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.

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