Normalization of $\mathbb{C}[x,y]/(y^2-x^5)$

Question

I wish to calculate the normalization of $R=\mathbb{C}[x,y]/(y^2-x^5)$. My first two steps were to define $s=y/x, t=y/x^2$ both of which are integral over $R$, as you can see from dividing the equation by $x^2$, $x^4$ respectively.

Now, I guess that these are all that I need, that is, that $R[t,s]$ is already integrally closed. However when calculating we get: $$R[t,s]\simeq\mathbb{C}[x,y,t,s]/(s^2-x^3,t^2-x,xs-y,x^2t-y)\simeq\mathbb{C}[t,s]/(s^2-t^6,t^2s-t^5).$$

But the latter is super reducible and thus for example not smooth and hence not normal (dimension 1). My question is:

1) Am I missing another generator for the integral closure?

2) Is my calculation of $R[t,s]$ correct?

3) How to calculate the integral closure in this case?

Answer 1

In the ring $R[t,s]$ you have

• $x=t^2$ (you seem to have figured this out), and consequently also
• $y=tx^2=t^5$, and also
• $s=y/x=xt=t^3$.

Therefore $R[t,s]\cong \Bbb{C}[t]$, which is integrally closed.

In other words, you have found the integral closure of $R$. You were missing the generator $xt-s$ from the ideal you are moding out.