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I'm trying to follow a text (Lang's Algebraic Number Theory) in which it fully determines an integral basis for quadratic fields (also seen here). Is there any easy or analogous way to determine one for cubic fields of the form $\mathbb Q(\sqrt[3]{a})$, where $a\in\mathbb Z$?

Can one also conclude (or stipulate various restrictions so) that $\mathcal O_K$ is a PID?

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2… – user27126 Dec 4 '12 at 3:09
$\mathcal{O}_K$ is not in general a PID when $K$ is quadratic, so there should be less hope for when $K$ is cubic. – Rankeya Dec 4 '12 at 4:06
If you know about the ideal class group, then $\mathcal{O}_K$ is a PID if and only if its class number (i.e. the order of the ideal class group) is 1. This wikipedia article has a list of some quadratic and cubic fields with class number 1:… – Rankeya Dec 4 '12 at 4:09
up vote 0 down vote accepted

You can find an "elementary" proof in example 4.3.6 of Murty and Esmonde, Problems in Algebraic Number Theory, here.

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Integral basis for ${\bf Q}(\root3\of a)$ is given in Theorem 7.3.2 of Alaca and Williams, Introductory Algebraic Number Theory:

Let $m$ be a cubefree integer. Set $m=hk^2$, where $h$ is squarefree, so that $k$ is squarefree and $(h,k)=1$. Set $\theta=m^{1/3}$ and $K={\bf Q}(\theta)$. Then an integral basis for $K$ is $$\eqalign{&\{{1,\theta,\theta^2/k\}},{\rm\ if\ }m^2\not\equiv1\pmod9,\cr&\{{1,\theta,(k^2\pm k^2\theta+\theta^2)/3k\}},{\rm\ if\ }m\equiv\pm1\pmod9.\cr}$$

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I do not have access to the text, but as far as I can gleam online, the proof to that theorem is left to the reader (!). – Dustin Tran Dec 4 '12 at 6:36
Yes, that's right. Considering how complex the statement is, you might want to reconsider asking for an easy way to determine an integral basis for these fields. – Gerry Myerson Dec 4 '12 at 6:43

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