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A known example of a number field that has no power basis is the field $\mathbb{Q}(\theta)$, where $\theta$ is a root of the polynomial $x^3-x^2-2x-8$. The discriminant of this polynomial is $-2012 =-503 . 4$ and also is the discriminant of the basis $(1, \theta, \theta^2)$. Now it appears that the basis $(1, \theta, (\theta+\theta^2)/2)$ has discriminant $503$ and that $(\theta+\theta^2)/2)$ is an algebraic integer(having minimal polynomial $x^3-3x^2-10x-8$ ).

If I look at the polynomial $x^3-8x-6$ I see that it has discriminant $1076=269 . 4$. If it has an integral basis with elements that has non integer coefficients with respect to the basis $(1, \theta, \theta^2)$ then these coefficients have to be multiples of $1/2$, but none of such elements around $0$ is an integer. Is there an algorithm to find such elements if they exist or is there a critirion the field for such elements to exist?

Marc Bogaerts
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Such an integral basis always exists. It can be found, for instance, by starting with a basis consisting of integral elements, looking at the discrimant and 'removing' squares from the discriminant by repeatedly replacing basis elements by 'better' basis elements. See, for instance, Computing Integral Basis by Cook for a practical guide.

Note that computer algebra systems generally include algorithms to compute an integral basis. For instance, in Magma, Basis(RingOfIntegers(K), K) will produce such a basis for a field K of an appropriate type.

For the cubic case, the case $\theta = \sqrt[3]{a}$ is answered here. Computing the basis is simple enough; the proofs seems to be hard.

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Magdiragdag
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  • Essential, in the paper you mention, is theorem 3.4 which shows that the number of tests to perform is finite. The proof of this theorem is rather simple and based on the rule for discriminants by a change of basis. But part of the question remains, is there a criterion for cubic extensions as there is for quadratic (d= 1 mod 4); – Marc Bogaerts Mar 27 '14 at 08:01
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    Ok, I thought your question was about the general case (only) and not about the specific cubic case. So, just to make sure I understand the (remainder of) your question: you want to know if there is a criterion to decide if $(1,\theta,\theta^2)$ is an integral basis (for a cubic extension)? – Magdiragdag Mar 27 '14 at 08:24
  • This is exactly my question. Why has the first polynomial (an example due to Dedekind) no powerbasis and the second does? How did Dedkind fell on this example, by pure coincidence? Are there many examples of this kind, or are the rather exceptional? – Marc Bogaerts Mar 27 '14 at 10:18