Genus of a smooth projective curve

Question

I was trying to prove that the genus $g(X)$ of a smooth projective complex plane curve $X=\{[x:y:z]\in \mathbb{P}^2 \vert F(x,y,z)=0\}$ of degree $d$ is equal to $$g(X)=\dfrac{(d-1)(d-2)}{2}.$$

My attempt was to take the standard projection $$\pi:\mathbb{P}^2 \to \mathbb{P}^1$$ $$[x :y : z] \to [x : z]$$ which has degree $d$ and then apply Riemann Hurwitz formula.

The problem is that I do not know how to explicitly write the multiplictity of this map at the ramification points(which are the ones for which $\dfrac{\partial F}{\partial y}=0$). In particular, I would like to say that the multiplicity of the map $\pi $ in a point $x \in \mathbb{P}^2$ is $I\left(x,F \cap \dfrac{\partial F}{\partial y}+1\right)$, but I'm stuck (the last I is the intersection multiplicity of the two algebraic curves).

Answer 1

One proof of this variant of the genus-degree formula (aka "Plücker's formula") is given on page 144 of Miranda's Algebraic Curves and Riemann Surfaces. Supposedly, another more geometric proof is sketched here.

Let me finish your proof with the help of Miranda. As you note, the proper non-constant holomorphic mapping $$\pi\colon X\rightarrow \mathbb{P}^1$$ $$[x:y: z]\mapsto [x:z]$$ has degree $d$, where $d$ is the degree of the homogeneous polynomial $F$. Thus, to calculate the topological genus of the compact Riemann surface $X$ (with the help of the Riemann-Hurwitz formula) all that is left to do is to find the non-negative integer $\sum_{p\in X}\big(\operatorname{mult}_p(\pi)-1\big)$. Here, $\operatorname{mult}_p(\pi)$ denotes the multiplicity of $\pi$ at $p\in X$.

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To do so, we employ the notion of a divisor on a Riemann surface. Namely, we use the following two examples of divisors on $X$. Firstly, we introduce a divisor whose degree is precisely the non-negative integer $\sum_{p\in X}\big(\operatorname{mult}_p(\pi)-1\big)$ which we wish to calculate.

Definition. The ramification divisor of the non-constant holomorphic mapping $\pi\colon X \rightarrow \mathbb{P}^1$ is the divisor on $X$ defined by $$R_\pi =\sum_{p\in X}\big(\operatorname{mult}_p(\pi)-1\big)\cdot p.$$

Secondly, we introduce a divisor on $X$ which will help us calculate the degree of the ramification divisor $R_\pi$.

Definition. Let $G$ a homogeneous polynomial that is not identically zero on $X$. Let $p\in X$. If $G(p)=0$, then find a homogeneous polynomial $H$ of the same degree as $G$ that does not vanish at $p$. In this case, let the integer $\operatorname{div}(G)(p)$ be the order of the meromorphic function $G/H$ at $p$. If $G(p)\neq 0$, then we set $\operatorname{div}(G)(p)=0$. One shows that the integer $\operatorname{div}(G)(p)$ is well-defined (i.e., independent of the choice of $H$) and calls the divisor $\operatorname{div}(G)\colon X \rightarrow \mathbb{Z}$ the intersection divisor of $G$ on $X$.

Next, we recall the following two results. Proofs can be found in Miranda's book. The first theorem gives a formula to calculate the degree of any intersection divisor on $X$.

Bezout's theorem (essentially Theorem 2.13 in Miranda's book). Let $G$ be a homogeneous polynomial which is not identically zero on $X$. Then the degree of the intersection divisor $\operatorname{div}(G)$ of $G$ on $X$ is equal to the product of the degrees of $G$ and $X$: $$\operatorname{deg}(\operatorname{div}(G))=\operatorname{deg}(G)\operatorname{deg}(X).$$

The second lemma states that the ramification divisor $R_\pi$ is equal to the intersection divisor of a certain homogeneous polynomial obtained by partial differentiation of $F$.

Lemma (Lemma 2.14 in Miranda's book). The polynomial $\partial F /\partial y$ is homogeneous of degree $d-1$ and the intersection divisor of $\partial F /\partial y$ of $X$ is equal to the ramification divisor $R_\pi$ of $\pi$ on $X$.

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We now have all the necessary ingredients to prove Plücker's formula. Namely, Bezout's theorem and the above lemma yield $$\sum_{p\in X}(\operatorname{mult}_p(\pi)-1)=d(d-1).$$ With the Riemann-Hurwitz formula, we thus have $$2g(X)-2=-2d+d(d-1).$$ Solving for $g(X)$ shows the claim $$g(X)=\frac{(d-1)(d-2)}{2}={d-1\choose 2}.$$