extension by localization

Let $R$ be a commutative ring and let $S$ be a non-empty 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  $0\notin S$.

• The mapping  $a\mapsto \frac{as}{s}$, where $s$ is any element of $S$, is well-defined 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.

Title	extension by localization
Canonical name	ExtensionByLocalization
Date of creation	2013-03-22 14:24:42
Last modified on	2013-03-22 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