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

Solution 1:

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.

Solution 2:

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.)