# Proving the class number of $\mathbb{Q}(\sqrt{-5})$ is 2 using Ireland-Rosen's bound

In this MO answer, Keith Conrad states that you can use the method of proof of the finiteness of the class number in Ireland & Rosen to prove that the class number $h_K=2$, when $K=\mathbb{Q}(\sqrt{-5})$. The point is to avoid using Minkowski's bound.

I would like to have some hints as to how to do this (since this is part of a homework question).

The outline of Ireland & Rosen's proof is the following (pp. 178-179):

Lemma: There exists a positive integer $M_K$ such that for all $\alpha,\beta\in \mathcal{O}_K$, $\beta\not=0$, there is an integer $t$, $1\leq t\leq M_K$ and an element $\omega \in \mathcal{O}_K$ such that $\lvert N(t\alpha-\omega \beta) \rvert < \lvert N(\beta)\rvert$.

If I understand the proof correctly, $M_K$ is as follows:

Let $\omega_1,\dots, \omega_n$ be an integral basis for $K$. Let $C=\prod_i \sum_j \lvert \sigma_i(\omega_j) \rvert$ where $\sigma_i$ are the $n$ $\mathbb{Q}$-monomorphisms $K\to \mathbb{C}$.

Let $m> \sqrt[n]{C}$ be an integer. Then we let $M_K=m^n$. $\square$

Now the finiteness of the class number follows, by proving that every non-zero ideal is equivalent to an ideal that contains $M_K!$. Since these ideals are in bijection with the ideals of $\mathcal{O}_K/M_K!\mathcal{O}_K$ which is a finite ring, there are finitely many ideal classes.

How to use this to prove that $h_K=2$ when $K=\mathbb{Q}(\sqrt{-5})$?

I'm getting $M_K=16$ by using the standard integral basis $\{1, \sqrt{-5}\}$: indeed, $C=(1+\sqrt{5})(1+\sqrt{5})\approx 10,4$, then $\sqrt{C}\approx 3,2$, thus we take $m=4$, whence $M_K=4^2=16$.

This $M_K$ does not seem useful...

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Any bound is useful. Just show that all ideals with norm less than 16 are either principal or equivalent to the prime ideal above 2. –  franz lemmermeyer Nov 2 '12 at 13:53
@franzlemmermeyer: but I don't see how this $M_K$ as in the lemma has something to do with norm of ideals. –  Bruno Stonek Nov 2 '12 at 14:04
What exactly is part of a homework question? More specifically, what is the homework question? –  KCd Nov 3 '12 at 21:45
@KCd: "Can you use the lemma and the bound just found to prove that $h_K=2$?" The part before asked to prove that $M_K=4$ using the method of proof of I&R, and asked if it could be reduced to 2 or 3. –  Bruno Stonek Nov 3 '12 at 21:54
If every ideal class contains an ideal $I$ such that $24 \in I$, then $I$ divides (24). Think about what that tells you about prime ideals that generate the ideal class group. –  KCd Nov 3 '12 at 22:42

Anyway, the approach I'm going to use involves the concept of the norm of an ideal, but maybe you can adapt it to what you have. If $A$ is an ideal of ${\cal O}_K$ then $N(A)$ is the cardinality of ${\cal O}_K/A$. If $t=[\sqrt{N(A)}]$ then there are $(t+1)^2$ distinct numbers of the form $b_1+b_2\sqrt{-5}$ with $0\le b_i\le t$. Now $(t+1)^2\gt N(A)$ so two of these numbers must be congruent mod $A$, so $A$ contains a nonzero number $\alpha=a_1+a_2\sqrt{-5}$ with $|a_i|\le t$. Then $N(\alpha)=a_1^2+5a_2^2\le6t^2\le6N(A)$.
I think the $6$ here may be an improvement on the $16$ you got, though I'm not sure I see where your $16$ comes from.
Anyway, from here you can show every nonzero ideal is equivalent to an ideal of norm at most $6$. Then you can find all the ldeals of norm at most $6$ --- there aren't that many of them --- and you can prove that all the non-principal ones are equivalent, and you're done.
Thank you for your answer. Unfortunately, I'm quite far from Sydney! I've added an explanation to my $M_K$ in the question. I understand this approach does not use the Minkowski bound, and I appreciate it, but I don't see how it has anything to do with Ireland & Rosen's proof... Could you please elaborate? –  Bruno Stonek Nov 2 '12 at 12:44