# Local normalization of algebraic curves

I am currently reading about the normalization theorem: Suppose $C$ is an irreducible plane algebraic curve, and let S be the set of singular points. Then there exists a compact Riemann surface $\hat C$ and a holomorphic map from the surface to $C$ such that the image of $\hat C$ is $C$, it's injective on the set of singular points, and the inverse image of the singular points is finite.

To prove this, we locally normalize each singular point by factoring it into irreducible local analytic curve components (using Weierstrass polynomials). Then for every curve component, we get can get a holomorphic, injective map from a disk at the origin of the complex plane onto the component, that is biholomorphic if we remove the origin from the disk and the origin from the curve component (where we have placed the singular point at the origin). Gluing all of these disks to the curve gives the normalization.

My question is: why do these curve components correspond exactly to the structure of the curve? The theorem just says that it factors and that a normalization is possible. Specifically:

1) If we have an ordinary double point, intuitively, it should factor into two analytic curve components. Why is this?

2) Why does the curve factor into a single irreducible component at smooth points?

The precise details can be found in "Introduction to Algebraic Curves" by Griffiths.

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When talking about complex manifolds of dimension one, one has the option of talking about curves or surfaces: but one should pick one and stick with it! :D – Mariano Suárez-Alvarez Feb 26 '12 at 2:57
You should explain how you are constructing the normalization so that one can use that construction to answer. – Mariano Suárez-Alvarez Feb 26 '12 at 3:00
Sadly, the accepted nomenclature is "Algebraic curve" and "Riemann surface." – Potato Feb 26 '12 at 3:00

Let's take an example for ordinary double points.
Consider the affine curve $C$ given by $y^2-x^2-x^3=0$ in $\mathbb C^2$.
It has an ordinary double point at the origin $O=(0,0)$.

Although $y^2-x^2-x^3$ is irreducible in $\mathbb C[x,y]$ (and even in the local ring $\mathbb C[x,y]_{(x,y)}$), the germ $y^2-x^2-x^3\in \mathcal O_{\mathbb C^2,O}=\mathbb C\lbrace x,y\rbrace$ is holomorphically reducible and factorizes as
$y^2-x^2-x^3=(y-xs(x))(y+xs(x))$ where
$s(x)=\sqrt {1-x}=1-\frac{1}{2}x-\frac{1}{8}x^2+...$ is a series of convergence radius $1$.

This means that in the normalization $\tilde C$ there will be two points $O_+,O_-\in \tilde C$ above $O\in C$ and that in suitable holomorphic (but not algebraic) charts the normalization at $O_+$ will look like $\; U\to C:z\mapsto (z,zs(z))$ and the normalization at $O_-$ like $\:U\to C:z\mapsto (z,-zs(z))$ (where $U$ is a disk of radius $1$).

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I see why this is true in this example, but why is it always the case that we get exactly two factors in $\mathbb{C}\{x,y\}$ for any ODP? – Potato Feb 26 '12 at 15:55
Dear @Potato, this is more or less included in the definition of "ordinary double point", which boils down to the condition that the equation of the curve starts with a non degenerate quadratic form $ax^2+bxy+cy^2$ or equivalently that the tangent cone at the singular point consists of two distinct lines. But I'll concede that there is some technical work to do if you want to give a complete proof. I recommend you look at Example 16.8 of Kunz's Introduction to Plane Algebraic Curves which is a classification of all double points of a plane curve. – Georges Elencwajg Feb 26 '12 at 17:10
Thanks! And could you give a brief reason why there is just one component at nonsingular points? It seems simple but I don't see it. – Potato Feb 26 '12 at 17:30
Dear @Potato, just use the implicit function theorem: near a nonsingular point $P$ the equation of the curve will be $y=0$ in a suitable holomorphic chart. Hence the normalization will be the local isomorphism $U\to C:z\mapsto (z,0)$ , where $U$ is a disk centered at $\tilde P$ ( the point above $P$) . The only local component of $\tilde C$ passing through $\tilde P$ is the disk $U$. – Georges Elencwajg Feb 26 '12 at 20:28
More specifically, why will it be irreducible in $\mathbb{C}\{x,y\}$? I don't see a priori why it couldn't factor. – Potato Feb 26 '12 at 20:28