Let $R$ be an integral domain, and $D\subset R$ be a multiplicatively closed subset such that $1\in D$ and $0\not\in D$ . Prove/disprove that there is a one-to-one correspondence between the ideals of the ring of fractions $D^{-1}R$ and the ideals $I$ of $R$ such that $I\cap D=\emptyset$.
I've shown that every ideal in $D^{-1}R$ is the extension of some ideal in $R$ and that an ideal $I\subset R$ is a contraction of some ideal in $D^{-1}R$ if and only if ($\forall d\in D, x\in R:dx\in I\implies x\in I$).
I think that this fact is false but I can't find a counterexample.
Consider $A = k[x, y]$ and $D = \{1, x, x^2, \ldots\}$. The ideals $(y)$ and $(xy)$ are distinct in $A$ and neither intersects $D$, but when we invert $x$ they become the same ideal.