Proof that $\mathbb Z[\sqrt{3}]$ is a Euclidean Domain

Solution 1:

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 $3(s - q)^2 \leq 3/4$.