# Find an integral basis of $\mathbb{Q}(\alpha)$ where $\alpha^3-\alpha-4=0$

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.

• You may find something useful starting on page 28 of ucl.ac.uk/~ucahmki/ant/3704notes.pdf – Gerry Myerson Feb 7 '13 at 22:38
• I think you should then examine eight numbers: $\Sigma _{i=0}^2 a_i \alpha ^i /2$, where $a_i$ are either $1$ or $0$. See if there are any algebraic integers among them; if so, replace the basis with one with a smaller discriminant, and do this again. If none of them are, then $\frac{1}{2}Z[\alpha]$ is the ring of integers. – awllower Feb 8 '13 at 6:15
• – Watson Jul 23 '16 at 15:34

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.
• 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. – PerelMan Sep 29 '19 at 4:36
• Why do you replace $α²$ by $(α+α²)/2$? – PerelMan Sep 29 '19 at 4:40