The study of solutions to polynomial equations over finite fields has a long
history in mathematics and is an interesting area of contemporary research. In
recent years the subject has found important applications in the modelling of
problems from applied mathematical fields such as signal analysis, system
theory, coding theory and cryptology. In this connection it is of interest to
know criteria for the existence of squares and other powers in arbitrary finite
fields. Making good use of polynomial division in polynomial rings over finite
fields, we have examined a classical criterion of Euler for squares in odd
prime fields, giving it a formulation which is apt for generalization to
arbitrary finite fields and powers. Our proof uses algebra rather than
classical number theory, which makes it convenient when presenting basic
methods of applied algebra in the classroom.Comment: 4 page

Aabrandt, Andreas

Hansen, Vagn Lundsgaard

English

arXiv

Andreas Aabrandt

Vagn Lundsgaard Hansen

Crossref

International Journal of Mathematical Education in Science and Technology

A note on powers in finite fields

The study of solutions to polynomial equations over finite fields has a long history in mathematics and is an interesting area of contemporary research. In recent years the subject has found important applications in the modelling of problems from applied mathematical fields such as signal analysis, system theory, coding theory and cryptology. In this connection it is of interest to know criteria for the existence of squares and other powers in arbitrary finite fields. Making good use of polynomial division in polynomial rings over finite fields, we have examined a classical criterion of Euler for squares in odd prime fields, giving it a formulation which is apt for generalization to arbitrary finite fields and powers. Our proof uses algebra rather than classical number theory, which makes it convenient when presenting basic methods of applied algebra in the classroom

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Downloaded from orbit.dtu.dk on: Dec 21, 2017A Note on Powers in Finite FieldsAabrandt, Andreas; Hansen, Vagn LundsgaardPublished in:International Journal of Mathematical Education in Science and TechnologyLink to article, DOI:10.1080/0020739X.2015.1129076Publication date:2016Document VersionPeer reviewed versionLink back to DTU OrbitCitation (APA):Aabrandt, A., & Hansen, V. L. (2016). A Note on Powers in Finite Fields. International Journal of MathematicalEducation in Science and Technology, 47(6), 987-991. DOI: 10.1080/0020739X.2015.1129076arXiv:1510.06243v1  [math.NT]  21 Oct 2015A NOTE ON POWERS IN FINITE FIELDSANDREAS AABRANDT AND VAGN LUNDSGAARD HANSENAbstract. The study of solutions to polynomial equations over finitefields has a long history in mathematics and is an interesting area ofcontemporary research. In recent years the subject has found importantapplications in the modelling of problems from applied mathematicalfields such as signal analysis, system theory, coding theory and cryptol-ogy. In this connection it is of interest to know criteria for the existenceof squares and other powers in arbitrary finite fields. Making good useof polynomial division in polynomial rings over finite fields, we haveexamined a classical criterion of Euler for squares in odd prime fields,giving it a formulation which is apt for generalization to arbitrary finitefields and powers. Our proof uses algebra rather than classical numbertheory, which makes it convenient when presenting basic methods ofapplied algebra in the classroom.Subject class: 11A15, 12C15Keywords: Finite fields, prime numbers, squares and powers in finite fieldsFor any prime number p we denote by Fp the field of residue classes ofintegers modulo p. And more generally, for any integer n ≥ 1, we denote byFq the finite field with q = pn elements. We identify Fp with the prime fieldof Fq by identifying 1 ∈ Fp with the identity element e ∈ Fq. Denote by F∗qthe group of nonzero elements in Fq.For the prime fields Fp, the following result was known to Euler; [1], [2].Theorem 1 (Euler’s Criterion). Let p be an odd prime, and let a ∈ F∗p.Then there exists an element x ∈ F∗p such that a = x2 in F∗p if and only ifa(p−1)/2 ≡ 1 (mod p).This criterion can be generalized to all finite fields as follows.Theorem 2 (Generalized Euler’s Criterion). Let q = pn for an odd prime pand an arbitrary integer n ≥ 1, and let a ∈ F∗q. Then there exists an elementx ∈ F∗q such that a = x2 in F∗q if and only ifa(q−1)/2 = 1.Proof. The finite field Fq is uniquely determined up to an isomorphism asthe splitting field for the polynomial f(x) = xq − x over Fp, and consists ofthe q roots of f(x); see [3].From this description follows easily that there exists an element x ∈ F∗qsuch that a = x2 ∈ F∗q if and only if the polynomial g(x) = x2−a splits intolinear factors and hence is a divisor in f(x) = xq − x.By polynomial division we getf(x) = xq − x = h(x)(x2 − a) + (a(q−1)/2 − 1)x,whereh(x) = xq−2 + axq−4 + a2xq−6 + · · ·+ a(q−3)/2x.12 ANDREAS AABRANDT AND VAGN LUNDSGAARD HANSENFrom the above follows immediately that g(x) is a divisor in f(x) andhence that there exists an element x ∈ F∗q such that x2 = a if and only ifa(q−1)/2 = 1.Example. For p = 3, the series of fields F3n for n ≥ 1, exhibits interestingphenomena for the existence of nontrivial squares, i.e. squares 6= 0, 1.The finite field F3 contains no nontrivial squares as can easily be seen bydirect computations, or, by using Euler’s criterion.The finite field F32 can be described as the polynomial ring F3[t] modulothe irreducible polynomial t2+1. The nine elements in F32 are then uniquelydescribed by the nine polynomials x = a0+a1t, for a0, a1 ∈ F3, and t2+1 = 0.It follows that t2 = −1 = 2 and (1+ 2t)2 = 1+4t+4t2 = t, showing that 2,t, and 2t are the three nontrivial squares in F32 . This is in accordance withthe generalized Euler’s criterion.Using the generalized Euler’s criterion, it can be proved by induction that2 is a square in the finite field F3n for an arbitrary integer n ≥ 1 if and onlyif n is even. The formula 2(3n−1)/2 = (2(3n−2−1)/2)9, which holds in F3, isuseful for the induction step.The fields F2 and F3 are the only prime fields without nontrivial squares.The number of nontrivial squares in an arbitrary finite field is given byTheorem 3. The finite field Fq, with q = pn for an odd prime p and aninteger n ≥ 1, containsq − 12− 1nontrivial squares.Proof. The multiplicative group F∗q is a cyclic group of order q−1 generatedby an element γ ∈ F∗q; see [3]. Every element in F∗q has then a uniquepresentation as a power γk of γ, where the exponent k is counted modulo q.The squaring homomorphism x2 : F∗q → F∗q maps the element a = γl ∈ F∗qinto c = γ2l ∈ F∗q. From this we conclude that c = γk is a square in F∗q ifand only if k is even modulo q. Hence there are equally many squares andnon-squares in F∗q. Now the theorem follows immediately. It is not easy to find a square root of an element in the finite field Fpnfor large primes p and integers n ≥ 1. Even using the generalized Euler’scriterion it is fairly complicated just to establish that an element has a squareroot without the use of a computer.Example. Consider the finite field F133 . Using the irreducible polynomialt3 + 2t+ 11 in the polynomial ring F13[t], the field can be described asF133 = F13[t]/(t3 + 2t+ 11).By the generalized Euler’s criterion we have(5 + t+ 8t2)(133−1)/2 = (5 + t+ 8t2)1098 = 1,showing that 5 + t+ 8t2 is a square. And indeed,(7 + t+ 2t2)2 = 5 + t+ 8t2.A NOTE ON POWERS IN FINITE FIELDS 3The results on squares in a finite field Fq can be extended to higher powers.This leads to our main result in Theorem 4. It should be mentioned thatTheorem 4 can be derived from the similar Proposition 7.1.2 in [2]. Ourproof follows, however, different lines and uses algebra rather than classicalnumber theory, which in our opinion makes the proof more transparent.Let r be a positive integer in the interval 2 ≤ r < q − 1. If r is not adivisor in the order q− 1 of the multiplicative group F∗q, then the rth powerhomomorphism xr : F∗q → F∗q is injective and hence an isomorphism in F∗q.If r is not a divisor in q − 1, every element in F∗q is therefore an rth power.In case r is a divisor in q − 1 we haveTheorem 4 (Extended Euler’s Criterion). Let q = pn for an odd prime pand an arbitrary integer n ≥ 1. Suppose the number r is a proper divisor inq − 1. Then it holds that(1) An element a ∈ F∗q is an rth power in F∗q if and only if a(q−1)/r = 1.(2) There are exactly q−1r different rth powers in F∗q.Proof. We exploit again that Fq is the splitting field for the polynomialf(x) = xq − x in Fp, and is generated by the q roots of f(x); see [3].The multiplicative group F∗q is a cyclic group of order q − 1. Choose agenerator γ ∈ F∗q. Then the powers γi, i = 1, . . . , q− 1, runs through all theelements in F∗q.Let r be a proper divisor in q − 1 and put k = q−1r .The rth power homomorphism xr : F∗q → F∗q maps γi into γir, showingthat the k elements aj = γjr, j = 1, . . . , k, are exactly the rth powers in F∗q.This proves part (2) of the theorem.For a ∈ F∗q, consider the polynomial g(x) = xr − a in Fq.By polynomial division in Fq[x] we getf(x) = xq − x = h(x)(xr − a) + (a(q−1)/r − 1)x,whereh(x) = xq−r + axq−2r + a2xq−3r + · · ·+ a(q−3)/rx.It follows that g(x) is a divisor in f(x) if and only if a(q−1)/r = 1.Sincef(x) = xq − x = x(x− γ1)(x− γ2) . . . (x− γq−1),it is clear that g(x) = xr − a is a divisor in f(x) if and only if it splitscompletely into linear factors, or in other words, if and only if a ∈ F∗q is therth power of r roots γi1 , . . . , γir ∈ F∗q of f(x).Combining information we conclude that a ∈ F∗q is an rth power in F∗q ifand only if a(q−1)/r = 1, thus proving part (1) of the theorem. Example. The finite field F52 can be described as the polynomial ring F5[t]modulo the irreducible polynomial t2 + t+ 1, i.e. F52 = F5[t]/(t2 + t+ 1).For each of the proper divisors r = 2, 3, 4, 6, 8, 12 in 24, which is the orderof the multiplicative group F∗52 , we can using the extended Euler’s criterionby hand determine which elements in the prime field F5 are rth powers in F∗52 .r = 2, 3, 64 ANDREAS AABRANDT AND VAGN LUNDSGAARD HANSENAll elements in F∗5 are rth powers in F∗52 .In fact: 2 = (1 + 2t)2 = 33 = (3 + t)6.3 = (3 + t)2 = 23 = (1 + 2t)6.4 = 22 = 43 = 26.r = 4The elements 1, 4 are 4-powers in F∗52 , and 2, 3 are not 4-powers in F∗52 .In fact: 4 = (1 + 2t)4.r = 8The elements 2, 3, 4 are not 8-powers in F∗52 .r = 12The elements 1, 4 are 12-powers in F∗52 , and 2, 3 are not 12-powers in F∗52 .In fact: 4 = (1 + 2t)12.Example. As a more complicated example, consider again the finite fieldF133 = F13[t]/(t3 + 2t+ 11).Making use of the extended Euler’s criterion we investigate the existenceof rth powers in F∗133 for r = 12, 61, 1098.The computation(5 + 7t)(133−1)/12 = 1,shows that 5 + 7t is a power of degree 12. In fact: (t2)12 = 5 + 7t.The computation(5 + 3t+ 7t2)(133−1)/61 = (5 + 3t+ 7t2)36 = 1,shows that 5 + 3t+ 7t2 is a power of degree 61. In fact:(6 + 2t)61 = 5 + 3t+ 7t2.The computation12(133−1)/1098 = 122 = 1,shows that 12 is a power of degree 1098. In fact: t1098 = 12.References[1] H. Davenport, The Higher Arithmetic. An Introduction to the Theory of Numbers,Dover Publications, Inc., New York, 1983.[2] K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory,Springer, 1982.[3] S. Lang, Algebra, Springer, 2005.Technical University of Denmark

A Note on Powers in Finite Fields

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