# Find the integral closure of an integral domain in its field of fractions [duplicate]

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.

## marked as duplicate by user26857, user147263, MathOverview, Zev Chonoles abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 2 '15 at 18:57

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 $k[x/y]$).
• 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. – Zhulin Li Jul 2 '15 at 8:19
• 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. – user26857 Jul 2 '15 at 8:21