On minimal ring extensions
Rahul Kumar, Atul Gaur
Abstract
Open-access reader
Rahul Kumar, Atul Gaur
Abstract
Open-access reader
Let $R$ be a commutative ring with identity. The ring $R\times R$ can be viewed as an extension of $R$ via the diagonal map $Δ: R \hookrightarrow R\times R$, given by $Δ(r) = (r, r)$ for all $r\in R$. It is shown that, for any $a, b\in R$, the extension $Δ(R)[(a,b)] \subset R\times R$ is a minimal ring extension if and only if the ideal $$ is a maximal ideal of $R$. A complete classification of maximal subrings of $R(+)R$ is also given. The minimal ring extension of a von Neumann regular ring $R$ is either a von Neumann regular ring or the idealization $R(+)R/\mathfrak{m}$ where $\mathfrak{m}\in \text{Max}(R)$. If $R\subset T$ is a minimal ring extension and $T$ is an integral domain, then $(R:T) = 0$ if and only if $R$ is a field and $T$ is a minimal field extension of $R$, or $R_J$ is a valuation ring of altitude one and $T_{J}$ is its quotient field.
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Let $R$ be a commutative ring with identity. The ring $R\times R$ can be viewed as an extension of $R$ via the diagonal map $Δ: R \hookrightarrow R\times R$, given by $Δ(r) = (r, r)$ for all $r\in R$. It is shown that, for any $a, b\in R$, the extension $Δ(R)[(a,b)] \subset R\times R$ is a minimal ring extension if and only if the ideal $$ is a maximal ideal of $R$. A complete classification of maximal subrings of $R(+)R$ is also given. The minimal ring extension of a von Neumann regular ring $R$ is either a von Neumann regular ring or the idealization $R(+)R/\mathfrak{m}$ where $\mathfrak{m}\in \text{Max}(R)$. If $R\subset T$ is a minimal ring extension and $T$ is an integral domain, then $(R:T) = 0$ if and only if $R$ is a field and $T$ is a minimal field extension of $R$, or $R_J$ is a valuation ring of altitude one and $T_{J}$ is its quotient field.
Key concepts: Integral domain, Quotient ring, Ring (chemistry), Mathematics, Maximal ideal, Principal ideal ring, Valuation ring, Quotient