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

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

Group representation point of view

A field extension Template:Nowrap 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 $\varphi :F[G]\to K$ is of form $\varphi (r)=r\beta$ for some $\beta \in K$. Since $\{1\cdot \sigma |\sigma \in G\}$ is a linear basis of F[G] over F, it follows easily that $\varphi$ is bijective iff $\beta$ generates a normal basis of K over F. The normal basis theorem therefore amounts to the statement saying that if Template:Nowrap is finite Galois extension, then $K\cong F[G]$ as a left $F[G]$-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: Let $F=\mathrm{G}\mathrm{F}(q)={𝔽}_q$ denote the field of q elements, where Template:Nowrap is a prime power, and let $K=\mathrm{G}\mathrm{F}(q^n)={𝔽}_{q^n}$ denote its extension field of degree Template:Nowrap. Here the Galois group is $G=\text{Gal}(K/F)=\{1,\mathrm{\Phi },{\mathrm{\Phi }}^2,\dots ,{\mathrm{\Phi }}^{n-1}\}$ with ${\mathrm{\Phi }}^n=1,$ a cyclic group generated by the q-power Frobenius automorphism $\mathrm{\Phi }(\alpha )={\alpha }^q,$with ${\mathrm{\Phi }}^n=1={\mathrm{I}\mathrm{d}}_K.$ Then there exists an element Template:Nowrap such that $$\{\beta ,\mathrm{\Phi }(\beta ),{\mathrm{\Phi }}^2(\beta ),\dots ,{\mathrm{\Phi }}^{n-1}(\beta )\}=\{\beta ,{\beta }^q,{\beta }^{q^2},\dots ,{\beta }^{q^{n-1}}\}$$ is a basis of K over F.

Proof for finite fields

In case the Galois group is cyclic as above, generated by $\mathrm{\Phi }$ with ${\mathrm{\Phi }}^n=1,$ 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 $\chi (h_1{h}_2)=\chi (h_1)\chi (h_2)$; then any distinct characters ${\chi }_1,{\chi }_2,\dots$ are linearly independent in the K-vector space of mappings. We apply this to the Galois group automorphisms ${\chi }_i={\mathrm{\Phi }}^i:K\to K,$ thought of as mappings from the multiplicative group $H=K^{\times }$. Now $K\cong F^n$as an F-vector space, so we may consider $\mathrm{\Phi }:F^n\to F^n$ as an element of the matrix algebra M_n(F); since its powers $1,\mathrm{\Phi },\dots ,{\mathrm{\Phi }}^{n-1}$ are linearly independent (over K and a fortiori over F), its minimal polynomial must have degree at least n, i.e. it must be $X^n-1$.

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

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

Example

Consider the field $K=\mathrm{G}\mathrm{F}(2^3)={𝔽}_8$ over $F=\mathrm{G}\mathrm{F}(2)={𝔽}_2$, with Frobenius automorphism $\mathrm{\Phi }(\alpha )={\alpha }^2$. 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 $$X^n-1=X^3-1=(X-1)(X^2+X+1)\in F[X]$$ means we have a direct sum of F[G]-modules (by the Chinese remainder theorem):$$K\cong \frac{F[X]}{(X^3-1)}\cong \frac{F[X]}{(X+1)}\oplus \frac{F[X]}{(X^2+X+1)}.$$ The first component is just $F\subset K$, while the second is isomorphic as an F[G]-module to ${𝔽}_{2^2}\cong {𝔽}_2[X]/(X^2+X+1)$ under the action $\mathrm{\Phi }\cdot X^i=X^{i+1}.$ (Thus $K\cong {𝔽}_2\oplus {𝔽}_4$ as F[G]-modules, but not as F-algebras.)

The elements $\beta \in K$ which can be used for a normal basis are precisely those outside either of the submodules, so that $(\mathrm{\Phi }+1)(\beta )\ne 0$ and $({\mathrm{\Phi }}^2+\mathrm{\Phi }+1)(\beta )\ne 0$. In terms of the G-orbits of K, which correspond to the irreducible factors of: $$t^{2^3}-t=t(t+1)(t^3+t+1)(t^3+t^2+1)\in F[t],$$ the elements of $F={𝔽}_2$ are the roots of $t(t+1)$, the nonzero elements of the submodule ${𝔽}_4$ are the roots of $t^3+t+1$, while the normal basis, which in this case is unique, is given by the roots of the remaining factor $t^3+t^2+1$.

By contrast, for the extension field $L=\mathrm{G}\mathrm{F}(2^4)={𝔽}_{16}$ in which Template:Nowrap is divisible by Template:Nowrap, we have the F[G]-module isomorphism $$L\cong {𝔽}_2[X]/(X^4-1)={𝔽}_2[X]/(X+1{)}^4.$$ Here the operator $\mathrm{\Phi }\cong X$ is not diagonalizable, the module L has nested submodules given by generalized eigenspaces of $\mathrm{\Phi }$, and the normal basis elements β are those outside the largest proper generalized eigenspace, the elements with $(\mathrm{\Phi }+1{)}^3(\beta )\ne 0$.

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 $K=\mathrm{G}\mathrm{F}(2^3)={𝔽}_8$ above, we may represent elements as bit-strings: $$\alpha =(a_2,a_1,a_0)=a_2{\mathrm{\Phi }}^2(\beta )+a_1\mathrm{\Phi }(\beta )+a_0\beta =a_2{\beta }^4+a_1{\beta }^2+a_0\beta ,$$ where the coefficients are bits $a_i\in \mathrm{G}\mathrm{F}(2)=\{0,1\}.$ Now we can square elements by doing a left circular shift, ${\alpha }^2=\mathrm{\Phi }(a_2,a_1,a_0)=(a_1,a_0,a_2)$, since squaring β⁴ gives Template:Nowrap. This makes the normal basis especially attractive for cryptosystems that utilize frequent squaring.

Proof for the case of infinite fields

Suppose $K/F$ is a finite Galois extension of the infinite field F. Let Template:Nowrap, $\text{Gal}(K/F)=G=\{{\sigma }_1...{\sigma }_n\}$, where ${\sigma }_1=\text{Id}$. By the primitive element theorem there exists $\alpha \in K$ such $i\ne j⟹{\sigma }_i(\alpha )\ne {\sigma }_j(\alpha )$ and $K=F[\alpha ]$. Let us write ${\alpha }_i={\sigma }_i(\alpha )$. $\alpha$'s (monic) minimal polynomial f over K is the irreducible degree n polynomial given by the formula $$\begin{array}{cc}f(X)& =\prod _{i=1}^n(X-{\alpha }_i)\end{array}$$ Since f is separable (it has simple roots) we may define $$\begin{array}{cc}g(X)& =\frac{f(X)}{(X-\alpha )f^{\prime }(\alpha )}\\ g_i(X)& =\frac{f(X)}{(X-{\alpha }_i)f^{\prime }({\alpha }_i)}={\sigma }_i(g(X)).\end{array}$$ In other words, $$\begin{array}{cc}{g}_i(X)& =\prod _{\begin{array}{c}1\le j\le n\\ j\ne i\end{array}}\frac{X-{\alpha }_j}{{\alpha }_i-{\alpha }_j}\\ g(X)& =g_1(X).\end{array}$$ Note that $g(\alpha )=1$ and $g_i(\alpha )=0$ for $i\ne 1$. Next, define an $n\times n$ matrix A of polynomials over K and a polynomial D by $$\begin{array}{cc}{A}_{ij}(X)& ={\sigma }_i({\sigma }_j(g(X))={\sigma }_i(g_j(X))\\ D(X)& =\det A(X).\end{array}$$ Observe that $A_{ij}(X)=g_k(X)$, where k is determined by ${\sigma }_k={\sigma }_i\cdot {\sigma }_j$; in particular $k=1$ iff ${\sigma }_i={\sigma }_j^{-1}$. It follows that $A(\alpha )$ is the permutation matrix corresponding to the permutation of G which sends each ${\sigma }_i$ to ${\sigma }_i^{-1}$. (We denote by $A(\alpha )$ the matrix obtained by evaluating $A(X)$ at $x=\alpha$.) Therefore, $D(\alpha )=\det A(\alpha )=\pm 1$. 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 $a\in F$ such that $D(a)\ne 0$. Define $$\begin{array}{cc}\beta & =g(a)\\ {\beta }_i& =g_i(a)={\sigma }_i(\beta ).\end{array}$$ We claim that $\{{\beta }_1,\dots ,{\beta }_n\}$ is a normal basis. We only have to show that ${\beta }_1,\dots ,{\beta }_n$ are linearly independent over F, so suppose ${\sum }_{j=1}^n{x}_j{\beta }_j=0$ for some $x_1...x_n\in F$. Applying the automorphism ${\sigma }_i$ yields ${\sum }_{j=1}^n{x}_j{\sigma }_i(g_j(a))=0$ for all i. In other words, $A(a)\cdot \bar{x}=\bar{0}$. Since $\det A(a)=D(a)\ne 0$, we conclude that $\bar{x}=\bar{0}$, which completes the proof.

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

Primitive normal basis

A primitive normal basis of an extension of finite fields Template:Nowrap is a normal basis for Template:Nowrap 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 extension of finite fields possesses a primitive normal basis, the case when F is a prime field having been settled by Harold Davenport.

Free elements

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

See also

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

References

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