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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?

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  • $\begingroup$ I think less than a week ago I returned a book to the library that would help me answer your question. It said something about how both Maple and Mathematica can understand multiple field extensions provided one doesn't get carried away. It also said about how a CAS can help you find a set of field extensions but it may or may not be optimal, and I believe the book also provided a means to determine if it is optimal or not. But this is all very fuzzy in my mind now. I can't remember the title of the book, but I do remember it was written some ten or twelve years ago... $\endgroup$ – Robert Soupe Aug 29 '14 at 2:25
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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 $1,2,4$).

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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 $\mathcal{O}_{K}$.

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