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The subring $\mathbb{Z}[\sqrt[3]{2}]\subset\mathbb{C}$ is a PID. I remember reading somewhere that the units in $\mathbb{Z}[\sqrt[3]{2}]$ are precisely the elements $\pm(1+\sqrt[3]{2}+(\sqrt[3]{2})^2)^n$ with $n\in\mathbb{Z}$, but I don't know how one would go about proving this. I guess if $\mathbb{Z}[\sqrt[3]{2}]$ were a euclidean ring, then one could try using the euclidean norm but I'm not sure whether this is the case or not (anyway, I don't have a euclidean norm in my hands to work with). Is there a quick way to determine the units of $\mathbb{Z}[\sqrt[3]{2}]$?

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A quick matrix representation of multiplication by $a+b\sqrt[3]2+c\sqrt[3]4$, $a,b,c\in\mathbb Z$, would have an inverse of the same form if and only if the determinant is $\pm 1$. This yields the equation: $a^3+2b^3+4c^3-6abc=\pm 1$. Not sure if that helps any. ($a^3+2b^3+4c^3-6abc$ is the norm for this ring.) – Thomas Andrews Nov 19 '12 at 20:49
This is an example given in… – Cocopuffs Nov 19 '12 at 21:05

Multiply your prospective unit by $1-\sqrt[3]2$ to show you have a unit.

The Dirichlet Unit Theorem tells you the rank of the unit group is 1. This may be more advanced than you want to use, and I would be interested in a special case argument for this case.

Then you need to work on showing that $\pm 1$ are the only units of finite order, and that you can reduce any other unit of form $a+b\sqrt[3]2+c\sqrt[3]4$ to one with a lower absolute value of $a$ until you get to $a = \pm1$ - by multiplying/dividing by the units you've already identified. After that there is some tidying up to do.

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