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 $\mathbb{Z}[\sqrt[3]{2}]$?
Units in $\mathbb{Z}[\sqrt[3]{2}]$ : $\pm(1+\sqrt[3]{2}+(\sqrt[3]{2})^2)^n$?
3
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ring-theory
principal-ideal-domains
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3This is an example given in http://planetmath.org/encyclopedia/UnitsOfRealCubicFieldsWithExactlyOneRealEmbedding.html – 2012-11-19
1 Answers
1
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.