Let $D$ be a domain, $S$ a submonoid of the multiplicative monoid of non-zero elements of $D$. Let $D_S$ be the subring of the field of fractions $F$ of $D$ consisting of elements $as^{-1}$, $a$ in $D$, $s$ in $S$.
Let $I'$ be a fractional ideal for $D_S$. Why $I'$ has the form $I_S$ for some fractional ideal $I$ in $D$?
Would you help me, please? Thank you in advance.
Let $R$ be a commutative ring with identity, and $S \subseteq R$ a multiplicative subset. There is a natural ring homomorphism $\phi:R \rightarrow S^{-1}R$. If $R$ is a domain, then all localizations are subrings of the field of fractions of $R$, and $\phi$ is the inclusion map. If $I$ is an ideal of $R$, then $S^{-1}I$ is defined to be the ideal in $S^{-1}R$ generated by the image of $I$ under $\phi$. It is easy to check that every element of $S^{-1}I$ is equal to $\frac{r}{s}$ for some $s \in S$ and $r \in I$.
Lemma: Every ideal of $S^{-1}R$ is of the form $S^{-1}I$ for some ideal $I$ of $R$.
Proof: Let $J$ be an ideal of $S^{-1}R$, and let $I$ be the ideal $\phi^{-1}(J)$ of $R$ (if $R$ is a domain, then $I$ is just $J \cap R$). I claim that $J = S^{-1}I$. Obviously $S^{-1}I \subseteq J$. Conversely if $\frac{r}{s} \in J$ for some $r \in R$, and $s \in S$, then $\phi(r) = \frac{r}{1} = \frac{s}{1} \frac{r}{s} \in J$. Thus $r \in \phi^{-1}(J) = I$, and so $\frac{r}{s} \in S^{-1}I$. $\blacksquare$
Now let $J$ be a fractional ideal of $D_S$. This means that there exists a nonzero $x \in F$ (in fact, in $D_S$) such that $xJ \subseteq D_S$. Then $xJ$ is an honest ideal of $D_S$, so by the lemma it is equal to $I_S$ for some ideal $I$ of $D$. Now $J = x^{-1}I_S$, which you can easily check is equal to $(x^{-1}I)_S$, the localization of the fractional ideal $x^{-1}I$ of $R$ at $S$.