4
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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 $n$. $\endgroup$ – Tim Dokchitser Nov 6 '13 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. $\endgroup$ – Stefan Keil Nov 8 '13 at 10:12
7
$\begingroup$

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.