Normal basis

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Short description: Mathematical theorem used in cryptography

In mathematics, specifically the algebraic theory of fields, a normal basis is a special kind of basis for Galois extensions of finite degree, characterised as forming a single orbit for the Galois group. The normal basis theorem states that any finite Galois extension of fields has a normal basis. In algebraic number theory, the study of the more refined question of the existence of a normal integral basis is part of Galois module theory.

Normal basis theorem

Let F⊆K be a Galois extension with Galois group G. The classical normal basis theorem states that there is an element β∈K such that {σ(β):σ∈G} forms a basis of K, considered as a vector space over F. That is, any element α∈K can be written uniquely as α=∑σ∈Gaσσ(β) for some coefficients aσ∈F.

A normal basis contrasts with a primitive element basis of the form {1,β,β2,…,βn−1}, where β∈K is an element whose minimal polynomial has degree n=[K:F].

Group representation point of view

A field extension K / F with Galois group G can be naturally viewed as a representation of the group G over the field F in which each automorphism is represented by itself; thus K is also a left module for the group algebra F[G]. Every homomorphism of left F[G]-modules ϕ:F[G]→K is of form ϕ(σ)=σ(β) for some β∈K. Since {σ:σ∈G} is a linear basis of F[G] over F, we see that ϕ is bijective iff β generates a normal basis of K over F.

The normal basis theorem therefore amounts to the statement saying that if K / F is finite Galois extension, then K≅F[G] as a left F[G]-module: K is isomorphic to the regular representation. Also, a given β generates a normal basis iff, when considering K as a G-representation, β does not lie in any proper subrepresentation.

Case of finite fields

For finite fields this can be stated as follows:[1] Let F=GF(q)=𝔽q denote the field of q elements, where q = pm is a prime power, and let K=GF(qn)=𝔽qn denote its extension field of degree n ≥ 1. Here the Galois group is G=Gal(K/F)={1,Φ,Φ2,…,Φn−1} with Φn=1, a cyclic group generated by the q-power Frobenius automorphism Φ(α)=αq,with Φn=1=IdK. Then there exists an element β ∈ K such that {β,Φ(β),Φ2(β),…,Φn−1(β)} = {β,βq,βq2,…,βqn−1} is a basis of K over F.

Proof for finite fields

In case the Galois group is cyclic as above, generated by Φ with Φn=1, the normal basis theorem follows from two basic facts. The first is the linear independence of characters: a multiplicative character is a mapping χ from a group H to a field K satisfying χ(h1h2)=χ(h1)χ(h2) and χ(1)=1; then any distinct characters χ1,χ2,… are linearly independent in the K-vector space of mappings H→K. We apply this to the Galois group automorphisms χi=Φi:K→K, thought of as mappings from the multiplicative group H=K×. Now K≅Fnas an F-vector space, so we may consider Φ:Fn→Fn as an element of Mn(F)⊂Mn(K); since its powers 1,Φ,…,Φn−1 are linearly independent over K in Mn(K) and a fortiori over F in Mn(F), its minimal polynomial must have degree at least n, i.e. it must be Xn−1.

The second basic fact is the classification of modules over a PID such as

F[X]

. Every such module

M

of finite dimension over F can be represented as

M≅⨁i=1kF[X](fi(X))

, where

fi(X)

are monic polynomials and

fi+1(X)

is a multiple of

fi(X)

, so that the highest

fk(X)

is the monic polynomial of smallest degree annihilating the module. For our cyclic Galois group

G

of order n, we have an

F

-algebra isomorphism

F[G]≅F[X](Xn−1)

taking the generator

Φ

to the variable

X

: this makes every

F[G]

-module into an

F[X]

-module. Consider

M=K

as an

F[X]

-module under

Xα=Φ(α)

: the minimal monic polynomial

fk(X)

annihilating

M

is the minimal polynomial of

Φ

, namely

fk(X)=Xn−1

. Since

dimF(M)=n,

we can only have

k=1

, and

K≅F[X](Xn−1)

as

F[X]

-modules (but this is not an isomorphism of rings). Thus we have an isomorphism of

F[G]

-modules

K≅F[X](Xn−1)≅F[G]

,

under which the basis

{1,X,X2,…,Xn−1}

corresponds to the basis

{1,Φ,Φ2,…,Φn−1}

of

F[G]

on the right side, and to a normal basis

{β,Φ(β),Φ2(β),…,Φn−1(β)}

of

K

on the left.

Note that this proof would also apply in characteristic zero for a cyclic Kummer extension.

Example

Consider the field K=GF(23)=𝔽8 over F=GF(2)=𝔽2, with Frobenius automorphism Φ(α)=α2. The proof above clarifies the choice of normal bases in terms of the structure of K as a representation of G (or F[G]-module). The irreducible factorization Xn−1 = X3−1 = (X−1)(X2+X+1) ∈ F[X] means we have a direct sum of F[G]-modules (by the Chinese remainder theorem):K ≅ F[X](X3−1) ≅ F[X](X+1)⊕F[X](X2+X+1). The first component is just F⊂K, while the second is isomorphic as an F[G]-module to 𝔽22≅𝔽2[X]/(X2+X+1) under the action Φ⋅Xi=Xi+1. (Thus K≅𝔽2⊕𝔽4 as F[G]-modules, but not as rings.)

The elements β∈K which can be used for a normal basis are precisely those outside either of the submodules, so that (Φ+1)(β)≠0 and (Φ2+Φ+1)(β)≠0. In terms of the G-orbits of K, which correspond to the irreducible factors in: t8−t = t(t+1)(t3+t+1)(t3+t2+1) ∈ F[t], the elements of the submodule F=𝔽2 are the roots of t(t+1); and the nonzero elements of the submodule 𝔽4 are the roots of t3+t+1; while the normal basis, which in this case is unique, is given by the roots of the remaining factor t3+t2+1.

By contrast, for the extension field L=GF(24)=𝔽16 in which n = 4 is divisible by p = 2, we have the F[G]-module isomorphism L ≅ 𝔽2[X]/(X4−1) = 𝔽2[X]/(X+1)4. Here the operator Φ≅X is not diagonalizable, the module L has nested submodules given by generalized eigenspaces of Φ, and the normal basis elements β are those outside the largest proper generalized eigenspace, the elements with (Φ+1)3(β)≠0.

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 K=GF(23)=𝔽8 above, we may represent elements as bit-strings: α = (a2,a1,a0) = a2Φ2(β)+a1Φ(β)+a0β = a2β4+a1β2+a0β, where the coefficients are bits ai∈GF(2)={0,1}. Now we can square elements by doing a left circular shift, α2=Φ(a2,a1,a0)=(a1,a0,a2), since squaring β4 gives β8 = β. This makes the normal basis especially attractive for cryptosystems that utilize frequent squaring.

Primitive normal basis

A primitive normal basis of an extension of finite fields K/F is a normal basis that is generated by a primitive element of K, that is a generator of the multiplicative group K×. (Note that this is a stronger definition of primitive element than mentioned above: one requires powers of the element to produce every non-zero element of K, not merely a basis.) Lenstra and Schoof (1987) proved that every extension of finite fields possesses a primitive normal basis, the case when F is a prime field having been settled by Harold Davenport.

Proof for the case of infinite fields

Suppose K/F is a finite Galois extension of the infinite field F. Let n=[K:F], G=Gal(K/F)={σ1...σn} with σ1=Id. By the primitive element theorem there exists α∈K such that K=F[α] and σi(α)≠σj(α) for i≠j; and write αi=σi(α). The minimal polynomial f of α over K is:f(X)=∏i=1n(X−αi),a degree n monic polynomial which is irreducible in K[X]. Since f is separable (having simple roots) we may define the Lagrange interpolation polynomials g1(X),…,gn(X) satisfying gi(αi)=1 and gi(αj)=0 for i≠j:g(X)= f(X)(X−α)f′(α) = ∏1≤j≤nj≠iX−αjα−αj,gi(X)= σi(g(X)) = f(X)(X−αi)f′(αi) = ∏1≤j≤nj≠iX−αjαi−αj.Next, define an n×n matrix of polynomials A=(Aij(X)) byAij(X)=σi(σj(g(X))=σi(gj(X)),and let D(X)=det⁡A(X). To see this a non-zero polynomial, observe that Aij(X)=gk(X), where k is determined by σk=σi⋅σj, and k=1 iff σi=σj−1; thus A(α) is the permutation matrix corresponding to the permutation of G which sends each σi to σi−1, and D(α)=det⁡A(α)=±1. Since D(X) is a non-zero polynomial, it has only a finite number of roots; and since we assumed F is infinite, we can find a∈F with D(a)≠0. Define β=g(a),βi=gi(a)=σi(β). We claim that {β1,…,βn} is a normal basis; i.e. that β1,…,βn are linearly independent over F. Suppose ∑j=1ncjβj=0 for some c→=(c1,...,cn)∈Fn; applying any σi gives ∑j=1ncjσi(gj(a))=0, so that A(a)⋅c→=0→. Since det⁡A(a)=D(a)≠0, we conclude that c→=0→, which completes the proof.

It is tempting to take a=α because D(α)≠0, but we need a∈F to conclude that σi(gj(a))=σi(gj(X))|X=σi(a)=σi(gj(X))|X=a is a matrix entry of A(a)=A(X)|x=a.

Free elements

If K / F is a Galois extension and x in K generates a normal basis over F, then x is free in K / F. If x has the property that for every subgroup H of the Galois group G, with fixed field KH, x is free for K / KH, then x is said to be completely free in K / F. Every Galois extension has a completely free element.[2]

See also

References

  1. ↑ Nader H. Bshouty; Gadiel Seroussi (1989), Generalizations of the normal basis theorem of finite fields, p. 1, https://hotcrp-vee2014.cs.technion.ac.il/users/wwwb/cgi-bin/tr-get.cgi/1989/CS/CS0578.pdf ; SIAM J. Discrete Math. 3 (1990), no. 3, 330–337.
  2. ↑ Dirk Hachenberger, Completely free elements, in Cohen & Niederreiter (1996) pp. 97–107 Zbl 0864.11066
  • Cohen, S.; Niederreiter, H., eds (1996). Finite Fields and Applications. Proceedings of the 3rd international conference, Glasgow, UK, July 11–14, 1995. London Mathematical Society Lecture Note Series. 233. Cambridge University Press. ISBN 978-0-521-56736-7. 
  • Lenstra, H.W. Jr; Schoof, R.J. (1987). "Primitive normal bases for finite fields". Mathematics of Computation 48 (177): 217–231. doi:10.2307/2007886. 
  • Menezes, Alfred J., ed (1993). Applications of finite fields. The Kluwer International Series in Engineering and Computer Science. 199. Boston: Kluwer Academic Publishers. ISBN 978-0792392828.