Class number of $\mathbb{Q}[\sqrt{2}]$

Question

Show that the class number of $\mathbb{Q}[\sqrt{2}]$ is $1$.

We set $M=(2/\pi)^s\sqrt{|d_K|}$ as in the proof of the finiteness of the class group. Since $s=0$, because there are no complex embeddings, and $d_K=8$, we get $M=2\sqrt{2}$. Again, by the proof of the finiteness of the ideal class group, we must find all ideals $I$ in $\mathbb{O}_K$ with norm $n(I)\leq 2\sqrt{2}$. Now, the possible values for the norm of such ideal are $1$ and $2$.

I'm not really convinced of the following:

1. Thus, the prime ideals dividing $I$ such that $n(I)\leq 2\sqrt{2}$ are the prime ideals dividing the ideal $(2)$.

2. But $2=(\sqrt{2})^2$. If we show that $(\sqrt{2})$ is prime, then all prime ideals with norm equal or less than $2\sqrt{2}$ are principal, so each ideal $I$ with $n(I)\leq 2\sqrt{2}$ is principal.

I get the last statement of 2. (if $I=p_1...p_r$ is the prime decomposition of $I$ and all $p_i$ are principal then $I$ is principal), but the rest is still cloudy.

I appreciate any help, and thanks in advance!

Answer 1

Let $K=\Bbb Q(\sqrt{d})$ with squarefree $d$. Then the absolute discriminant is given by $$ |d_K|= \begin{cases} 4d & \text{ for } d\equiv 2,3 \mod 4, \\ d & \text{ for } d\equiv 1 \hspace{0.484cm} \mod 4. \end{cases} $$ The Minkowski bound is given by $$ B_K = \frac{n!}{n^n}\left( \frac{4}{\pi}\right)^s\sqrt{|d_K|}. $$ In the quadratic case we have $n=2$ and $s=0$ for $d>0$, $s=1$ for $d<0$. The Minkowski bound for $\Bbb Q(\sqrt{2})$ is given by $$ B_K=\sqrt{2}<2. $$ Hence the class number is equal to $1$.

Edit: This argument works for $d=-7,-3,-2,-1,2,3,5,13$. The converse is not true, because, say, for $d=-11$ the Minkowski bound is $\frac{2}{\pi}\sqrt{11}>2$, but still this number field has class number $1$.