Inmathematics, the(field) norm is a particular mapping defined infield theory, which maps elements of a largerfield into asubfield.

LetK be a field andL afiniteextension (and hence analgebraic extension) ofK.

The fieldL is then afinite-dimensionalvector space overK.

Multiplication byα, an element ofL,

${\displaystyle m_{\alpha }:L\to L}$

${\displaystyle m_{\alpha }(x)=\alpha x}$,

is aK-linear transformation of this vector space into itself.

Thenorm,N_L/K(α), is defined as thedeterminant of this linear transformation.^[1]

IfL/K is aGalois extension, one may compute the norm ofα ∈L as the product of all theGalois conjugates ofα:

${\displaystyle {\mathrm{N}}_{L/K}(\alpha )=\prod _{\sigma \in \mathrm{Gal}(L/K)}\sigma (\alpha ),}$

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

For a general field extensionL/K, and nonzeroα inL, letσ₁(α), ..., σ_n(α) be theroots of theminimal polynomial ofα overK (roots listed with multiplicity and lying in some extension field ofL); then

${\displaystyle {\mathrm{N}}_{L/K}(\alpha )=(\prod _{j=1}^n{\sigma }_j(\alpha ){)}^{[L:K(\alpha )]}}$.

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

One of the basic examples of norms comes fromquadratic field extensions${\displaystyle \mathbb{Q}(\sqrt{a})/\mathbb{Q}}$ where${\displaystyle a}$ is asquare-free integer.

Then, the multiplication map by${\displaystyle \sqrt{a}}$ on an element${\displaystyle x+y\cdot \sqrt{a}}$ is

${\displaystyle \sqrt{a}\cdot (x+y\cdot \sqrt{a})=y\cdot a+x\cdot \sqrt{a}.}$

The element${\displaystyle x+y\cdot \sqrt{a}}$ can be represented by the vector

${\displaystyle \left[\begin{array}{c}x\\ y\end{array}\right],}$

since there is a direct sum decomposition${\displaystyle \mathbb{Q}(\sqrt{a})=\mathbb{Q}\oplus \mathbb{Q}\cdot \sqrt{a}}$ as a${\displaystyle \mathbb{Q}}$-vector space.

Thematrix of${\displaystyle m_{\sqrt{a}}}$ is then

and the norm is${\displaystyle N_{\mathbb{Q}(\sqrt{a})/\mathbb{Q}}(\sqrt{a})=-a}$, since it is the determinant of this matrix.

Consider thenumber field${\displaystyle K=\mathbb{Q}(\sqrt{2})}$.

The Galois group of${\displaystyle K}$ over${\displaystyle \mathbb{Q}}$ has order${\displaystyle d=2}$ and is generated by the element which sends${\displaystyle \sqrt{2}}$ to${\displaystyle -\sqrt{2}}$. So the norm of${\displaystyle 1+\sqrt{2}}$ is:

${\displaystyle (1+\sqrt{2})(1-\sqrt{2})=-1.}$

The field norm can also be obtained without the Galois group.

Fix a${\displaystyle \mathbb{Q}}$-basis of${\displaystyle \mathbb{Q}(\sqrt{2})}$, say:

${\displaystyle \{1,\sqrt{2}\}}$.

Then multiplication by the number${\displaystyle 1+\sqrt{2}}$ sends

${\displaystyle 1}$ to${\displaystyle 1+\sqrt{2}}$ and

${\displaystyle \sqrt{2}}$ to${\displaystyle 2+\sqrt{2}}$.

So the determinant of "multiplying by${\displaystyle 1+\sqrt{2}}$" is the determinant of the matrix which sends the vector

${\displaystyle \left[\begin{array}{c}1\\ 0\end{array}\right]}$ (corresponding to the first basis element, i.e., 1) to${\displaystyle \left[\begin{array}{c}1\\ 1\end{array}\right]}$,

${\displaystyle \left[\begin{array}{c}0\\ 1\end{array}\right]}$ (corresponding to the second basis element, i.e.,${\displaystyle \sqrt{2}}$) to${\displaystyle \left[\begin{array}{c}2\\ 1\end{array}\right]}$,

viz.:

The determinant of this matrix is −1.

Another easy class of examples comes from field extensions of the form${\displaystyle \mathbb{Q}(\sqrt[p]{a})/\mathbb{Q}}$ where the prime factorization of${\displaystyle a\in \mathbb{Q}}$ contains no${\displaystyle p}$-th powers, for${\displaystyle p}$ a fixed odd prime.

The multiplication map by${\displaystyle \sqrt[p]{a}}$ of an element is

giving the matrix

The determinant gives the norm

${\displaystyle N_{\mathbb{Q}(\sqrt[p]{a})/\mathbb{Q}}(\sqrt[p]{a})=(-1{)}^{p-1}a=a.}$

The field norm from thecomplex numbers to thereal numbers sends

x +iy

to

x² +y²,

because the Galois group of${\displaystyle \mathbb{C}}$ over${\displaystyle \mathbb{R}}$ has two elements,

• theidentity element and
• complex conjugation,

and taking the product yields(x +iy)(x −iy) =x² +y².

LetL = GF(qⁿ) be a finite extension of afinite fieldK = GF(q).

SinceL/K is a Galois extension, ifα is inL, then the norm ofα is the product of all the Galois conjugates ofα, i.e.^[3]

${\displaystyle {\mathrm{N}}_{L/K}(\alpha )=\alpha \cdot {\alpha }^q\cdot {\alpha }^{q^2}\cdots {\alpha }^{q^{n-1}}={\alpha }^{(q^n-1)/(q-1)}.}$

In this setting we have the additional properties,^[4]

• ${\displaystyle \mathrm{\forall }\alpha \in L,{\mathrm{N}}_{L/K}({\alpha }^q)={\mathrm{N}}_{L/K}(\alpha )}$
• ${\displaystyle \mathrm{\forall }a\in K,{\mathrm{N}}_{L/K}(a)=a^n.}$

Several properties of the norm function hold for any finite extension.^[5][6]

The normN_L/K :L* →K* is agroup homomorphism from the multiplicative group ofL to the multiplicative group ofK, that is

${\displaystyle {\mathrm{N}}_{L/K}(\alpha \beta )={\mathrm{N}}_{L/K}(\alpha ){\mathrm{N}}_{L/K}(\beta )\text{ for all }\alpha ,\beta \in L^{\ast }.}$

Furthermore, if${\displaystyle a\in K}$:

${\displaystyle {\mathrm{N}}_{L/K}(a\alpha )=a^{[L:K]}{\mathrm{N}}_{L/K}(\alpha )\text{ for all }\alpha \in L,}$ and

${\displaystyle {\mathrm{N}}_{L/K}(a)=a^{[L:K]}.}$

Additionally, the norm behaves well intowers of fields:

ifM is a finite extension ofL, then the norm fromM toK is just the composition of the norm fromM toL with the norm fromL toK, i.e.

${\displaystyle {\mathrm{N}}_{M/K}={\mathrm{N}}_{L/K}\circ {\mathrm{N}}_{M/L}.}$

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 is

${\displaystyle N_{L/K}(\alpha )=N_{K(\alpha )/K}(\alpha {)}^{[L:K(\alpha )]}}$^[6]

For example, for${\displaystyle \alpha =\sqrt{2}}$ in the field extension${\displaystyle L=\mathbb{Q}(\sqrt{2},{\zeta }_3),K=\mathbb{Q}}$, the norm of${\displaystyle \alpha }$ is

since the degree of the field extension${\displaystyle L/K(\alpha )}$ is${\displaystyle 2}$.

For${\displaystyle {\mathcal{O}}_K}$ thering of integers of analgebraic number field${\displaystyle K}$, an element${\displaystyle \alpha \in {\mathcal{O}}_K}$ is aunit if and only if${\displaystyle N_{K/\mathbb{Q}}(\alpha )=\pm 1}$.

For instance

${\displaystyle N_{\mathbb{Q}({\zeta }_3)/\mathbb{Q}}({\zeta }_3)=1}$

where

${\displaystyle {\zeta }_3^3=1}$.

Thus, any number field${\displaystyle K}$ whose ring of integers${\displaystyle {\mathcal{O}}_K}$ contains${\displaystyle {\zeta }_3}$ has it as a unit.

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

Inalgebraic number theory one defines also norms forideals. This is done in such a way that ifI is a nonzero ideal ofO_K, the ring of integers of the number fieldK,N(I) is the number of residue classes in${\displaystyle O_K/I}$ – i.e. the cardinality of thisfinite ring. Hence thisideal norm is always a positive integer.

WhenI is aprincipal idealαO_K thenN(I) is equal to theabsolute value of the norm toQ ofα, forα analgebraic integer.

• Field trace
• Ideal norm
• Norm form

• Lidl, Rudolf;Niederreiter, Harald (1997) [1983],Finite Fields, Encyclopedia of Mathematics and its Applications, vol. 20 (Second ed.),Cambridge University Press,ISBN 0-521-39231-4,Zbl 0866.11069
• Mullen, Gary L.; Panario, Daniel (2013),Handbook of Finite Fields, CRC Press,ISBN 978-1-4398-7378-6
• Roman, Steven (2006),Field theory,Graduate Texts in Mathematics, vol. 158 (Second ed.), Springer, Chapter 8,ISBN 978-0-387-27677-9,Zbl 1172.12001
• Rotman, Joseph J. (2002),Advanced Modern Algebra, Prentice Hall,ISBN 978-0-13-087868-7

• Algebraic number theory

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• Short description matches Wikidata