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



Function field of the Jacobian of genus 2 curve over $\mathbb{F}_q$
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 $H \times H$.)


