Field norm

From HandWiki

In mathematics, the (field) norm is a particular mapping defined in field theory, which maps elements of a larger field into a subfield.

Formal definition

Let K be a field and L a finite extension (and hence an algebraic extension) of K. The field L is then a finite dimensional vector space over K. Multiplication by α, an element of L,

[math]\displaystyle{ m_\alpha:L\to L \text{ given by } m_\alpha (x) = \alpha x }[/math],

is a K-linear transformation of this vector space into itself. The norm, NL/K(α), is defined as the determinant of this linear transformation.[1]

If L/K is a Galois extension, one may compute the norm of α ∈ L as the product of all the Galois conjugates of α:

[math]\displaystyle{ \operatorname{N}_{L/K}(\alpha)=\prod_{\sigma\in\operatorname{Gal}(L/K)} \sigma(\alpha) }[/math],

where Gal(L/K) denotes the Galois group of L/K.[2] (Note that there may be a repetition in the terms of the product)

For a general field extension L/K, and nonzero α in L, let σ1(α), ..., σn(α) be the roots of the minimal polynomial of α over K (roots listed with multiplicity and lying in some extension field of L); then

[math]\displaystyle{ \operatorname{N}_{L/K}(\alpha)=\left (\prod_{j=1}^n\sigma_j(\alpha) \right )^{[L:K(\alpha)]} }[/math].

If L/K is separable, then each root appears only once in the product (though the exponent, the degree [L:K(α)], may still be greater than 1).

Example

The field norm from the complex numbers to the real numbers sends

x + iy

to

x2 + y2,

because the Galois group of [math]\displaystyle{ \C }[/math] over [math]\displaystyle{ \R }[/math] has two elements, the identity element and complex conjugation, and taking the product yields (x + iy)(xiy) = x2 + y2.

In this example the norm was the square of the usual Euclidean distance norm in [math]\displaystyle{ \C }[/math]. In general, the field norm is very different from the usual distance norm. We will illustrate that with an example where the field norm can be negative. Consider the number field [math]\displaystyle{ K=\Q(\sqrt{2}) }[/math]. The Galois group of [math]\displaystyle{ K }[/math] over [math]\displaystyle{ \Q }[/math] has order d = 2 and is generated by the element which sends [math]\displaystyle{ \sqrt{2} }[/math] to [math]\displaystyle{ -\sqrt{2} }[/math]. So the norm of [math]\displaystyle{ 1+\sqrt{2} }[/math] is:

[math]\displaystyle{ (1+\sqrt{2})(1-\sqrt{2}) = -1. }[/math]

The field norm can also be obtained without the Galois group. Fix a [math]\displaystyle{ \Q }[/math]-basis of [math]\displaystyle{ \Q(\sqrt{2}) }[/math], say [math]\displaystyle{ \{1,\sqrt{2}\} }[/math]: then multiplication by the number [math]\displaystyle{ 1+\sqrt{2} }[/math] sends 1 to [math]\displaystyle{ 1+\sqrt{2} }[/math] and [math]\displaystyle{ \sqrt{2} }[/math] to [math]\displaystyle{ 2+\sqrt{2} }[/math]. So the determinant of "multiplying by [math]\displaystyle{ 1+\sqrt{2} }[/math]" is the determinant of the matrix which sends the vector [math]\displaystyle{ (1,0)^\mathrm{T} }[/math] (corresponding to the first basis element, i.e. 1) to [math]\displaystyle{ (1,1)^\mathrm{T} }[/math] and the vector [math]\displaystyle{ (0,1)^\mathrm{T} }[/math] (which represents the second basis element [math]\displaystyle{ \sqrt{2} }[/math]) to [math]\displaystyle{ (2,1)^\mathrm{T} }[/math], viz.:

[math]\displaystyle{ \begin{bmatrix}1 & 2 \\1 & 1 \end{bmatrix}. }[/math]

The determinant of this matrix is −1.

Properties of the norm

Several properties of the norm function hold for any finite extension.[3]

The norm NL/K : L* → K* is a group homomorphism from the multiplicative group of L to the multiplicative group of K, that is

[math]\displaystyle{ \operatorname{N}_{L/K}(\alpha \beta) = \operatorname{N}_{L/K}(\alpha) \operatorname{N}_{L/K}(\beta) \text{ for all }\alpha, \beta \in L^* }[/math].

Furthermore, if a in K:

[math]\displaystyle{ \operatorname{N}_{L/K}(a \alpha) = a^{[L:K]} \operatorname{N}_{L/K}(\alpha) \text{ for all }\alpha \in L }[/math].

If aK then [math]\displaystyle{ \operatorname{N}_{L/K}(a) = a^{[L:K]}. }[/math]

Additionally, the norm behaves well in towers of fields: if M is a finite extension of L, then the norm from M to K is just the composition of the norm from M to L with the norm from L to K, i.e.

[math]\displaystyle{ \operatorname{N}_{M/K}=\operatorname{N}_{L/K}\circ\operatorname{N}_{M/L} }[/math].

Finite fields

Let L = GF(qn) be a finite extension of a finite field K = GF(q). Since L/K is a Galois extension, if α is in L, then the norm of α is the product of all the Galois conjugates of α, i.e.[4]

[math]\displaystyle{ \operatorname{N}_{L/K}(\alpha)=\alpha \cdot \alpha^q \cdot \alpha^{q^2} \cdots \alpha^{q^{n-1}} = \alpha^{(q^n - 1)/(q-1)} }[/math].

In this setting we have the additional properties,[5]

  • [math]\displaystyle{ \operatorname{N}_{L/K}(\alpha^q) = \operatorname{N}_{L/K}(\alpha) \text{ for all } \alpha \in L }[/math]
  • [math]\displaystyle{ \text{for any }a \in K, \text{ we have } \operatorname{N}_{L/K}(a) = a^n }[/math].

Further properties

The norm of an algebraic integer is again an integer, because it is equal (up to sign) to the constant term of the characteristic polynomial.

In algebraic number theory one defines also norms for ideals. This is done in such a way that if I is a nonzero ideal of OK, the ring of integers of the number field K, N(I) is the number of residue classes in [math]\displaystyle{ O_K / I }[/math] – i.e. the cardinality of this finite ring. Hence this ideal norm is always a positive integer. When I is a principal ideal αOK then N(I) is equal to the absolute value of the norm to Q of α, for α an algebraic integer.

See also

Notes

References