Field norm

Examples

Quadratic field extensions

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

The matrix of ${\displaystyle m_{\sqrt{a}}}$ is then

${\displaystyle m_{\sqrt{a}}=\left[\begin{array}{cc}0& a\\ 1& 0\end{array}\right]}$

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

Norm of Q(√2)

Consider the number 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.:

${\displaystyle \left[\begin{array}{cc}1& 2\\ 1& 1\end{array}\right].}$

The determinant of this matrix is −1.

p-th root field extensions

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

${\displaystyle \begin{array}{cc}{m}_{\sqrt[p]{a}}(x)& =\sqrt[p]{a}\cdot (a_0+a_1\sqrt[p]{a}+a_2\sqrt[p]{a^2}+\cdots +a_{p-1}\sqrt[p]{a^{p-1}})\\ & =a_0\sqrt[p]{a}+a_1\sqrt[p]{a^2}+a_2\sqrt[p]{a^3}+\cdots +a_{p-1}a\end{array}}$

giving the matrix

${\displaystyle \left[\begin{array}{ccccc}0& 0& \cdots & 0& a\\ 1& 0& \cdots & 0& 0\\ 0& 1& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & 1& 0\end{array}\right]}$

The determinant gives the norm

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

Complex numbers over the reals

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

x + iy

to

x² + y²,

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

• the identity element and
• complex conjugation,

and taking the product yields (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.

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

In this setting we have the additional properties,

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

Properties of the norm

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

Group homomorphism

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

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

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.

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

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 is

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

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

${\displaystyle \begin{array}{cc}{N}_{\mathbb{Q}(\sqrt{2},{\zeta }_3)/\mathbb{Q}}(\sqrt{2})& =N_{\mathbb{Q}(\sqrt{2})/\mathbb{Q}}(\sqrt{2}{)}^{[\mathbb{Q}(\sqrt{2},{\zeta }_3):\mathbb{Q}(\sqrt{2})]}\\ & =(-2{)}^2\\ & =4\end{array}}$

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

Detection of units

For ${\displaystyle {\mathcal{O}}_K}$ the ring of integers of an algebraic number field ${\displaystyle K}$, an element ${\displaystyle \alpha \in {\mathcal{O}}_K}$ is a unit 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.

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 ${\displaystyle O_K/I}$ – 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.