Theorem. If $\vartheta$ is an algebraic number of degree $n$ with minimal polynomial $f(x)$ , then the discriminant of the number $\vartheta$ , i.e. the discriminant $\Delta(1,\,\vartheta,\,\ldots,\,\vartheta^{n-1})$ , is $$d(\vartheta) = (-1)^\frac{n(n-1)}{2}\mbox{N}(f'(\vartheta)),$$ where N means the absolute norm.

Proof. Let the algebraic conjugates of the number $\vartheta$ , i.e. all complex zeroes of $f(x)$ , be $\vartheta_1 = \vartheta,\,\vartheta_2,\,\ldots,\,\vartheta_n$ . If $f(x) = x^n+a_1x^{n-1}+\ldots+a_n$ , we have $$f'(\vartheta) = n\vartheta^{n-1}+(n-1)a_1\vartheta^{n-2}+\ldots+2a_{n-2}\vartheta+a_{n-1} \in \mathbb{Q}(\vartheta).$$ The norm of $f'(\vartheta)$ in $\mathbb{Q}(\vartheta)/\mathbb{Q}$ is the product of all conjugates $[f'(\vartheta)]^{(i)}$ of $f'(\vartheta)$ , which is $$\mbox{N}(f'(\vartheta)) = [f'(\vartheta)]^{(1)}[f'(\vartheta)]^{(2)}\cdots[f'(\vartheta)]^{(n)} = f'(\vartheta_1)f'(\vartheta_2)\cdots f'(\vartheta_n).$$ 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 $$f'(x) = \sum_{\nu=1}^n(x-\vartheta_1)\cdots(x-\vartheta_{\nu-1})\,(x-\vartheta_{\nu+1})\cdots(x-\vartheta_n).$$ Substituting $x = \vartheta_\nu$ gives simply $$f'(\vartheta_\nu) = \prod_{j\neq\nu}(\vartheta_\nu-\vartheta_j)\quad \mbox{for\;\;} \nu = 1,\,\ldots,\,n.$$ Multiplying these equations we obtain $$\mbox{N}(f'(\vartheta)) = \prod_{\nu=1}^nf'(\vartheta_\nu) = \prod_{i\neq j}(\vartheta_i-\vartheta_j).$$ The discriminant of $\vartheta$ is same as the discriminant of the equation $f(x) = 0$ . Therefore $$d(\vartheta) = \left[\prod_{i<j}(\vartheta_i-\vartheta_j)\right]^2,$$ 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 $$d(\vartheta) = \left[\prod_{i<j}(\vartheta_i-\vartheta_j)\right]\cdot(-1)^\frac{n(n-1)}{2}\left[\prod_{j<i}(\vartheta_i-\vartheta_j)\right] = (-1)^\frac{n(n-1)}{2}\prod_{i\neq j}(\vartheta_i-\vartheta_j) = (-1)^\frac{n(n-1)}{2}\mbox{N}(f'(\vartheta)).$$