- Finite field
In

abstract algebra , a**finite field**or**Galois field**(so named in honor ofÉvariste Galois ) is a field that contains only finitely many elements. Finite fields are important innumber theory ,algebraic geometry ,Galois theory ,cryptography , andcoding theory . The finite fields are completely known.**Classification**There is a unique field of order "p"

^{"n"}for every prime "p" and every positive integer "n", up toisomorphism .In detail, the finite fields are classified as follows Harv|Jacobson|1985|p=287:

* The**order**, or number of elements, of a finite field is of the form "p"^{"n"}, where "p" is a prime number called the**characteristic**of the field, and "n" is a positive integer.

* For everyprime number "p" and positive integer "n", there exists a finite field with "p"^{"n"}elements.

* Any two finite fields with the same number of elements areisomorphic . 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

**F**_{"p""n"}. Another notation is**GF**("p"^{"n"}), where the letters "GF" stand for "Galois field".**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**/"p**"Z**. 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**Z**_{"p"}, but within some areas of mathematics, particularly number theory, this may cause confusion because the same notation**Z**_{"p"}is used for the ring of p-adic integers.Two isomorphic constructions of the field with 4 elements are (

**Z**/2**Z**) ["T"] /("T"^{2}+"T"+1) and**Z**["φ"] /(2**Z**), where "φ" = $frac\{-1\; +\; sqrt\{5\{2\}$. A field with 8 elements is (**Z**/2**Z**) ["T"] /("T"^{3}+T+1). Two isomorphic constructions of the field with 9 elements are (**Z**/3**Z**) ["T"] /("T"^{2}+1) and**Z**[i] /(3**Z**). There is "no" field with 6 elements, because 6 is not aprime power .Even though all fields of size "p" are isomorphic to

**Z**/"p**"Z**, for "n" ≥ 2 the ring**Z**/"p"^{"n"}**Z**(the ring of integers modulo "p"^{"n"}) is "not" a field. The element "p" (mod "p"^{"n"}) is nonzero and has no multiplicative inverse. By comparison with the ring**Z**/4**Z**of size 4, the underlying additive group of the field (**Z**/2**Z**) ["T"] /("T"^{2}+"T"+1) of size 4 is not cyclic but rather is isomorphic to theKlein four-group , (**Z**/2**Z**)^{2}.**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**/"p**"Z**, hence the field has "p"^{"n"}elements. A field of order "p" exists, because**F**_{"p"}=**Z**/p**Z**is a field, where primality is required for the nonzero elements to have multiplicative inverses.For any prime power "q" = "p"

^{"n"},**F**_{"q"}is thesplitting field of the polynomial "f"("T") = "T"^{"q"}− "T" over**F**_{"p"}. This field exists and is unique up to isomorphism by theconstruction 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 "T"^{"q"}− "T" is separable over**F**_{"p"}: 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 itsmultiplicative 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**/"p**"Z**. "F" is avector space over**Z**/"p**"Z**, and it must have finite dimension over**Z**/"p**"Z**. Call the dimension "n", so each element of "F" is specified uniquely by "n" coordinates in**Z**/"p**"Z**. There are "p" possibilities for each coordinate, with no dependencies among different coordinates, so the number of elements in "F" is "p"^{"n"}. This proves the first statement, and does a little more: it shows that, additively, "F" is a direct sum of copies of**Z**/"p**"Z**.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**/"p**"Z**(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" = "p"

^{"n"}for any prime "p" and positive integer "n", is more involved. We again give two arguments.The case "n" = 1 is easy: take

**F**_{"p"}=**Z**/"p**"Z**.For general "n", inside

**F**_{"p"}["T"] consider the polynomial "f"("T") = "T"^{"q"}− "T". It is possible to construct a field "F" (called thesplitting field of "f" over**F**_{"p"}), which contains**F**_{"p"}and which is large enough for "f"("T") to split completely into linear factors::"f"("T") = ("T"−"r"_{1})("T"−"r"_{2})⋯("T"−"r"_{"q"}) in "F" ["T"] . The existence of splitting fields in general is discussed inconstruction of splitting fields . These "q" roots are distinct, because "T"^{"q"}− "T" is a polynomial of degree "q" which has no repeated roots in "F": its derivative is "qT"^{"q"−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" = { "r"_{1}, …, "r"_{"q"}} = { roots of the equation "T"^{"q"}= "T" }one sees that "R" "itself forms a field", as follows. Both 0 and 1 are in "R", because 0^{"q"}= 0 and 1^{"q"}= 1. If "r" and "s" are in "R", then:("r"+"s")^{"q"}= "r"^{"q"}+ "s"^{"q"}= "r" + "s"so that "r"+"s" is in "R". The first equality above follows from thebinomial 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"}= "r"^{"q"}"s"^{"q"}= "rs"and: ("r"^{−1})^{"q"}= ("r"^{"q"})^{−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" = "p"

^{"n"}exists, we just sketch the ideas. We will give a combinatorial argument that a monic irreducible "f"("T") of degree "n" exists in**F**_{"p"}["T"] . Then the quotient ring**F**_{"p"}["T"] / ("f"("T")) is a field of size "q". Because "T"^{"q"}− "T" has no repeated irreducible factors (it is a separable polynomial in**F**_{"p"}["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**F**_{"p"}["T"] is a factor precisely when its degree divides "n". Writing N_{"p"}("d") for the number of monic irreducibles of degree "d" in**F**_{"p"}["T"] , computing the degree of the irreducible factorization of "T"^{"q"}− "T" shows "q" = "p"^{"n"}is the sum of "dN"_{"p"}("d") over all "d" dividing "n". This holds for all "n", so by Moebius inversion one can get a formula for "N"_{"p"}("n") for all "n", and a simple lower bound estimate using this formula shows "N"_{"p"}("n") is positive. Thus a (monic) irreducible of degree "n" in**F**_{"p"}["T"] exists, for any "n".**Uniqueness**Finally the uniqueness statement: a field of size "q" = "p"

^{"n"}is the splitting field of "T"^{q}- "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 "p"^{"n"}are isomorphic to each other in "n" ways (because a field of size "p"^{"n"}is isomorphic to itself in "n" ways, from Galois theory for finite fields).**Explicitly constructing finite fields***Given a*prime power "q" = "p"^{"n"}, we may explicitly construct a finite field with "q" elements as follows. Select a monicirreducible polynomial "f"("T") of degree "n" in**F**_{"p"}["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**F**_{"p"}.) Then**F**_{"p"}["T"] /("f"("T")) is a field of size "q". Here,**F**_{"p"}["T"] denotes the ring of allpolynomial s in "T" with coefficients in**F**_{"p"}, ("f"("T")) denotes the ideal generated by "f"("T"), and the quotient is meant in the sense ofquotient ring s — the set of polynomials in "T" with coefficients in**F**_{"p"}modulo ("f"("T")).**Examples***The polynomial "f"("T") = "T"*^{ 2}+ "T" + 1 is irreducible over**Z**/2**Z**, and (**Z**/2**Z**) ["T"] / ("T"^{2}+"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 "t"^{2}+ "t" + 1 = 0. In fact, since we are working over**Z**/2**Z**(that is, in characteristic 2), we may write this as "t"^{2}= "t" + 1. (This follows because −1 = 1 in**Z**/2**Z**!) Then, for example, to determine "t"^{3}, we calculate: "t"^{3}= "t"("t"^{2}) = "t"("t"+1) = "t"^{2}+"t" = "t"+1+"t" = 2t + 1 = 1, so "t"^{3}= 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**/3**Z**. The field (**Z**/3**Z**) ["T"] /("T"^{ 3}+ "T"^{ 2}+ "T" + 2) has size 27. Its elements have the form "at"^{2}+ "bt" + "c" where "a", "b", and "c" lie in**Z**/3**Z**and the multiplication is defined by "t"^{ 3}+ "t"^{ 2}+ "t" + 2 = 0, or by rearranging this equation, "t"^{3}= 2"t"^{2}+ 2"t" + 1.**Properties and facts****Frobenius automorphisms***If "F" is a finite field with "q" = "p"*^{"n"}elements (where "p" is prime), then*:"x"*^{"q"}= "x"*for all "x" in "F". Furthermore, the map**:"f" : "F" → "F"**defined by**:"f"("x") = "x"*^{"p"}*is*bijective and ahomomorphism , and is therefore anautomorphism . It is called theFrobenius automorphism , afterFerdinand Georg Frobenius .*The Frobenius automorphism of a field of size "p"*^{"n"}has order "n", and thecyclic group it generates is the full group of automorphisms of the field.**Algebraic closure***Finite fields are not algebraically closed: the polynomial:$f(T)=1+prod\_\{alpha\; in\; F\}left(T-alpha\; ight)$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***F**_{"p""n"}contains a copy of**F**_{"p""m"}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**F**_{"p""m"}.*The*direct limit of this system is a field, and is analgebraic closure of**F**_{"p"}(or indeed of**F**_{"p""n"}for any "n"), denoted $armathbf\{F\}\_p$. This field is infinite, as it is algebraically closed, or more simply because it contains a subfield of size "p"^{"n"}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 "p"th power), so the Frobenius map defines an automorphism of $armathbf\{F\}\_p$, which carries all subfields back to themselves. Unlike in the case of finite fields, the Frobenius automorphism on the algebraic closure of***F**_{"p"}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 "p"th power map. However, the iterates of the "p"th 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 theprofinite integers (direct product of the "p"-adic integers over all primes "p", with theproduct topology ).*The field***F**_{"p""n"}can be recovered as the fixed points of the "n"th iterate of the Frobenius map.*If we actually construct our finite fields in such a fashion that***F**_{"p""n"}is contained in**F**_{"p""m"}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.**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 finitedivision ring s are commutative, hence finite fields. The result holds even if we relax associativity and consideralternative ring s, by theArtin-Zorn theorem .**Multiplicative structure****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 ).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", "x"*^{2}, ..., "x"^{"q"-2}}.*The primitive element "x" is not unique (unless "q" = 2 or 3): the set of generators has size $varphi(q-1)$ where $phi$ is*Euler's totient function . If we fix a generator, then for any non-zero element "a" in "F"_{"q"}, there is a unique integer "n" with*:0 ≤ "n" ≤ "q" − 2**such that**:"a" = "x"*^{"n"}.*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 field of size "q" satisfies "a"*^{"q"}= "a". When "q" is prime, this is justFermat's little theorem , which states that "a"^{"p"}≡ "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 "a"*^{"q"−1}= 1 for any nonzero "a" in the field. Then "a"^{"q"}= "a" and this holds for 0 as well.**Applications***Discrete exponentiation, also known as calculating "a" = "x"*^{"n"}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 thediscrete logarithm "n" given "a" and "x" is known, and this has many applications in cryptography, such as theDiffie-Hellman protocol.*Finite fields also find applications in*coding theory : many codes are constructed as subspaces ofvector space s 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 , theWeil conjectures , and in fact was the motivation for Grothendieck's development of modern algebraic geometry.**Some small finite fields****F**_{2}:*+ | 0 1 × | 0 1 --+---- --+---- 0 | 0 1 0 | 0 0 1 | 1 0 1 | 0 1***F**_{3}:*+ | 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***F**_{4}:*+ | 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***See also*****Finite field arithmetic

*Quasi-finite field

*Trigonometry in Galois fields

*Field with one element **References*****

***External links**** [**http://mathworld.wolfram.com/FiniteField.html Finite Fields*] at Wolfram research.

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