You are here
Homeextension by localization
Primary tabs
extension by localization
Let $R$ be a commutative ring and let $S$ be a nonempty multiplicative subset of $R$. Then the localisation 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 $0\notin S$.

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 $S^{{1}}R$.

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.
Mathematics Subject Classification
13B30 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections