Supercommuting maps on unital algebras with idempotents

1.   Introduction

Let $\mathcal{A}$ be an associative algebra over $R$, a commutative ring with unity. By $Z(\mathcal{A})$, we denote the center of $\mathcal{A}$. Set $[x, y] = xy-yx$ and $x\circ y = xy+yx$.

A linear map $f$ on $\mathcal{A}$ is called a commuting map if $[f(x), x] = 0$ for all $x\in\mathcal{A}$. It is clear that $f(x) = \lambda x+\tau(x)$ is a commuting map, where $\lambda\in Z(\mathcal{A})$ and $\tau:\mathcal{A}\rightarrow Z(\mathcal{A})$, which is said to be a proper commuting map.

In 1991, Brešar ^[1] proved that a commuting map on noncommutative Lie ideals of prime rings is always proper. In 1993, Brešsar ^[2] discussed centralizing mappings and derivations in prime rings. In the same year, Brešsar ^[3] discussed commuting traces of biadditive mappings, commutativity-preserving mappings and Lie mappings. In 2020, Jia and Xiao ^[4] discussed commuting maps on certain incidence algebras. Results related to commuting maps are discussed in ^[5,6,7].

It should be mentioned that the study of commuting maps on rings initiated the theory of functional identities on rings (see ^[8] for details).

An associative algebra $\mathcal{A}$ is said to be a superalgebra if $\mathcal{A}$ is the direct sum of two $R$-submodules $\mathcal{A}_0$ and $\mathcal{A}_1$ such that $\mathcal{A}_i\mathcal{A}_j\subseteq \mathcal{A}_{i+j}$ (modulo $2$). We call $\mathcal{A}_0$ the even part and $\mathcal{A}_1$ the odd part of $\mathcal{A}$. Elements in $\mathcal{H} = \mathcal{A}_0\bigcup \mathcal{A}_1$ are called homogeneous, and we write $|a| = i$ to mean $a\in \mathcal{A}_i$. For $a, b\in \mathcal{H}$, the supercommutator of $a$ and $b$ is defined to be

It is clear that $[a, b]_s = a\circ b$ if both $a$ and $b$ are odd, and $[a, b]_s = [a, b]$ if either $a$ or $b$ is even. The definition can be extended linearly to arbitrary $a, b\in\mathcal{A}$.

Let $\mathcal{A} = \mathcal{A}_0\bigoplus \mathcal{A}_1$ be a superalgebra. A linear map $f: \mathcal{A}\rightarrow \mathcal{A}$ is said to be supercommuting if

for all $x\in\mathcal{A}$.

In 2002, Beidar, Chen, Fong, and Ke ^[9] discussed graded polynomial identities with an antiautomorphism. In 2003, Beidar, Bresǎr, and Chebotar ^[10] discussed Jordan superhomomorphisms on superalgebras. In 2008, Wang ^[11] discussed skew-supercommuting maps in superalgebras. In 2009, Wang ^[12] gave a description of supercentralizing superautomorphisms on prime superalgebras. In the same year, Lee and Wang ^[13] gave a description of supercommuting maps of prime superalgebras. In 2017, Fan and Dai ^[14] investigated Super-biderivations on Lie superalgebras. In 2019, Cheng and Sun ^[15] discussed Super-biderivations and linear supercommuting maps on the super-BMS3 algebras.

Let $\mathcal{A}$ be a unital algebra with an idempotent $e\neq 0, 1$. Let $f$ denote the idempotent $1-e$. In this case, $\mathcal{A}$ can be represented in the so-called Peirce decomposition form

where $e\mathcal{A}e$ and $f\mathcal{A}f$ are subalgebras with unitary elements $e$ and $f$, respectively, $e\mathcal{A}f$ is an $(e\mathcal{A}e, f\mathcal{A}f)$-bimodule and $f\mathcal{A}e$ is an $(f\mathcal{A}f, e\mathcal{A}e)$-bimodule, which is said to be a generalized matrix algebra (see ^[16] for details).

For brevity, we set

It is clear that

In 2012, Benkovič and Širovnik ^[17] defined the following useful condition:

Some examples of unital algebras with nontrivial idempotents having the property (1.1) are triangular algebras, matrix algebras, and prime (and hence in particular simple) algebras with nontrivial idempotents (see ^[17]).

In 2010, Xiao and Wei ^[16] initiated the study of commuting mappings of generalized matrix algebras, which generalized a typical result on commuting maps of triangular algebras (see ^[18] for details). In 2019, Li, Wei, and Fošner ^[19] discussed $k$-commuting mappings of generalized matrix algebras, which generalized a result on $k$-commuting maps of triangular algebras (see ^[20] for details). In 2002, Du and Wang ^[21] gave a description of Lie derivations of generalized matrix algebras. In 2018, Benkovič ^[22] discussed generalized Lie derivations of unital algebras with idempotents. Additional results on mappings of generalized matrix algebras can be found in ^[23,24,25].

In 2024, Ghahramani and Zadeh ^[26] considered $\mathcal{A}$ as a superalgebra by:

where

They determined the class of generalized matrix algebras for which every Lie superderivation is proper (see [26, Theorem 5.1]. As a consequence, they gave some descriptions of Lie superderivations on both matrix algebras and triangular algebras (see ^[26] for details).

Recently, Chen ^[27] discussed Jordan superderivations of unital algebras with idempotents. As a consequence, she gave some descriptions of Jordan superderivations of matrix algebras and triangular algebras.

In the present paper, we give a description of supercommuting maps of unital algebras with idempotents. As a consequence, we give some descriptions of supercommuting maps of matrix algebras and triangular algebras.

We organize the paper as follows: In Section $2$, we give preliminaries and the definition of proper supercommuting maps. In Section $3$, we give a description of supercommuting maps of degree $0$ on unital algebras. In Section $4$, we give a description of supercommuting maps of degree $1$ on unital algebras. In Section $5$, we give the main result of the paper. In Section $6$, we give a description of supercommuting maps on matrix algebras. As a consequence, we prove that every supercommuting map on matrix algebras over a $2$-torsion free unital algebra is supercentral. In the last section, we prove that every supercommuting map on triangular algebras is a commuting map.

2.   Preliminaries

Let $\mathcal{A} = \mathcal{A}_0\bigoplus \mathcal{A}_1$ be a superalgebra. The supercenter of $\mathcal{A}$ is the set

We set

It is easy to check that $Z(\mathcal{A}) = Z(\mathcal{A})_0\bigoplus Z(\mathcal{A})_1$ is a graded subalgebra of $\mathcal{A}$ (see [4, Section 2] for details). It is clear that $Z(\mathcal{A})_0\subseteq Z(\mathcal{A})_s$.

We begin with the following definition.

Definition 2.1. Let $\mathcal{A} = \mathcal{A}_0\bigoplus \mathcal{A}_1$ be a superalgebra. We call a linear map $f: \mathcal{A}\rightarrow \mathcal{A}$ a proper supercommuting map if

for all $x\in \mathcal{A}$, where $\lambda\in Z(\mathcal{A})_0$ with $2\lambda \mathcal{A}_1^2 = \{0\}$, and $\tau:\mathcal{A}\rightarrow Z(\mathcal{A})_s$ is a linear map. In particular, if $f(x)\in Z(\mathcal{A})_s$ for all $x\in\mathcal{A}$, we call $f$ a supercentral map.

The following result shows that a proper supercommuting map is a supercommuting map.

Lemma 2.1. Let $\mathcal{A} = \mathcal{A}_0\bigoplus \mathcal{A}_1$ be a superalgebra. Then

for all $x\in \mathcal{A}$, is a supercommuting map of $\mathcal{A}$, where $\lambda\in Z(\mathcal{A})_0$ with $2\lambda \mathcal{A}_1^2 = \{0\}$, and $\tau:\mathcal{A}\rightarrow Z(\mathcal{A})_s$ is a linear map.

Proof. For any $x = x_0+x_1\in \mathcal{A}$ we get

We obtain that $f$ is a supercommuting map. □

Definition 2.2. Let $\mathcal{A} = \mathcal{A}_0\bigoplus \mathcal{A}_1$ be a superalgebra. A supercommuting map $f$ on $\mathcal{A}$ is said to be a supercommuting map of degree $0$ if $f(\mathcal{A}_0)\subseteq \mathcal{A}_0$ and $f(\mathcal{A}_1)\subseteq \mathcal{A}_1$. A supercommuting map $f$ of $\mathcal{A}$ is said to be a supercommuting map of degree $1$ if $f(\mathcal{A}_0)\subseteq \mathcal{A}_1$ and $f(\mathcal{A}_1)\subseteq \mathcal{A}_0$.

The following result shows that a supercommuting map is the sum of a supercommuting map of degree $0$ and a supercommuting map of degree $1$.

Lemma 2.2. Let $\mathcal{A} = \mathcal{A}_0\bigoplus\mathcal{A}_1$ be a superalgebra. Let $f$ be a supercommuting map of $\mathcal{A}$. Then

where $f_0$ is a supercommuting map of degree $0$ on $\mathcal{A}$ and $f_1$ is a supercommuting map of degree $1$ on $\mathcal{A}$.

Proof. For $i = 0$ or $1$, let $\pi_i$ be the canonical projection of $\mathcal{A}$. We set

It is easy to check that $f_i$ is a linear map of $\mathcal{A}$ and $f = f_0+f_1$, where $i = 0, 1$. Moreover, $f_0(\mathcal{A}_0)\subseteq \mathcal{A}_0$, $f_0(\mathcal{A}_1)\subseteq \mathcal{A}_1$, $f_1(\mathcal{A}_0)\subseteq \mathcal{A}_1$, and $f_1(\mathcal{A}_1)\subseteq \mathcal{A}_0$. We now claim that both $f_0$ and $f_1$ are supercommuting maps on $\mathcal{A}$.

For $i = 0, 1$, and any $x = x_0+x_1\in\mathcal{A}$, we have

Since $[f_0(x_i), x_i]_s$ is even and $[f_1(x_i), x_i]_s$ is odd, we obtain that

It follows from (2.1) and the linearity of $f_i$, where $i = 0, 1$, that

and

For $i = 0, 1$, we get from (2.1), (2.2), and (2.3) that

This implies that $f_i$ is a supercommuting map of degree $i$. The proof of the result is complete. □

Form now on we always assume that $\mathcal{A}$ is a unital algebra with nontrivial idempotents.

The following result is essentially the same as [26, Lemma 2.1].

Lemma 2.3.

We define two natural projection $\pi_A:\mathcal{A}\rightarrow A$ and $\pi_B:\mathcal{A}\rightarrow B$ by

The following result is essentially the same as [26, Lemma 2.3].

Lemma 2.4. Let $\mathcal{A}$ be a unital algebra with nontrivial idempotents having the property (1.1). Then

Furthermore, $\pi_A(Z(\mathcal{A}))\subseteq Z(A)$, $\pi_B(Z(\mathcal{A}))\subseteq Z(B)$, and there exists a unique isomorphism $\varphi$ from $\pi_A(Z(\mathcal{A}))$ to $\pi_B(Z(\mathcal{A}))$ such that $am = m\varphi(a)$, $mb = \varphi^{-1}(b)m$, $na = \varphi(a)n$, and $bn = n\varphi^{-1}(b)$ for all $m\in M, n\in N$.

3.   Supercommuting maps of degree $0$

We begin with the structure of supercommuting maps of degree $0$.

Lemma 3.1. Let $f_0$ be a supercommuting map of degree $0$ on $\mathcal{A}$. Then

for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$, where $\alpha_1:A\rightarrow A$, $\alpha_4:B\rightarrow Z(A)$, $\beta_1:A\rightarrow Z(B)$, and $\beta_4:B\rightarrow B$ are linear maps satisfying the following conditions:

(i) $\alpha_1$ and $\beta_4$ are commuting mappings of $A$ and $B$, respectively. In particular, $\alpha_1(1)\in Z(A), \beta_4(1)\in Z(B)$;

(ii) $\alpha_1(a)m-m\beta_1(a) = a(\alpha_1(1)m-m\beta_1(1))$, $\beta_1(a)n-n\alpha_1(a) = (n\alpha_1(1)-\beta_1(1)n)a$;

(iii) $\alpha_4(b)m-m\beta_4(b) = (m\beta_1(1)-\alpha_1(1)m)b$, $\beta_4(b)n-n\alpha_4(b) = b(n\alpha_1(1)-\beta_1(1)n)$;

(iv) $2\alpha_1(1)mn = 2m\beta_1(1)n$ and $2n\alpha_1(1)m = 2\beta_1(1)nm$.

Proof. Since $f_0(\mathcal{A}_0)\subseteq \mathcal{A}_0$ and $f_0(\mathcal{A}_1)\subseteq \mathcal{A}_1$, we can write

for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$, where $\alpha_1:A\rightarrow A$, $\alpha_4:B\rightarrow A$, $\mu_2:M\rightarrow M$, $\mu_3:N\rightarrow M$, $\nu_2:M\rightarrow N$, $\nu_3:N\rightarrow N$, $\beta_1:A\rightarrow B$, and $\beta_4:B\rightarrow B$ are linear maps.

Linearizing $[f_0(X), X]_s = 0$ leads to

for all $X, Y\in\mathcal{A}$. For any $m\in M$, taking $X = 1_A$ and $Y = m$ in (3.2) yields

That is

This implies that $v_2(m) = 0$ and $\mu_2(m) = \alpha_1(1)m-m\beta_1(1)$ for all $m\in M$. Similarly, if we choose $X = 1_A$ and $Y = n$ in (3.2), then we arrive at $\mu_3(n) = 0$ and $\nu_3(n) = n\alpha_1(1)-\beta_1(1)n$ for all $n\in N$. Therefore (3.1) becomes

for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$. For any $a\in A$ and $b\in B$, taking $X = a$ and $Y = b$ into (3.3) yields

That is

Then $[\alpha_4(b), a] = 0$ and $[\beta_1(a), b] = 0$ for all $a\in A$ and $b\in B$. This implies that $\alpha_4(B)\subseteq Z(A)$ and $\beta_1(A)\subseteq Z(B)$. By (3.3) we obtain

Then $[\alpha_1(a), a] = 0$ for all $a\in A$ and $[\beta_4(b), b] = 0$ for all $b\in B$. This implies that $\alpha_1$ and $\beta_4$ are commuting mappings of $A$ and $B$, respectively. It is easy to check that $\alpha_1(1)\in Z(A)$ and $\beta_4(1)\in Z(B)$. This proves the statement (ⅰ).

By (3.3) we get

and

Note that

The above three relations imply that $\alpha_1(a)m-m\beta_1(a) = a(\alpha_1(1)m-m\beta_1(1))$ and $\beta_1(a)n-n\alpha_1(a) = (n\alpha_1(1)-\beta_1(1)n)a$ for all $a\in A$, $m\in M$, and $n\in N$. This proves the statement (ii). Similarly, taking $X = b$ and $Y = m+n$ in (3.2) we can obtain that $\alpha_4(b)m-m\beta_4(b) = (m\beta_1(1)-\alpha_1(1)m)b$ and $\beta_4(b)n-n\alpha_4(b) = b(n\alpha_1(1)-\beta_1(1)n)$ for all $b\in B$, $m\in M$, and $n\in N$. This proves the statement (ⅲ).

Since $[f_0(m+n), m+n]_s = 0$ for all $m\in M$, $n\in N$, we get from (3.3) that

That is

for all $m\in M$ and $n\in N$. This implies that

for all $m\in M$, $n\in N$. This proves the statement (ⅳ). We complete the proof of the result. □

The idea of proving the following result is taken from [18, Lemma 1].

Lemma 3.2. Let $\mathcal{A}$ be a unital algebra with nontrivial idempotents having the property (1.1). Let $f_0$ be a supercommuting map of degree $0$ on $\mathcal{A}$. With notations as above, then $\beta_1^{-1}(\pi_B(Z(\mathcal{A})))$ and $\alpha_4^{-1}(\pi_A(Z(\mathcal{A})))$ are ideals of $A$ and $B$, respectively. Furthermore, $[A, A]\subseteq \beta_1^{-1}(\pi_B(Z(\mathcal{A})))$ and $[B, B]\subseteq\alpha_4^{-1}(\pi_A(Z(\mathcal{A})))$.

Proof. We prove the part of the statement related to $\beta_1$. The part related to $\alpha_4$ can be proved analogously. For any $a, a'\in A$, $m\in M$, and $n\in N$ we get from Lemma 3.1(ⅱ) that

From (3.4) and (3.5), we have

and from (3.6) and (3.7), we have

Taking the difference of (3.8) and (3.9), we have

For any $a, a'\in A$ and $n\in N$, we get from Lemma 3.1(ⅱ) that

From (3.11) and (3.12), we have

and from (3.13) and (3.14), we have

Taking the difference of (3.15) and (3.16), we have

Since $\alpha_1$ is commuting map of $A$, we get that $[a, \alpha_1(a')] = [\alpha_1(a), a']$. Thus, we get from (3.17) that

In view of Lemma 2.4 we get from both (3.10) and (3.18) that $\beta_1([a, a'])\in \pi_B(Z(\mathcal{A}))$. Hence $[A, A]\subseteq\beta_1^{-1}(\pi_B(Z(\mathcal{A})))$.

Suppose that $a\in \beta_1^{-1}(\pi_B(Z(\mathcal{A})))$. From both (3.8) and (3.15) we have

By Lemma 2.4 we get that $\beta_1(a'a)\in \pi_B(Z(\mathcal{A})))$. Hence $a'a\in \beta_1^{-1}(\pi_B(Z(\mathcal{A})))$. Similarly, from both (3.9) and (3.16) we have

By Lemma 2.4 we get that $\beta_1(aa')\in \pi_B(Z(\mathcal{A})))$. Hence $aa'\in \beta_1^{-1}(\pi_B(Z(\mathcal{A})))$. As a result, $\beta_1^{-1}(\pi_B(Z(\mathcal{A})))$ is an ideal of $A$ containing $[A, A]$. This proves the result. □

Now we obtain necessary and sufficient conditions for a supercommuting map of degree $0$ on $\mathcal{A}$ to be proper. The idea of proving the following result is taken from [18, Theorem 1].

Lemma 3.3. Let $\mathcal{A}$ be a unital algebra with nontrivial idempotents having the property (1.1). Let $f_0$ be a supercommuting map of degree $0$ on $\mathcal{A}$ such that

Then, the following three conditions are equivalent:

(i) $f_0$ is proper;

(ii) $\beta_1(A)\subseteq\pi_B(Z(\mathcal{A}))$ and $\alpha_4(B)\subseteq\pi_A(Z(\mathcal{A}))$;

(iii) $\alpha_1(1)\in \pi_A(Z(\mathcal{A}))$ and $\beta_1(1)\in \pi_B(Z(\mathcal{A}))$.

Proof. (ⅱ)$\Rightarrow$ (ⅲ). $\beta_1(1)\in \beta_1(A)\subseteq\pi_B(Z(\mathcal{A}))$. Taking $b = 1$ in Lemma 3.1(ⅲ), we get

for all $m\in M$ and $n\in N$. By Lemma 2.4 we get that $\alpha_1(1)\in \pi_A(Z(\mathcal{A}))$.

(ⅲ) $\Rightarrow$ (ⅱ). Since $\beta_1(1)\in \pi_B(Z(\mathcal{A}))$, the ideal $\beta_1^{-1}(\pi_B(Z(\mathcal{A})))$ of $A$ contains $1$. Hence $A = \beta_1^{-1}(\pi_B(Z(\mathcal{A})))$. We have that $\beta_1(A)\subseteq\pi_B(Z(\mathcal{A}))$. By Lemma 3.1(ⅲ), we have $\alpha_4(b)m-m\beta_4(b) = (m\beta_1(1)-\alpha_1(1)m)b$, which implies

for all $m\in M$ and $b\in B$. By Lemma 3.1(ⅲ) again, we have $\beta_4(b)n-n\alpha_4(b) = b(n\alpha_1(1)-\beta_1(1)n)$, which implies

for all $n\in N$ and $b\in B$. In view of Lemma 2.4 we get from both (3.19) and (3.20) that $\alpha_4(b)\in \pi_A(Z(\mathcal{A}))$ for all $b\in B$.

(ⅲ) $\Rightarrow$ (ⅰ). We set

for all $X\in \mathcal{A}$, where $\lambda = \alpha_1(1)-\varphi^{-1}(\beta_1(1))+\varphi(\alpha_1(1))-\beta_1(1)\in Z(\mathcal{A})$. We claim that $\tau(\mathcal{A})\subseteq Z(\mathcal{A})$. Indeed, we have

By Lemma 3.1(ⅱ) we have

for all $a\in A$, $m\in M$, and $n\in N$. By Lemma 2.4 we get that

for all $a\in A$. Similarly, we get from Lemma 3.1(ⅲ) that

for all $m\in M$, $n\in N$, and $b\in B$. By Lemma 2.4 we get that

for all $b\in B$. We obtain that $\tau(a+m+n+b)\in Z(\mathcal{A})$ for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$ as desired. We next claim that $2\lambda \mathcal{A}_1^2 = \{0\}$.

For any $m\in M$ and $n\in N$, by Lemma 3.1(ⅳ) we have

and

It follows that

for all $m, m'\in M$, $n, n'\in N$. This implies that $2\lambda \mathcal{A}_1^2 = \{0\}$.

(ⅰ) $\Rightarrow$ (ⅲ). Suppose that $f_0(X) = \lambda X+\tau(X)$ for all $X\in \mathcal{A}$, where $\lambda\in Z(\mathcal{A})$ with $2\lambda \mathcal{A}_1^2 = \{0\}$ and $\tau:\mathcal{A}\rightarrow Z(\mathcal{A})$ is a linear map. For any $m\in M$ and $n\in N$, we have

By Lemma 3.1 we get that

We get from the last relation that

This implies that

By Lemma 2.4 we get that $\alpha_1(1)-\pi_A(\lambda)\in \pi_A(Z(\mathcal{A}))$ and $\beta_1(1)\in \pi_B(Z(\mathcal{A}))$. Hence, $\alpha_1(1)\in \pi_A(Z(\mathcal{A}))$ and $\beta_1(1)\in \pi_B(Z(\mathcal{A}))$ as desired. The proof of the result is complete. □

We now give sufficient conditions for every supercommuting map of degree $0$ on $\mathcal{A}$ to be proper. The idea of proving the following result is taken from [18, Theorem 2].

Theorem 3.1. Let $\mathcal{A}$ be a unital algebra with nontrivial idempotents having the property (1.1). Suppose that the following two conditions are satisfied:

(i) $Z(B) = \pi_B(Z(\mathcal{A}))$, or $A = [A, A]$;

(ii) $Z(A) = \pi_A(Z(\mathcal{A}))$, or $B = [B, B]$.

Then every supercommuting map of degree $0$ on $\mathcal{A}$ is proper.

Proof. Let $f_0$ be a supercommuting mapping of degree $0$ on $\mathcal{A}$. With notations as above, we note that $\alpha_4(B)\subseteq Z(A)$ and $\beta_1(A)\subseteq Z(B)$. By the condition (ⅰ) we note that either $Z(B) = \pi_B(Z(\mathcal{A}))$ or $A = [A, A]$. Suppose first that $Z(B) = \pi_B(Z(\mathcal{A}))$. We get that $\beta_1(A)\subseteq\pi_B(Z(\mathcal{A}))$. Suppose next that $A = [A, A]$. In view of Lemma 3.2 we note that $[A, A]\subseteq \beta_1^{-1}(\pi_B(Z(\mathcal{A})))$. This implies that $\beta_1(A)\subseteq\pi_B(Z(\mathcal{A}))$.

By the condition (ⅱ) we note that either $Z(A) = \pi_A(Z(\mathcal{A}))$ or $B = [B, B]$. Suppose first that $Z(A) = \pi_A(Z(\mathcal{A}))$. We get that $\alpha_4(B)\subseteq\pi_A(Z(\mathcal{A}))$. Suppose next that $B = [B, B]$. In view of Lemma 3.2 we note that $[B, B]\subseteq \alpha_4^{-1}(\pi_A(Z(\mathcal{A})))$. We obtain that $\alpha_4(B)\subseteq\pi_A(Z(\mathcal{A}))$. By Lemma 3.3 we obtain that $f_0$ is proper. This proves the result. □

4.   Supercommuting maps of degree $1$

We first give the structure of supercommuting map of degree $1$ on $\mathcal{A}$.

Lemma 4.1. Let $\mathcal{A}$ be a unital algebra with nontrivial idempotents. Let $f_1$ be a commuting mapping of degree $1$ on $\mathcal{A}$. Then $f_1$ is of the form

for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$, where $\alpha_2:M\rightarrow Z(A)$, $\alpha_3:N\rightarrow Z(A)$, $\beta_2:M\rightarrow Z(B)$, and $\beta_3:N\rightarrow Z(B)$ are linear maps satisfying the following conditions:

(i) $(\alpha_2(m)+\alpha_3(n))m = m(\beta_2(m)+\beta_3(n))$;

(ii) $n(\alpha_2(m)+\alpha_3(n)) = (\beta_2(m)+\beta_3(n))n$

for all $m\in M$, $n\in N$.

Proof. Note that $f_1(\mathcal{A}_0)\subseteq \mathcal{A}_1$ and $f_1(\mathcal{A}_1)\subseteq \mathcal{A}_0$. So $f_1$ is of the form

for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$, where $\alpha_2:M\rightarrow A$, $\alpha_3:N\rightarrow A$, $\mu_1:A\rightarrow M$, $\mu_4:B\rightarrow M$, $\nu_1:A\rightarrow N$, $\nu_4:B\rightarrow N$, $\beta_2:M\rightarrow B$, and $\beta_3:N\rightarrow B$ are linear maps. Note that

We get that

This implies that $\mu_1(1) = 0 = \nu_1(1)$. Linearizing $[f_1(X), X]_s = 0$ leads to

for all $X, Y\in \mathcal{A}$. For any $a\in A$ and $b\in B$, taking $X = a+b$ and $Y = 1_A$ into (4.2) yields

That is

This implies that $\mu_1(a)+\mu_4(b) = 0$ and $\nu_1(a)+\nu_4(b) = 0$ for all $a\in A$ and $b\in B$. We get that $\mu_1 = \mu_4 = 0$ and $\nu_1 = \nu_4 = 0$. Thus, the relation (4.1) becomes

for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$. For any $a\in A$ and $b\in B$, $m\in M$, and $n\in N$, taking $X = a+b$ and $Y = m+n$ in (4.3) yields

It follows from (4.3) that

This implies that $[\alpha_2(m)+\alpha_3(n), a] = 0$ and $[\beta_2(m)+\beta_3(n), b] = 0$ for all $a\in A$ and $b\in B$, $m\in M$, and $n\in N$. We get that $[\alpha_2(m), a] = 0$, $[\alpha_3(n), a] = 0$, $[\beta_2(m), b] = 0$, and $[\beta_3(n), b] = 0$ for all $a\in A$ and $b\in B$, $m\in M$, and $n\in N$. That is, $\alpha_2(m), \alpha_3(n)\in Z(A)$ and $\beta_2(m), \beta_3(n)\in Z(B)$ for all $m\in M$ and $n\in N$.

For any $m\in M$ and $n\in N$, we have that

It follows from (4.3) that

This implies that $(\alpha_2(m)+\alpha_3(n))m = m(\beta_2(m)+\beta_3(n))$ and $n(\alpha_2(m)+\alpha_3(n)) = (\beta_2(m)+\beta_3(n))n$ for all $m\in M$ and $n\in N$. We complete the proof of the result. □

We now give a sufficient condition for every supercommuting map of degree $1$ on $\mathcal{A}$ to be supercentral.

Theorem 4.1. Let $f_1$ be a supercommuting mapping of degree $1$ on $\mathcal{A}$. Suppose that there exists $Y_1\in\mathcal{A}_1$ such that

Then $f_1(\mathcal{A})\subseteq Z(\mathcal{A})$.

Proof. We set

By Lemma 4.1 we note that

for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$, where $\alpha_2:M\rightarrow Z(A)$, $\alpha_3:N\rightarrow Z(A)$, $\beta_2:M\rightarrow Z(B)$, and $\beta_3:N\rightarrow Z(B)$. Moreover,

for all $m\in M$ and $n\in N$. We set

for all $m\in M$ and $n\in N$. It is easy to check that

It follows that $\delta(m, n)\in Z(\mathcal{A}_0)$ for all $m\in M$, $n\in N$. Since

for all $a\in A$, $m\in M$, $n\in N$, and $b\in B$ we get from (4.4) that

for all $m\in M$ and $n\in N$. In particular, $[\delta(m_0, n_0), Y_1] = 0$. By assumption we have that $\delta(m_0, n_0)\in Z(\mathcal{A})$. Note that

for all $m\in M$ and $n\in N$. It is clear that

for all $m\in M$ and $n\in N$. We get from (4.5) that

for all $m\in M$ and $n\in N$. This implies that

for all $m\in M$ and $n\in N$. By assumption again we obtain that $\delta(m, n)\in Z(\mathcal{A})$ for all $m\in M$ and $n\in N$. It follows from (4.4) that $f_1(\mathcal{A})\subseteq Z(\mathcal{A})$. This proves the result. □

5.   The main results

We are in a position to give the main result of the paper.

Theorem 5.1. Let $\mathcal{A}$ be a unital algebra with nontrivial idempotents satisfying (1.1). Suppose that

(i) $Z(B) = \pi_B(Z(\mathcal{A}))$, or $A = [A, A]$;

(ii) $Z(A) = \pi_A(Z(\mathcal{A}))$, or $B = [B, B]$;

(iii) there exists $Y_1\in \mathcal{A}_1$ such that

Then, every supercommuting map of $\mathcal{A}$ is proper.

Proof. Let $f$ is a supercommuting map of $\mathcal{A}$. By Lemma 2.2 we have that $f = f_0+f_1$, where $f_i$ is a supercommuting map of degree $i$ on $\mathcal{A}$, $i = 0, 1$. By Theorem 3.1 we have that

for all $X\in \mathcal{A}$, where $\lambda\in Z(\mathcal{A})$ with $2\lambda \mathcal{A}_1^2 = \{0\}$ and $\tau_0:\mathcal{A}\rightarrow Z(\mathcal{A})$ is a linear map. By Theorem 4.1 we have that $f_1(\mathcal{A})\subseteq Z(\mathcal{A})$. We set $\tau = \tau_0+f_1$. Then

for all $X\in \mathcal{A}$. This proves the result. □

6.   Supercommuting maps of matrix algebras

Let $S$ be a unital algebra over $R$. Let $M_n(S)$ be the set of all $n\times n$ matrices over $S$, where $n\geq 2$. We can view $M_n(S)$ as a unital algebra with nontrivial idempotents:

where $A = M_s(S)$, $M = M_{s, t}(S)$, $N = M_{t, s}(S)$, $B = M_{t}(S)$, where $s+t = n$. Thus, $M_n(S)$ is a generalized matrix algebra (see ^[16] for details). We set

It is easy to check that $M_n(S) = M_n(S)_0\bigoplus M_n(S)_1$ is a superalgebra (see ^[26] for details).

Applying Theorem 3.1, we give a description of supercommuting maps of matrix algebras, which is different from the result on commuting maps of matrix algebras (see [16, Corollary 4.1] for details).

Theorem 6.1. Let $S$ be a unital algebra. Let $M_n(S)$ be the $n\times n$ matrix algebra. Then every supercommuting map on $M_n(R)$ is the form $X\rightarrow \lambda X+\tau(X)$, where $\lambda\in Z(S)$ with $2\lambda = 0$, and $\tau:M_n(S)\rightarrow Z(S)\cdot 1$ is a linear map.

Proof. Note that $M_n(S)$ satisfies the condition (1.1) (see [17, Section 1]). It is easy to check that $Z(M_n(S)) = Z(S)\cdot 1$. This implies that the conditions (ⅰ) and (ⅱ) in Theorem 5.1 are satisfied. We now claim that the condition (ⅲ) in Theorem 5.1 is satisfied.

We set

It is clear that $Z(S)\cdot 1\subseteq \mathcal{W}$. Note that

For any $X\in \mathcal{W}$, we may assume that

where $\lambda_1, \lambda_2\in Z(S)$. We have that

This implies that $(\lambda_1-\lambda_2)e_{1, s+1} = 0$. Hence, $\lambda_1 = \lambda_2$. It follows that $X\in Z(S)\cdot 1$. Thus, $\mathcal{W}\subseteq Z(S)\cdot 1$. Hence, $\mathcal{W} = Z(S)\cdot 1$, as desired. We obtain that $M_n(S)$ satisfies all conditions in Theorem 5.1.

Let $f$ be a supercommuting map of $M_n(S)$. By Theorem 5.1 we have that there exist $\lambda\in Z(S)$ with $2\lambda M_n(S)_1^2 = 0$, and a supercentral map $\tau:M_n(S)\rightarrow Z(S)\cdot 1$ such that

for all $X\in M_n(S)$. It suffices to prove that $2\lambda = 0$. It is clear that $e_{1, s+1}+e_{s+1, 1}\in M_{n}(S)_1$. It follows that

Hence, $2\lambda = 0$. This proves the result. □

As a consequence of theorem 6.1 we have the following interesting result.

Corollary 6.1. Let $S$ be a $2$-torsion free unital algebra. Let $M_n(S)$ be the $n\times n$ matrix algebra with $n\geq 2$. Then every supercommuting map on $M_n(R)$ is supercentral.

Proof. Let $f$ is a supercommuting map on $M_n(S)$. By Theorem 6.1 we have that $f(X) = \lambda X+\tau(X)$, where $\lambda\in Z(S)$ with $2\lambda = 0$, and $\tau:M_n(S)\rightarrow Z(S)\cdot 1$. Since $S$ is $2$-torsion free we get that $\lambda = 0$. This implies that $f(M_n(S))\subseteq Z(S)\cdot 1$. This proves the result. □

We conclude the section with an example, which implies that a commuting map is not a supercommuting map in general.

Example 6.1. Let $S$ be a $2$-torsion free unital algebra. Let $M_n(S)$ be the $n\times n$ matrix algebra with $n\geq 2$. We defined a linear map $f:M_n(S)\rightarrow M_n(S)$ by

for all $x\in M_n(S)$. Then $f$ is a commuting map on $M_n(S)$, but $f$ is not a supercommuting map on $M_n(S)$.

Proof. It is clear that $f$ is a commuting map. Assume that $f$ is a supercommuting map. By Corollary 6.1 we get that $f(x) = x\in Z(S)\cdot 1$ for all $x\in M_n(S)$. In particular, we have that

which is a contradiction. Hence, $f$ is not a supercommuting map. □

7.   Supercommuting maps of triangular algebras

Let $\mathcal{A}$ be a unital algebra with a nontrivial idempotent $e$ satisfying (1.1). If $f\mathcal{A}e = 0$, we have that

where $M$ is a faithful $(A, B)$-bimodule. In this case, $\mathcal{A}$ is said to be a triangular algebra (see ^[18,28] for details). Upper triangular matrix algebras and nest algebras are the two usual examples of triangular algebras (see ^[18] for details).

In 2001, Cheung ^[18] initiated the study of commuting maps on triangular algebra. He determined the class of triangular algebras for which every commuting map is proper (see [18, Theorem 8]. In 2003, Cheung ^[28] gave a description of Lie derivations of triangular algebras. In 2012, Du and Wang ^[20] discussed $k$-commuting maps of triangular algebras, where $k\geq 1$.

We set

Note that $\mathcal{A} = \mathcal{A}_0\bigoplus\mathcal{A}_1$ is a superalgebra (see ^[26] for details).

We begin with the following result, which shows that a supercommutator is a commutator in triangular algebras.

Proposition 7.. Let $\mathcal{A}$ be a triangular algebra. Then $[x, y]_s = [x, y]$ for all $x, y\in \mathcal{A}$.

Proof. For any $a, a'\in e\mathcal{A}e$, $m, m'\in e\mathcal{A}f$, and $b, b'\in f\mathcal{A}f$, we note that

We get that

We obtain that $[x, y]_s = [x, y]$ for all $x, y\in \mathcal{A}$. This proves the result. □

As a consequence of Proposition 7.1, we have the following interesting result.

Corollary 7.1. Every supercommuting map is the same as a commuting map on triangular algebras.

8.   Conclusions

Let $\mathcal{A}$ be a unital algebra with nontrivial idempotents. We consider $\mathcal{A}$ as a generalized matrix algebra according to Ghahramani and Zadeh's method. We give a description of supercommuting maps on generalized matrix algebras. As a consequence, we give a description of supercommuting maps on matrix algebras, which is different from the result on commuting maps of matrix algebras. We finally prove that every supercommuting map is the same as a commuting map on triangular algebras.

Author contributions

Yingyu Luo: Writing-riginal draft, Funding acquisition; Yu Wang: Validation, Writing-review & editing.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The first author is supported by Scientific Research Foundation of Jilin Province Education Department (JJKH20241000KJ) and Doctoral research start-up fund project of Changchun Normal University.

Conflict of interest

The authors declare no conflict of interest.