# Without using Heegner-Stark-Baker, $\mathbb{Q}(\sqrt{-11})$ has class number $1$.

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.

• math.blogoverflow.com/2014/08/23/… May 17, 2015 at 20:16
• I think with a smidgen of cleverness one can show that this ring of integers is (norm-) Euclidean, for example. May 17, 2015 at 20:21
• Have you heard of the Minkowski Bound? May 17, 2015 at 21:15
• How about Bachman Turner Overdrive? May 17, 2015 at 21:24
• Minkowski bound is $2\sqrt{11}/\pi\approx 2.11143$, and you have checked ideals of norm two, so looks like you're done? Read this set of examples from Planetmath. May 18, 2015 at 9:09

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 $z\in \mathbb{Z}[\frac{1 + \sqrt{-11}}2]$.

• +1 I wanted to see what the picture looks like. I took the liberty of adding an image to your answer. Feel free to remove it, if you think it makes it too easy it follow up on your hint. May 18, 2015 at 9:46
• @JyrkiLahtonen A nice picture. Thanks a lot for helping me with improving my answer! May 18, 2015 at 9:48
• Why downvote? Any error? May 20, 2015 at 9:27

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 $\mathbb{Z}\left[\tfrac{1 + \sqrt{-19}}{2}\right]$.