On Ideals and Multiplicative (Generalized) - $(\Phi,\Phi)$ - Derivations

S. Lalitha1,2, S. Sreenivasulu1 and A. Mallikarjun Reddy2


1Department of Mathematics, GDC (M), S. K. University, Ananthapuramu, Andhra Pradesh, India.
2Department of Mathematics, S. K. University, Ananthapuramu, Andhra Pradesh, India.

Abstract: Let $P$ be a prime ring. $I$ is a nonzero ideal of $P$. $\Phi$ is an automorphism on P. A mapping $M:P \to P$ is called Multiplicative (generalized) $(\Phi, \Phi)$-derivation if there exist a map $d: P\to P$ such that $M(a, b)=M(a) \Phi(b)+\Phi(a) d(b)$ holds for all $a, b \in P$. The objective of the present paper is to study the following identities (i). If $M(ab)+M(a) M(b)=0$ for all $a, b \in I$ then $\Phi(I)[M(a), M(b)]=0$ for all $a \in I$ (ii). Let $M_{1}$ and $M_{2}$ be two multiplicative (generalized)-$(\Phi, \Phi)$ derivations on P associated with the maps $d_{1}$ and $d_{2}$ on P respectively. If $M_{1}(a b)=\Phi(b) \circ M_{2}(a)$ for all $a, b \in I$ then $R$ is abelian or commutative or $\Phi(I)\left[\Phi(I), M_{2}(I)\right]=0$ (iii). If $M_{1}(ab)=\left[\Phi(b), M_{2}(a)\right]$ for all $a, b \in I$ then either $\Phi(I)\left[\Phi(I), M_{2}(I)\right]=(0)$ or $R$ is commutative.
Keywords: Primering, Ideal, Multiplicative (generalized) derivation, Multiplicative (generalized)-$(\Phi, \Phi)$-derivation.


Cite this article as: S. Lalitha, S. Sreenivasulu and A. Mallikarjun Reddy, On Ideals and Multiplicative (Generalized) - $(\Phi,\Phi)$ - Derivations, Int. J. Math. And Appl., vol. 9, no. 4, 2021, pp. 53-57.

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