11
$\begingroup$

Let $R_d$ be the ring defined as $R_d=\left \{ x+y\omega : x,y\in \mathbb{Z} \right\}$, where $$\omega = \begin{cases} \sqrt{d}, & \text{if } \quad d \not \equiv 1\mod 4 \\ \frac{1+\sqrt{d}}{2}, & \text{if } \quad d\equiv 1\mod 4. \end{cases}$$

It has been proven that $R_d$ is Euclidean for several positive values of $d$.

Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?

Thank you.

$\endgroup$
12
$\begingroup$

Define the norm on $\mathbb Z[\sqrt 3]$ to be $N(a + b \sqrt 3) = \vert a^2 - 3 b^2 \vert$.

Let $\alpha, \beta \in \mathbb Z[\sqrt 3]$ with $\beta \neq 0$.

Say $\alpha = a + b \sqrt 3$ and $\beta = c + d \sqrt 3$.

Notice that \begin{align*} \frac\alpha\beta &= \frac{a + b \sqrt 3}{c + d \sqrt 3} \cdot \frac{c - d \sqrt 3}{c - d \sqrt 3} \\ &= \frac{ac - 3bd}{c^2 - 3d^2} + \frac{-ad + bc}{c^2 - 3d^2} \sqrt 3 \\ &= r + s\sqrt 3 \end{align*}

where $r = \displaystyle \frac{ac - 3bd}{c^2 - 3d^2}$ and $s = \displaystyle \frac{-ad + bc}{c^2 - 3d^2}$.

Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $\vert r - p \vert \leq 1/2$ and $\vert s - q \vert \leq 1/2$.

We want to show that $\alpha = (p + q\sqrt 3) \beta + \gamma$ for some $\gamma \in \mathbb Z[\sqrt 3]$ such that either $\gamma = 0$ or $N(\gamma) < N(\beta)$. (We'll show the latter holds always.)

Define $\theta := (r - p) + (s - q)\sqrt 3$ and define $\gamma = \beta \cdot \theta \in \mathbb Z[\sqrt 3]$ and observe that \begin{align*} \gamma &= \beta \cdot \theta\\ &= \beta ( (r - p) + (s - q)\sqrt 3)\\ &= \beta (r + s\sqrt 3) - \beta(p + q\sqrt 3) \\ &= \beta \cdot\frac\alpha\beta - \beta (p + q\sqrt 3) \\ &= \alpha - \beta (p + q\sqrt 3) \end{align*}

Hence we have $\alpha = \beta(p + q\sqrt 3) + \gamma$.

Finally notice that \begin{align*} N(\gamma) &= N(\beta \cdot \theta) \\ &= N(\beta) \cdot N(\theta) \\ &= N(\beta) \cdot \vert (r - p)^2 - 3 (s - q)^2 \vert \\ &\leq N(\beta) \cdot \max\{ (r - p)^2, 3(s - q)^2\} \\ & \leq\frac34 N(\beta)\\ &< N(\beta) \end{align*}

The key here was that $\vert (r - p)^2 - 3 (s - q)^2 \vert \leq \max\{ (r - p)^2, 3(s - q)^2\}$ since $(r - p)^2, 3(s - q)^2 \geq 0$ and then we use that $(r - p)^2 \leq 1/4$ and $3(s - q)^2 \leq 3/4$.

$\endgroup$
  • 1
    $\begingroup$ When you're proving that $N(\gamma) < N(\beta)$, you say that $N(\theta) = |(p-r)^2 - 3(q-s)^2|$. However, $N$ was only defined for elements in $\Bbb{Z}[\sqrt{3}]$. You showed that $(\beta)(\theta) \in \Bbb{Z}[\sqrt{3}]$. But how does this imply that $\theta \in \Bbb{Z}[\sqrt{3}]$? $\endgroup$ – Artus Mar 12 '15 at 14:05
  • 2
    $\begingroup$ That's a very good question! We can just think of the norm $N$ on $\mathbb Z[\sqrt 3]$ as being the norm on $\mathbb Q[\sqrt 3]$, restricted to $\mathbb Z[\sqrt 3]$. $\endgroup$ – Robert Cardona Mar 12 '15 at 15:13
  • $\begingroup$ why is |p - r |< 1/2 and |q- s| < 1/2 $\endgroup$ – user10024395 Apr 23 '15 at 9:33
  • $\begingroup$ @user136266, because we define $p, q$ to be the closest integers to $r, s$, respectively. Say $p = \frac14$, the closest integer is $0$ and the difference is less than $\frac12$, if $p = \frac34$, then the closest integer is $1$. If $p = \frac12$, then pick either $0$ or $1$ and the claim holds (less than or equal to). $\endgroup$ – Robert Cardona Apr 23 '15 at 14:21
  • $\begingroup$ @RobertCardona how is 1/4 and 1/2 integers? these are not integers. i think you meant r but even so why must r be only these 3 possibilities?If i have a, c = 1 and b,d = 0. Then r = 0 and so u can't find an integer p such that p-r < 1/2. $\endgroup$ – user10024395 Apr 24 '15 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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