Finite Fields

Size of Finite Fields

Lemma

The characteristic of any field is either zero or a prime.

Proof Assume a field has characteristic neither zero nor a prime, say . Then , where is the multiplicative identity of and is the additive identity of . Let be the smallest prime divisor of . Then, we have , which contradicts the assumption that the characteristic of is neither zero nor a prime.

Lemma

A finite field must have characteristic for some prime and its order is for some .

Proof Let be a finite field. The characteristic of a finite field cannot be zero, so it must be a prime . Hence, contains the finite subfield . Note that is finite, thus a finite-dimensional vector space over . Let be the dimension of as a vector space over . Then, the order of is .

e.g. Consider . Suppose is a root of in . Then, we have . And is also a root of .

Warning

Do not confuse the field with the ring , which isn’t a field. Indeed, is a field if and only if is prime. (See proposition)

Properties of Finite Fields & The Frobenius Automorphism

Frobenius Automorphism

The Frobenius automorphism of a field of characteristic is the map defined by for all .

Note that this is NOT a well-defined homomorphism for rings. But it is a well-defined homomorphism for finite fields, which we will show next.

Proof First, we show that is a homomorphism. Let , then we have where the second equality follows from the binomial theorem, all coefficients in the middle are divisible by , thus vanish. Multiplication is preserved trivially: Thus, is a homomorphism. Next, we show that is bijective. Since is a homomorphism, it is injective, thus automatically surjective by the proposition.

Theorem

Let be a prime integer, and let be a positive power of . Then

1. The elements of a order field are exactly fixed points of , where is the corresponding Frobenius automorphism. In other words, is the smallest integer such that .
2. Any finite field of order is isomorphic to for some polynomial . Equivalently, is a finite extension of , and we can write , where is a root of an irreducible polynomial of degree .
3. Let be a field of order . The multiplicative group of nonzero elements of is a cyclic group of order .
4. There exists a field of order , and all fields of order are isomorphic.
5. If is a field of order and is a field of order , then there is a homomorphism if and only if . In this case, it is an isomorphism if and only if .

Proof We shall prove each statement one by one:

1. Let be a field of order . Note that the group has order . Therefore the order of any element divides by the Lagrange’s Theorem, so , which means . Observe that is also a fixed point of , so we have for all . We shall show the “in other words” part in (3).
3. Since is a finite Abelian group, by the structure theorem, we can write where and . So is the exponent of (i.e. the smallest positive integer such that ), and for all . That is, the polynomial has roots in , so its degree, has to be greater than or equal to . On the other hand, by (1), we know that for all , so (since is the exponent), which means , thus . Therefore, is cyclic.
4. We know that the elements of a field of order will be roots of the polynomial . There exists a field extension of in which this polynomial splits completely by the proposition. We claim that the set of all its roots in is a field of order . To see this, we need to check that has no multiple roots, as well as this set is indeed a field. Note that the derivative of is in a field of characteristic . Thus, the polynomial has no multiple roots. Moreover, clearly, and are roots of . Let and be the roots of , then we have which means is also a root of . So the set of roots of is closed under addition. The same holds for multiplication clearly, so the set of roots of is indeed a field. We now show that two fields and of order must be isomorphic. Let and , where and are roots of some irreducible polynomials of degree respectively. We may assume that and both and are not linear, otherwise naturally. Note that is also a root of , so . Since splits completely in , there is some element that is also a root of , so the map that will be an injective homomorphism. Thus it is an isomorphism since and are finite fields of the same size.
5. Based on (4), we can assume without loss of generality that and are field extensions of , and it suffices to show that is a subfield of iff . If is a subfield, then by multiplicative property, we have , so , which means . Conversely, if , then we can write for some integer . For any element , is a root of in , so it satisfies , thus . Therefore, .