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

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!

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 $$1$$.
• You may have rendered the bound for $\mathbb Q[\sqrt{\color{blue}{-2}}]$ in your last equation. Also the discriminant is $8$ here, not $2$, though that is still small enough to render the bound less than $2$. Anyway please check. Jan 12, 2022 at 18:13
• @OscarLanzi Yes, you were right, I was in the line with $d=-2$. Corrected. Jan 12, 2022 at 19:15
• Although $(2/\pi)\sqrt{11}>2$, it is still less than $3$, and then $(2)$ is inert therefore prime with the discriminant $\equiv5\bmod 8$. So the Minkowski bound still leads to class 1 for $\mathbb Q[\sqrt{-11}]$. Also works for $\mathbb Q[\sqrt{-19}]$. Jan 13, 2022 at 1:42
• @OscarLanzi Yes, right, but the converse of the implication $B_K<2 \Longrightarrow h_K=1$ is already wrong for $d=-11$. Jan 14, 2022 at 16:12