# How do I find an integral basis, given a basis consisting of algebraic integers?

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?

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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.
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 8:24