When is this statement true $\frac{a}{s} \in S^{-1}I \implies a \in I$?

Question

If $R$ is a ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset of $R$.

We know that if $a \in I$ then $\frac{a}{s} \in S^{-1}I$. But the converse isn't true always. If $\frac{a}{s} \in S^{-1}P$ then $a \in P$ if $P$ is a prime ideal and $P \cap S= \emptyset$.

My question is that, if we know that $I \cap S= \emptyset$, is there another property that we can have on the ideal $I$ so that this statement $\frac{a}{s} \in S^{-1}I \implies a \in I$ is true? (I mean other than $I$ being prime.)

Answer 1

If $q$ is primary and $S \cap q=\varnothing$ then, $\frac{a}{s} \in S^{-1}q \implies a \in q$

Answer 2

$I \subset R$ with $I\cap S = \emptyset$ satisfies your property if and only if whenever $sa\in I$ with $s\in S$ and $a\in R$, we have that $a\in I$.

To prove it is sufficient is the same proof as in the prime case. To prove it is necessary, suppose there is some $a\in R-I$ and $s\in S$ with $sa \in I$. Then $a/1 = (sa)/s \in S^{-1}I$ but $a\notin I$.