Let $D$ be a domain, S a submonoid of the multiplicative monoid of non-zero elements of $D$, $F$ its field of fractions.
Put $D_S$ the subring of the field of fractions $F$ consisting of the elements $as^{-1}$, $a$ in $D$, $s$ in $S$.
Let $I$ be a fractional ideal for $D$ and put $I_S$ the set of the elements $bs^{-1}$, $b$ in $I$, $s$ in $S$.
I want to show that $I_S$ is a $D_S$-fractional ideal of $F$.
- It is clear that $I_S$ is a $D_S$-submodule of $F$.
- $I_S$ is diffrent from zero.
- If $a$ is a non-zero element of $D$ such that $aI\subset$$D$, it follows that $aI_S\subset$$D_S$.
- I am not understanding why $F$ is also the field of fractions of $D_S$?
Would you help me, please?
Consider the embedding $f : D \hookrightarrow F$. Then $f(S) \subseteq F^\times$ so by the universal property of localization, $f$ factors through $D_S$:
$$ D \to D_S \to F. $$
Since $D \to D_S$ and $D \to F$ are injections, so is $D_S \to F$. We can view an injection as a subring so let us consider this as
$$ D \subseteq D_S \subseteq F. $$
The fraction field of $D_S$ is the smallest field containing $D_S$. So we have $D \subseteq D_S \subseteq \operatorname{Frac}(D_S)$. Hence $\operatorname{Frac}(D_S)$ is a field containing $D$. It follows that $F \subseteq \operatorname{Frac}(D_S)$ since $F$ is the smallest field containing $D$. And because $F$ is a field containing $D_S$, $\operatorname{Frac}(D_S) \subseteq F$ since $\operatorname{Frac}(D_S)$ is the smallest field containing $D_S$. Thus $F = \operatorname{Frac}(D_S)$.