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I'm currently reading through Ravi Vakil's notes on Algebraic Geometry. I've been having trouble grasping some things conceptually though and I hope that you can help me.

For an elliptic curve (E,p) (where p is a k-valued point), we have that $\mathcal{O}_E(2p)$ has dimension 2, so we have a section that vanishes to order 2 there according to Ravi. OK, I'm fine with that, although I have to admit I'm not personally, entirely convinced. Isn't it true that all the global sections should be constant, so the other section of $\mathcal{O}_E(2p)$ (the one that is not constant) should have a pole of order 2 at p? How can it vanish then?

Now, for the second part, we have that $\mathcal{O}_E(2p)$ defines a hyperelliptic covering with 4 branch points. Why do these branch points occur from our sections? And further, If q is another branchpoint of this hyperelliptic covering, then $\mathcal{O}_E(2p) \cong \mathcal{O}_E(2q)$. I do understand that to do this, I should show that there's a section of $\mathcal{O}_E(2p)$ that vanishes at q of order 2. But how can this arise from our two sections of $\mathcal{O}_E(2p)$, one of which has divisor 2p? What goes wrong with the construction if q is not a branch point?

Many questions, but I hope that you can help me.

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    Dear Dedalus, There are a lot of inaccuracies in your question: E.g. $\mathcal O_E(2p)$ is an invertible sheaf on $E$, so it doesn't have a dimension. (It has a locally free rank, which is one in this case, since it is an invertible sheaf.) Also, it is certainly *not* the case that all the global sections of $\mathcal O_E(2p)$ are constant. (You yourself seem to acknowledge this when you speak of the non-constant section in the subsequent clause. You are correct though that the non-constant global sections have poles of order two at $p$.) I suggest that you clarify your undestanding ...2012-04-30
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    ... of some of these foundational points before trying to understand the particular aspects of the situation under discussion. On a different note, did you try writing down an actual elliptic curve, say $y^2 = x^3 -x$, taking $p$ to be the point at infinity, and work through the discussion/construction in this case? Regards,2012-04-30
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    Hi Matt, Thanks for your wise input and I apologize for the inaccuracies. What I mean with $\mathcal{O}_E(2p)$ having dimension 2, I meant that $h^0(C,\mathcal{O}_E(2p)) = 2$. I realize that the wording is incorrect here. I didn't mean that $\mathcal{O}_E(2p)$ only had constant sections, but that $\mathcal{O}_E$ only had constant sections. Now, I understand that $\mathcal{O}_E(2P)$ is an invertible sheaf, but what does it mean to say that a section that vanishes at p of order 2? Is it just that 1, under the local trivialization will have that as a corresponding divisor?2012-04-30
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    Dear Dedalus, Everything in your comment is correct. In particular, the section $1$ of $\mathcal O_E$, when regarded as a section of $\mathcal O_E(2p)$, vanishes to order $2$ at $p$ (just as you wrote). But this is not the section that is going to give a finite map to $\mathbb P^1$; for that you need a non-constant rational function, and so a non-constant section of $\mathcal O_E(2p)$. Regards,2012-04-30
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    Matt: Thank you so much for finally clearing the thing up with that 1, under the local trivialization, vanishes to order 2. This had been confusing me for a long time. So, the one that gives my my finite map should be s, with a pole of order 2 at p? If that is the case, I still have trouble seeing how we from linear combinations of these 2 can get the other branch points, and if we can do so, why we couldn't do the same for any point.2012-04-30
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    Dear Dedalus, Did you try looking at a concrete example? Regards,2012-05-01
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    Dear Matt, I believe I tried to check with an example.But probably I'm misunderstanding something. Let us take y^2=x^3-x, so that p would be at infinity. Then, we have that the fraction field will be k(x)[y]/(y^2-x^3-x). So, if we look at say (x-1) and wanna find the divisors of zeros and poles, we first see that if we localize at $(x-1,y)$, y will be a local uniformizer, and since (x-1) = y^2/(x+1)x, it will have a zero of order 2 there (?) and for infinity, we do a similar analysis, and see that it has a pole of order 2? Am I at all correct or totally wrong, or somewhere in between?2012-05-01
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    Dear Dedalus, That's correct. But now think of $x-1$ as a map from $E$ to $\mathbb P^1$, and compute the branch points. Regards,2012-05-01
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    Dear Dedalus, if I am looking at the correct document, the notes by Ravi Vakil are huge, at 656 pages. Do you mind including in your original post specific information about the chapter, section and/or pages that you are looking at in the notes? That may help others provide additional feedback.2012-05-01

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