Normal basis theorem
Let
be a Galois extension with Galois group
. The classical normal basis theorem states that there is an element
such that
forms a basis of
, considered as a vector space over
. That is, any element
can be written uniquely as
for some coefficients
.
A normal basis contrasts with a primitive element basis of the form
, where
is an element whose minimal polynomial has degree
.
Case of finite fields
For finite fields this can be stated as follows: Let
denote the field of q elements, where q = pm is a prime power, and let
denote its extension field of degree n ≥ 1. Here the Galois group is
with
a cyclic group generated by the q-power Frobenius automorphism
with
Then there exists an element β ∈ K such that
is a basis of K over F.
Proof for finite fields
In case the Galois group is cyclic as above, generated by
with
the normal basis theorem follows from two basic facts. The first is the linear independence of characters: a multiplicative character is a mapping
from a group
to a field
satisfying
and
; then any distinct characters
are linearly independent in the
-vector space of mappings
. We apply this to the Galois group automorphisms
thought of as mappings from the multiplicative group
. Now
as an F-vector space, so we may consider
as an element of
; since its powers
are linearly independent over
in
and a fortiori over
in
, its minimal polynomial must have degree at least n, i.e. it must be
.
The second basic fact is the classification of modules over a PID such as
. Every such module
of finite dimension over F can be represented as
, where
are monic polynomials and
is a multiple of
, so that the highest
is the monic polynomial of smallest degree annihilating the module. For our cyclic Galois group
of order n, we have an
-algebra isomorphism
taking the generator
to the variable
: this makes every
-module into an
-module. Consider
as an
-module under
: the minimal monic polynomial
annihilating
is the minimal polynomial of
, namely
. Since
we can only have
, and
as
-modules (but this is not an isomorphism of rings). Thus we have an isomorphism of
-modules
,
under which the basis
corresponds to the basis
of
on the right side, and to a normal basis
of
on the left.
Note that this proof would also apply in characteristic zero for a cyclic Kummer extension.
Example
Consider the field
over
, with Frobenius automorphism
. The proof above clarifies the choice of normal bases in terms of the structure of K as a representation of G (or
-module). The irreducible factorization
means we have a direct sum of F[G]-modules (by the Chinese remainder theorem):
The first component is just
, while the second is isomorphic as an
-module to
under the action
(Thus
as
-modules, but not as rings.)
The elements
which can be used for a normal basis are precisely those outside either of the submodules, so that
and
. In terms of the
-orbits of
, which correspond to the irreducible factors in:
the elements of the submodule
are the roots of
; and the nonzero elements of the submodule
are the roots of
; while the normal basis, which in this case is unique, is given by the roots of the remaining factor
.
By contrast, for the extension field
in which n = 4 is divisible by p = 2, we have the
-module isomorphism
Here the operator
is not diagonalizable, the module
has nested submodules given by generalized eigenspaces of
, and the normal basis elements β are those outside the largest proper generalized eigenspace, the elements with
.
Application to cryptography
The normal basis is frequently used in cryptographic applications based on the discrete logarithm problem, such as elliptic curve cryptography, since arithmetic using a normal basis is typically more computationally efficient than using other bases.
For example, in the field
above, we may represent elements as bit-strings:
where the coefficients are bits
Now we can square elements by doing a left circular shift,
, since squaring β4 gives β8 = β. This makes the normal basis especially attractive for cryptosystems that utilize frequent squaring.
Proof for the case of infinite fields
Suppose
is a finite Galois extension of the infinite field F. Let
,
with
. By the primitive element theorem there exists
such that
and
for
; and write
. The minimal polynomial f of
over K is:
a degree n monic polynomial which is irreducible in
.
Since f is separable (having simple roots) we may define the Lagrange interpolation polynomials
satisfying
and
for
:
Next, define an
matrix of polynomials
by
and let
. To see this a non-zero polynomial, observe that
, where k is determined by
, and
iff
; thus
is the permutation matrix corresponding to the permutation of G which sends each
to
, and
. Since
is a non-zero polynomial, it has only a finite number of roots; and since we assumed F is infinite, we can find
with
. Define
We claim that
is a normal basis; i.e. that
are linearly independent over F. Suppose
for some
; applying any
gives
, so that
. Since
, we conclude that
, which completes the proof.
It is tempting to take
because
, but we need
to conclude that
is a matrix entry of
.