Let $\vartheta$ be an algebraic integer of degree $n$ . 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 $j$ , $$-\frac{d_i}{2} < a_{ij} \leqq \frac{d_i}{2}.$$ 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 $d$ .