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.

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 ${\displaystyle n=[K:F]}$.

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.

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

$${\displaystyle \{\beta ,\Phi (\beta ),\Phi ^{2}(\beta ),\ldots ,\Phi ^{n-1}(\beta )\}\ =\ \{\beta ,\beta ^{q},\beta ^{q^{2}},\ldots ,\beta ^{q^{n-1}}\!\}}$$

is a basis of K over F.

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.

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 ${\displaystyle \sigma (h(a))}$.

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.

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 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]

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