discriminant of algebraic number

Theorem.  If ϑ is an algebraic numberMathworldPlanetmath of degree n with minimal polynomialPlanetmathPlanetmath f(x), then the of the number ϑ, i.e. the discriminantMathworldPlanetmathPlanetmathPlanetmath Δ(1,ϑ,,ϑn-1),  is


where N means the absolute norm.

Proof. Let the algebraic conjugates of the number ϑ, i.e. all complex zeroes of f(x),  be  ϑ1=ϑ,ϑ2,,ϑn.  If  f(x)=xn+a1xn-1++an,  we have


The norm (http://planetmath.org/AbsoluteNorm) of f(ϑ) in (ϑ)/ is the product of all http://planetmath.org/node/12046(ϑ)-conjugates [f(ϑ)](i) of f(ϑ), which is


On the other side, the polynonomial f(x) in its linear factors is


whence its derivative may be written


Substituting  x=ϑν  gives simply

f(ϑν)=jν(ϑν-ϑj)for ν=1,,n.

Multiplying these equations we obtain


The discriminant of ϑ is same as the discriminant of the equation  f(x)=0.  Therefore


where the number of the factors in the brackets is  (n-1)+(n-2)++1=(n-1)n2.  Thus we obtain the asserted result

Title discriminant of algebraic number
Canonical name DiscriminantOfAlgebraicNumber
Date of creation 2013-03-22 17:49:59
Last modified on 2013-03-22 17:49:59
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 10
Author pahio (2872)
Entry type Theorem
Classification msc 11R29
Related topic Discriminant
Related topic DerivativeOfPolynomial
Defines discriminant of number