canonical basis

Let $\vartheta$ be an algebraic integer of degree (http://planetmath.org/ExtensionField) $n$.  The algebraic number field $\mathbb{Q}(\vartheta )$ has always an integral basis of the form

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

where the $a_{ij}$’s and $d_i$’s are rational integers such that

$$d_2\mid d_3\mid d_4\mid \mathrm{\dots }\mid d_n,$$

i.e.

$$d_i\mid d_{i+1}\mathit{\hspace{1em}}\forall i=2,\mathrm{\hspace{0.17em}3},\mathrm{\dots },n-1.$$

The integral basis  ${\omega }_1,{\omega }_2,\mathrm{\dots },{\omega }_n$ is called a canonical basis of the number field.

Remark.  The integers $a_{ij}$ can be reduced so that for all $i$ and $j$,

Then one may speak of an adjusted canonical basis.  In the case of a quadratic number field $\mathbb{Q}(\sqrt{d})$ with $d\equiv 1(\text{mod}\mathrm{\hspace{0.17em}4})$  we have (see the examples of ring of integers of a number field)

$${\omega }_1=1,{\omega }_2=\frac{1+\sqrt{d}}{2}.$$

The discriminant of this basis is $d$.

Title	canonical basis
Canonical name	CanonicalBasis
Date of creation	2015-02-06 13:12:19
Last modified on	2015-02-06 13:12:19
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	14
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11R04
Related topic	MinimalityOfIntegralBasis
Related topic	ExamplesOfRingOfIntegersOfANumberField
Related topic	ConditionForPowerBasis
Related topic	IntegralBasisOfQuadraticField
Related topic	CanonicalFormOfElementOfNumberField
Defines	canonical basis
Defines	canonical basis of a number field
Defines	adjusted canonical basis