# 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

$\displaystyle\omega_{1}=1,$
$\displaystyle\omega_{2}=\frac{a_{21}\!+\!\vartheta}{d_{2}},$
$\displaystyle\omega_{3}=\frac{a_{31}\!+\!a_{32}\vartheta\!+\!\vartheta^{2}}{d_% {3}},$
$\vdots\,\qquad\vdots\,\qquad\vdots$
$\displaystyle\omega_{n}=\frac{a_{n1}\!+\!a_{n2}\vartheta\!+\ldots+\!a_{n,n-1}% \vartheta^{n-2}\!+\!\vartheta^{n-1}}{d_{n}}$,

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

 $d_{2}\mid d_{3}\mid d_{4}\mid\ldots\mid d_{n},$

i.e.

 $d_{i}\mid d_{i+1}\quad\forall\,i=2,\,3,\,\ldots,\,n\!-\!1.$

The integral basis  $\omega_{1},\,\omega_{2},\,\ldots,\,\omega_{n}$ is called a canonical basis of the number field.

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

 $-\frac{d_{i}}{2}

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

 $\omega_{1}=1,\quad\omega_{2}=\frac{1\!+\!\sqrt{d}}{2}.$

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