# 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 $\mathbb{Q}(\vartheta)$.

Title condition for power basis ConditionForPowerBasis 2013-03-22 17:49:56 2013-03-22 17:49:56 pahio (2872) pahio (2872) 9 pahio (2872) Theorem msc 11R04 IntegralBasis PowerBasis CanonicalBasis PropertiesOfDiscriminantInAlgebraicNumberField