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I'm writing a survey paper about Weierstrass subgroup of plane quartic curves. I found that Weierstrass subgroup of genus $g$ hyper-elliptic curve is just 2-torsion subgroup of its Jacobian, namely $J(C)[2]$. I want to find a proof of this fact, but I can't. I know that there are exactly $2g+2$ Weierstrass points on hyper-elliptic curves, and these are fixed points of a hyper elliptic involution. So I just want to know why the fixed points of the hyper-elliptic involution generates 2-torsion subgroup of $J(C)$.

Also, there is one more question : why inflection points (flexes) corresponds to Weierstrass points in case of smooth (plane) quartic curve? I proved that every flexes (resp. hyper-flexes) are Weierstrass point of weight 1 (resp. weight 2) by constructing meromorphic functions on a curve $C$ which has order 3 (resp. order 4) pole at a given flex using equation of tangent lines. (For example, if $p$ is hyper-flex and $L_{p}$ is linear form defining target lines at $p$, then $L/L_{p}$ ($L$ is a linear form which defines a line does not contain $p$) is meromorphic function which has a pole only at $p$ with order 4. But I can't prove the converse, why all the Weierstrass points should be flexes? Thanks.

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  • $\begingroup$ If this question (math.stackexchange.com/questions/2060970/…) is true, than I have an answer for second question. Anyone can help? $\endgroup$ – Seewoo Lee Dec 16 '16 at 8:37
  • $\begingroup$ Do you see why $J(C)[2]$ is what you claim it is when $C$ is an elliptic curve? $\endgroup$ – Ben Lim Dec 22 '16 at 3:05
  • $\begingroup$ @BenjaLim Yes, since there are exactly 4 fixed points of involution which are just 2 torsion points. $\endgroup$ – Seewoo Lee Dec 26 '16 at 6:57

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