# Holomorphic mappings, Plücker's formula (Mistake in Rick Miranda's book?)

In Rick Miranda's book "Algebraic Curves and Riemann Surfaces", in order to prove Plücker's formula for smooth projective plane curves, he first defines a projective plane curve by the formula $X:F(x,y,z)=0$, where of course $F$ is a homogeneous polynomial. Now, he defines the map $\pi:X\to\mathbb{P}^2$ such that $[x:y:z]\mapsto[x:z]$, and uses properties of this map to prove the formula. This may be a really dumb question with an obvious answer, but is this function $\pi$ really defined on all the curve $X$? What if $F(x,y,z)=x+z$; then $\pi[0:1:0]$ wouldn't be defined.

The proof in the book uses the ramification divisor of $\pi$, and so necessarily makes use of the fact that $\pi$ is well defined.

If anyone can clarify this problem for me, I would greatly appreciate it.

Thanks!

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The proof starts with the words " Let $X$ be a smooth projective curve...". And smooth is the magic word that solves your problem. Indeed there is a proposition stating that a rational (or meromorphic) map from a smooth curve to projective space is actually regular (or holomorphic).
In your example you have, for a point $[x:y:z]\in X$ different from $P=[0:1:0]$, the formula $\pi [x:y:z]=[x:z]=[x:-x]=[1:-1]$, taking into account that $x+z=0$ on $X$. So it is easy to extend $\pi$ regularly through $P$, since $\pi$ is actually constant. You just define $\pi(P)=\pi( [0:1:0])=[1:-1]$ .
Remark The proposition mentioned above (on regularity of rational maps) is not difficult. If you take a coordinate $t$ for your curve near the troublesome point $P$, locally at $P \;$ the map is $t\mapsto [t^au(t):t^bv(t)]$ (where I have assumed that the target is $\mathbb P^1$) with $a,b\in \mathbb N$ and $u,v\in\mathcal O_{X,P}^\ast$ (units of $\mathcal O_{X,P})$ . If $a>b$ the map is just $t\mapsto [t^au(t):t^bv(t)]=[t^{a-b}u(t):v(t)]$ and can clearly be extended regularly by sending $P$ to $[0:v(0)]=[0:1]$ . And similarly if $a\leq b$ . And similarly if the target is $\mathbb P^N$.