# The identifications of $R$ in its ring of fractions $S^{-1}R$

If $R$ is an integral domain, we can identify the elements $r\in R$ as elements $rs/s$ of the ring of fractions $S^{-1}R$. In this way, we can identify $r\in R$ as $r/1_R$. I've seen in somewhere that we can have a more general case identifying $r$ as $r/u$, where $u$ is a unit, is that true?

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The relevant result: $\tau: R \to S^{-1}R \ , \ r\mapsto r/1$ is injective if $S$ doesn't contain zero-divisors of $R.$

Proof: Suppose $r/1 = r'/1.$ By definition, this is true if and only if there exists $s\in S$ such that $s(r\cdot 1 - r'\cdot 1)= s( r-r')=0.$ But $S$ has no zero divisors, so we must have $r=r'.$

You can modify this proof to show that the map $\sigma: R \to S^{-1}R \ , \ r \mapsto r/u$ (where $u$ is a unit of $R$ contained in $S$) is injective if $S$ doesn't contain zero-divisors of $R.$

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using your argument it seems that we can use any nonzero element of $S$ instead of 1. Am I wrong? – user42912 Jun 9 '13 at 5:24
@user42912 No you are correct. – Ragib Zaman Jun 9 '13 at 6:56