condition for power basis

Lemma. If $K$ is an algebraic number field of degree (http://planetmath.org/Degree) $n$ and the elements $\\alpha_{1},\\,\\alpha_{2},\\,\\ldots,\\,\\alpha_{n}$ of $K$ can be expressed as linear combinations

\\displaystyle\\begin{cases}\\alpha_{1}=c_{11}\\beta_{1}+c_{12}\\beta_{2}+\\ldots+c_%{1n}\\beta_{n}\\\\\\alpha_{2}=c_{21}\\beta_{1}+c_{22}\\beta_{2}+\\ldots+c_{2n}\\beta_{n}\\\\\\qquad\\cdots\\\\\\alpha_{n}=c_{n1}\\beta_{1}+c_{n2}\\beta_{2}+\\ldots+c_{nn}\\beta_{n}\\end{cases}

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 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 (http://planetmath.org/DegreeOfAnAlgebraicNumber) $n$ . 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

d=\\Delta(\\omega_{1},\\,\\omega_{2},\\,\\ldots,\\,\\omega_{n})=\\left|\\begin{array}[]{%cccc}1&0&\\ldots&0\\\\\\frac{a_{21}}{d_{2}}&\\frac{1}{d_{2}}&\\ddots&0\\\\\\vdots&\\vdots&\\ddots&0\\\\\\frac{a_{n1}}{d_{n}}&\\frac{a_{n2}}{d_{n}}&\\ldots&\\frac{1}{d_{n}}\\end{array}%\\right|^{2}\\Delta(1,\\,\\vartheta,\\,\\ldots,\\,\\vartheta^{n-1})=\\frac{d(\\vartheta)%}{(d_{2}d_{3}\\cdots d_{n})^{2}}.

Thus $(d_{2}d_{3}\\cdots d_{n})^{2}d=d(\\vartheta)$ , and since $d(\\vartheta)$ is assumed to be square-free, we have $(d_{2}d_{3}\\cdots d_{n})^{2}=1$ , and accordingly $d(\\vartheta)$ equals the discriminant of the field (http://planetmath.org/MinimalityOfIntegralBasis). This implies (see minimality of integral basis) that the numbers $1,\\,\\vartheta,\\,\\ldots,\\,\\vartheta^{n-1}$ form an integral basis of the field $\\mathbb{Q}(\\vartheta)$ .