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...

  • $\begingroup$ 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. $\endgroup$ – franz lemmermeyer Nov 2 '12 at 13:53
  • $\begingroup$ @franzlemmermeyer: but I don't see how this $M_K$ as in the lemma has something to do with norm of ideals. $\endgroup$ – Bruno Stonek Nov 2 '12 at 14:04
  • $\begingroup$ What exactly is part of a homework question? More specifically, what is the homework question? $\endgroup$ – KCd Nov 3 '12 at 21:45
  • $\begingroup$ @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. $\endgroup$ – Bruno Stonek Nov 3 '12 at 21:54
  • $\begingroup$ 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. $\endgroup$ – KCd Nov 3 '12 at 22:42

Ha. I'm going to lecture on exactly this next week. If you're in Sydney, drop on in.

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.

  • $\begingroup$ 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? $\endgroup$ – Bruno Stonek Nov 2 '12 at 12:44

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