canonical basisCanonicalBasis2005-06-17 16:59:322008-02-21 08:10:45Theoremintegral basiscanonical basiscanonical basis of a number fieldadjusted canonical basis% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}Let $\vartheta$ be an algebraic integer of \PMlinkname{degree}{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 {\em canonical basis} of the number field.
\textbf{Remark.}\, The integers $a_{ij}$ can be reduced so that for all $i$ and $j$,
$$-\frac{d_i}{2} < a_{ij} \leqq \frac{d_i}{2}.$$
Then one may speak of an {\em 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 $d$.