condition for power basis

Lemma.  If K is an algebraic number fieldMathworldPlanetmath of degree ( n and the elements α1,α2,,αn of K can be expressed as linear combinationsMathworldPlanetmath


of the elements β1,β2,,βn of K with rational coefficients cij, then the discriminantsMathworldPlanetmathPlanetmathPlanetmath of αi and βj are by the equation


Theorem.  Let ϑ be an algebraic integerMathworldPlanetmath of degree ( n.  The set  {1,ϑ,,ϑn-1}  is an integral basis of (ϑ) if the discriminant  d(ϑ):=Δ(1,ϑ,,ϑn-1)  is square-free.

Proof.  The adjusted canonical basis


of (ϑ) is an integral basis, where d2,d3,,dn are integers.  Its discriminant is the fundamental number d of the field.  By the lemma, we obtain


Thus  (d2d3dn)2d=d(ϑ),  and since d(ϑ) is assumed to be square-free, we have (d2d3dn)2=1,  and accordingly  d(ϑ) equals the discriminant of the field (  This implies (see minimality of integral basis) that the numbers 1,ϑ,,ϑn-1 form an integral basis of the field (ϑ).

Title condition for power basis
Canonical name ConditionForPowerBasis
Date of creation 2013-03-22 17:49:56
Last modified on 2013-03-22 17:49:56
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 9
Author pahio (2872)
Entry type Theorem
Classification msc 11R04
Related topic IntegralBasis
Related topic PowerBasis
Related topic CanonicalBasis
Related topic PropertiesOfDiscriminantInAlgebraicNumberField