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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