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...
 A: 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. 
A: 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.)
