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

• Discriminant of a number field
• Discriminant of an algebraic integer

See also

• Absolute norm of an ideal of the ring of integers of a number field
• Discriminant of a number field

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Footnotes

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