# Degree of the projection map of projective plane curve to projective line

Let $$\mathbb{P}^1$$ be the complex projective line and $$\mathbb{P}^2$$ be the complex projective plane. Let $$\mathcal{C} \subset \mathbb{P}^2$$ be a compact projective plane curve, i.e. the zero locus of a homogeneous polynomial $$F \in \mathbb{C}[X,Y,Z]$$ of degree $$d$$. We consider the holomorphic map between Riemann surfaces $$\pi : \mathcal{C} \to \mathbb{P}^1, \, [x:y:z] \mapsto [x:z]$$. I want to show that the degree of this map is equal to $$d$$, as it is stated in "Algebraic Curves and Riemann Surfaces" by R. Miranda (Plücker's formula, page 144) and in this post (without further details) : Genus of a smooth projective curve.

My attempt is the following. We take $$p = [x_0,z_0] \in \mathbb{P}^1$$ such that it is not the image by $$\pi$$ of a ramification point (it is possible as ramifications points are a finite subset of $$\mathcal{C}$$, the latter being compact). Then the degree of $$\pi$$ is just the number of preimages of $$p$$. Without loss of generalities we suppose $$x_0 \neq 0$$ so we can rewrite $$p = [1:w_0]$$. Now a preimage of $$p$$ is of the form $$[1:y:w_0]$$ and such that $$F(1,y,w_0) = 0$$. If $$f(Y) = F(1,Y,w_0) \in \mathbb{C}[Y]$$ we see that preimages of $$p$$ are in bijection with roots of $$f$$.

So I'd like to show that $$f$$ has $$d$$ distinct roots, but I don't see how to prove this (and I'm not 100% sure it is the right statement). Any help is welcome !

• Does this answer your question? Plucker's Formula proof Commented Jun 4, 2023 at 19:00
• @M.C. Oh yes thank you I had not seen that post ! Commented Jun 5, 2023 at 12:02

I helps me a lot to think about this more geometrically: You can embed $$\mathbb P^1$$ as a line $$L$$ into $$\mathbb P^2$$ by $$\mathbb P^1 \ni [x:z] \mapsto [x:0:z] \in L \subset \mathbb P^2.$$ With that embedding in mind, you may think of your map $$\pi: \mathbb P^2 \setminus \{[0:1:0]\} \to \mathbb P^1$$ as the projection from the point $$[0:1:0]$$ to $$L$$. This means that the image point $$\pi([x:y:z])\in L$$ is the point obtained by intersecting the line $$N$$, which connects $$[0:1:0]$$ and $$[x:y:z]$$, with $$L$$. As $$N$$ is a line, it intersects $$C$$ in $$d$$ points, counted with multiplicitiy.