discriminant of algebraic number

Theorem.  If $\vartheta$ is an algebraic number of degree $n$ with minimal polynomial $f(x)$, then the of the number $\vartheta$, i.e. the discriminant $\mathrm{\Delta }(1,\vartheta ,\mathrm{\dots },{\vartheta }^{n-1})$,  is

$$d(\vartheta )={(-1)}^{\frac{n(n-1)}{2}}\text{N}(f^{\prime }(\vartheta )),$$

where N means the absolute norm.

Proof. Let the algebraic conjugates of the number $\vartheta$, i.e. all complex zeroes of $f(x)$,  be  ${\vartheta }_1=\vartheta ,{\vartheta }_2,\mathrm{\dots },{\vartheta }_n$.  If  $f(x)=x^n+a_1{x}^{n-1}+\mathrm{\dots }+a_n$,  we have

$$f^{\prime }(\vartheta )=n{\vartheta }^{n-1}+(n-1)a_1{\vartheta }^{n-2}+\mathrm{\dots }+2a_{n-2}\vartheta +a_{n-1}\in \mathbb{Q}(\vartheta ).$$

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

$$\text{N}(f^{\prime }(\vartheta ))={[f^{\prime }(\vartheta )]}^{(1)}{[f^{\prime }(\vartheta )]}^{(2)}\mathrm{\cdots }{[f^{\prime }(\vartheta )]}^{(n)}=f^{\prime }({\vartheta }_1)f^{\prime }({\vartheta }_2)\mathrm{\cdots }{f}^{\prime }({\vartheta }_n).$$

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

$$f(x)=(x-{\vartheta }_1)(x-{\vartheta }_2)\mathrm{\cdots }(x-{\vartheta }_n),$$

whence its derivative may be written

$$f^{\prime }(x)=\sum _{\nu =1}^n(x-{\vartheta }_1)\mathrm{\cdots }(x-{\vartheta }_{\nu -1})(x-{\vartheta }_{\nu +1})\mathrm{\cdots }(x-{\vartheta }_n).$$

Substituting  $x={\vartheta }_{\nu }$  gives simply

$$f^{\prime }({\vartheta }_{\nu })=\prod _{j\ne \nu }({\vartheta }_{\nu }-{\vartheta }_j)\mathit{\hspace{1em}}\text{for }\nu =1,\mathrm{\dots },n.$$

Multiplying these equations we obtain

$$\text{N}(f^{\prime }(\vartheta ))=\prod _{\nu =1}^n{f}^{\prime }({\vartheta }_{\nu })=\prod _{i\ne j}({\vartheta }_i-{\vartheta }_j).$$

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

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