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Let $X=V(X^3-Y^2)$ the cuspidal cubic over an algebraically closed field $k$.

This variety $X$ has a singularity at $(0,0)$ and this fact is reflected in the coordinate ring as $A=k[X,Y]/(X^3-Y^2)=k[x,y]$ being not normal. The normalization of this ring (i.e the integral closure in the fraction field) is given by $A[\frac{y}{x}]$ as is proved here.

I want to write $A[\frac{y}{x}]$ as a quotient of the polinomial ring $A[Z]$ by an ideal $I$ in order to find an embedding of $\text{spec}(A[\frac{y}{x}])$ in $\mathbb{A}^3$ (given by $V(I)$) and a natural map $\text{spec}(A[\frac{y}{x}])\rightarrow \text{spec}(A)=X$ induced by $A\rightarrow A[Z]/I$.

For this I need to compute the kernel of the map $A[Z]\rightarrow A[\frac{y}{x}]$ given by $z\mapsto \frac{y}{x}$ but I can't do this. I have noticed that this kernel contains the ideal $(Zx-y,Zy-x^2)$ but I think this is not the entire kernel because the variaty $V(X^3-Y^2,ZX-Y,ZY-X^2)$ is still singular at 0.

Does someone has any idea in how to proceed here?

Thanks in advance.

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    $\begingroup$ The equation of integrality for $y/x$ over $A$, $Z- x^2$, needs to be in the kernel. $\endgroup$
    – Youngsu
    May 24, 2018 at 4:58
  • $\begingroup$ This post may be helpful. $\endgroup$
    – Viktor Vaughn
    May 24, 2018 at 5:50

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