Let K be an algebraic number field. Let $\alpha \in$ K. Let $\beta$ be conjugate of $\alpha$ relative to K . Prove that $D(\alpha)=D(\beta)$.
$D(\alpha)$:= Let K be algebraic number field of degree n . Let $\alpha \in$ K. Then we define the discriminant $D(\alpha)$ of $\alpha$ by $D(\alpha)$=$D(1,\alpha,\alpha^2,...,\alpha^{n-1})$
or if $\alpha \in$ K , $D(\alpha)=\prod(\alpha^{(i)}-\alpha^{(j)})^2$ where $\alpha^{(1)}=\alpha,\alpha^{(2)},...,\alpha^{(n)}$ are the conjugates of $\alpha$ with respect to K .
I could not do this question. Can you help me please? .
Since $\alpha$ conjugate of $\beta$ is an equivalence relation. We have that the set of all conjugates of $\alpha$ equals the set of all conjugates of $\beta$. Thus your formula is such that it's equal on any two representatives of that set!