discriminant in algebraic number field

Let us consider the elements ${\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n$ of an algebraic number field$\mathbb{Q}(\vartheta )$ of degree (http://planetmath.org/NumberField) $n$.  Let${\vartheta }_1=\vartheta ,{\vartheta }_2,\mathrm{\dots },{\vartheta }_n$ be the algebraic conjugates of the primitive element $\vartheta$ and

$${\alpha }_i=r_i(\vartheta )\mathit{\hspace{1em}}(i=\mathrm{\hspace{0.33em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots },n)$$

the canonical forms of the elements ${\alpha }_i$.  Then the $\mathbb{Q}(\vartheta )$-conjugates (http://planetmath.org/CharacteristicPolynomialOfAlgebraicNumber) of those elements are

$${\alpha }_i^{(j)}=r_i({\vartheta }_j).$$

Using these, one can define the discriminant  $\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)$ of the elenents ${\alpha }_i$ as

$$\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n):=det{(r_i({\vartheta }_j))}^2=det{\left({\alpha }_i^{(j)}\right)}^2,$$

i.e.

Basing on the properties of determinants, one sees at once that the discriminant is of the numbers ${\alpha }_i$.  The entry independence of characteristic polynomial on primitive element allows to see that the discriminant also does not depend on the used primitive element of the field.  Moreover, the method for multiplying the determinants enables to convert the discriminant into the form

where S is the trace function defined in $\mathbb{Q}(\vartheta )$; therefore the discriminant is always a rational number (and an integer if every ${\alpha }_i$ is an algebraic integer of the field).  Cf. the parent entry (http://planetmath.org/DiscriminantOfANumberField).