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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 $K$?

user26857
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user444824
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1 Answers1

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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...

Angina Seng
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  • $N(5 + \sqrt{23}) = 5^2 - 23 \times 1^2 = 2$. I'm not sure how that helps though. – user444824 May 09 '17 at 16:11
  • Does it help to find ideals of norm 2? – Angina Seng May 09 '17 at 16:43
  • What can you say about the factorisation into prime ideals of such an ideal? – Tim.ev May 11 '17 at 12:01
  • All the subtlety of a nuclear bomb: it looks like $22=2×11=(\sqrt{23}+1)(\sqrt{23}-1)$ is nonunique. But $2=\color{blue}{(5+\sqrt{23})}(5-\sqrt{23})$, $11=(-1)(9+2\sqrt{23})\color{blue}{(9-2\sqrt{23})}$, and the product of the blue factors is $(-1)(1+\sqrt{23})$. This suggests that in reality ... . – Oscar Lanzi Jun 01 '22 at 01:17