Abstract Algebra – Proving $\mathbb{Z}[\sqrt{6}]$ is a Euclidean Domain

Question

I'm attempting to modify the proof the $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain to prove a similar result for $\mathbb{Z}[\sqrt{6}]$. The idea is to prove that $\mathbb{Q}[\sqrt{6}]$ is Euclidean which will then give the result for $\mathbb{Z}[\sqrt{6}]$. According to Dummit and Foote's Abstract Algebra this should have norm $d(a+b\sqrt{6})=|a^2-6b^2|$. The result for $\mathbb{Z}[\sqrt{2}]$ relies on the fact that for any element $x$ in $\mathbb{Q}[\sqrt{2}]$ there is an element $x'$ in $\mathbb{Z}[\sqrt{2}]$ with $d(x-x')<1$, which is then used to show that for $x=qy+r$ we have
$$
d(r)=d(x-qy)=d\left(y\left(\frac{x}{y}-q\right)\right)=d(y)d\left(\frac{x}{y}-q\right)<d(y)
$$
But in this case, I'm having a hard time showing that we can find an element with $d(x-x')<1$ for $x\in\mathbb{Q}[\sqrt{6}], x'\in\mathbb{Z}[\sqrt{6}]$. Consider $x=a+b\sqrt{6}\in\mathbb{Q}[\sqrt{6}]$. The furthest this element can be away from any $x'=a'+b'\sqrt{6}\in\mathbb{Z}[\sqrt{6}]$ is when $|a-a'|=\frac{1}{2}=|b-b'| \Rightarrow$
$$
d(x-x')=\left|\left(\frac{1}{2}\right)^2-6\left(\frac{1}{2}\right)^2\right|=\frac{5}{4}>1
$$
What can I do to get around this issue?

Answer 1

Alright, this is from the viewpoint of what is called "covering radius" for positive quadratic forms. You have a quadratic form $f(x,y) = x^2 - 6 y^2.$ Now, with any real point $(a,b)$ in the plane, you want to find an integer point $(m,n)$ such that $$|f(a-m, b-n)| < 1. $$

Now, what is the set, centered around $(0,0)$ with $| x^2 - 6 y^2| < 1?$ It is a strange starfish shape with four arms, all along lines of slope $\pm \frac{1}{\sqrt 6}.$ The starfish "centered at" some integer point $(m,n)$ is $$ |(x -m)^2 - 6 (y - n)^2 | < 1. $$

All that is needed to show your inequality is that a finite set of such starfish covers the ordinary unit square with corners at $(0,0),(1,0),(1,1),(0,1).$

Well, the square $0 \leq x \leq 1, \; \; 0 \leq y \leq 1 $ is covered by eight starfish centered at $$ (0,0),(0,1), (1,0),(1,1), (-1,0),(-1,1),(2,0),(2,1). $$ The four starfish centered at the corners of the square itself cover all except a funny curvy diamond shape around $(\frac{1}{2}, \frac{1}{2} ),$ the vertices of the diamond being $$ \left(\frac{1}{2}, \frac{1}{2} \sqrt{\frac{5}{6}} \right), \; \left(\frac{1}{2}, 1 - \frac{1}{2} \sqrt{\frac{5}{6}} \right), \; \left(\frac{1}{\sqrt2}, \frac{1}{2} \right), \; \left( 1 -\frac{1}{\sqrt2}, \frac{1}{2} \right). $$ Next, out of the four remaining squares, the (open) starfish centered at $(-1,0)$ covers the entire closed rectangle $$ \frac{1}{5} \leq x \leq \frac{1}{2}, \; \; \frac{1}{2} \leq y \leq \frac{3}{5}, $$ so you can see that the four final starfish cover the closed rectangle $$ \frac{1}{5} \leq x \leq \frac{4}{5}, \; \; \frac{2}{5} \leq y \leq \frac{3}{5}, $$ thus entirely covering the missing diamond shape.

Earlier I had a solution with $20$ points, this is easier. Still, I suggest you have a computer draw the intersections of these sets with the square I indicate. I did this with a calculator and some graph paper, but I have special eyes.

September 2016: by computer, a bit much. Maybe if I show just the first four starfish, the central diamond that is not yet covered should be more clear

Note that the four boundary curves of the starfish centered at the integer point $(m,n)$ can be parametrized by a variable $t$ as $$ \left( m + \cosh t, \; n + \frac{ \;\sinh t \;}{\sqrt 6} \right), $$ $$ \left( m - \cosh t, \; n + \frac{ \;\sinh t \;}{\sqrt 6} \right), $$ $$ \left( m + \sinh t, \; n + \frac{ \;\cosh t \;}{\sqrt 6} \right), $$ $$ \left( m + \sinh t, \; n - \frac{ \;\cosh t \;}{\sqrt 6} \right). $$

EDITTT: this was first proved by Perron in 1932, with a better method by Oppenheim in 1934. See page 11 in survey.pdf at LEMMERMEYER