Let $X$ be a smooth projective curve of genus 1, let $P_0\in X$ and consider the linear system $|2P_0|$. By Riemman-Roch $l(2P_0)=2$. I understand why this linear system is base-point free, and it defines a morphism \begin{align*} f:X &\to\mathbb{P}^1\\ P &\mapsto [f_0(P):f_1(P)],\end{align*} 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.
Why is this a degree 2 morphism?
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algebraic-geometry
1 Answers
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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 $\deg f=\# f^{-1}(Q)=2$.
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0Thank you! I can see why the image of the embedding of function fields $f^*:K(\mathbb{P}^1)\to K(X)$ is $K(x)$, but why is $K(X)=K(x,y)$ ? – 2017-01-11
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0It is a legitimate question. Let me assume a bit more and edit an adequate answer. – 2017-01-11
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0Are you assuming the field to be algebraically closed? I think when you're counting points in the fibre, you need to count both the ramification of the map at each point and the degree of each point. Thus if $C_1$ has no degree $1$ points, it will never be true that the number of points in the fibre of a degree $2$ map is $2$. – 2017-01-12
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0Ah, that's true, I missed that $P_0$ was degree one. I'm still slightly concerned about the "All but finitely many points" and saying that the two zeroes will be distinct generically. What if $C_1$ only has finitely many rational points? – 2017-01-12
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0It is good point. – 2017-01-12
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0As long as $K$ is the full constant field of $K(C)$ (which is the case), you can safely go the the algebraic closure, as the degree $[K(C):K(x)]$ will not change. – 2017-01-12