On Robbins example of a continued fraction expansion for a quartic power series over F 13

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1 Journal of Number Theory 28 (2008) On Robbins example of a continued fraction expansion for a quartic power series over F 3 Alain Lasjaunias C.N.R.S.-UMR 5465, Université Bordeaux I, Talence 33405, France Received 24 November 2006 Available online 23 March 2007 Communicated by David Goss Abstract The continued fraction expansion for a quartic power series over the finite field F 3 was conjectured first in [W. Mills, D. Robbins, Continued fractions for certain algebraic power series, J. Number Theory 23 (986) ] and later in a more precise way in [W. Buc, D. Robbins, The continued fraction of an algebraic power series satisfying a quartic equation, J. Number Theory 50 (995) ]. Here this conjecture is proved by describing the continued fraction expansion for a large family of algebraic power series over a finite field Elsevier Inc. All rights reserved. MSC: J70; T55 Keywords: Continued fractions; Fields of power series; Finite fields. Introduction In this note we discuss continued fractions in the function fields case. For a general presentation of this subject, the reader may consult [S]. The examples which are considered here belong to a particular class of algebraic power series called hyperquadratic. More references concerning this class of elements as well as comments on the particular quartic equation considered first by Mills and Robbins and then by Buc and Robbins can be found in [BL]. In this note we consider power series over a prime field F p where p is an odd prime. We consider an indeterminate T, the ring of polynomials F p [T ] and the field of rational functions F p (T ). If T is a fixed address: alain.lasjaunias@math.u-bordeaux.fr X/$ see front matter 2007 Elsevier Inc. All rights reserved. doi:0.06/j.jnt

2 0 A. Lasjaunias / Journal of Number Theory 28 (2008) 09 5 real number greater than one, we consider the ultrametric absolute value defined on F p (T ) by P/Q = T deg(p ) deg(q). The completion of this field is the field of power series in /T over F p, which will here be denoted by F(p). Ifα F(p) and α 0, we have α = 0 u T, where 0 Z, u F p,u 0 0 and α = T 0. We introduce the following subset of F(p): F(p) + = { α F(p) with α T }. We also now that each irrational element α of F(p) can be expanded as an infinite continued fraction. This will be denoted by α =[a,...,a n,...], where the a i F p [T ] are the partial quotients. As usual the tail of the expansion, [a n,a n+,...], called the complete quotient, is denoted by α n (α = α). Finally the numerator and the denominator of the convergent [a,a 2,...,a n ] are denoted by x n and y n. These polynomials, called continuants, are both defined by the same recursive relation: K n = a n K n + K n 2 for n 2, with the initials conditions x 0 = and x = a for the numerator, while the initial conditions are y 0 = 0 and y = for the denominator. 2. Results The main result, Theorem below, is based upon the study developed in [L], Section 4. We recall the notations which were introduced there and which we have slightly modified here. For integers and i 2 we introduce the rational numbers u i, = j<i/2 (2j)(2 2j)/ j<(i+)/2 (2j )(2 2j + ) and also θ = ( ) ( /2j). j Given the integer, we fix a prime number p with p 2 +. Thus it is clear that the rational numbers u i, and θ can be reduced modulo p. Therefore in this note these numbers will also be considered as elements of F p. We introduce the following pair of polynomials: P (T ) = ( T 2 ) and Q (T ) = T 0 ( x 2 ) dx. Again these polynomials can either be considered as elements of Q[T ] or, by reduction modulo p, as elements of F p [T ]. Note that the existence of Q (T ) is ensured by the condition p 2 +. We recall that the origin of the numbers u i, is to be found in the following continued fraction expansion, P (T )/Q (T ) = [ (2 )T,...,(2 2i + )u ( )i i, T,...,( 2 + )u 2, T ],

3 A. Lasjaunias / Journal of Number Theory 28 (2008) 09 5 which holds in Q(T ) as well as in F p (T ) by reduction modulo p. Note also the lin between θ and Q which is given by the formula: 2θ Q () =. For an integer l and an integer n, we define f(n)= (2 + )n + l 2. We define the sequence (i(n)) n in the following way: i(n) = if n/ f ( N ) and i ( f(n) ) = i(n) +. Finally we introduce the sequence (A i ) i of polynomials in F p [T ] defined recursively by A = T and A i+ = [ A p i /P ] for i (here the square bracets denote the integer part, i.e., the polynomial part). We have the following result. Theorem. Let p be an odd prime number. Let be an integer with 2 <p. Let l be an integer. Let (λ,λ 2,...,λ l ) be a l-tuple in (F p )l. Let (ɛ,ɛ 2 ) (F p )2. Let α be the infinite continued fraction α =[a,...,a l,α l+ ] F(p) defined by a i (T ) = λ i T for i l and α p = ɛ P α l+ + ɛ 2 Q. Then α is the unique root in F(p) + of the algebraic equation y l x p+ x l x p + (ɛ P y l ɛ 2 Q y l )x ɛ P x l + ɛ 2 Q x l = 0. (E) Set formally δ i =[2θ λ i,...,2θ λ,ɛ 2 ] for i l. We assume that and Then we have δ i F p for i l (H ) δ l = ɛ (θ u 2, ɛ 2 ). (H 2 ) a n = λ n A i(n), where λ n F p for n. (I ) Let (δ n ) n be the sequence defined from the l-tuple (δ,...,δ l ) by the recurrence relations and (D i ),for i 2, δ f(n) = ɛ ( )n θ δ n for n (D 0 ) δ f(n)+i = ɛ ( )n+i 2θ i(u i, δ n ) ( )i for n. Then the sequence (λ n ) n is defined from the l-tuple (λ,...,λ l ) by the recurrence relation λ f(n) = ɛ ( )n λ n for n (L 0 )

4 2 A. Lasjaunias / Journal of Number Theory 28 (2008) 09 5 and (L i ),for i 2, λ f(n)+i = ɛ ( )n+i (2 2i + )(u i, δ n ) ( )i for n. Remar. It is interesting to observe that, in the extremal case p = 2 +, the sequence of polynomials A i is constant and we have A i (T ) = T for i. Consequently when the conditions (H ) and (H 2 ) are satisfied all the partial quotients of the expansion are linear. A first example was given in [MR], p. 400, and such continued fractions have been studied in a different approach in [LR] and [LR2]. There we started from the algebraic equation (E) but we imposed a special form to this equation, this was apparently artificial and constrained us to tae l p. Now we can turn to the conjecture made by Buc, Mills and Robbins [MR, p. 404], [BR, p. 342]. Theorem 2. Let α be the unique root in F(3) of the algebraic equation x 4 + x 2 Tx+ = 0. Let (A i ) i be the sequence of polynomials in F 3 [T ] defined recursively by Let U and V be the following vectors: and A = T and A i+ = [ A 3 i /(T 2 ) 4] for i. U = (u,...,u 8 ) = (7, 0, 5, 2, 9,,, 5) (F 3 ) 8 V = (v 0,...,v 8 ) = (, 0,,, 2, 5,, 2, 6) (F 3 ) 9. Let u F 69 with u 2 = 8. Then we have the continued fraction expansion with α =[0,a,...,a n,...] a n = u ( )n+ λ n A v9 (8n 2)+(uT ) for n, where λ n F 3 (here v 9 (m) denotes the largest power of 9 dividing m). If (δ n ) n is the sequence in F 3 defined by the recurrence relations δ 9n 2+i = v i δ ( )i n for 0 i 8 and n with the initial conditions (δ,...,δ 6 ) = (, 2, 5,, 2, 6), then the sequence (λ n ) n is defined by the recurrence relation λ 9n 2 = λ n for n

5 A. Lasjaunias / Journal of Number Theory 28 (2008) with the initial conditions (λ,...,λ 6 ) = (5, 2, 9,,, 5) and by 3. Proofs λ 9n 2+i = u i δ ( )i n for i 8 and n. Proof of Theorem. The existence of the continued fraction α F(p) + satisfying the given definition and the fact that it is the only root of equation (E) follows from Theorem [L]. Now we use Proposition 4.6 of [L]. Hence we now that there exists N N { }depending on (λ,...,λ l ) and (ɛ,ɛ 2 ) such that a n = λ n A i(n) with λ n F p for n f(n). (I ) The sequence (λ n ) n f(n) is defined in the following way: and (L i ),for i 2, λ f(n) = ɛ ( )n λ n for n N (L 0 ) λ f(n)+i = ɛ ( )n+i (2 2i + )(u i, δ n ) ( )i for n N, where the sequence (δ n ) n N is defined recursively by δ n = 2θ i(n) λ n + (δ n u 2, ) for n N (DL) with the initial condition δ 0 = (u 2, ɛ 2 ). Note that the frame of Proposition 4.6 is more general than here. Indeed here the base field is supposed to be prime, it follows that the Frobenius isomorphism is the identity on F p and this implies a simplification in (L 0 ) and (DL). Moreover the formulas given here are adapted to our new notations. Both sequences (λ n ) and (δ n ) are depending on each other and the process of their definition can be carried on as long as δ n 0. If this process terminates then N N,wehaveδ N = 0 and the continued fraction expansion is described by (I) but only up to a certain ran f(n). This is what happens if the l-tuple (λ,...,λ l ) is taen arbitrarily. We need to prove that under conditions (H ) and (H 2 ) we have N = and also that the sequence (δ n ) satisfies the formulas (D 0 ) to (D 2 ) stated in this theorem. First we observe that (H ) simply says that N>l. Indeed for n l we have i(n) = and (DL) becomes δ n =[2θ λ n,...,2θ λ,ɛ 2 ]. We will prove that δ n 0forn l + by showing that the equalities (D i ) for 0 i 2 hold for n. We introduce the following equalities δ f(n) = ɛ ( )n θ δ n for n. (H 2 ) n We observe that (H 2 ) is simply (H 2 ) and thus it is assumed to be true. Now we shall prove that for n we have the following implications: (H 2 ) n (D 0 ) n, () (D i ) n (D i+ ) n for 0 i 2, (2) (D 2 ) n (H 2 ) n+. (3)

6 4 A. Lasjaunias / Journal of Number Theory 28 (2008) 09 5 This will prove by induction on n that, for 0 i 2, (D i ) hold for n. To establish these implications we use the following equalities which are immediately derived from the definitions: and 4 2 θ 2 u 2, = (4) u, =, u i+, = u i, ( i(2 i) ) ( ) i for i 2. (5) To prove () we start from (DL) at the ran f(n). Since i(f(n)) = i(n) +, we have δ f(n) = 2θ i(n)+ λ f(n) + (u 2, δ f(n) ) and, by (L 0 ) and (DL) at the ran n, this becomes δ f(n) = ( δ n (δ n u 2, ) ) ɛ ( )n θ + (u 2, δ f(n) ). Now using (H 2 ) n we obtain immediately δ f(n) = δ n ɛ ( )n θ which is (D 0 ) n. To prove (2) we start from (DL) at the ran f(n)+ i + for0 i 2. Since i(f(n) + i + ) =, we have δ f(n)+i+ = 2θ λ f(n)+i+ + (u 2, δ f(n)+i ). Applying (L i+ ) and (D i ) n and using (4) and (5), we obtain without difficulties (D i+ ) n. Finally the last implication (3) is obtained very simply with (4) δ f(n+) = δ f(n)+2 = ɛ ( )n 4 2 θ u 2, δ n = ɛ ( )n θ δ n. So the proof is complete. Remar. It is clear that condition (H ) is necessary to have N =, but condition (H 2 ) may not be so. Indeed this condition implies a simple form for the sequence (δ n ) n showing that it never vanishes. We say that the expansion is perfect if (I) holds for n (i.e.,n = ). So the expansion could perhaps be perfect without δ n having the form given in the theorem. Indeed it is interesting to notice that more complex forms for this sequence have been pointed out when the base field is not prime and in the particular case p = 2 + (see [LR2], p. 562). Now we turn to the second theorem. Proof of Theorem 2. Let us consider the infinite continued fraction β =[b,...,b 6,β 7 ] F(3) defined by (b,...,b 6 ) = (5T,2T,9T,T,T,5T) and β 3 = P 4 β 7 4Q 4. This element β is the only solution in F(3) + of the algebraic equation (E) ( T 5 + 3T 3 + T ) x 4 + ( 8T 6 + 2T 4 + 3T 2 + ) x 3 + ( 5T 2 + ) x + 7T = 0.

7 A. Lasjaunias / Journal of Number Theory 28 (2008) It is easy to chec that conditions (H ) and (H 2 ) of Theorem are satisfied. Here = 4, θ 4 = 2 and u 8,4 = 3. Moreover l = 6 and (ɛ,ɛ 2 ) = (, 4). We have the resulting 6-tuple (δ,...,δ 6 ) = (, 2, 5,, 2, 6). Thus the expansion for β is perfect and, according to Theorem, we have b n = λ n A i(n), where λ n F 3 for n. Here we have f(n)= 9n 2. It is elementary to chec that the sequence v 9 (8n 2) + satisfies the same relations as i(n), and therefore here we have i(n) = v 9 (8n 2) + forn. By adapting the formulas given in Theorem for the sequences (λ n ) n and (δ n ) n, we obtain immediately those which are stated in this theorem. Now we turn to the quartic equation which can be written as x = /T + (x 2 + x 4 )/T. Hence by iteration, we see that it has a root α in F(3) with α = T. We put γ(t)= u α (u T). Then it follows that γ is solution of the algebraic equation B(x) = x 4 5Tx 3 + 5x 2 = 0 and γ F(3) +. In order to have the continued fraction expansion for α as described in the theorem, we only need to prove that /α(t ) = uβ(ut ) or equivalently that γ = β. Thus it remains to prove that γ satisfies equation (E). This will be shown if the polynomial A on the left side of equation (E) is divisible by the polynomial B. A straightforward calculation shows that A = BC with C = ( T 5 + 3T 3 + T ) x 0 + ( T 4 + 6T 2 + ) x 9 + ( 2T 3 + 2T ) x 8 + ( 5T 4 + 6T ) x 7 + ( 0T 3 + 2T ) x 6 + 2T 2 x 5 + ( 2T 3 + 5T ) x 4 + ( 0T ) x 3 + 4Tx 2 + ( 8T ) x + 6T. So the proof is complete. References [BL] A. Bluher, A. Lasjaunias, Hyperquadratic power series of degree four, Acta Arith. 24 (2006) [BR] W. Buc, D. Robbins, The continued fraction of an algebraic power series satisfying a quartic equation, J. Number Theory 50 (995) [L] A. Lasjaunias, Continued fractions for hyperquadratic power series over a finite field, Finite Fields Appl. 4 (2) (2008) [LR] A. Lasjaunias, J.-J. Ruch, Flat power series over a finite field, J. Number Theory 95 (2002) [LR2] A. Lasjaunias, J.-J. Ruch, On a family of sequences defined recursively in F q (II), Finite Fields Appl. 0 (2004) [MR] W. Mills, D. Robbins, Continued fractions for certain algebraic power series, J. Number Theory 23 (986) [S] W. Schmidt, On continued fractions and Diophantine approximation in power series fields, Acta Arith. 95 (2000)