# Miranda - Classification of Curves of Genus 4

I'm trying to understand pag. 207 of Miranda's book "Algebraic Curves and Riemann Surfaces" where it is explained which homogenous polynomial equations define a Riemann Surface of Genus Four embedded in $$\mathbb{CP}^3$$ by the canonical map. We will call $$X$$ the embedded Riemann Surface. We recall that the degree of this curve is $$6$$.

Miranda first observes that there exists at least one quadratic polynomial $$F$$ vanishing on $$X$$. I'm stuck with understanding that in fact there no exists $$F_1$$ quadratic homogenous polynomial such that $$F$$ and $$F_1$$ are linearly independent.

The book proceeds this way: it is taken a general hyperplane $$H \simeq \mathbb{CP}^2$$ (i.e. an hyperplane such that the intersection with $$X$$ consists of exactly $$6$$ points.) and the author observes that restricting $$F$$ and $$F_1$$ to $$H$$ the intersection

$$\{F=0\} \cap \{F_1=0\} \cap \{H=0\}$$

consists of at most $$4$$ points because of Bézout's theorem. In this way it immediately follows the conclusion.

I'm having some trouble in understanding why we can in fact use Bézout theorem. In fact in order to apply Bézout Theorem I think that one should prove first that the restriction of $$F$$ and $$F_1$$ to $$H$$ which are homogenous polynomials of degree $$2$$ in $$3$$ variables of $$H$$ have no non-trivial common factor.

I have proved that $$F$$ and $$F_1$$ are irreducible if it can help. Could someone give me some suggestions?

Thank you

• If polynomials $F$ and $F_1$ have a common factor on a general plane, they have a common factor on $\mathbb{P}^3$, and then the curve $C$ should be contained in a plane, – Sasha Feb 3 at 9:49
• Thanks. I tried also this way but I don't understand why it follows from "having no common factor on a general hyperplane" that they have a common factor in $\mathbb{P}^3$. I think that in general, taken two irreducible linearly indipendent polynomials - for example $x_0^2+x_3^2$ and $x_0^2$ - and restricting them to an hyperplane - for example $x_3=0$ - it doesn't follow that having a common factor on the plane implies having a common factor in $\mathbb{P}^3$. Maybe I am making some confusion on the concept of general hyperplane:it is true that I can find a general hyperplane that works? – ghiroz Feb 3 at 10:06
• In your example the hyperplane is not general. To be true for a "general" hyperplane means to be true for an open subset of the set of all hyperplanes. – Sasha Feb 3 at 10:15