Normalization of a variety I'm currently in a number theory course and this question popped up. As I'm not super familiar with algebraic geometry, I was wondering if my reasoning is correct:

Show that $\mathbb{C}[X,Y]/(Y^2 - X^2 - X^3)$ is a one-dimensional noetherian ring which is also an integral domain. Determine the normalization of this ring. 

Now, my strategy was to let $t = X/Y$ and define a map from $\mathbb{C}[X,Y] \to \mathbb{C}[t]$ where $X \mapsto t^2 -1$ and $Y \mapsto t(t^2 -1)$. Since this map has kernel $(Y^2 - X^2 - X^3)$, the two rings are isomorphic. Thus, since $\mathbb{C}[t]$ is noetherian and an integral domain, so is the quotient ring. However, I'm not sure how to determine its normalization, or really how to determine the normalization of $\mathbb{C}[X,Y]/(f)$ in general, where $f \in \mathbb{C}[X,Y]$. Could someone explain how normalization works over rings defined in such a way? By normalization I mean the integral closure of $\mathbb{C}[X,Y]/(f)$.
 A: I don't think that your homomorphism is surjective. Although, you seem to be in the right direction. Let $\mathbb{C}[\bar{x}, \bar{y}]$ denote the ring $\mathbb{C}[x,y]/\langle y^2-x^2-x^3\rangle$, where $\bar{y}$ and $\bar{x}$ are the images of $y$ and $x$ in $\mathbb{C}[x,y]/\langle y^2-x^2-x^3\rangle$. Notice that $\frac{\bar{y}}{\bar{x}}$ is integral over $\mathbb{C}[\bar{x},\bar{y}]$ since it satisfies the monic polynomial $f(z) = z^2 - \bar{x} - 1$. Thus the integral closure of this ring must contain $\mathbb{C}[t]$. As $\mathbb{C}[t]$ is a P.I.D, it is integrally closed. As a result, the integral closure of $\mathbb{C}[\bar{x},\bar{y}]$ is isomorphic to $\mathbb{C}[t]$. 
A: Here's how you do it geometrically.
$y^2 = x^2 + x^3$ is a nodal cubic, like so:

Take any line $\ell$ in the plane not passing through the node at the origin, and project onto this line from the node. In other words, associate to each point $P \in C$ the point $Q$ of intersection of $\ell$ and the line through $O$ and $P$.  (This association is well-defined because a cubic in one variable has three roots: a double root at the node, and one more somewhere else on the curve.)

This is a birational map between your curve and the line (which is smooth), so the line is a normalization of your curve.
