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 be a Galois extension with Galois group . The classical **normal basis theorem** states that there is an element such that forms a basis of *K*, considered as a vector space over *F*. That is, any element can be written uniquely as for some elements

A normal basis contrasts with a primitive element basis of the form , where is an element whose minimal polynomial has degree .

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 is of form for some . Since is a linear basis of *F*[*G*] over *F*, it follows easily that is bijective iff 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 as left -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 denote the field of *q* elements, where *q* = *p*^{m} is a prime power, and let denote its extension field of degree *n* ≥ 1. Here the Galois group is with a cyclic group generated by the *q*-power Frobenius automorphism with Then there exists an element *β* ∈ *K* such that

is a basis of

In case the Galois group is cyclic as above, generated by with 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 ; then any distinct characters are linearly independent in the *K*-vector space of mappings. We apply this to the Galois group automorphisms thought of as mappings from the multiplicative group . Now as an *F*-vector space, so we may consider as an element of the matrix algebra M_{n}(*F*); since its powers are linearly independent (over *K* and a fortiori over *F*), its minimal polynomial must have degree at least *n*, i.e. it must be .

The second basic fact is the classification of finitely generated modules over a PID such as . Every such module *M* can be represented as , where may be chosen so that they are monic polynomials or zero and is a multiple of . is the monic polynomial of smallest degree annihilating the module, or zero if no such non-zero polynomial exists. In the first case , in the second case . In our case of cyclic *G* of size *n* generated by we have an *F*-algebra isomorphism where *X* corresponds to , so every -module may be viewed as an -module with multiplication by *X* being multiplication by . In case of *K* this means , so the monic polynomial of smallest degree annihilating *K* is the minimal polynomial of . Since *K* is a finite dimensional *F*-space, the representation above is possible with . Since we can only have , and as *F*[*X*]-modules. (Note this is an isomorphism of *F*-linear spaces, but *not* of rings or *F*-algebras.) This gives isomorphism of -modules that we talked about above, and under it the basis on the right side corresponds to a normal basis of *K* on the left.

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

Consider the field over , with Frobenius automorphism . 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

means we have a direct sum of

The first component is just , while the second is isomorphic as an

The elements which can be used for a normal basis are precisely those outside either of the submodules, so that and . In terms of the *G*-orbits of *K*, which correspond to the irreducible factors of:

the elements of are the roots of , the nonzero elements of the submodule are the roots of , while the normal basis, which in this case is unique, is given by the roots of the remaining factor .

By contrast, for the extension field in which *n* = 4 is divisible by *p* = 2, we have the *F*[*G*]-module isomorphism

Here the operator is not diagonalizable, the module

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 above, we may represent elements as bit-strings:

where the coefficients are bits Now we can square elements by doing a left circular shift, , since squaring

Suppose is a finite Galois extension of the infinite field *F*. Let [*K* : *F*] = *n*, , where . By the primitive element theorem there exists such and . Let us write . 's (monic) minimal polynomial *f* over *K* is the irreducible degree *n* polynomial given by the formula

Since

In other words,

Note that and for . Next, define an matrix

Observe that , where

We claim that is a normal basis. We only have to show that are linearly independent over

It is tempting to take because . But this is impermissible because we used the fact that to conclude that for any *F*-automorphism and polynomial over the value of the polynomial at *a* equals .

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]}