Finite field

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In abstract algebra, a finite field or Galois field (so named in honor of Évariste Galois) is a field that contains only finitely many elements. Finite fields are important in number theory, algebraic geometry, Galois theory, cryptography, and coding theory. The finite fields are completely known.

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[edit] Classification

There is a unique field of order pn for every prime p and every positive integer n, up to isomorphism.

In detail, the finite fields are classified as follows (Jacobson 1985, p. 287):

  • The order, or number of elements, of a finite field is of the form pn, where p is a prime number called the characteristic of the field, and n is a positive integer.
  • For every prime number p and positive integer n, there exists a finite field with pn elements.
  • Any two finite fields with the same number of elements are isomorphic. That is, under some renaming of the elements of one of these, both its addition and multiplication tables become identical to the corresponding tables of the other one.

This classification justifies using a naming scheme for finite fields that specifies only the order of the field. One notation for a finite field is Fpn. Another notation is GF(pn), where the letters "GF" stand for "Galois field".

[edit] Examples

First we consider fields where the size is prime, i.e., n = 1. An example of such a finite field is the ring Z/pZ. It is a finite field with p elements, usually labelled 0, 1, 2, …, p−1, where arithmetic is performed modulo p. It is also sometimes denoted Zp, but within some areas of mathematics, particularly number theory, this may cause confusion because the same notation Zp is used for the ring of p-adic integers.

Two isomorphic constructions of the field with 4 elements are (Z/2Z)[T]/(T2+T+1) and Z[φ]/(2Z), where φ = (1 + 51/2)/2. A field with 8 elements is (Z/2Z)[T]/(T3+T+1). Two isomorphic constructions of the field with 9 elements are (Z/3Z)[T]/(T2+1) and Z[i]/(3Z). There is no field with 6 elements, because 6 is not a prime power.

Even though all fields of size p are isomorphic to Z/pZ, for n ≥ 2 the ring Z/pnZ (the ring of integers modulo pn) is not a field. The element p (mod pn) is nonzero and has no multiplicative inverse. By comparison with the ring Z/4Z of size 4, the underlying additive group of the field (Z/2Z)[T]/(T2+T+1) of size 4 is not cyclic but rather is isomorphic to the Klein four-group, (Z/2Z)2.

[edit] Proof outline

The characteristic of a finite field is a prime p (since a field has no zero divisors), and the field is a vector space of some finite dimension, say n, over Z/pZ, hence the field has pn elements. A field of order p exists, because Fp = Z/pZ is a field, where primality is required for the nonzero elements to have multiplicative inverses.

For any prime power q = pn, Fq is the splitting field of the polynomial f(T) = Tq − T over Fp. This field exists and is unique up to isomorphism by the construction of splitting fields. The roots are closed under field operations, so the splitting field is exactly the q roots of this polynomial, which are distinct because the polynomial Tq − T is separable over Fp: its derivative is −1, which has no roots.

[edit] Detailed proof of the classification

[edit] Order

We give two proofs that a finite field has prime-power order.

For the first proof, let F be a finite field. Write its additive identity as 0 and its multiplicative identity as 1. The characteristic of F is a prime number p as the characteristic of a finite ring is positive and must be prime or else the ring would have zero divisors. The p distinct elements 0, 1, 2, …, p−1 (where 2 means 1+1, etc.) form a subfield of F that is isomorphic to Z/pZ. F is a vector space over Z/pZ, and it must have finite dimension over Z/pZ. Call the dimension n, so each element of F is specified uniquely by n coordinates in Z/pZ. There are p possibilities for each coordinate, with no dependencies among different coordinates, so the number of elements in F is pn. This proves the first statement, and does a little more: it shows that, additively, F is a direct sum of copies of Z/pZ.

For the second proof, which is longer than the one above, we look more closely at the additive structure of a finite field. When F is a finite field and a and b are any two nonzero elements of F, the function f(x) = (b/a)x on F is an additive automorphism which sends a to b. (It certainly is not multiplicative too, in general!) So F is, under addition, a finite abelian group in which any two nonidentity elements are linked by an automorphism. Let's show that for any nontrivial finite abelian group A where any two nonzero elements are linked by an automorphism of A, the size of A must be a prime power. Let p be a prime factor of the size of A. By Cauchy's theorem, there is an element a of A of order p. Since we are assuming for every nonzero b in A there is an automorphism f of A such that f(a) = b, b must have order p as well. Hence all nonzero elements in A have order p. If q were any prime dividing the size of A, by Cauchy's theorem there is an element in A of order q, and since we have shown all nonzero elements have order p it follows that q = p. Thus p is the only prime factor of the size of A, so A has order equal to a power of p.

Remark: In that group-theoretic argument, one could remove the assumption that A is abelian and directly show A has to be abelian. That is, if G is a nontrivial finite group in which all nonidentity elements are linked by an automorphism, G must be an abelian group of p-power order for some prime p. The prime-power order argument goes as above, and once we know G is a p-group we appeal once again to the automorphism-linking condition, as follows. Since G is a nontrivial finite p-group, it has a nontrivial center. Pick a nonidentity element g in the center. For any h in G, there is an automorphism of G sending g to h, so h has to be in the center too. An automorphism of a group preserves the center. Therefore all elements of G are in its center, so G is abelian.

We can go further with this and show A has to be a direct sum of cyclic groups of order p. From the classification of finite abelian p-groups, A is a direct sum of cyclic groups of p-power order. Since all nonzero elements of A have order p, the cyclic groups in such a direct sum decomposition can't have order larger than p, so they all have order p. Returning to the motivating application where A is F as an additive group, we have recovered the fact that F is a direct sum of copies of Z/pZ (cyclic group of order p).


Now the first proof, using linear algebra, is a lot shorter and is the standard argument found in (nearly) all textbooks that treat finite fields. The second proof is interesting because it gets the same result by working much more heavily with the additive structure of a finite field. Of course we had to use the multiplicative structure somewhere (after all, not all finite rings have prime-power order), and it was used right at the start: multiplication by b/a on F sends a to b. The second proof is actually the one which was used in E. H. Moore's 1903 paper which (for the first time) classified all finite fields.

[edit] Existence

The proof of the second statement, concerning the existence of a finite field of size q = pn for any prime p and positive integer n, is more involved. We again give two arguments.

The case n = 1 is easy: take Fp = Z/pZ.

For general n, inside Fp[T] consider the polynomial f(T) = Tq − T. It is possible to construct a field F (called the splitting field of f over Fp), which contains Fp and which is large enough for f(T) to split completely into linear factors:

f(T) = (T−r1)(T−r2)⋯(T−rq)

in F[T]. The existence of splitting fields in general is discussed in construction of splitting fields. These q roots are distinct, because Tq − T is a polynomial of degree q which has no repeated roots in F: its derivative is qTq−1 − 1, which is −1 (because q = 0 in F) and therefore the derivative has no roots in common with f(T). Furthermore, setting R to be the set of these roots,

R = { r1, …, rq } = { roots of the equation Tq = T }

one sees that R itself forms a field, as follows. Both 0 and 1 are in R, because 0q = 0 and 1q = 1. If r and s are in R, then

(r+s)q = rq + sq = r + s

so that r+s is in R. The first equality above follows from the binomial theorem and the fact that F has characteristic p. Therefore R is closed under addition. Similarly, R is closed under multiplication and taking inverses, because

(rs)q = rq sq = rs

and

(r−1)q = (rq)−1 = r−1.

Therefore R is a field with q elements, proving the second statement.


For the second proof that a field of size q = pn exists, we just sketch the ideas. We will give a combinatorial argument that a monic irreducible f(T) of degree n exists in Fp[T]. Then the quotient ring Fp[T] / (f(T)) is a field of size q. Because Tq − T has no repeated irreducible factors (it is a separable polynomial in Fp[T]), it is a product of distinct monic irreducibles. We ask: which monic irreducibles occur in the factorization? Using some group theory, one can show that a monic irreducible in Fp[T] is a factor precisely when its degree divides n. Writing Np(d) for the number of monic irreducibles of degree d in Fp[T], computing the degree of the irreducible factorization of Tq − T shows q = pn is the sum of dNp(d) over all d dividing n. This holds for all n, so by Moebius inversion one can get a formula for Np(n) for all n, and a simple lower bound estimate using this formula shows Np(n) is positive. Thus a (monic) irreducible of degree n in Fp[T] exists, for any n.

[edit] Uniqueness

Finally the uniqueness statement: a field of size q = pn is the splitting field of Tq - T over its subfield of size p, and for any field K, two splitting fields of a polynomial in K[T] are unique up to isomorphism over K. That is, the two splitting fields are isomorphic by an isomorphism extending the identification of the copies of K inside the two splitting fields. Since a field of size p can be embedded in a field of characteristic p in only one way (the multiplicative identity 1 in the field is unique, then 2 = 1 + 1, and so on up to p - 1), the condition of two fields of size q being isomorphic over their subfields of size p is the same as just being isomorphic fields.

Warning: it is not the case that two finite fields of the same size are isomorphic in a unique way, unless the fields have size p. Two fields of size pn are isomorphic to each other in n ways (because a field of size pn is isomorphic to itself in n ways, from Galois theory for finite fields).

[edit] Explicitly constructing finite fields

Given a prime power q = pn, we may explicitly construct a finite field with q elements as follows. Select a monic irreducible polynomial f(T) of degree n in Fp[T]. (Such a polynomial is guaranteed to exist, once we know that a finite field of size q exists: just take the minimal polynomial of any primitive element for that field over the subfield Fp.) Then Fp[T]/(f(T)) is a field of size q. Here, Fp[T] denotes the ring of all polynomials in T with coefficients in Fp, (f(T)) denotes the ideal generated by f(T), and the quotient is meant in the sense of quotient rings — the set of polynomials in T with coefficients in Fp modulo (f(T)).

[edit] Examples

The polynomial f(T) = T 2 + T + 1 is irreducible over Z/2Z, and (Z/2Z)[T] / (T2+T+1) has size 4. Its elements can be written as the set {0, 1, t, t+1} where the multiplication is carried out by using the relation t2 + t + 1 = 0. In fact, since we are working over Z/2Z (that is, in characteristic 2), we may write this as t2 = t + 1. (This follows because −1 = 1 in Z/2Z!) Then, for example, to determine t3, we calculate: t3 = t(t2) = t(t+1) = t2+t = t+1+t = 2t + 1 = 1, so t3 = 1.

In order to find the multiplicative inverse of t in this field, we have to find a polynomial p(T) such that T * p(T) = 1 modulo T 2 + T + 1. The polynomial p(T) = T + 1 works, and hence 1/t = t + 1.

To construct a field of size 27, we could start for example with the irreducible polynomial T 3 + T 2 + T + 2 over Z/3Z. The field (Z/3Z)[T]/(T 3 + T 2 + T + 2) has size 27. Its elements have the form at2 + bt + c where a, b, and c lie in Z/3Z and the multiplication is defined by t 3 + t 2 + t + 2 = 0, or by rearranging this equation, t3 = 2t2 + 2t + 1.

[edit] Properties and facts

[edit] Frobenius automorphisms

If F is a finite field with q = pn elements (where p is prime), then

xq = x

for all x in F. Furthermore, the map

f : F → F

defined by

f(x) = xp

is bijective and a homomorphism, and is therefore an automorphism. It is called the Frobenius automorphism, after Ferdinand Georg Frobenius.

The Frobenius automorphism of a field of size pn has order n, and the cyclic group it generates is the full group of automorphisms of the field.

[edit] Algebraic closure

Finite fields are not algebraically closed: the polynomial

f(T)=1+\prod_{\alpha \in F}\left(T-\alpha\right)

has no roots over F, as f(α) = 1 for all α in F. However, for each prime p there is an algebraic closure of any finite field of characteristic p, as below.

[edit] Containment

The field Fpn contains a copy of Fpm if and only if m divides n. "Only if" is because the larger field is a vector space over the smaller field, of some finite dimension, say d, so it must have size (pm)d = pmd, so m divides n. "If" is because there exist irreducible polynomials of every degree over Fpm.

The direct limit of this system is a field, and is an algebraic closure of Fp (or indeed of Fpn for any n), denoted \bar\mathbf{F}_p. This field is infinite, as it is algebraically closed, or more simply because it contains a subfield of size pn for all n.

The inclusions commute with the Frobenius map, as it is defined the same way on each field (its still just the function raising to the pth power), so the Frobenius map defines an automorphism of \bar\mathbf{F}_p, which carries all subfields back to themselves. Unlike in the case of finite fields, the Frobenius automorphism on the algebraic closure of Fp has infinite order (no iterate of it is the identity function on the whole field), and it does not generate the full group of automorphisms of this field. That is, there are automorphisms of the algebraic closure which are not iterates of the pth power map. However, the iterates of the pth power map do form a dense subgroup of the automorphism group in the Krull topology. Algebraically, this corresponds to the additive group Z being dense in the profinite integers (direct product of the p-adic integers over all primes p, with the product topology).

The field Fpn can be recovered as the fixed points of the nth iterate of the Frobenius map.

If we actually construct our finite fields in such a fashion that Fpn is contained in Fpm whenever n divides m, then this direct limit can be constructed as the union of all these fields. Even if we do not construct our fields this way, we can still speak of the algebraic closure, but some more delicacy is required in its construction.

[edit] Finite division rings are fields

A division ring is a generalization of field, which are not assumed commutative. There are no non-commutative finite division rings: Wedderburn's (little) theorem states that all finite division rings are commutative, hence finite fields. The result holds even if we relax associativity and consider alternative rings, by the Artin-Zorn theorem.

[edit] Multiplicative structure

[edit] Cyclic

The multiplicative group of every finite field is cyclic, a special case of a theorem mentioned in the article about fields (see Field (mathematics)#Some first theorems).

This means that if F is a finite field with q elements, then there exists an element x in F, called "a generator for the group of units" such that

F = { 0, 1, x, x2, ..., xq-2 }.

The element x is not unique (unless q = 2 or 3): the set of generators has size \varphi(q-1) where φ is Euler's totient function. If we fix a generator, then for any non-zero element a in Fq, there is a unique integer n with

0 ≤ n ≤ q − 2

such that

a = xn.

The value of n for a given a is called the discrete log of a (in the given field, to base x).

[edit] Analog of Fermat's little theorem

Every element of a field of size q satisfies aq = a. When q is prime, this is just Fermat's little theorem, which states that ap ≡ a (mod p) for any integer a and prime p.

The general statement for any finite field follows because the non-zero elements in a field of size q form a group under multiplication of order q−1, so by Lagrange's theorem aq−1 = 1 for any nonzero a in the field. Then aq = a and this holds for 0 as well.

[edit] Applications

Discrete exponentiation, also known as calculating a = xn from x and n, can be computed quickly using techniques of fast exponentiation such as binary exponentiation, which takes only O(log n) field operations. No fast way of computing the discrete logarithm n given a and x is known, and this has many applications in cryptography, such as the Diffie-Hellman protocol.

Finite fields also find applications in coding theory: many codes are constructed as subspaces of vector spaces over finite fields.

Within number theory, the significance of finite fields is their role in the definition of the Frobenius element (or, more accurately, Frobenius conjugacy class) attached to a prime ideal in a Galois extension of number fields, which in turn is needed to make sense of Artin L-functions of representations of the Galois group, the non-abelian generalization of Dirichlet L-functions.

Counting solutions to equations over finite fields leads into deep questions in algebraic geometry, the Weil conjectures, and in fact was the motivation for Grothendieck's development of modern algebraic geometry.

[edit] Some small finite fields

F2:

 + | 0 1        × | 0 1
 --+----        --+----
 0 | 0 1        0 | 0 0
 1 | 1 0        1 | 0 1

F3:

 + | 0 1 2       × | 0 1 2
 --+------       --+------
 0 | 0 1 2       0 | 0 0 0
 1 | 1 2 0       1 | 0 1 2
 2 | 2 0 1       2 | 0 2 1

F4:

 + | 0 1 A B       × | 0 1 A B
 --+--------       --+--------
 0 | 0 1 A B       0 | 0 0 0 0
 1 | 1 0 B A       1 | 0 1 A B
 A | A B 0 1       A | 0 A B 1
 B | B A 1 0       B | 0 B 1 A

[edit] See also

[edit] References

[edit] External links

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