Separable extension

Discriminant of a separable extension

Let be a finite separable extension of degree , be the Algebraic closure of , and be the distinct embeddings of into . For some elements , the discriminant is defined as1

where

For we then define .

Properties

  2. iff are linearly dependent over .
Proof of 1–2

By linearity of the embeddings we see that linearly dependent give a singular matrix and therefore a zero discriminant.

Conversely, suppose form a -basis for . By the primitive element theorem, for some , so also form a -basis, so there exists a change of basis such that

whence also

for each . Thus and . It therefore suffices to show that .

Since is a Vandermonde matrix, the corresponding Vandermonde determinant is nonzero since each is distinct by separability — since these are precisely the roots of the minimal polynomial . It follows as required.

Special cases

See also

#state/develop | #lang/en | #SemBr

Footnotes

  1. 2022. Algebraic number theory course notes, p. 23 ↩