Euclidean domain How to prove that $\mathcal{O}_{\sqrt[3]{3}}$ is an euclidean domain? I heard that one should prove the following but why it is enough?
For any $ a,b,c\in\mathbb{R}$, prove that there are $ x,y,z\in\mathbb{R}$ such that $ x-a,y-b,z-c\in\mathbb{Z}$ and that
$$-1\leq x^3+3y^3+9z^3-9xyz\leq 1.$$
 A: As Alex mentioned, the proof that you have in mind amounts to showing that your cubic field is Euclidean with respect to the norm. In fact it is known that there are only three pure cubic fields $\rm\ Q(\sqrt[3] m)\ $ that are norm Euclidean, viz. $\rm\ \mathbb Q(\sqrt[3] 2),\ \mathbb Q(\sqrt[3] 3),\ \mathbb Q(\sqrt[3] {10})\:.\ $ You can find the proofs in Cioffari: The Euclidean Condition in Pure Cubic and Complex Quartic Fields.
A: Note that the ring of integers of $\mathbb{Q}(\sqrt[3]{3})$ is $\mathbb{Z}[\sqrt[3]{3}]$ with basis $1,\sqrt[3]{3},\sqrt[3]{9}$ over $\mathbb{Z}$. You want to show that the norm, defined by
\begin{eqnarray*}
N(a+b\sqrt[3]{3} + c\sqrt[3]{9}) & = & (a+b\sqrt[3]{3} + c\sqrt[3]{9})(a+\zeta_3b\sqrt[3]{3} + \zeta_3^2c\sqrt[3]{9})(a+\zeta_3^2b\sqrt[3]{3} + \zeta_3c\sqrt[3]{9})\\
& = & a^3 + 3b^3 + 9c^3 - 9abc
\end{eqnarray*}
gives in fact a Euclidean norm upon taking absolute values, where $\zeta_3$ is a fixed primitive cube root of unity. Now, the first (and maybe the main) thing you might wonder about is where I pulled this norm out. For that you have to know a little bit of Galois theory. The basic idea is that from the perspective of $\mathbb{Q}$, the elements $\sqrt[3]{3}$ and $\zeta_3\sqrt[3]{3}$ are indistinguishable: they are both just some roots of the polynomial $x^3-3$ and can be thought of as mirror images of each other. Essentially, the factors that I am multiplying are like "mirror images" of my given element of $\mathbb{Z}[\sqrt[3]{3}]$.
Now, the strategy is exactly the same as in the case of some quadratic rings like $\mathbb{Z}[\sqrt{2}]$: given $\alpha = u+v\sqrt[3]{3} + w\sqrt[3]{9}$ and
$\beta = f+g\sqrt[3]{3} + h\sqrt[3]{9}\in \mathbb{Z}[\sqrt[3]{3}]$, you want to show that there exist $p$ and $r$ in that ring such that
\begin{eqnarray*}\alpha = p\beta + r,
\end{eqnarray*}
where either $r=0$ or $N(r)<N(\beta)$. Dividing by $\beta$, this becomes equivalent to showing that given any $\alpha/\beta=a+b\sqrt[3]{3} + c\sqrt[3]{9}$ in the field of fraction $\mathbb{Q}(\sqrt[3]{3})$ of $\mathbb{Z}[\sqrt[3]{3}]$, there exists $r/\beta\in \mathbb{Q}(\sqrt[3]{3})$ such that the difference is in the ring of integers itself (since the difference is $p$) and $|N(r/\beta)|<1$. That is the statement you posted.
