On the period length of vector sequences generated by polynomials modulo prime powers Текст научной статьи по специальности «Математика»

ПРИКЛАДНАЯ ДИСКРЕТНАЯ МАТЕМАТИКА

2016 Теоретические основы прикладной дискретной математики № 1(31)

UDC 511.176 DOI 10.17223/20710410/31/5

ON THE PERIOD LENGTH OF VECTOR SEQUENCES GENERATED BY POLYNOMIALS MODULO PRIME POWERS

N. G. Parvatov

National Research Tomsk State University, Tomsk, Russia

We give an upper bound on the period length for vector sequences defined recursively by systems of multivariate polynomials with coefficients in the ring of integers modulo a prime power.

Keywords: recurrence sequences, vector sequences, period length, polynomial functions, polynomial permutations, finite rings.

Introduction

Let n and m be positive integers, p be a prime number, and f1,..., fn be polynomials in n variables with integer coefficients. Consider a recurrence sequence

f0(x) + pmZn, f1 (x) + pmZn, f2(x) + pmZn,...,

where x E Zn, f (x) = (f1(x),...,fn(x)), f0(x) = x, and fk(x) = f (fk-1 (x)) for all positive k. Denote it by s(f,m,x). The sequence s(f,m,x) is said to be purely periodic if there exists a positive integer d such that fd(x) = x mod pmZn. In this case, the smallest d is called the period of s(f,m,x) and is denoted by t(f,m,x).

Further, the function on Zn/pmZn induced by f is denoted by [f ]m. Clearly, this function is a permutation iff the sequence s(f,m,x) is purely periodic for all x E Zn.

Permutations induced by polynomials modulo prime powers are considered in [1-3]. They are characterized in [1]. Transitive polynomial permutations are described in [1, 2]. The cycle structure of permutations induced by univariate polynomials over Galois rings is investigated in [3]. In this paper, we extend this result to polynomials in several variables over the ring of integers modulo pm. Namely, we derive an upper bound on the period length t(f,m,x) under the condition that the sequence s(f,m,y) is purely periodic for each y E x + pZn.

This paper is organized as follows. In section 1, we formulate Theorem 1. This theorem gives an upper bound on the value of t(f, m, x). In section 2, we prove auxiliary Lemmas 1 and 2. In section 3, we prove the theorem.

1. Main results

We begin with some notation. Let Mn be the ring of (n x n)-matrices over Z with the identity matrix E. For a matrix A, let det(A) denote its determinant. If det(A) = 0 mod pZ, then there exists a positive integer k such that Ak = E mod pMn. The smallest integer with this property is denoted by ordp(A). By definition, put

Jf (x)

( dfl (x) ■ dx\ dfn dx\

\ dfl (x) ■ V dxn dfn dxn

(x)

\

and JJ(x) = Jf (/0(x)) ■ ■ ■ Jf (/T-1(x)) for a positive integer t. The matrix Jf (x) is called the Jacobi matrix and the determinant det( Jf (x)) is called the Jacobian of the function / at the point x.

The aim of this paper is to prove the following result.

Theorem 1. Let x be a tuple in Zn and m be a positive integer such that m > 1. Suppose the sequence s(/, 1,x) is purely periodic and t1 = t(/, 1,x); then the following statements hold.

1) If the sequence s(/, m, y) is purely periodic for every y E x + pZn, then

det(Jf (x)) ^ 0 mod pZ.

2) If detJT1 (x)) ^ 0 mod pZ and y E x + pZn, then the sequence s(/, m, y) is purely periodic and the following relation holds:

t(/, m, y) | ti ■ pm-1 ■ ordp(JT1 (x)).

3) If det(JT1 (x)) ^ 0 mod pZ and det(JT1 (x) — E) ^ 0 mod pZ, then, for every y E x + pZn, the following relation holds:

t(/, m, y) | ti ■ pm-2 ■ ordpJ (x)).

We will prove Theorem 1 in section 3.

Remark 1. We have ordp(A) ^ pn — 1 for each A E M„ such that det(A) ^ 0 mod pZ. Indeed, ordp(A) is equal to the period of the sequence of nonzero polynomials

x0 mod mA(x), x1 mod mA(x), x2 mod mA(x), ...

from the ring Z/pZ [x], where m^(x) is the minimal polynomial of the matrix A over the field Z/pZ. Since degmA ^ n, there are less than pn distinct polynomials here.) Thus, we obtain

t(/, m, y) ^ T1 ■ pk (pn — 1) ^ pn ■ pk(pn — 1),

where k = m — 1 in conditions of statement 2 and k = m — 2 in conditions of statement 3 in Theorem 1.

Remark 2. Let / be given by /(z) = z ■ A for all z E Zn, where A E Mn and det(A) ^ 0 mod pZ. In this case, s(/, m,x) is the congruential sequence

x + pmZn, x ■ A + pmZn, x ■ A2 + pmZn, ...

In conditions of statement 2, we have t1 | ordp(A) and JT1 (x) = AT1. Hence,

t (/, m, y) ^ T1 ■ pm-1 ■ ordp(AT1) = pm-1 ■ ordp (A) ^ pm-1(pn — 1).

In [4], this bound is proved and congruential sequences of period pm-1(pn — 1) are constructed.

Remark 3. Let expp(Mn) denote the exponent of the multiplicative group of the ring Mn/pMn. Suppose that [/]m is a permutation of order t(/, m). Then we have

t(/, m) | t(/, 1) ■ pk ■ expp(Mn),

where k = m — 1 in conditions of statement 2 and k = m — 2 in conditions of statement 3. The value of expp(Mn) is determined in [5, 6].

To prove Theorem 1, we need two auxiliary lemmas.

2. Two Lemmas

We use the notation U(J, k) = E + J + ... + Jk-1.

Lemma 1. Let Z,k,T, T1 be positive integers and x,y,z,w be tuples in Zn such that x = y mod pZn. Suppose the sequence s(/, 1,x) is purely periodic and t(/, 1,x) | T1. Then the following statements hold.

1) /k(y + p1 z) = /k(y) + p1 z ■ Jk(x) mod p1+1Zn

2) If /T (y) = y + p^w and T1 | t, then

/kT(y + p1 z) = y + pzw ■ U(JT1 (x)CT, k) + p1 z ■ JJ1 (x)kCT mod p1+1Zn,

where a = t/t^. Proof. It is well known (see, for example, [1]) that

/(y + p'z) = /(y) + p1 z ■ Jf (y) mod p1+1Zn.

Using this formula, we get

/2(y + p*z) = /(/(y) + p*z ■ Jf (y)) = /2(y) + p1 z ■ Jf (y) ■ Jf (/(y)) = = /2(y)+ p*z ■ J2(y) mod p1+1 Zn;

p3/ , z \ _ (• ( f2(„\ , t2{„,\\ — r3/ \ , z j2 („ 1 ( f2i

/3(y + p'z) = /(/2(y) + p1 z ■ J2(y)) = /3(y) + p1 z ■ J2(y) ■ Jf (/2(y)) =

= /3(y) + p'z ■ J?(y) mod p1+1 Zn

/k(y + p*z) = /k(y) + p*z ■ Jk(y) mod p1+1Zn. Here, take Jk(x) in place of Jk(y). We claim that this replacing is correct. Indeed, since x = y, /(x) = /(y), ..., /k-1(x) = /k-1(y) modpZn,

we have

Jf(x) = Jf(y), Jf(/(x)) = Jf(/(y)), ..., Jf(/k-1(x)) = Jf(/k-1 (y)) modpMn.

Hence, Jk(y) = J^(x) mod pMn and p1 z ■ J^(y) = pzz ■ J^(x) mod p1+1Zn. This proves the statement 1. Let us prove the statement 2. Note that the sequence

Jf (x) mod pMn, Jf (/(x)) mod pMn, Jf (/2(x)) mod pMn, ...

is purely periodic and its period divides T1. Hence, JJ(x) = JJ1 (x)CT mod pZn. Using the statement 1, we get

/T(y + p1 z) = /T (y) + p1 z ■ JJ1 (x)CT = y + p1 w + p1 z ■ JJ1 (x)CT = = y + pzw ■ U(JJ1 (x)CT, 1) + p^z ■ JJ1 (x)CT mod p1+1Zn.

In the same manner, we can see that

/2t(y + pzz) = y + p1 w ■ U(JJ1 (x)CT, 2) + p1 z ■ JJ1 (x)2CT mod p1+1Zn /3t(y + pzz) = y + p1 w ■ U(JJ1 (x)CT, 3) + p1 z ■ JJ1 (x)3CT mod p1+1Zn

/kT(y + p1 z) = y + pzw ■ U(JJ1 (x)CT, k) + p1 z ■ JJ1 (x)kCT mod p1+1Zn. This completes the proof. ■

Lemma 2. Let r be a positive integer. Suppose J E Mn and det(J) = 0 mod pZ. Then the following statements hold.

1) U(J,p • ordp(J) • r) = 0 mod pMn.

2) If det(J - E) = 0 mod pZ, then U(J, ordp(J) • r) = 0 mod pMn. Proof. Clearly, if i = j mod ordp(J), then J% = Jj mod pMn. Hence,

U (J,p • ordp( J) • r) = p • r • U (J, ordp( J)) = 0 mod pMn

and statement 1 holds. Further, for every positive integer k we have

(J - E)U(J, k) = Jk - E.

For k = ordp(J) • r, this gives

(J - E)U(J, ordp(J) • r) = 0 mod pMn.

If det(J - E) = 0 mod pZ, then the matrix J - E is invertible modulo pMn. In this case, U(J, ordp(J) • r) = 0 mod pMn. ■

3. Proof of Theorem 1

Suppose that, for every y E x + pZn, the sequence s(f,m,y) is purely periodic; then the sequence s(f, 2,y) is purely periodic too. We may choose a positive integer k such that the relation t(f, 2,y) | kT1 holds for each y E x + pZn. This means that

fkT1 (x + pz) = x + pz mod p2Zn

for all z E Zn. At the same time, by statement 2 of Lemma 1, we have

fkT1 (x + pz) = x + pw • U(J}1 (x), k) + pz • J}1 (x)k mod p2Zn,

where rpw = fT1 (x) - x. If we take z = 0, we have rpw • U(J}1 (x),k) = 0 modp2Zn and

fkT1 (x + pz) = x + pz • J}1 (x)k = x + pz mod p2Zn

for all z E Zn. This implies that

pz • Jf1 (x)k = pz mod p2Zn and z • Jff1 (x)k = z mod pZn

for all z. Hence, Jf (x)k = E mod pMn and (detJfP (x)))k = 1 mod pZ. Thus, det(Jf (x)) = = 0 mod pZ. We have proved the first statement of Theorem 1. Assume det(J^ (x)) = 0 mod pZ and y E x + pZn. Let

j-i

1 ti • /

Tl

Ti • pl-1 • ordp(JT1 (x)), if detJ (x) - E) = 0 mod pZ, Ti • pl-2 • ordpJ1 (x)), if det(JT1 (x) - E) = 0 mod pZ

for all l ^ 2. Suppose inductively that the following relation holds:

fTl (y) = y mod plZn,

where l ^ 1. Then using Lemma 1, we obtain

fTl+1 (y) = y + pw • U J1 (xf, k) mod pl+1Zn,

where pw = fTl (y) - y, a = t1/t1, and k = t1+1/t1 . For l = 1, we have a =1 and

i p • ordp(Jf(x)), for det(Jf (x) - E) = 0 mod pZ, = | 1 • ordpJ (x)), for detJ (x) - E) = 0 mod pZ.

For l ^ 2, we have ordp( J1 (x)) | a and p | k.

Using Lemma 2, we get U(J1 (x),k) = 0 modpMn and fTl+1 (y) = y modplZn for all

l ^ 1. Thus, for every l ^ 1, the sequence s(f,l,y) is purely periodic and t(f,l,y) | Tl.

We take l = m to complete the proof. ■

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