Computing the class group of a quadratic function field

Question

I am asking for a reference in which I can find tools to answer questions like the following: Let $K=\mathbb{F}_q(X)$ be a rational function field over the finite field with $q$ elements. Let $E/K$ be a quadratic extension. How can I compute the class groups of $K$?

I'm not interested in the general case, but in the special case where $E = K(\sqrt{D})$, where $D$ is an irreducible polynomial of even degree $\deg D=2g+2$ and $g,D$ may vary (that is, statistical results are also welcomed).

Edit. I've been asked to be more specific in my question: Given a finite field $\mathbb{F}_q$ and a prime $\ell$ (say $\ell=3$ to be concrete), can is there always an hyperelliptic curve $y^2=D(x)$ with $D$ irreducible of degree $2g+2$ such that the class group has a order divisible by $\ell$?

Answer 1

The answer to the precise question is yes. See Theorem 1.2 of:

"Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields," https://arxiv.org/pdf/0912.0325.pdf.

This implies that a positive proportion of imaginary quadratic extensions of $\mathbb{F}_q(x)$ has class number divisible by $\ell$, at least for $q$ large and $q \not \equiv 1 \bmod \ell$.

The corresponding infinitude is easier and was known previously in more generality (see the discussions following Theorem 1.2).

Answer 2

This is more of a too-long comment than an answer.

The specific question in the comments, "can I find for every finite field, say of characteristic neq 2, an elliptic curve $y^2 = D(x)$, with $D$ of degree 4 and irreducible such that the number of points is even" has a nice positive answer: yes, for any irreducible $D$. And the same trick applies to produce hyperelliptic examples in every odd genus.

Let $g \ge 1$ be odd, and let $F$ be an irreducible polynomial of degree $2g+2$ over $\mathbb{F}_q$ with $q$ odd (I will use $F$, rather than $D$, for the polynomial, so that I can use $D$ for divisors). We define a genus-$g$ hyperelliptic curve $ H: y^2 = F(x) $ over $\mathbb{F}_q$. Let $D_\infty$ be the (degree 2) divisor at infinity on $H$.

The polynomial $F$ factors over $\mathbb{F}_{q^2}$ as $F = GG'$ where $G$ and $G'$ are irreducible polynomials of degree $g+1$ over $\mathbb{F}_{q^2}$, conjugate over $\mathbb{F}_q$. Let $\{\alpha_1,\ldots,\alpha_{g+1}\}$ be the roots of $G$ and $\{\alpha_1',\ldots,\alpha_{g+1}'\}$ be the roots of $G'$. Since $g$ is odd, we can define divisors $D := \sum_{i=1}^{g+1}(\alpha_i,0) - ((g+1)/2)D_\infty$ and $D' := \sum_{i=1}^{g+1}(\alpha_i',0) - ((g+1)/2)D_\infty$ on $H$, both of degree $0$ and both defined over $\mathbb{F}_{q^2}$. Now $D$ and $D'$ are not principal (they correspond to the ideals $(G(x),y)$ and $G'(x),y)$, but by construction $2D = (G)$ and $2D' = (G')$, so $[D]$ and $[D']$ are 2-torsion divisor classes exchanged by Galois. But in fact they are the same class, because $D + D' = (y)$; so $[D] = [D']$ is an $\mathbb{F}_q$-rational divisor class of order 2, and therefore the divisor class group of $H$ has even order.