basis of ideal in algebraic number field


Theorem.  Let 𝒪K be the maximal orderMathworldPlanetmathPlanetmath of the algebraic number fieldMathworldPlanetmath K of degree n.  Every ideal 𝔞 of 𝒪K has a basis, i.e. there are in 𝔞 the linearly independentMathworldPlanetmath numbers  α1,α2,,αn  such that the numbers

m1α1+m2α2++mnαn,

where the mi’s run all rational integers, form precisely all numbers of 𝔞.  One has also

𝔞=(α1,α2,,αn),

i.e. the basis of the ideal can be taken for the system of generatorsPlanetmathPlanetmathPlanetmath of the ideal.

Since  {α1,α2,,αn}  is a basis of the field extension K/, any element of 𝔞 is uniquely expressible in the form m1α1+m2α2++mnαn.

It may be proven that all bases of an ideal 𝔞 have the same discriminantMathworldPlanetmathPlanetmathPlanetmath Δ(α1,α2,,αn), which is an integer; it is called the discriminant of the ideal.  The discriminant of the ideal has the minimality property, that if β1,β2,,βn are some elements of 𝔞, then

Δ(β1,β2,,βn)Δ(α1,α2,,αn)orΔ(β1,β2,,βn)=0

But if  Δ(β1,β2,,βn)=Δ(α1,α2,,αn),  then also the βi’s form a basis of the ideal 𝔞.

Example.  The integers of the quadratic field (2) are l+m2 with  l,m.  Determine a basis {α1,α2}  and the discriminant of the ideal a)  (6-62, 9+62),  b) (1-22).

a) The ideal may be seen to be the principal idealMathworldPlanetmathPlanetmathPlanetmathPlanetmath (3), since the both generators are of the form  (l+m2)3  and on the other side,  3=0(6-62)+(3-22)(9+62).  Accordingly, any element of the ideal are of the form

(m1+m22)3=m13+m232

where m1 and m2 are rational integers.  Thus we can infer that

α1=3,α2=32

is a basis of the ideal concerned.  So its discriminant is

Δ(α1,α2)=|3323-32|2=648.

b) All elements of the ideal (1-22) have the form

α=(a+b2)(1-22)=(a-4b)+(b-2a)2with a,b. (1)

Especially the rational integers of the ideal satisfyb-2a=0,  when  b=2a  and thus  α=a-42a=-7a.  This means that in the presentationMathworldPlanetmathα=m1α1+m2α2  we can assume α1 to be 7.  Now the rational portion a-4b in the form (1) of α should be splitted into two parts so that the first would be always divisible by 7 and the second by b-2a, i.e.  a-4b=7m1+(b-2a)x; this equation may be written also as

(2x+1)a-(x+4)b=7m1.

By experimenting, one finds the simplest value  x=3,  another would be  x=10.  The first of these yields

α=7(a-b)+(b-2a)(3+2)=m17+m2(3+2),

i.e. we have the basis

α1=7,α2=3+2.

The second alternative  x=10  similarly would give

α1=7,α2=10+2.

For both alternatives,  Δ(α1,α2)=392.

Title basis of ideal in algebraic number field
Canonical name BasisOfIdealInAlgebraicNumberField
Date of creation 2013-03-22 17:51:15
Last modified on 2013-03-22 17:51:15
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 9
Author pahio (2872)
Entry type Theorem
Classification msc 12F05
Classification msc 11R04
Classification msc 06B10
Synonym basis of ideal in number field
Related topic IntegralBasis
Related topic IdealNorm
Related topic AlgebraicNumberTheory
Defines basis of ideal
Defines ideal basis
Defines discriminant of the ideal