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I am reading the following paper: https://arxiv.org/abs/1101.4792.

In lemma 4, they prove: If $P_1,\ldots,P_6$ are the Weierstrass points of a genus 2 hyperelliptic curve $H$, then every element of the $2$-torsion subgroup $J[2]$ of the Jacobian variety $J$ of $H$ contains a unique pair of divisors $\{P_i-P_j,P_j-P_i\}$.

They say that the Riemann-Roch theorem implies the following facts:

  • the points $\{P_i-P_j,P_j-P_i\}$ are nonzero,
  • and they are distinct. In fact, in the following discussion, they claim that $\{P_i-P_j,P_j-P_i\}+\{P_k-P_l,P_l-P_k\}=$ the divisor belonging to the two remaining points, if $i,j,k,l$ are all distinct.

I understand the rest of the proof, but not how two derive these two facts. I understand the statement of the Riemann-Roch theorem as it can be found in Silverman's book "The Arithmetic of Elliptic Curves".

My best guess is that it has something to do with the definition of Weierstrass points using the gap sequence like on Wikipedia: https://en.wikipedia.org/wiki/Weierstrass_point (for me, the Weierstrass points are the fixed points of the hyperelliptic involution).

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  • $\begingroup$ You really threw me with the wrong genus. You mean genus=$2$. I’d expect that a hyperelliptic of genus four would have $10$ W-points. $\endgroup$ – Lubin Jun 14 '17 at 0:23
  • $\begingroup$ Ohhh yes, stupid mistake! Thanks for pointing that out. $\endgroup$ – mainomai Jun 14 '17 at 9:00
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I’m sorry that nobody else has given you an answer here. I’ll try, but there’s a good chance that anybody else would have given a better than what you see below. The little algebraic geometry I learned, I learned in an earlier geological era, so my treatment will be laughably old-fashioned.

Let’s try to use Riemann-Roch. Let’s call the constant field $\kappa$, and the hyperelliptic function field, $K$. Let’s choose one Weierstrass point, and label it $I$, it’s going to be special. You know that there’s one new element of the function field $K$ with a double pole at $I$, let’s call it $\xi\in K$, and $\xi^2$ has pole of order four at $I$. Genus two implies that there’s a new function with pole of order $5$ at $I$, call it $\eta$, which is not in the $\kappa$-span of $\{1,\xi,\xi^2\}$. From there on, for every order of pole $6\le n\le9$, there’s a new function with pole of order $n$, they’re $\xi^3$, $\eta\xi$, $\xi^4$, and $\eta\xi^2$, respectively. But with pole of order ten, there are two new functions, $\xi^5$ and $\eta^2$, so there must be a $\kappa$-linear relation $$ b\eta^2+a_1\eta\xi^2+a_3\eta\xi+a_5\eta =c\xi^5+a_2\xi^4+a_4\xi^3+a_6\xi^2+a_8\xi+a_{10}\,, $$ where $bc\ne0$, that is, $\eta^2$ and $\xi^5$ definitely must appear in the relation of linear dependence. It’s an old story, but by choosing $\lambda_1,\lambda_3,\lambda_5$ correctly, you can set $\eta=\eta'+\lambda_1\xi^2+\lambda_3\xi+\lambda_5$, still in the same linear space, still with a pole of order $5$ at $I$, so that $b{\eta'}^2=c\xi^5+a'_2\xi^4+a'_4\xi^3+a'_6\xi^2+a'_8\xi+a'_{10}\,$. In other words, we have $$ {\eta'}^2=\mu\prod_{i=1}^5(\xi-\rho_i)\,,\quad\rho_i\in\kappa, \mu\in\kappa^\times $$ at least if $\kappa$ is algebraically closed (and you can treat the general case very easily). Let’s call $Q(\xi)$ that quintic polynomial in $\xi$, for convenience.

Now, in this model of your curve, the hyperelliptic involution acts on a point $(\alpha,\beta)$ to change the sign of $\beta$, and the five other Weierstrass points are $P_i=(\rho_i,0)$, and we can call $I=P_6$. At this point, since the divisor of $\xi-\rho_i$ is $2P_i-2I$, you see why $P_i-P_j$ gives a point of order two on the Jacobian, and for the fact about the “two remaining points”, that comes from the fact that the divisor of $\eta'$ is $-5I+\sum_1^5P_i$.

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  • $\begingroup$ Thank you very much! This treatment was actually just right for me. $\endgroup$ – mainomai Jun 14 '17 at 23:42

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