Field norm

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

${\displaystyle {\begin{bmatrix}x\\y\end{bmatrix}},}$

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

${\displaystyle m_{\sqrt {a}}={\begin{bmatrix}0&a\\1&0\end{bmatrix}}}$

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

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

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

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

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

viz.:

${\displaystyle {\begin{bmatrix}1&2\\1&1\end{bmatrix}}.}$

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

${\displaystyle {\begin{aligned}m_{\sqrt[{p}]{a}}(x)&={\sqrt[{p}]{a}}\cdot (a_{0}+a_{1}{\sqrt[{p}]{a}}+a_{2}{\sqrt[{p}]{a^{2}}}+\cdots +a_{p-1}{\sqrt[{p}]{a^{p-1}}})\\&=a_{0}{\sqrt[{p}]{a}}+a_{1}{\sqrt[{p}]{a^{2}}}+a_{2}{\sqrt[{p}]{a^{3}}}+\cdots +a_{p-1}a\end{aligned}}}$

giving the matrix

${\displaystyle {\begin{bmatrix}0&0&\cdots &0&a\\1&0&\cdots &0&0\\0&1&\cdots &0&0\\\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&\cdots &1&0\end{bmatrix}}}$

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 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.

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.[3]

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

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

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

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

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

Furthermore, if a in K:

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

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

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.

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

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

${\displaystyle {\begin{aligned}N_{\mathbb {Q} ({\sqrt {2}},\zeta _{3})/\mathbb {Q} }({\sqrt {2}})&=N_{\mathbb {Q} ({\sqrt {2}})/\mathbb {Q} }({\sqrt {2}})^{[\mathbb {Q} ({\sqrt {2}},\zeta _{3}):\mathbb {Q} ({\sqrt {2}})]}\\&=(-2)^{2}\\&=4\end{aligned}}}$

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

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.