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