So far I have only seen extensions of $\mathbb{Q}$ with "trivial" integral basis. Meaning that the integral basis is the most natural one e.g. the integral basis for $\mathbb{Q}(\sqrt[3]{2})$ is just $\{1, \sqrt[3]{2}, \sqrt[3]{2}^2 \}$ (I know this is not easily proven). So is there a extension which integral basis is really hard to intuitively see?
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Do you mean that the extension is not 'mongenic', meaning that $\mathcal{O}_K$ is not of the form $\mathbb{Z}[\alpha]$ for some $\alpha\in\mathcal{O}_K$? – Alex Youcis Nov 20 '14 at 08:37
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@AlexYoucis Well I think that's what I'm looking for. Never heard of that term though. – NPHA Nov 20 '14 at 08:45
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Not sure if it's exactly the same but I'm looking for a extensions which integral basis is really hard to find (I Don't really 'care' if its easily proven to be the integral basis or not). – NPHA Nov 20 '14 at 08:51
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See also http://math.stackexchange.com/questions/727985/how-do-i-find-an-integral-basis-given-a-basis-consisting-of-algebraic-integers. – lhf Nov 20 '14 at 11:19
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Considering pure cubic number fields shows already that we do not always have such a nice integral basis. Suppose that $K=\mathbb{Q}(\sqrt[3]{d})$ with $d=ab^2$ cubefree with coprime squarefree integers $a$ and $b$.
Theorem If $a^2\not\equiv b^2(9)$, then $K$ is monogenic if and only if the Diophantine equation $ax^3+by^3=1$ has integer solutions. If $a^2\equiv b^2(9)$, then $K$ is monogenic if and only if the Diophantine equation $ax^3+by^3=9$ has integer solutions.
In particular, take any prime $p\equiv 1(9)$, which can be represented by $7x^2+3xy+9y^2$ with integers $x,y$. Then $K=\mathbb{Q}(\sqrt[3]{p})$ is not monogenic, i.e., the ring of integers $\mathcal{O}_K$ is not of the easy form $\mathbb{Z}[\alpha]$ for some $\alpha\in \mathcal{O}_K$. For example, $p=19$ can be represented with $x=y=1$ in this form, so that $K=\mathbb{Q}(\sqrt[3]{19})$ is not monogenic. So can you "intuively see" the integral basis here ?
P.S. As to the question, why this is hard in general, see the discussion here.
Dietrich Burde
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