Integer basis for $\mathbb{Z}_{\mathbb Q(\alpha)}$

Question

So we are studying integral field and i am stuck on this problem.

Let $f(X) = X^3 -X -12 \in \mathbb{Q}[X]$ and $\alpha \in \bar{\mathbb{Q}}$ be a zero of $f(X)$.

I already showed that $f(X)$ is irreducible but i am struggling to show what the integer basis of $\mathbb{Z}_{\mathbb Q(\alpha)}$ is. Any help is really appreciated.

Answer 1

Well, I do disagree with @DavidPopović, but I had the benefit of a rather primitive symbolic calculation package. My method was not at all as systematic as David’s: first I looked locally at the prime $2$, where you see that $f$ has a root and a quadratic factor. By playing around with the factor, I found that (at $2$) its roots were in the quadratic unramified extension of $\Bbb Q_2$, so that (always mistrusting my package) I thought that $\Bbb Q(\alpha)$ should be unramified at two, so that the factor $4$ in the preliminary discriminant of $-3884$ should be spurious. By further messing about, I found that the $\Bbb Q$-minimal polynomial for $\frac12(\alpha+\alpha^2)$ was $x^3-x^2-9x-18$.

This led me to try for $\bigl\lbrace1,\alpha,\frac12(\alpha+\alpha^2)\bigr\rbrace$ as integral basis. Using this $\Bbb Q$-basis, I found that the matrix of the trace pairing was $$ \begin{pmatrix} 3&0&1\\ 0&2&19\\ 1&19&19 \end{pmatrix}\,, $$ whose determinant is surenough $-971$. So my choice is an integral basis for $\Bbb Q(\alpha)$.

Answer 2

I will interpret the question as finding the integral basis of the ring of integers $\mathcal{O_K}$.

Firstly, the discriminant of the polynomial is $\text{Disc}(f) = -3884 = -2^2 \cdot 971$, which is non-zero, but not square-free. This means that $1, \alpha, \alpha^2$ is a rational basis of $\mathcal{O}_K$, but not necessarily an integral basis.

Let $\beta_1, \beta_2, \beta_3$ be an integral basis and $A$ a matrix (with integral coefficients!) such that $A(\beta_1, \beta_2, \beta_3)^T = (1, \alpha, \alpha^2)^T$. Then $ (\det A)^2 \text{Disc}(\beta_1, \beta_2, \beta_3) = \text{Disc}(1, \alpha, \alpha^2)$ and so $$\lvert \det A \rvert= [\mathcal{O}_K : \mathbb{Z}[\alpha]]= 1 \text{ or }2.$$

This is because $2$ is the only square that divides the discriminant of alphas. At this point we get the idea that we can try to see if any of $\alpha/2$ or $\alpha^2/2$ or $(\alpha^2 + \alpha)/2$ or similar expressions are in fact algebraic integers. If we manage to show that, the $\mathbb{Z}$-module $\mathbb{Z}[\alpha]$ will have index $2$ in the new module, which means that we have found an integral basis.

It turns out that $\frac{1}{2} (\alpha + \alpha^2)$ is in fact a root of $X^3-X^2-9X-18$ so an algebraic integer and $1, \alpha, \frac{1}{2}(\alpha+\alpha^2)$ is an integral basis.

If we were unsuccessful, we would have to adopt a general approach. Let $a+b \alpha +c\alpha^2 \in \mathcal{O}_K$ where $a, b, c \in \frac{1}{2}\mathbb{Z}$. By calculating the trace, which has to be an integer, we get that $3a+2c \in \mathbb{Z}$, so $a \in \mathbb{Z}$. There are only $4$ remaining cases based on the parity of $2b, 2c$ and we could check whether these are algebraic integers (for example by calculating traces, norms and if all else fails by brute force calculation).