Rings of fractions

Question

If $R$ is a ring, $I$ is an ideal of $R$ such that $I\ne R$, and $S$ a multiplicatively closed set of $R$, and we have that this property is true, $$\frac{a}{s} \in S^{-1}I \implies a\in I$$ Can we prove that $S\cap I= \varnothing$ ??

If yes, how is it done? Any help is appreciated...

Answer 1

Suppose $s\in S\cap I$. Then $$ \frac{1}{1}=\frac{s}{s}\in S^{-1}I $$ and therefore $S^{-1}I=S^{-1}R$. It follows that $I=R$, because for every $a\in R$, we have $a/s\in S^{-1}I$, so $a\in I$.

Answer 2

Since $I \ne R \implies S^{-1}I \ne S^{-1}R$

Assume that $S \cap I \ne \varnothing \implies \exists s \in S \cap I$

But $1_{S^{-1}R}=\dfrac{1_R}{1_R}=\dfrac{s}{s} \in S^{-1}I$

$\implies S^{-1}I=S^{-1}R$ #contradiction.

Thus, $S \cap I=\varnothing$

Is my proof correct?