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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 $$-1\leq x^3+3y^3+9z^3-9xyz\leq 1.$$

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What is your background? E.g. do you know how you would prove that $\mathbb{Z}(\sqrt{2})$ is a Euclidean domain? – Alex B. Nov 23 '10 at 15:34
I have read the definition and managed to prove some ring to be Euclidean, like $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[i]$. I have never thought why $\mathbb{Z}(\sqrt{2})$ is Euclidean. – Jaska Nov 23 '10 at 15:46
Sorry, I did mean square brackets. Just wanted to know, where to start with my answer. – Alex B. Nov 23 '10 at 15:54
up vote 2 down vote accepted

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.

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OK, I understood. Now, how one proves the existence on such $x,y,z$? – Jaska Nov 23 '10 at 16:35

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.

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It looks like I got the idea of the proof. However, I'm unable to perceive why the problem looks so hard that it requires a computer. – Jaska Nov 23 '10 at 17:10
Jaska, just to give you a vague idea of just how hard the problem is: we still don't know whether infinitely many real quadratic rings of integers are Euclidean domains. Computation and certain heuristics suggest that about 3/4 of them are PIDs and then the generalised Riemann hypothesis would imply that they are Euclidean (although not Norm-Euclidean). But all this is wide open! – Alex B. Nov 23 '10 at 17:38
@Jaska: Generally such results require computational assistance. It may be that this case is amenable to hand computation but I do not recall off the top of my head if this is so. Try googling Cioffari's paper to look for citations mentioning simplifications. – Bill Dubuque Nov 23 '10 at 17:45

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