Discriminant (of a field basis)

Let $L/K$ be a finite field extension of degree $n$ , and let $\mathrm{Tr}_{L/K}$ be the field trace. For an $n$ -tuple $\mathbf{b}=(b_1,\dots,b_n)$ in $L$ that is a $K$ -basis of $L$ , the discriminant of $\mathbf{b}$ (relative to $L/K$ ) is

\mathrm{disc}_{L/K}(\mathbf{b}) \;:=\; \det\!\big(\mathrm{Tr}_{L/K}(b_i b_j)\big)_{1\le i,j\le n}\in K.

Remarks

If $L/K$ is separable (see separable extension), then $\mathrm{disc}_{L/K}(\mathbf{b})\neq 0$ , and one can also express it using $K$ -embeddings $\sigma_1,\dots,\sigma_n:L\hookrightarrow \Omega$ into a common overfield $\Omega$ :

\mathrm{disc}_{L/K}(\mathbf{b}) \;=\; \det(\sigma_i(b_j))_{i,j}^2.

Thus separability is equivalent to nondegeneracy of the trace pairing, and hence to the nonvanishing of the discriminant of every basis.

Examples

Quadratic basis. Let $L=K(\sqrt{d})$ with $\mathrm{char}(K)\neq 2$ and basis $(1,\sqrt{d})$ . Using $\mathrm{Tr}_{L/K}(1)=2$ , $\mathrm{Tr}_{L/K}(\sqrt{d})=0$ , $\mathrm{Tr}_{L/K}(d)=2d$ ,

\mathrm{disc}_{L/K}(1,\sqrt{d})=\det\begin{pmatrix}2&0\\0&2d\end{pmatrix}=4d.

Power basis in a simple extension. If $L=K(\alpha)$ with $[L:K]=n$ , the “power basis” $(1,\alpha,\dots,\alpha^{n-1})$ has discriminant

$\mathrm{disc}_{L/K}(1,\alpha,\dots,\alpha^{n-1})=\det(\mathrm{Tr}_{L/K}(\alpha^{i+j}))_{0\le i,j\le n-1}$ , which can be computed from the minimal polynomial of $\alpha$ in concrete cases.

Finite fields. For $L=\mathbb{F}_{q^n}$ over $K=\mathbb{F}_q$ , every $K$ -basis has nonzero discriminant because finite fields are perfect.