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 $S^{-1}I=S^{-1}R$ .