Characterization of $(\alpha, \beta)$ Jordan bi-derivations in prime rings

1.   Introduction

We consider $\mathfrak{S}$ as an associative ring throughout the paper, and $\mathfrak{Z}(\mathfrak{S})$ its center, unless otherwise mentioned. For all $k, l \in \mathfrak{S}$, if $k \mathfrak{S} l = \{0\}\; (k \mathfrak{S} k = \{0\})$ implies that $k = 0$ or $l = 0\; (k = 0)$, then $\mathfrak{S}$ is called a prime ring $\text{(semiprime ring)}$. For all $k, l \in \mathfrak{S}$, the representations $[k, l] = kl - lk$ and $k \circ l = kl + lk$ are termed as the commutator (Lie product) and the skew-commutator (Jordan product), respectively. For all $a \in \mathfrak{S}$, $\mathfrak{S}$ is called an $n$-torsion free ring if $na = 0$ implies $a = 0$. A map $g$ is called a centralizing (resp. commuting) map on $\mathfrak{S}$ if $[g(k), l] \in \mathfrak{Z}(\mathfrak{S})$ (resp. $[g(k), l] = 0) \; \forall$ $k, l \in \mathfrak{S}$. An additive map $f: \mathfrak{S} \rightarrow \mathfrak{S}$ satisfying $f(kl) = f(k)l + kf(l)$ (resp. $f(k^2) = f(k)k + kf(k))$ is termed as a derivation (resp. Jordan derivation) $\forall\; k, l \in \mathfrak{S}$. Clearly we can see that every derivation is a Jordan derivation, however the converse is not true in general. In 1957, Herstein [1, Theorem 4.1] showed that the Jordan derivation $d$ of ring $\mathfrak{S}$ is a derivation if $d$ satisfies $d(klk) = d(k)lk + kd(l)k + kld(k)$ for all $k, l \in \mathfrak{S}$, where $\mathfrak{S}$ is a prime ring which is not a commutative integral domain.

For a prime ring $\mathfrak{S}$, an additive map $d$ from $\mathfrak{S}$ to $\mathfrak{S}$ is said to be a $\sigma$-derivation (resp. Jordan $\sigma$-derivation) with associated endomorphism $\sigma$ if $d(kl) = d(k)l + \sigma(k)d(l)$ for all $k, l \in \mathfrak{S}$ (resp. $d(k^2) = d(k)k + \sigma(k)d(k)$). In 2015, Lee ^[2] proved that, for a non-commutative prime ring $\mathfrak{S}$, every Jordan $\sigma$-derivation is a $\sigma$-derivation if it satisfies the condition $d(klk) = d(k)lk + \sigma(k)d(l)k + \sigma(kl)d(k)$ for all $k, l \in S.$

If a map $d$ from $\mathfrak{S} \times \mathfrak{S}$ to $\mathfrak{S}$ is additive in both arguments, then it is called bi-additive and $d$ is said to be symmetric if $d(k, l) = d(l, k)$. Let $d$ be a symmetric bi-additive map satisfying the condition $d(kl, s) = d(k, s)l + kd(l, s)$ (resp. $d(k^2, s) = d(k, s)k + kd(k, s)$) for all $k, l, s \in \mathfrak{S}$. Then $d$ is said to be a symmetric bi-derivation (resp. symmetric Jordan bi-derivation). In 1980 ^[3], Maksa was the first to introduce the idea of symmetric bi-derivation, and after that Vukman ^[4] proved some results related to symmetric bi-derivation using prime and semiprime rings. In 2017, Abdioǧlu and Lee ^[5] studied a basic functional identity in the presence of bi-additive maps on a non-commutative prime ring $\mathfrak{S}$. Precisely, they have proved: "Let $\mathfrak{S}$ be a non-commutative prime ring and let $\mathfrak{B}: \mathfrak{S} \times \mathfrak{S} \rightarrow \mathfrak{Q}_{ml}(\mathfrak{S})$ be a bi-additive map. Suppose that $[\mathfrak{B}(x, y), [x, y]] = 0$ for all $x, y \in \mathfrak{S}$. Then there exist $\lambda \in \mathfrak{C}$ and a bi-additive map $\beta: \mathfrak{S} \times \mathfrak{S} \rightarrow \mathfrak{C}$ such that $\mathfrak{B}(x, y) = \lambda[x, y] + \beta(x, y)$ for all $x, y \in \mathfrak{S}$". We can also find some related results [2, Theorem 2.1] and their application in order to investigate Jordan $\sigma$-derivation.

Motivated by all these mentioned results, in this paper we defined symmetric ($\alpha, \beta$) Jordan bi-derivation in prime rings. It is a map $\mathfrak{D}$ from $\mathfrak{S} \times \mathfrak{S}$ to $\mathfrak{S}$ defined by $\mathfrak{D}(k^2, s) = \mathfrak{D}(k, s)\alpha(k) + \beta(k) \mathfrak{D}(k, s)$ for all $k, s \in \mathfrak{S}$. Any symmetric ($\alpha, \beta$) Jordan bi-derivation is symmetric ($\alpha, \beta$) bi-derivation, but the converse is not true in general. In this article, we find the answer under which the map symmetric ($\alpha, \beta$) Jordan bi-derivation becomes an ($\alpha, \beta$) bi-derivation in prime rings $\mathfrak{S}$. Also, we characterize the symmetric ($\alpha, \beta$) Jordan bi-derivation on any rings $\mathfrak{S}$.

2.   Preliminaries

Lemma 2.1. [2,Theorem 2.1] Let $\mathfrak{S}$ be a non-commutative prime ring and $f : \mathfrak{S} \times \mathfrak{S} \rightarrow \mathfrak{Q}_{mr}(\mathfrak{S})$ be a bi-additive map. If $f(k, l)[k, l] = 0$ for all $k, l \in \mathfrak{S}$, then $f = 0$.

3.   Results

Theorem 3.1. Let $\mathfrak{D}$ be a symmetric $(\alpha, \beta)$ Jordan bi-derivation on a ring $\mathfrak{S}$ with char$(\mathfrak{S}) \ne 2$. Consequently, the following claims are true for all $k, l, s \in \mathfrak{S}$.

(1) $\mathfrak{D}(kl + lk, s) = \mathfrak{D}(k, s)\alpha(l) + \mathfrak{D}(l, s)\alpha(k) + \beta(l) \mathfrak{D}(k, s) + \beta(k) \mathfrak{D}(l, s).$

(2) $\mathfrak{D}(klk, s) = \mathfrak{D}(k, s)\alpha(l)\alpha(k) + \beta(k)\beta(l) \mathfrak{D}(k, s) + \beta(k) \mathfrak{D}(l, s)\alpha(k)$.

(3) $\mathfrak{D}(klr + rlk, s) = \mathfrak{D}(k, s)\alpha(l)\alpha(r) + \beta(k) \mathfrak{D}(l, s)\alpha(r) + \mathfrak{D}(r, s)\alpha(l)\alpha(k)\nonumber + \beta(r) \mathfrak{D}(l, s)\alpha(k) + \beta(r)\beta(l) \mathfrak{D}(k, s) + \beta(k)\beta(l) \mathfrak{D}(r, s)$.

(4) $\mathfrak{D}(klr + lrk, s) = \mathfrak{D}(lr, s)\alpha(k) + \mathfrak{D}(k, s)\alpha(l)\alpha(r) + \beta(k) \mathfrak{D}(lr, s) + \beta(l)\beta(r) \mathfrak{D}(k, s)$.

(5) $\mathfrak{D}(l^2, s)\alpha(k) + \beta(k) \mathfrak{D}(l^2, s) = \mathfrak{D}(l, s)\alpha(l)\alpha(k) + \beta(k) \mathfrak{D}(l, s)\alpha(l) + \beta(l) \mathfrak{D}(l, s)\alpha(k) + \beta(k) \mathfrak{D}(l, s)\alpha(l) + \beta(k)\beta(l) \mathfrak{D}(l, s)$.

Proof. ${(1)}$ In the definition of symmetric $(\alpha, \beta)$ Jordan bi-derivation, substituting $k$ by $k + l$, we get

Also, we have

From (3.1) and (3.2), we conclude that

(2) Replacing $l$ by $kl + lk$ in (3.3), we obtain

Using Theorem 3.1 $(1)$ in the right hand side of (3.4), after simplifying we get

Now, solving the left side of (3.4) and using (3.3), we have

From (3.5) and (3.6), we see that

Since char$(\mathfrak{S}) \ne 2$, then from the last relation we get the required result:

(3) Substituting $k$ by $k + r$ in (2) for all $r \in \mathfrak{S}$, we have

Now, solving the left side of (3.8), we obtain

Now, solving the right side of (3.8), we obtain

From (3.9) and (3.10), we find that

This is the required result.

(4) Substituting $k$ by $k + lr$ in the definition of $(\alpha, \beta)$ Jordan bi-derivation, we get

Solving the left side of (3.12) and using the definition of $(\alpha, \beta)$ Jordan bi-derivation, we see that

Now, solving the right side of (3.12) and using the definition of $(\alpha, \beta)$ Jordan bi-derivation, we find that

Comparing (3.13) and (3.14), we have

We are done.

(5) Subtracting (3.15) from (3.11), we get

In particular, for $r = l$, (3.16) reduces to

which implies that

So, the last equality is done. Hence, with this proof our theorem is completed. □

Definition 3.2. Let $\mathfrak{S}$ be a ring with char$(\mathfrak{S}) \ne 2$. We define the symbol

for all $k, l, s \in \mathfrak{S}$.

Now we present a theorem regarding $k^{l}$ under the action of symmetric $(\alpha, \beta)$ Jordan bi-derivation.

Theorem 3.3. Let $\mathfrak{D}$ be a symmetric $(\alpha, \beta)$ Jordan bi-derivation on a ring $\mathfrak{S}$. Consequently, the following claims are true for all $k, k_i, l, l_i, s \in \mathfrak{S}$, for $i \in \{1, 2\}$.

(1) $k^{l} + l^{k} = 0$.>

(2) $k^{l_1 + l_2} = k^{l_1} + k^{l_2}$.

(3) $(k_1 +k_2)^{l} = k_1^{l} + k_2^{l}$.

Proof. ${(1)}$ Let $k, l, s \in \mathfrak{S}$. Then, from Definition 3.2, we have

The above relation can be re-written as

Using (2) of Theorem 3.1 in (3.20), we get

Solving (3.21), we get

(2) Let $k, l_1, l_2, s \in \mathfrak{S}$. Then, from Definition 3.2, we have

The above relation can be re-written as

Using Definition 3.2 in (3.23), we find that

(3) Let $k_1, k_2, l, s \in \mathfrak{S}$. Then, from the Definition 3.2, we have

The above relation can be re-written as

Using Definition 3.2 in (3.26), we see that

□

Since we know that every symmetric $(\alpha, \beta)$ bi-derivation of $\mathfrak{S}$ is always a symmetric $(\alpha, \beta)$ Jordan bi-derivation of $\mathfrak{S}$, the converse need not be true. So, in our next theorem we will show that a symmetric $(\alpha, \beta)$ Jordan bi-derivation of $\mathfrak{S}$ becomes a symmetric $(\alpha, \beta)$ bi-derivation of $\mathfrak{S}$ by imposing certain conditions on ring $\mathfrak{S}$.

Theorem 3.4. Let $\mathfrak{S}$ be a non-commutative prime ring with automorphisms $\alpha$ and $\beta$. Any symmetric $(\alpha, \beta)$ Jordan bi-derivation $\mathfrak{D}$ of $\mathfrak{S}$ which satisfies the properties (2) and (3) of Theorem 3.1 (it is trivial if char ($\mathfrak{S}) \ne$ 2), $\mathfrak{D}$ is a symmetric $(\alpha, \beta)$ bi-derivation of $\mathfrak{S}$.

Proof. Substituting $r$ by $kl$ in (3) of Theorem 3.1, we get

We solve this equation in two parts. In the first part we solve the right side of (3.28), and in second we solve the left side of (3.28). Solving right side, we get

Now, solving the left side of (3.28), we find that

Using the definition of $(\alpha, \beta)$ Jordan bi-derivation and (2) of Theorem 3.1 in (3.30), we see that

The above relation can be re-written as

Comparing (3.29) and (3.32), we find that

The above equation can be re-arranged as

This implies that

By using Lemma 2.1 in (3.34), we obtain

That is,

So, the above equation is a symmetric $(\alpha, \beta)$ bi-derivation of $\mathfrak{S}$, and we are done. □

4.   Conclusions

In this paper, we focused on symmetric $(\alpha, \beta)$ Jordan biderivation $\mathfrak{D}$ on a prime ring $\mathfrak{S}$, where $\alpha$ and $\beta$ are automorphisms of $\mathfrak{S}$. In Theorem 3.1, we have defined certain properties of $(\alpha, \beta)$ Jordan biderivation $\mathfrak{D}$ on a ring $\mathfrak{S}$. Finally, we have provided the conditions under which the symmetric $(\alpha, \beta)$ Jordan biderivation $\mathfrak{D}$ transforms into a symmetric $(\alpha, \beta)$ biderivation on a prime ring $\mathfrak{S}$.

Use of AI tools declaration

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

Acknowledgments

The authors are thankful to the anonymous referees for their valuable comments and suggestions which have improved the manuscript immensely.

The second author extend their appreciation to Princess Nourah Bint Abdulrahman University (PNU), Riyadh, Saudi Arabia, for funding this research under the Researchers Supporting Project, No. PNURSP2024R231. Also, third author was very thankful to the DST-SERB (project MATRICS, File No. MTR/2022/000153) for their necessary support and facility.

Conflict of interest

The authors have no competing interests to declare that are relevant to the content of this article.