Normal basis

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 ${\displaystyle F\subset K}$ be a Galois extension with Galois group ${\displaystyle G}$. The classical normal basis theorem states that there is an element ${\displaystyle \beta \in K}$ such that ${\displaystyle \{g(\beta ):g\in G\}}$ forms a basis of K, considered as a vector space over F. That is, any element ${\displaystyle \alpha \in K}$ can be written uniquely as ${\textstyle \alpha =\sum _{g\in G}a_{g}\,g(\beta )}$ for some elements ${\displaystyle a_{g}\in F.}$

A normal basis contrasts with a primitive element basis of the form ${\displaystyle \{1,\beta ,\beta ^{2},\ldots ,\beta ^{n-1}\}}$, where ${\displaystyle \beta \in K}$ is an element whose minimal polynomial has degree ${\displaystyle n=[K:F]}$.

Group representation point of view[]

A field extension ${\displaystyle K/F}$ 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. Representations of G over the field F can be viewed as left modules for the group algebra ${\displaystyle F[G]}$. Every homomorphism of left ${\displaystyle F[G]}$-modules ${\displaystyle \phi :F[G]\rightarrow K}$ is of form ${\displaystyle \phi (r)=r\beta }$ for some ${\displaystyle \beta \in K}$. Since ${\displaystyle \{1\cdot \sigma |\sigma \in G\}}$ is a linear basis of ${\displaystyle F[G]}$ over F, it follows easily that ${\displaystyle \phi }$ is bijective iff ${\displaystyle \beta }$ generates a normal basis of K over F. The normal basis theorem therefore amounts to the statement saying that if ${\displaystyle K/F}$ is finite Galois extension, then ${\displaystyle K\cong F[G]}$ as left ${\displaystyle F[G]}$-module. In terms of representations of G over F, this means that K is isomorphic to the regular representation.

Case of finite fields[]

For finite fields this can be stated as follows:^[1] Let ${\displaystyle F=\mathrm {GF} (q)=\mathbb {F} _{q}}$ denote the field of q elements, where q = p^m is a prime power, and let ${\displaystyle K=\mathrm {GF} (q^{n})=\mathbb {F} _{q^{n}}}$ denote its extension field of degree n ≥ 1. Here the Galois group is ${\displaystyle G={\text{Gal}}(K/F)=\{1,\Phi ,\Phi ^{2},\ldots ,\Phi ^{n-1}\}}$ with ${\displaystyle \Phi ^{n}=1,}$ a cyclic group generated by the q-power Frobenius automorphism ${\displaystyle \Phi (\alpha )=\alpha ^{q},}$with ${\displaystyle \Phi ^{n}=1=\mathrm {Id} _{K}.}$ 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 ${\displaystyle \Phi }$ with ${\displaystyle \Phi ^{n}=1,}$ 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 ${\displaystyle \chi (h_{1}h_{2})=\chi (h_{1})\chi (h_{2})}$; then any distinct characters ${\displaystyle \chi _{1},\chi _{2},\ldots }$ are linearly independent in the K-vector space of mappings. We apply this to the Galois group automorphisms ${\displaystyle \chi _{i}=\Phi ^{i}:K\to K,}$ thought of as mappings from the multiplicative group ${\displaystyle H=K^{\times }}$. Now ${\displaystyle K\cong F^{n}}$as an F-vector space, so we may consider ${\displaystyle \Phi :F^{n}\to F^{n}}$ as an element of the matrix algebra ${\displaystyle M_{n}(F);}$ since its powers ${\displaystyle 1,\Phi ,\ldots ,\Phi ^{n-1}}$ are linearly independent (over K and a fortiori over F), its minimal polynomial must have degree at least n, i.e. it must be ${\displaystyle X^{n}-1}$.

The second basic fact is the classification of finitely generated modules over a PID such as ${\displaystyle F[X]}$. Every such module M can be represented as ${\textstyle M\cong \bigoplus _{i=1}^{k}{\frac {F[X]}{(f_{i}(X))}}}$, where ${\displaystyle f_{i}(X)}$ may be chosen so that they are monic polynomials or zero and ${\displaystyle f_{i+1}(X)}$ is a multiple of ${\displaystyle f_{i}(X)}$. ${\displaystyle f_{k}(X)}$ is the monic polynomial of smallest degree annihilating the module, or zero if no such non-zero polynomial exists. In the first case ${\textstyle \dim _{F}M=\sum _{i=1}^{k}\deg f_{i}}$, in the second case ${\displaystyle \dim _{F}M=\infty }$. In our case of cyclic G of size n generated by ${\displaystyle \Phi }$ we have an F-algebra isomorphism ${\textstyle F[G]\cong {\frac {F[X]}{(X^{n}-1)}}}$ where X corresponds to ${\displaystyle 1\cdot \Phi }$, so every ${\displaystyle F[G]}$-module may be viewed as an ${\displaystyle F[X]}$-module with multiplication by X being multiplication by ${\displaystyle 1\cdot \Phi }$. In case of K this means ${\displaystyle X\alpha =\Phi (\alpha )}$, so the monic polynomial of smallest degree annihilating K is the minimal polynomial of ${\displaystyle \Phi }$. Since K is a finite dimensional F-space, the representation above is possible with ${\displaystyle f_{k}(X)=X^{n}-1}$. Since ${\displaystyle \dim _{F}(K)=n,}$ we can only have ${\displaystyle k=1}$, and ${\textstyle K\cong {\frac {F[X]}{(X^{n}{-}\,1)}}}$ as ${\displaystyle F[X]}$-modules. (Note this is an isomorphism of F-linear spaces, but not of rings or F-algebras!) This gives isomorphism of ${\displaystyle F[G]}$-modules ${\displaystyle K\cong F[G]}$ that we talked about above, and under it the basis ${\displaystyle \{1,X,X^{2},\ldots ,X^{n-1}\}}$ on the right side corresponds to a normal basis ${\displaystyle \{\beta ,\Phi (\beta ),\Phi ^{2}(\beta ),\ldots ,\Phi ^{n-1}(\beta )\}}$ of K on the left.

Note that this proof would also apply in the case of a cyclic Kummer extension.

Example[]

Consider the field ${\displaystyle K=\mathrm {GF} (2^{3})=\mathbb {F} _{8}}$ over ${\displaystyle F=\mathrm {GF} (2)=\mathbb {F} _{2}}$, with Frobenius automorphism ${\displaystyle \Phi (\alpha )=\alpha ^{2}}$. 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

${\displaystyle X^{n}-1\ =\ X^{3}-1\ =\ (X{+}1)(X^{2}{+}X{+}1)\ \in \ F[X]}$

means we have a direct sum of F[G]-modules (by the Chinese remainder theorem):

${\displaystyle K\ \cong \ {\frac {F[X]}{(X^{3}{-}\,1)}}\ \cong \ {\frac {F[X]}{(X{+}1)}}\oplus {\frac {F[X]}{(X^{2}{+}X{+}1)}}.}$

The first component is just ${\displaystyle F\subset K}$, while the second is isomorphic as an F[G]-module to ${\displaystyle \mathbb {F} _{2^{2}}\cong \mathbb {F} _{2}[X]/(X^{2}{+}X{+}1)}$ under the action ${\displaystyle \Phi \cdot X^{i}=X^{i+1}.}$ (Thus ${\displaystyle K\cong \mathbb {F} _{2}\oplus \mathbb {F} _{4}}$ as F[G]-modules, but not as F-algebras.)

The elements ${\displaystyle \beta \in K}$ which can be used for a normal basis are precisely those outside either of the submodules, so that ${\displaystyle (\Phi {+}1)(\beta )\neq 0}$ and ${\displaystyle (\Phi ^{2}{+}\Phi {+}1)(\beta )\neq 0}$. In terms of the G-orbits of K, which correspond to the irreducible factors of:

the elements of ${\displaystyle F=\mathbb {F} _{2}}$ are the roots of ${\displaystyle t(t{+}1)}$, the nonzero elements of the submodule ${\displaystyle \mathbb {F} _{4}}$ are the roots of ${\displaystyle t^{3}+t+1}$, while the normal basis, which in this case is unique, is given by the roots of the remaining factor ${\displaystyle t^{3}{+}t^{2}{+}1}$.

By contrast, for the extension field ${\displaystyle L=\mathrm {GF} (2^{4})=\mathbb {F} _{16}}$ in which n = 4 is divisible by p = 2, we have the F[G]-module isomorphism

Here the operator ${\displaystyle \Phi \cong X}$ is not diagonalizable, the module L has nested submodules given by generalized eigenspaces of ${\displaystyle \Phi }$, and the normal basis elements β are those outside the largest proper generalized eigenspace, the elements with ${\displaystyle (\Phi {+}1)^{3}(\beta )\neq 0}$.

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 ${\displaystyle K=\mathrm {GF} (2^{3})=\mathbb {F} _{8}}$ above, we may represent elements as bit-strings:

where the coefficients are bits ${\displaystyle a_{i}\in \mathrm {GF} (2)=\{0,1\}.}$ Now we can square elements by doing a left circular shift, ${\displaystyle \alpha ^{2}=\Phi (a_{2},a_{1},a_{0})=(a_{1},a_{0},a_{2})}$, 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 ${\displaystyle K/F}$ is finite Galois extension of infinite field F. Let ${\displaystyle [K:F]=n}$, ${\displaystyle {\text{Gal}}(K/F)=G=\{\sigma _{1}...\sigma _{n}\}}$, where ${\displaystyle \sigma _{1}={\text{Id}}}$. By primitive element theorem there exist ${\displaystyle \alpha \in K}$ such that ${\displaystyle K=F[\alpha ]}$. Let f be the minimal monic polynomial of ${\displaystyle \alpha }$. Then f is irreducible monic polynomial of degree n over F/ Denote ${\displaystyle \alpha _{i}=\sigma _{i}\alpha }$. Since f is of degree n, we have ${\displaystyle \alpha _{i}\neq \alpha _{j}}$ for ${\displaystyle i\neq j}$. Denote

In other words, we have Note that ${\displaystyle g(\alpha )=1}$ and ${\displaystyle g_{i}(\alpha )=0}$ for ${\displaystyle i\neq 1}$. Next, define ${\displaystyle n\times n}$ matrix A of polynomials over K and polynomial D by Observe that ${\displaystyle A_{ij}(X)=g_{k}(X)}$, where k is determined by ${\displaystyle \sigma _{k}=\sigma _{i}\cdot \sigma _{j}}$, in particular ${\displaystyle k=1}$ iff ${\displaystyle \sigma _{i}=\sigma _{j}^{-1}}$. It follows that ${\displaystyle A(\alpha )}$ is permutation matrix corresponding to the permutation of G which sends each ${\displaystyle \sigma _{i}}$ to ${\displaystyle \sigma _{i}^{-1}}$. (We denote by ${\displaystyle A(\alpha )}$ matrix elements of which are values of elements of ${\displaystyle A(X)}$ at ${\displaystyle \alpha }$.)Therefore, we have ${\displaystyle D(\alpha )=\det A(\alpha )=\pm 1}$. We see that D is a non-zero polynomial, therefore it may have only finite number of roots. Since we assume F is infinite, we can find ${\displaystyle a\in F}$ such that ${\displaystyle D(a)\neq 0}$. Define We claim that ${\displaystyle \{\beta _{1},\ldots ,\beta _{n}\}}$ is a normal basis. We only have to show that ${\displaystyle \beta _{1},\ldots ,\beta _{n}}$ are linearly independent over F, so suppose ${\textstyle \sum _{j=1}^{n}x_{j}\beta _{j}=0}$ for some ${\displaystyle x_{1}...x_{n}\in F}$. Applying automorphism ${\displaystyle \sigma _{i}}$ we get ${\textstyle \sum _{j=1}^{n}x_{j}\sigma _{i}(g_{j}(a))=0}$ for all i. In other words, ${\displaystyle A(a)\cdot {\overline {x}}={\overline {0}}}$. Since ${\displaystyle \det A(a)=D(a)\neq 0}$,we conclude ${\displaystyle {\overline {x}}={\overline {0}}}$, which completes the proof.

Note that we have made use of the fact that ${\displaystyle a\in F}$, so for any F-automorphism ${\displaystyle \sigma }$ and polynomial ${\displaystyle h(X)}$ over ${\displaystyle K}$ value of the polynomial ${\displaystyle \sigma (h(X))}$ at ${\displaystyle a}$ equals ${\displaystyle \sigma (h(a))}$. Therefore, we could not have simply taken ${\displaystyle a=\alpha }$.

Primitive normal basis[]

A primitive normal basis of an extension of finite fields E/F is a normal basis for E/F that is generated by a primitive element of E, that is a generator of the multiplicative group ${\displaystyle K^{\times }}$. (Note that this is a more restrictive definition of primitive element than that mentioned above after the general Normal Basis Theorem: 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 finite field extension possesses a primitive normal basis, the case when F is a prime field having been settled by Harold Davenport.

Free elements[]

If K/F is a Galois extension and x in E 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 K^H, x is free for K/K^H, then x is said to be completely free in K/F. Every Galois extension has a completely free element.^[2]

See also[]

• Dual basis in a field extension
• Polynomial basis
• Zech's logarithm

References[]

• 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. Zbl 0851.00052.
• 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. JSTOR 2007886. Zbl 0615.12023.
• 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. Zbl 0779.11059.