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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.

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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.

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  • $\begingroup$ 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 $\mathcal{O}([0:1])$? $\endgroup$
    – oleout
    Nov 13, 2022 at 12:35
  • $\begingroup$ 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). $\endgroup$
    – Omri Zemer
    Nov 13, 2022 at 15:05
  • $\begingroup$ Thanks again, but what is $\mathcal{O}([0:1])$? $\endgroup$
    – oleout
    Nov 13, 2022 at 16:45
  • $\begingroup$ 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). $\endgroup$
    – Omri Zemer
    Nov 14, 2022 at 8:50

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