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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.

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    Maybe instead of "I am trying to build my intuition" you meant "I am trying to develop more experience"?2012-09-02

2 Answers 2

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I don't guarantee that this is free of mistakes.

To compute the class group, we need to factor the ideals $(p)$ for primes $2 \le p \le 31$.

Case: $p = 2$. We have $(2) = (\sqrt{2})^2$ and $\mathcal{O}_L/(\sqrt{2}) \cong \mathbb{F}_2[\alpha]/(\alpha^2 - 1)$, so $(\sqrt{2}) = (\sqrt{2}, \alpha - 1)^2$. Hence

$(\sqrt{2}) = A^2$

where $A = (\sqrt{2}, \alpha - 1)$ is prime of norm $2$ and has order dividing $2$ in the class group.

The remaining primes are odd, and when working $\bmod p$ for $p$ odd we have $\mathcal{O}_L/(p) \cong \mathbb{F}_p[\sqrt{2}, \sqrt{-13}]$, so things simplify a little.

Case: $p = 13$. We have $(13) = (\sqrt{-13})^2$ and $\mathcal{O}_L/(\sqrt{-13}) \cong \mathbb{F}_{13}[\sqrt{2}]$ (note that $\bmod \sqrt{-13}$ we have $2 \alpha = \sqrt{2}$), which is $\mathbb{F}_{13^2}$. Hence $(\sqrt{-13})$ is prime and trivial in the class group.

The remaining primes do not ramify.

Case: $p = 7, 17, 31$. These are the primes for which $2, -13, -26$ are all quadratic residues. We can write them all up to sign as $a^2 - 2b^2$; consequently, $(p) = (a + b \sqrt{2})(a - b \sqrt{2})$, and $\mathcal{O}_L/(a + b \sqrt{2}) \cong \mathbb{F}_p[\sqrt{-13}] \cong \mathbb{F}_p^2$, so

$(a + b \sqrt{2}) = B_{1, p} B_{2, p}$

where the $B_{i, p}$ are distinct primes of norm $p$ and inverses in the class group. Similarly, $(a - \sqrt{2} b) = B_{3, p} B_{4, p}$.

Case: $p = 23$. We have $(23) = (3 + 4 \sqrt{2})(3 - 4 \sqrt{2})$ and $\mathcal{O}_L/(3 + 4 \sqrt{2}) \cong \mathbb{F}_{23}[\sqrt{-13}] \cong \mathbb{F}_{23^2}$. Hence the ideals $(3 \pm 4 \sqrt{2})$ are prime and trivial in the class group.

Case: $p = 29$. We have $(29) = (4 + \sqrt{-13})(4 - \sqrt{-13})$ and $\mathcal{O}_L/(4 + \sqrt{-13}) \cong \mathbb{F}_{29}[\sqrt{2}] \cong \mathbb{F}_{29^2}$. Hence the ideals $(4 \pm \sqrt{-13})$ are prime and trivial in the class group.

Case: $p = 3, 5, 11, 19$. These are the remaining primes for which $2$ is not a quadratic residue and exactly one of $-13, -26$ is. In this case $\mathbb{F}_p[\sqrt{2}, \sqrt{-13}] \cong \mathbb{F}_{p^2} \times \mathbb{F}_{p^2}$ and

$(p) = C_{1, p} C_{2, p}$

where the $C_{i, p}$ are distinct prime of norm $p^2$.

In summary, the class group is generated by the following prime ideals:

  • $A$ (norm $2$, order dividing $2$),
  • $B_{1, 7}, B_{3, 7}, B_{1, 17}, B_{3, 17}, B_{1, 31}, B_{3, 31}$ (norms $7, 7, 17, 17, 31, 31$),
  • $C_{1, 3}, C_{1, 5}, C_{1, 11}, C_{1, 19}$ (norms $3^2, 5^2, 11^2, 19^2$).

To find relations between these elements we will compute norms of elements of $\mathbb{Z}[\sqrt{2}, \alpha]$:

  • $N(\sqrt{2} + \alpha) = 11^2$, hence $C_{1, 11}$ is trivial.
  • $N(\sqrt{2} + \sqrt{-13}) = 3^2 \cdot 5^2$, hence $C_{1, 3} = C_{1, 5}^{\pm}$.
  • $N(3 + \sqrt{-26}) = 5^2 \cdot 7^2$, hence $C_{1, 5}$ is some product of the $B_{i, 7}$s.
  • $N(14 + \sqrt{-13}) = 11^2 \cdot 19^2$, hence $C_{1, 19}$ is trivial.
  • $N(1 + \alpha) = 2 \cdot 31$, hence $B_{1, 31} = A$. Conjugating, we find that $B_{3, 31} = A$.
  • $N(3 + \alpha) = 2 \cdot 7 \cdot 17$, hence $A$ is some product of a $B_{i, 7}$ and a $B_{j, 17}$. Conjugating, we obtain another such product.
  • $N(8 + \alpha) = 17^3$. We compute that $8 + \alpha$ is not divisible by $1 \pm 3 \sqrt{2}$, hence $B_{i, 17}$ has order dividing $3$. Together with the above, we conclude that $A = B_{i, 7}^3$.
  • $N(\alpha) = 7^2$. We will use this below.

So the class group is generated by $B_{1, 7}$ and $B_{3, 7}$, both of which cube to $A$, which has order $2$. Now, the prime factorization of $(\alpha)$ is either (WLOG) $B_{1, 7}^2$ or $B_{1, 7} B_{3, 7}$. In the first case, $A = B_{i, 7}$ in the class group, so $A B_{i, 7}$ is trivial in the class group. It follows that some element of $\mathcal{O}_K$ has norm $14$, but using the tower property of the norm, the norm of any element of $\mathcal{O}_K$ has the form $a^2 + 26b^2$ ($a, b \in \mathbb{Z}$), and $14$ is not of this form.

Hence $(\alpha) = B_{1, 7} B_{3, 7}$, from which it follows that the class group is generated by $B_{1, 7}$, which has order $6$. Again using the tower property, there are no elements of norm $2$, so $A$ is nontrivial and $B_{1, 7}$ has order either $2$ or $6$. We ruled out the first possibility above, so:

The class group is cyclic of order $6$. It is generated by any prime lying over $7$.

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    @KCd - Although I accepted Qiaochu's answer, I would definitely be interested in your alternative solution based on the $\mathbb{Z}[\sqrt{2}]$-module structure of $\mathcal{O}_K$, whenever you have time.2012-09-04
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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?

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    +1 Wow Qiaochu, thanks for putting in all this work. Your two answers together is exactly the type of thing I was hoping for. I wanted to see the watch completely disassembled so I could examine all the gears.2012-09-03