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.
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 $[\mathfrak{p}']=[\mathfrak{p}]^{-1}=[\mathfrak{p}]^2$.