Why a particular ring of integers is not generated by a single element

Question

It says here in the Sage documentation that the ring of integers in the number field obtained from $$f(x) = x^3 + x^2 - 2x + 8$$ is not generated by a single element. How would one go about showing that this is the case?

Answer 1

Let $\alpha$ be a root of $f(x) = x^{3} + x^{2} - 2x + 8$. We will show that $\mathcal{O}_{K} = \mathbb{Z}[\alpha, \beta]$, where $\beta = (\alpha + \alpha^{2})/2$. Granting this, assume that $\mathbb{Z}[\vartheta]$ for some $\vartheta = a + b\alpha + c\beta$. Since $\{1, \vartheta, \vartheta^{2}\}$ is an integral basis iff $\{1, (\vartheta + 1), (\vartheta + 1)^{2}\}$ is an integral basis iff $\{1, (\vartheta - a), (\vartheta - a)^{2}\}$ is an integral basis, we may assume that $a = 0$. We find that (writing $\alpha^{2}, \alpha\beta, \beta^{2}$ in terms of $\alpha, \beta$) $$(b\alpha + c\beta)^{2} = -8bc - 6c^{2} + \left(2bc - b^{2} - \frac{c^{2}}{2}\right)\alpha - 2b^{2}\beta,$$ so that the change of basis matrix is given by $$A = \begin{bmatrix} 1 & 0 & -8bc - 6c^{2}\\ 0 & b & 2bc - b^{2} - \frac{c^{2}}{2}\\ 0 & c & -2b^{2} \end{bmatrix}.$$ If $\mathbb{Z}[\vartheta]$ is indeed an integral basis, then this matrix must have determinant $\pm 1$. As $$\pm 1 = \det A = -2b^{3} - 2bc^{2} + b^{2}c + \frac{c^{3}}{2},$$ we must have $c$ even. But then the determinant is even, a contradiction, so that no such $\vartheta$ exists.

We now show that $\mathcal{O}_{K} = \mathbb{Z}[\alpha, \beta]$. Note that the linear transformation of multiplication by $a + b\alpha + c\alpha^{2}$ is given by $$L = \begin{bmatrix} a & -8c & -8(b - c)\\ b & 2c + a & 2b - 10c\\ c & b - c & a - b + 3c \end{bmatrix}.$$ Consider $\mathcal{O} = \mathbb{Z}[\alpha]$. First, $f$ is indeed the minimal polynomial since it is irreducible mod 3 (no roots). Then the discriminant of $\{1, \alpha, \alpha^{2}\}$ is $$d(1, \alpha, \alpha^{2}) = -N\Big(f'(\alpha)\Big) = -N(3\alpha^{2} + 2\alpha - 2) = -\rm{det}\begin{bmatrix} -2 & -24 & 8\\ 2 & 4 & -26\\ 3 & -1 & 7 \end{bmatrix} = -4 \cdot 523.$$ As a result, since $2^{2} \mid d$, if $\mathcal{O}_{K} \not= \mathcal{O}$, (by a theorem) there exists an algebraic integer of the form $$\frac{1}{2}(\lambda_{1} + \lambda_{2}\alpha + \lambda_{3}\alpha^{2}),$$ where $\lambda_{i} \in \{0, 1\}$.

If such an algebraic integer $\gamma$ of this form exists, we have $$\rm{Tr}\, \gamma = \rm{tr}\, L = \frac{3\lambda_{1} - \lambda_{2} + 5\lambda_{3}}{2} \in \mathbb{Z}.$$ Examining the parity of the numberator, we see that for $\gamma \not= 0$, we must have exactly two $\lambda_{i} = 1$. Thus, the possibilities for $\gamma$ are $$\frac{1 + \alpha}{2}, \frac{1 + \alpha^{2}}{2}, \frac{\alpha + \alpha^{2}}{2}.$$ However, for the first two possibilities, $$N(\gamma) = \rm{det} \begin{bmatrix} \frac{1}{2} & 0 & -4\\ \frac{1}{2} & \frac{1}{2} & 1\\ 0 & \frac{1}{2} & 0 \end{bmatrix} = -\frac{5}{4} \notin \mathbb{Z}, \qquad N(\gamma) = \rm{det} \begin{bmatrix} \frac{1}{2} & -2 & 4\\ 0 & \frac{3}{2} & -5\\ \frac{1}{2} & -\frac{1}{2} & 2 \end{bmatrix} = \frac{9}{4} \notin \mathbb{Z}.$$ For the third possibility, which we denoted $\beta$ earlier, the norm is indeed an integer and we further find (upon examining $\beta, \beta^{2}, \beta^{3}$) that $$\beta^{3} - \beta^{2} + 6\beta - 8 = 0,$$ so that $\beta \in \mathbb{O}_{K}$ and $\mathbb{Z}[\alpha, \beta] \in \mathcal{O}_{K}$.

Now, $\mathbb{Q} \oplus \mathbb{Q}\alpha \oplus \mathbb{Q}\alpha^{2} \approx \mathbb{Q} \oplus \mathbb{Q}\alpha \oplus \mathbb{Q}\beta$ and we have a change of basis matrix ($\{1, \alpha, \alpha^{2}\} \rightarrow \{1, \alpha, \beta\}$) $$B = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & -1\\ 0 & 0 & 2 \end{bmatrix}, \qquad \rm{det} B = 2.$$ Hence, $d(1, \alpha, \beta) = d(1, \alpha, \alpha^{2})/2^{2} = 523$. Since $523$ is squarefree, $\{1, \alpha, \beta\}$ is a basis for $\mathcal{O}_{K}$.

Answer 2

Here is a relatively short proof:

If $O_K=\mathbb Z[\alpha]$ and $f$ is the minimal polynomial of $\alpha$, then we can find the factorization of $(2)$ to prime ideals in $O_K$ by factoring $\bar f\in\mathbb F_2[x]$ to irreducible polynomials ($\bar f$ denotes the reduction of $f\in\mathbb Z[x]$ mod $2$); if $\bar f =\prod h_i^{e_i}$ then $(2)=\prod P_i^{e_i}$. Whatever the cubic polynomial $f$ is, we cannot get $(2)=P_1P_2P_3$, as that would mean that $\bar f$ has $3$ different roots in $\mathbb F_2$, which has only two elements.

However, in reality $(2)=P_1P_2P_3$. It is equivalent to $g(x):=x^3+x^2-2x+8$ having $3$ roots in $\mathbb Q_2$, and indeed $g(1)\equiv0$ mod $8$ and $g(2)\equiv g(4)\equiv0$ mod $16$, while $g'(1)\equiv 1$ mod $2$ and $g'(2)\equiv g'(4)\equiv 2$ mod $4$, so Hensel's lemma (i.e. Newton's method for computing roots) will give us $3$ roots (close to $1,2,4$).