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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.


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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 $\mathbb P^N$.

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