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So let $(E,p)$ be an elliptic curve over a field $k$ with a choice of $k$-valued point $p$. Then by Riemann-Roch, there are two global sections of $\mathcal{O}_{E}(2p)$ which gives a double cover of $\mathbb{P}^{1}$ and I let this morphism be called $\phi$. Then one of the sections which give the map to $\mathbb{P}^{1}$ vanishes at $p$ of order $2$, meaning that the preimage of $0$ is one point, so $p$ is a branch point. Write this section as $\sigma_{p}$.

Then by Riemann-Hurwitz, there are three other branch points, so we pick $q$ to be one of them.

Why do we have that $2p\sim 2q$ as divisors (reference [Hartshorne IV Example 4.8.2], [Vakil, Ex 19.9.A])?

The only way I can think of is that by the same argument, $\mathcal{O}_{E}(2q)$ also has 2 sections, one of which vanishes at $q$ of order 2. Write this as $\sigma_{q}$.

Now, I am very tempted to say that hence $2p-2q$ is principal because we can take $\sigma_{p}/\sigma_{q}$. But I think this proof is missing some crucial steps because what if I take a non-branched point say $z$ of the morphism $\phi$ and run the whole argument, I will arrive at the same conclusion... which looks to me absurd...

What should be the correct argument in this case?

A more elaborate explanation of my doubt

I should be more precise with my question: I was thinking of what happens if instead we take $\mathcal{O}_{E}(2z)$. I am afraid of a contradiction here because we already know that $z$ is not a branched point of $E$, and as you have said any sections $\tau$ that vanishes at $z$ must vanish at $z'$ also. Taking into account that $\deg div(\tau)=2$, you can't have $\tau$ that vanishes at $z$ of order 2.

On the other hand, if we choose $(E,z)$ as a elliptic curve instead of $p$, by Riemann-Roch, we have $\mathcal{O}_{E}(2z)$ having two global sections? one vanishing at $z$ at two points again. Are we talking about two different elliptic curves, or two different coverings $E\rightarrow\mathbb{P}^1$?

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  • $\begingroup$ I think it is very helpful to draw $E$ as a plane cubic curve, with $p$ and $q$ being two of the intersection points with the $x$-axis, and divisors in the linear system being intersections of $E$ with vertical lines. $\endgroup$ – Relapsarian Jun 18 '15 at 14:40
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If you take a non branched point $z$, you can find a global section that vanishes simply at $z$ but will also vanish at another point $z' \neq z$ (because $\phi^{-1}(\phi(z)) = \{z,z'\}$). Then $2p \sim z + z'$. There is no absurdity there.

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  • $\begingroup$ Thanks for your response. Probably I should be more precise with my question: I was thinking of what happens if instead we take $\mathcal{O}_{E}(2z)$. I am afraid of a contradiction here because we already know that $z$ is not a branched point of $E$, and as you have said any sections $\tau$ that vanishes at $z$ must vanish at $z'$ also. Taking into account that $\deg\div\tau=2$, you can't have $\tau$ that vanishes at $z$ of order 2. $\endgroup$ – enoughsaid05 Jun 18 '15 at 15:15
  • $\begingroup$ ... On the other hand, if $z$ is taken to be unbranched, RR would imply that $\mathcal{O}_{E}(2z)$ will have two sections, one vanish at order 2 at $z$. Are we talking about the same covering $E\rightarrow \mathbb{P}^1$ in this case? $\endgroup$ – enoughsaid05 Jun 18 '15 at 15:16
  • $\begingroup$ Oh if you change the line bundle, it changes the map into $\Bbb P^1$, so the branch points of the maps will be different. "a branch point of $E$" has no meaning. It's a branch point of a map. $\endgroup$ – mercio Jun 18 '15 at 15:58
  • $\begingroup$ I see! Now it makes more sense. But I still have more questions: for example if the point $p$ gives the covering $E\rightarrow \mathbb{P}^{1}$, and we have branch points $q$, $r$, and $\infty$. I pick one of them, say $q$ and I use this to define a mapping $(E,q)\rightarrow\mathbb{P}^{1}$, how will it differ from $(E,p)\rightarrow\mathbb{P}^{1}$. Is there obvious way of seeing that they should be the same (or not), and why it is it different from $(E,z)\rightarrow\mathbb{P}^{1}$? $\endgroup$ – enoughsaid05 Jun 19 '15 at 19:00
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    $\begingroup$ Well the divisors $2p$ and $2q$ will give equivalent line bundles, so they define the same mappings up to an automorphism of $\Bbb P^1$ $\endgroup$ – mercio Jun 19 '15 at 19:02

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