Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.