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Lemma. If $K$ is an algebraic number field of degree $n$ and the elements $\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n$ of $K$ can be expressed as linear combinations
of the elements $\beta_1,\,\beta_2,\,\ldots,\,\beta_n$ of $K$ with rational coefficients $c_{ij}$ , then the discriminants of $\alpha_i$ and $\beta_j$ are related by the equation $$\Delta(\alpha_1,\,\alpha_2,\,\ldots,\,\alpha_n) = \det(c_{ij})^2\cdot\Delta(\beta_1,\,\beta_2,\,\ldots,\,\beta_n).\\$$
Theorem. Let $\vartheta$ be an algebraic integer of degree $n$ . The set $\{1,\,\vartheta,\,\ldots,\,\vartheta^{n-1}\}$ is an integral basis of $\mathbb{Q}(\vartheta)$ if the discriminant $d(\vartheta) := \Delta(1,\,\vartheta,\,\ldots,\,\vartheta^{n-1})$ is square-free.
Proof. The adjusted canonical basis
$\displaystyle\omega_1 = 1,$
$\displaystyle\omega_2 = \frac{a_{21}\!+\!\vartheta}{d_2},$
$\displaystyle\omega_3 = \frac{a_{31}\!+\!a_{32}\vartheta\!+\!\vartheta^2}{d_3},$
$\vdots\,\qquad\vdots\,\qquad\vdots$
$\displaystyle\omega_n = \frac{a_{n1}\!+\!a_{n2}\vartheta\!+\ldots+\!a_{n,n-1}\vartheta^{n-2}\!+\!\vartheta^{n-1}}{d_n}$
of $\mathbb{Q}(\vartheta)$ is an integral basis, where $d_2,\,d_3,\,\ldots,\,d_n$ are integers. Its discriminant is the fundamental number $d$ of the field. By the lemma, we obtain
Thus $(d_2d_3\cdots d_n)^2d = d(\vartheta)$ , and since $d(\vartheta)$ is assumed to be square-free, we have $(d_2d_3\cdots d_n)^2 = 1$ , and accordingly $d(\vartheta)$ equals the discriminant of the field. This implies (see minimality of integral basis) that the numbers $1,\,\vartheta,\,\ldots,\,\vartheta^{n-1}$ form an integral basis of the field $\mathbb{Q}(\vartheta)$ .
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