# Conditions for S^-1I=S^-1R (Ring of quotients)

Conditions for $S^{-1}I=S^{-1}R$:
Let $S$ be a multiplicative subset of a commutative ring $R$ with identity and let $I$ be an ideal of $R$. Then $S^{-1}I=S^{-1}R$ if and only if $S\cap I\neq\varnothing$.

Proof
(H pg 146) $(\implies)$ Assume $S^{-1}I=S^{-1}R$. Consider the ring homomorphism $\displaystyle \phi_S: R\to S^{-1}R$ given by $r\mapsto rs/s$ (for any $s\in S$). Then $\phi_S^{-1}(S^{-1}I)=R$ hence $\phi_S(1_R)=a/s$ for some $a\in I$, $s\in S$. Since $\phi_S(1_R)=1_Rs/s$, we have $s_1(s^2-as)=0$ for some $s_1\in S$, i.e.\ $s^2s_1=ass_1$. But $s^2s_1\in S$ and $ass_1\in I$ imply $S\cap I\neq\varnothing$. $(\impliedby)$ If $s\in S\cap I$, then $1_{S^{-1}R}=s/s\in S^{-1}I$. Note that for any $r/s\in S^{-1}R$, $\displaystyle (\frac{r}{s})(\frac{s}{s})=\frac{rs}{s^2}=\frac{r}{s}\in S^{-1}I$ since $rs\in I$ and $s^2\in S$. Thus $S^{-1}I=S^{-1}R$. ## Author: mathtuition88

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