Let $V$ be the curve defined by $x^3+x^2-y^2$, and let $\mathbb{C}[V]$ be its coordinate ring. Let $\alpha = \frac{\bar{y}}{\bar{x}}$. I want to prove that for every maximal ideal $M \subseteq \mathbb{C}[V]$, $M \neq (\bar{x},\bar{y})$, there exists a maximal ideal $N \subseteq \mathbb{C}[V][\alpha]$ such that $M = \mathbb{C}[V] \cap N$ and $\mathbb{C}[V]_M = \mathbb{C}[V][\alpha]_N$ (i.e. the localizations with respect to $M$ and $N$, respectively).
For this, I was able to show that $\mathbb{C}[V][\alpha]$ is the integral closure of $\mathbb{C}[V]$ in $\mathbb{C}(V)$, and that $\mathbb{C}[V][\alpha]$ is isomorphic to a polynomial ring in one variable. Also, I'm not entirely sure where the hypothesis that $M \neq (\bar{x},\bar{y})$ would be necessary.
Maybe a classification of the maximal ideals of $\mathbb{C}[V]$ could be given?
Thanks in advance.
Best Answer
By the nullstellensatz, the maximal ideals of $\Bbb C[V]$ are of the form $(x-a,y-b)$ with $b^2-a^3-a^2=0$, and the only maximal ideal with $a=0$ is the ideal $(x,y)$. For any such ideal $M$ with $a\neq 0$, we see that $\frac{y}{x}$ is already an element of the local ring $\Bbb C[V]_M$ because $x\notin (x-a,y-b)$. So this immediately shows that if we let $N=(x-a,y-b)\subset \Bbb C[V][\alpha]$, we have that $N\cap \Bbb C[V] = M$ and $\Bbb C[V]_M=\Bbb C[V][\alpha]_N$.
It should be clear what goes wrong with this argument in the case $M=(x,y)$: $x\in M$, so $\frac{y}{x}\notin \Bbb C[V]_M$. For a more conceptual explanation of why things are different here, the map of varieties which corresponds to the map on coordinate rings $\Bbb C[V]\to \Bbb C[V][\alpha]$ is a resolution of singularities of $V$, which is an isomorphism on the smooth locus of $V$ (the complement of $M=(x,y)$) and thus induces isomorphisms on local rings. The only place where something different happens is at the singular point, given by $M=(0,0)$.