# discriminant of algebraic number

Proof. Let the algebraic conjugates of the number $\vartheta$, i.e. all complex zeroes of $f(x)$,  be  $\vartheta_{1}=\vartheta,\,\vartheta_{2},\,\ldots,\,\vartheta_{n}$.  If  $f(x)=x^{n}+a_{1}x^{n-1}+\ldots+a_{n}$,  we have

 $f^{\prime}(\vartheta)=n\vartheta^{n-1}+(n-1)a_{1}\vartheta^{n-2}+\ldots+2a_{n-% 2}\vartheta+a_{n-1}\in\mathbb{Q}(\vartheta).$

The norm (http://planetmath.org/AbsoluteNorm) of $f^{\prime}(\vartheta)$ in $\mathbb{Q}(\vartheta)/\mathbb{Q}$ is the product of all http://planetmath.org/node/12046$\mathbb{Q}(\vartheta)$-conjugates $[f^{\prime}(\vartheta)]^{(i)}$ of $f^{\prime}(\vartheta)$, which is

 $\mbox{N}(f^{\prime}(\vartheta))=[f^{\prime}(\vartheta)]^{(1)}[f^{\prime}(% \vartheta)]^{(2)}\cdots[f^{\prime}(\vartheta)]^{(n)}=f^{\prime}(\vartheta_{1})% f^{\prime}(\vartheta_{2})\cdots f^{\prime}(\vartheta_{n}).$

On the other side, the polynonomial $f(x)$ in its linear factors is

 $f(x)=(x-\vartheta_{1})(x-\vartheta_{2})\cdots(x-\vartheta_{n}),$

whence its derivative may be written

 $f^{\prime}(x)=\sum_{\nu=1}^{n}(x-\vartheta_{1})\cdots(x-\vartheta_{\nu-1})\,(x% -\vartheta_{\nu+1})\cdots(x-\vartheta_{n}).$

Substituting  $x=\vartheta_{\nu}$  gives simply

 $f^{\prime}(\vartheta_{\nu})=\prod_{j\neq\nu}(\vartheta_{\nu}-\vartheta_{j})% \quad\mbox{for\;\;}\nu=1,\,\ldots,\,n.$

Multiplying these equations we obtain

 $\mbox{N}(f^{\prime}(\vartheta))=\prod_{\nu=1}^{n}f^{\prime}(\vartheta_{\nu})=% \prod_{i\neq j}(\vartheta_{i}-\vartheta_{j}).$

The discriminant of $\vartheta$ is same as the discriminant of the equation  $f(x)=0$.  Therefore

 $d(\vartheta)=\left[\prod_{i

where the number of the factors in the brackets is  $(n-1)+(n-2)+\ldots+1=\frac{(n-1)n}{2}$.  Thus we obtain the asserted result

 $d(\vartheta)=\left[\prod_{i
Title discriminant of algebraic number DiscriminantOfAlgebraicNumber 2013-03-22 17:49:59 2013-03-22 17:49:59 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 11R29 Discriminant DerivativeOfPolynomial discriminant of number