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Let $X = \{F=0\} \subset \mathbb{P}^2$ be a projective plane curve and let $\pi : X \to \mathbb{P}^1$ be defined by $\pi [x:y:z] \to [x:y]$.

I'm trying to understand why the following is true:

The holomorphic map $\pi$ is ramified at $p \in X$ Iff $\frac{\partial F}{\partial z} (p)=0.$

So far I understand the case where $X$ is an affine curve in $\mathbb{C}^2$ and the map $\pi$ projects to $\mathbb{C}$. Here the same condition condition applies as i've been able to prove.

A projective plane curve $X$ is locally just an affine curve so the same condition must apply but the projection here is confusing since there are two coordinates. Could someone carefully outline the argument needed here to conclude the projective case from the affine case?

user286568
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  • Even more is true: The intersection divisor of $\frac{\partial F}{\partial z}$ is equal to the ramification divisor of $\pi$. A proof can be found on pages 143-144 of Miranda's Algebraic Curves and Riemann Surfaces. The statement you quote is equivalent to the weaker claim that both divisors have the same support. – Max Demirdilek Feb 14 '23 at 13:10

1 Answers1

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You better assume that the curve $X$ is smooth: else the morphism $\pi$ might not be defined everywhere and moreover ramification would not be a very clear concept.
Your morphism $\pi:X\to \mathbb P^1$ is best seen as the projection of the curve $X$ from the point $O=(0:0:1)$ to the line at infinity $z=0$.
Indeed that line at infinity is a copy of $\mathbb P^1$ with coordinates $(x:y:0)$ .
Now, the morphism $\pi$ is ramified at $P=(a:b:c)\in X$ iff the line $\overline {OP}$ is tangent to $X$ at $P$.
But the tangent line at $P$ to $X$ (remember that $X$ is assumed smooth) has equation $$\frac{\partial F}{ \partial x}(P)\cdot x+ \frac{\partial F}{ \partial y}(P)\cdot y+\frac{\partial F}{ \partial z}(P) \cdot z=0 $$ That tangent line will pass through $O=(0:0:1)$ iff $\frac{\partial F}{ \partial x}(P)\cdot 0+ \frac{\partial F}{ \partial y}(P)\cdot 0+\frac{\partial F}{ \partial z}(P) \cdot 1=0 $ i.e. iff $$\frac{\partial F}{ \partial z}(P)=0$$ which is thus the required condition for $P\in X$ to be a ramification point for $\pi$.
[Notice the little trick of replacing the condition the line $\overline {OP}$ is tangent to $X$ by the condition the tangent line to $X$ at $P$ goes through $O$ ]

Georges Elencwajg
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  • This is v. pretty (+1)... In the context it is perfectly clear, so it doesn't much matter, but you are using the symbol $X$ as a coordinate when it also denotes the curve – peter a g Nov 04 '15 at 02:20
  • Dear @peter a g, you are perfectly right: thank you for your friendly and judicious comment. I have now replaced the upper case letters by lower case letters. – Georges Elencwajg Nov 04 '15 at 08:20