# Understanding a genus 2 projective curve

I guess it's convenient if our base field is algebraically closed of characteristic zero, let $$k$$ denote such a field. I read a paper that gave a basic background of genus 2 curve $$C$$, which is supposed to be easy so I am trying to understand it better.

...the canonical bundle $$K_C$$ of $$C$$ has degree $$2$$ and $$h^0(C,K_C)=2$$. That is, the corresponding complete linear system is a $$g_2^1$$. We therefore have a map $$x:C \rightarrow \mathbb{P}^1$$ which is ramified at $$6$$ points by the Riemann-Hurwitz formula, and the function field of $$C$$ is a quadratic extension of $$k(x)$$.

I know that the degree of $$K_C$$ is given by $$2g(C) - 2 = 2$$ and I suppose $$h^0(C,K_C)=2$$ because by Serre duality, we have $$h^0(C,K_C) = h^1(C,\mathcal{O}_C) = g(C) = 2$$. Since $$\dim K_C = h^0(C,K_C)-1 = 1$$, the linear system $$|K_C|$$ corresponds to an morphism of $$C$$ into $$\mathbb{P}^1$$. But why does it follow that this morphism, which we denote as $$x$$, has degree $$2$$?

Also, by the Riemann-Hurwitz formula (following notation in Hartshorne), we have $$R= 2g(C)-2 - 2(2g(\mathbb{P}^1)-2)) = 2 - 2(-2) = 6.$$ I believe that since every ramified point can have a ramification index of at most $$2$$ (because $$\deg x = 2$$), we must have $$6$$ ramification points, am I right?

Finally, what does the notation $$k(x)$$ even mean? Here $$x$$ denotes given map.

More generally, suppose $$\mathscr{L}$$ is a line bundle of degree $$d$$ on an irreducible regular curve $$C$$, and $$s_1, s_2$$ are two global sections without a common zero, then the map $$\pi: C\to \mathbb{P}^1$$ given by $$[s_1:s_2]$$ is a map of degree $$d$$ (this is Exercise 19.4.E in Vakil's The Rising Sea book, for example).

The proof goes as follows. If the map is not finite and hence has degree 0, then $$s_1, s_2$$ are linearly dependent so cannot vanish. This implies $$\mathscr{L}$$ is trivial and $$d=0$$.

Now assume $$\deg \pi>0$$. The fiber over $$[0:1]\in \mathbb{P}^1$$ is a closed subscheme of $$C$$, a divisor $$D$$ (it's an effective cartier divisor, because $$\pi$$ is dominant) with degree $$\deg \pi$$ (according to this question). Now write

$$\mathscr{L}\cong \pi^*\mathscr{O}(1)\cong \pi^*\mathscr{O}([0:1])\cong \mathscr{O}_C(D)$$

(the last equation the non-trivial one: it follows because $$D=\pi^*([0:1])$$ and $$[0:1]$$ is locally cut by a non-zero-divisor)

This implies $$d=\deg \mathscr{L}=\deg D = \deg \pi$$ which finishes the proof. The second equality follows from the general fact that on a regular curve $$C$$ $$\deg \mathscr{O}_C(D)=\deg D$$ where $$D$$ is any divisor.

Also, you are correct about each ramified point having ramification index at most $$2$$ (again by this).

For the last point, I guess that $$k(x)$$ is just the field of rational functions over $$k$$ in a single variable, the $$x$$ has nothing to do with the map $$x$$. This field is the function field of $$\mathbb{P}^1$$, and the morphism $$x$$ embeds it into $$k(C)$$. The resulting field extension has degree 2, equal to the degree of the morphism.

• Thank you for your answer. When you say "If the map is not finite and hence has degree $0$...", does it mean that a positive degree map is finite? Also, what does it mean when you say $$\pi^*\mathcal{O}(1) \cong \pi^*\mathcal{O}([0:1])?$$ What is $\mathcal{O}([0:1])$? What is the relation between the invertible sheaf $\mathcal{O}(1)$ and $\mathcal{O}([0:1])$? Nov 13, 2022 at 12:35
• For a map $\pi : X\to Y$ between irreducible varieties of the same dimension, the degree is defined as follows: It is 0 if $\pi$ is not dominant, and otherwise it is defined as the degree of the field extension $k(X)/k(Y)$. In the latter case, and if further $X, Y$ are regular curves over a field, then $\pi$ must be finite (this is proved in section 18.4.4 in Vakil's The Rising Sea, for example). Nov 13, 2022 at 15:05
• Thanks again, but what is $\mathcal{O}([0:1])$? Nov 13, 2022 at 16:45
• Oh sorry, forgot to write that in my first comment. $[0:1]$ is a point $p\in \mathbb{P}^1$, so we can define the line bundle $\mathscr{O}(p)$ (by treating $p$ as a divisor). Nov 14, 2022 at 8:50