The Problem: Let $F=\mathbb{Q}(\sqrt{-3})$ be a quadratic field with associated integer ring $\mathcal{O}=\mathbb{Z}[\omega]=\{a+b\omega\mid a, b\in\mathbb{Z}\}$ where $\omega=\frac{1+\sqrt{-3}}{2}$, and field norm $N(x+y\sqrt{-3})=x^2+3y^2$ ($x, y\in\mathbb{Q}$). Show that $\mathcal{O}$ is a Euclidean Domain.
[Hint: Prove that every element of $F$ differs from an element in $\mathcal{O}$ by an element whose norm is at most $(1+|D|^2)/(16|D|)$, which is less than $1$ for the value of $-3$].
Source: Abstract Algebra $\mathit{3^{rd}}$ edition by Dummit and Foote.
My Attempt: I failed to make use of the hint, so I tried the following:
Suppose $\alpha, \beta\in\mathcal{O}$ and $\beta\neq0$. Suppose $\alpha=a+b\omega$ and $\beta=c+d\omega$, where $a, b, c, d\in\mathbb{Z}$. Then $\frac{\alpha}{\beta}=r+s\omega$ for some $r, s\in\mathbb{Q}$. Choose $p, q\in\mathbb{Z}$ such that $|r-p|, |s-q|\leq\frac{1}{2}$. Let $\theta=(r-p)+(s-q)\omega$, then let $\gamma=\beta\theta$. Then $\gamma\in\mathcal{O}$ and $\alpha=(p+q\omega)\beta+\gamma$. Note $N(\theta)=(r-p)^2+3(s-q)^2\leq\frac{1}{4}+\frac{3}{4}=1$; so $N(\gamma)=N(\theta)N(\beta)\mathbb{\leq} N(\beta)$. Screeching Halt.
Obviously my approach failed because the last inequality is not strict, as required by the definition of Euclidean Domain. Any help would be greatly appreciated.
In Dummit and Foote, the case when $D=-3$ is a subcase of $D=-3,-7,-11.$ The following proof is supposed to work for all these values of $D.$
Let $D=-3,-7,-11.$ Then $D\equiv 1\pmod 4,$ so the ring of integers of the quadratic field $\Bbb{Q}(\sqrt{D})$ is $\Bbb{Z}\left[{1+\sqrt{D}\over2}\right].$ Let $\alpha = a+b {1+\sqrt{D}\over2}$ for some $a, b \in \Bbb{Z}.$ As $\alpha$ can be written as $\alpha = \left(a+\frac{b}{2}\right)+\frac{b}{2}\sqrt{D},$ we see that $\Bbb{Z}\left[{1+\sqrt{D}\over2}\right] \subset \Bbb{Q}(\sqrt{D}).$
Now, take any nonzero $\beta = c + d {1+\sqrt{D}\over2} \in \Bbb{Z} \left[{1 + \sqrt{D} \over 2}\right].$ Since $\Bbb{Q}(\sqrt{D})$ is a field, in $\Bbb{Q}(\sqrt{D})$ we have \begin{align*} \alpha\beta^{-1} &= \frac{\left(a+\frac{b}{2}\right)+\frac{b}{2}\sqrt{D}}{\left(c+\frac{d}{2}\right)+\frac{d}{2}\sqrt{D}}\\ &= \frac{\left(\left(a+\frac{b}{2}\right)+\frac{b}{2}\sqrt{D}\right)\left(\left(c+\frac{d}{2}\right)-\frac{d}{2}\sqrt{D}\right)}{\left(c+\frac{d}{2}\right)^2-\left(\frac{d}{2}\sqrt{D}\right)^2}\\ &= r + s\sqrt{D}, \end{align*} where $$r = \frac{2ac+bc+ad+2bd\frac{1-D}{4}}{2\left(c^2 + cd + d^2 \frac{1-D}{4}\right)} \text{ and } s = \frac{bc-ad}{2\left(c^2+cd+\frac{1-D}{4}d^2\right)}.$$
If we can find some $\theta \in \Bbb{Z}\left[{1+\sqrt{D}\over2}\right]$ such that $\alpha = \theta\beta+\gamma$ where $N(\gamma) < N(\beta),$ then we’re done. Suppose there is such an element $\theta.$ Write $\theta = u+v\sqrt{D} \in \Bbb{Q}(\sqrt{D}).$ Then $$\gamma = \alpha - \theta\beta = \left(\alpha\beta^{-1}-\theta\right) \beta.$$ Since the norm $N$ is multiplicative, $$N(\gamma)=N\left(\alpha\beta^{-1}-\theta\right) N(\beta).$$ Our goal now is to show that $N\left(\alpha\beta^{-1}-\theta\right) < 1$.
Note that since $D$ is negative, $$\alpha\beta^{-1} = r+s\sqrt{D} = r+s\sqrt{|D|}i \text{ and } \theta = u+v\sqrt{D}=u+v\sqrt{|D|}i,$$ two points in the complex plane. Therefore, \begin{align*} N(\alpha\beta^{-1}-\theta) &= N\left((r-u)+(s-v)\sqrt{|D|}i\right)\\ &= (r-u)^2+(s-v)^2|D|. \end{align*}
Take any two integers $x, y \in \Bbb{Z}$. Denoting ${1+\sqrt{D}\over2}=ω$, there are elements $x+yω$, $x+(y+1)ω$, $(x+1)+yω$, $(x+1)+(y+1)ω$ in the ring of integers $\Bbb{Z}\left[{1+\sqrt{D}\over2}\right]=\Bbb{Z}[ω]$. These elements form a “rectangle” of minimal square (with sides of minimal possible lengths $1$ and $ω$) in the plane $\Bbb{Z}[ω]$, i.e. there cannot be any “rectangular” inside it in the plane $\Bbb{Z}[ω]$. Thus, $\Bbb{Z}[ω]$ can be viewed as a lattice of such rectangles of minimal square.
As before, \begin{align*} x + y \omega &= \left(x+\frac{y}{2}\right)+\frac{y}{2}\sqrt{|D|}i,\\x+(y+1) \omega &= \left(x+\frac{y}{2}+\frac{1}{2}\right)+\left(\frac{y}{2}\sqrt{|D|}+\frac{1}{2}\sqrt{|D|}\right)i\\ (x+1) + y \omega &= \left(x+\frac{y}{2}+1\right)+\frac{y}{2}\sqrt{|D|}i,\\(x+1) + (y+1) \omega &= \left(x+\frac{y}{2}+1+\frac{1}{2}\right)+\left(\frac{y}{2}\sqrt{|D|}+\frac{1}{2}\sqrt{|D|}\right)i. \end{align*}
Plotting these points in $\Bbb{C}$ produces the following parallelogram, where the measurements in the picture are calculated from the coordinates above:
So, a rectangle of minimal square in $\Bbb{Z}[\omega]$ is a parallelogram in $\Bbb{C}.$ Hence, since we view the plane $\Bbb{Z}[\omega]$ as a lattice of minimal-square rectangles, the complex plane can inherently be viewed as a lattice of such parallelograms. Note that, by construction, this lattice covering the entire $\Bbb{C}$ is comprised of the parallelograms of minimal possible square such that their vertices belong to $\Bbb{Z}[\omega].$
Then, our point $r+s\sqrt{|D|}i$ with $r,s∈\Bbb{Q}$ lies in some parallelogram in $\Bbb{C}$. Furthermore, if we connect the vertices $x+(y+1)ω$ and $(x+1)+yω$ (as in the picture above), we get that our point $r+s\sqrt{|D|}i$ lies in some isosceles triangle in $\Bbb{C}$ (by the way, when $D=-3$ this triangle is equilateral). Then the maximal possible distance from $r+s\sqrt{|D|}i$ to the nearest vertex (which is also a point in the plane $\Bbb{Z}[ω]$) is achieved when $r+s\sqrt{|D|}i$ is equidistant from all the vertices of the isosceles triangle it lies in. Let’s denote this maximal possible distance as $l$. Then, using the picture above and the Pythagorean Theorem, we can get the following relation: $\sqrt{l^2-\left(\frac{1}{2}\right)^2}+l=\frac{\sqrt{|D|}}{2} ⇔ \sqrt{l^2-\frac{1}{4}}=\frac{\sqrt{|D|}}{2}-l ⇔ l^2-\frac{1}{4}=\left(\frac{\sqrt{|D|}}{2}-l\right)^2 ⇔ l^2-\frac{1}{4}=\frac{|D|}{4}-l\sqrt{|D|}+l^2 ⇔ l\sqrt{|D|}=\frac{1}{4}+\frac{|D|}{4} ⇔ l\sqrt{|D|}=\frac{1+|D|}{4} ⇔ l=\frac{1+|D|}{4\sqrt{|D|}}$
Let $r+s\sqrt{|D|}i$ be located this way inside the isosceles triangle (the worst possible case). Then, if $u+v\sqrt{D}$ (see 2.) is the nearest vertex, then, as was shown in 2., the norm $N\left(\left(r+s\sqrt{|D|}i\right)-\left(u+v\sqrt{|D|}i\right)\right)=l^2=\left(\frac{1+|D|}{4\sqrt{|D|}}\right)^2=\frac{(1+|D|)^2}{16|D|}$.
If $D=-3$, then $\frac{(1+|D|)^2}{16|D|}=\frac{(1+3)^2}{16\cdot3}=\frac{16}{48}<1$
If $D=-7$, then $\frac{(1+|D|)^2}{16|D|}=\frac{(1+7)^2}{16\cdot7}=\frac{64}{112}<1$
If $D=-11$, then $\frac{(1+|D|)^2}{16|D|}=\frac{(1+11)^2}{16\cdot11}=\frac{144}{176}<1$
So, for $D=-3,-7,-11$ we have (see 2.) that $N(γ)=N\left(\left(r+s\sqrt{D}\right)-\left(u+v\sqrt{D}\right)\right)N(β)<1\cdot{N(β)}=N(β)$, i.e., $N(γ)<N(β)$. Hence we can write $α=\left(u+v\sqrt{D}\right)β+γ$ with $N(γ)<N(β)$, where $u+v\sqrt{D}=\left(p+\frac{q}{2}\right)+\frac{q}{2}\sqrt{D}=p+q{1+\sqrt{D}\over2}∈\Bbb{Z}\left[{1+\sqrt{D}\over2}\right]$ (see 2.) so that $γ=α-\left(u+v\sqrt{D}\right)β∈\Bbb{Z}\left[{1+\sqrt{D}\over2}\right]$. Therefore, for any $α∈\Bbb{Z}\left[{1+\sqrt{D}\over2}\right]$ and any nonzero $β∈\Bbb{Z}\left[{1+\sqrt{D}\over2}\right]$ there are $μ=p+q{1+\sqrt{D}\over2}∈\Bbb{Z}\left[{1+\sqrt{D}\over2}\right]$ and $γ∈\left[{1+\sqrt{D}\over2}\right]$ such that $α=μβ+γ$ with $N(γ)<N(β)$.
Thus, $\Bbb{Z}\left[{1+\sqrt{D}\over2}\right]$ is an Euclidean Domain for $D=-3,-7,-11$.