Prove that $\mathcal{O}_3$ and $\mathcal{O}_7$ are euclidean domains

Question

For a non-square integer $d$ such that $d \equiv 1 \mod 4,$ we define the set $$\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}.$$

Prove that $\mathcal{O}_3$ and $\mathcal{O}_7$ are euclidean domains (draw their lattices in the complex plane).

In my class we drew a general lattice for $\mathbb{Z}[i]$:

and then proved $\mathbb{Z}[i]$ is a euclidean domain: Consider $\dfrac{b}{a} \in \mathbb{C}.$ Let $q \in \mathbb{Z}[i]$ be the closest Gaussian integer to $\dfrac{b}{a}.$ Then the norm function $N \left(\dfrac{b} {a}-q \right) \leqslant \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{1}{2}.$ Set $r := b -aq \in \mathbb{Z}[i].$ Then $N(r) = N(b - aq) = N \left(a \left(\dfrac{b}{a} -q \right) \right) = N(a)N \left(\dfrac{b}{a} - q \right) \leqslant \dfrac{1}{2} N(a) < N(a).$

First off, how do we draw these lattices? I am not entirely sure how to draw $\mathbb{Z}[\sqrt{-2}]$ or $\mathbb{Z}[\sqrt{3}]$ since I don't fully understand how the values $\dfrac{1}{\sqrt{2}}$ and $1$ are chosen in the example above. Would the same methods work for drawing $\mathcal{O}_3$ and $\mathcal{O}_7$ in the complex plane? I'm not really sure where to start proving that they are both euclidean domains.

Answer 1

Using the norm $N((a+b\sqrt{3})/2):= \frac{1}{4}|a^2-3b^2|$ I drew the fundamental domain $N(z) < 1$ and the corresponding hexagonal lattice. These look like Eisenstein integers $\mathbb{Z}[e^{2\pi i /3}]$ except the discriminant is positive instead of negative.

Your goal is to show any of the blue dots can be translated onto the green area using lattice vectors. The fact is you get infinitely many as you go up the tentacles of the hyperbola, but they are all related by a factor of units in this ring.

To the compute the units in this ring note that $\sqrt{3} = [1, \overline{1,2}] $ is the continued fraction expansion and from you compute the units by solving Pell's equation.

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The corresponding situation for $\sqrt{7}$ is a bit more tricky. The diagram suggests you get a Euclidean norm as before - $N((a+b\sqrt{7})/2):= \frac{1}{4}|a^2-7b^2|$ - by a narrower margin.

This trick works for discriminants up to around $13$ and there are norm Euclidean rings up to $d = 51$ and $73$. The proof is not obvious.

Answer 2

I'm sorry, I didn't notice that $O_3$ and $O_7$ are no domains, less Euclidean domains. We need $d\equiv 1\mod 4$, which is not true for $3$ and $7$. In fact, for any $p=4k+3$ we have $$ \left(\frac{1+\sqrt{p}}{2}\right)^2=\frac{1}{4}(p+1+2 \sqrt{p})=k+1+\frac 12\sqrt{p}\notin O_p. $$ so the question about $O_3$ and $O_7$ is settled. If you take $p=5$ or $p=13$, then the following works:

Let $p=5$. Your lattice consists of the points $(a/2,b/2)$ in $\Bbb{R}^2$ with $a,b$ integer and $a\equiv b\mod 2$. Then you can show that for any point $(x,y)$ in $\Bbb{R}^2$, (hence the same holds for points in $\Bbb{Q}^2$), you can find a point in the lattice with $|x-a/2|\le 1/2$ and $|y-b/2|\le 1/4$ (Draw the picture).

Now, if you are working in $O_5$, you take the multiplicative function $$ N((a+b\sqrt{5})/2):=|a^2-5b^2|/4, $$ defined in all $\Bbb{Q}[\sqrt{5}]$. If you divide $(a+b\sqrt{5})/2$ by $(c+d\sqrt{5})/2$, then you find $$ \frac{a+b\sqrt{5}}{c+d\sqrt{5}}=\frac{ac+5bd}{c^2-5d^2}+\frac{bc-ad}{c^2-5d^2}\sqrt{5}=x+y\sqrt{5} $$ in $\Bbb{Q}[\sqrt{5}]$. Now you take $(e/2,f/2)$ in your lattice with $|x-e/2|\le 1/2$ and $|y-f/2|\le 1/4$.

Then $$ N(x-e/2+(y-f/2)\sqrt{5})=|(x-e/2)^2-5(y-f/2)^2|\le (x-e/2)^2+5(y-f/2)^2 $$ $$ \le 1/4+5/16<1. $$

Hence, you can write $$ (a+b\sqrt{5})/2=((c+d\sqrt{5})/2)((e+f\sqrt{5})/2)+r, $$ with $r=(a+b\sqrt{5})/2-((c+d\sqrt{5})/2)((e+f\sqrt{5})/2)$ and $$ N(r)=N\left(\frac{(a+b\sqrt{5})/2}{(c+d\sqrt{5})/2}-(e+f\sqrt{5})/2\right)N((c+d\sqrt{5})/2). $$

But $$ N\left(\frac{(a+b\sqrt{5})/2}{(c+d\sqrt{5})/2}-(e+f\sqrt{5})/2\right)= N(x-e/2+(y-f/2)\sqrt{5})<1, $$ hence $N(r)<N((c+d\sqrt{5})/2)$. This shows that $O_5$ is an Euclidean domain with Euclidean function $N$.

For $O_{13}$ the lattice is exactly the same, the Euclidean function is $$ N((a+b\sqrt{13})/2):=|a^2-13b^2|/4, $$ and all works exactly the same way, once you have established that for any point $(x,y)$ in $\Bbb{R}^2$, (hence the same holds for points in $\Bbb{Q}^2$), you can find a point $(e/2,f/2)$ in the lattice with $$ N(x-e/2+(y-f/2)\sqrt{13})=|(x-e/2)^2-13(y-f/2)^2|<1. $$ For this it suffices to show that the translations of the elipse $x^2+13y^2<1$ to every point of the lattice cover all of $\Bbb{R}^2$. For this you need to know that $\frac{\sqrt{3}}{2\sqrt{13}}+\frac{1}{\sqrt{13}}>\frac 12.$

${\bf{Edit:}}$ I'just noticed that $O_3$ and $O_7$ are not closed ounder multiplication, so I rewrote it for $O_5$ and $O_{13}$.