Why is this a degree 2 morphism?

Question

Let $X$ be a smooth projective curve of genus 1, let $P_0\in X$ and consider the linear system $|2P_0|$. By Riemann-Roch $l(2P_0)=2$. I understand why this linear system is base-point free, and it defines a morphism $$\begin{array}{rccc} f\colon&X&\longrightarrow&\mathbb{P}^1\\ &P &\mapsto &[f_0(P):f_1(P)],\end{array}$$ where $\{f_0,f_1\}$ is any basis of the $K$-vector space $L(2P_0)$. Hartshorne [IV, §4] says it is a degree $2$ morphism, why is that? I would appreciate an answer as elementary as possible.

Answer 1

Choose $x \in L(2P_0)$ such that $\{1,x\}$ is a basis for $ L(2P_0)$, and consider the map $f: X \longrightarrow \mathbb{P}^1$, with $P\mapsto [1,x(P)]$. We will use the following fact:

If $f:\mathcal{C}_1 \longrightarrow \mathcal{C}_2$ is a nonconstant map of smooth curves, then for all but finitely many points $Q\in \mathcal{C}_2$ $$\deg f=\# f^{-1}(Q). $$

Now, let $Q=[1:\alpha] \in \mathbb{P}^1$ be a generic point. If $P_1,P_2,\cdots,P_n \in X$ are such that $x(P_1)=x(P_2)=\cdots=x(P_n)=\alpha $, then $P_1,P_2,\cdots,P_n$ are zeros of $(x-\alpha)$. However, the fact that $P_0$ is the only pole (a double pole) of $x$ gives that $P_0$ will be the only pole (a double pole) of $x-\alpha $. This implies that $x-\alpha$ has only two zeros. Generically the two zeros will be distinct, and so $\deg f=\# f^{-1}(Q)=2$.

Answer 2

We dicuss this more scheme-theorically.

Let $L=\mathscr{O}(2P_0)$. Actually we get a basis $\{x,1\}\in\Gamma(X,L)$. So if we let $f_0$ defines $P_0$ locally (over $V\ni P_0$), then $x|_{V}=f_0^{-2}$, as $x$ has pole of order 2 only at $P_0$. This defines $f:X\to\mathbb{P}^1$ (classically, $P\mapsto[x(P):1]$) as follows: Let cordinates of $\mathbb{P}^1$ is $x_0,x_1$. Let $U_0,U_1$ are non-vanising sets of $x$ and $1$.

First $k[x_1/x_0]\to\Gamma(U_0,\mathscr{O}_{U_0})$ is $x_1/x_0\mapsto 1/x$. Second $k[x_0/x_1]\to\Gamma(U_1,\mathscr{O}_{U_1})$ is $x_0/x_1\mapsto x$.

Next we need to find the image of $P_0$ in $\mathbb{P}^1$.

(a) Actually $x$ not vanish for all but $P_0$, as near $P_0$, $x=f_0^{-2}$. Using the fact that $\mathfrak{m}_{P_0}=(f_0)$. So if $x_{P_0}\in\mathfrak{m}_{P_0}L_{P_0}$, then $x_{P_0}=uf_0^k\cdot f_0^{-2}$ where $u$ be a unit and $k>0$. But this is impossible! So $x_{P_0}\notin\mathfrak{m}_{P_0}L_{P_0}$. So $U_0=X$!

(b) Now as (a), we get $1\in\mathfrak{m}_{P_0}L_{P_0}$! So $U_1=X-\{P_0\}$!

SO there is only $P_0$ maps to $Q=[1:0]\in\mathbb{P}^1$.

Now using the proposition II.6.9, we get $\deg(f)=v_{P_0}(t_Q)$. Now $t_Q=x_1/x_0$ mapping to the $x$ in $\mathscr{O}_{X,P_0}$ with $t_{P_0}=f_0$! As $x$ locally is $f_0^{-2}$, we get $v_{P_0}(t_Q)=2$, hence $\deg(f)=2$.