# Contradictory results when computing the Ideal class group of $\mathbb{Q}(\sqrt{-7})$

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?

• Minkowski bound say we have to check ideals of norm $\le 2$, that is $(1)$ and the possible prime ideals above $(2)$. Indeed $(2) =(\frac{1+\sqrt{-7}}{2})^2$ is a product of principal ideals so the ideal class group is trivial. – reuns Dec 13 '18 at 0:12
• I see now, I was trying to factor in $Z[\sqrt{-7}]$ but the ring of integers is larger than that by @RicardoBuring. – TrostAft Dec 13 '18 at 0:14

The ring of integers of $$\mathbb{Q}(\sqrt{-7})$$ is not $$\mathbb{Z}[\sqrt{-7}]$$ but bigger, because $$-7 \equiv 1 \pmod 4$$.
• Oh of course, it's actually $\mathbb{Z}[ (1 + \sqrt{-7})/2 ]$, and then we immediately see it. That was crazy of me, thanks. – TrostAft Dec 13 '18 at 0:12