Let $E/K$ be a separable field extension of degree $n$, let $A$ be a Dedekind Domain which quotient field is $K$, and let $B$ be the integral closure of $A$ in $E$. Then we have that the ideal generated by the discriminants $D_{E/K}(\alpha_1, \cdots , \alpha_n)$ where, $\alpha_1, \cdots , \alpha_n$ runs over all integral bases of $B$, is the discriminant ideal of $E/K$.
Ok, now, we need $E/K$ be separable because if it is not separable, then the trace is always $0$ and thus the discriminants will be zero. But, why do we need $A$ to be a Dedekind Domain? Is it not enought to be just an integral Domain?
Thanks and sorry for my English