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Yapay Zeka Hızlı Bakış & Çıkarımlar
AI Insight⚡ TL;DR(1 Cümlede Hızlı Çıkarım)
We show that $R\{X\}$ is a Prüfer ring if and only if $R$ is a von Neumann regular ring, and so if $R\{X\}$ satisfies one of the Prüfer conditions, it satisfies all of them.
📌 3 Maddede Bu Makale
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Amaç & Hipotez:Let $R$ be a commutative ring with unity and $W=\{f(X)\in R[X]:f(0)=1\}$.
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Temel Bulgular & Veri:We define $R\{X\}=W^{-1}R[X]$.
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Sonuç & Katkı:We show that $R\{X\}$ is a Prüfer ring if and only if $R$ is a von Neumann regular ring, and so if $R\{X\}$ satisfies one of the Prüfer conditions, it satisfies all of them.
🔬Tür:Research Article
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler:Prime ideal · localization · over ring
Abstract
Let $R$ be a commutative ring with unity and $W=\{f(X)\in R[X]:f(0)=1\}$. We define $R\{X\}=W^{-1}R[X]$. We show that the maximal ideals of $R\{X\} $ are of the form $W^{-1}(M,X)$ where $M$ is a maximal ideal of $R$, and so if $R$ is finite dimensional, then $\dim R\{X\}=\dim R[X]$. We show that $R\{X\}$ is a Prüfer ring if and only if $R$ is a von Neumann regular ring, and so if $R\{X\}$ satisfies one of the Prüfer conditions, it satisfies all of them.
Anahtar kelimeler
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Bu Makaleye Atıf Yap
Abuosba, E., Al-azaizeh, M. (2025). The ring $R\{X\}$. International Electronic Journal of Algebra, Volume: 37 Issue: 37. https://doi.org/10.24330/ieja.1478925