Classgroup of $\mathbb{Q}(\sqrt{2},\sqrt{-13})$ How would you compute the classgroup of the biquadratic number field $\mathbb{Q}(\sqrt{2},\sqrt{-13})$?
I would prefer a method as "from scratch" as possible. Please avoid, if possible, quoting theorems that relate the desired classgroup to those of the quadratic subfields. My objective is to "see" the classgroup in the ring of integers of $\mathbb{Q}(\sqrt{2},\sqrt{-13})$ "as clearly as possible." Apologies for the vagueries here. I am trying to build my intuition, so I cannot be entirely precise with my question. Thanks in advance.
 A: There is probably a reason people don't do these kinds of things from scratch... 
Let $\alpha = \frac{\sqrt{2} + \sqrt{-26}}{2}$ and let $K = \mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt{2}, \sqrt{-13})$. The order $\mathbb{Z}[\sqrt{2}, \alpha]$ has integral basis $1, \sqrt{2}, \alpha, \sqrt{-13}$ and discriminant $2^8 \cdot 13^2$, so it has index dividing $2^4 \cdot 13$ in the ring of integers. Write an element of the ring of integers as
$$\frac{a + b \sqrt{2} + c \alpha + d \sqrt{-13}}{2^4 \cdot 13}$$
where $a, b, c, d$ are integers. Computing the trace down to all three quadratic subfields shows that $a, d$ must be divisible by $2^3 \cdot 13$, that $2b + c$ must be divisible by $2^4 \cdot 13$, and that $c$ must be divisible by $2^4 \cdot 13$, hence $b$ must be divisible by $2^3 \cdot 13$. Up to addition of an element of $\mathbb{Z}[\sqrt{2}, \alpha]$ we may therefore write an element of the ring of integers as
$$\frac{e + f \sqrt{2} + g \sqrt{-13}}{2}$$
where $e, f, g \in \{ 0, 1 \}$. Multiplying by $\alpha$ and simplifying we conclude that we may take $e = g$, which gives $4$ remaining cases. The case $e = f = g = 0$ can be ignored, the cases $e = g = 0, f = 1$ and $e = g = 1, f = 0$ are straightforward to rule out, and the remaining one is
$$\frac{1 + \sqrt{2} + \sqrt{-13}}{2}$$
which can be ruled out by computing its norm down to $\mathbb{Q}(\sqrt{2})$. Hence in fact $\mathbb{Z}[\sqrt{2}, \alpha] = \mathcal{O}_K$. The Minkowski bound is
$$\sqrt{2^8 \cdot 13^2} \left( \frac{4}{\pi} \right)^2 \frac{4!}{4^4} \approx 31.6$$
so the class group is generated by the primes over $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31$. Computing the class group from here seems like a fairly hard slog and I am not convinced there is anything to be gained from it. Can you be more precise about what you want to see? 
