# Let $K = \mathbb{Q}(\sqrt{23})$. Compute the class number of $K$.

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 $K$?

• Just a quick comment: in general, if $p$ is a positive prime, $\mathbb Q(\sqrt p)$ is quite likely UFD. – Robert Soupe May 9 '17 at 17:27

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...
• $N(5 + \sqrt{23}) = 5^2 - 23 \times 1^2 = 2$. I'm not sure how that helps though. – user444824 May 9 '17 at 16:11