TITLE: Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.
QUESTION [13 upvotes]: Prove that $\mathbb{Q}(\sqrt{-11})$ is of class number $1$.

I have found that the ideal $(2)$ of the integer ring $\mathbb{Z}[(1 + \sqrt{-11})/2]$ of $\mathbb{Q}(\sqrt{-11})$ is a prime ideal. But I am not sure where to go from there. Any help would be greatly appreciated.

REPLY [6 votes]: The Euclidean algorithm says for any $a,b \in \mathcal{O}_K$ there is a quotient and remainder $q,r \in \mathcal{O}_K$ with $N(r)<N(q)$ such that $a=bq+r$.
Another way of thinking about it is to divide both side by $b$ and say
$$ \frac{a}{b}=q+   \frac{r}{b}$$
So that every rational number can be represented as the sum of a lattice point $q \in \mathcal{O}_K$ and element of unit disk $N( \frac{r}{b})<1$.
Once we prove the existence of the Euclidean algorithm for $K=\mathbb{Q}(\sqrt{-11})$  then unique factorization into primes follows.  Every Euclidean domain is a principal ideal domain.

The norm for the ring of integers  $\mathbb{Z}\left[\frac{1 + \sqrt{-11}}{2}\right]$ is a binary quadratic form over $\mathbb{Z}$. 
$$\big|\big|x + \tfrac{1 + \sqrt{-11}}{2}y \big|\big|^2 = x^2 + xy + 3y^2 \hspace{0.25in}\text{and}\hspace{0.25in}
\big|\big| \tfrac{m + n\sqrt{-11}}{2} \big|\big|^2 = \frac{1}{4}(m^2 + 11n^2 )
$$
The "unit circle" for this norm is actually an ellipse
with $ Area(\bigcirc)= \frac{4\pi}{11} >1 = Area(\square) $   
The oval area is just slightly bigger than the fundamental area of the lattice, so you are fine by Minkowski.

Check it out! The Minkowski sum of the integer lattice $\mathbb{Z}\left[\tfrac{1 + \sqrt{-11}}{2}\right]$  and the unit circle $||z|| < 1$ covers the entire plane:


This is no longer true for $\mathbb{Z}\left[\tfrac{1 + \sqrt{-19}}{2}\right]$.