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Let $\alpha = \sqrt[3]{2}$, $K = \mathbb{Q}(\alpha)$. We want to show that $O_K = \mathbb{Z}[\alpha]$. That is to say that if $a + b\alpha + c\alpha^2$ is in $O_K$ then $a,b,c$ are in $\mathbb{Z}$.

Since disc$(K) = -2^23^3$ we only need to show that $2$ and $3$ don't appear in the denominator of $a,b$ and $c$ (or equivalently that $2$ and $3$ don't divide $(O_K:\mathbb{Z}[\alpha])$).

In Keith Conrad's note : he proves this by showing that the ring of integers of $\mathbb{Q}_2(\alpha)$ (resp. $\mathbb{Q}_3(\alpha)$) is $\mathbb{Z}_2[\alpha]$ (resp. $\mathbb{Z}_3[\alpha])$).

Although I should probably know why this implies the result I can't seem to be able to figure it out. Can someone explain (the most generality would be the best) why it true ? (a reference is also fine)

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It's just about the embedding of fields $\mathbb Q(\alpha)\to\mathbb Q_p(\alpha)$ which certainly sends integers to integers (for $p=2$ and $3$ the polynomial $x^3-2$ is irreducible in $\mathbb Q_p[x]$; if not, some care is needed in the definition of $\mathbb Q_p(\alpha)$, i.e. all irreducible factors have to be considered). So $a,b,c$ are in $\mathbb Z_p$ for $p=2,3$, that's all. – user8268 Oct 12 '13 at 20:06
I'm a little confused with what he did. The method he used to show that the ring of integers of the two $p$-adic fields is what he wants, works equally well to just directly show that $\mathcal{O}_K=\mathbb{Z}[\sqrt[3]{2}]$. Indeed, $K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{2}+1)$. Since the min polys of $\sqrt[3]{2}$ and $\sqrt[3]{2}+1$ are $2$-eisenstein and $3$-eisenstein respectively, and the degree of our extension is $2$, you may conclude that $2^2\mid d_K$ and $3^2\mid d_K$ respectively. Since the discriminant of $\mathbb{Z}[\alpha]$ differs from $d_K$ by a square – Alex Youcis Oct 12 '13 at 20:46
you can easily conclude that $\text{disc}(\mathbb{Z}[\alpha])=d_K$ so that $\mathcal{O}_K=\mathbb{Z}[\alpha]$. I don't see why he needed to move to completions. – Alex Youcis Oct 12 '13 at 20:47
Thanks for your comments. Alex your remark seems correct to me (and definetly simpler), maybe we are missing something. – Zorba le Grec Oct 12 '13 at 21:07

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