Show that $h(Q(\sqrt{10})) = 2$
use this theorem
Theorem: Let $K = Q(\sqrt{d})$ where $d$ not in $(0; 1)$ is a square free integer. If $W:=\{\sqrt{d} \text{ whenever } d \equiv 2\text{ or }3 \bmod{4};\}$ or $W:=\{1+ \sqrt{d}/2\text{ whenever }d \equiv 1 \bmod{4} \}$ then $OK = Z[W]$.
I am trying to work it out and this is my solution could you please help with using the theorem:
Note that this has discriminant $4 * 10 = 40$. The degree of this over $Q$ is $n = 2$, and there are $s = 2$ real embeddings and $t = 0$ imaginary pairs of embeddings.
So, every class of ideals contains one with norm at most
$(4/\pi)^0 * (2!/2^2) * \sqrt{40} = \sqrt{10}$
(which is just above 3).
Hence, we need to factorize the primes at most 3.
Since $10 \not\equiv 1 \bmod{4}$, we know that the number ring is $Z[\sqrt{10}]$, and $t^2 - 10$ is the minimal polynomial for $\sqrt{10}$ over $Q$.
(i) Working mod $3$, $t^2 - 10 = t^2 - 1 = (t + 1)(t - 1)$ (mod 3).
$\Rightarrow \rm<3\rm> = \rm< 3, \sqrt{10} + 1 \rm> * \rm< 3, \sqrt{10} - 1 \rm >$
(ii) Working mod 2, we have $t^2 - 10 \equiv t^2$ (mod 2). $\Rightarrow \rm<2 \rm> = \rm <2, \sqrt{10} \rm >^2 $
Suppose that $\rm<2, \sqrt{10}\rm>$ were principal; that is, $\rm< 2, \sqrt{10} \rm> = \rm <a + b\sqrt{10}\rm >$ for some $a, b$ in $\mathbb{Z}$. So, $N(\rm <a + b\sqrt{10} \rm>^2) = N( \rm <2 \rm>) = 4$. $==> N(\rm <a + b\sqrt{10} \rm>) = 2$ $==> a^2 - 10b^2 = 2 or -2$, neither of which have solutions (try reducing mod 10). So, we have a contradiction, and $p := \rm <2, √10 \rm>$ is not principal.
Next, check that $\rm <2, √10 \rm> * \rm <3, \sqrt{10} + 1 \rm> = \rm <-2 + \sqrt{10} \rm>$, and so $[\rm <-2 + \sqrt{10} \rm >] = [p]^{-1}$. (We have a similar result for the other prime ideal above 3.)
That means that every class of fractional ideals either contains a principal ideal or $p$; thus the class group equals $\{[O], [p]\}$. So, $h = 2$, as required.
