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prove that $\{1,\sqrt[3]{3},(\sqrt[3]{3})^2 \}$ is integral basis for $\Bbb{Q}(\sqrt{3^3})$

my aim is to show that $\Bbb{Q}_K=\Bbb{Z}+\sqrt[3]{3}\Bbb{Z}+(\sqrt[3]{3})^2\Bbb{Z}$

$\supseteq$ : it is clear

$\subseteq$ : let $\alpha \in \Bbb{Q}_K$ i want to show $\alpha=z_1+z_2\sqrt[3] {3}+z_3(\sqrt[3]{3})^2$ and $z_1,z_2,z_3 \in \Bbb{Z}$

is there any short way to show this ? Thanks a lot.

Seirios
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izaag
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    IMVHO your notation is a bit funny/inconsistent. If I guess that you want to look at the ring of integers $\mathcal{O}_K$ of the field $K=\Bbb{Q}(\root3\of3)$? A good start is usually to look at the traces as integral elements have integral traces. If you calculate $\mathrm{tr}((\root3\of3)^i\alpha)$ for $i=0,1,2$ you will learn that $3z_i, i=1,2,3,$ are all in $\Bbb{Z}$. As $\mathcal{O}_K$ is a ring that (using the other inclusion) leaves you with the task of checking out the 27 possible combinations $z_i={0,1/3,2/3}, i=1,2,3.$ – Jyrki Lahtonen May 25 '14 at 13:15
  • @izaag: Related: http://math.stackexchange.com/questions/99913/easy-way-to-show-that-mathbbz-sqrt32-is-the-ring-of-integers-of-mat – Watson Aug 26 '16 at 08:25

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