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I am trying to find the ring of integers of $\mathbb{Q}(\sqrt[3]{6})$. Unfortunately, the discriminant of the basis $\{1, \theta, \theta^2\}$, where $\theta = \sqrt[3]{6}$, is equal to $2^2 (-3)^5$ (so it's not square-free). How can I proceed in this case to find the ring of integers? I could take the prime $2$, and analyze the expressions of the form $(a + b\theta + c\theta^2)/2$ for binary values of $a,b,c$, but maybe there is a more direct approach.

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$Disc(x^3-6)= -27(-6)^2=3^a2^b$

Since $v_3(6^{1/3})=v_3(6)/3=1/3$ then $\Bbb{Q}_3(6^{1/3})/\Bbb{Q}_3$ is totally ramified of degree $3$ with uniformizer $6^{1/3}$

Similarly $\Bbb{Q}_2(6^{1/3})/\Bbb{Q}_2$ is totally ramified with uniformizer $6^{1/3}$

The uniformizers of the ramified completions are in $\Bbb{Z}[6^{1/3}]$ thus it is a Dedekind domain ie. $$O_{\Bbb{Q}(6^{1/3})} = \Bbb{Z}[6^{1/3}]$$ For $p\ne 2,3$ then $(p)$ is a product of distinct prime ideals of $\Bbb{Z}[6^{1/3}]$, for $p=2,3$ then $(p)=(p,6^{1/3})^3$.

The more elementary solution : check that $(p)=(p,6^{1/3})^3$ for $p=2,3$, since the unramified prime ideals are easily shown to be inversible, this implies every prime ideal is inversible, thus it is a Dedekind domain, thus it is integrally closed ($=O_K$). The point is that (in Dedekind domains) a prime ideal $P$ becomes principal $=(\pi)$ in $(O_K-P)^{-1} O_K$, the uniformizer is $\pi$, from which we obtain a discrete valuation and a $p$-adic completion.

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  • $\begingroup$ I do not understand this solution; in my class we have not talk about uniformizer. Do you know a less technical approach different that the one I mentioned? $\endgroup$ – Philomeno Dec 6 '19 at 3:03
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    $\begingroup$ In general, if the minimal polynomial of $\theta$ is Eisenstein with respect to the prime $p$, then the index of $\Bbb Z[\theta]$ in the ring of integers is coprime to $p$. If you know about extensions of the field of $p$-adic numbers this is straightforward, but maybe is less easy to prove from first principles. @Philomeno $\endgroup$ – Angina Seng Dec 6 '19 at 3:09
  • $\begingroup$ @LordSharktheUnknown I feel better already =). $\endgroup$ – Philomeno Dec 6 '19 at 3:17
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If you are interested, a complete determination of the ring of integers $R$ of a pure cubic field $\mathbf Q(\sqrt [3] m)$, where $m$ is a cube free integer written as $m=hk^2$, withe $h,k$ coprime and square free, can be found in chap.3 of D.A. Marcus' book "Number Fields".

In the general case of a number field $\mathbf Q(\alpha)$, where $\alpha$ is an algebraic integer of degree $n$, thm.13 therein states that $R$ admits an integral basis of the form {$1, f_1(\alpha)/d_1,..., f_{n-1}(\alpha))/d_{n-1}$}, where the $d_i \in \mathbf Z$ , with $d_1|d_2|...|d_{n-1}$, are uniquely determined, and the $f_i \in \mathbf Z[X]$ are monic of degree $i$. In the particular case of $\alpha=\sqrt [3] m$, everything becomes explicit. The (long) exercise 41 op. cit. shows that one can take $f_1(\alpha)/d_1=\alpha$, and $f_2(\alpha)/d_2=\alpha^2/k$ if $m\neq \pm 1$ (mod $9$), $\alpha^2\pm k^2\alpha+k^2/3k$ if $m\equiv \pm 1$ (mod $9$) .

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  • $\begingroup$ For $h,k$ coprime squarefree the same argument as in my answer shows that $\Bbb{Z}[3^{-1},(hk^2)^{1/3},(h^2k)^{1/3}]$ is a Dedekind domain, thus it remains to find the uniformizer of $\Bbb{Q}_3((hk^2)^{1/3})$ depending on $hk^2\bmod 9$. $\endgroup$ – reuns Dec 8 '19 at 10:16
  • $\begingroup$ Sure, but @Philomeno says that he's not familiar with local methods. $\endgroup$ – nguyen quang do Dec 8 '19 at 10:21

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