Let $R$ be a commutative ring with identity $1_R$, $0 \not \in S\subseteq R$ be a multiplicative set, and $I\subseteq R$ be an ideal of $R$.Consider the ring of quotients $S^{-1}I$.
I was trying to prove a theorem, and I proved almost everything, but stuck at showing that if $\frac{b}{s} \in I $, then $b \in I$, so my question is
Does $\frac{b}{s} \in S^{-1} I$ directly implies that $b \in I$ ?
The definition only says that for $i \in I, s' \in S$ satisfying $\frac{b}{s} = \frac{i}{s'}$, i.e $\exists s_1 \in S \quad s.t \quad s_1 (s'b - si) = 0,$, the equivalence class $b/s \in S^{-1}I$. but since $R$ is not necessarily an integral domain, I couldn't derive anything useful from this relation.
Edit:
In the original theorem, $I$ and $b$ has some other properties, but what I want to understand in this question is that whether I need to look for other thing, or I'm done but just I haven't realized it.
No, this is false. For example, let $\newcommand{\Z}{\mathbb{Z}}R = \Z$, $P = 3\Z$, so $S = \Z \setminus 3\Z$, and $I = 6\Z$. Then $$ I_P = S^{-1} I = \left\{\frac{6m}{n} : m,n \in \Z, 3 \nmid n\right\} \, . $$ Since $2 \in S$ then $\frac{3}{1} = \frac{6}{2} \in I_P$, but $3 \notin I$.