condition for power basis

Lemma.  If $K$ is an algebraic number field of degree (http://planetmath.org/Degree) $n$ and the elements ${\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n$ of $K$ can be expressed as linear combinations

$\{\begin{array}{cc}{\alpha }_1=c_{11}{\beta }_1+c_{12}{\beta }_2+\mathrm{\dots }+c_{1n}{\beta }_n\hfill & \\ {\alpha }_2=c_{21}{\beta }_1+c_{22}{\beta }_2+\mathrm{\dots }+c_{2n}{\beta }_n\hfill & \\ \mathrm{\cdots }\hfill & \\ {\alpha }_n=c_{n1}{\beta }_1+c_{n2}{\beta }_2+\mathrm{\dots }+c_{nn}{\beta }_n\hfill & \end{array}$

of the elements ${\beta }_1,{\beta }_2,\mathrm{\dots },{\beta }_n$ of $K$ with rational coefficients $c_{ij}$, then the discriminants of ${\alpha }_i$ and ${\beta }_j$ are by the equation

$$\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)=det{(c_{ij})}^2\cdot \mathrm{\Delta }({\beta }_1,{\beta }_2,\mathrm{\dots },{\beta }_n).$$

Theorem.  Let $\vartheta$ be an algebraic integer of degree (http://planetmath.org/DegreeOfAnAlgebraicNumber) $n$.  The set  $\{1,\vartheta ,\mathrm{\dots },{\vartheta }^{n-1}\}$  is an integral basis of $\mathbb{Q}(\vartheta )$ if the discriminant  $d(\vartheta ):=\mathrm{\Delta }(1,\vartheta ,\mathrm{\dots },{\vartheta }^{n-1})$  is square-free.

Proof.  The adjusted canonical basis

${\omega }_1=1,$
${\omega }_2={\displaystyle \frac{a_{21}+\vartheta }{d_2}},$
${\omega }_3={\displaystyle \frac{a_{31}+a_{32}\vartheta +{\vartheta }^2}{d_3}},$
$\mathrm{⋮}\mathit{\hspace{1em}\hspace{1em} }\mathrm{⋮}\mathit{\hspace{1em}\hspace{1em} }\mathrm{⋮}$
${\omega }_n={\displaystyle \frac{a_{n1}+a_{n2}\vartheta +\mathrm{\dots }+a_{n,n-1}{\vartheta }^{n-2}+{\vartheta }^{n-1}}{d_n}}$

of $\mathbb{Q}(\vartheta )$ is an integral basis, where $d_2,d_3,\mathrm{\dots },d_n$ are integers.  Its discriminant is the fundamental number $d$ of the field.  By the lemma, we obtain

Thus  ${(d_2{d}_3\mathrm{\cdots }{d}_n)}^{2}d=d(\vartheta )$,  and since $d(\vartheta )$ is assumed to be square-free, we have ${(d_2{d}_3\mathrm{\cdots }{d}_n)}^2=1$,  and accordingly  $d(\vartheta )$ equals the discriminant of the field (http://planetmath.org/MinimalityOfIntegralBasis).  This implies (see minimality of integral basis) that the numbers $1,\vartheta ,\mathrm{\dots },{\vartheta }^{n-1}$ form an integral basis of the field $\mathbb{Q}(\vartheta )$.

Title	condition for power basis
Canonical name	ConditionForPowerBasis
Date of creation	2013-03-22 17:49:56
Last modified on	2013-03-22 17:49:56
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	9
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11R04
Related topic	IntegralBasis
Related topic	PowerBasis
Related topic	CanonicalBasis
Related topic	PropertiesOfDiscriminantInAlgebraicNumberField