Fabien Durand
email: fabien.durand@u-picardie.fr

F R Éd

Éric Paccaut
email: frederic.paccaut@u-picardie.fr

MINIMAL POLYNOMIAL DYNAMICS ON THE SET OF 3-ADIC INTEGERS

coefficients, that have all their orbits dense in the set of 3-adic integers Z 3 .

Introduction

The study of dynamical systems on Z p or Q p is now developing a lot [START_REF] Anashin | Uniformly distributed sequences of p-adic integers II[END_REF][START_REF] Bryk | Measurable dynamics of simple p-adic polynomials[END_REF][START_REF] Coelho | Ergodicity of p-adic multiplications and the distribution of Fibonacci numbers[END_REF][START_REF] Fan | p-adic repellers in are subshifts of finite type[END_REF][START_REF] Fan | Strict ergodicity of affine dynamical systems on Zp[END_REF]. It is sometimes guided by or leading to applications in physics, cognitive science and cryptography [START_REF] Anashin | Uniformly distributed sequences in computer algebra or how to construct program generators of random numbers[END_REF][START_REF] Khrennikov | P -adic deterministic and random dynamics[END_REF]. For example, in the theory of pseudo random number generators, it is usefull to have a map, defined on the integers, giving rise to large cycles modulo n for a given integer n. When n is a power of a prime, good candidates are minimal maps in the set of p-adic integers (that is maps with all their orbits dense). Namely, these minimal maps, whenever polynomial, have only one cycle with maximal length modulo p n for every n. The point is: how to find such a minimal map or decide if a given polynomial map is minimal ? There is a complete characterization of minimal polynomial maps (with a different vocabulary) for p = 2 in [START_REF] Larin | Transitive polynomial transformations of residue class rings[END_REF] and necessary conditions on the Mahler coefficients of 1-Lipschitz maps to be minimal are given in [START_REF] Anashin | Uniformly distributed sequences of p-adic integers II[END_REF]. Complete results also exist for affine maps (see [START_REF] Coelho | Ergodicity of p-adic multiplications and the distribution of Fibonacci numbers[END_REF] and [START_REF] Fan | Strict ergodicity of affine dynamical systems on Zp[END_REF] where the dynamics are also studied when non minimal) and for polynomials of degree 2 for all prime p [START_REF] Knuth | The art of computer programming[END_REF]. But a complete description of minimal polynomials in Z p in terms of their coefficients seems to be a much harder task. Here we give this description for p = 3. First, we give definitions concerning p-adic analysis and dynamical systems. In Section 3, we develop the main properties of the dynamics of a compatible map (a map that preserves congruences) on the set of p-adic integers. In particular, we prove that an onto compatible map is minimal if and only if it is conjugated to the odometer in base p. Section 4 is devoted to the characterization of minimality for polynomials. In this section, we have reorganized ideas of [START_REF] Desjardins | On the structure of polynomial mappings modulo an odd prime power[END_REF] and rewritten a proof from [START_REF] Larin | Transitive polynomial transformations of residue class rings[END_REF] so as to develop a strategy to deal with Z 2 and Z 3 . We hope this strategy will also apply to Z p for p > 3.

2. Some definitions 2.1. p-adic integers and compatible maps. Let p be a prime number. We endow Z/p n Z with the discrete topology and n∈N Z/p n Z with the product topology. Let ϕ n : Z/p n+1 Z → Z/p n Z, n ∈ N, be the canonical homomorphisms. We call Z p the projective limit of (Z/p n Z, ϕ n ). This limit can be seen as 2000 Mathematics Subject Classification: Primary 37E99, Secondary 11S85

1 Z p = (x n ) n ∈ n∈N Z/p n Z; ϕ n (x n+1 ) = x n
which is a compact subset of n∈N Z/p n Z (see [START_REF] Robert | A Course in p-adic Analysis[END_REF]). The canonical projections are

π n : Z p → Z/p n Z. We of course have ϕ n • π n+1 = π n . Let x = (x n ) ∈ Z p and y = (y n ) ∈ Z p . Then (x n + y n ) n and (x n .y n ) n
are also elements of Z p ; this defines the addition x + y and the multiplication x.y in Z p . It can be checked that with these operations Z p is a topological ring. As a topological space Z p , is a Cantor set (a compact metric space with no isolated point and a countable base of its topology consisting of clopen sets). More precisely, one can prove that it is homeomorphic to n∈N {0, 1, . . . , p -1} endowed with the infinite product of the discrete topologies. Through this homeomorphism, a point x = (x n ) n in Z p can be represented by

x = x ′ 0 + x ′ 1 p + x ′ 2 p 2 + . . . with x ′ i ∈ {0, 1, . . . , p -1} and x i = x ′ 0 + x ′ 1 p + . . . + x ′ i p i .
In this way, Z/p n Z can be viewed as a subset of Z p :

Z/p n Z = {x ∈ Z p , ∀i n, x ′ i = 0}.
Thereofore, when no confusion is possible, we will consider an element x ∈ Z/p n Z as an element

x = x ′ 0 + x ′ 1 p + x ′ 2 p 2 + . . . + x ′ i p i of Z p with i ≤ n -1. By definition, |x| p = p -min{i,x ′ i =0} is the p-adic norm of x (with |0| p = 0). Each set x + p k Z p (x ∈ Z p , k ∈ N)
is a clopen ball of radius p -k and for each k, Z p is the union of p k balls of radius p -k . The Haar measure, the unique translation invariant probability measure on Z p , gives measure p -k to any ball of radius p -k . This measure will be denoted by µ p . The following lemma will be extensively used in this paper. The proof is left to the reader.

Lemma 1. Let f : Z p → Z p be a function. Then, the following are equivalent.

(1) For every n, if π n (x) = π n (y) then π n (f (x)) = π n (f (y));

(2) For every x and y in Z p , |f (y)f (x)| p ≤ |y -x| p ;

(3) For every n and all x ∈ Z p , f (x + p n Z p ) ⊂ f (x) + p n Z p ; (4) For every n, there exists a unique map f /n : Z/p n Z → Z/p n Z satisfying

(2.1) π n • f = f /n • π n .
Moreover we have the following other commuting relation:

(2.2) f /n • ϕ n = ϕ n • f /n+1 . Z p f ✲ Z p Z/p n+1 Z f /n+1 ✲ π n + 1 ✲ Z/p n+1 Z ✛ π n + 1 Z/p n Z π n ❄ f /n ✲ ✛ ϕ n Z/p n Z π n ❄ ϕ n ✲ Definition 2.
A compatible map f : Z p → Z p is a map satisfying one of the assertions of Lemma 1.

Note that compatible maps are exactly 1-Lipschitz maps. This terminology can be found in [START_REF] Anashin | Uniformly distributed sequences of p-adic integers II[END_REF]. In the sequel, whenever f is a compatible map, f /n will always denote the map defined in Lemma 1.

Proposition 3. Any polynomial whose coefficients belong to Z p is compatible.

2.2. Dynamical systems. Let X be a compact metric space and T : X → X be a homeomorphism. The couple (X, T ) is called a topological dynamical system. Let B be the Borel σ-algebra of X and µ a probability measure defined on B.

A set A ∈ B is called invariant by T if T -1 A = A. We say that µ is invariant by T if, for every set A ∈ B, µ(T -1 A) = µ(A). The dynamical stystem (X, T ) is called ergodic for µ if for every invariant set A ∈ B, µ(A) = 0 or µ(X \ A) = 0. It is called minimal if A = ∅ or A = X

Dynamics of compatible maps in Z p

In this section we give dynamical properties of compatible maps. These are known results but we give the proofs for the sake of completeness and a better understanding of the next section.

Proposition 4. Let f : Z p → Z p be a compatible map. The following are equivalent :

(1) f is onto;

(2) f is an isometry;

(3) for all n ∈ N, f /n : Z/p n Z → Z/p n Z (defined in Lemma 1) is bijective;

(4) f preserves the Haar measure.

Proof. (1) =⇒ (2) : Let f be onto. Let us show that for all x ∈ Z p and all n ∈ N, f (x + p n Z p ) = f (x) + p n Z p (we already have one inclusion because of the compatibility of f ). By contradiction, suppose it is not true : there exist

x ∈ Z p , n ∈ N, and z ∈ Z p with z ∈ f (x)+p n Z p \f (x+p n Z p ). Then f (x)+p n Z p = z +p n Z p .
As f is onto, there exists y ∈ Z p with z = f (y) and y / ∈ x + p n Z p . This implies (x + p n Z p ) ∩ (y + p n Z p ) = ∅. We can write

Z p = (x + p n Z p ) ∪ (y + p n Z p ) ∪ i∈{0,...,p n -1},i =x,i =y (i + p n Z p )
and, using the compatibility of f ,

f (Z p ) = Z p ⊂ (z + p n Z p ) ∪ i∈{0,...,p n -1},i =x,i =y (f (i) + p n Z p )
which means that at least one ball of radius p -n is lacking in 4) : Let us prove that for all n ∈ N, any i ∈ Z/p n Z, we have

Z p . (2) =⇒ (3) : As Z/p n Z is finite, it suffices to show that f /n is one-to-one. Let a, b be two different elements of Z/p n Z such that f /n (a) = f /n (b). We can consider a and b as elements of Z p . This implies that f (a) -f (b) ∈ p n Z p . Consequently, 1 p n-1 ≤ |a -b| p = |f (a) -f (b)| p ≤ 1 p n , a contradiction. Hence f /n is bijective. (3) =⇒ (
f -1 (i + p n Z p ) = f /n -1 (i) + p n Z p
where f /n -1 (i) is reduced to one point as f /n is bijective. This will do the job as this will imply

µ p (f -1 (i + p n Z p )) = µ p (f /n -1 (i) + p n Z p ) = 1 p n = µ p (i + p n Z p ) First inclusion : using compatibility, f (f /n -1 (i) + p n Z p ) ⊂ f (f /n -1 (i)) + p n Z p = i + p n Z p and so f /n -1 (i) + p n Z p ⊂ f -1 (f (f /n -1 (i) + p n Z p )) ⊂ f -1 (i + p n Z p ). For the other inclusion, let x ∈ f -1 (i + p n Z p ). We have f (x) ∈ i + p n Z p . Let x n = π n (x). Using compatibility, we have f (x) ∈ f (x n + p n Z p ) ⊂ f /n (x n ) + p n Z p .
Hence f (x) is a point common to the two balls with the same radius f /n (x n )+ p n Z p and i + p n Z p . This implies that both balls have the same center, so i = f /n (x n ) and x ∈ f /n -1 (i) + p n Z p . ( 4) =⇒ (1) : By contradiction, suppose that there exists x ∈ Z p \ f (Z p ). Let us show there exists n such that x + p n Z p ⊂ Z p \ f (Z p ). If it is not the case, for all n ∈ N there exists x n ∈ (x + p n Z p ) ∩ f (Z p ) and y n with x n = f (y n ) and |f (y n ) -x| p ≤ p -n . By compacity one can suppose that (y n ) converges to some y ∈ Z p and, thus, that x = f (y). This contradicts our assumption. To end remark that µ

p (f (Z p )) = µ p (f -1 (f (Z p ))) = µ p (Z p ) = 1 which is absurd because a ball x + p n Z p (which has a measure p -n ) is lacking in f (Z p ).
Now we deal with minimality of maps in Z p . Let T be the translation in Z p :

T : Z p → Z p x → x + 1 It is sometimes called odometers in base (p n ) n ([10]).
Lemma 5. Let n ∈ N and g : Z/nZ → Z/nZ. Then g is minimal if and only if:

∀x ∈ Z/nZ, g n (x) = x, and , ∀x ∈ Z/nZ, g k (x) = x =⇒ k ≡ 0 [n].
Proof. The proof is easy and left to the reader.

A cycle for a map h :

Z/p n Z → Z/p n Z is a finite sequence (x k ) 0≤k≤K-1 such that for every i ∈ {0, . . . , K -2}, x i+1 = h(x i ), h(x K-1 ) = x 0 and #{x k ; 0 ≤ k ≤ K -1} = K. It is a full-cycle for h if K = p n .
Theorem 6. Let f : Z p → Z p be an onto compatible map. The five following propositions are equivalent:

(1) f is minimal;

(2) for all n, f /n has a full-cycle in Z/p n Z;

(3) f is conjugate to the translation T (x) = x + 1 on Z p ;

(4) f is uniquely ergodic;

(5) f is ergodic for the Haar measure.

Proof.

(1) =⇒ ( 2): First note that as f is onto, for every n, f /n is bijective. By contradiction, suppose that f /n does not have a full-cycle in Z/p n Z for some n.

Then, as f /n is bijective, it has several cycles and there exists a proper subset A of Z/p n Z such that f /n (A) = A. Then A + p n Z p is a non empty closed invariant proper subset of Z p and f is not minimal.

(2) =⇒ (3): We need to find a conjugacy Ψ :

Z p → Z p such that Ψ • f = T • Ψ = Ψ + 1.
Actually, we will look for a compatible conjugacy Ψ. Hence we only need to define maps

ψ n : Z/p n Z → Z/p n Z such that ψ n • ϕ n = ϕ n • ψ n+1 and ψ n • f /n = T /n • ψ n . Indeed, defining Ψ n = π -1 n • ψ n • π n
for any n, one can check that (Ψ n ) n uniformly converges to a conjugacy Ψ we are looking for. Let us fix n ∈ N. As f /n : Z/p n Z → Z/p n Z is onto and has only one cycle, it has a full-cycle, and thus (Z/p n Z, f /n ) is minimal. Consequently, it is sufficient to define ψ n on the orbit of 0, which is a dense orbit. Notice that in this case, dense orbit only means that the orbit goes through each point of Z/p n Z. For all n ∈ N and k ∈ {1, . . . , p n -1} we set

ψ n (0) = 0 and ψ n (f k /n (0)) = k. Hence, ψ n • f /n = T /n • ψ n for all n ∈ N.
Let us now check the compatibility condition

ψ n • ϕ n = ϕ n • ψ n+1 , that is to say, if x = (x n ) n∈N ∈ Z p : ∀n ∈ N, ψ n+1 (x n+1 ) ≡ ψ n (x n ) [p n ].
As x n and x n+1 belong to, respectively, Z/p n Z and Z/p n+1 Z, there exist 0 ≤ k n < p n and 0 ≤ k n+1 < p n+1 such that x n = f kn /n (0) and 4): By definition of the Haar measure. (4) =⇒ (5): Suppose that f is uniquely ergodic. Using Proposition 4, f being an isometry, the Haar measure µ p is invariant. Thus it is the only invariant ergodic probability measure.

x n+1 = f kn+1 /n+1 (0). Consequently, f kn+1 /n+1 (0) ≡ f kn /n (0) [p n ] because ϕ n (x n+1 ) = x n . Moreover, f being compatible, f kn+1 /n+1 (0) ≡ f kn+1 /n (0) [p n ]. This implies f kn+1 /n (0) ≡ f kn /n (0) [p n ] and, using Lemma 5, k n+1 -k n = 0 in Z/p n Z which is equivalent to ψ n+1 (f kn+1 /n+1 (0)) ≡ ψ n (f kn /n (0)) [p n ], that is to say ψ n+1 (x n+1 ) ≡ ψ n (x n ) [p n ]. (3) =⇒ (
(5) =⇒ (1): Let x, y in Z p and ǫ > 0. It suffices to show that for some k, |f k (x) -y| p < ǫ. Let n such that p -n < ε/2. As f is ergodic for the Haar measure, there

exists some k > 0 such that µ p (f k (x + p n Z p ) ∩ (y + p n Z p )) > 0. As f is compatible f k (x + p n Z p ) ⊂ f k (x) + p n Z p and thus (f k (x) + p n Z p ) ∩ (y + p n Z p ) = ∅.
Therefore these two balls coincide and f k (x) and y lie in the same ball of radius p -n . Then |f k (x) -y| p ≤ 2p -n ≤ ε. This concludes the proof.

It is well known that minimal isometries on topological groups are conjugate to group rotations [START_REF] Kurka | Topological and symbolic dynamics, Cours spécialisés[END_REF]. Here, we obtain a little more, namely we prove that the conjugacy goes from Z p to Z p .

Minimality for polynomials

4.1. Some preliminary results. In this section, we are interested in polynomials f : Z p → Z p with coefficients in Z p . For the sequel, we need some definitions inspired by [START_REF] Desjardins | On the structure of polynomial mappings modulo an odd prime power[END_REF]. Let f : Z p → Z p be a polynomial. Let g n = f p n . Suppose there exists a finite sequence x 0 , . . . , x p n -1 in Z p so that f /n has a full cycle π n (x 0 ), . . . , π n (x p n -1 ) in Z/p n Z. Then, g n (x 0 ) belongs to x 0 + p n Z p . Consequently gn(x0)-x0 p n belongs to Z p . Using Taylor's expansion, for n 1, we get for all z ∈ Z p :

g n (x 0 + p n z) ∈ x 0 + p n g n (x 0 ) -x 0 p n + p n (g n ) ′ (x 0 )z + p 2n Z p . (4.1)
We set

α n (x 0 ) = (g n ) ′ (x 0 ), β n (x 0 ) = g n (x 0 ) -x 0 p n and Φ n (x 0 ) : Z p → Z p z → α n (x 0 )z + β n (x 0 ).
We have (4.2)

g n (x 0 + p n z) ∈ x 0 + p n Φ n (x 0 )(z) + p 2n Z p .
Then, Proof. The proof is easy and left to the reader.

p n β n (f (x 0 )) = g n (f (x 0 )) -f (x 0 ) = f (g n (x 0 )) -f (x 0 ) ≡ f (x 0 + p n Φ n (x 0 )(0)) -f (x 0 ) ≡ p n Φ n (x 0 )(0)f ′ (x 0 ) [p 2n ] Hence β n (f (x 0 )) ≡ β n (x 0 )f ′ (x 0 ) [p n ].
Lemma 8. Let f : Z p → Z p be a polynomial and n 1 be such that (Z/p n Z, f /n ) is minimal. Then, the following are equivalent :

(1) (Z/p n+1 Z, f /n+1 ) is minimal;

(2) For all x ∈ Z p , f p n (x)x ∈ p n+1 Z p and f p n ′ (x) ∈ 1 + pZ p ;

(3) There exists x ∈ Z p such that f p n (x)-x ∈ p n+1 Z p and f p n ′ (x) ∈ 1+pZ p .

Proof

f p n ′ (x) ∈ 1 + pZ p . Let y ∈ Z p . Then, (f i /n (y)) 0≤i≤p n -1 is a full cycle in Z/p n Z. Hence, 1 ≡ f p n ′ (x) ≡ p n -1 i=0 f ′ f i (x) ≡ p n -1 i=0 f ′ (i) ≡ p n -1 i=0 f ′ f i (y) ≡ f p n ′ (y) [p n ].
We conclude using (4.3).

Proposition 9. Let f : Z p → Z p be a polynomial, then (Z p , f ) is minimal if and only if (Z/p δ Z, f /δ ) is minimal, where δ = 2 if p > 3 and δ = 3 if p ∈ {2, 3}.
Proof. The necessary condition is due to Theorem 6. Suppose (Z/p δ Z, f /δ ) is minimal, i.e f /δ : Z/p δ Z → Z/p δ Z has one full-cycle. From Theorem 6, it is sufficient to prove that, for any n, f /n : Z/p n Z → Z/p n Z has a full-cycle. Remark that, as the number of cycles in Z/p n Z is non decreasing with n, this implies that f /n has only one cycle for all n ≤ δ.

Let us proceed by induction on n for the other cases. Suppose (Z/p n Z, f /n ) is minimal. From Lemma 8 it suffices to prove that β n (0) ∈ pZ p and α n (0) ∈ 1 + pZ p . As δ is greater than 2, Lemma 8 implies that α n-1 (0) belongs to 1+pZ p and β n-1 (0) does not belong to pZ p . We set α k = α k (0), β k = β k (0) and Φ k = Φ k (0) for all k. In order to conclude, we establish relations between (α n , β n ) and (α n-1 , β n-1 ). Remark that g n+1 = g p n so, using (4.2),

α n = (g p n-1 ) ′ (0) = p-1 i=0 g ′ n-1 (g i n-1 (0)) ∈ g ′ n-1 (0) p + p n-1 Z p ⊂ 1 + pZ p . (4.4)
For β n the situation is not so easy. We need to consider three different cases. Let us first consider the case p = 3 and n 3. We have

β n = g n (0) p n = g p n-1 (0) p n , g p n-1 (0) ∈ Φ p n-1 (0)p n-1 + p 2(n-1) Z p and Φ p n-1 (0) = β n-1 1 + α n-1 + . . . + α p-1 n-1 ∈ β n-1 p + p 2 Z p . Hence β n ∈ β n-1 + pZ p , which conclude the proof for p = 3.
Let us now consider the case p = 2 and n 3. Then n-2 1 and (Z/p n-2 Z, f /n-2 ) is minimal. From Lemma 8 we obtain α n-2 = 1+pz for some z ∈ Z p . Consequently, proceeding as in (4.4),

α n-1 = α n-2 α n-2 (g n-2 (0)) = α n-2 α n-2 + p n-2 z ′ g ′′ n-2 (0) = 1 + 2pz + p 2 z 2 + α n-2 p n-2 z ′ g ′′ n-2 (0) ∈ 1 + p 2 Z p , for some z ′ ∈ Z p , because g ′′ n-2 (
x) belongs to pZ p for all x ∈ Z p . Then, doing as in the previous case, we deduce that β n ∈ β n-1 + pZ p . Let us end with the case p 5 and n 2. Note that what we did in the case p = 3 also holds here. Hence we just have to consider the case n = 2 and thus to prove that β 2 ∈ β 1 + pZ p . We set γ = g ′′ 1 (0)/2 (as p 5, it belongs to Z p ). By induction, as observed in [START_REF] Desjardins | On the structure of polynomial mappings modulo an odd prime power[END_REF], we can prove that

g i 1 (0) ≡ pβ 1 i-1 j=0 α j 1 + p 2 γβ 2 1 i-2 j=0 α i-2-j 1 1 + α 1 + • • • + α j 1 2 [p 3 ], hence g p 1 (0) ≡ p 2 β 1 + p 2 γβ 2 1 (p -1)p(2p -1) 6 = p 2 β 1 [p 3 ]. (4.5)
This achieves the proof.

Remark 10. The value of δ obtained in Proposition 9 cannot be smaller. For p = 2, let f (x) = 1 + 3x + 2x 3 . The map f /2 has a full cycle 0,1,2,3 but (Z 2 , f ) is not minimal : f /3 has a cycle 0,1,6,3. For p = 3, let f (x) = 1+4x+4x 3 +2x 5 . The map f /2 has a full cycle 0,1,2,6,7,5,3,4,8 but (Z 3 , f ) is not minimal, f /3 has a cycle 0,1,11,15,7,23,3,13,17.

Remark 11. Let f be a polynomial with f (0) = a 0 . If f is minimal, then a 0 ≡ 0 [p], otherwise 0 is a fixed point modulo p for f . Let g(x) = 1 a0 f (a 0 x). Then g(0) = 1 and (Z p , f ) and (Z p , g) are conjugate. Therefore, f is minimal if and only if g is minimal. Hence we will restrict in the following to polynomials f with f (0) = 1.

4.2.

Characterization for p = 2. Using the previous ideas, we can characterize the polynomials that are minimal in Z 2 . The result is not new (see for example [START_REF] Larin | Transitive polynomial transformations of residue class rings[END_REF]) but the method described here is interesting as it will also apply to

Z 3 . Let f (x) = a d x d + a d-1 x d-1 + • • • + a 1 x + 1 be a polynomial of degree d. We set A 0 = i∈2Z,i =0 a i , A 1 = i∈1+2Z a i .
Lemma 12. The dynamical system (Z/2Z, f /1 ) is minimal if and only if

A 0 + A 1 ∈ 1 + 2Z 2 . Proof. If comes from the fact that f /1 (0) = 1 and f /1 (1) = 1 + A 0 + A 1 . Hence f /1 is minimal if and only if 1 + A 0 + A 1 ≡ 0 [2]. Lemma 13. The dynamical system (Z/4Z, f /2 ) is minimal if and only if    a 1 ≡ 1 [2] A 1 ≡ 1 [2] A 0 + A 1 ≡ 1 [4]
Proof. Suppose (Z/4Z, f /2 ) minimal, then (Z/2Z, f /1 ) is also minimal, therefore A 0 + A 1 ≡ 1 [START_REF] Anashin | Uniformly distributed sequences in computer algebra or how to construct program generators of random numbers[END_REF] (see Lemma 12). In view of Lemma 8, (Z/4Z, f /2 ) is minimal if and only if [START_REF] Anashin | Uniformly distributed sequences in computer algebra or how to construct program generators of random numbers[END_REF], which is equivalent to a 1 ∈ 1 + 2Z 2 and A 1 ∈ 1 + 2Z 2 . Secondly, f 2 (0) = 1 + A 0 + A 1 . We thus get immediately 1 + A 0 + A 1 ≡ 0 [START_REF] Coelho | Ergodicity of p-adic multiplications and the distribution of Fibonacci numbers[END_REF] and consequently A 0 + A 1 ≡ 1 [START_REF] Coelho | Ergodicity of p-adic multiplications and the distribution of Fibonacci numbers[END_REF].

A 0 + A 1 ≡ 1 [2] , (f 2 ) ′ (0) ∈ 1 + 2Z 2 and f 2 (0) ∈ 2Z 2 \ 4Z 2 First, (f 2 ) ′ (0) ≡ f ′ (0)f ′ (1) ≡ a 1 i 1 ia i ≡ a 1 A 1
Theorem 14. The dynamical system (Z 2 , f ) is minimal if and only if

       a 1 ≡ 1 [2] A 1 ≡ 1 [2] A 0 + A 1 ≡ 1 [4] 2a 2 + a 1 A 1 ≡ 1 [4]
Proof. Lemma 8 implies that (Z 2 , f ) is minimal if and only if

(Z/4Z, f /2 ) is minimal, (f 4 ) ′ (0) ∈ 1 + 2Z 2 and f 4 (0) ∈ 4Z 2 \ 8Z 2 .
First, (f 4 ) ′ (0) ≡ ((f 2 ) ′ (0)) 2 [2] so this condition has already been checked. Secondly, using Taylor's expansion, we get

f 2 (pz) ≡ p(β 1 + α 1 z) + p 2 z 2 (f 2 ) ′′ (0) 2 [8]
and then

f 4 (0) ≡ 2β 1 (1 + (f 2 ) ′ (0)) + 4β 2 1 (f 2 ) ′′ (0) 2 [8]
with

β 1 ≡ 1[2] and 1 + (f 2 ) ′ (0) ≡ 0 [2]. The condition f 4 (0) ∈ 4Z 2 \ 8Z 2 is therefore equivalent to (f 2 ) ′ (0) + (f 2 ) ′′ (0) ≡ 1 [4]. Using a 1 ≡ 1 [2], (f 2 ) ′ (0) ≡ a 1 i 1 ia i ≡ a 1   i 0 a 4i+1 + 2 i 0 a 4i+2 - i 0 a 4i+3   [4] ≡ a 1 i 0 a 4i+1 + 2 i 0 a 4i+2 -a 1 i 0 a 4i+3 [4]. Moreover, as i 2 i(i -1)a i ≡ 0 [2], a 2 1 ≡ 1 [2] and i 1 ia i ≡ A 1 ≡ 1 [2]: (f 2 ) ′′ (0) = f ′′ (0)f ′ (1) + (f ′ (0)) 2 f ′′ (1) = 2a 2 i 1 ia i + a 2 1 i 2 i(i -1)a i ≡ 2a 2 + i 2 i(i -1)a i ≡ i 3 i(i -1)a i [4] ≡ 2 i 1 a 4i+2 + 2 i 0 a 4i+3 [4].
This gives, using 2 -

a 1 ≡ a 1 [4] (f 2 ) ′ (0) + (f 2 ) ′′ (0) ≡ 2a 2 + a 1 i 0 a 4i+1 + (2 -a 1 ) i 0 a 4i+3 [4] ≡ 2a 2 + a 1 A 1 [4].
This achieves the proof.

These conditions are easily proved to be equivalent to those given in [START_REF] Larin | Transitive polynomial transformations of residue class rings[END_REF]:

   a 1 ≡ 1 [2] A 1 -a 1 ≡ 2a 2 [4] A 0 -a 2 ≡ a 1 + a 2 -1 [4]. Corollary 15. Let f (x) = a d x d + a d-1 x d-1 + • • • + a 1
x + a 0 be a polynomial of degree d. The dynamical system (Z 2 , f ) is minimal if and only if

       a 1 ≡ 1 [2] A ′ 1 ≡ 1 [2] A ′ 0 + A ′ 1 ≡ 1 [4] 2a 2 a 0 + a 1 A ′ 1 ≡ 1 [4]
where

A ′ 0 = i∈2Z,i =0 a i a i-1 0 and A ′ 1 = i∈1+2Z a i a i-1 0 . 4.3. Characterization for p = 3. Let f (x) = a d x d + a d-1 x d-1 + • • • + a 1 x + 1 be a polynomial of degree d. We set A 0 = i∈2Z,i =0 a i , A 1 = i∈1+2Z a i , D 0 = i∈2Z,i =0 ia i , D 1 = i∈1+2Z ia i .
Lemma 16. The dynamical system (Z/3Z, f /1 ) is minimal if and only if

A 0 ∈ 3Z 3 and A 1 ∈ 1 + 3Z 3 .
Proof. Suppose f is minimal. We remark that if x = 0 then f /1 (x) = 1 + A 0 + A 1 x and f /1 (0) = 1. Hence A 0 + A 1 x ≡ 0 [START_REF] Bryk | Measurable dynamics of simple p-adic polynomials[END_REF] for all x ≡ 0 [START_REF] Bryk | Measurable dynamics of simple p-adic polynomials[END_REF]. Then, A 0 ≡ 0 [START_REF] Bryk | Measurable dynamics of simple p-adic polynomials[END_REF] and A 1 ≡ 1 [START_REF] Bryk | Measurable dynamics of simple p-adic polynomials[END_REF]. Consequently (Z/3Z, f /1 ) is minimal. The reciprocal is as easy to establish.

Lemma 17. The dynamical system (Z/9Z, f /2 ) is minimal if and only if A 0 ∈ 3Z 3 , A 1 ∈ 1 + 3Z 3 and f fulfils one of the following conditions:

(1) D 0 ∈ 3Z 3 , D 1 ∈ 2 + 3Z 3 , a 1 ∈ 1 + 3Z 3 and A 1 + 5 ∈ 9Z 3 ; (2) D 0 ∈ 3Z 3 , D 1 ∈ 1 + 3Z 3 , a 1 ∈ 1 + 3Z 3 and A 0 + 6 ∈ 9Z 3 ; (3) D 1 ∈ 3Z 3 , D 0 ∈ 1 + 3Z 3 , a 1 ∈ 2 + 3Z 3 and A 1 + 5 ∈ 9Z 3 ; (4) D 1 ∈ 3Z 3 , D 0 ∈ 2 + 3Z 3 , a 1 ∈ 2 + 3Z 3 and A 0 + 6 ∈ 9Z 3 ;
Proof. In Lemma 8 we prove (Z/9Z, f /2 ) is minimal if and only if

(Z/3Z, f /1 ) is minimal, (f 3 ) ′ (0) ∈ 1 + 3Z 3 and f 3 (0) ∈ 3Z 3 \ 9Z 3 . Suppose (Z/9Z, f /2 ) is minimal. From Lemma 16, A 0 ∈ 3Z 3 and A 1 ∈ 1 + 3Z 3 . We easily check that (f 3 ) ′ (0) ≡ f ′ (0)f ′ (1)f ′ (2) ≡ a 1 (D 2 1 -D 2 0 ) [3]. Hence either D 0 ∈ 3Z 3 and D 1 ∈ 3Z 3 or D 1 ∈ 3Z 3 and D 0 ∈ 3Z 3 . Moreover we have f (0) = 1, f 2 (0) = 1 + A 0 + A 1 and f 3 (0) ≡1 + d i=1 a i (1 + A 0 + A 1 ) i ≡ 1 + d i=1 a i (1 + A 1 ) i + iA 0 (1 + A 1 ) i-1 [9] ≡1 + d i=1 a i 2 i + i2 i-1 (A 1 -1) + iA 0 2 i-1 [9] ≡1 + d i=1 a i (-1) i + 3i(-1) i-1 + i(-1) i-1 (A 0 + A 1 -1) [9] ≡1 + A 0 -A 1 + (A 0 + A 1 + 2)(D 1 -D 0 ) [9]. Consequently, (Z/9Z, f /2 ) is minimal if and only if A 0 ∈ 3Z 3 , A 1 ∈ 1 + 3Z 3 , a 1 (D 2 1 -D 2 0 ) ∈ 1 + 3Z 3 and 1 + A 0 -A 1 + (A 0 + A 1 + 2)(D 1 -D 0 ) ∈ 3Z 3 \ 9Z 3 . Suppose D 0 ∈ 3Z 3 . Then a 1 ∈ 1 + 3Z 3 and f 3 (0) ≡ 3 + 2A 0 [9] if D 1 ∈ 1 + 3Z 3 , 5 + A 1 [9] if D 1 ∈ 2 + 3Z 3 . Suppose D 1 ∈ 3Z 3 . Then a 1 ∈ 2 + 3Z 3 and f 3 (0) ≡ -1 -2A 1 [9] if D 0 ∈ 1 + 3Z 3 , -6 -A 0 [9] if D 0 ∈ 2 + 3Z 3 .
This achieves the proof.

Theorem 18. The dynamical system (Z 3 , f ) is minimal if and only if A 0 ∈ 3Z 3 , A 1 ∈ 1 + 3Z 3 and f fulfils one of the conditions (1), ( 2), (3) or (4):

D 0 [3] D 1 [3] a 1 [3]
(1) 0 2 1 A 1 + 5 ≡ 0 [9] A 1 + 5 ≡ 3a 2 + 3 j 0 a 5+6j [START_REF] Knuth | The art of computer programming[END_REF] (2) 0 1 1 A 0 + 6 ≡ 0 [9] A 0 + 6 ≡ 6a 2 + 3 j 0 a 2+6j [START_REF] Knuth | The art of computer programming[END_REF] (3) 1 0 2 A 1 + 5 ≡ 0 [9] A 1 + 5 ≡ 6a 2 + 3 j 0 a 5+6j [START_REF] Knuth | The art of computer programming[END_REF] (4) 2 0 2 A 0 + 6 ≡ 0 [9] A 0 + 6 ≡ 3a 2 + 3 j 0 a 2+6j [START_REF] Knuth | The art of computer programming[END_REF] Proof. From Lemma 8 and Proposition 9, we know that (Z 3 , f ) is minimal if and only if (Z/9Z, f /2 ) is minimal, f 9 ′ (0) ∈ 1 + 3Z 3 and f 9 (0) ∈ 3 3 Z 3 . From Lemma 8, as (Z/9Z, f /2 ) is minimal, f 3 ′ (0) ∈ 1 + 3Z 3 . The calculus (4.4) shows that f 9 ′ (0) ∈ 1 + 3Z 3 . Hence, using Lemma 17, (Z 3 , f ) is minimal if and only if f fulfils Conditions (1), ( 2), ( 3), (4), of Lemma 17, and f 9 (0) ∈ 3 3 Z 3 . But from (4.5) we see that

f 9 (0) ≡ 3 2 β 1 + 3 2 γβ 2 1 5 = 3 2 (β 1 -γ) [3 3 ],
where β 1 = f 3 (0) 3 and γ = (f 3 ) ′′ (0) 2 . Thus, f 9 (0) ∈ 3 3 Z 3 if and only if β 1 ≡ γ [START_REF] Bryk | Measurable dynamics of simple p-adic polynomials[END_REF]. That is to say

A ′ 0 = i∈2Z,i =0 a i a i-1 0 , A ′ 1 = i∈1+2Z a i a i-1 0 , D ′ 0 = i∈2Z,i =0 ia i a i-1 0 , D ′ 1 = i∈1+2Z ia i a i-1 0 .
Corollary 20. Let f : Z 3 → Z 3 be a polynomial defined by f (x) = a 5 x 5 + a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + 1. Then, (Z 3 , f ) is minimal if and only f satisfies one of the following conditions 4.4. Discussion. We had planned to address the question of minimality of polynomial dynamics in Z p for any prime p. We have no idea if this is tractable or not to get a general condition for all prime numbers p > 3. We have obtained sufficient conditions for any p, but they are far from necessary, as we have seen performing simulations. In some sense, it is not surprising since giving conditions on the coefficients to ensure bijectivity of polynomials is already known to be a difficult problem [START_REF] Lidl | Finite fields[END_REF].

(4. 3 ) 7 .

 37 Lemma Let a, b ∈ Z/pZ, and h : Z/pZ → Z/pZ t → b + at then (Z/pZ, h) is minimal if and only if a = 1 and b = 0.

a 1 [ 3 ] a 2 [ 3 ] a 3 [ 3 ] a 4 [ 3 ] a 5 1 + a 3 + a 5 ≡ 7 2 + a 4 ≡ 0 or 6 1 + a 3 + a 5 ≡ 7 2 +

 132333435135724613572 a 4 ≡ 0[START_REF] Knuth | The art of computer programming[END_REF]

  . Suppose (Z/p n+1 Z, f /n+1 ) is minimal. Let x ∈ Z p . If f p n (x)x belongs to p n+1 Z p , then f /n+1 has a cycle of length p n , which is not possible. From (4.2) we see that, (Z/p n+1 Z, f /n+1 ) being minimal, Φ n (x) /1 : Z/pZ → Z/pZ should be minimal. Lemma 7 implies f p n ′ (x) ∈ 1 + pZ p . Hence (1) implies[START_REF] Anashin | Uniformly distributed sequences in computer algebra or how to construct program generators of random numbers[END_REF]. Suppose[START_REF] Anashin | Uniformly distributed sequences in computer algebra or how to construct program generators of random numbers[END_REF]. Then, Φ n (x) /1 is minimal for all x ∈ Z p . We deduce with (4.2) that (Z/p n+1 Z, f /n+1 ) is minimal : (2) implies[START_REF] Anashin | Uniformly distributed sequences of p-adic integers II[END_REF]. It is trivial that (2) gives[START_REF] Bryk | Measurable dynamics of simple p-adic polynomials[END_REF]. Let us show that (3) implies[START_REF] Anashin | Uniformly distributed sequences in computer algebra or how to construct program generators of random numbers[END_REF]. Suppose there exists x ∈ Z p such that f p n (x)x ∈ p n+1 Z p and

Acknowledgement

We would like to thank the mathematical department of Vaxjö University, Sweden, where part of this work has been initiated, for its hospitality. We are grateful to Prof. Anashin, Prof. Fan and Prof. Khrennikov for fruitful discussions

2f 3 (0) ≡ 3 f 3 ′′ (0) [START_REF] Knuth | The art of computer programming[END_REF].

We already computed f 3 (0) in the proof of Lemma 17. Let us compute f 3 ′′ (0) :

Moreover, we easily obtain that

This gives:

which gives the conditions : Condition (1) Lemma 17 A 1 + 5 ≡ 3a 2 + 3 j 0 a 5+6j [START_REF] Knuth | The art of computer programming[END_REF] (2) Lemma 17 A 0 + 6 ≡ 6a 2 + 3 j 0 a 2+6j [START_REF] Knuth | The art of computer programming[END_REF] (3) Lemma 17 A 1 + 5 ≡ 6a 2 + 3 j 0 a 5+6j [START_REF] Knuth | The art of computer programming[END_REF] (4) Lemma 17 A 0 + 6 ≡ 3a 2 + 3 j 0 a 2+6j [START_REF] Knuth | The art of computer programming[END_REF] This achieves the proof.

3 and f fulfils one of the Conditions (1), ( 2), (3) or (4):

A ′ 1 + 5 ≡ 6a 2 a 0 + 3 j 0 a 5+6j a 4+6j 0 [START_REF] Knuth | The art of computer programming[END_REF] (4) 2 0 2 A ′ 0 + 6 ≡ 0 [9] A ′ 0 + 6 ≡ 3a 2 a 0 + 3 j 0 a 2+6j a 1+6j 0 [START_REF] Knuth | The art of computer programming[END_REF] where