Let $R$ be a prime ring and $I$ be a nonzero ideal of $R.$ A mapping $d:R\rightarrow R$ is called a multiplicative semiderivation if there exists a function $g:R\rightarrow R$ such that (i) $d(xy)=d(x)g(y)+xd(y)=d(x)y+g(x)d(y)$ and (ii) $d(g(x))=g(d(x))$ hold for all $x,y\in R.$ In the present paper, we shall prove that $[x,d(x)]=0,$ for all $x\in I$ if any of the followings holds: i) $d(xy)\pm xy\in Z,$ ii) $d(xy)\pm yx\in Z,$ iii) $d(x)d(y)\pm xy\in Z,$ iv) $d(xy)\pm d(x)d(y)\in Z,$ viii) $d(xy)\pm d(y)d(x)\in Z,$ for all $x,y\in I.$ Also, we show that $R$ must be commutative if $d(I)\subseteq Z.$
📌 3 Key Takeaways
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Objective & Hypothesis:Let $R$ be a prime ring and $I$ be a nonzero ideal of $R.$ A mapping $d:R\rightarrow R$ is called a multiplicative semiderivation if there exists a function $g:R\rightarrow R$ such that (i) $d(xy)=d(x)g(y)+xd(y)=d(x)y+g(x)d(y)$ and (ii) $d(g(x))=g(d(x))$ hold for all $x,y\in R.$ In the present paper, we shall prove that $[x,d(x)]=0,$ for all $x\in I$ if any of the followings holds: i) $d(xy)\pm xy\in Z,$ ii) $d(xy)\pm yx\in Z,$ iii) $d(x)d(y)\pm xy\in Z,$ iv) $d(xy)\pm d(x)d(y)\in Z,$ viii) $d(xy)\pm d(y)d(x)\in Z,$ for all $x,y\in I.$ Also, we show that $R$ must be commutative if $d(I)\subseteq Z.$
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Key Findings & Data:Let $R$ be a prime ring and $I$ be a nonzero ideal of $R.$ A mapping $d:R\rightarrow R$ is called a multiplicative semiderivation if there exists a function $g:R\rightarrow R$ such that (i) $d(xy)=d(x)g(y)+xd(y)=d(x)y+g(x)d(y)$ and (ii) $d(g(x))=g(d(x))$ hold for all $x,y\in R.$ In the present paper, we shall prove that $[x,d(x)]=0,$ for all $x\in I$ if any of the followings holds: i) $d(xy)\pm xy\in Z,$ ii) $d(xy)\pm yx\in Z,$ iii) $d(x)d(y)\pm xy\in Z,$ iv) $d(xy)\pm d(x)d(y)\in Z,$ viii) $d(xy)\pm d(y)d(x)\in Z,$ for all $x,y\in I.$ Also, we show that $R$ must be commutative if $d(I)\subseteq Z.$
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Conclusion & Impact:Let $R$ be a prime ring and $I$ be a nonzero ideal of $R.$ A mapping $d:R\rightarrow R$ is called a multiplicative semiderivation if there exists a function $g:R\rightarrow R$ such that (i) $d(xy)=d(x)g(y)+xd(y)=d(x)y+g(x)d(y)$ and (ii) $d(g(x))=g(d(x))$ hold for all $x,y\in R.$ In the present paper, we shall prove that $[x,d(x)]=0,$ for all $x\in I$ if any of the followings holds: i) $d(xy)\pm xy\in Z,$ ii) $d(xy)\pm yx\in Z,$ iii) $d(x)d(y)\pm xy\in Z,$ iv) $d(xy)\pm d(x)d(y)\in Z,$ viii) $d(xy)\pm d(y)d(x)\in Z,$ for all $x,y\in I.$ Also, we show that $R$ must be commutative if $d(I)\subseteq Z.$
🔬Design:Research Article
📍Scope:Comprehensive Sample
⚙️Variables:multiplicative semiderivation · prime rings · semiderivation
Abstract
Let $R$ be a prime ring and $I$ be a nonzero ideal of $R.$ A mapping $d:R\rightarrow R$ is called a multiplicative semiderivation if there exists a function $g:R\rightarrow R$ such that (i) $d(xy)=d(x)g(y)+xd(y)=d(x)y+g(x)d(y)$ and (ii) $d(g(x))=g(d(x))$ hold for all $x,y\in R.$ In the present paper, we shall prove that $[x,d(x)]=0,$ for all $x\in I$ if any of the followings holds: i) $d(xy)\pm xy\in Z,$ ii) $d(xy)\pm yx\in Z,$ iii) $d(x)d(y)\pm xy\in Z,$ iv) $d(xy)\pm d(x)d(y)\in Z,$ viii) $d(xy)\pm d(y)d(x)\in Z,$ for all $x,y\in I.$ Also, we show that $R$ must be commutative if $d(I)\subseteq Z.$
Gölbaşı, Ö., Bedir, Z. (2022). Some identities involving multiplicative semiderivations on ideals. Hacettepe Journal of Mathematics and Statistics, Cilt: 50 - Sayı: 4, 963–969. https://ajindex.com/articles/529480