I seem to have arrived at some contradictory results in computing this group, would you mind helping me resolve this?
By Sage + internet I find that it should be true that this class group is trivial.
We know that Minkowski's bound for this group is:
$$
M_k = \frac{1}{2}\frac{\pi}{4}\sqrt{7*4} \approx 2.08.
$$
Therefore the only prime we need to check is $2 \mathcal{O}_k$. Notice that $x^2 + 7 \equiv_2 x^2 + 1 \equiv_2 (x+1)^2$, which implies that it is totally ramified. So then we have to consider $\mathfrak{p}_2 = (2, \sqrt{-7}+1)$. We know that $\mathfrak{p}_2^2 = (2)$, which implies that we know that the order of the ideal class group is less than or equal to 2. So then we consider $(2, \sqrt{-7}+1)$ and want to show that this is principle (otherwise sage/internet is wrong).
Suppose we have some element $z$ such that $(z) = (2, 1 + \sqrt{-7})$. Then it must be true that $N(z) \mid N(2)$ and $N(z) \mid N(1+\sqrt{-7}) \implies N(z) \mid 4, 8 \implies N(z)$ is one of $1, 2, 4$. However, since $N(a+b\sqrt{-7}) = a^2 + 7b^2$, we see that it is impossible for it to take on the values $2$ and $4$, therefore if such $z$ exists it must be $1$.
However, by this answer on stack exchange, such a $z$ is not possible!
Therefore I must conclude that the class group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, but this is clearly not true.
Could you point out my error?
Best Answer
The ring of integers of $\mathbb{Q}(\sqrt{-7})$ is not $\mathbb{Z}[\sqrt{-7}]$ but bigger, because $-7 \equiv 1 \pmod 4$.