Class Group of $\mathbb Q(\sqrt{-35})$ As an exercise I am trying to compute the class group of $\mathbb Q(\sqrt{-35})$.
We have $-35\equiv 1$ mod $4$, so the Minkowski bound is $\frac{4}{\pi}\frac12 \sqrt{35}<\frac23\cdot 6=4$. So we only need to look at the prime numbers $2$ and $3$.
$-35\equiv 5$ mod $8$, so $2$ is inert. Also, $-35\equiv 1$ mod $3$, so $3$ splits, i.e. $(3)=Q\overline Q$ with $Q=(3,1+\sqrt{-35})$, $\overline Q=(3,1-\sqrt{-35})$. The ideals $Q,\overline Q$ are not principal, because there are no solutions to $x^2+35y^2=12$, i.e. no elements of norm 3.
Now we know that there are at most $3$ elements (or do we?), namely $(1),Q,\overline Q$. Mathematica tells me that the class number is $2$, so $Q$ and $\overline Q$ must be in the same equivalence class and $Q^2$ has to be a principal ideal. But how can I show this? 
 A: Note that the ring of integers is $\mathbb Z[(1+\sqrt{-35})/2]$.
If you compute $(3, 1 + \sqrt{-35})^2$, you get $$(9,3 + 3\sqrt{-35}, -34 + 2 \sqrt{-35} ) = (9, 1 + \sqrt{-35}) = ( \dfrac{1-\sqrt{-35}}{2} \dfrac{1 + \sqrt{-35}}{2}, 2 \dfrac{1+\sqrt{-35}}{2}) = ((1 + \sqrt{-35})/2 )$$
(because 
$\dfrac{1-\sqrt{-35}}{2}$ and $2$ are coprime, and so generate the ideal $1$).

I find the computation a bit easier by phrasing the factorization of $(3)$
in the following alternative way:
$(3) = (3, (1 + \sqrt{-35})/2) (3,(1- \sqrt{-35})/2)$, and
$$(3,(1+\sqrt{-35})/2)^2 = (9, 3(1+\sqrt{-35})/2,(-17+\sqrt{-35})/2) = ( (1+\sqrt{-35})/2 )$$ is principal.

As a consistency check, note that $ (9) = (3) (3) = Q \overline{Q} Q \overline{Q}
= Q^2 \overline{Q}^2,$ but also $9  = ( (1+\sqrt{-35})/2) ( ( 1 - \sqrt{-35})/2),$ so we must have $Q^2$ equal to one of $( (1 \pm \sqrt{-35})/2).$
A: Let $\alpha=\frac{1+\sqrt{-35}}{2}$, then in fact, $(3)=Q\overline{Q}=(3,\alpha)(3,\alpha+2)$. In the class group then, $[Q]+[\overline{Q}]=[\mathcal{O}_K]$, so $[Q]$ and $[\overline{Q}]$ are inverses. So the class group is generated by $Q$, and we just need to show that $Q^2$ is principal.
The minimal polynomial of $\alpha$ is $f(x)=x^2-x+9$, and recall that $f(\beta)=\text{Nm}(\beta-\alpha)$. In particular, $f(0)=9=3^2$. The element $(\alpha)$ is sent to $0$ under the evaluation $\alpha\mapsto 0$, corresponding to the ideal $Q$, so $(\alpha)=Q^2$, and the order of $[Q]$ divides $2$. You've already shown that $Q$ can't be principal, so we're done.
