In mathematics, the (field) norm is a particular mapping defined in field theory, which maps elements of a larger field into a subfield.
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\alpha\colonL\toL
m\alpha(x)=\alphax
The norm, NL/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 Galois conjugates of α:
\operatorname{N}L/K(\alpha)=\prod\sigma\in\operatorname{Gal(L/K)}\sigma(\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
\operatorname{N}L/K(\alpha)=\left
n\sigma | |
(\prod | |
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).
One of the basic examples of norms comes from quadratic field extensions
\Q(\sqrt{a})/\Q
a
Then, the multiplication map by
\sqrt{a}
x+y ⋅ \sqrt{a}
\sqrt{a} ⋅ (x+y ⋅ \sqrt{a})=y ⋅ a+x ⋅ \sqrt{a}.
x+y ⋅ \sqrt{a}
\begin{bmatrix}x\ y\end{bmatrix},
\Q(\sqrt{a})=\Q ⊕ \Q ⋅ \sqrt{a}
\Q
The matrix of
m\sqrt{a}
m\sqrt{a
N\Q(\sqrt{a)/\Q}(\sqrt{a})=-a
K=\Q(\sqrt{2})
The Galois group of
K
\Q
d=2
\sqrt{2}
-\sqrt{2}
1+\sqrt{2}
(1+\sqrt{2})(1-\sqrt{2})=-1.
The field norm can also be obtained without the Galois group.
Fix a
\Q
\Q(\sqrt{2})
\{1,\sqrt{2}\}
Then multiplication by the number
1+\sqrt{2}
1 to
1+\sqrt{2}
\sqrt{2}
2+\sqrt{2}
So the determinant of "multiplying by
1+\sqrt{2}
\begin{bmatrix}1\ 0\end{bmatrix}
\begin{bmatrix}1\ 1\end{bmatrix}
\begin{bmatrix}0\ 1\end{bmatrix}
\sqrt{2}
\begin{bmatrix}2\ 1\end{bmatrix}
\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
Q(\sqrt[p]{a})/Q
a\inQ
p
p
The multiplication map by
\sqrt[p]{a}
giving the matrix}(x) &= \sqrt[p] \cdot (a_0 + a_1\sqrt[p] + a_2\sqrt[p] + \cdots + a_\sqrt[p])\\&= a_0\sqrt[p] + a_1\sqrt[p] + a_2\sqrt[p] + \cdots + a_a \end\begin{align} m\sqrt[p]{a
The determinant gives the norm\begin{bmatrix} 0&0& … &0&a\\ 1&0& … &0&0\\ 0&1& … &0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0& … &1&0 \end{bmatrix}
NQ(\sqrt[p]{a)/Q
The field norm from the complex numbers to the real numbers sends
to
,
because the Galois group of
\Complex
\R
and taking the product yields .
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.
\operatorname{N}L/K(\alpha)=\alpha ⋅ \alphaq ⋅
q2 | |
\alpha |
…
qn-1 | |
\alpha |
=
(qn-1)/(q-1) | |
\alpha |
.
In this setting we have the additional properties,
\forall\alpha\inL, \operatorname{N}L/K(\alphaq)=\operatorname{N}L/K(\alpha)
\foralla\inK, \operatorname{N}L/K(a)=an.
Several properties of the norm function hold for any finite extension.
The norm N : 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)forall\alpha,\beta\inL*.
\operatorname{N}L/K(a\alpha)=a[L:K]\operatorname{N}L/K(\alpha)forall\alpha\inL.
If a ∈ K then
\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.
\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
For example, for[1]NL/K(\alpha)=NK(\alpha)/K(\alpha)[L:K(\alpha)]
\alpha=\sqrt{2}
L=Q(\sqrt{2},\zeta3),K=Q
\alpha
since the degree of the field extension}(\sqrt) &= N_(\sqrt)^\\&= (-2)^\\&= 4\end\begin{align} NQ(\sqrt{2,\zeta3)/Q
L/K(\alpha)
2
For
l{O}K
K
\alpha\inl{O}K
NK/Q(\alpha)=\pm1
For instance
N | |
Q(\zeta3)/Q |
(\zeta3)=1
3 | |
\zeta | |
3 |
=1
Thus, any number field
K
l{O}K
\zeta3
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 OK, the ring of integers of the number field K, N(I) is the number of residue classes in
OK/I
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.