3, $(\tau _A,0)$ is a central map which vanishes on all commutators. Now, $[\tau _A(eae), m] =0 = [\tau _A(e'ae'), m]$ for all $a \in A$ and $m \in M$ (which follows from Lemma 4.11 and the condition (2)). Therefore, by Theorem 4.4, $A < imes M$ has the generalized Lie derivation property. $\square$