A finite field (also called a Galois field) is a field that has finitely many elements. The number of elements in a finite field is sometimes called the order of the field. We will present some basic facts about finite fields.
1 Size of a finite field
The characteristic of is positive because otherwise the additive subgroup generated by would be an infinite subset of . Accordingly, the prime subfield of is isomorphic to the field of integers mod . The integer is prime since otherwise would have zero divisors. Since the field is an –dimensional vector space over for some finite , it is set–isomorphic to and thus has cardinality . ∎
2 Existence of finite fields
Now that we know every finite field has elements, it is natural to ask which of these actually arise as cardinalities of finite fields. It turns out that for each prime and each natural number , there is essentially exactly one finite field of size .
In any field with elements, the equation is satisfied by all elements of .
For each prime and each natural number , there exists a finite field of cardinality , and any two such are isomorphic.
For , the finite field has elements, and any two such are isomorphic by the map sending to .
In general, the polynomial has derivative and thus is separable over . We claim that the splitting field of this polynomial is a finite field of size . The field certainly contains the set of roots of . However, the set is closed under the field operations, so is itself a field. Since splitting fields are minimal by definition, the containment means that . Finally, has elements since is separable, so is a field of size .
For the uniqueness part, any other field of size contains a subfield isomorphic to . Moreover, equals the splitting field of the polynomial over , since by Lemma 2.1 every element of is a root of this polynomial, and all possible roots of the polynomial are accounted for in this way. By the uniqueness of splitting fields up to isomorphism, the two fields and are isomorphic. ∎
Note: The proof of Theorem 2.2 given here, while standard because of its efficiency, relies on more abstract algebra than is strictly necessary. The reader may find a more concrete presentation of this and many other results about finite fields in [1, Ch. 7].
Every finite field is a normal extension of its prime subfield .
This follows from the fact that field extensions obtained from splitting fields are normal extensions. ∎
3 Units in a finite field
Henceforth, in light of Theorem 2.2, we will write for the unique (up to isomorphism) finite field of cardinality . A fundamental step in the investigation of finite fields is the observation that their multiplicative groups are cyclic:
The multiplicative group consisting of nonzero elements of the finite field is a cyclic group.
We begin with the formula
where denotes the Euler totient function. It is proved as follows. For every divisor of , the cyclic group of size has exactly one cyclic subgroup of size . Let be the subset of consisting of elements of which have the maximum possible order (http://planetmath.org/OrderGroup) of . Since every element of has maximal order in the subgroup of that it generates, we see that the sets partition the set , so that
We now prove the theorem. Let , and for each divisor of , let be the number of elements of of order . We claim that is either zero or . Indeed, if it is nonzero, then let be an element of order , and let be the subgroup of generated by . Then has size and every element of is a root of the polynomial . But this polynomial cannot have more than roots in a field, so every root of must be an element of . In particular, every element of order must be in already, and we see that only has elements of order .
We have proved that for all . If were 0, then we would have
which is impossible since the first sum must equal (because every element of has order equal to some divisor of ). ∎
4 Automorphisms of a finite field
Observe that, since a splitting field for over contains all the roots of , it follows that the field contains a subfield isomorphic to . We will show later (Theorem 4.2) that this is the only way that extensions of finite fields can arise. For now we will construct the Galois group of the field extension , which is normal by Corollary 2.3.
The Galois group of the field extension is a cyclic group of size generated by the power Frobenius map .
The fact that is an element of , and that is the identity on , is obvious. Since the extension is normal and of degree , the group must have size , and we will be done if we can show that , for , are distinct elements of .
It is enough to show that none of , for , is the identity map on , for then we will have shown that is of order exactly equal to . But, if any such were the identity map, then the polynomial would have distinct roots in , which is impossible in a field since . ∎
We can now use the Galois correspondence between subgroups of the Galois group and intermediate fields of a field extension to immediately classify all the intermediate fields in the extension .
The field extension contains exactly one intermediate field isomorphic to , for each divisor of , and no others. In particular, the subfields of are precisely the fields for .
By the fundamental theorem of Galois theory, each intermediate field of corresponds to a subgroup of . The latter is a cyclic group of order , so its subgroups are exactly the cyclic groups generated by , one for each . The fixed field of is the set of roots of , which forms a subfield of isomorphic to , so the result follows.
The subfields of can be obtained by applying the above considerations to the extension . ∎
|Date of creation||2013-03-22 12:37:50|
|Last modified on||2013-03-22 12:37:50|
|Last modified by||yark (2760)|