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