extension by localization
Let $R$ be a commutative ring and let $S$ be a nonempty multiplicative subset of $R$. Then the localisation (http://planetmath.org/Localization^{}) of $R$ at $S$ gives the commutative ring ${S}^{1}R$ but, generally, it has no subring isomorphic^{} to $R$. Formally, ${S}^{1}R$ consists of all elements $\frac{a}{s}$ ($a\in R$, $s\in S$). Therefore, ${S}^{1}R$ is called also a ring of quotients of $R$. If $0\in S$, then ${S}^{1}R=\{0\}$; we assume now that $0\notin S$.

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The mapping $a\mapsto \frac{as}{s}$, where $s$ is any element of $S$, is welldefined and a homomorphism^{} from $R$ to ${S}^{1}R$. All elements of $S$ are mapped to units of ${S}^{1}R$.

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If, especially, $S$ contains no zero divisors^{} of the ring $R$, then the above mapping is an isomorphism from $R$ to a certain subring of ${S}^{1}R$, and we may think that ${S}^{1}R\supseteq R$. In this case, the ring of fractions of $R$ is an extension ring of $R$; this concerns of course the case that $R$ is an integral domain^{}. But if $R$ is a finite ring, then ${S}^{1}R=R$, and no proper extension is obtained.
Title  extension by localization 
Canonical name  ExtensionByLocalization 
Date of creation  20130322 14:24:42 
Last modified on  20130322 14:24:42 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  15 
Author  pahio (2872) 
Entry type  Definition 
Classification  msc 13B30 
Synonym  ring extension by localization 
Related topic  TotalRingOfFractions 
Related topic  ClassicalRingOfQuotients 
Related topic  FiniteRingHasNoProperOverrings 
Defines  ring of fractions 
Defines  ring of quotients 