TITLE: Proving the class number of $\mathbb{Q}(\sqrt{-5})$ is 2 using Ireland-Rosen's bound
QUESTION [10 upvotes]: 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...

REPLY [2 votes]: We can take $M_K=2$, solely using elementary means.
Using the multiplicativity of norm, rewrite $N(t\alpha - \omega \beta) < N(\beta)$ as $N(t \tfrac{\alpha}{\beta} - \omega)<1$. In other words, we want to show that, for all $\gamma \in K$, there is $1 \leq t \leq M_K$ and $\omega \in R_K$ such that $N(t \gamma-\omega)<1$. I claim that this holds for $M_K=2$, where $R_K$ is $\mathbb{Z}[\sqrt{-D}]$ for $D \in \{1,2,3,4,5,6 \}$ or $\mathbb{Z}\left[\tfrac{1}{2} + \sqrt{-D} \right]$ for $D \in \{ 0.75, 1.75, 2.75, 3.75, 4.75, 5.75 \}$. For shorthand, put $\zeta$ to be either $\sqrt{-D}$ or $\tfrac{1}{2} + \sqrt{-D}$ accordingly.
Let $\gamma = x+iy$. Subtracting off an integer multiple of $\zeta$, we may assume that $|y| < \tfrac{\sqrt{D}}{2}$. WLOG, we may assume $y \geq 0$.
Case 1: $0 \leq y <\tfrac{\sqrt{3}}{2}$. In this case, subtracting off integers, we may assume that $|x|<\tfrac{1}{2}$. So
$$N(x+iy) < \left(\tfrac{1}{2} \right)^2 + \left
(\tfrac{\sqrt{3}}{2} \right)^2 < 1.$$
Case 2: $\tfrac{\sqrt{3}}{2} \leq y < \tfrac{\sqrt{D}}{2}$. (Note that this case can only occur if $D \geq 3$.) Then $\sqrt{3} \leq 2 y < \sqrt{D}$ so $|2y - \sqrt{D}| < \sqrt{D}-\sqrt{3} \leq \sqrt{6} - \sqrt{3}$. In other words, the imaginary part of $2(x+iy) - \zeta$ is at most $\sqrt{6}-\sqrt{3}$. Subtracting off an appropriate integer, we get an number of the form $2(x+iy) - \omega$ with imaginary part at most $\sqrt{6} - \sqrt{3}$ and real part at most $1/2$. We compute
$$(1/2)^2 + (\sqrt{6} - \sqrt{3})^2 \approx 0.76 < 1$$
so we win. $\square$
I like this argument because not only can it do $\mathbb{Z}[\sqrt{-5}]$, it can also do $\mathbb{Z}\left[ \tfrac{1+\sqrt{-19}}{2} \right]$, which is a PID that is not Euclidean. (Note that this is the $D=4.75$ case, so it morally should be easier than the $D=5$ case.)