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Let X be a genus 2 curve with affine equation y^2 = f(x), where f is a polynomial of degree 6. Write $P_1, ..., P_6$ for the points on X(C) with y=0. Then why every $P_i-P_j$ is a 2-torsion points in Jacobian $Pic^0(X)$ and they are distinct? (Since there are 16 2-torsion points, we find all of them.)

My professor says one can prove that $P_i-P_j$ is a 2-torsion points by some argument involving divisor, and he also mentions one can prove $P_i-P_j$ are distinct by using the fact that a genus 2 curve has one degree-2 map to $\mathbb P^1$.

I just have no idea, so I am not able to say more. Could any one help me?

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The fibers $f^*(Q)=Q'+Q''$ of your degree-two morphism $f:X\to \mathbb P^1$ exactly decribe the complete linear system of effective canonical divisors $\vert K_X\vert $ of $X$ and these fibers are linearly equivalent divisors as $Q$ runs through $\mathbb P^1$.
In particular $f^*(P_i)=2P_i\equiv f^*(P_j)=2P_j$ for $1\leq i,j\leq 6$, so that $2(P_i-P_j)\equiv 0$ or equivalently $2(P_i-P_j)=0\in \operatorname {Pic^0(X)}$, just as desired.

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  • $\begingroup$ Thanks! So could you please explain more why $P_i-P_j$ is distinct for all i,j? $\endgroup$ – user198206 Nov 21 '15 at 14:58
  • $\begingroup$ On any smooth projective curve of genus $g\geq 1$ two different points cannot be linearly equivalent. $\endgroup$ – Georges Elencwajg Nov 21 '15 at 17:50

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