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Normal basis

Summary

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 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 F[G]. Every homomorphism of left 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 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 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 = pm 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 βK such that ${\displaystyle \{\beta ,\Phi (\beta ),\Phi ^{2}(\beta ),\ldots ,\Phi ^{n-1}(\beta )\}\ =\ \{\beta ,\beta ^{q},\beta ^{q^{2}},\ldots ,\beta ^{q^{n-1}}\!\}}$  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 Mn(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 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: ${\displaystyle t^{2^{3}}-t\ =\ t(t{+}1)\left(t^{3}+t+1\right)\left(t^{3}+t^{2}+1\right)\ \in \ F[t],}$  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 ${\displaystyle L\ \cong \ \mathbb {F} _{2}[X]/(X^{4}{-}1)\ =\ \mathbb {F} _{2}[X]/(X{+}1)^{4}.}$  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: ${\displaystyle \alpha \ =\ (a_{2},a_{1},a_{0})\ =\ a_{2}\Phi ^{2}(\beta )+a_{1}\Phi (\beta )+a_{0}\beta \ =\ a_{2}\beta ^{4}+a_{1}\beta ^{2}+a_{0}\beta ,}$  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 a finite Galois extension of the infinite field F. Let [K : F] = n, ${\displaystyle {\text{Gal}}(K/F)=G=\{\sigma _{1}...\sigma _{n}\}}$ , where ${\displaystyle \sigma _{1}={\text{Id}}}$ . By the primitive element theorem there exists ${\displaystyle \alpha \in K}$  such ${\displaystyle i\neq j\implies \sigma _{i}(\alpha )\neq \sigma _{j}(\alpha )}$  and ${\displaystyle K=F[\alpha ]}$ . Let us write ${\displaystyle \alpha _{i}=\sigma _{i}(\alpha )}$ . ${\displaystyle \alpha }$ 's (monic) minimal polynomial f over K is the irreducible degree n polynomial given by the formula {\displaystyle {\begin{aligned}f(X)&=\prod _{i=1}^{n}(X-\alpha _{i})\end{aligned}}}  Since f is separable (it has simple roots) we may define {\displaystyle {\begin{aligned}g(X)&=\ {\frac {f(X)}{(X-\alpha )f'(\alpha )}}\\g_{i}(X)&=\ {\frac {f(X)}{(X-\alpha _{i})f'(\alpha _{i})}}=\ \sigma _{i}(g(X)).\end{aligned}}}  In other words, {\displaystyle {\begin{aligned}g_{i}(X)&=\prod _{\begin{array}{c}1\leq j\leq n\\j\neq i\end{array}}{\frac {X-\alpha _{j}}{\alpha _{i}-\alpha _{j}}}\\g(X)&=g_{1}(X).\end{aligned}}}  Note that ${\displaystyle g(\alpha )=1}$  and ${\displaystyle g_{i}(\alpha )=0}$  for ${\displaystyle i\neq 1}$ . Next, define an ${\displaystyle n\times n}$  matrix A of polynomials over K and a polynomial D by {\displaystyle {\begin{aligned}A_{ij}(X)&=\sigma _{i}(\sigma _{j}(g(X))=\sigma _{i}(g_{j}(X))\\D(X)&=\det A(X).\end{aligned}}}  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 the permutation matrix corresponding to the permutation of G which sends each ${\displaystyle \sigma _{i}}$  to ${\displaystyle \sigma _{i}^{-1}}$ . (We denote by ${\displaystyle A(\alpha )}$  the matrix obtained by evaluating ${\displaystyle A(X)}$  at ${\displaystyle x=\alpha }$ .) Therefore, ${\displaystyle D(\alpha )=\det A(\alpha )=\pm 1}$ . We see that D is a non-zero polynomial, and therefore it has only a finite number of roots. Since we assumed F is infinite, we can find ${\displaystyle a\in F}$  such that ${\displaystyle D(a)\neq 0}$ . Define {\displaystyle {\begin{aligned}\beta &=g(a)\\\beta _{i}&=g_{i}(a)=\sigma _{i}(\beta ).\end{aligned}}}  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 the automorphism ${\displaystyle \sigma _{i}}$  yields ${\textstyle \sum _{j=1}^{n}x_{j}\sigma _{i}(g_{j}(a))=0}$  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 that ${\displaystyle {\overline {x}}={\overline {0}}}$ , which completes the proof.

It is tempting to take ${\displaystyle a=\alpha }$  because ${\displaystyle D(\alpha )\neq 0}$ . But this is impermissible because we used the fact that ${\displaystyle a\in F}$  to conclude that for any F-automorphism ${\displaystyle \sigma }$  and polynomial ${\displaystyle h(X)}$  over ${\displaystyle K}$  the value of the polynomial ${\displaystyle \sigma (h(X))}$  at a equals ${\displaystyle \sigma (h(a))}$ .

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 K×. (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 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]