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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 2.2012-11-02
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    @franzlemmermeyer: but I don't see how this $M_K$ as in the lemma has something to do with norm of ideals.2012-11-02
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    What exactly is part of a homework question? More specifically, what is the homework question?2012-11-03
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    @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 3.2012-11-03
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    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.2012-11-03

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