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Let $F = \mathbb{Q}(\sqrt{-3})$ be a quadratic field with associated integer ring $$\mathcal{O} = \mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right] = \left\{a+b\frac{1+\sqrt{-3}}{2} : a,b \in \mathbb{Z} \right\}$$ and field norm $N$ defined by $N(a+b\sqrt{-3}) = a^2 + 3b^2$. Prove that $\mathcal{O}$ is a Euclidean Domain with respect to $N$. Prove that every element of $F$ differs from an element of $\mathcal{O}$ by an element whose norm is at most $1/3$ (this is what I'm stuck on).

Following almost the exact steps to this proof (Proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain) I write $a = \beta q + r$ and get that $N(r) < \left[\left(\frac{1}{2}\right)^2 + 3 \left(\frac{1}{2}\right)^2\right] *N(\beta) = 1*N(\beta)$, but I'm not sure how I'm supposed to get a bound of $1/3$ instead of $1$.

user20354139
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  • Suggestion: draw a picture of the lattice in theomplex plaine of points $a+ b\omega$ where $\omega = (1 + \sqrt{-3})/2$. Find a vertex near enough to the complex quotient. – Ethan Bolker Feb 01 '18 at 00:05
  • See the answer at https://math.stackexchange.com/questions/23086 – Watson Nov 26 '18 at 21:24

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