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This question already has an answer here:

Let $k$ be a field and let $R = k[x,y]/(x^2-y^2+y^3)$. Note that $R$ is an integral domain. Let $F$ be the field of fractions of $R$. How to determine the integral closure of $R$ in $F$?

I have no idea how this integral closure looks like. But I find that $F = k(\overline{x}/\overline{y})$, because $\overline{y} = -(\overline{x}/\overline{y})^2 + 1$ and hence $\overline{x} = -(\overline{x}/\overline{y})^3 + (\overline{x}/\overline{y})$.

Also, a general method for find the integral closure of an integral domain in its field of fractions is strongly desirable. Many thanks.

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marked as duplicate by user26857, user147263, MathOverview, Zev Chonoles abstract-algebra Jul 2 '15 at 18:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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One can uses a parametrization of $x^2-y^2+y^3=0$: $x=t^3-t$, and $y=1-t^2$. This shows that $k[x,y]/(x^2-y^2+y^3)\simeq k[t^3-t,t^2-1]$. But the integral closure of $k[t^3-t,t^2-1]$ is $k[t]$, so you can conclude that the integral closure of $k[x,y]/(x^2-y^2+y^3)$ is isomorphic to $k[t]$ (or, if you like, equals $k[x/y]$).

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    $\begingroup$ Could you please explain why the integral closure of $k[t^3-t,t^2-1]$ is $k[t]$? I can only see that $k[t]$ is contained in its integral closure. BTW, do you have any reference for the method of parametrization? Thank you very much. $\endgroup$ – Zhulin Li Jul 2 '15 at 8:19
  • $\begingroup$ The ring extension $k[t^3-t,t^2-1]\subset k[t]$ is integral, both rings have the same field of fractions, that is, $k(t)$ and $k[t]$ is integrally closed. $\endgroup$ – user26857 Jul 2 '15 at 8:21
  • $\begingroup$ Vakil's book on the foundations of algebraic geometry gives plenty of examples of parametrizations. $\endgroup$ – user26857 Jul 2 '15 at 8:23
  • $\begingroup$ Do you mean this one?math.stanford.edu/~vakil/216blog/FOAGjun1113public.pdf $\endgroup$ – Zhulin Li Jul 2 '15 at 8:59

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