discriminant of algebraic number
Theorem. If ϑ is an algebraic number of degree n with minimal polynomial
f(x), then the of the number ϑ, i.e. the discriminant
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∈ℚ(ϑ). |
The norm (http://planetmath.org/AbsoluteNorm) of f′(ϑ) in ℚ(ϑ)/ℚ is the product of all http://planetmath.org/node/12046ℚ(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)=n∑ν=1(x-ϑ1)⋯(x-ϑν-1)(x-ϑν+1)⋯(x-ϑn). |
Substituting x=ϑν gives simply
f′(ϑν)=∏j≠ν(ϑν-ϑj) |
Multiplying these equations we obtain
The discriminant of is same as the discriminant of the equation . Therefore
where the number of the factors in the brackets is . 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 |