Discriminant of a field extension

Remarks

If $c_1, \ldots, c_n$ is another basis and $c = Pb$ with $P \in GL_n(K)$, then $d(c) = \det(P)^2\, d(b)$, so the class modulo squares is well defined. If $L/K$ is separable with embeddings $\sigma_1, \ldots, \sigma_n$ into an algebraic closure, then $d(b) = \det\left(\sigma_i(b_j)\right)^2$. The discriminant of a basis is nonzero if and only if $L/K$ is separable. If $L = K(\alpha)$ and $f$ is the minimal polynomial of $\alpha$, then $d(1, \alpha, \ldots, \alpha^{n-1})$ is the discriminant of $f$. For a number field $L$ over $\Q$, the discriminant of a $\Z$-basis of the ring of integers is an integer independent of the basis, the discriminant of the number field.

Example: $L = \Q(i)$, $K = \Q$, basis $1, i$. Check: $\mathrm{Tr}(1) = 2$, $\mathrm{Tr}(i) = 0$ and $\mathrm{Tr}(i^2) = -2$, so $d(1, i) = \det\begin{pmatrix} 2 & 0 \\ 0 & -2 \end{pmatrix} = -4$, the discriminant of $x^2 + 1$.

Non-example: for $K = \F_p(t)$ and $L = K(s)$ with $s^p = t$, every trace is $0$, so the discriminant of every basis is $0$; the extension is not separable.