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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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    Sadly, the accepted nomenclature is "Algebraic curve" and "Riemann sur$f$ace."2012-02-26

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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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    Ah, duh. I need to think about this a bit more. Thank you for the help.2012-02-26