Given a positive integer $n$, is there an algorithm (or even better a closed formula) that provides me with a hyperelliptic curve $C/\mathbb Q$, such that its Jacobian $J:=Jac(C)$ possesses a $\mathbb Q$-rational torsion point $P$ of exact order $n$?

The genus of $C$ may very well vary with different values of $n$.

If it makes the situation simpler, please feel free to assume that $n$ is prime.

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    $\begingroup$ Hello Stefan! The curve $y^2+f(x)y=x^{2g+1}$ for $n=2g+1$ odd, and $y^2+2(x^{g+1}+f(x))y=-x^{2g+2}$ for $n=2g+2$ even has div$(y)=n(0,0)-n(\infty)$, so $(0,0)-(\infty)$ is in both cases a rational divisor of degree $n$. $\endgroup$ Commented Nov 6, 2013 at 19:39
  • $\begingroup$ Dear Tim, thanks for your comment! It is exactly what I was looking for. And apparently it wasn't too hard to answer, but if you like to repost it as an answer I can accept it. $\endgroup$ Commented Nov 8, 2013 at 10:12

1 Answer 1


One can take $$ \begin{array}{llllll} y^2 + (a_g x^g+...+a_0) y &=& x^{2g+1} &&& \text{ ($n=2g+1$ odd)} \cr y^2 + (2cx_{g+1} +a_g x^g+...+a_0) y &=& -c^2x^{2g+2} &&& \text{ ($n=2g+2$ even).} \cr \end{array} $$ The divisor of the $y$ function is $$ \text{div}(y) = n(0,0)-n(\infty), $$ and so $D=(0,0)-(\infty)$ is an $n$-torsion point on the Jacobian.

Note that in both cases, completing the square on the left gives an equation $y^2=$ polynomial of degree $2g+1$, so the curve has a unique point at infinity. Also, the order of $D$ is exactly $n$, for otherwise $y$ would be (up to a constant) a power of a rational function which has a point of order $<n$ at $\infty$ and no other poles, and all such functions are polynomials in $x$.

P.S. We needed such `interesting' torsion points to construct elements of $K_2$ on hyperelliptic curves


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