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 Maddede Bu Makale
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Amaç & Hipotez: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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Temel Bulgular & Veri: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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Sonuç & Katkı: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 $d(I)\subseteq Z.$
🔬Tür:Research Article
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler: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 $d(I)\subseteq Z.$
Golbasi, O., Bedir, Z. (2021). Some identities involving multiplicative semiderivations on ideals. Hacettepe Journal of Mathematics and Statistics, Volume: 50 Issue: 4. https://doi.org/10.15672/hujms.650600