1 Introduction and preliminaries

Let \({\mathcal {R}}\) be a ring and let n be a positive integer. A mapping \(\Delta : {\mathcal {R}} \rightarrow {\mathcal {R}}\) is called a derivation of degree n or \(\{n\}\)-derivation if \(\Delta (xy) = \Delta (x) y^n + x^n \Delta (y)\) holds for all \(x, y \in {\mathcal {R}}\). Also, \(\Delta\) is called a Jordan derivation of degree n or Jordan \(\{n\}\)-derivation if \(\Delta (x^2) = \Delta (x) x^n + x^n \Delta (x)\) holds for all \(x \in {\mathcal {R}}\). In this paper, we provide an example of a Jordan derivation of degree n which is not a derivation of degree n.

By getting the idea from cubic derivations and quadratic derivations, we define a derivation of degree n from an algebra into a module. Before stating the results of this article, let us recall some basic definitions and set the notations which we use in what follows. An algebra \({\mathcal {A}}\) is called a domain if \({\mathcal {A}} \ne \{0\}\), and \(a = 0\) or \(b = 0\), whenever \(ab = 0\). A commutative domain is called an integral domain. Recall that the Jacobson radical of an algebra \({\mathcal {A}}\) is the intersection of all primitive ideals of \({\mathcal {A}}\) which is denoted by \(rad({\mathcal {A}})\). An algebra \({\mathcal {A}}\) is called semisimple if \(rad({\mathcal {A}}) = \{0\}\). A nonzero linear functional \(\varphi\) on an algebra \({\mathcal {A}}\) is called a character if \(\varphi (ab) = \varphi (a) \varphi (b)\) for every \(a,b \in {\mathcal {A}}\). The set of all characters on \({\mathcal {A}}\) is denoted by \(\Phi _{\mathcal {A}}\) and is called the character space of \({\mathcal {A}}\). We know that \(\ker \varphi\) is a maximal ideal of \({\mathcal {A}}\) for every \(\varphi \in \Phi _{\mathcal {A}}\) (see [4, Proposition 1.3.37]).

Let \({\mathcal {A}}\) be a complex algebra and let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule. Recall that a linear mapping \(\delta :{\mathcal {A}} \rightarrow {\mathcal {M}}\) is called a derivation if it satisfies the Leibnitz’s rule \(\delta (ab) = \delta (a)b + a\delta (b)\) for all \(a, b \in {\mathcal {A}}\). In [5], Eshaghi Gordji et al. introduced the concept of a cubic derivation. A mapping \(D: {\mathcal {A}} \rightarrow {\mathcal {M}}\) is called a cubic derivation if D is a cubic homogeneous mapping, that is \(D(\lambda a) = \lambda ^3 D(a)\) (\(\lambda \in {\mathbb {C}}\), \(a \in {\mathcal {A}}\)), and \(D(ab) = D(a) b^3 + a^3 D(b)\) for all \(a, b \in {\mathcal {A}}\). Also, a mapping \(d:{\mathcal {A}} \rightarrow {\mathcal {M}}\) is called a quadratic derivation if d is a quadratic homogeneous mapping, that is \(d(\lambda a) = \lambda ^2 d(a)\) (\(\lambda \in {\mathbb {C}}\), \(a \in {\mathcal {A}}\)), and \(d(ab) = d(a) b^2 + a^2 d(b)\) for all \(a, b \in {\mathcal {A}}\). The most papers to date have been focused on investigating stability of cubic derivations and quadratic derivations, see, e.g. [1, 5, 6, 9, 13, 17], and references therein.

In this paper, by getting the idea from the notions of cubic derivation and quadratic derivation, we define the notion of derivation of degree n on algebras, where n is a positive integer. In what follows, let \({\mathcal {A}}\) be a complex algebra, let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule and let n be a positive integer. A mapping \(\Delta : {\mathcal {A}} \rightarrow {\mathcal {M}}\) is called a derivation of degree n or \(\{n\}\)-derivation if it satisfies both the equations \(\Delta (ab) = \Delta (a) b^n + a^n \Delta (b)\) and \(\Delta (\lambda a) = \lambda ^n \Delta (a)\) for all \(a, b \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\).

Now let us to give a background about the image of derivations. The image of derivations has a fairly long history and so far, many authors have studied the image of derivations, see, e.g. [2, 3, 7, 10,11,12, 14,15,16] and references therein. As a pioneering work, Singer and Wermer [14] achieved a fundamental result which started investigation into the image of derivations on Banach algebras. The so-called Singer-Wermer theorem, which is a classical theorem of complex Banach algebra theory, states that every continuous derivation on a commutative Banach algebra maps the algebra into its Jacobson radical, and Thomas [15] proved that the Singer-Wermer theorem remains true without assuming the continuity of the derivation.

One of our aims in this research is to prove some results similar to Singer- Wermer theorem and Thomas theorem for derivations of degree n. In this regard, we first prove the following theorem which has been motivated by [7]:

Let \({\mathcal {A}}\) be a unital integral domain and let \(\Delta : {\mathcal {A}} \rightarrow {\mathcal {A}}\) be an \(\{n\}\)-derivation such that its rank is at most one. Then \(\Delta\) is identically zero. Using this result, it is proved that if \({\mathcal {A}}\) is a unital algebra and \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) is an \(\{n\}\)-derivation such that \(\Delta (a) - \Delta (b) \in ker \varphi\) whenever \(a - b \in ker \varphi\) for every \(a, b \in {\mathcal {A}}\) and every \(\varphi \in \Phi _{\mathcal {A}}\), then \(\Delta ({\mathcal {A}}) \subseteq \bigcap _{ \varphi \in \Phi _{\mathcal {A}}} \ker \varphi\). If \({\mathcal {A}}\) is also commutative, then \(\Delta ({\mathcal {A}}) \subseteq rad({\mathcal {A}})\). In this regard, we provide an example of an \(\{n\}\)-derivation on an algebra \({\mathfrak {A}}\) mapping the algebra into the intersection of all characters of \({\mathfrak {A}}\). In addition, we prove that if \({\mathcal {A}}\) is a unital, commutative Banach algebra and \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) is an additive \(\{n\}\)-derivation, then \(\Delta ({\mathcal {A}}) \subseteq rad({\mathcal {A}})\). As another result in this regard, we prove that every \(\{n\}\)-derivation on finite dimensional algebras is identically zero under certain conditions. Indeed, we establish the following result. Let m be a positive integer and let \({\mathcal {A}}\) be an m-dimensional unital algebra with the basis \({\mathfrak {B}} = \{{\mathfrak {b}}_1,\ldots , {\mathfrak {b}}_m\}\). Furthermore, suppose that for every integer k, \(1\le k \le m\), an ideal \({\mathfrak {X}}_{k}\) generated by \({\mathfrak {B}}-\{{\mathfrak {b}}_k\}\) is a proper subset of \({\mathcal {A}}\). If \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) is an \(\{n\}\)-derivation such that \(\Delta (a) - \Delta (b) \in {\mathfrak {X}}_k\) whenever \(a - b \in {\mathfrak {X}}_k\) for every \(a, b \in {\mathcal {A}}\) and \(1\le k \le m\), then \(\Delta\) is identically zero.

Another objective of this paper is to characterize \(\{n\}\)-derivations, \(\{n\}\)-generalized derivations and \(\{n\}\)-ternary derivations on algebras. First, we introduce these notions. A mapping \(f:{\mathcal {A}} \rightarrow {\mathcal {M}}\) is called a generalized derivation of degree n or an \(\{n\}\)-generalized derivation if there exists a mapping \(d:{\mathcal {A}} \rightarrow {\mathcal {M}}\) such that

$$\begin{aligned}&f(ab) = f(a) b^n + a^n d(b), \\&f(\lambda a) = \lambda ^n f(a), \end{aligned}$$

for all \(a, b \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). In this case, d is called an associated mapping of f.

A ternary derivation of degree n is defined as follows. A ternary derivation of degree n or an \(\{n\}\)-ternary derivation is a triple of mappings \((d_1, d_2, d_3)\) from \({\mathcal {A}}\) into \({\mathcal {M}}\) such that

$$\begin{aligned}&d_1(ab) = d_2(a)b^n + a^n d_3(b), \\&d_1(\lambda a) = \lambda ^n d_1(a), \end{aligned}$$

for all \(a, b \in {\mathcal {A}}\), \(\lambda \in {\mathbb {C}}\).

For instance, we establish the result below concerning the characterization of \(\{n\}\)-generalized derivations. Let \({\mathcal {A}}\) be a unital algebra with the identity element \({{\textbf {e}}}\), let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule and let \(f:{\mathcal {A}} \rightarrow {\mathcal {M}}\) be an additive generalized \(\{n\}\)-derivation with an associated mapping \(d:{\mathcal {A}} \rightarrow {\mathcal {M}}\) such that \(d(2 {{\textbf {e}}}) = 2 d({{\textbf {e}}})\). Then either f is a nonzero linear generalized derivation with the associated linear derivation d or f and d are identically zero.

A theorem similar to the above result is presented for the \(\{n\}\)-ternary derivations.

2 Definitions and examples

In this section, without further mention, \({{\textbf {e}}}\) denotes the identity of any unital ring or algebra. We begin this section with the following definition.

Definition 1

Let \({\mathcal {R}}\) be a ring and let n be a positive integer. A mapping \(\Delta : {\mathcal {R}} \rightarrow {\mathcal {R}}\) is called a derivation of degree n if

$$\begin{aligned} \Delta (xy) = \Delta (x) y^n + x^n \Delta (y) \end{aligned}$$

holds for all \(x, y \in {\mathcal {R}}\). Also, \(\Delta\) is called a Jordan derivation of degree n if it satisfies

$$\begin{aligned} \Delta (x^2) = \Delta (x) x^n + x^n \Delta (x) \end{aligned}$$

for all \(x \in {\mathcal {R}}\).

Obviously, if \(\Delta\) is a Jordan derivation of degree n on \({\mathcal {R}}\), then \(\Delta (0) = 0\). Also, if \({\mathcal {R}}\) is unital with the identity element \({{\textbf {e}}}\), then \(\Delta ({{\textbf {e}}}) = 0\). It is clear that every derivation of degree n is a Jordan derivation of degree n, but the converse is, in general, not true. In the following, we present a Jordan derivation of degree n which is not a derivation of degree n.

Example 2

Let \({\mathcal {R}}\) be a ring such that \(x^{4} = 0\) for all \(x \in {\mathcal {R}}\), but the product of some nonzero elements of \({\mathcal {R}}\) is nonzero. Let

$$\begin{aligned} {{\mathfrak {R}}} = \left\{ \left[ \begin{array}{ccc} 0 &{} x &{} y\\ 0 &{} 0 &{} x\\ 0 &{} 0 &{} 0 \end{array}\right] \ : \ x, y \in {\mathcal {R}}\right\} \end{aligned}$$

Define \(\Delta : {{\mathfrak {R}}} \rightarrow {\mathfrak {R}}\) by

$$\begin{aligned} \Delta \left( \left[ \begin{array}{ccc} 0 &{} x &{} y\\ 0 &{} 0 &{} x\\ 0 &{} 0 &{} 0 \end{array}\right] \right) = \left[ \begin{array}{ccc} 0 &{} 0 &{} y^2\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array}\right] . \end{aligned}$$

For any \(A = \left[ \begin{array}{ccc} 0 &{} x &{} y\\ 0 &{} 0 &{} x\\ 0 &{} 0 &{} 0 \end{array}\right] \in {\mathfrak {R}}\), we have

$$\begin{aligned} \Delta (A^2) = \Delta \left( \left[ \begin{array}{ccc} 0 &{} 0 &{} x^2\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array}\right] \right) = \left[ \begin{array}{ccc} 0 &{} 0 &{} x^4\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array}\right] = 0. \end{aligned}$$

A straightforward verification shows that

$$\begin{aligned} \Delta (A)A^n + A^n \Delta (A) = 0, \end{aligned}$$

for all \(A \in {\mathfrak {R}}\) and all \(n \in {\mathbb {N}}\). We see that \(\Delta\) is a Jordan derivation of degree n for any \(n \in {\mathbb {N}}\). Also, it is easy to see that \(\Delta (A)B^n + A^n \Delta (B) = 0\) for all \(A, B \in {\mathfrak {R}}\) and all \(n \in {\mathbb {N}}\), but \(\Delta (AB) \ne 0\) for some \(A, B \in {\mathfrak {R}}\). It means that \(\Delta\) is not a derivation of degree n for all \(n \in {\mathbb {N}}\).

In the rest of this article, we consider derivations of degree n from algebras into modules as follows.

Definition 3

Let \({\mathcal {A}}\) be a complex algebra, let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule and let n be a positive integer. A mapping \(\Delta : {\mathcal {A}} \rightarrow {\mathcal {M}}\) is called a derivation of degree n if it satisfies both of the following equations:

$$\begin{aligned} \Delta (ab)&= \Delta (a) b^n + a^n \Delta (b), \\ \Delta (\lambda a)&= \lambda ^n \Delta (a), \end{aligned}$$

for all \(a, b \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\).

Example 4

Let \({\mathcal {A}}\) an algebra, let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule, let n be a positive integer and let \(x_0\) be an element of \({\mathcal {M}}\) satisfying

$$\begin{aligned} x_0 \big [(ab)^n - a^n b^n \big ] = \big [(ab)^n - a^n b^n \big ]x_0 \end{aligned}$$

for all \(a, b \in {\mathcal {A}}\). Define a mapping \(\Delta : {\mathcal {A}} \rightarrow {\mathcal {M}}\) by \(\Delta (a) = a^n x_0 - x_0 a^n\) for any \(a \in {\mathcal {A}}\). It is routine to see that \(\Delta (ab) = \Delta (a)b^n + a^n \Delta (b)\) and \(\Delta (\lambda a) = \lambda ^n \Delta (a)\) for all \(a, b \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). This means that \(\Delta\) is an \(\{n\}\)-derivation. We call such mapping inner derivation of degree n or inner \(\{n\}\)-derivation.

Example 5

Let \({\mathcal {A}}\) be a commutative algebra, let n be an arbitrary positive integer and let

$$\begin{aligned} {\mathfrak {A}} = \left\{ \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} c &{} 0\\ 0 &{} 0 &{} e \end{array}\right] \ : \ a, b, c, e \in {\mathcal {A}}\right\} \end{aligned}$$

It is clear that \({\mathfrak {A}}\) is a non-commutative algebra. Define \(\Delta : {\mathfrak {A}} \rightarrow {\mathfrak {A}}\) by

$$\begin{aligned} \Delta \left( \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} c &{} 0\\ 0 &{} 0 &{} e \end{array}\right] \right) = \left[ \begin{array}{ccc} 0 &{} 0 &{} b^n\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array}\right] . \end{aligned}$$

It is easy to see that for any \(A = \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} c &{} 0\\ 0 &{} 0 &{} e \end{array}\right] \in {\mathfrak {A}}\) and any \(k \in {\mathbb {N}}\), we have

$$\begin{aligned} A^k = \left[ \begin{array}{ccc} 0 &{} ac^{k - 1} &{} be^k\\ 0 &{} c^k &{} 0\\ 0 &{} 0 &{} e^k \end{array}\right] \end{aligned}$$

One can easily get that \(\Delta (AB) = \Delta (A) B^n + A^n \Delta (B)\) and \(\Delta (\lambda A) = \lambda ^n \Delta (A)\) for all \(A, B \in {\mathfrak {A}}\) and all \(\lambda \in {\mathbb {C}}\), which means that \(\Delta\) is a derivation of degree n on \({\mathfrak {A}}\).

Example 6

Let \({\mathcal {A}}\) be an algebra, let n be an arbitrary positive integer and let

$$\begin{aligned} {\mathfrak {A}} = \left\{ \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} 0 &{} c\\ 0 &{} 0 &{} 0 \end{array}\right] \ : \ a, b, c \in {\mathcal {A}}\right\} \end{aligned}$$

Define \(\Delta : {\mathfrak {A}} \rightarrow {\mathfrak {A}}\) by

$$\begin{aligned} \Delta \left( \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} 0 &{} c\\ 0 &{} 0 &{} 0 \end{array}\right] \right) = \left[ \begin{array}{ccc} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} c^n\\ 0 &{} 0 &{} 0 \end{array}\right] . \end{aligned}$$

It is straightforward to see that \(\Delta (AB) = \Delta (A) B^n + A^n \Delta (B)\) and \(\Delta (\lambda A) = \lambda ^n \Delta (A)\) for all \(A, B \in {\mathfrak {A}}\) and all \(\lambda \in {\mathbb {C}}\), which means that \(\Delta\) is a derivation of degree n on \({\mathfrak {A}}\).

Definition 7

Let \({\mathcal {A}}\) be an algebra and let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule. A mapping \(f:{\mathcal {A}} \rightarrow {\mathcal {M}}\) is called a generalized derivation of degree n or an \(\{n\}\)-generalized derivation if there exists a mapping \(d:{\mathcal {A}} \rightarrow {\mathcal {M}}\) such that

$$\begin{aligned} f(ab)&= f(a) b^n + a^n d(b), \\ f(\lambda a)&= \lambda ^n f(a), \end{aligned}$$

for all \(a, b \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). In this case, d is called an associated map of f.

Example 8

Let \({\mathcal {A}}\) an algebra, let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule, let n be a positive integer and let \(x_0\) and \(y_0\) be two elements of \({\mathcal {M}}\) satisfying

$$\begin{aligned} y_0 \big [(ab)^n - a^n b^n \big ] = \big [(ab)^n - a^n b^n \big ]x_0, \end{aligned}$$

for all \(a, b \in {\mathcal {A}}\). Define the mappings \(f, d: {\mathcal {A}} \rightarrow {\mathcal {M}}\) by \(f(a) = a^n x_0 - y_0 a^n\) and \(d(a) = a^n x_0 - x_0 a^n\) for any \(a \in {\mathcal {A}}\). It is routine to see that \(f(ab) = f(a)b^n + a^n d(b)\) and \(f(\lambda a) = \lambda ^n f(a)\) for all \(a, b \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). This means that f is an \(\{n\}\)-generalized derivation with the associated mapping d. We call such mapping inner generalized derivation of degree n or inner \(\{n\}\)-generalized derivation.

In the following, we define a ternary derivation of degree \(\{n\}\).

Definition 9

Let \({\mathcal {A}}\) be an algebra and let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule. A ternary derivation of degree n or an \(\{n\}\)-ternary derivation is a triple of mappings \((d_1, d_2, d_3)\) from \({\mathcal {A}}\) into \({\mathcal {M}}\) such that

$$\begin{aligned} d_1(ab)&= d_2(a)b^n + a^n d_3(b), \\ d_1(\lambda a)&= \lambda ^n d_1(a), \end{aligned}$$

for all \(a, b \in {\mathcal {A}}\), \(\lambda \in {\mathbb {C}}\).

3 Results and proofs

Let \({\mathcal {A}}\) and \({\mathcal {B}}\) be two algebras over a field \(\mathbb {F}\). Throughout this section, a mapping \(D:{\mathcal {A}} \rightarrow {\mathcal {B}}\) is called a rank-one mapping if there exist a nonzero element \({\mathfrak {b}}\) of \({\mathcal {B}}\) and a functional \(\mu :{\mathcal {A}} \rightarrow \mathbb {F}\) such that \(D(a) = \mu (a){\mathfrak {b}}\) for all \(a \in {\mathcal {A}}\).

We begin our results with the following theorem.

Theorem 10

Let \({\mathcal {A}}\) be a unital integral domain and let \(\Delta : {\mathcal {A}} \rightarrow {\mathcal {A}}\) be a derivation of degree n such that its rank is at most one. Then \(\Delta\) is identically zero.

Proof

Let \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) be a derivation of degree n such that its rank is at most one. We are going to show that \(\Delta ({\mathcal {A}}) = \{0\}\). Suppose that \(\Delta\) is a rank-one mapping. So there exist a nonzero element \({\mathfrak {c}}\) of \({\mathcal {A}}\) and a functional \(\mu :{\mathcal {A}} \rightarrow {\mathbb {C}}\) such that \(\Delta (a) = \mu (a){\mathfrak {c}}\) for all \(a \in {\mathcal {A}}\). To obtain a contradiction, suppose there exists a nonzero element \({\mathfrak {a}} \in {\mathcal {A}}\) such that \(\Delta ({\mathfrak {a}}) \ne 0\). It is clear that \(\mu ({\mathfrak {a}}) \ne 0\). We observe two cases for \(\Delta ({\mathfrak {c}})\).

Case 1.:

\(\Delta ({\mathfrak {c}}) = 0\). In this case, we have \(\mu ({\mathfrak {c}}) {\mathfrak {c}} = 0\) and it implies that \(\mu ({\mathfrak {c}}) = 0\). We have the following expressions:

$$\begin{aligned} \mu ({\mathfrak {a}}^2) {\mathfrak {c}}&= \Delta ({\mathfrak {a}}^2) \\&= \Delta ({\mathfrak {a}}) {\mathfrak {a}}^n + {\mathfrak {a}}^n \Delta ({\mathfrak {a}}) \\&= 2 {\mathfrak {a}}^n \Delta ({\mathfrak {a}}) \\&= 2 {\mathfrak {a}}^n \mu ({\mathfrak {a}}) {\mathfrak {c}} \\&= 2 \mu ({\mathfrak {a}}) {\mathfrak {a}}^n {\mathfrak {c}}. \end{aligned}$$

Since we are assuming that \(\Delta ({\mathfrak {c}}) = 0\), we have

$$\begin{aligned} 0&= (\mu ({\mathfrak {a}}^2))^n \Delta ({\mathfrak {c}}) = \Delta (\mu ({\mathfrak {a}}^2){\mathfrak {c}}) = \Delta (2 \mu ({\mathfrak {a}}){\mathfrak {a}}^n {\mathfrak {c}}) \\&= 2^n (\mu ({\mathfrak {a}}))^n [\Delta ({\mathfrak {a}}^n) {\mathfrak {c}}^n + {\mathfrak {a}}^{n^2} \Delta ({\mathfrak {c}})] \\&= 2^n (\mu ({\mathfrak {a}}))^n \Delta ({\mathfrak {a}}^n) {\mathfrak {c}}^n \end{aligned}$$

Since \({\mathcal {A}}\) is a domain and \(\mu ({\mathfrak {a}})\) and \({\mathfrak {c}}\) are nonzero, we get that \(\Delta ({\mathfrak {a}}^n) = 0\). Using induction, for any \(m \in {\mathbb {N}}\), one can easily prove that

$$\begin{aligned} \Delta (a^m) = \Sigma _{k = 1}^{m}a^{(k - 1)n}\Delta (a)a^{(m - k)n} \end{aligned}$$

in which \(a^0 = {{\textbf {e}}}\). So we have

$$\begin{aligned} 0&= \Delta ({\mathfrak {a}}^n) = \Delta ({\mathfrak {a}}^{n - 1}{\mathfrak {a}}) \\&= \Delta ({\mathfrak {a}}^{n - 1}){\mathfrak {a}}^n + {\mathfrak {a}}^{n(n - 1)}\Delta ({\mathfrak {a}}) \\&= \Big [\Sigma _{k = 1}^{n - 1}{\mathfrak {a}}^{(k - 1)n}\Delta ({\mathfrak {a}}){\mathfrak {a}}^{(n- 1 - k)n}\Big ] {\mathfrak {a}}^n + {\mathfrak {a}}^{n(n - 1)}\Delta ({\mathfrak {a}}) \\&= \Sigma _{k = 1}^{n - 1}\Big [\Delta ({\mathfrak {a}}){\mathfrak {a}}^{n^2 - n}\Big ] + {\mathfrak {a}}^{n^2 - n}\Delta ({\mathfrak {a}}) \\&= n \Delta ({\mathfrak {a}}) {\mathfrak {a}}^{n^2 - n}, \end{aligned}$$

which implies that \(\Delta ({\mathfrak {a}}) = 0\), a contradiction.

Case 2.:

\(\Delta ({\mathfrak {c}}) \ne 0\). In this case, we have \(\mu ({\mathfrak {c}}) \ne 0\). Now look at the following statements:

$$\begin{aligned} \mu ({\mathfrak {c}}^2){\mathfrak {c}} = \Delta ({\mathfrak {c}}^2) = \Delta ({\mathfrak {c}}){\mathfrak {c}}^n + {\mathfrak {c}}^n \Delta ({\mathfrak {c}}) = 2 {\mathfrak {c}}^{n}\Delta ({\mathfrak {c}}) = 2 \mu ({\mathfrak {c}}){\mathfrak {c}}^{n + 1} \end{aligned}$$
(1)

If \(\mu ({\mathfrak {c}}^2) = 0\), then it follows from (1) that either \(\mu ({\mathfrak {c}}) = 0\) or \({\mathfrak {c}} = 0\), and we know that both of them are nonzero. So \(\mu ({\mathfrak {c}}^2) \ne 0\). Putting \(\frac{\mu ({\mathfrak {c}}^2)}{2 \mu ({\mathfrak {c}})} = \alpha\) in (1), we have \({\mathfrak {c}}({\mathfrak {c}}^n - \alpha {{\textbf {e}}}) = {\mathfrak {c}}^{n+1} - \alpha {\mathfrak {c}} = 0\). In view of this assumption that \({\mathcal {A}}\) is a domain, we infer that \({\mathfrak {c}} = 0\), a contradiction, or \({\mathfrak {c}}^n = \alpha {{\textbf {e}}}\). So we have

$$\begin{aligned} 0&= \alpha ^n \Delta ({{\textbf {e}}}) = \Delta (\alpha {{\textbf {e}}}) = \Delta ({\mathfrak {c}}^n) = \Delta ({\mathfrak {c}}^{n - 1}{\mathfrak {c}})\\&= \Delta ({\mathfrak {c}}^{n - 1}){\mathfrak {c}}^n + {\mathfrak {c}}^{n(n - 1)}\Delta ({\mathfrak {c}}) \\&= \Big [\Sigma _{k = 1}^{n - 1}{\mathfrak {c}}^{(k - 1)n}\Delta ({\mathfrak {c}}){\mathfrak {c}}^{(n- 1 - k)n}\Big ] {\mathfrak {c}}^n + {\mathfrak {c}}^{n(n - 1)}\Delta ({\mathfrak {c}}) \\&= \Sigma _{k = 1}^{n - 1}\Big [\Delta ({\mathfrak {c}}){\mathfrak {c}}^{n^2 - n}\Big ] + {\mathfrak {c}}^{n^2 - n}\Delta ({\mathfrak {c}}) \\&= n \Delta ({\mathfrak {c}}) {\mathfrak {c}}^{n^2 - n}. \end{aligned}$$

Reusing the assumption that \({\mathcal {A}}\) is a domain, we get that \({\mathfrak {c}} = 0\) or \(\Delta ({\mathfrak {c}}) = 0\), which these are contradictions. It is observed that both Cases 1 and 2 lead to contradictions. Therefore, there is no element \({\mathfrak {a}}\) of \({\mathcal {A}}\) such that \(\Delta ({\mathfrak {a}}) \ne 0\), and consequently, \(\Delta\) must be zero. \(\square\)

In the following, we provide some examples that show that the conditions of Theorem 10 are not superfluous.

Example 11

  1. (i)

    Let n be a positive number. Define \(\Delta : {\mathbb {R}} \rightarrow {\mathbb {R}}\) by

    $$\begin{aligned} \Delta (a) = \left\{ \begin{array}{c l} a^{n} \ln (\mid a\mid ) &{} \ \hbox { }\ a \ne 0,\\ 0 &{} \hbox { }\ a = 0. \end{array} \right. \end{aligned}$$

    One can easily check that \(\Delta (ab) = \Delta (a)b^n + a^n \Delta (b)\) for all \(a, b \in {\mathbb {R}}\) and also it is clear that the rank of \(\Delta\) is at most one, but we observe that \(\Delta (\alpha a) \ne \alpha ^n \Delta (a)\) for some \(\alpha , a \in {\mathbb {R}}\). We see that \(\Delta\) is a nonzero mapping.

  2. (ii)

    In Example 5, considering \({\mathcal {A}} = {\mathbb {C}}\), we see that \(\Delta : {\mathfrak {A}} \rightarrow {\mathfrak {A}}\) defined by

    $$\begin{aligned} \Delta \left( \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} c &{} 0\\ 0 &{} 0 &{} e \end{array}\right] \right) = \left[ \begin{array}{ccc} 0 &{} 0 &{} b^n\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array}\right] = b^n \left[ \begin{array}{ccc} 0 &{} 0 &{} 1\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array}\right] \end{aligned}$$

    is a nonzero, rank one derivation of degree n. Note that \({\mathfrak {A}}\) is not an integral domain.

In the following theorem, we present some conditions under which every derivation of degree n on an algebra maps the algebra into its Jacobson radical.

Theorem 12

Let \({\mathcal {A}}\) be a unital algebra and let \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) be a derivation of degree n such that \(\Delta (a) - \Delta (b) \in ker \varphi\) whenever \(a - b \in ker \varphi\) for every \(a, b \in {\mathcal {A}}\) and every \(\varphi \in \Phi _{\mathcal {A}}\). In this case, \(\Delta ({\mathcal {A}}) \subseteq \bigcap _{ \varphi \in \Phi _{\mathcal {A}}} \ker \varphi\). If \({\mathcal {A}}\) is also commutative, then \(\Delta ({\mathcal {A}}) \subseteq rad({\mathcal {A}})\).

Proof

Let \(\varphi\) be an arbitrary character on \({\mathcal {A}}\). We define a mapping \(\Omega :\frac{{\mathcal {A}}}{\ker \varphi } \rightarrow \frac{{\mathcal {A}}}{\ker \varphi }\) by \(\Omega (a + \ker \varphi ) = \Delta (a) + \ker \varphi\) for every \(a \in {\mathcal {A}}\). \(\Omega\) is a derivation of degree n on the algebra \(\frac{{\mathcal {A}}}{\ker \varphi }\). It is clear that the algebra \(\frac{{\mathcal {A}}}{\ker \varphi }\) is a unital, integral domain and it follows from [4, Proposition 1.3.37] that \(\dim (\frac{{\mathcal {A}}}{\ker \varphi }) = 1\). So the rank of \(\Omega\) is at most one. Now, Theorem 10 yields that \(\Omega\) is identically zero, and it means that \(\Delta ({\mathcal {A}}) \subseteq \ker \varphi\). Since we are assuming \(\varphi\) is an arbitrary element of \(\Phi _{\mathcal {A}}\), \(\Delta ({\mathcal {A}}) \subseteq \bigcap _{ \varphi \in \Phi _{\mathcal {A}}} \ker \varphi\). It is obvious that if \({\mathcal {A}}\) is commutative, then \(\bigcap _{ \varphi \in \Phi _{\mathcal {A}}} \ker \varphi = rad({\mathcal {A}})\) (see [4]). Hence, we deduce that \(\Delta ({\mathcal {A}}) \subseteq rad({\mathcal {A}})\). \(\square\)

An immediate corollary of the previous theorem is as follows:

Corollary 13

Let \({\mathcal {A}}\) be a unital algebra such that \(\bigcap _{ \varphi \in \Phi _{\mathcal {A}}} \ker \varphi = \{0\}\) and let \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) be a derivation of degree n such that \(\Delta (a) - \Delta (b) \in ker \varphi\) whenever \(a - b \in ker \varphi\) for every \(a, b \in {\mathcal {A}}\) and every \(\varphi \in \Phi _{\mathcal {A}}\). Then \(\Delta\) is identically zero.

Proof

According to [8, Proposition 2.10], the algebra \({\mathcal {A}}\) is commutative and semisimple. Now the previous theorem gives the result. \(\square\)

Remark 14

In this remark, we show that the image of derivation of degree n presented in Example 5 is contained in \(\bigcap _{ \varphi \in \Phi _{\mathfrak {A}}} \ker \varphi\). Let \({\mathcal {A}}\) be a unital commutative Banach algebra and let

$$\begin{aligned} {\mathfrak {A}} = \left\{ \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} c &{} 0\\ 0 &{} 0 &{} e \end{array}\right] \ : \ a, b, c, e \in {\mathcal {A}}\right\} \end{aligned}$$

Note that \({\mathfrak {A}}\) is a non-commutative algebra. Since \({\mathcal {A}}\) is a unital commutative Banach algebra, it follows from [4, Theorem 2.3.1] that its character space is a non-empty set, i.e. \(\Phi _{\mathcal {A}} \ne \phi\). Let \(\varphi\) be a character of \({\mathcal {A}}\). We define \(\theta _\varphi :{\mathfrak {A}} \rightarrow {\mathbb {C}}\) by \(\theta _\varphi \left( \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} c &{} 0\\ 0 &{} 0 &{} e \end{array}\right] \right) = \varphi (c)\). It is clear that \(\theta _\varphi\) is a character on \({\mathfrak {A}}\) and it is easy to see that

$$\begin{aligned} ker (\theta _\varphi ) = \left[ \begin{array}{ccc} 0 &{} {\mathcal {A}} &{} {\mathcal {A}}\\ 0 &{} ker(\varphi ) &{} 0\\ 0 &{} 0 &{} {\mathcal {A}} \end{array}\right] = \left\{ \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} x &{} 0\\ 0 &{} 0 &{} e \end{array}\right] \ : \ a, b, e \in {\mathcal {A}}, x \in \ker (\varphi )\right\} . \end{aligned}$$

Also, if we define \(\theta _\varphi :{\mathfrak {A}} \rightarrow {\mathbb {C}}\) by

\(\theta _\varphi \left( \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} c &{} 0\\ 0 &{} 0 &{} e \end{array}\right] \right) = \varphi (e)\), then we deduce that \(\theta _\varphi\) is a character on \({\mathfrak {A}}\). It is easy to see that

$$\begin{aligned} ker (\theta _\varphi ) = \left[ \begin{array}{ccc} 0 &{} {\mathcal {A}} &{} {\mathcal {A}}\\ 0 &{} {\mathcal {A}} &{} 0\\ 0 &{} 0 &{} ker(\varphi ) \end{array}\right] = \left\{ \left[ \begin{array}{ccc} 0 &{} a &{} b\\ 0 &{} c &{} 0\\ 0 &{} 0 &{} z \end{array}\right] \ : \ a, b, c \in {\mathcal {A}}, z \in \ker (\varphi )\right\} . \end{aligned}$$

Therefore, \(\Phi _{\mathfrak {A}} = \{\theta _\varphi \ : \ \varphi \in \Phi _{\mathcal {A}}\}\). It is observed that \(\Delta ({\mathfrak {A}}) \subseteq \bigcap _{\varphi \in \Phi _{\mathfrak {A}}} \ker \theta _{\varphi }\).

In the next theorem, we prove that every derivation of degree n on a unital finite-dimensional algebra is identically zero under certain conditions. Let m be a positive integer and let \({\mathcal {A}}\) be an m-dimensional unital algebra with the basis \({\mathfrak {B}} = \{{\mathfrak {b}}_1, {\mathfrak {b}}_2,\ldots , {\mathfrak {b}}_m\}\).

Theorem 15

Suppose that for every integer k, \(1\le k \le m\), an ideal \({\mathfrak {X}}_{k}\) generated by \({\mathfrak {B}}-\{{\mathfrak {b}}_k\}\) is a proper subset of \({\mathcal {A}}\). Let \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) be a derivation of degree n such that \(\Delta (a) - \Delta (b) \in {\mathfrak {X}}_k\) whenever \(a - b \in {\mathfrak {X}}_k\) for every \(a, b \in {\mathcal {A}}\) and every \(1\le k \le m\). Then \(\Delta\) is identically zero.

Proof

It is clear that \(\dim (\frac{{\mathcal {A}}}{{\mathfrak {X}}_k}) = 1\) for every \(k \in \{1, ..., m\}\). We show that \({\mathfrak {X}}_k\) is a maximal ideal of \({\mathcal {A}}\) for each \(k \in \{1, ..., m\}\). If \({\mathfrak {X}}_k\) is not a maximal ideal of \({\mathcal {A}}\) for some k, \(1\le k \le m\), then there exists a maximal ideal \(\mathfrak {M}_{k}\) of \({\mathcal {A}}\) such that \({\mathfrak {X}}_k \subset \mathfrak {M}_{k} \subset {\mathcal {A}}\), and so \(m - 1 = \dim ({\mathfrak {X}}_k)< \dim (\mathfrak {M}_{k}) < m\), a contradiction. Hence, every \({\mathfrak {X}}_k\) is a maximal ideal of \({\mathcal {A}}\). Moreover, it follows from Proposition 1.3.37 and Corollary 1.4.38 of [4] that for every maximal ideal \({\mathfrak {X}}_k\) (\(1 \le k \le m\)) there exists a character \(\varphi _k \in \Phi _{\mathcal {A}}\) such \({\mathfrak {X}}_k = ker \varphi _k\). So the algebra \(\frac{{\mathcal {A}}}{{\mathfrak {X}}_k}\) is an integral domain. Now Theorem 10 yields that \(\Omega :\frac{{\mathcal {A}}}{{\mathfrak {X}}_k} \rightarrow \frac{{\mathcal {A}}}{{\mathfrak {X}}_k}\) defined by \(\Omega (a + {\mathfrak {X}}_k) = \Delta (a) + {\mathfrak {X}}_k\), which is a derivation of degree n, is identically zero. This means that \(\Delta ({\mathcal {A}}) \subseteq {\mathfrak {X}}_k\), for every \(k \in \{1, ..., m\}\), and so \(\Delta ({\mathcal {A}}) \subseteq \bigcap _{k = 1}^{n}{\mathfrak {X}}_k\). Now suppose that there is an element \({\mathfrak {a}}\) of \({\mathcal {A}}\) such that \(\Delta ({\mathfrak {a}}) \ne 0\). Since \({\mathfrak {B}} = \{{\mathfrak {b}}_1, ..., {\mathfrak {b}}_m\}\) is a basis for \({\mathcal {A}}\), there exist the complex numbers \(\mu _{i_{j}}\), and the elements \({\mathfrak {b}}_{i_j}\) of \({\mathfrak {B}}\) such that

$$\begin{aligned} \Delta ({\mathfrak {a}}) = \sum _{j = 1}^{r}\mu _{i_{j}} {\mathfrak {b}}_{i_j} = \mu _{i_{1}}{\mathfrak {b}}_{i_1} + \mu _{i_{2}}{\mathfrak {b}}_{i_2} + ... + \mu _{i_{r}}{\mathfrak {b}}_{i_r}, \ \ \ (r \le m). \end{aligned}$$

We know that \(\Delta ({\mathcal {A}}) \subseteq {\mathfrak {X}}_k\) for every \(k \in \{1, ... , m\}\). So we can assume that \(\Delta ({\mathcal {A}}) \subseteq {\mathfrak {X}}_{i_1} = {\mathfrak {B}} - \{{\mathfrak {b}}_{i_1}\}\). Thus, we have

$$\begin{aligned} \Delta ({\mathfrak {a}}) = \mu _{i_{1}}{\mathfrak {b}}_{i_1} + \mu _{i_{2}}{\mathfrak {b}}_{i_2} + ... + \mu _{i_{r}}{\mathfrak {b}}_{i_r} \in {\mathfrak {X}}_{i_1}. \end{aligned}$$

The previous equation asserts that \({\mathfrak {b}}_{i_1} \in {\mathfrak {X}}_{i_1}\), which is a contradiction. This contradiction proves our claim. \(\square\)

In the following, we are going to characterize \(\{n\}\)-derivations, \(\{n\}\)-generalized derivations and \(\{n\}\)-ternary derivations on algebras under certain conditions.

Theorem 16

Let \({\mathcal {A}}\) be a unital algebra, let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule and let \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {M}}\) be an additive \(\{n\}\)-derivation. Then either \(\Delta\) is a nonzero linear derivation or \(\Delta\) is identically zero.

Proof

Since \(\Delta\) is an additive mapping, \(\Delta (a(b + c)) = \Delta (ab) + \Delta (ac)\) for all \(a, b, c \in {\mathcal {A}}\). We have

$$\begin{aligned} \Delta (a(b + c)) = \Delta (a)(b +c)^n + a^n \Delta (b) + a^n \Delta (c). \end{aligned}$$
(2)

Also, we have

$$\begin{aligned} \Delta (ab) + \Delta (ac) = \Delta (a)b^n + a^n \Delta (b) + \Delta (a)c^n + a^n \Delta (c). \end{aligned}$$
(3)

Comparing (2) and (3), we get that

$$\begin{aligned} \Delta (a) \big [(b + c)^n - b^n - c^n \big ] = 0, \quad \ \ (a, b, c \in {\mathcal {A}}). \end{aligned}$$
(4)

Putting \(b = c = {{\textbf {e}}}\) in (4), we arrive at

$$\begin{aligned} (2^n - 2)\Delta (a)= 0, \ \ \ \ \ \ \ \ \ \ \ \ (a \in {\mathcal {A}}). \end{aligned}$$

It follows from the previous equation that either \(n = 1\), which means that \(\Delta\) is a nonzero linear derivation from \({\mathcal {A}}\) into \({\mathcal {M}}\) or \(\Delta\) is identically zero. By the way, in both cases \(\Delta\) is a derivation on \({\mathcal {A}}\). \(\square\)

Corollary 17

Let \({\mathcal {A}}\) be a unital, commutative Banach algebra and let \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) be an additive \(\{n\}\)-derivation for some \(n \in {\mathbb {N}}\). Then \(\Delta ({\mathcal {A}}) \subseteq rad({\mathcal {A}})\).

Proof

It follows from the previous theorem that \(\Delta\) is a derivation and now [15, Theorem 4.4] yields the required result. \(\square\)

Theorem 18

Let \({\mathcal {A}}\) be a unital algebra, let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule and let \(f:{\mathcal {A}} \rightarrow {\mathcal {M}}\) be a generalized \(\{n\}\)-derivation with an associated mapping \(d:{\mathcal {A}} \rightarrow {\mathcal {M}}\). Then d is an \(\{n\}\)-derivation if and only if \(f({{\textbf {e}}}) \big [(bc)^n - b^n c^n \big ] = 0\) for all \(b, c \in {\mathcal {A}}\).

Proof

For every \(a, b, c \in {\mathcal {A}}\), we have

$$\begin{aligned} f(abc) = f(a)(bc)^n + a^n d(bc). \end{aligned}$$

On the other hand, we have

$$\begin{aligned} f(abc) = f(ab)c^n + (ab)^n d(c) = f(a)b^n c^c + a^n d(b)c^n + (ab)^n d(c). \end{aligned}$$

Comparing the last two equations, we get that

$$\begin{aligned} f(a) \big [(bc)^n - b^n c^n \big ] = a^n \big [d(b)c^n - d(bc) \big ] + (ab)^n d(c). \end{aligned}$$
(5)

Putting \(a = {{\textbf {e}}}\) in (5), we have

$$\begin{aligned} f({{\textbf {e}}}) \big [(bc)^n - b^n c^n \big ] = d(b)c^n - d(bc) + b^n d(c). \end{aligned}$$

If follows from the previous equation that \(f({{\textbf {e}}}) \big [(bc)^n - b^n c^n \big ] = 0\) if and only if \(d(bc) = d(b)c^n + b^n d(c)\) for all \(b, c \in {\mathcal {A}}\). We know that \(f(\lambda a) = \lambda ^n f(a)\) for all \(a \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). Hence, for any \(a, b \in {\mathcal {A}}\) and any \(\lambda \in {\mathbb {C}}\), we have the following statements:

$$\begin{aligned} f(a)(\lambda b)^n + a^n d(\lambda b) = f( a \lambda b) = \lambda ^n f(a)b^n + \lambda ^n a^n d(b), \end{aligned}$$

which implies that \(a^n d(\lambda b) = \lambda ^n a^n d(b)\). Putting \(a = {{\textbf {e}}}\) in the previous equation, we get that \(d(\lambda b) = \lambda ^n d(b)\) for all \(b \in {\mathcal {A}}\). This means that d is an \(\{n\}\)-derivation. \(\square\)

Theorem 19

Let \({\mathcal {A}}\) be a unital algebra, let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule and let \(f:{\mathcal {A}} \rightarrow {\mathcal {M}}\) be an additive generalized \(\{n\}\)-derivation with an associated mapping \(d:{\mathcal {A}} \rightarrow {\mathcal {M}}\) such that \(d(2 {{\textbf {e}}}) = 2 d({{\textbf {e}}})\). Then either f is a nonzero linear generalized derivation with the associated linear derivation d or f and d are identically zero.

Proof

Since f is an additive mapping, \(f(a(b + c)) = f(ab) + f(ac)\) for all \(a, b, c \in {\mathcal {A}}\). We have

$$\begin{aligned} f(a(b + c)) = f(a)(b +c)^n + a^n d(b +c). \end{aligned}$$
(6)

Also, we have

$$\begin{aligned} f(ab) + f(ac) = f(a)b^n + a^n d(b) + f(a)c^n + a^n d(c). \end{aligned}$$
(7)

Comparing (6) and (7), we get that

$$\begin{aligned} f(a) \big [(b + c)^n - b^n - c^n \big ] = a^n \big [d(b) + d(c) - d(b + c) \big ], \ \ (a, b, c \in {\mathcal {A}}). \end{aligned}$$
(8)

Setting \(b = c = {{\textbf {e}}}\) in (8) and using the assumption that \(d(2 {{\textbf {e}}} ) = 2 d({{\textbf {e}}})\), we arrive at

$$\begin{aligned} (2^n - 2)f(a)= 0, \ \ \ \ \ \ \ \ \ \ \ \ (a \in {\mathcal {A}}). \end{aligned}$$
(9)

We consider the following two cases:

Case 1.:

\(2^n - 2 = 0\). Then \(n = 1\) and this means that f is a linear generalized derivation with an associated mapping \(d:{\mathcal {A}} \rightarrow {\mathcal {M}}\). Now we show that d is a linear derivation. Since \(n = 1\), it follows from (8) that

$$\begin{aligned} 0 = a \big [d(b) + d(c) - d(b + c) \big ], \ \ (a, b, c \in {\mathcal {A}}). \end{aligned}$$
(10)

Putting \(a = {{\textbf {e}}}\) in (10), we see that d is an additive mapping. Also, note that \(f( \lambda a) = \lambda ^n f(a) = \lambda f(a)\) for all \(a \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). Similar to the proof of Theorem 18, one can easily show that \(d(\lambda a) = \lambda d(a)\) for all \(a \in {\mathcal {A}}\) and we leave it to the interested reader. So d is a linear derivation.

Case 2.:

\(2^n - 2 \ne 0\). It follows from (9) that f is identically zero. This fact with \(f(ab) = f(a)b^n + a^n d(b)\) imply that \(a^n d(b) = 0\) for all \(a, b \in {\mathcal {A}}\). Putting \(a = {{\textbf {e}}}\) in the previous equation, we infer that d is identically zero. By the way, in both above-mentioned cases f is a generalized derivation with an associated derivation d on \({\mathcal {A}}\). \(\square\)

In the following, we present a characterization of \(\{n\}\)-ternary derivations on algebras.

Theorem 20

Let \({\mathcal {A}}\) be a unital algebra, let \({\mathcal {M}}\) be an \({\mathcal {A}}\)-bimodule and let \((d_1, d_2, d_3):{\mathcal {A}} \rightarrow {\mathcal {M}}\) be an \(\{n\}\)-ternary derivation. Let \(d_3(2 {{\textbf {e}}}) = 2 d_3({{\textbf {e}}})\) or \(d_2(2 {{\textbf {e}}}) = 2 d_2({{\textbf {e}}})\). If \(d_1\) is an additive mapping, then either all the mappings \(d_1\), \(d_2\) and \(d_3\) are linear and \((d_1, d_2, d_3)\) is a ternary derivation on \({\mathcal {A}}\) or \(d_1 = d_2 = d_3 = 0\).

Proof

Suppose that \(d_3(2{{\textbf {e}}}) = 2 d_3({{\textbf {e}}})\). Let a, b, c be arbitrary elements of \({\mathcal {A}}\). We have the following expressions:

$$\begin{aligned} d_1(a(b + c)) = d_2(a) (b + c)^n + a^n d_3(b + c). \end{aligned}$$
(11)

On the other hand, we have

$$\begin{aligned} d_1(a(b + c))&= d_1(ab) + d_1(ac) \\&= d_2(a)b^n + a^n d_3(b) + d_2(a)c^n + a^n d_3(c) \\&= d_2(a)(b^n + c^n) + a^n(d_3(b) + d_3(c)), \end{aligned}$$

which means that

$$\begin{aligned} d_1(a(b + c)) = d_2(a)(b^n + c^n) + a^n(d_3(b) + d_3(c)). \end{aligned}$$
(12)

Comparing (11) and (12), we get that

$$\begin{aligned} d_2(a)\big [(b + c)^n - b^n - c^n\big ] = a^n\big [d_3(b) + d_3(c) - d_3(b + c)\big ]. \end{aligned}$$
(13)

Putting \(b = c = {{\textbf {e}}}\) in (13) and using the assumption that \(d_3(2{{\textbf {e}}}) = 2 d_3({{\textbf {e}}})\), we get that

$$\begin{aligned} (2^n - 2) d_2(a) = 0 \ \ for \ all \ a \in {\mathcal {A}}. \end{aligned}$$
(14)

We have two cases concerning \(2^n - 2\) as follows:

Case 1.:

\(2^n - 2 = 0\). So \(n = 1\) and it follows from (13) that

$$\begin{aligned} 0 = a \big [d_3(b) + d_3(c) - d_3(b + c)\big ]. \end{aligned}$$
(15)

Setting \(a = {{\textbf {e}}}\) in (15), we see that \(d_3\) is an additive mapping. We know that \(d_1(\lambda a) = \lambda ^n d_1(a) = \lambda d_1(a)\) for all \(a \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). Hence, for any \(a, b \in {\mathcal {A}}\) and any \(\lambda \in {\mathbb {C}}\), we have the following statements:

$$\begin{aligned} d_2(a)(\lambda b) + a d_3(\lambda b) = d_1( a \lambda b) = \lambda d_2(a)b + \lambda a d_3(b), \end{aligned}$$

which implies that \(a d_3(\lambda b) = \lambda a d_3(b)\). Putting \(a = {{\textbf {e}}}\) in the previous equation, we get that \(d_3(\lambda b) = \lambda d_3(b)\) for all \(b \in {\mathcal {A}}\). This means that \(d_3\) is a linear mapping. Similarly, we can show that \(d_2\) is a linear mapping. Hence, \((d_1, d_2, d_3)\) is a ternary derivation on \({\mathcal {A}}\).

Case 2.:

\(2^n - 2 \ne 0\). Then equation (14) yields that \(d_2\) must be zero. Considering this case and using \(d_1(ab) = d_2(a)b^n + a^n d_3(b)\) for all \(a, b \in {\mathcal {A}}\), we get that

$$\begin{aligned} d_1(ab) = a^n d_3(b) \ \ \ for \ all \ a, b \in {\mathcal {A}}. \end{aligned}$$
(16)

We know that \(d_1\) is an additive mapping. So we have \(d_1((b+c)a) = d_1(ba) + d_1(ca)\) for all \(a, b , c \in {\mathcal {A}}\). This equation along with (16) imply that

$$\begin{aligned} \big [(b + c)^n - b^n - c^n \big ]d_3(a) = 0, \ \ for \ all \ a, b , c \in {\mathcal {A}}. \end{aligned}$$
(17)

Putting \(b = c = {{\textbf {e}}}\) in (17) and considering the assumption that \(2^n - 2 \ne 0\), we infer that \(d_3 = 0\) and it follows from (16) that so is \(d_1\). Therefore, \(d_1\), \(d_2\) and \(d_3\) are zero. Reasoning like above, we obtain the required result if we assume that \(d_2(2 {{\textbf {e}}}) = 2 d_2({{\textbf {e}}})\). Note, however, that in both above-mentioned cases, \((d_1, d_2, d_3)\) is a ternary derivation. \(\square\)

In the next theorem, we present a characterization of \(\{n\}\)-generalized derivations using some functional equations.

Theorem 21

Let \({\mathcal {A}}\) be a unital algebra, let n be a positive integer, and let \(d_1, d_2, d_3:{\mathcal {A}} \rightarrow {\mathcal {A}}\) be mappings satisfying

$$\begin{aligned} d_1(ab)&= d_2(a)b^n + a^n d_3(b) = d_3(a)b^n + a^n d_2(b) \end{aligned}$$
(18)
$$\begin{aligned} d_1(\lambda a)&= \lambda ^n d_1(a) \end{aligned}$$
(19)

for all \(a,b \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). Furthermore, assume that \(d_i({{\textbf {e}}}) \big [a^n b^n - (ab)^n \big ] = 0\) for all \(a, b \in {\mathcal {A}}\) and \(i \in \{2, 3\}\). Then there exists an \(\{n\}\)-derivation \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) such that \(d_1, d_2\) and \(d_3\) are \(\{n\}\)-generalized derivations with the associated \(\{n\}\)-derivation \(\Delta\).

Proof

Putting \(b = {{\textbf {e}}}\) in (18), we obtain

$$\begin{aligned} d_1(a) = d_2(a) + a^n d_3({{\textbf {e}}}) = d_3(a) + a^n d_2({{\textbf {e}}}), \end{aligned}$$
(20)

and taking \(a = {{\textbf {e}}}\) in (18), we see that

$$\begin{aligned} d_1(b) = d_2({{\textbf {e}}}) b^n + d_3(b) = d_3({{\textbf {e}}}) b^n + d_2(b). \end{aligned}$$
(21)

Comparing (20) and (21), we get that

$$\begin{aligned} d_i({{\textbf {e}}}) a^n = a^n d_i({{\textbf {e}}}) \end{aligned}$$
(22)

for all \(a \in {\mathcal {A}}\) and \(i \in \{1, 2, 3\}\). It follows from (20) and (22) that

$$\begin{aligned} d_3(a) = d_2(a) + \left( d_3({{\textbf {e}}}) - d_2({{\textbf {e}}})\right) a^n = d_2(a) + a^n \left( d_3({{\textbf {e}}}) - d_2({{\textbf {e}}})\right) , \end{aligned}$$

for all \(a \in {\mathcal {A}}\). Using (20), we have

$$\begin{aligned} d_2(a) b^n + a^n d_3(b) = d_1(ab) = d_2(ab) + d_3({{\textbf {e}}})(ab)^n \end{aligned}$$

and so

$$\begin{aligned} d_2(ab)&= d_2(a)b^n + a^n d_3(b) - d_3({{\textbf {e}}})(ab)^n \\&= d_2(a)b^n + a^n \big [d_2(b) + (d_3({{\textbf {e}}}) - d_2({{\textbf {e}}}))b^n \big ] - d_3({{\textbf {e}}})(ab)^n \\&= d_2(a)b^n + a^n d_2(b) - d_2({{\textbf {e}}}) a^n b^n \end{aligned}$$

We define \(\Delta :{\mathcal {A}} \rightarrow {\mathcal {A}}\) by \(\Delta (a) = d_2(a) - d_2({{\textbf {e}}})a^n\). So by (22) and the assumption that \(d_i({{\textbf {e}}}) \big [a^n b^n - (ab)^n \big ] = 0\) for all \(a, b \in {\mathcal {A}}\) and \(i \in \{2, 3\}\), we have the following expressions:

$$\begin{aligned} \Delta (ab)&= d_2(ab) - d_2({{\textbf {e}}})(ab)^n \\&= d_2(a)b^n + a^n d_2(b) - d_2({{\textbf {e}}})a^n b^n - d_2({{\textbf {e}}}) (ab)^n\\&= \big [d_2(a)- d_2({{\textbf {e}}})a^n \big ]b^n + a^n \big [d_2(b)- d_2({{\textbf {e}}}) b^n\big ]\\&= \Delta (a)b^n + a^n \Delta (b), \end{aligned}$$

which means that

$$\begin{aligned} \Delta (ab) = \Delta (a)b^n + a^n \Delta (b), \ \ \ \ for \ all \ a, b \in {\mathcal {A}}. \end{aligned}$$

Our next task is to show that \(\Delta (\lambda a) = \lambda ^n \Delta (a)\) for all \(a \in {\mathcal {A}}\) and \(\lambda \in {\mathbb {C}}\). Before that, we prove that \(d_2(\lambda a) = \lambda ^n d_2(a)\) for all \(a \in {\mathcal {A}}\) and \(\lambda \in {\mathbb {C}}\). We know that \(d_1(\lambda a) = \lambda ^n d_1(a)\) for all \(a \in {\mathcal {A}}\) and \(\lambda \in {\mathbb {C}}\). So we have

$$\begin{aligned} d_1(\lambda ab) = \lambda ^n d_1(ab) = \lambda ^n d_2(a)b^n + \lambda ^n a^n d_3(b) \end{aligned}$$

and on the other hand

$$\begin{aligned} d_1(\lambda ab) = d_2(\lambda a)b^n + \lambda ^n a^n d_3(b) \end{aligned}$$

for all \(a, b \in {\mathcal {A}}\) and all \(\lambda \in {\mathbb {C}}\). By comparing these two equations related to \(d_1(\lambda ab)\), we deduce that \(\lambda ^n d_2(a)b^n = d_2(\lambda a)b^n\). Putting \(b = {{\textbf {e}}}\) in the previous equation, we get that \(d_2(\lambda a) = \lambda ^n d_2(a)\) for all \(a \in {\mathcal {A}}\) and \(\lambda \in {\mathbb {C}}\). Consequently, \(\Delta (\lambda a) = \lambda ^n \Delta (a)\) for all \(a \in {\mathcal {A}}\) and \(\lambda \in {\mathbb {C}}\). So \(\Delta\) is an \(\{n\}\)-derivation. Using this fact, we have

$$\begin{aligned} d_2(ab)&= \Delta (ab) + d_2({{\textbf {e}}})(ab)^n \\&= \Delta (a)b^n + a^n \Delta (b) + d_2({{\textbf {e}}})a^n b^n\\&= (\Delta (a) + d_2({{\textbf {e}}})a^n)b^n + a^n \Delta (b) \\&= d_2(a)b^n + a^n \Delta (b)\\&= \Delta (a)b^n + a^n d_2(b), \end{aligned}$$

which means that

$$\begin{aligned} d_2(ab) = d_2(a)b^n + a^n \Delta (b) = \Delta (a)b^n + a^n d_2(b), \ \ \ for \ all \ a, b \in {\mathcal {A}}. \end{aligned}$$

So \(d_2\) is an \(\{n\}\)-generalized derivation with the associated \(\{n\}\)derivation \(\Delta\). Using a similar argument, one can easily show that

$$\begin{aligned} d_3(ab) = d_3(a) b^n + a^n d_3(b) - d_3({{\textbf {e}}}) (ab)^n, \ \ \ \ for \ all \ a, b \in {\mathcal {A}}. \end{aligned}$$

By defining \(\delta : {\mathcal {A}} \rightarrow {\mathcal {A}}\) by \(\delta (a) = d_3(a) - d_3({{\textbf {e}}})a^n\) and by reasoning like the mapping \(d_2\), it is observed that \(d_3\) is an \(\{n\}\)-generalized derivation with the associated \(\{n\}\)-derivation \(\delta\). In the following, we show that \(\delta = \Delta\). We know that \(\Delta (a) = d_2(a) - d_2({{\textbf {e}}})a^n\) and it follows from (21) that \(d_2(a) = d_3(a) + a^n d_2({{\textbf {e}}}) - d_3({{\textbf {e}}})a^n\) for all \(a \in {\mathcal {A}}\). So we have

$$\begin{aligned} \Delta (a)&= d_2(a) - d_2({{\textbf {e}}})a^n \\&= d_3(a) + a^n d_2({{\textbf {e}}}) - d_3({{\textbf {e}}})a^n - d_2({{\textbf {e}}}) a^n \\&= d_3(a) - d_3({{\textbf {e}}})a^n \\&= \delta (a) \end{aligned}$$

for all \(a \in {\mathcal {A}}\). Hence, both \(d_2\) and \(d_3\) are \(\{n\}\)-generalized derivations with the associated \(\{n\}\)-derivation \(\Delta\). We are now ready to show that \(d_1\) is also an \(\{n\}\)-generalized derivation with the associated \(\{n\}\)-derivation \(\Delta\). We know that \(d_1(a) = d_2(a) + d_3({{\textbf {e}}}) a^n\) and \(d_2(a) = \Delta (a) + d_2({{\textbf {e}}})a^n\) for all \(a \in {\mathcal {A}}\). Hence, we have

$$\begin{aligned} d_1(a) = \Delta (a) + d_2({{\textbf {e}}}) a^n + d_3({{\textbf {e}}}) a^n = \Delta (a) + d_1({{\textbf {e}}})a^n, \end{aligned}$$

which means that \(d_1\) is an \(\{n\}\)-generalized derivation with the associated \(\{n\}\)-derivation \(\Delta\), as required. \(\square\)

We conclude this paper with the following questions.

Question 22

Let \({\mathcal {A}}\) be an algebra or ring, let \(n > 1\) be a positive integer, and let \(\Delta : {\mathcal {A}} \rightarrow {\mathcal {A}}\) be a mapping such that \(\Delta (a^2) = \Delta (a)a^n + a^n \Delta (a)\) holds for all \(a \in {\mathcal {A}}\). Under what conditions we have \(\Delta (ab) = \Delta (a)b^n + a^n \Delta (b)\) for all \(a, b \in {\mathcal {A}}\)?

Question 23

Let \({\mathcal {A}}\) be a unital algebra or ring, let \(n > 1\) be a positive integer, and let \(\Delta : {\mathcal {A}} \rightarrow {\mathcal {A}}\) be a mapping satisfying

$$\begin{aligned} \Delta (a^m) = \Sigma _{k = 1}^{m}a^{(k - 1)n}\Delta (a)a^{(m - k)n} \end{aligned}$$

in which \(a^0 = {{\textbf {e}}}\), for all \(a \in {\mathcal {A}}\) and for some positive integer m. Under what conditions we have \(\Delta (ab) = \Delta (a)b^n + a^n \Delta (b)\) for all \(a, b \in {\mathcal {A}}\)?