TITLE: Normalization of a variety
QUESTION [6 upvotes]: 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 $\mathbb{C}[X,Y]/(f)$.

REPLY [6 votes]: 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.