# Finite field

In abstract algebra, a finite field or Galois field (so named in honor of Évariste Galois) is a field that contains a finite number of elements. Finite fields are important in number theory, algebraic geometry, Galois theory, cryptography, coding theory and quantum error correction.[1] The finite fields are classified by size; there is exactly one finite field up to isomorphism of size pk for each prime p and positive integer k. Each finite field of size q is the splitting field of the polynomial xqx, and thus the fixed field of the Frobenius endomorphism which takes x to xq. Similarly, the multiplicative group of the field is a cyclic group. Wedderburn's little theorem states that the Brauer group of a finite field is trivial, so that every finite division ring is a finite field. Finite fields have applications in many areas of mathematics and computer science, including coding theory, LFSRs, modular representation theory, and the groups of Lie type. Finite fields are an active area of research, including recent results on the Kakeya conjecture and open problems on the size of the smallest primitive root.

Finite fields appear in the following chain of class inclusions:

Commutative ringsintegral domainsintegrally closed domainsunique factorization domainsprincipal ideal domainsEuclidean domainsfieldsfinite fields.

## Classification

The finite fields are classified as follows (Jacobson 2009, §4.13,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 $\mathbb{F}_{p^n}$ or Fpn. Another notation is GF(pn), where the letters "GF" stand for "Galois field".

### Examples

First we consider fields where the size is prime, i.e., n = 1. Such a field is also called a Prime field. An example of such a finite field is the ring Z/pZ, the set of integers 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. However this ring Z/pZ is a field because it contains a multiplicative inverse for each element N other than zero (an integer that, multiplied by the element modulo p yields 1), and it has a finite number of elements (p), making it a finite field.

Next we consider fields where the size is not prime, but is a prime power, i.e., n > 1.

Two isomorphic constructions of the field with 4 elements are (Z/2Z)[T]/(T2+T+1) and Z[φ]/(2Z[φ]), where φ = $\frac{-1 + \sqrt{5}}{2}$. Here (Z/2Z)[T] is the polynomial ring of Z/2Z and (Z/2Z)[T]/(T2+T+1) are the equivalence classes of these polynomials modulo T2+T+1. Roughly T2+T+1=0 so that T2=T+1 (since −1=1 in Z/2Z) and hence the elements of (Z/2Z)[T]/(T2+T+1) are the polynomials of degree up to 1 with coefficients in Z/2Z, i.e. the set {0, 1, T, T+1 } (see below for more details). Notice that (Z/2Z)[T]/(T2+1) is not a field since it admits a zero divisor (T+1)2=T2+1=0 (since we work in Z/2Z where 2=0).

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[i]).

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.

A prime power field with p=2 is also called a binary field.

Finally, we consider fields where the size is not a prime power. As it turns out, none exists. For example, there is no field with 6 elements, because 6 is not a prime power. Each and every pair of operations on a set of 6 elements fails to satisfy the mathematical definition of a field.

### 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) = TqT over Fp. This field exists and is unique up to isomorphism by the construction of splitting fields. The set of roots is a field, the fixed field of the nth iterate of the Frobenius endomorphism, so the splitting field is exactly the q roots of this polynomial, which are distinct because the polynomial TqT is separable over Fp: its derivative is −1, which has no roots.

### Detailed proof of the classification

#### 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, and all the elements can be deduced by induction) form a subfield Fp of F that is isomorphic to Z/pZ. F is a vector space over Z/pZ, and it must have a 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. (Certainly, it is not multiplicative either, 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 denotes 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 since any automorphism of a group preserves the center. Therefore all elements of G are in the 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.

#### 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) = TqT. It is possible to construct a field F (called the splitting field of f(T) over Fp), which contains Fp and which is large enough for f(T) to split completely into linear factors:

f(T) = (Tr1)(Tr2)⋯(Trq)

in F[T]. The existence of splitting fields in general is discussed in construction of splitting fields. These q roots are distinct, because TqT 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 by induction on n for q = pn, from the binomial theorem, the fact that for n = 1 all binomial coefficients except first and last are divisible by p, 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 TqT 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 TqT 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.

#### Uniqueness

Finally the uniqueness statement: a field of size q = pn is the splitting field of TqT 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).

## 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)).

### 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.

### A simple representation of Fp2

Consider the numbers of the form
$a+b\sqrt c$ ,
where $a$ and $b$ are integers in $Z_p$ and $c$ is a (quadratic) nonresidue, meaning $c$ reduced modulo $p$ is not the square of any integer in $Z_p$.

Notice that the sum and product of any two is done in the obvious way.

$(a+b\sqrt c)+(j+k\sqrt c) =(a+j)+(b+k)\sqrt c$, and

$(a+b\sqrt c)(j+k\sqrt c) =(aj+c\,bk)+(ak+bj)\sqrt c$

A quadratic residue is, on the other hand, an integer whose value reduced modulo $p$ equals the square of some integer in $Z_p$. There are exactly $(p-1)/2$ quadratic residues and nonresidues each, for any prime other than $2$.

The Quadratic Residue Multiplication Rule states:

(i)   The product of two quadratic residues modulo $p$ is a quadratic residue.
(ii)  The product of a residue and nonresidue is a nonresidue.

This leads to the conclusion, in our case when $c$ is a nonresidue, that $a^2 -c\,b^2 \ne 0$,
provided that $a$ and $b$ are not both $0$.

Now, $(a+b\sqrt c)(a - b\sqrt c)/(a^2 - c\,b^2)$
is justified, when $a+b\sqrt c$ is not zero, and equals
($a^2 - c\,b^2)/(a^2 - c\,b^2)$ $=1.$

This means the numbers $a+b\sqrt c$, as described at the start, are a field and so $F_{p^2}$.

### 2×2 matrices in integers modulo p

Consider a 2×2 matrix A over the integers modulo p, where p is a prime number, of the form:

$A =\begin{bmatrix} 0 & b \\ 1 & 1 \end{bmatrix}$

The set of matrices, let's call it F,

{ uI + vA },

with I the 2×2 identity matrix and u, v elements of Zp forms an additive group of p2 elements. It is easily seen to be closed with repect to multiplication since A satisfies its characteristic polynomial equation   x2 − x − b = 0. Simple inspection of a product shows multiplication is associative, commutative, and distributive. It is only left to determine the conditions on the integer b so that every nonzero element of F has an inverse (with respect to multiplication) in F. All the elements of F of the kind  x I + A  have an inverse (verified by direct multiplication and making use of   A2 − A − b I = 0).

(x2 + x − b)−1
((x + 1) I − A)

provided that x2 + x − b is not 0 for any x in Zp. The number of b for which x2 + x − b is irreducible over Zp is quite numerous, being (p − 1)/2. It is only left to observe that  v((u/v) I + A)  =  u I + v A.

Determining for which b that A is a primitive element for F p2  is not as easy to analyze. For p, not too large for brute calculation of all the 2x2 matrices B of integers in Zp such that B p2−1 = I and B k does not equal I, for k = 1, 2, ..., p2−2, showed many cases. Such a matrix B is a primitive element for Fp2 . To see this notice that B p2−2 = B −1 and so, if B k = B r for 0 < r < k < p2−1, then B k − r = I, which is assumed not to be the case. As noted before, since B satisfies it's characteristic equation, the powers of B must lie in the span of  u I + v B. Since there are p2−1 of them distinct together with 0 they are Fp2.

p b for which A is a primitive element
2: 1
3: 1
5: 3
7: 4
11: 3 4
13: 11
17: 7 10 14
19: 5 16 17
23: 4 9 16
29: 15 21 26
31: 7 9 14 18 19
37: 15 17 22 24 32
41: 7 13 17 29
43: 9 15 17 23 31 40
47: 8 14 17 18 21 27 32 34
53: 5 8 18 21 32 33 34 39 48
59: 3 16 17 19 25 27 35 57
61: 7 17 26 43 44 55 59
67: 10 17 21 26 35 36 49 55
71: 3 8 10 15 16 24 38 43 49 50 60
73: 5 11 13 28 29 31 33 39 40 44 47 58 60 62
79: 4 9 13 19 40 49 50 51 73 76
83: 21 25 31 33 37 51 59 64 75 77 81
89: 15 29 35 41 46 48 65 70 74 82 83
97: 7 10 14 17 29 37 39 41 58 76 80 82 83 84 87 92
p PEs/p^4 PEs p^4
2: 12.50% 2 16
3: 14.81% 12 81
5: 12.80% 80 625
7: 13.99% 336 2401
11: 12.02% 1760 14641
13: 13.11% 3744 28561
17: 15.63% 13056 83521
19: 12.60% 16416 130321
23: 14.47% 40480 279841
29: 11.02% 77952 707281
31: 12.89% 119040 923521
37: 15.35% 287712 1874161
41: 11.14% 314880 2825761
43: 12.68% 433440 3418801
47: 15.60% 761024 4879681

PEs are the 2×2 matrices in integers modulo p that are primitive elements

p4 is the total number of 2×2 matrices in integers modulo p

More generally let A be a 2x2 matrix of integers in Zp with characteristic polynomial x2 + b x + c. Then F, the span of { uI + vA }, u, v members of Zp, is a field if and only if x2 + b x + c is irreducible over Zp. Assume A is independent of I over Zp.

proof:

F contains 0 and I and is closed with respect to associative, commutative, and distributive addition and multiplication. Consider any element of F of the sort  xI + A  and consider the multiplication with the element  (x − b)I − A .

(xI + A) ((x − b)I − A)   =   x2 I − A2  − b x I − b A
=   x2 I + (b A + c I)  − b x I − b A  =  (x2 − b x + c) I

The polynomial x2 − b x + c is the same as x2 + b x + c after substituting −x for x so their irreducibilities are the same. In the case that x2 + b x + c is irreducible (x2 − b x + c)−1 ((x − b)I − A) is a multiplicative inverse for  xI + A  and F is a field. In the case that  x2 − b x + c  has a root x in Zp then  xI + A  is a divisor of 0 and F is not a field.

### n×n matrices in integers modulo p

To clarify notation it is the usual convention when P(x) = u0 + u1 + , ..., + ukxk is a polynomial
and A is a matrix P(A) = u0I + u1 + , ..., + ukAk.

A monic polynomial P(x) is said to be minimal for A if P(A) = 0 and there is no nonzero polynomial
of less degree for which A is a zero.

The characteristic polynomial of a n×n matrix A is given by P(x) = det( x I − A),
and has degree n.

By the Cayley–Hamilton Theorem a matrix A satisfies its characteristic equation, that is P(A) = 0. If A has the characteristic polynomial P(x) =  a0 + a1 + a2x2 , ...,  + an−1xn−1  + xn ,
then it satisfies the relation  An =( a0I + a1A a2A2 , ...,  + an−1An−1 ).

Expressions of the kind    u0I + u1 + u2A2 , ...,  + un−1An−1    are closed with respect to multiplication besides addition. Since powers of a matrix commute with one another as do polynomials the multiplication will be commutative, as well as associative and distributive (these last two being properties of matrix algebra).

If P(x) is irreducible then P(x) is also minimal for A. To see this suppose Q(x) is of positive degree
less than n and the minimal polynomial for A instead. By the division algorithm
P(x) = T(x) Q(x) + R(x), for some polynomials T(x) and R(x) with the degree of R(x)
less than the degree of Q(x). R(x) can not be the zero polynomial since P(x) is irreducible.
But, Q(A) = 0, and P(A) = T(A) Q(A) + R(A) = 0, meaning R(A) = 0 too,
which would be that Q(x) is not minimal for A.

Theorem

Let A be a n×n matrix of integers in Zp with characteristic polynomial P(x). Then F, the span of
{ u0I + u1 + u2A2 , ...,  + un−1An−1 }, with u0, u1, ...,  un−1  members of Zp, is a field
with pn elements if and only if P(x) is irreducible over Zp.

proof:
First the necessity of the condition on P(x) is shown. If P(x) is not irreducible then there exists
nontrivial polynomials Q(x) and T(x) such that P(A) = T(A) Q(A) = 0,
in which case F has divisors of zero and is not a field.

Now, assume P(x) is irreducible. Then for u ∈ Zp, by the Remainder Theorem
P(x) = T(x) (x − u) + P(u), for some polynomial T(x) of degree n − 1.
Then P(A) = T(A) (A − u I) + P(u) I = 0.. Since P(x) is irreducible, P(u) ≠ 0 and it follows
−(P(u))−1T(A) (A − u I) = I. and so (A − u I)−1 exists and is in F.

Now, let our induction hypothesis be that for any polynomial R(x) =  u0 + u1 + , ...,  + ukxk,
for
k < m < n, that (R(A))−1 exists and is in F. Consider Q(x) =  u0 + u1 + , ...,  + umxm.
By the division algorithm P(x) = T(x) Q(x) + R(x), for some polynomials T(x) and R(x)
with the degree of R(x) less than m.

So   P(A) = T(A) Q(A) + R(A) = 0 ,    −(R(A))−1 T(A) Q(A)  = I , and
(Q(A))−1 = −(R(A))−1 T(A) exists and is in F.
The induction proceeds until m = n − 1, and it is seen that F is a field.

Remark: In the above argument the indeterminant x can replace A, in the first place. The polynomials with their coefficents in Zp having xn resolved by an irreducible monic polynomial of degree n, in the same manner, is a field.

## Properties and facts

Finite fields cannot be ordered: in an ordered field the elements 0 < 1 < 1 + 1 < 1 + 1 + 1 < … are all different, so that an ordered field necessarily contains infinitely many elements.

### Frobenius automorphisms

If F is a finite field with q = pn elements, then

xq = x

for all x in F (see Analog of Fermat's little theorem below). Furthermore, the map

f : FF

defined by

f(x) = xp

is bijective and a homomorphism, and is therefore an automorphism on the field F which fixes the subfield with p elements. It is called the Frobenius automorphism, after Ferdinand Georg Frobenius. The fact that the Frobenius map is surjective implies that a finite field is perfect.

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.

### 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.

### 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 $(p^m)^d=p^{md}$, 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 (it is 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.

### Irreducibility of polynomials

If F is a finite field, a polynomial f(X) with coefficients in F is said to be irreducible over F if and only if f(X) is irreducible as an element of the polynomial ring over F (that is, in F[X]). Note that since the polynomial ring F[X] is a unique factorization domain, a polynomial f(X) is irreducible if and only if it is prime as an element of F[X].

There are several fundamental questions one can ask about irreducible polynomials over a given finite field. Firstly, is it possible to give an explicit formula, in the variables q and n, that yields the number of irreducible polynomials over Fq of degree n? Note that since there are only finitely many polynomials of a given degree n over the finite field Fq, there can be only finitely many such irreducible polynomials. However, while little theory is required to compute the number of polynomials of degree n over Fq (there are precisely qn(q−1) such polynomials), it is not immediately obvious how to compute the number of irreducible polynomials of degree n over q.

Secondly, is it possible to describe an algorithm that may be used to decide whether a given polynomial over Fq is irreducible? In fact, there exists two such (known) algorithms: the Berlekamp algorithm and the Cantor–Zassenhaus algorithm. Furthermore, these algorithms do much more than merely decide whether a given polynomial is irreducible; they may also be implemented to explicitly compute the irreducible factors of f.

#### Number of monic irreducible polynomials of a given degree over a finite field

If Fq denotes the finite field of order q, then the number N of monic irreducible polynomials of degree n over Fq is given by:[2]

$N(q,n)=\frac{1}{n}\sum_{d|n} \mu(d)q^{\frac{n}{d}},$

where μ is the Möbius function. By the above formula, the number of irreducible polynomials of degree n over Fq is given by $(q-1)N(q,n)$. A (slightly simpler) lower bound on N also exists, and is given by:

$N\geq\frac{1}{n} \left(q^n-\sum_{p|n, \; p \text{ prime }} q^{\frac{n}{p}}\right).$

### Wedderburn's little theorem

A division ring is a generalization of field. Division rings 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.

## Multiplicative structure

### Cyclic

The multiplicative group of every finite field is cyclic, a special case of a theorem mentioned in Fields. A generator for the multiplicative group is a primitive element.

This means that if F is a finite field with q elements, then there exists an element x in F such that

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

The primitive element x is not unique (unless q = 2 or 3): the set of generators has size $\varphi(q-1)$ where $\varphi$ 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 ≤ nq − 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).

### Analog of Fermat's little theorem

Every element of a finite field of size q satisfies aq = a. When q is prime, this is just Fermat's little theorem, which states that apa (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.

### Roots of unity

Let $n$ be a positive integer, a $n$-th root of unity in a finite field $\mathbb{F}$ is a solution of the equation $x^n = 1$, a $n$-th primitive root of unity is a solution of the equation $x^n = 1$ that is not a solution of the equation $x^m = 1$ for any positive integer $m < n$. Unlike the $n$-th roots of unity in $\mathbb{C}$, the number of $n$-th roots of unity in $\mathbb{F}$ may be less than $n$. Let $q := |\mathbb{F}|$, then the number of $n$-th roots of unity in $\mathbb{F}$ is $\gcd(n, q - 1)$. If $n \nmid q - 1$ then $\mathbb{F}$ has no primitive $n$-roots of unity, while if $n \mid q - 1$ then the number of primitive $n$-th roots of unity in $\mathbb{F}$ is $\varphi(n)$, where $\varphi(\cdot)$ is the Euler totient function.

## 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.

## Some small finite fields

F2:

+ 0 1
0 0 1
1 1 0
× 0 1
0 0 0
1 0 1

F3:

+ 0 1 2
0 0 1 2
1 1 2 0
2 2 0 1
× 0 1 2
0 0 0 0
1 0 1 2
2 0 2 1

F4:

+ 0 1 A B
0 0 1 A B
1 1 0 B A
A A B 0 1
B B A 1 0
× 0 1 A B
0 0 0 0 0
1 0 1 A B
A 0 A B 1
B 0 B 1 A
Field of 8 elements represented as matrices
integers are modulo 2

element (0)     element (1)     element (2)     element (3)

0  0  0         1  0  0         0  1  0         0  0  1
0  0  0         0  1  0         0  0  1         1  1  0
0  0  0         0  0  1         1  1  0         0  1  1

element (4)     element (5)     element (6)     element (7)

1  1  0         0  1  1         1  1  1         1  0  1
0  1  1         1  1  1         1  0  1         1  0  0
1  1  1         1  0  1         1  0  0         0  1  0

+/  (0) (1) (2) (3) (4) (5) (6) (7)
(0)  0   1   2   3   4   5   6   7
(1)  1   0   4   7   2   6   5   3
(2)  2   4   0   5   1   3   7   6
(3)  3   7   5   0   6   2   4   1
(4)  4   2   1   6   0   7   3   5
(5)  5   6   3   2   7   0   1   4
(6)  6   5   7   4   3   1   0   2
(7)  7   3   6   1   5   4   2   0

x/  (0) (1) (2) (3) (4) (5) (6) (7)
(0)  0   0   0   0   0   0   0   0
(1)  0   1   2   3   4   5   6   7
(2)  0   2   3   4   5   6   7   1
(3)  0   3   4   5   6   7   1   2
(4)  0   4   5   6   7   1   2   3
(5)  0   5   6   7   1   2   3   4
(6)  0   6   7   1   2   3   4   5
(7)  0   7   1   2   3   4   5   6

_________________________________________________________

Field of 9 elements represented as matrices
integers are modulo 3

element (0)     element (1)     element (2)

0  0            1  0            0  1
0  0            0  1            1  1

element (3)     element (4)     element (5)

1  1            1  2            2  0
1  2            2  0            0  2

element (6)     element (7)     element (8)

0  2            2  2            2  1
2  2            2  1            1  0

+/  (0) (1) (2) (3) (4) (5) (6) (7) (8)
(0)  0   1   2   3   4   5   6   7   8
(1)  1   5   3   8   7   0   4   6   2
(2)  2   3   6   4   1   8   0   5   7
(3)  3   8   4   7   5   2   1   0   6
(4)  4   7   1   5   8   6   3   2   0
(5)  5   0   8   2   6   1   7   4   3
(6)  6   4   0   1   3   7   2   8   5
(7)  7   6   5   0   2   4   8   3   1
(8)  8   2   7   6   0   3   5   1   4

x/  (0) (1) (2) (3) (4) (5) (6) (7) (8)
(0)  0   0   0   0   0   0   0   0   0
(1)  0   1   2   3   4   5   6   7   8
(2)  0   2   3   4   5   6   7   8   1
(3)  0   3   4   5   6   7   8   1   2
(4)  0   4   5   6   7   8   1   2   3
(5)  0   5   6   7   8   1   2   3   4
(6)  0   6   7   8   1   2   3   4   5
(7)  0   7   8   1   2   3   4   5   6
(8)  0   8   1   2   3   4   5   6   7



_________________________________________________________

F16 is represented by the polynomials a + b x + c x2 + d x3.
a, b, c, and d are integers modulo 2
The polynomials are generated by the powers of x using the rule

x4 = 1 + x

e ( 0)        e ( 1)        e ( 2)        e ( 3)
[ 0  0  0  0] [ 1  0  0  0] [ 0  1  0  0] [ 0  0  1  0]

e ( 4)        e ( 5)        e ( 6)        e ( 7)
[ 0  0  0  1] [ 1  1  0  0] [ 0  1  1  0] [ 0  0  1  1]

e ( 8)        e ( 9)        e (10)        e (11)
[ 1  1  0  1] [ 1  0  1  0] [ 0  1  0  1] [ 1  1  1  0]

e (12)        e (13)        e (14)        e (15)
[ 0  1  1  1] [ 1  1  1  1] [ 1  0  1  1] [ 1  0  0  1]

+/   0_ 1_ 2_ 3_ 4_ 5_ 6_ 7_ 8_ 9_10_11_12_13_14_15_
0_  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
1_  1  0  5  9 15  2 11 14 10  3  8  6 13 12  7  4
2_  2  5  0  6 10  1  3 12 15 11  4  9  7 14 13  8
3_  3  9  6  0  7 11  2  4 13  1 12  5 10  8 15 14
4_  4 15 10  7  0  8 12  3  5 14  2 13  6 11  9  1
5_  5  2  1 11  8  0  9 13  4  6 15  3 14  7 12 10
6_  6 11  3  2 12  9  0 10 14  5  7  1  4 15  8 13
7_  7 14 12  4  3 13 10  0 11 15  6  8  2  5  1  9
8_  8 10 15 13  5  4 14 11  0 12  1  7  9  3  6  2
9_  9  3 11  1 14  6  5 15 12  0 13  2  8 10  4  7
10_ 10  8  4 12  2 15  7  6  1 13  0 14  3  9 11  5
11_ 11  6  9  5 13  3  1  8  7  2 14  0 15  4 10 12
12_ 12 13  7 10  6 14  4  2  9  8  3 15  0  1  5 11
13_ 13 12 14  8 11  7 15  5  3 10  9  4  1  0  2  6
14_ 14  7 13 15  9 12  8  1  6  4 11 10  5  2  0  3
15_ 15  4  8 14  1 10 13  9  2  7  5 12 11  6  3  0

x/   0_ 1_ 2_ 3_ 4_ 5_ 6_ 7_ 8_ 9_10_11_12_13_14_15_
0_  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
1_  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
2_  0  2  3  4  5  6  7  8  9 10 11 12 13 14 15  1
3_  0  3  4  5  6  7  8  9 10 11 12 13 14 15  1  2
4_  0  4  5  6  7  8  9 10 11 12 13 14 15  1  2  3
5_  0  5  6  7  8  9 10 11 12 13 14 15  1  2  3  4
6_  0  6  7  8  9 10 11 12 13 14 15  1  2  3  4  5
7_  0  7  8  9 10 11 12 13 14 15  1  2  3  4  5  6
8_  0  8  9 10 11 12 13 14 15  1  2  3  4  5  6  7
9_  0  9 10 11 12 13 14 15  1  2  3  4  5  6  7  8
10_  0 10 11 12 13 14 15  1  2  3  4  5  6  7  8  9
11_  0 11 12 13 14 15  1  2  3  4  5  6  7  8  9 10
12_  0 12 13 14 15  1  2  3  4  5  6  7  8  9 10 11
13_  0 13 14 15  1  2  3  4  5  6  7  8  9 10 11 12
14_  0 14 15  1  2  3  4  5  6  7  8  9 10 11 12 13
15_  0 15  1  2  3  4  5  6  7  8  9 10 11 12 13 14


_________________________________________________________

F25 represented by the numbers a + b√2, a and b are integers modulo 5
generated by powers of  2 + √2

 e ( 0) e ( 1) e ( 2) e ( 3) e ( 4) 0 + 0√2 1 + 0√2 2 + 1√2 1 + 4√2 0 + 4√2 e ( 5) e ( 6) e ( 7) e ( 8) e ( 9) 3 + 3√2 2 + 4√2 2 + 0√2 4 + 2√2 2 + 3√2 e (10) e (11) e (12) e (13) e (14) 0 + 3√2 1 + 1√2 4 + 3√2 4 + 0√2 3 + 4√2 e (15) e (16) e (17) e (18) e (19) 4 + 1√2 0 + 1√2 2 + 2√2 3 + 1√2 3 + 0√2 e (20) e (21) e (22) e (23) e (24) 1 + 3√2 3 + 2√2 0 + 2√2 4 + 4√2 1 + 2√2
+/   0_ 1_ 2_ 3_ 4_ 5_ 6_ 7_ 8_ 9_10_11_12_13_14_15_16_17_18_19_20_21_22_23_24_
0_  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
1_  1  7 18  6  3 12 14 19 22  5 20  2 10  0 23 16 11 21 15 13  9  8 24  4 17
2_  2 18  8 19  7  4 13 15 20 23  6 21  3 11  0 24 17 12 22 16 14 10  9  1  5
3_  3  6 19  9 20  8  5 14 16 21 24  7 22  4 12  0  1 18 13 23 17 15 11 10  2
4_  4  3  7 20 10 21  9  6 15 17 22  1  8 23  5 13  0  2 19 14 24 18 16 12 11
5_  5 12  4  8 21 11 22 10  7 16 18 23  2  9 24  6 14  0  3 20 15  1 19 17 13
6_  6 14 13  5  9 22 12 23 11  8 17 19 24  3 10  1  7 15  0  4 21 16  2 20 18
7_  7 19 15 14  6 10 23 13 24 12  9 18 20  1  4 11  2  8 16  0  5 22 17  3 21
8_  8 22 20 16 15  7 11 24 14  1 13 10 19 21  2  5 12  3  9 17  0  6 23 18  4
9_  9  5 23 21 17 16  8 12  1 15  2 14 11 20 22  3  6 13  4 10 18  0  7 24 19
10_ 10 20  6 24 22 18 17  9 13  2 16  3 15 12 21 23  4  7 14  5 11 19  0  8  1
11_ 11  2 21  7  1 23 19 18 10 14  3 17  4 16 13 22 24  5  8 15  6 12 20  0  9
12_ 12 10  3 22  8  2 24 20 19 11 15  4 18  5 17 14 23  1  6  9 16  7 13 21  0
13_ 13  0 11  4 23  9  3  1 21 20 12 16  5 19  6 18 15 24  2  7 10 17  8 14 22
14_ 14 23  0 12  5 24 10  4  2 22 21 13 17  6 20  7 19 16  1  3  8 11 18  9 15
15_ 15 16 24  0 13  6  1 11  5  3 23 22 14 18  7 21  8 20 17  2  4  9 12 19 10
16_ 16 11 17  1  0 14  7  2 12  6  4 24 23 15 19  8 22  9 21 18  3  5 10 13 20
17_ 17 21 12 18  2  0 15  8  3 13  7  5  1 24 16 20  9 23 10 22 19  4  6 11 14
18_ 18 15 22 13 19  3  0 16  9  4 14  8  6  2  1 17 21 10 24 11 23 20  5  7 12
19_ 19 13 16 23 14 20  4  0 17 10  5 15  9  7  3  2 18 22 11  1 12 24 21  6  8
20_ 20  9 14 17 24 15 21  5  0 18 11  6 16 10  8  4  3 19 23 12  2 13  1 22  7
21_ 21  8 10 15 18  1 16 22  6  0 19 12  7 17 11  9  5  4 20 24 13  3 14  2 23
22_ 22 24  9 11 16 19  2 17 23  7  0 20 13  8 18 12 10  6  5 21  1 14  4 15  3
23_ 23  4  1 10 12 17 20  3 18 24  8  0 21 14  9 19 13 11  7  6 22  2 15  5 16
24_ 24 17  5  2 11 13 18 21  4 19  1  9  0 22 15 10 20 14 12  8  7 23  3 16  6

x/   0_ 1_ 2_ 3_ 4_ 5_ 6_ 7_ 8_ 9_10_11_12_13_14_15_16_17_18_19_20_21_22_23_24_
0_  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
1_  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
2_  0  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1
3_  0  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2
4_  0  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3
5_  0  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4
6_  0  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5
7_  0  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6
8_  0  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7
9_  0  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8
10_  0 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9
11_  0 11 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10
12_  0 12 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11
13_  0 13 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12
14_  0 14 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13
15_  0 15 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14
16_  0 16 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
17_  0 17 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
18_  0 18 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17
19_  0 19 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18
20_  0 20 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19
21_  0 21 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
22_  0 22 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21
23_  0 23 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22
24_  0 24  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23