Computing the fiber over $(0,0)$ for the normalization $\operatorname{Spec} \Bbb C[t]\to\operatorname{Spec} \Bbb C[x,y]/(y^2-x^2(x+1))$ Let $R=\mathbb{C}[x,y]/(y^2 -x^2 (x+1))$. Setting $t:=y/x$, we can show that $R[t]=\mathbb{C}[t]$ and use it to conclude that $R[t]$ is the normalization of $R.$
Now consider the corresponding normalization map $\phi:\operatorname{Spec}(\mathbb{C}[t])\rightarrow \operatorname{Spec}(R)$. My goal is to show that the fiber of $(x,y)\in \operatorname{Spec}(R)$ has exactly two points.
My idea was to make use of the fact that there is a natural bijection between $\phi^{-1}((x,y))$ and $\operatorname{Spec}(k((x,y))\otimes_R R[t])$, where $k((x,y)):=R_{(x,y)}/(x,y)_{(x,y)}$ is the residue field of $(x,y)$.
But I'm currently stuck on computing $\operatorname{Spec}(k((x,y))\otimes_R R[t])$. So any hint/help will be extremely useful.
Also, is there a different approach to prove the same result?
Thanks in advance.
 A: First, you can simplify $k((x,y))$ a little - for a maximal ideal $m\subset R$, $R_m/m_m$ is the same as $R/m$. So we're looking to understand what $R[t]\otimes_R R/(x,y)\cong R[t]/(x,y)$ is (ref for the last simplification).
Next, we need to do a little better with understanding what $R[t]$ is. Let's play around a little with $t$: as $t=\frac{y}{x}$, we get that $t^2=\frac{y^2}{x^2}$. But as $y^2=x^2(x+1)$, we actually have $t^2=x+1$. So we can write $x=t^2-1$ and $y=xt=t^3-t$, so it turns out that $R[t]\cong k[t]$ and the map $R\to k[t]$ is given by sending $x\mapsto t^2-1$ and $y\mapsto t^3-t$. This puts us down to considering $k[t]/(t^2-1,t^3-t)=k[t]/(t^2-1)$. As long as the characteristic of your field is not 2, $t^2-1$ splits in to distinct linear factors and so we find that $k[t]/(t^2-1)\cong k^2$ as rings. The spectrum of this ring is two points, and you're done.
A: The question is stated in geometric language, and probably the other answer is what you are looking for, but let me try a purely algebraic approach.
Since $t=y/x$ and $y^2=x^2(x+1)$ in $R$, we get $x=t^2-1$ and $y=t(t^2-1)$, so $R=\mathbb C[t^2-1,t(t^2-1)]$ and $S:=R[t]=\mathbb C[t]$. The maximal ideals of $S$ are of the form $fS$ with $f$ an irreducible polynomial (equivalently, $\deg f=1$). We are looking for the maximal ideals lying over $(t^2-1,t(t^2-1))$, that is, $fS\cap R=(t^2-1,t(t^2-1))$. In particular, $t^2-1\in fS$, and this shows that $f=t\pm1$.
