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

d(ϑ)=(-1)n(n-1)2N(f(ϑ)),

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

f(ϑ)=nϑn-1+(n-1)a1ϑn-2++2an-2ϑ+an-1(ϑ).

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

N(f(ϑ))=[f(ϑ)](1)[f(ϑ)](2)[f(ϑ)](n)=f(ϑ1)f(ϑ2)f(ϑn).

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

f(x)=(x-ϑ1)(x-ϑ2)(x-ϑn),

whence its derivative may be written

f(x)=ν=1n(x-ϑ1)(x-ϑν-1)(x-ϑν+1)(x-ϑn).

Substituting  x=ϑν  gives simply

f(ϑν)=jν(ϑν-ϑj)for ν=1,,n.

Multiplying these equations we obtain

N(f(ϑ))=ν=1nf(ϑν)=ij(ϑi-ϑj).

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

d(ϑ)=[i<j(ϑi-ϑj)]2,

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

d(ϑ)=[i<j(ϑi-ϑj)](-1)n(n-1)2[j<i(ϑi-ϑj)]=(-1)n(n-1)2ij(ϑi-ϑj)=(-1)n(n-1)2N(f(ϑ)).
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