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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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The class groups are often very random. You may try sage or Magma for the job. Hands down computation is usually very time consuming unless you have a good bound. –  Bombyx mori Sep 1 '12 at 8:54
Along the lines of what user32240 writes, I don't think seeing a worked example is going to build intuition. Even for quadratic fields, you can't stare at one and guess the structure of the class group from intuition. –  KCd Sep 1 '12 at 21:58
Maybe instead of "I am trying to build my intuition" you meant "I am trying to develop more experience"? –  KCd Sep 2 '12 at 2:51

2 Answers 2

up vote 8 down vote accepted

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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First of all, using PARI to find out what the answer should be (my crystal ball is out at the shop), the class number is 6. I agree with you that the class group is generated by the primes lying over 2 and 7 (I calculated it myself by hand in another way), but it can't be the case that $B_{1,7}$ and $B_{3,7}$ have order dividing 2 since $h=6$. I can show by hand that $A$ is not principal, so the ideal class of $A$ has order 2. –  KCd Sep 2 '12 at 3:02
Although $(\alpha)$ is a product of primes lying over 7 and has norm $7^2$, the claim that a prime over 7 has order dividing 2 in the class group need not follow: you are seeking the lowest power of the prime that is principal, and that need not be related to the exponent of 7 in the norm of $(\alpha)$. –  KCd Sep 2 '12 at 3:48
@KCd: thanks for the correction! I'd forgotten to get back to this; I can finish the computation of the class group from here, but I'm not sure what there is to be gained from this computation. –  Qiaochu Yuan Sep 2 '12 at 3:59
What is to gained for the OP, I think, is seeing an instance of how this kind of thing can be carried out in an actual example. Another typo: you write near the end "any element of ${\mathcal O}_K$ has the form $a^2 + 26b^2$, but you meant "the norm of any element of...". –  KCd Sep 2 '12 at 19:08
In looking over your work, I see that you didn't exploit the ${\mathbf Z}[\sqrt{2}]$-module structure as much as I did. At some point I'll get around to posting my solution to this problem. –  KCd Sep 2 '12 at 19:16

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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Will it to be easier to treat this by decomposition groups? I have not touched AN for a while so I just ask. –  Bombyx mori Sep 1 '12 at 8:31
@Qiaochu: Is there a reason you chose to write ‘order’ instead of just ‘ring’? Someone mistakenly tried to change it... –  Zhen Lin Sep 1 '12 at 9:02
+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. –  Ben Blum-Smith Sep 3 '12 at 20:33

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