I've been struggling in this problem:
If $m$ is a maximal ideal of $R:=\mathbb{C}[x,y]/(y^2-x^2(x+1))$, when is $R_m$ integrally closed (normal)?
Well, I know that $t=y/x$ is integral over R, but isn't in R, what makes it not normal. Moreover, $R[t]\simeq \mathbb{C}[x]$ is the integral closure of R in $Frac(R)$ and for $\overline{m}=\overline{(x-a,y-b)}$ such that $(y^2-x^2(x+1))\subseteq m$ (maximals of R) and $S=R-m$, we have $S^{-1}\mathbb{C}[x]$ as the integral closure of $R_m$ in $S^{-1}Frac(R)$. So the question is:
Is it true that $S^{-1}Frac(R)\simeq Frac(R_m)$?
If yes, then I can advance by noticing that when $a\neq 0\Leftrightarrow x\not \in m \Leftrightarrow x \in S$, then $t=y/x\in R_m$ and $R_m=R_m[t]\simeq S^{-1}R[t]\simeq S^{-1}\mathbb{C}[x]$, that is, $R_m$ is normal. But:
Is it true that $R_m[t]\simeq S^{-1}R[t]$?
I also know that for $a=0, b\neq0$, $y\in S$ and $t$ is invertible in $R_m$.
Then what happens with $R_m[t]$? Perhaps $R_m[t]\simeq S^{-1}\mathbb{C}[x]\simeq \mathbb{C}$?
If the answer to this question implies on $R_m$ is normal, we're ok. Now, when $a=b=0$, we have $x,y\not \in S$, then $r=(y/s)/(x/s)\in Frac(R_m)$ is integral over $R_m$, because $r^2-\frac{(x+1)}{1}=0$ and $r\not \in R_m$. Thus $R_m$ isn't normal.
Can you point out what I missed and my mistakes?
Thanks in advance.
P.S.: I asked another question around this problem, but focusing on singularities, which is not the point here.
