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Construction of Irreducible Polynomials Over Finite Fields

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Construction of Irreducible Polynomials Over Finite Fields U.U.D.M. Project Report 2014:17 Construction of Irreducible Polynomials over Finite Fields Gustav Hammarhjelm Examensarbete i matematik, 15 hp Handledare och examinator: Karl-Heinz Fieseler Maj 2014 Department of Mathematics Uppsala University Construction of irreducible polynomials over finite fields Gustav Hammarhjelm May 22, 2014 Contents 1 Introduction 3 2 Basic results on finite fields 4 2.1 The reciprocal of a polynomial . 7 2.2 The Mobius¨ inversion formula . 8 3 Finding irreducible polynomials (examples) 9 4 Sequences of irreducible polynomials 12 4.1 The Q-transformation and the trace . 12 4.2 Sequences of irreducible polynomials over finite fields of characteristic 2 15 4.3 Sequences of irreducible polynomials over finite fields of odd characeristic 18 4.4 The polynomial xqn+1 − 1.......................... 23 References 25 1 Abstract In this paper we investigate some results on the construction of irreducible poly- nomials over finite fields. Basic results on finite fields are introduced and proved. Several theorems proving irreducibility of certain polynomials over finite fields are presented and proved. Two theorems on the construction of special sequences of irreducible polynomials over finite fields are investigated in detail. Acknowledgements I would like to thank my supervisor Karl-Heinz Fieseler for guidance, inspiration and insightful comments. I would also like to thank my family for their support. 2 1 Introduction The concept of a prime number is well known. The properties that make prime numbers interesting include (but are not limited to) the fact that a prime number does not admit any non-trivial factorization in integers and that if a prime number divides a product of numbers, it necessarily divides one of the factors. The first quality is what defines an irreducible element in any unital ring: Definition 1.1. Let R be a commutative ring with unity and let r 2 R. A non-zero, non- unit r is said to be irreducible if r = ab for a; b 2 R implies a is a unit or b is a unit. If one is challenged to find, explicitly, infinite sequences of distinct irreducible ele- ments of a ring one can have various outcomes: In the ring Z the irreducible elements are ±p where p is any prime number. As of today, as far as I know, nobody has come up with an explicit infinite sequence of distinct prime numbers. The challenge turns out to be a rather modest one in some rings. For instance in Q[x], the polynomial ring over the field of rational numbers, it is very easy to explicitly define sequences of irreducible elements, e.g. the sequence xn − 2 where n is a non-zero natural number, using Eisenstein’s criterion. In this text we will consider the setting when R is the polynomial ring Fq[x] over a finite field Fq. A non-constant polynomial f (x) of Fq[x] is called irreducible over Fq if f (x) = g(x)h(x) for polynomials g(x); h(x) 2 Fq[x] implies g(x) or h(x) is a unit, i.e. g(x) or h(x) is in Fq, according to the definition of irreducibility. We will show that it is indeed possible (but requires more work than in Q[x]) to generate infinite sequences of irreducible elements of strictly increasing degrees over Fq[x] for various finite fields Fq[x]. The existence of such sequences are not only valuable for recreational purposes, they may also be used for applications in mathematics. Indeed, one important role of irre- ducible polynomials is that one can explicitly construct fields using irreducible polyno- mials through factor rings. If one wants to make explicit calculations in say a finite field, it is often required to find an irreducible polynomial, in order to get information of the structure of the field. This is important for applications of field theory, for instance error correcting codes. In this text we shall, after presenting some auxiliary results, investigate some ways of recognizing irreducible polynomials over finite fields. In the last part, we carefully investigate a theorem on the construction of infinite sequences of irreducible polynomials of increasing degree over finite fields. 3 2 Basic results on finite fields Firstly, some notation that will be used in the text. If F and K are fields F > K expresses that F is a field extension of K (or K is a subfield of F). If the extension is finite, then [F : K] denotes the dimension of F over K, when F is considered a vector space over K. If α1; : : : ; αn are algebraic over F then F(α1; : : : ; αn) is the extension of F obtained by adjoining α1; : : : ; αn to F. F[x] denotes the polynomial ring over F and Fq denotes the ∗ finite field of q elements, Fq its multiplicative group. Some fundamental results of algebra shall be used frequently, but will not be proved here, for instance that there is a finite field of pn elements for each prime p and each positive natural number n, unique up to isomorphism, as well as the tower law for finite extensions and that the multiplicative group of a finite field is cyclic. Theorem 2.1. Let F be a finite field of characteristic p. Then n n n n n n (a + b)p = ap + bp ; (a − b)p = ap − bp for a; b 2 F; n 2 N>0. Proof. By the binomial theorem for commutative rings p ! X p (a + b)p = akbp−k k k=0 p p pn pn−1 p where each p j k for each 0 < k < p so (a + b) = a + b. Now (a + b) = ((a + b) ) and the first result follows by induction. For the second result n n n n (a − b)p = (a + (−b))p = ap + (−b)p : Now if p is odd, (−1)pn = −1, if p is even −1 = 1 so in either case we have obtained the other result. Theorem 2.2. Let Fq be a finite field and let f 2 Fq[x] be irreducible over Fq, deg f = n. Then the splitting field of f is Fqn . Furthermore, if α is a zero of f , then the other zeros of f are given by αq; : : : ; αqn−1 . Proof. The theorem is trivial if n = 1 so assume n > 1. Let α be a zero in the splitting field of f , α , 0 since f (x) irreducible. [Fq(α): Fq] = n so Fq(α) Fqn . Now, suppose Pn k Pn k f (x) = k=0 ak x , so that f (α) = k=0 akα = 0. By theorem 2.1, for 0 < i < n i 0 n 1q n BX C X i i i 0 = B a αkC = aq αkq = f (αq ); @B k AC k k=0 k=0 q qi q j since ak = ak, as ak 2 Fq. It remains to show that α = α , 0 ≤ i; j < n implies i = j (so that we really have obtained n distinct zeros of f ), until we, with clear conscience, may declare Fq(α) Fqn the splitting field of f . To this end, we use the fact that an irreducible polynomial f (x) of degree m over a qn finite field Fq divides x − x if and only if m j n. If m j n then Fqm < Fqn and as Fqn consists of the zeros of xqn − x each zero of f (x) is a zero of xqn − x so f (x) divides this polynomial. 4 qn Conversely, if f (x) j x − x and β is a zero of f (x) in Fqm , we have the equality qn Fqm = F(β) since f (x) is irreducible of degree m, then α is a zero of x − x as well and thus α 2 Fqn . Therefore we have Fq < Fqm < Fqn and m j n by the tower law of finite field extensions. i j Now, for a contradiction, assume αq = αq , 0 ≤ i; j < n. Then, since α , 0 we have i j i j−i j−i i αq = αq () αq (q −1) = 1 () (αq −1)q = 1 by raising the right hand side to the power qn−i and multiplying with α we get αq j−i = α q j−i−1 j−i since α 2 Fqn . Thus, α is a zero of x − x and so m j j − i with 0 < j − i < m which is absurd. Remark 2.3. We have seen in the proof of the last theorem that an irreducible polynomial over a finite field of degree m must have m distinct zeros. With this information we can deduce that polynomials of certain forms are never irreducible. p Let Fq be a field of characteristic p and consider the polynomial x +a for some a 2 Fq. p p p p p Let α be a zero of x + a = 0 with α 2 Fqp . Then (x − α) = x − α = x + a and we see that the only zero of xp + a = 0 is α and since p > 1 the polynomial xp + a must be reducible over Fq since if it would be irreducible, it would have p distinct zeros. Definition 2.4. Let F be a field and K be a subfield of F. An automorphism σ of F is an automorphism of F over K if σ(a) = a for all a 2 K. Theorem 2.5. Let Fq and Fqm , m > 1 be finite fields. Then the automorphisms of Fqm over qi Fq are precisely σi; i = 1;:::; m where σi(α) = α for all α 2 Fqm .
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