# 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