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AI Summary & Key Takeaways
AI Insight⚡ TL;DR(Too Long; Didn't Read)
Commutativity of R has also been investigated under different sets of constraints on integral exponents.
📌 3 Key Takeaways
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Objective & Hypothesis:In the present paper, it is shown that if R is a left ( resp. right) s-unital ring satisfying $[f(y^mx^ry^s) pm x^ty, x]$ = 0 (resp. $[f(y^mx^ry^s) pm yx^t, x]$ = 0), where m, r, s, t are fixed non-negative integers and $f(lambda)$ is a polynomial in $lambda^2{bf Z}[lambda]$, then R is commutative.
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Key Findings & Data:In the present paper, it is shown that if R is a left ( resp. right) s-unital ring satisfying $[f(y^mx^ry^s) pm x^ty, x]$ = 0 (resp. $[f(y^mx^ry^s) pm yx^t, x]$ = 0), where m, r, s, t are fixed non-negative integers and $f(lambda)$ is a polynomial in $lambda^2{bf Z}[lambda]$, then R is commutative.
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Conclusion & Impact:Commutativity of R has also been investigated under different sets of constraints on integral exponents.
🔬Design:Research Article
📍Scope:Comprehensive Sample
⚙️Variables:Domain Parameters
Abstract
In the present paper, it is shown that if R is a left ( resp. right) s-unital ring satisfying $[f(y^mx^ry^s) pm x^ty, x]$ = 0 (resp. $[f(y^mx^ry^s) pm yx^t, x]$ = 0), where m, r, s, t are fixed non-negative integers and $f(lambda)$ is a polynomial in $lambda^2{bf Z}[lambda]$, then R is commutative. Commutativity of R has also been investigated under different sets of constraints on integral exponents.
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Cite This Article
KHAN, A. M. (2022). Some commutativity results for S-unital rings. Turkish Journal of Mathematics, Cilt: 24 - Sayı: 2, 165–172. https://ajindex.com/articles/608938