Discriminant of a number field

Remarks

The value does not depend on the integral basis or on the order of the embeddings. Two integral bases differ by a matrix in $\mathrm{GL}_n(\Z)$, whose determinant is $\pm 1$. Reordering the rows changes only the sign of the determinant. An equivalent form is $d_K = \det\big(\operatorname{Tr}_{K/\Q}(\omega_i \omega_j)\big)$, where the field trace $\operatorname{Tr}_{K/\Q}(\alpha)$ is the trace of the $\Q$-linear map $x \mapsto \alpha x$ on $K$. The field trace of an algebraic integer is an integer, so $d_K$ is an integer. It is never $0$, and its sign is $(-1)^{r_2}$, where $r_2$ is the number of pairs of complex conjugate embeddings.

Example: for $K = \Q(i)$ the basis $1, i$ of $\Z[i]$ gives $\det \begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix} = -2i$, so $d_K = -4$. More generally, for a squarefree integer $d \ne 1$, the discriminant of $\Q(\sqrt{d})$ is $d$ if $d \equiv 1 \pmod 4$ and $4d$ otherwise.

Non-example: the basis $1, 2i$ gives $\det \begin{pmatrix} 1 & 2i \\ 1 & -2i \end{pmatrix}^2 = (-4i)^2 = -16$. This is not $d_{\Q(i)}$, because $1, 2i$ is a basis of the subring $\Z[2i]$ and not of the whole ring of integers $\Z[i]$.