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.


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 $\deg f=\# f^{-1}(Q)=2$.

  • $\begingroup$ Thank 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 $K(X)=K(x,y)$ ? $\endgroup$ – Marco Flores Jan 11 '17 at 16:48
  • $\begingroup$ It is a legitimate question. Let me assume a bit more and edit an adequate answer. $\endgroup$ – MathChat Jan 11 '17 at 21:02
  • $\begingroup$ Are 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 $2$. $\endgroup$ – Tom Oldfield Jan 12 '17 at 19:07
  • $\begingroup$ Ah, 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? $\endgroup$ – Tom Oldfield Jan 12 '17 at 19:58
  • $\begingroup$ It is good point. $\endgroup$ – MathChat Jan 12 '17 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.