Discriminant of a separable extension

Let $𝐿 : 𝐾$ be a finite separable extension of degree $𝑛$ , $―― 𝐾$ be the Algebraic closure of $𝐾$ , and ${𝜎 𝑖} 𝑛 𝑖 = 1$ be the distinct embeddings of $𝐿$ into $―― 𝐾$ . For some elements ${𝛼 𝑖} 𝑛 𝑖 = 1 \subset 𝐿$ , the discriminant is defined as¹

Δ 𝐿 : 𝐾 (𝛼 1, \dots, 𝛼 𝑛) = d e t 𝑇 (𝛼 1, \dots, 𝛼 𝑛) 2

where

𝑇 (𝛼 1, \dots, 𝛼 𝑛) = ⎡ ⎢ ⎢ ⎣ 𝜎 1 (𝛼 1) \dots 𝜎 1 (𝛼 𝑛) ⋮ ⋱ ⋮ 𝜎 𝑛 (𝛼 1) \dots 𝜎 𝑛 (𝛼 𝑛) ⎤ ⎥ ⎥ ⎦ .

For $𝛼 \in 𝐿$ we then define $Δ 𝐿 : 𝐾 (𝛼) = Δ 𝐿 : 𝐾 (1, 𝛼, \dots, 𝛼 𝑛 - 1)$ .

Properties

$Δ 𝐿 : 𝐾 (𝛼 1, \dots, 𝛼 𝑛) \in 𝐾$
$Δ 𝐿 : 𝐾 (𝛼 1, \dots, 𝛼 𝑛) = 0$ iff $𝛼 1, \dots, 𝛼 𝑛$ are linearly dependent over $𝐾$ .

Proof of 1–2

By linearity of the embeddings ${𝜎 𝑖} 𝑛 𝑖 = 1$ we see that linearly dependent $𝛼 1, \dots, 𝛼 𝑛$ give a singular matrix and therefore a zero discriminant.

Conversely, suppose ${𝛼 𝑖} 𝑛 𝑖 = 1$ form a $𝐾$ -basis for $𝐿$ . By the primitive element theorem, $𝐿 = 𝐾 (𝜗)$ for some $𝜗 \in 𝐿$ , so ${𝜗 𝑖 - 1} 𝑛 𝑖 = 1$ also form a $𝐾$ -basis, so there exists a change of basis $𝑃 \in G L 𝑛 (𝐾)$ such that

⎡ ⎢ ⎢ ⎣ 𝛼 1 ⋮ 𝛼 𝑛 ⎤ ⎥ ⎥ ⎦ = 𝑃 ⎡ ⎢ ⎢ ⎣ 1 ⋮ 𝜗 𝑛 - 1 ⎤ ⎥ ⎥ ⎦

whence also

⎡ ⎢ ⎢ ⎣ 𝜎 𝑖 𝛼 1 ⋮ 𝜎 𝑖 𝛼 𝑛 ⎤ ⎥ ⎥ ⎦ = 𝑃 ⎡ ⎢ ⎢ ⎣ 1 ⋮ 𝜗 𝑛 - 1 ⎤ ⎥ ⎥ ⎦

for each $𝑖$ . Thus $𝑇 (𝛼 1, \dots, 𝛼 𝑛) = 𝑇 (𝜗) 𝑃 𝖳$ and $Δ (𝛼 1, \dots, 𝛼 𝑛) = (d e t 𝑃) 2 Δ (𝜗)$ . It therefore suffices to show that $0 \neq Δ (𝜗) \in 𝐾$ .

Since $𝑇 (𝜗)$ is a Vandermonde matrix, the corresponding Vandermonde determinant is nonzero since each $𝜎 𝑖 (𝜗)$ is distinct by separability — since these are precisely the roots of the minimal polynomial $𝑚 𝜗 (𝑥) \in 𝐾 [𝑥]$ . It follows $Δ \in 𝐾$ as required.

Discriminant of a separable extension

Properties

Special cases

See also

Discriminant of a separable extension

Properties

Special cases

See also

Footnotes