Let $R=\mathbb{C}[x,y,z]/(y^3-x^2z)$ and let $Q$ be the quotient field of $R$. Find integral closure of $R$ in $Q$.

I found that $$R \cong \mathbb{C}[s,st,st^3] \subset \mathbb{C}[s,t],\quad (st^2)^2-(st)(st^3)=0$$ so I expect that the integral closure of $R$ is $$\displaystyle \mathbb{C}[s,st,st^3][st^2]=R\left[\frac{\overline{x}\overline{z}}{\overline{y}}\right]$$ How to prove it? Thanks.

  • $\begingroup$ Show that $\mathbb C[s,st,st^2,st^3]$ is integrally closed. This is isomorphic to $\mathbb C[x^3,x^2y,xy^2,y^3]$ which is known as being normal. $\endgroup$ – user26857 Aug 19 '16 at 5:20
  • $\begingroup$ One way is to realize the integral closure you wrote as the toric variety coming from a cone generated by $e_2, 3e_1 -e_2$ in $\mathbb{Z}^2$. Such varieties are normal. $\endgroup$ – Youngsu Aug 21 '16 at 20:30

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