2
$\begingroup$

I have been trying to build the function field of the jacobian of a genus 2 smooth curve over a finite field, but I am having problems making it explicit, I need to work with another curve with points in that field.

Let $H$ be a smooth curve of genus 2 over $\mathbb{F}_q$ defined by the equation $y^2 = f(x)$ where $deg(f(x))=5$. what I want to build is $\mathbb{F}=\mathbb{F}_q(Jac(H))$.

What I did so far is:

Take $(a,b)=p\in H$ (not Weierstrass) and take $g(x)=x-a\in \mathbb{F}_q[H]$.

Consider the basic Zariski open

$U=D_g=\lbrace P\in H : g(P)\neq 0\rbrace$

Now we have that

$U=Spec\space \mathbb{F}_q[H]_g=Spec\space \mathcal{O}_H(D_g)$

where

$R_g=\mathbb{F}_q[H]_g=\lbrace \frac{h}{(x-a)^r}: h\in \mathbb{F}_q[H], r\in\mathbb{Z}^{+}\rbrace$

is the localization at the multiplicative set $\lbrace 1,g,g^2,..,\rbrace$

I know what $Sym^{2}(H)=H\times H/S_2$ (unordered pair of points, $S_2$ is acting in the coordinate functions) is birational to $Jac(H)$ and that if $V\subset H$ is an open (quasi-affine variety) then $\mathbb{F}_q(V)=\mathbb{F}_q(H)$ (have the same function field) so I think

$\mathbb{F}=Quot(R_g\otimes_{\mathbb{F}_q} R_g/<x\otimes y-y\otimes x>)$

I think I need to do the calculation of the symmetric square of algebras explicit, I would like to know if this is the best way and any hints if there are, or corrections.

I was thinking using basic symmetric polynomials because I know that all the elements of $\mathbb{F}_q[H]$ are of the form $p(x)+q(x)y$ (because $y^2=f(x)$) but I get lost in the generalisation and in the tensor product.

Thank you

$\endgroup$
  • $\begingroup$ You should perhaps look at the book by Cassels and Flynn: Prolegomena to a middlebrow arithmetic of curves of genus 2. They give a fairly explicit description of the Jacobian that works in all characteristics except 2. $\endgroup$ – Michael Stoll Feb 26 '15 at 16:46
  • $\begingroup$ Yes, but in this case what I am interested in is in the function field of the Jacobian variety, I will check it again too $\endgroup$ – Eduardo R. Duarte Feb 26 '15 at 16:48
4
$\begingroup$

You get an open subset of the Jacobian by looking at points "in general position", i.e., points represented by divisors of the form $(P)+(P')-2(\infty)$, where $\infty$ denotes the point at infinity, such that $P, P' \neq \infty$ and $P$ and $P'$ are not images of each other under the hyperelliptic involution. Such points can be specified uniquely by their Mumford representation $(a(x),b(x))$, where $a(x) = x^2 + a_1 x + a_0$ and $b(x) = b_1 x + b_0$ are such that the $x$-coordinates of $P = (\xi,\eta)$ and $P' = (\xi',\eta')$ are the roots of $a(x)$ and $\eta = b(\xi)$, $\eta' = b(\xi')$; more precisely, $y = b(x)$ is the line through $P$ and $P'$ (tangent to the curve when $P = P'$). The condition that the points are on the curve translates into $$ f(x) \equiv b(x)^2 \bmod a(x).$$ This can be expanded into a pair of equations that $a_0, a_1, b_0, b_1$ have to satisfy and gives you an affine variety in $\mathbb A^4$ that is birational to the Jacobian (and therefore a quite explicit construction of its function field).

(We are actually considering an open subset of the symmetric square of the curve, since we are looking at certain unordered pairs of points on it. The coefficients $a_0 = \xi\xi', a_1 = -(\xi+\xi'), b_0, b_1$ are invariants of the action of $S_2$ on the corresponding open subset of $H \times H$.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ (1) The $y$-coordinates are taken into account via the polynomial $b(x)$. Concretely, $b_1 = (\eta'-\eta)/(\xi'-\xi)$ and $b_0 = (\eta\xi'-\eta'\xi)/(\xi'-\xi)$. (2) $x^2 + a_1 x + a_0 = (x-\xi)(x-\xi') = x^2 - (\xi+\xi') x + \xi\xi'$. $\endgroup$ – Michael Stoll Feb 26 '15 at 19:04
  • $\begingroup$ Thank you Michael, Is awesome to represent elements on the Jacobian with the pair of polynomials :) Great Answer $\endgroup$ – Eduardo R. Duarte Feb 26 '15 at 19:21

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.