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Field norm

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Field norm
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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,
m_alphacolon Lto L m_alpha (x) = alpha x,
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.{{harvnb|Rotman|2002|loc=p. 940}}If L/K is a Galois extension, one may compute the norm of αL as the product of all the Galois conjugates of α:
operatorname{N}_{L/K}(alpha)=prod_{sigmainoperatorname{Gal}(L/K)} sigma(alpha),
where Gal(L/K) denotes the Galois group of L/K.{{harvnb|Rotman|2002|loc=p. 943}} (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 σ{{sub|1}}(α), ..., σ{{sub|n}}(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
operatorname{N}_{L/K}(alpha)=left (prod_{j=1}^nsigma_j(alpha) right )^{[L:K(alpha)]}.
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).

Examples

Quadratic field extensions

One of the basic examples of norms comes from quadratic field extensions Q(sqrt{a})/Q where a is a square-free integer.Then, the multiplication map by sqrt{a} on an element x + y cdot sqrt{a} is
sqrt{a}cdot (x + ycdotsqrt{a}) = y cdot a + x cdot sqrt{a}.
The element x + y cdot sqrt{a} can be represented by the vector
begin{bmatrix}x yend{bmatrix},
since there is a direct sum decomposition Q(sqrt{a}) = Qoplus Qcdotsqrt{a} as a Q-vector space.The matrix of m_sqrt{a} is then
m_{sqrt{a}} = begin{bmatrix}
1 & 0end{bmatrix}and the norm is N_{Q(sqrt{a})/Q}(sqrt{a}) = -a, since it is the determinant of this matrix.

Norm of Q(√2)

Consider the number field K=Q(sqrt{2}).The Galois group of K over Q has order d = 2 and is generated by the element which sends sqrt{2} to -sqrt{2}. So the norm of 1+sqrt{2} is:
(1+sqrt{2})(1-sqrt{2}) = -1.
The field norm can also be obtained without the Galois group.Fix a Q-basis of Q(sqrt{2}), say:
{1,sqrt{2}}.
Then multiplication by the number 1+sqrt{2} sends
1 to 1+sqrt{2} and sqrt{2} to 2+sqrt{2}.
So the determinant of "multiplying by 1+sqrt{2}" is the determinant of the matrix which sends the vector
begin{bmatrix}1 0end{bmatrix} (corresponding to the first basis element, i.e., 1) to begin{bmatrix}1 1end{bmatrix}, begin{bmatrix}0 1end{bmatrix} (corresponding to the second basis element, i.e., sqrt{2}) to begin{bmatrix}2 1end{bmatrix},
viz.:
begin{bmatrix}1 & 2 1 & 1 end{bmatrix}.
The determinant of this matrix is −1.

p-th root field extensions

Another easy class of examples comes from field extensions of the form mathbb{Q}(sqrt[p]{a})/mathbb{Q} where the prime factorization of a in mathbb{Q} contains no p-th powers, for p a fixed odd prime.The multiplication map by sqrt[p]{a} of an element isbegin{align}m_{sqrt[p]{a}}(x) &= sqrt[p]{a} cdot (a_0 + a_1sqrt[p]{a} + a_2sqrt[p]{a^2} + cdots + a_{p-1}sqrt[p]{a^{p-1}} )&= a_0sqrt[p]{a} + a_1sqrt[p]{a^2} + a_2sqrt[p]{a^3} + cdots + a_{p-1}a end{align}giving the matrixbegin{bmatrix}1 & 0 & cdots & 0 & 0 vdots & vdots & ddots & vdots & vdots end{bmatrix}The determinant gives the norm
N_{mathbb{Q}(sqrt[p]{a})/mathbb{Q}}(sqrt[p]{a}) = (-1)^{p-1} a = a.

Complex numbers over the reals

The field norm from the complex numbers to the real numbers sends
{{nowrap|x + iy}}
to
{{nowrap|x2 + y2}},
because the Galois group of Complex over R has two elements,
  • the identity element and
  • complex conjugation,
and taking the product yields {{nowrap|1=(x + iy)(x − iy) = x2 + y2}}.

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.{{harvnb|Lidl|Niederreiter|1997|loc=p. 57}}
operatorname{N}_{L/K}(alpha)=alpha cdot alpha^q cdot alpha^{q^2} cdots alpha^{q^{n-1}} = alpha^{(q^n - 1)/(q-1)}.
In this setting we have the additional properties,{{harvnb|Mullen|Panario|2013|loc=p. 21}}
  • forall alpha in L, quad operatorname{N}_{L/K}(alpha^q) = operatorname{N}_{L/K}(alpha)
  • forall a in K, quad operatorname{N}_{L/K}(a) = a^n.

Properties of the norm

Several properties of the norm function hold for any finite extension.{{harvnb|Roman|2006|p=151}}

Group homomorphism

The norm N{{sub|L/K}} : L* → K* is a group homomorphism from the multiplicative group of L to the multiplicative group of K, that is
operatorname{N}_{L/K}(alpha beta) = operatorname{N}_{L/K}(alpha) operatorname{N}_{L/K}(beta) text{ for all }alpha, beta in L^*.
Furthermore, if a in K:
operatorname{N}_{L/K}(a alpha) = a^{[L:K]} operatorname{N}_{L/K}(alpha) text{ for all }alpha in L.
If a ∈ K then operatorname{N}_{L/K}(a) = a^{[L:K]}.

Composition with field extensions

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.
operatorname{N}_{M/K}=operatorname{N}_{L/K}circoperatorname{N}_{M/L}.

Reduction of the norm

The norm of an element in an arbitrary field extension can be reduced to an easier computation if the degree of the field extension is already known. This isN_{L/K}(alpha) = N_{K(alpha)/K}(alpha)^{[L:K(alpha)]}BOOK, Oggier,weblink Introduction to Algebraic Number Theory, 15, 2020-03-28, 2014-10-23,weblink" title="web.archive.org/web/20141023023935weblink">weblink dead, For example, for alpha = sqrt{2} in the field extension L = mathbb{Q}(sqrt{2},zeta_3), K =mathbb{Q}, the norm of alpha isbegin{align}N_{mathbb{Q}(sqrt{2},zeta_3)/mathbb{Q}}(sqrt{2}) &= N_{mathbb{Q}(sqrt{2})/mathbb{Q}}(sqrt{2})^{[mathbb{Q}(sqrt{2},zeta_3):mathbb{Q}(sqrt{2})]}&= (-2)^{2}&= 4end{align}since the degree of the field extension L/K(alpha) is 2.

Detection of units

For mathcal{O}_K the ring of integers of an algebraic number field K, an element alpha in mathcal{O}_K is a unit if and only if N_{K/mathbb{Q}}(alpha) = pm 1.For instance
N_{mathbb{Q}(zeta_3)/mathbb{Q}}(zeta_3) = 1
where
zeta_3^3 = 1.
Thus, any number field K whose ring of integers mathcal{O}_K contains zeta_3 has it as a unit.

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 O'K, the ring of integers of the number field K, N'(I'') is the number of residue classes in O_K / I â€“ 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

{{reflist|3}}

References

  • {{citation | first1=Rudolf | last1=Lidl | first2=Harald | last2=Niederreiter | author2-link=Harald Niederreiter | title=Finite Fields | series=Encyclopedia of Mathematics and its Applications | volume=20 | year=1997 | orig-year=1983 | edition=Second | publisher=Cambridge University Press | isbn=0-521-39231-4 | zbl=0866.11069 | url-access=registration | url=https://archive.org/details/finitefields0000lidl_a8r3 }}
  • {{citation|first1=Gary L.|last1=Mullen|first2=Daniel|last2=Panario|title=Handbook of Finite Fields|year=2013|publisher=CRC Press|isbn=978-1-4398-7378-6}}
  • {{citation | last=Roman | first=Steven | title=Field theory | edition=Second | year=2006 | publisher=Springer | series=Graduate Texts in Mathematics | volume=158 | at=Chapter 8 | isbn=978-0-387-27677-9 | zbl=1172.12001 }}
  • {{citation|first=Joseph J.|last=Rotman|title=Advanced Modern Algebra|year=2002|publisher=Prentice Hall|isbn=978-0-13-087868-7}}


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