Localization of an integral domain and fields of fractions Is it true that every localization of an integral domain is isomorphic to a subring of its field of fractions? How are the localizations of an integral domain related to its field of fractions? Is there a handy criterion to tell whether a subring of an integral domain has the same field of fractions as the overring? I've read that $k[x^2,x^3]\subset k[x]$ ($k$ a field, algebraically closed if this matters) have the same field of fractions, how come? I am very curious about these things.
 A: Answering your questions:


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*Yes. A localization of a domain $R$ with respect to some multiplicatively closed subset $S$ of $R$ is a subring of $K$ (the field of fractions of $R$) consisting of the elements $m/s$, with $m \in R$ and $s \in S$.

*If you take $S = R\setminus\{0\}$, the localization of $R$ with respect the set $S$ is the field of fractions $K$ of $R$.

*The field of fractions of a domain $R$ is the smallest (with relation to inclusion) field that contains $R$. So, two domains have the same field of fractions if the smallest fields that contains each one are the same.

A: Let $D$ be a domain an let $K(D)$ be its fraction field. Let $S$ be a multiplicative subset of $D$.


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*Is $S^{-1}D$ a subring of $K(D)$? Yes, to prove it remembre that injectivity is a local property of morphism.(We obtain that EVERY localization of $D$ is a subring of $K(D)$)

*Have $k[x^2,x^3]$ and $k[x]$ the same fraction field? Yes: We need just to show that $x \in K(k[x^2,x^3])$, but you can obtain $x$ as the class of $x^3/x^2$.

*Are there some handly criterions to show that $A \subset D$ is such that $K(A)=K(D)$? Well,seriously I don't know but i think there are not general results. It works certainly if $A$ and $D$ have the same integral closure.

