TITLE: Let $K = \mathbb{Q}(\sqrt{23})$. Compute the class number of $K$.
QUESTION [6 upvotes]: Let $K = \mathbb{Q}(\sqrt{23})$. Compute the class number of $K$.

Here's what I know:
As $d = 23 \not\equiv 1$ (mod $4$) we have $\Delta_K = 4d = 92$.
The Minkowski bound is $M_K = \sqrt{23} < 5$ so we only need to check the ideals generated by the primes 2 and 3.
With the minimal polynomial $f =X^2 -23$ we find that in $\mathbb{F}_2$ this factorizes as $X^2 + 1 = (X + 1)^2$, meaning $(2) = (2, 1 + \sqrt{23})^2$.
In $\mathbb{F}_3$, $f$ is irreducible and therefore $(3)$ is prime. 
This is where I get stuck. How do I use these facts to find the class number of $K$?

REPLY [3 votes]: So every ideal class contains an ideal of norm $1$, $2$, $3$ or $4$.
There aren't any of norm $3$. What of norm $2$? What's the norm of
$5+\sqrt{23}$? The ideals of norm $4$ are products of norm $2$ ideals,
so if norm $2$ ideals are principal...