Class group of $\mathbb{Q}(\sqrt{7})$

Question

Find the class group of $\mathbb{Q}(\sqrt{7})$.

$\textit{Hint}$: notice that $2=(3+\sqrt{7})(3-\sqrt{7})$ and that $-1+\sqrt{7}=(2+\sqrt{7})(3-\sqrt{7})$

Here's what I've done: the Minkowski bound is $\sqrt{7}\approx 2.65$, so we need to analyse $2\mathcal{O}$. Since $x^2-7=(x+1)^2(\text{mod }2)$, then $(2)=\mathfrak{p}^2$, where $\mathfrak{p}=(2, 1+\sqrt{7})$. $[\mathfrak{p}]$ has order $\leq 2$ because $[\mathfrak{p}]^2=[(2)]=e$. If $\mathfrak{p}=(a+b\sqrt{7})$ for some $a+b\sqrt{7}\in\mathcal{O}$, then $(2)=\mathfrak{p}^2=((a+b\sqrt{7})^2)$, so $2=u(a+b\sqrt{7})^2$ for some unit $u$. The only units in $\mathbb{Z}[\sqrt{7}]$ are $(8+3\sqrt{7})^n$ for some $n\in\mathbb{Z}$. Obviously $n$ cannot be even, because that would mean $2=w^2$ for some $w\in\mathbb{Z}[\sqrt{7}]$ (absurd), so we must have $2(8+3\sqrt{7})=w^2$ for some $w\in\mathbb{Z}[\sqrt{7}]$, which is also impossible by simple verification. Therefore $[\mathfrak{p}]\neq e$ and $Cl_{\mathbb{Q}(\sqrt{7})}=\mathbb{Z}/2\mathbb{Z}$.

That took a little bit of work, and I suppose the tip could be useful, but I dont see how.

Answer 1

Maybe I’ve done these computations so many times that I don’t remember where the pitfalls are. But I would argue this way:

You know that $(2)=\mathfrak p^2$, and you know that $2=(3+\sqrt7\,)(3-\sqrt7\,)$. It only remains to show that $3+\sqrt7$ and $3-\sqrt7$ generate the same ideal, which must then be $\mathfrak p$. But: $$ \frac{3+\sqrt7}{3-\sqrt7}=\frac{(3+\sqrt7\,)^2}2=8+3\sqrt7\,, $$ a unit. That seems to do it.