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I'm trying to show $\mathbb{Z}[\sqrt{-31}]$ has three distinct ideal classes, but something keeps going wrong and I can only find the identity and one other class.

I found $|d_{K}| = 31$ because $-31 \equiv 1 \bmod 4$, and using Minkowski's Bound I computed $\frac{1}{2} \frac{4}{\pi} \sqrt{31} < 4$, so I need only check primes 2 and 3 (I feel like this is where I may be missing something).

$\langle 2 \rangle$ splits into $\left\langle 2, \frac{1 \pm \sqrt{-31}}{2} \right\rangle$ which I can show is not principal by considering the norm of an arbitrary element needs to equal $2$ $\implies$ $N(\alpha) = N\left(\frac{a + b\sqrt{-31}}{2}\right) = \frac{a^{2} + 31b^{2}}{4} \neq 2$.

My problems is with $\langle 3 \rangle$ because it is prime in $\mathcal{O}_{K}$ due to $-31$ not being congruent to a square modulo $3$. Thus $\langle 3 \rangle$ is equivalent to the identity in the ideal class group, therefore the ideal class group has only two classes.

Can anyone see where I am making a mistake? Your comments would be much appreciated. Thanks for reading.

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    $\begingroup$ I think the two ideals $(2,\frac{1\pm\sqrt{-31}}{2})$ belong to different classes? Or maybe I miss some element $\alpha\in\mathcal O_K$ such that $(2,\frac{1+\sqrt{-31}}{2})\cdot(\alpha)=(2,\frac{1-\sqrt{-31}}{2})$? $\endgroup$
    – awllower
    Nov 12, 2017 at 17:10
  • $\begingroup$ Thank you! I never considered that. I kept thinking the prime factors had to be in the same class. $\endgroup$ Nov 12, 2017 at 17:31

1 Answer 1

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As awllower points out, $\mathfrak{p}=(2,\frac12(1+\sqrt{-31}))$ and $\mathfrak{p}'=(2,\frac12(1-\sqrt{-31}))$ are in different ideal classes. Clearly $\mathfrak{p}\mathfrak{p}'=(2)$ so $[\mathfrak{p}]$ and $[\mathfrak{p}']$ are inverse classes. Also as there is no norm $2$ element in the ring of integers, each is non-principal. But $(\frac12(1+\sqrt{-31}))=\mathfrak{p}^3$ as this ideal has norm $8$ and cannot be divisible by $\mathfrak{p}'$. Thus $[\mathfrak{p}]$ is an ideal class of order $3$ in the classgroup and $[\mathfrak{p}']=[\mathfrak{p}]^{-1}=[\mathfrak{p}]^2$.

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  • $\begingroup$ Awesome, thank you very much. I kept thinking the two factors were in the same class. This makes a lot of sense though. $\endgroup$ Nov 12, 2017 at 17:32
  • $\begingroup$ Could you explain why it is that $p'$ doesn't divide $p^3$? (not sure how to type a gothic p) $\endgroup$ Nov 12, 2017 at 17:49
  • $\begingroup$ @celticburns, Well $\mathfrak p$ certainly does, and if $\mathfrak p'$ did so too then so would $(2)=\mathfrak p\mathfrak p'$. $\endgroup$ Nov 12, 2017 at 17:57
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    $\begingroup$ @JackD'Aurizio Looks like you listed four there.... $a^2+ab+8b^2$ and $a^2-ab+8b^2$ are properly equivalent ... the latter is not reduced by the usual definition. $2a^2+ab+4b^2$ and $2a^2-ab+4b^2$ are not properly equivalent, and neither are equivalent to $a^2+ab+8b^2$. Both are reduced. So of the four forms you list, three are reduced. In general I would counsel against believing Wolfram's web pages; there are errors on the web-page cited. In particular it classifies the wrong class as positive definite. $\endgroup$ Nov 12, 2017 at 18:19
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    $\begingroup$ @JackD'Aurizio In case of interest, I recommend Buell, Binary Quadratic Forms. He does a good job of relating forms and quadratic fields. Also, while he is consistent with older texts, he does very well on forms with odd $ab$ coefficient. For this specific question, recommend Hudson and Williams on class number three, zakuski.utsa.edu/~jagy/Hudson_Williams_1991.pdf $\endgroup$
    – Will Jagy
    Nov 12, 2017 at 19:23

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