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Let $K=\mathbb{Q}(\alpha)$ where $\alpha$ has minimal polynomial $X^3-X-4$. Find an integral basis for $K$.

I have calculated the discriminant of the minimal polynomial is $-2^2 \times 107$, so the ring of algebraic integers is contained in $\frac{1}{2}\mathbb{Z}[\alpha]$. But I don't know how to then find an integral basis.

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As in the link of Gerry, one is supposed to check if there are any algebraic integers among the seven :$\Sigma_{i=0}^2 a_i \alpha^i/2$, where $a_i$ are either $0$ or $1$, and not all of them are $0$.
Now, after some computations(some minutes maybe), one finds that the only one among them which is an algebraic integer is: $(\alpha+\alpha²)/2$, satisfying the irreducible polynomial: $x^3-x²-3x-2$.
Replace $\alpha²$ by $(\alpha+\alpha²)/2$ in the basis, one then finds that the discriminant becomes $-107$ by the transition formula. Hence this is an integral basis, as required.
P.S. Since this is subject to a certain amount of calculations, of which I think quite tediously, it is quite expectable that some errors penetrated in the above arguments. Thus, if there are some mistakes, please tell me. Thanks in advance.

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    $\begingroup$ Why did you choose the prime $p=2$ to apply 111 Theorem in the link? $X^3-X-4$ doesn't satisfy Eisenstein criterion at 2. $\endgroup$ – PerelMan Sep 29 '19 at 4:36
  • $\begingroup$ Why do you replace $α²$ by $(α+α²)/2$? $\endgroup$ – PerelMan Sep 29 '19 at 4:40
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    $\begingroup$ Indeed the theorem mentioned does not apply. Thanks for pointing out the error. But I think perhaps the approach is valid: take a basis, compute the discriminant, and then for the square factors of the discriminant, divide them out and see if there are algebraic integers. Repeat this process until we reach an integral basis. $\endgroup$ – awllower Oct 5 '19 at 3:03

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