Field norm

 ( Algebraic Number Theory ) 

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

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Formal definition Let K be a field and L a finite field 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,

$m\_\backslash alpha\backslash colon\; L\backslash to\; L$

$m\_\backslash alpha\; (x)\; =\; \backslash alpha\; x$,

is a K-linear transformation of this vector space into itself.

The norm, N _{L/ K}( α), is defined as the determinant of this linear transformation.

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

$\backslash operatorname\{N\}\_\{L/K\}(\backslash alpha)\; =\; \backslash !\; \backslash prod\_\{\backslash sigma\backslash in\backslash operatorname\{Gal\}(L/K)\}\; \backslash !\backslash !\; \backslash sigma(\backslash alpha),$

where Gal( L/ K) denotes the Galois group of L/ K. (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( α), ..., σ( 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

$\backslash operatorname\{N\}\_\{L/K\}(\backslash alpha)=\backslash biggl(\backslash prod\_\{j=1\}^n\backslash sigma\_j(\backslash alpha)\; \backslash biggr)^\{L:K(\backslash alpha)\}$.

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).

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Examples

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Quadratic field extensions One of the basic examples of norms comes from quadratic field extensions $\backslash Q(\backslash sqrt\{a\})/\backslash Q$ where $a$ is a square-free integer.

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

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

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

$\backslash begin\{bmatrix\}x\; \backslash \backslash \; y\backslash end\{bmatrix\},$

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

The matrix of $m\_\backslash sqrt\{a\}$ is then

$m\_\{\backslash sqrt\{a\}\}\; =\; \backslash begin\{bmatrix\}$

0 & a \\ 1 & 0 \end{bmatrix} and the norm is $N\_\{\backslash Q(\backslash sqrt\{a\})/\backslash Q\}(\backslash sqrt\{a\})\; =\; -a$, since it is the determinant of this matrix.

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Norm of Q(√2) Consider the number field $K=\backslash Q(\backslash sqrt\{2\})$.

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

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

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

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

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

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

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

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

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

$\backslash begin\{bmatrix\}1\; \backslash \backslash \; 0\backslash end\{bmatrix\}$ (corresponding to the first basis element, i.e., 1) to $\backslash begin\{bmatrix\}1\; \backslash \backslash \; 1\backslash end\{bmatrix\}$,

$\backslash begin\{bmatrix\}0\; \backslash \backslash \; 1\backslash end\{bmatrix\}$ (corresponding to the second basis element, i.e., $\backslash sqrt\{2\}$) to $\backslash begin\{bmatrix\}2\; \backslash \backslash \; 1\backslash end\{bmatrix\}$,

viz.:

The determinant of this matrix is −1.

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

The multiplication map by $\backslash sqrtp\{a\}$ of an element is

giving the matrix

The determinant gives the norm

$N\_\{\backslash mathbb\{Q\}(\backslash sqrtp\{a\})/\backslash mathbb\{Q\}\}(\backslash sqrtp\{a\})\; =\; (-1)^\{p-1\}\; a\; =\; a.$

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Complex numbers over the reals The field norm from the to the sends

to

,

because the Galois group of $\backslash Complex$ over $\backslash R$ has two elements,

• the identity element and
• complex conjugation,

and taking the product yields .

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Finite fields Let L = GF( q ⁿ) 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.

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

In this setting we have the additional properties,

• $\backslash forall\; \backslash alpha\; \backslash in\; L,\; \backslash quad\; \backslash operatorname\{N\}\_\{L/K\}(\backslash alpha^q)\; =\; \backslash operatorname\{N\}\_\{L/K\}(\backslash alpha)$
• $\backslash forall\; a\; \backslash in\; K,\; \backslash quad\; \backslash operatorname\{N\}\_\{L/K\}(a)\; =\; a^n.$

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Properties of the norm Several properties of the norm function hold for any finite extension.

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Group homomorphism

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

$\backslash operatorname\{N\}\_\{L/K\}(\backslash alpha\; \backslash beta)\; =\; \backslash operatorname\{N\}\_\{L/K\}(\backslash alpha)\; \backslash operatorname\{N\}\_\{L/K\}(\backslash beta)\; \backslash text\{\; for\; all\; \}\backslash alpha,\; \backslash beta\; \backslash in\; L^*.$

Furthermore, if a in K:

$\backslash operatorname\{N\}\_\{L/K\}(a\; \backslash alpha)\; =\; a^\{L:K\}\; \backslash operatorname\{N\}\_\{L/K\}(\backslash alpha)\; \backslash text\{\; for\; all\; \}\backslash alpha\; \backslash in\; L.$

If a ∈ K then $\backslash operatorname\{N\}\_\{L/K\}(a)\; =\; a^\{L:K\}.$

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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.

$\backslash operatorname\{N\}\_\{M/K\}=\backslash operatorname\{N\}\_\{L/K\}\backslash circ\backslash operatorname\{N\}\_\{M/L\}.$

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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 is

$N\_\{L/K\}(\backslash alpha)\; =\; N\_\{K(\backslash alpha)/K\}(\backslash alpha)^\{L:K(\backslash alpha)\}$

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

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

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Detection of units

For $\backslash mathcal\{O\}\_K$ the ring of integers of an algebraic number field $K$, an element $\backslash alpha\; \backslash in\; \backslash mathcal\{O\}\_K$ is a unit if and only if $N\_\{K/\backslash mathbb\{Q\}\}(\backslash alpha)\; =\; \backslash pm\; 1$.

For instance

$N\_\{\backslash mathbb\{Q\}(\backslash zeta\_3)/\backslash mathbb\{Q\}\}(\backslash zeta\_3)\; =\; 1$

where

$\backslash zeta\_3^3\; =\; 1$.

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

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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 $O\_K\; /\; I$ – i.e. the cardinality of this finite ring. Hence this ideal norm is always a positive integer.

When I is a principal ideal αO_K then N( I) is equal to the absolute value of the norm to Q of α, for α an algebraic integer.

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See also

• Field trace
• Ideal norm
• Norm form

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Notes

Categories: Algebraic Number Theory

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