Let $\mathcal S$ be a set of monic degree $2$ polynomials over a finite field
and let $C$ be the compositional semigroup generated by $\mathcal S$. In this
paper we establish a necessary and sufficient condition for $C$ to be
consisting entirely of irreducible polynomials. The condition we deduce depends
on the finite data encoded in a certain graph uniquely determined by the
generating set $\mathcal S$. Using this machinery we are able both to show
examples of semigroups of irreducible polynomials generated by two degree $2$
polynomials and to give some non-existence results for some of these sets in
infinitely many prime fields satisfying certain arithmetic conditions

A Batra

A Ostafe

D Gomez

GL Mullen

J ZurGathen von

O Ahmadi

R Barbulescu

R Jones

R Lidl

S Gao

V Shoup

English

arXiv

Let S be a set of monic degree 2 polynomials over a finite field and let C be the compositional semigroup generated by S. In this paper we establish a necessary and sufficient condition for C to be consisting entirely of irreducible polynomials. The condition we deduce depends on the finite data encoded in a certain graph uniquely determined by the generating set S. Using this machinery we are able both to show examples of semigroups of irreducible polynomials generated by two degree 2 polynomials and to give some non-existence results for some of these sets in infinitely many prime fields satisfying certain arithmetic conditions

Ferraguti, Andrea

Micheli, Giacomo

Schnyder, Reto

Archivio istituzionale della ricerca - Università di Brescia

On Sets of Irreducible Polynomials Closed by Composition

Archivio istituzionale della Ricerca - Scuola Normale Superiore

arXiv:1604.05223v1  [math.NT]  18 Apr 2016ON SETS OF IRREDUCIBLE POLYNOMIALS CLOSED BYCOMPOSITIONANDREA FERRAGUTI, GIACOMO MICHELI, AND RETO SCHNYDERAbstract. Let S be a set of monic degree 2 polynomials over a finitefield and let C be the compositional semigroup generated by S. In thispaper we establish a necessary and sufficient condition for C to be consist-ing entirely of irreducible polynomials. The condition we deduce dependson the finite data encoded in a certain graph uniquely determined by thegenerating set S. Using this machinery we are able both to show exam-ples of semigroups of irreducible polynomials generated by two degree 2polynomials and to give some non-existence results for some of these setsin infinitely many prime fields satisfying certain arithmetic conditions.1. IntroductionSince irreducible polynomials play a fundamental role in applications andin the whole theory of finite fields (see for example [1, 2, 3, 4, 5, 6]), relatedquestions have a long history (see for example [7, 8, 9, 10, 11, 12, 13]). Inthis paper we specialize on irreducibility questions regarding compositionalsemigroups of polynomials. This kind of question has been addressed in thespecific case of semigroups generated by a single quadratic polynomial, see forexample in [6, 5, 11, 10], for analogous results related to additive polynomials,see [14, 15]. It is worth mentioning that one of these results [11, Lemma 2.5]has been recently used in [16] by the first and the second author of the presentpaper to prove [12, Conjecture 1.2].Throughout the paper, q will be an odd prime power, Fq[x] the univariatepolynomial ring over the finite field Fq and Irr(Fq[x]) the set of irreduciblepolynomials in Fq[x]. Let us give an example which motivates this paper.For a prime p congruent to 1 modulo 4, we can fix in Fp[x] two quadraticpolynomials f = (x − a)2 + a and g = (x − a − 1)2 + a such that both a anda + 1 are non-squares in Fp. One can experimentally check that any possiblecomposition of a sequence of f ’s and g’s is irreducible (for a concrete example,take q = 13, (x− 5)2 + 5 and g = (x − 6)2 + 5). Let us denote the set of suchcompositions by C. A couple of observations are now necessary:2010 Mathematics Subject Classification. 11T06.Key words and phrases. Finite Fields; Irreducible Polynomials; Semigroups; Graphs.The Second Author is thankful to SNSF grant number 161757.12 A. FERRAGUTI, G. MICHELI, AND R. SCHNYDER• In principle, it is unclear whether a finite number of irreducibility checkswill ensure that C is a subset of Irr(Fq[x]).• The fact that C ⊆ Irr(Fq[x]) is indeed pretty unlikely to happen bychance, as the density of degree 2n monic irreducible polynomials overFq is roughly 1/2n. Thus, if C satisfies this property, one reasonablyexpects that there must be an algebraic reason for that.We address these issues by giving a necessary and sufficient condition for thesemigroup C ⊂ Fq[x] to be contained in Irr(Fq[x]). In addition, this condition isalgebraic and can be checked by performing only a finite amount of computationover Fq, answering both points above.In Section 2 we describe the criterion (Theorem 2.4 and Corollary 2.5) andprovide a non-trivial example (Example 2.7) of a compositional semigroup inFq[x] contained in Irr(Fq[x]) and generated by two polynomials.In Section 3 we show the non-existence of such C whenever q is a primecongruent to 3 modulo 4 and the generating polynomials are of a certain form(Proposition 3.2). Example 3.3 shows that these conditions are indeed sharp.2. A general criterionIn order to state our main result, we first need the following definition, whichdescribes how to build a finite graph encoding only the useful (to our purposes)information contained in the generating set of the semigroup.Definition 2.1. Let q be an odd prime power, Fq the finite field of order qand S a subset of Fq[x]. We denote by GS the directed multigraph defined asfollows:• the set of nodes of GS is Fq;• for any node a ∈ Fq and any polynomial f ∈ S, there is a directed edgea → f(a). We label that edge with f .Before stating the next definition, we recall that for any monic polynomialf of degree 2 there exist unique pair (af , bf ) ∈ F2q such that f = (x−af )2− bf .Definition 2.2. Let S be a subset of Fq[x] consisting of monic polynomials ofdegree 2. We call the set DS := {−bf | f ∈ S} ⊆ Fq, the S-distinguished set ofFq.The following result is just an inductive extension of the classical Capelli’sLemma.Lemma 2.3 (Recursive Capelli’s Lemma). Let K be a field and f1, . . . , fl bea set of irreducible polynomials in K[X ]. The polynomial f1(f2(· · · (fl) · · · )) isON SETS OF IRREDUCIBLE POLYNOMIALS CLOSED BY COMPOSITION 3irreducible if and only if the following conditions are satisfiedf1is irreducible over K[X ]f2 − α1 is irreducible over K(α1)[X ] for a root α1 of f1f3 − α2 is irreducible over K(α1, α2)[X ] for a root α2 of f2 − α1· · ·fl − αl−1 is irreducible over K(α1, . . . , αl−1)[X ] for a root αl−1 of fl−1 − αl−2Proof. Given Capelli’s Lemma [11, Lemma 2.4], the proof is straightforwardby induction. We are now ready to state and prove the main theorem.Theorem 2.4. Let S be a set of generators for a compositional semigroupC ⊆ Fq[x]. Suppose that S consists of polynomials of degree 2. Then we havethat C ⊆ Irr(Fq[x]) if and only if no element of −DS = {bf | f ∈ S} ⊆ Fq isa square and in GS there is no path of positive length from a node of DS to asquare of Fq.Proof. It is clear that C contains a reducible polynomial of degree 2 if and onlyif one element of −DS is a square. Thus we can assume that S consists onlyof irreducible polynomials.We now show that in GS there is a path of positive length from a node of DSto a square if and only if C contains a reducible polynomial of degree greateror equal than 4.First, suppose that the composition f1f2 · · · fl+1 is a reducible polynomialof minimal degree, with fi ∈ S and fi = (x − ai)2 − bi, for i ∈ {1, . . . , l + 1}and l ≥ 1. Whenever β is not a square in Fq, we denote by√β a root ofthe polynomial T 2 − β in the algebraic closure of Fq. By Capelli’s Lemmaapplied to the composition of f1 · · · fl and by the minimality of the degree off1f2 · · · fl+1, we have that the following elements are not squares in their fieldof definition:β0 := b1 ∈ Fq,β1 := b2 + a1 +√β0 ∈ Fq2β2 = b3 + a2 +√β1 ∈ Fq22. . .βl−1 := bl + al−1 +√βl−2 ∈ Fq2l−1 .On the other hand, βl = bl+1 + al +√βl−1 ∈ Fq2l is necessarily a square.For j < i, let us denote by N ji : Fq2i −→ Fq2j the usual norm map. We claimthat the Fq-norm N0l : Fq2l −→ Fq maps βl to f1(· · · fl(−bl+1) · · · ), and thisdefines a path in GS from −bl to a square. This can be easily seen by first4 A. FERRAGUTI, G. MICHELI, AND R. SCHNYDERdecomposing N1l :N1l = N12 ◦N23 ◦ . . . N l−1land then by directly computing N12 ◦N23 ◦ . . . N l−1l (βl). It is important indeedthat β0, β1, . . . , βl−1 are not squares, as the computation above only gives thedesired result when (√βi)q2i= −√βi.Conversely, suppose that in GS there is a path to a square s. Choose such apath of minimal length, starting at some −bf in the distinguished set, for somef ∈ S. Consider now the composition associated to this path: ifs = f1f2 · · · fl(−bf),set fl+1 = f and let g := f1f2 · · · fl+1 ∈ Fq[x]. One can construct the βi’s asbefore, i.e. β0 = b1 and for i ∈ {1, . . . , l}, βi = bi+1 + ai +√βi−1. We cansuppose that the βi’s for i < l are all non-squares as otherwise, by taking thesmallest d such that βd is square, we find a composition f1f2 · · · fd+1 that isreducible by Recursive Capelli’s Lemma, and then we are done.As all the βi’s, for i < l, can be supposed to be non-squares, we have asabove that N0l (βl) = f1f2 · · · fl(−bl+1) = s, which we have assumed to be asquare. Now, recall that an element of a finite field is a square if and onlyif its norm is a square: this shows that g is reducible by Recursive Capelli’sLemma. The reader should observe that this theorem generalizes [11, Proposition2.3]. It is useful to mention the following corollary, which is immediate.Corollary 2.5. Let S be a set of irreducible degree two polynomials and Cdefined as in Theorem 2.4. Then C ⊆ Irr(Fq[x]) if and only if there is no pathof positive length from a node of DS to a square of Fq.Proof. It is enough to observe that whenever S ⊆ Irr(Fq[x]) then −DS consistsof non-squares. Remark 2.6. Given that C is generated by degree 2 polynomials, it is easy toobserve that the datum of S is equivalent to the datum of C.The following example shows a way to find examples of semigroups containedin Irr(Fq[x]) when q ≡ 1 mod 4.Example 2.7. Let q ≡ 1 mod 4 be a prime power, and let a ∈ Fq such that botha and b = a+1 are non-squares. Define f = (x− a)2 + a and g = (x− b)2 + a.In this situation, we have DS = {a}, and by assumption, −a, a and b are allnon-squares. Since f(a) = g(b) = a and f(b) = g(a) = b, all paths in GSstarting from a end in a non-square, and the conditions of Theorem 2.4 aresatisfied. Figure 1 shows the relevant part of the graph GS . The reader shouldobserve that this is indeed the example mentioned in the introduction.ON SETS OF IRREDUCIBLE POLYNOMIALS CLOSED BY COMPOSITION 5a bggf fFigure 1. The nodes of GS reachable from DS .3. The case p ≡ 3 mod 4Whenever q = p is a prime congruent to 3 modulo 4, we have the followingnon-existence results.Lemma 3.1. Let p ≡ −1 mod 8 be a prime, and let f = x2−b be a polynomialin Fp[x]. Let C be the semigroup generated by f . Then C contains a reduciblepolynomial.Proof. Assume for contradiction that C ⊂ Irr(Fp[x]). First note that if b is asquare, then f is reducible, so we can assume that b is not a square, and thus−b is a square. Consider the set of iterates T = {f(−b), f2(−b), . . .} ⊆ Fp. ByCorollary 2.5, C contains only irreducible polynomials if and only if T containsonly nonsquares. So assume that this condition holds. Since T is finite, thereexist k < m ∈ N>0 such that fm(−b) = fk(−b). Choose k to be minimal. Nowthere are two cases: if k > 1, then there exist two distinct elements u, v ∈ Tsuch that u2 − b = v2 − b. Thus, u = −v, which implies that one betweenu and v is a square, a contradiction. If on the other hand k = 1, then wehave fm(−b) = f(−b) = b2 − b, and so fm−1(−b) is either −b or b. It can’tbe −b, since that is a square, so we must have fm−1(−b) = b ∈ T . Settingu = fm−2(−b), we get that u2 − b = b and so u2 = 2b, which is a contradictionbecause 2 is a square in Fp and consequently 2b is not. Proposition 3.2. Let p ≡ 3 mod 4 be a prime. Let f = x2−bf and g = x2−bgbe polynomials in Fp[x] with bf , bg distinct non-squares. Let S = {f, g} and letC be the semigroup generated by S. Then C contains a reducible polynomial.Proof. Let GS be the graph attached to S as in Definition 2.1. Let G′S bethe induced subgraph consisting of all nodes of GS that are reachable by somepath of positive length starting from −bf or −bg. That is, the edges of G′S arejust the edges of GS starting and ending at a node in G′S. From now on, whenwe speak of nodes and edges, we will always be referring to nodes and edgesin G′S. We call an edge from u to v an f -edge if it comes from the relationf(u) = v, while we call it a g-edge if it comes from g(u) = v. Since bf and bgare assumed nonsquare, we have by Corollary 2.5 that C contains a reduciblepolynomial if and only if at least one of the nodes of G′Sis a square. In thefollowing, we assume for contradiction that G′S consists only of non-squares.6 A. FERRAGUTI, G. MICHELI, AND R. SCHNYDERLet us observe the following: suppose that there exists a node v of G′Swhich is the target of two f -edges. By definition, this means that there existtwo distinct nodes u, u′ ∈ G′Ssuch that u2 − bf = u′2 − bf = v. This impliesthat u′ = −u, and thus one between u and u′ is a square, since −1 is not asquare in Fp. This contradicts our assumption. By symmetry, the same appliesto g-edges.By the argument above, we see that every node is the target of at most onef -edge and one g-edge, and by counting edges that it is indeed exactly one ofeach.Now, consider the sum∑v∈G′S(f(v)− g(v)).On one hand, each node u ∈ G′Sappears exactly once as f(v) and once as g(v′)for some v, v′ ∈ G′S, so the sum is zero. On the other hand, it clearly holdsthat f(v)− g(v) = bg − bf for all v. Letting n be the number of nodes in G′S ,we get the equation0 = n(bg − bf) in Fp.Since bf 6= bg by hypothesis, we must have p | n. This is impossible however,since G′Sis not empty and consists only of nonsquares, so 1 ≤ n ≤ p−12. The fact that the polynomials of Proposition 3.2 don’t have a linear termis of crucial importance. Let us see why by giving an explicit example of asemigroup of irreducible polynomials in Fp[x] for which Proposition 3.2 doesnot apply (but p ≡ 3 mod 4).Example 3.3. Let us fix p = 7 andf = (x− 1)2 − 5 = x2 + 5x+ 3 ∈ F7[x]g = (x− 4)2 − 5 = x2 + 6x+ 4 ∈ F7[x].The set S = {f, g} has distinguished set DS = {−5} and graph as in Figure 2.Since 5 is not a square, and we only look at paths of positive length, the final−53−1fgfgfgFigure 2. The nodes of GS reachable from −5.ON SETS OF IRREDUCIBLE POLYNOMIALS CLOSED BY COMPOSITION 7claim follows by checking that 3 and −1 are not squares modulo 7.References[1] Rudolf Lidl and Harald Niederreiter. Finite fields, volume 20. Cambridgeuniversity press, 1997.[2] Gary L. Mullen and Daniel Panario. Handbook of finite fields. CRC Press,2013.[3] Michael O. Rabin et al. Fingerprinting by random polynomials. Center forResearch in Computing Techn., Aiken Computation Laboratory, Univ.,1981.[4] Razvan Barbulescu, Pierrick Gaudry, Antoine Joux, and EmmanuelThomé. A heuristic quasi-polynomial algorithm for discrete logarithmin finite fields of small characteristic. In Advances in Cryptology–EUROCRYPT 2014, pages 1–16. Springer, 2014.[5] Alina Ostafe and Igor E. Shparlinski. On the length of critical orbits ofstable quadratic polynomials. Proceedings of the American MathematicalSociety, 138(8):2653–2656, 2010.[6] Omran Ahmadi, Florian Luca, Alina Ostafe, and Igor E. Shparlinski. 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Irreducible trinomials over finite fields. Mathe-matics of Computation, 72(244):1987–2000, 2003.[14] Anjula Batra and Patrick Morton. Algebraic dynamics of polynomial mapson the algebraic closure of a finite field, i. JOURNAL OF MATHEMAT-ICS, 24(2), 1994.[15] Anjula Batra and Patrick Morton. Algebraic dynamics of polynomial mapson the algebraic closure of a finite field, ii. Rocky Mountain J. Math.,8 A. FERRAGUTI, G. MICHELI, AND R. SCHNYDER24(3):905–932, 09 1994.[16] Andrea Ferraguti and Giacomo Micheli. On the existence of infinite, non-trivial F -sets. arXiv preprint arXiv:1602.06608, 2016.Institute of Mathematics, University of Zurich, Winterthurerstrasse 190, 8057Zurich, Switzerland,E-mail address: andrea.ferraguti@math.uzh.chMathematical Institute, University of Oxford, Woodstock Rd,, Oxford OX26GG, United KingdomE-mail address: giacomo.micheli@maths.ox.ac.ukInstitute of Mathematics, University of Zurich, Winterthurerstrasse 190, 8057Zurich, Switzerland,E-mail address: reto.schnyder@math.uzh.ch

On sets of irreducible polynomials closed by composition

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