# Show $\mathbb{Z}[\sqrt{6}]$ is a Euclidean domain

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?

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Note that if there is $x'$ in your lattice such that $|a-a'|=\frac12 = |b-b'|$, then there is also another point $x''$ in the lattice such that $|b-b''|=\frac12$ and $|a-a''| = \frac32$. And then $d(x-x'')$ is...

As you see, to get a point in the lattice close to your given $x$, the trick is not to minimise $|a-a'|^2$ and $|b-b'|^2$ separately, but to minimise a suitable weighted difference. Now if you have found a lattice point such that $6|b-b'|^2 <1$, then what you did works just fine, right? If not, then as you have noted, you can find a point such that $1\leq 6|b-b'|^2 \leq \frac64=\frac 32$. Then how do you choose $a$? The first paragraph should give you an idea.

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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.

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

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