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| In [[abstract algebra]], a '''finite field''' or '''Galois field''' (so named in honor of [[Évariste Galois]]) is a [[field (mathematics)|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]].<ref>A. R. Calderbank, E. M. Reins, P. W. Shor, and N. J. A. Sloane, Quantum error correction via codes over GF(4), IEEE Trans. Inform. Theory 44 (1998), 1369–1387; [http://arxiv.org/abs/quant-ph/9605005 arXiv:quant-ph/9605005]</ref> The finite fields are classified by size; there is exactly one finite field [[up to]] [[isomorphism]] of size ''p''<sup>''k''</sup> for each prime ''p'' and positive integer ''k''. Each finite field of size ''q'' is the [[splitting field]] of the polynomial {{nowrap|''x''<sup>''q''</sup> − ''x''}}, and thus the [[fixed field]] of the [[Frobenius endomorphism]] which takes ''x'' to ''x''<sup>''q''</sup>. Similarly, the [[group of units|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]], [[linear feedback shift register]]s (LFSRs), [[modular representation theory]], and the [[groups of Lie type]]. Finite fields are an active area of research, including recent results on the [[Kakeya set#Kakeya sets in vector spaces over finite fields|Kakeya conjecture]] and open problems on the size of the smallest [[Primitive root modulo n|primitive root]].
| | Apple has confirmed it will buy headphone maker and music-streaming service provider Beats Electronics.<br><br>The deal is worth a total of $3bn (�1.8bn), and is thought to be Apple's largest acquisition to date.<br> |
| | As part of the acquisition, Beats co-founders Jimmy Iovine and Dr Dre will join the technology firm<br> |
| | Apple boss Tim Cook said the deal would allow the firm to "continue to create the most innovative music products and services in the world"<br><br> |
| | Continue reading the main story "Start Quote The Beats deal is totally in character for Apple: everyone is puzzled<br> |
| | End Quote Benedict Evans Technology analyst In a statement, Apple said it is paying an initial $2.6bn (�1.6bn) for Beats, and approximately $400m (�239m) "that will vest over tim<br>. |
| | Beats was founded in 2008 by music producer Jimmy Iovine and hip-hop star Dr Dre and until recently was best known for its headphone<br><br> |
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| Finite fields appear in the following chain of [[subclass (set theory)|class inclusions]]:
| | It started a subscription-based music streaming service earlier this ye<br>. |
| | Apple has its own iTunes store, the world's largest music download service, and launched iTunes Radio last ye<br>. |
| | But despite having been an early pioneer of digital music, the Californian firm has been facing increased competition from subscription services such as Spotify, Pandora and Rd<br>. |
| | Continue reading the main story Analysis Leo Kelion Technology desk editor There had been speculation that [http://www.humitech.co.nz/images/beats-by-dre.asp beats music] Apple might drastically cut its offer price for Beats or pull the deal altogether, after a video of Dr Dre describing himself as hip-hop's first billionaire leaked onli<br><br> |
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| : '''[[Commutative ring]]s''' ⊃ '''[[integral domain]]s''' ⊃ '''[[integrally closed domain]]s''' ⊃ '''[[unique factorization domain]]s''' ⊃ '''[[principal ideal domain]]s''' ⊃ '''[[Euclidean domain]]s''' ⊃ '''[[field (mathematics)|field]]s''' ⊃ '''finite fields'''.
| | But the deal has now been confirmed for only $200m less than what had originally been report<br>. |
| | The press release announcing the tie-up mentions Beats' "premium sound entertainment," but there has been criticism of its headphones' sound quali<br>. |
| | However, there's no doubting their popularity or the skill of Beats' co-founders Jimmy Iovine and Dr Dre in building up the bra<br><br> |
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| == Classification ==
| | It's questionable whether Apple's co-founder Steve Jobs would have been willing to share the limelight with two such outspoken personaliti<br>. |
| The finite fields are classified as follows {{Harv|Jacobson|2009|loc=§4.13,p. 287}}:
| | But the challenge for the more reserved Tim Cook is to bring the two firms' very different cultures together, and to make the most of the Beats brand, as he delivers on a promise to launch new products this ye<br>. |
| * The '''order''', or number of elements, of a finite field is of the form ''p''<sup>''n''</sup>, where ''p'' is a prime number called the '''[[characteristic (algebra)|characteristic]]''' of the field, and ''n'' is a positive integer.
| | However, Beats' music service only has about 250,000 paying subscribers, [http://www.adobe.com/cfusion/search/index.cfm?term=&compared&loc=en_us&siteSection=home compared] with Spotify's 10 milli<br><br> |
| * For every [[prime number]] ''p'' and positive integer ''n'', there exists a finite field with ''p''<sup>''n''</sup> elements.
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| * 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 table]]s become identical to the corresponding tables of the other one.
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| 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 <math> \mathbb{F}_{p^n} </math> or '''F'''<sub>''p''<sup>''n''</sup></sub>. Another notation is '''GF'''(''p''<sup>''n''</sup>), where the letters "GF" stand for "Galois field".
| | 'It's the people' The deal with Beats also marks a departure for Apple, which has a reputation for developing new products in-house, rather than buying up smaller firms - a method preferred by rivals Goog<br>. |
| | But in an interview with the New York Times, Mr Cook hinted that Beats' founders may have been part of the attracti<br>. |
| | "Could Eddy's team [Eddy Cue, Apple executive in charge of iTunes] have built a subscription service? Of course," he told the newspap<br><br> |
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| === Examples ===
| | "You don't build everything yourself. It's not one thing that excites us here. It's the people. It's the servic<br>" |
| First we consider fields where the size is prime, i.e., ''n'' = 1. Such a field is called a [[prime field]], and is [[canonical (disambiguation)|canonically]] isomorphic to the [[Ring (mathematics)|ring]] '''Z'''/''p'''''Z''', the set of [[integers modulo n|integers modulo p]]. It is also sometimes denoted '''Z'''<sub>''p''</sub>, but within some areas of mathematics, particularly number theory, this may cause confusion because the same notation '''Z'''<sub>''p''</sub> is used for the ring of [[p-adic number|p-adic integers]]. However this ring '''Z'''/''p'''''Z''' 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.
| | "These guys are really unique," Mr Cook added. "It's like finding the precise grain of sand on the beach. They're rare and very hard to fin<br>" |
| | The Apple boss also said that Dr Dre and Mr Iovine would be coming up with "products you haven't thought of ye<br><br> |
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| Next we consider fields where the size is not prime, but is a prime power, i.e., ''n'' > 1.
| | Please turn on JavaScript. Media requires JavaScript to pl<br>. |
| | In a video posted online, Dr Dre is shown celebrating the deal before it was even done, as Michelle Fleury repo<br>s |
| | 'Everyone is puzzled' However some analysts remain baffled by the acquisiti<br>. |
| | Technology writer Benedict Evans tweeted: "If you think Apple's lost it, Beats deal is confirmation. If you don't, it's� perplexing. Few really convincing rational<br><br> |
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| Two isomorphic constructions of the field with 4 elements are <math>(\mathbb{Z}/2\mathbb{Z})[T]/(T^2+T+1)=\mathbb{Z}[T]/\langle 2,T^2+T+1\rangle</math> and <math>\mathbb{Z}[\varphi]/(2\mathbb{Z}[\varphi])=\mathbb{Z}[T]/\langle 2,T^2+T-1\rangle</math>, where <math>\varphi=\frac{-1 + \sqrt{5}}{2}</math>. Here ('''Z'''/2'''Z''')[''T''] is the [[polynomial ring]] of '''Z'''/2'''Z''' and ('''Z'''/2'''Z''')[''T'']/(''T''<sup>2</sup>+''T''+1) are the equivalence classes of these polynomials modulo ''T''<sup>2</sup>+''T''+1.
| | Beats co-founder Jimmy Iovine, who made his name as chairman of the Interscope Geffen A&M record label, welcomed the d<br>l. |
| Roughly ''T''<sup>2</sup>+''T''+1=0 so that ''T''<sup>2</sup>=''T''+1 (since −1=1 in '''Z'''/2'''Z''') and hence the elements of ('''Z'''/2'''Z''')[''T'']/(''T''<sup>2</sup>+''T''+1) are the polynomials of degree up to 1 with coefficients in '''Z'''/2'''Z''', i.e. the set {0, 1, ''T'', ''T''+1 } (see below for more details).
| | "I've always known in my heart that Beats belonged with Apple," he s<br>d. |
| Notice that ('''Z'''/2'''Z''')[''T'']/(''T''<sup>2</sup>+1) is not a field since it admits a zero divisor (''T''+1)<sup>2</sup>=''T''<sup>2</sup>+1=0 (since we work
| | Apple released this picture of Beats founders Jimmy Iovine and Dr Dre posing with Mr Cook and Mr Cue "The idea when we started the company was inspired by Apple's unmatched ability to marry culture and technolo<br><br> |
| in '''Z'''/2'''Z''' where 2=0).
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| A field with 8 elements is ('''Z'''/2'''Z''')[''T'']/(''T''<sup>3</sup>+T+1).
| | Apple analyst Jim Dalrymple hinted that it may be Mr Iovine's talent that the technology firm is af<br>r. |
| | Writing on his website The Loop, Mr Dalrymple said: "In my opinion, Jimmy is going to play an important role going forw<br>d. |
| | "Maybe not that you always see, but he'll be the<br>". |
| | As well as headphones, Beats sells earphones and portable speakers, and has even developed partnerships with carmakers and [http://www.humitech.co.nz/images/beats-by-dre.asp beats music] computer manufacturers to include its BeatsAudio software technology in their produc<br><br> |
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| Two isomorphic constructions of the field with 9 elements are ('''Z'''/3'''Z''')[''T'']/(''T''<sup>2</sup>+1) and '''Z'''[i]/(3'''Z'''[i]).
| | The company's flagship products have also received several celebrity endorsements, with stars including Lady Gaga, Lil Wayne and Nicki Minaj designing their own customised Beats headpho<br>s. |
| | | Subject to regulatory approvals, Apple said it expects the deal to close in the fourth financial quarter of this year. |
| Even though all fields of size ''p'' are isomorphic to '''Z'''/''p'''''Z''',
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| for ''n'' ≥ 2 the ring '''Z'''/''p''<sup>''n''</sup>'''Z''' (the [[ring (algebra)|ring]] of integers modulo ''p''<sup>''n''</sup>) is ''not'' a field. The element ''p'' (mod ''p''<sup>''n''</sup>) 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''<sup>2</sup>+''T''+1) of size 4 is not cyclic but rather is isomorphic to the [[Klein four-group]], ('''Z'''/2'''Z''')<sup>2</sup>.
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| A prime power field with ''p''=2 is also called a binary field.
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| 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 (mathematics)|field]].
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| === Proof outline ===
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| The [[Characteristic (algebra)|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''<sup>''n''</sup> elements. A field of order ''p'' exists, because '''F'''<sub>''p''</sub> = '''Z'''/p'''Z''' is a field, where primality is required for the nonzero elements to have multiplicative inverses.
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| For any prime power ''q'' = ''p''<sup>''n''</sup>, '''F'''<sub>''q''</sub> is the [[splitting field]] of the polynomial ''f''(''T'') = ''T''<sup>''q''</sup> − ''T'' over '''F'''<sub>''p''</sub>. This field exists and is unique up to isomorphism by the [[Splitting field#Constructing splitting fields|construction of splitting fields]]. The set of roots is a field, the fixed field of the ''n''th iterate of the [[Frobenius endomorphism]], so the splitting field is exactly the ''q'' roots of this polynomial, which are distinct because the polynomial ''T''<sup>''q''</sup> − ''T'' is separable over '''F'''<sub>''p''</sub>: its derivative is −1, which has no roots.
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| === Detailed proof of the classification ===
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| ====Order====
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| We give two proofs that a finite field has prime-power order.
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| 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 ''F''<sub>''p''</sub> of ''F'' that is isomorphic to '''Z'''/''p'''''Z'''. ''F'' is a [[vector space]] over '''Z'''/''p'''''Z''', and it must have a finite [[dimension (vector space)|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''<sup>''n''</sup>. 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'''.
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| 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.
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| 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''.
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| 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.
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| 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'').
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| 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.
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| ====Existence====
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| The proof of the second statement, concerning the existence of a finite field of size ''q'' = ''p''<sup>''n''</sup> for any prime ''p'' and positive integer ''n'', is more involved. We again give two arguments.
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| The case ''n'' = 1 is easy: take '''F'''<sub>''p''</sub> = '''Z'''/''p'''''Z'''.
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| For general ''n'', inside '''F'''<sub>''p''</sub>[''T''] consider the polynomial ''f''(''T'') = ''T''<sup>''q''</sup> − ''T''. It is possible to construct a field ''F'' (called the [[splitting field]] of ''f''(''T'') over '''F'''<sub>''p''</sub>), which contains '''F'''<sub>''p''</sub> and which is large enough for ''f''(''T'') to split completely into linear factors:
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| :''f''(''T'') = (''T''−''r''<sub>1</sub>)(''T''−''r''<sub>2</sub>)⋯(''T''−''r''<sub>''q''</sub>)
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| in ''F''[''T'']. The existence of splitting fields in general is discussed in [[construction of splitting fields]]. These ''q'' roots are distinct, because ''T''<sup>''q''</sup> − ''T'' is a polynomial of degree ''q'' which has no [[multiple root|repeated root]]s in ''F'': its [[formal derivative|derivative]] is ''qT''<sup>''q''−1</sup> − 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,
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| : ''R'' = { ''r''<sub>1</sub>, ..., ''r''<sub>''q''</sub> } = { roots of the equation ''T''<sup>''q''</sup> = ''T'' }
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| one sees that ''R'' ''itself forms a field'', as follows. Both 0 and 1 are in ''R'', because 0<sup>''q''</sup> = 0 and 1<sup>''q''</sup> = 1. If ''r'' and ''s'' are in ''R'', then
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| :(''r''+''s'')<sup>''q''</sup> = ''r''<sup>''q''</sup> + ''s''<sup>''q''</sup> = ''r'' + ''s''
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| so that ''r''+''s'' is in ''R''. The first equality above follows by induction on ''n'' for ''q'' = ''p''<sup>''n''</sup>, 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
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| : (''rs'')<sup>''q''</sup> = ''r''<sup>''q''</sup> ''s''<sup>''q''</sup> = ''rs''
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| and
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| : (''r''<sup>−1</sup>)<sup>''q''</sup> = (''r''<sup>''q''</sup>)<sup>−1</sup> = ''r''<sup>−1</sup>.
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| Therefore ''R'' is a field with ''q'' elements, proving the second statement.
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| For the second proof that a field of size ''q'' = ''p''<sup>''n''</sup> exists, we just sketch the ideas. We will give a combinatorial argument that a [[Monic polynomial|monic]] irreducible ''f''(''T'') of degree ''n'' exists in '''F'''<sub>''p''</sub>[''T'']. Then the quotient ring '''F'''<sub>''p''</sub>[''T''] / (''f''(''T'')) is a field of size ''q''. Because ''T''<sup>''q''</sup> − ''T'' has no repeated irreducible factors (it is a separable polynomial in '''F'''<sub>''p''</sub>[''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'''<sub>''p''</sub>[''T''] is a factor precisely when its degree divides ''n''. Writing N<sub>''p''</sub>(''d'') for the number of monic irreducibles of degree ''d'' in '''F'''<sub>''p''</sub>[''T''], computing the degree of the irreducible factorization of ''T''<sup>''q''</sup> − ''T'' shows ''q'' = ''p''<sup>''n''</sup> is the sum of ''dN''<sub>''p''</sub>(''d'') over all ''d'' dividing ''n''. This holds for all ''n'', so by Moebius inversion one can get a formula for ''N''<sub>''p''</sub>(''n'') for all ''n'', and a simple lower bound estimate using this formula shows ''N''<sub>''p''</sub>(''n'') is positive. Thus a (monic) irreducible of degree ''n'' in '''F'''<sub>''p''</sub>[''T''] exists, for any ''n''.
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| ====Uniqueness====
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| Finally the uniqueness statement: a field of size ''q'' = ''p''<sup>''n''</sup> is the splitting field of ''T''<sup>q</sup> − ''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.
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| 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.
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| 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''<sup>''n''</sup> are isomorphic to each other in ''n'' ways (because a field of size ''p''<sup>''n''</sup> is isomorphic to itself in ''n'' ways, from Galois theory for finite fields).
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| == Explicitly constructing finite fields ==
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| Given a [[prime power]] ''q'' = ''p''<sup>''n''</sup>, we may explicitly construct a finite field with ''q'' elements as follows. Select a [[Monic polynomial|monic]] [[irreducible polynomial]] ''f''(''T'') of degree ''n'' in '''F'''<sub>''p''</sub>[''T'']. (Such a polynomial is guaranteed to exist, once we know that a finite field of size ''q'' exists: just take the [[Minimal polynomial (field theory)|minimal polynomial]] of any primitive element for that field over the subfield '''F'''<sub>''p''</sub>.) Then '''F'''<sub>''p''</sub>[''T'']/(''f''(''T'')) is a field of size ''q''. Here,
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| '''F'''<sub>''p''</sub>[''T''] denotes the [[ring (algebra)|ring]] of all [[polynomial]]s in ''T'' with coefficients in '''F'''<sub>''p''</sub>, (''f''(''T'')) denotes the [[ideal (ring theory)|ideal]] generated by ''f''(''T''), and the quotient is meant in the sense of [[quotient ring]]s — the set of polynomials in ''T'' with coefficients in '''F'''<sub>''p''</sub> modulo (''f''(''T'')).
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| === Examples ===
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| The polynomial ''f''(''T'') = ''T''<sup> 2</sup> + ''T'' + 1 is irreducible over '''Z'''/2'''Z''', and ('''Z'''/2'''Z''')[''T''] / (''T''<sup>2</sup>+''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''<sup>2</sup> + ''t'' + 1 = 0. In fact, since we are working over '''Z'''/2'''Z''' (that is, in characteristic 2), we may write this as ''t''<sup>2</sup> = ''t'' + 1. (This follows because −1 = 1 in '''Z'''/2'''Z''') Then, for example, to determine ''t''<sup>3</sup>, we calculate: ''t''<sup>3</sup> = ''t''(''t''<sup>2</sup>) = ''t''(''t''+1) = ''t''<sup>2</sup>+''t'' = ''t''+1+''t'' = 2t + 1 = 1, so ''t''<sup>3</sup> = 1.
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| 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''<sup> 2</sup> + ''T'' + 1. The polynomial ''p''(''T'') = ''T'' + 1 works, and hence 1/''t'' = ''t'' + 1.
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| To construct a field of size 27, we could start for example with the irreducible polynomial ''T''<sup> 3</sup> + ''T''<sup> 2</sup> + ''T'' + 2 over '''Z'''/3'''Z'''. The field ('''Z'''/3'''Z''')[''T'']/(''T''<sup> 3</sup> + ''T''<sup> 2</sup> + ''T'' + 2) has size 27. Its elements have the form ''at''<sup>2</sup> + ''bt'' + ''c'' where ''a'', ''b'', and ''c'' lie in '''Z'''/3'''Z''' and the multiplication is defined by ''t''<sup> 3</sup> + ''t''<sup> 2</sup> + ''t'' + 2 = 0, or by rearranging this equation, ''t''<sup>3</sup> = 2''t''<sup>2</sup> + 2''t'' + 1.
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| ===A simple representation of ''F''<sub>''p''<sup>2</sup></sub>===
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| <p style="line-height:175%">
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| Consider the numbers of the form<br> <math>a+b\sqrt c</math> ,<br>
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| where <math>a</math> and <math>b</math> are integers in <math>\mathbb{Z}_p</math> and <math>c</math>
| |
| is a (quadratic) '''''nonresidue''''', meaning <math>c</math> reduced modulo <math>p</math>
| |
| is not the square of any integer in <math>\mathbb{Z}_p</math>.
| |
| </p>
| |
| | |
| Notice that the sum and product of any two is done in the obvious way.
| |
| | |
| <math>(a+b\sqrt c)+(j+k\sqrt c)
| |
| =(a+j)+(b+k)\sqrt c</math>, and
| |
| | |
| <math>(a+b\sqrt c)(j+k\sqrt c)
| |
| =(aj+c\,bk)+(ak+bj)\sqrt c</math>
| |
| | |
| A '''''quadratic residue''''' is, on the other hand, an integer whose value reduced
| |
| modulo <math>p</math> equals the square of some integer in <math>\mathbb{Z}_p</math>. There are exactly
| |
| <math>(p-1)/2</math> nonzero quadratic residues and nonresidues each, for any prime other than
| |
| <math>2</math>.
| |
| | |
| The '''''Quadratic Residue Multiplication Rule''''' states:
| |
| | |
| (i) The product of two quadratic residues modulo <math>p</math> is a quadratic residue.<br>
| |
| (ii) The product of a residue and nonresidue is a nonresidue.<br>
| |
| (iii) The product of two quadratic nonresidues is a quadratic residue.
| |
| | |
| This leads to the conclusion, in our case when <math>c</math> is a nonresidue, that
| |
| <math>a^2 -c\,b^2 \ne 0</math>,
| |
| <br>provided that <math>a</math> and <math>b</math> are not both <math>0</math>.
| |
| | |
| Now, <math>(a+b\sqrt c)(a - b\sqrt c)/(a^2 - c\,b^2)</math>
| |
| <br>is justified, when <math>a+b\sqrt c</math> is not zero, and equals
| |
| <br>
| |
| (<math>a^2 - c\,b^2)/(a^2 - c\,b^2)</math>
| |
| <math>=1.</math>
| |
| | |
| This means the numbers <math>a+b\sqrt c</math>, as described at the start,
| |
| are a field and so <math>F_{p^2}</math>.
| |
| | |
| ===2×2 matrices in integers modulo ''p''===
| |
| Consider a 2×2 matrix '''A''' over the [[integers modulo n|integers modulo '''p''']], where '''p''' is a prime number, of the form:
| |
| | |
| :<math>A =\begin{bmatrix} 0 & b \\ 1 & 1 \end{bmatrix}</math>
| |
| | |
| The set of matrices, let's call it '''F''',
| |
| | |
| :<span style="font-size:140%;">'''{ uI + vA }''',</span>
| |
| | |
| with '''I''' the '''2×2''' identity matrix and '''u, v''' elements of
| |
| '''Z<sub>p</sub>''' forms an additive group of '''p'''<sup>2</sup>
| |
| elements. It is easily seen to be closed with respect to multiplication since '''A'''
| |
| satisfies its characteristic polynomial equation
| |
| '''x'''<sup>2</sup>''' − 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
| |
| '''A'''<sup>2</sup>''' − A − b '''I''' = 0''').
| |
| | |
| : '''(x'''<sup>2</sup>''' + x − b)'''<sup>−1</sup>
| |
| : '''((x + 1) I − A)'''
| |
| | |
| <p style="line-height:180%">
| |
| provided that '''x'''<sup>2</sup>''' + x − b'''
| |
| is not '''0''' for any '''x''' in '''Z<sub>p</sub>'''. The number of '''b''' for which
| |
| '''x'''<sup>2</sup>''' + x − b''' is irreducible over
| |
| '''Z<sub>p</sub>''' is quite numerous, being '''(p − 1)/2'''. It is only left to observe
| |
| that '''v((u/v) I + A)''' = '''u I + v A'''.
| |
| </p>
| |
| | |
| <p style="line-height:180%">
| |
| Determining for which '''b''' that '''A''' is a ''primitive element'' for '''F'''
| |
| <sub>p<sup>2</sup></sub>
| |
| is not as easy to analyze. For '''p''', not too large for brute calculation
| |
| of all the 2x2 matrices '''B''' of integers in '''Z<sub>p</sub>'''
| |
| such that '''B''' <sup>p<sup>2</sup>−1</sup> = '''I''' and
| |
| '''B''' <sup>k</sup> does not equal '''I''',
| |
| for k = 1, 2, ..., p<sup>2</sup>−2, showed many cases.
| |
| Such a matrix '''B''' is a primitive element for
| |
| '''F'''<sub>p<sup>2</sup></sub> .
| |
| To see this notice that
| |
| '''B''' <sup>p<sup>2</sup>−2</sup> = '''B''' <sup>−1</sup>
| |
| and so, if '''B''' <sup>k</sup> = '''B''' <sup>r</sup> for
| |
| 0 < r < k < p<sup>2</sup>−1, then
| |
| '''B''' <sup>k − r</sup> = '''I''', which is assumed not to be the case.
| |
| As noted before, since '''B''' satisfies its characteristic equation, the powers of
| |
| '''B''' must lie in the span of '''u I + v B'''. Since there are
| |
| p<sup>2</sup>−1 of them distinct together with '''0''' they are
| |
| '''F'''<sub>p<sup>2</sup></sub>.
| |
| </p>
| |
| {{col-begin}}
| |
| {{col-break}}
| |
| {| class="wikitable"
| |
| ! ''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
| |
| |}
| |
| {{col-break}}
| |
| {| class="wikitable" style="text-align:right"
| |
| ! 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
| |
| | |
| ''p''<sup>4</sup> is the total number of 2×2 matrices in integers modulo ''p''
| |
| {{col-end}}
| |
| | |
| More generally let '''A''' be a 2x2 matrix of integers in
| |
| '''Z'''<sub>p</sub> with characteristic polynomial
| |
| '''x'''<sup>2</sup>''' + b x + c'''. Then '''F''', the
| |
| span of '''{ uI + vA }''', u, v members of '''Z'''<sub>p</sub>, is a
| |
| ''field'' if and only if
| |
| '''x'''<sup>2</sup>''' + b x + c'''
| |
| is irreducible over '''Z'''<sub>p</sub>.
| |
| Assume '''A''' is independent of '''I''' over '''Z'''<sub>p</sub>.
| |
| | |
| 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''') =
| |
| '''x'''<sup>2</sup> I − '''A'''<sup>2</sup>
| |
| ''' − b x I − b A'''<br> =
| |
| '''x'''<sup>2</sup>''' I + (b A + c I)'''
| |
| ''' − b x I − b A''' =
| |
| '''(x'''<sup>2</sup>''' − b x + c) I'''
| |
| | |
| The polynomial '''x'''<sup>2</sup>''' − b x + c''' is the
| |
| same as '''x'''<sup>2</sup>''' + b x + c''' after
| |
| substituting '''−x''' for '''x''' so their irreducibilities are the same. In the case that
| |
| '''x'''<sup>2</sup>''' + b x + c''' is irreducible
| |
| '''(x'''<sup>2</sup>''' − b x + c)'''<sup>−1</sup>
| |
| ('''(x − b)I − A''') is a multiplicative inverse for
| |
| '''xI + A''' and '''F''' is a ''field''. In the case that
| |
| '''x'''<sup>2</sup>''' − b x + c''' has a
| |
| root '''x''' in '''Z'''<sub>p</sub> 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) = u<sub>0</sub> + u<sub>1</sub>x
| |
| + , ..., + u<sub>k</sub>x<sup>k</sup>'''
| |
| is a polynomial<br> and '''A''' is a ''matrix'' '''P(A) = u<sub>0</sub>I + u<sub>1</sub>A
| |
| + , ..., + u<sub>k</sub>A<sup>k</sup>'''.
| |
| | |
| 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) ='''
| |
| ''' a<sub>0</sub> + a<sub>1</sub>x
| |
| + a<sub>2</sub>x<sup>2</sup> , ...,
| |
| + a<sub>n−1</sub>x<sup>n−1</sup> + x<sup>n</sup> ,'''<br>
| |
| then it satisfies the relation '''A<sup>n</sup> ='''
| |
| −
| |
| '''( a<sub>0</sub>I + ''a''<sub>1</sub>''A'' + ''a''<sub>2</sub>''A''<sup>2</sup> , ...,
| |
| + ''a''<sub>''n''−1</sub>A<sup>''n''−1</sup> ).'''
| |
| | |
| Expressions of the kind ''' u<sub>0</sub>I + u<sub>1</sub>A + u<sub>2</sub>A<sup>2</sup> , ..., + u<sub>n−1</sub>A<sup>n−1</sup> ''' 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''<br>
| |
| '''P(x) = T(x) Q(x) + R(x),''' for some polynomials '''T(x)''' and '''R(x)''' with the degree of '''R(x)'''<br> 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,<br>which would be that '''Q(x)''' is not minimal for '''A'''.
| |
| | |
| ;Theorem
| |
| | |
| Let '''A''' be a '''n×n''' matrix of integers in '''Z'''<sub>p</sub> with characteristic polynomial '''P(x)'''. Then '''F''', the span of {''' u<sub>0</sub>I + u<sub>1</sub>A + u<sub>2</sub>A<sup>2</sup> , ...,
| |
| + u<sub>n−1</sub>A<sup>n−1</sup> }''', with
| |
| '''u<sub>0</sub>, u<sub>1</sub>, ..., u<sub>n−1</sub> ''' members of '''Z'''<sub>p</sub>, is a ''field'' with '''p'''<sup>n</sup> elements if and only if '''P(x)''' is irreducible over '''Z'''<sub>p</sub>.
| |
| | |
| ;Proof
| |
| | |
| First, the necessity of the condition on '''P(x)''' is shown.
| |
| If '''P(x)''' is not irreducible, then there exist 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 ∈ '''Z'''<sub>p</sub>, by the ''Remainder Theorem''<br>
| |
| '''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 <br>
| |
| '''−(P(u))<sup>−1</sup>T(A) (A − u I) = I'''.
| |
| and so '''(A − u I)<sup>−1</sup>'''
| |
| exists and is in '''F'''.
| |
| | |
| Now, let an induction hypothesis be that for any polynomial '''R(x)''' =
| |
| ''' u<sub>0</sub> + u<sub>1</sub>x
| |
| + , ..., + u<sub>k</sub>x<sup>k</sup>''', for '''k''' < '''m''' < '''n''', that '''(R(A))<sup>−1</sup>''' exists and is in '''F'''.
| |
| | |
| Consider '''Q(x)''' =
| |
| ''' u<sub>0</sub> + u<sub>1</sub>x
| |
| + , ...,
| |
| + u<sub>m</sub>x<sup>m</sup>'''.
| |
| | |
| By the ''division algorithm''
| |
| '''P(x) = T(x) Q(x) + R(x),''' for some polynomials
| |
| '''T(x)''' and '''R(x)'''<br> with the degree of '''R(x)''' less than '''m'''.
| |
| | |
| So, '''P(A) = T(A) Q(A) + R(A) = 0 ,'''
| |
| '''−(R(A))<sup>−1</sup> T(A) Q(A) = I''' , and
| |
| | |
| '''(Q(A))<sup>−1</sup> = −(R(A))<sup>−1</sup> T(A)''' exists and is in '''F'''.<br>
| |
| 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 coefficients in '''Z<sub>p</sub>''' having '''x<sup>n</sup>''' resolved by an ''irreducible'' monic polynomial of degree '''n''', in the same manner, is a ''field''.
| |
| | |
| == Properties and facts ==
| |
| Finite fields cannot be [[ordered field|ordered]]: in an ordered field the elements 0 < 1 < {{nowrap|1 + 1}} < {{nowrap|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'' = ''p''<sup>''n''</sup> elements, then
| |
| | |
| :''x''<sup>''q''</sup> = ''x''
| |
| | |
| for all ''x'' in ''F'' (see ''[[Finite field#Analog of Fermat.27s little theorem|Analog of Fermat's little theorem]]'' below). Furthermore, the map
| |
| | |
| :''f'' : ''F'' → ''F''
| |
| | |
| defined by
| |
| | |
| :''f''(''x'') = ''x''<sup>''p''</sup>
| |
| | |
| 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 field|perfect]].
| |
| | |
| The Frobenius automorphism of a field of size ''p''<sup>''n''</sup> has order ''n'', and the [[cyclic group]] it generates is the full [[group (mathematics)|group]] of automorphisms of the field.
| |
| | |
| ===Algebraic closure===
| |
| Finite fields are not algebraically closed: the polynomial
| |
| :<math>f(T)=1+\prod_{\alpha \in F}\left(T-\alpha\right)</math>
| |
| 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'''<sub>''p''<sup>''n''</sup></sub> contains a copy of '''F'''<sub>''p''<sup>''m''</sup></sub> [[if and only if]] ''m'' [[divisor|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 <math>(p^m)^d=p^{md}</math>, so ''m'' divides ''n''. "If" is because there exist irreducible polynomials of every degree over '''F'''<sub>''p''<sup>''m''</sup></sub>.
| |
| | |
| The [[direct limit]] of this system is a field, and is an [[algebraic closure]] of '''F'''<sub>''p''</sub> (or indeed of '''F'''<sub>''p''<sup>''n''</sup></sub> for any ''n''), denoted <math>\bar{\mathbf{F}}_p</math>. This field is infinite, as it is algebraically closed, or more simply because it contains a subfield of size ''p''<sup>''n''</sup> 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 ''p''th power), so the Frobenius map defines an automorphism of <math>\bar{\mathbf{F}}_p</math>, which carries all subfields back to themselves. Unlike in the case of finite fields, the Frobenius automorphism on the algebraic closure of '''F'''<sub>''p''</sub> 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 the [[profinite integers]] (direct product of the ''p''-adic integers over all primes ''p'', with the [[product topology]]).
| |
| | |
| The field '''F'''<sub>''p''<sup>''n''</sup></sub> 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'''<sub>''p''<sup>''n''</sup></sub> is contained in '''F'''<sub>''p''<sup>''m''</sup></sub> whenever ''n'' divides ''m'', then this direct limit can be constructed as the [[union (set theory)|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 '''F'''<sub>''q''</sub> of degree ''n''? Note that since there are only finitely many ''polynomials'' of a given degree ''n'' over the finite field '''F'''<sub>''q''</sub>, 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 '''F'''<sub>''q''</sub> (there are precisely ''q''<sup>''n''</sup>(''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 '''F'''<sub>''q''</sub> 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 '''F'''<sub>''q''</sub> denotes the finite field of order ''q'', then the number ''N'' of ''monic irreducible polynomials'' of degree ''n'' over '''F'''<sub>''q''</sub> is given by:<ref>{{harvnb|Jacobson|2009|loc=§4.13}}</ref>
| |
| | |
| :<math>N(q,n)=\frac{1}{n}\sum_{d|n} \mu(d)q^{\frac{n}{d}},</math>
| |
| | |
| where μ is the [[Möbius function]]. By the above formula, the number of ''irreducible polynomials'' of degree ''n'' over '''F'''<sub>''q''</sub> is given by <math>(q-1)N(q,n)</math>. A (slightly simpler) lower bound on ''N'' also exists, and is given by:
| |
| | |
| :<math>N\geq\frac{1}{n} \left(q^n-\sum_{p|n, \; p \text{ prime }} q^{\frac{n}{p}}\right).</math>
| |
| | |
| ====Algorithm for computing the irreducible factors of a given polynomial over a finite field====
| |
| {{Main|Berlekamp algorithm|Cantor–Zassenhaus algorithm}}
| |
| | |
| ===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 ring]]s are commutative, hence finite fields. The result holds even if we relax associativity and consider [[alternative ring]]s, by the [[Artin–Zorn theorem]].
| |
| | |
| == Multiplicative structure ==
| |
| | |
| ===Cyclic===
| |
| The multiplicative [[group (mathematics)|group]] of every finite field is cyclic, a special case of [[field (mathematics)#Some first theorems|a theorem mentioned in ''Fields'']].
| |
| A generator for the multiplicative group is a ''[[primitive element (finite field)|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''<sup>2</sup>, ..., ''x''<sup>''q''-2</sup> }.
| |
| | |
| The primitive element ''x'' is not unique (unless ''q'' = 2 or 3): the set of generators has size <math>\varphi(q-1)</math> where <math>\varphi</math> is [[Euler's totient function]]. If we fix a generator, then for any non-zero element ''a'' in ''F''<sub>''q''</sub>, there is a unique integer ''n'' with
| |
| | |
| :0 ≤ ''n'' ≤ ''q'' − 2
| |
| | |
| such that
| |
| | |
| :''a'' = ''x''<sup>''n''</sup>.
| |
| | |
| The value of ''n'' for a given ''a'' is called the ''[[discrete logarithm|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 ''a''<sup>''q''</sup> = ''a''. When ''q'' is prime, this is just [[Fermat's little theorem]], which states that ''a''<sup>''p''</sup> ≡ ''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 (group theory)|Lagrange's theorem]] ''a''<sup>''q''−1</sup> = 1 for any nonzero ''a'' in the field. Then ''a''<sup>''q''</sup> = ''a'' and this holds for 0 as well.
| |
| | |
| === Roots of unity ===
| |
| Let <math>n</math> be a positive integer, a <math>n</math>-th ''root of unity'' in a finite field <math>\mathbb{F}</math> is a solution of the equation <math>x^n = 1</math>, a <math>n</math>-th ''primitive root of unity'' is a solution of the equation <math>x^n = 1</math> that is not a solution of the equation <math>x^m = 1</math> for any positive integer <math>m < n</math>. Unlike the <math>n</math>-th [[roots of unity]] in <math>\mathbb{C}</math>, the number of <math>n</math>-th roots of unity in <math>\mathbb{F}</math> may be less than <math>n</math>. If <math>\mathbb{F}</math> has <math>q</math> elements, then the number of <math>n</math>-th roots of unity in <math>\mathbb{F}</math> is <math>\gcd(n, q - 1)</math>. If <math>n \nmid q - 1</math> then <math>\mathbb{F}</math> has no primitive <math>n</math>-roots of unity, while if <math>n \mid q - 1</math> then the number of primitive <math>n</math>-th roots of unity in <math>\mathbb{F}</math> is <math>\varphi(n)</math>, where <math>\varphi(\cdot)</math> is the [[Euler totient function]].
| |
| | |
| == Applications ==
| |
| Discrete exponentiation, also known as calculating ''a'' = ''x''<sup>''n''</sup> from ''x'' and ''n'', can be computed quickly using techniques of [[Exponentiation#Efficiently computing a power|fast exponentiation]] such as
| |
| [[Exponentiation by squaring|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 [[Discrete logarithm problem#Cryptography|applications in cryptography]], such as the [[Diffie-Hellman]] protocol.
| |
| | |
| Finite fields also find applications in [[coding theory]]: many codes are constructed as [[linear subspace|subspace]]s of [[vector 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]], the [[Weil conjectures]], and in fact was the motivation for Grothendieck's development of modern algebraic geometry.
| |
| | |
| == Some small finite fields ==
| |
| | |
| === [[GF(2)|F<sub>2</sub>]] ===
| |
| {|
| |
| |-
| |
| |
| |
| {| class="wikitable" style="text-align: center; width: 81px; height: 81px;"
| |
| |-
| |
| ! + !! 0 !! 1
| |
| |-
| |
| ! 0
| |
| | 0 || 1
| |
| |-
| |
| ! 1
| |
| | 1 || 0
| |
| |}
| |
| |
| |
| {| class="wikitable" style="text-align: center; width: 81px; height: 81px;"
| |
| |-
| |
| ! × !! 0 !! 1
| |
| |-
| |
| ! 0
| |
| | 0 || 0
| |
| |-
| |
| ! 1
| |
| | 0 || 1
| |
| |}
| |
| |}
| |
| | |
| === F<sub>3</sub> ===
| |
| {|
| |
| |-
| |
| |
| |
| {| class="wikitable" style="text-align: center; width: 108px; height: 108px;"
| |
| |-
| |
| ! + !! 0 !! 1 !! 2
| |
| |-
| |
| ! 0
| |
| | 0 || 1 || 2
| |
| |-
| |
| ! 1
| |
| | 1 || 2 || 0
| |
| |-
| |
| ! 2
| |
| | 2 || 0 || 1
| |
| |}
| |
| |
| |
| {| class="wikitable" style="text-align: center; width: 108px; height: 108px;"
| |
| |-
| |
| ! × !! 0 !! 1 !! 2
| |
| |-
| |
| ! 0
| |
| | 0 || 0 || 0
| |
| |-
| |
| ! 1
| |
| | 0 || 1 || 2
| |
| |-
| |
| ! 2
| |
| | 0 || 2 || 1
| |
| |}
| |
| |}
| |
| | |
| === F<sub>4</sub> ===
| |
| {|
| |
| |-
| |
| |
| |
| {| class="wikitable" style="text-align: center; width: 135px; height: 135px;"
| |
| |-
| |
| ! + !! 0 !! 1 !! A !! B
| |
| |-
| |
| ! 0
| |
| | 0 || 1 || A || B
| |
| |-
| |
| ! 1
| |
| | 1 || 0 || B || A
| |
| |-
| |
| ! A
| |
| | A || B || 0 || 1
| |
| |-
| |
| ! B
| |
| | B || A || 1 || 0
| |
| |}
| |
| |
| |
| {| class="wikitable" style="text-align: center; width: 135px; height: 135px;"
| |
| |-
| |
| ! × !! 0 !! 1 !! A !! B
| |
| |-
| |
| ! 0
| |
| | 0 || 0 || 0 || 0
| |
| |-
| |
| ! 1
| |
| | 0 || 1 || A || B
| |
| |-
| |
| ! A
| |
| | 0 || A || B || 1
| |
| |-
| |
| ! B
| |
| | 0 || B || 1 || A
| |
| |}
| |
| |}
| |
| | |
| === F<sub>8</sub> ===
| |
| <pre>
| |
| 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
| |
| </pre>
| |
| | |
| === F<sub>9</sub> ===
| |
| <pre>
| |
| 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
| |
| | |
| </pre>
| |
| | |
| === F<sub>16</sub>===
| |
| | |
| '''F'''<sub>16</sub> is represented by the polynomials
| |
| '''a + b x + c x'''<sup>2</sup>''' + d x'''<sup>3</sup>.<br>
| |
| '''a''', '''b''', '''c''', and '''d''' are integers modulo '''2'''<br>
| |
| The polynomials are generated by the powers of '''x''' using the rule
| |
| | |
| '''x'''<sup>4</sup>''' = 1 + x'''.
| |
| | |
| <pre>
| |
| 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
| |
| </pre>
| |
| | |
| === F<sub>25</sub>===
| |
| | |
| '''F'''<sub>25</sub> represented by the numbers
| |
| '''a + b√2''',
| |
| '''a''' and '''b''' are integers modulo '''5'''<br>
| |
| generated by powers of '''2 + √2'''
| |
| | |
| {| cellspacing="10" cellpadding="0"
| |
| |- align="center"
| |
| | e ( 0) || e ( 1) || e ( 2) || e ( 3) || e ( 4)
| |
| |- align="center"
| |
| | 0 + 0√2 || 1 + 0√2 || 2 + 1√2 || 1 + 4√2 || 0 + 4√2
| |
| |- align="center"
| |
| | e ( 5) ||e ( 6) || e ( 7) || e ( 8) || e ( 9)
| |
| |- align="center"
| |
| | 3 + 3√2 || 2 + 4√2 || 2 + 0√2 || 4 + 2√2 || 2 + 3√2
| |
| |- align="center"
| |
| | e (10) || e (11) || e (12) || e (13) || e (14)
| |
| |- align="center"
| |
| | 0 + 3√2 || 1 + 1√2 || 4 + 3√2 || 4 + 0√2 || 3 + 4√2
| |
| |- align="center"
| |
| | e (15) || e (16) || e (17) || e (18) || e (19)
| |
| |- align="center"
| |
| | 4 + 1√2 || 0 + 1√2 || 2 + 2√2 || 3 + 1√2 || 3 + 0√2
| |
| |- align="center"
| |
| | e (20) || e (21) || e (22) || e (23) || e (24)
| |
| |- align="center"
| |
| | 1 + 3√2 || 3 + 2√2 || 0 + 2√2 || 4 + 4√2 || 1 + 2√2
| |
| |}
| |
| | |
| {| class="wikitable" style="text-align: center;"
| |
| ! + !! 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
| |
| |}
| |
| | |
| {| class="wikitable" style="text-align: center;"
| |
| |-
| |
| ! × !! 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
| |
| |}
| |
| | |
| == See also ==
| |
| * [[Finite field arithmetic]]
| |
| * [[Quasi-finite field]]
| |
| * [[Trigonometry in Galois fields]]
| |
| * [[Field with one element]]
| |
| * [[Finite ring]]
| |
| | |
| ==Notes==
| |
| {{Reflist}}
| |
| | |
| == References ==
| |
| *{{citation
| |
| | last=Jacobson
| |
| | first=Nathan
| |
| | author-link=Nathan Jacobson
| |
| | title=Basic algebra I
| |
| | year=2009
| |
| | edition=Second
| |
| | publisher=Dover Publications
| |
| | isbn=978-0-486-47189-1
| |
| | origyear=1985
| |
| }}
| |
| * {{Citation | last=Lidl | first=Rudolf | last2=Niederreiter | first2=Harald | title=Finite Fields | edition=2nd | year=1997 | publisher=[[Cambridge University Press]] | isbn=0-521-39231-4 }}
| |
| | |
| == External links ==
| |
| * [http://mathworld.wolfram.com/FiniteField.html Finite Fields] at Wolfram research.
| |
| | |
| {{DEFAULTSORT:Finite Field}}
| |
| [[Category:Finite fields| ]]
| |
| [[Category:Field theory]]
| |
| | |
| {{Link GA|fr}}
| |