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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.
Prove that $\mathbb{Z}[\frac{1 + \sqrt{-11}}2]$ is an ED, which is just alike the case of $\mathbb Z[\sqrt{-1}]$. The core is that for any $a+b\sqrt{-11}\in\mathbb{Q}(\sqrt{-11})$, there exists an element $z\in\mathbb{Z}[\frac{1 + \sqrt{-11}}2]$ such that, if we write $a+b\sqrt{-11}=z+r+t\sqrt{-11}$, then $\vert r\vert\leqslant\frac12,\ \vert t\vert\leqslant\frac14$, and consequently, $N(r+t\sqrt{-11})<1$.
With the help of this lemma, for any two elemnts $x,y\in\mathbb{Z}[\frac{1 + \sqrt{-11}}2]$ with $y\neq0$, we can find an $z\in\mathbb{Z}[\frac{1 + \sqrt{-11}}2]$ such that $\frac xy=z+r+t\sqrt{-11}$ and $N(r+t\sqrt{-11})<1$, or equivalently, $ x=yz+y(r+t\sqrt{-11})$ and $N(y(r+t\sqrt{-11}))<N(y)$ (note also that $y(r+t\sqrt{-11})=x-yz\in\mathbb{Z}[\frac{1 + \sqrt{-11}}2]$). So Euclidean algorithm is applicable.
The figure has some elements of $\mathbb{Z}[\frac{1 + \sqrt{-11}}2]$ (dots) and the above regions (grey rectangles) centered around some points $z\in \mathbb{Z}[\frac{1 + \sqrt{-11}}2]$.
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 $\mathbb{Z}\left[\tfrac{1 + \sqrt{-19}}{2}\right]$.
