ON FIXED DIVISORS OF THE MINIMAL POLYNOMIALS OVER Z OF ALGEBRAIC NUMBERS

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ON FIXED DIVISORS OF THE MINIMAL POLYNOMIALS OVER Z OF ALGEBRAIC

NUMBERS

M Ayad, A Bayad, O Kihel

To cite this version:

M Ayad, A Bayad, O Kihel. ON FIXED DIVISORS OF THE MINIMAL POLYNOMIALS OVER Z OF ALGEBRAIC NUMBERS. Hardy-Ramanujan Journal, Hardy-Ramanujan Society, 2020. �hal- 03029800�

ALGEBRAIC NUMBERS

M. AYAD, A. BAYAD AND O. KIHEL

ABSTRACT. LetKbe a number field of degreen,Abe its ring of integers, andA_n(resp.K_n) be the set of elements ofA(resp.K) which are primitive overQ. For anyγ∈K_n, letF_γ(x) be the unique irreducible polynomial inZ[x], such that its leading coefficient is positive andF_γ(γ) =0. Leti(γ) =gcd_x∈ZF_γ(x),i(K) =lcmθ∈A_ni(θ)and ˆı(K) =lcmγ∈K_ni(γ). For anyγ∈K_n, there exists a unique pair(θ,d), whereθ∈A_nanddis a positive integer such thatγ=θ/dandθ6≡0 (mod p)for any prime divisorpofd. In this paper, among other results, we show thati(K)and ˆı(K)have the same prime factors and we study the possible values ofν_p(d)whenp|i(γ). We introduce and study a new invariant ofKdefined using ν_p(d), whenγdescribesK_n. In the last theorem of this paper, we establish a generalisation of a theorem of MacCluer.

1. Introduction

Let K be a number field of degreen≥2 andAbe its ring of integers. Denote byA_n (resp. K_n) be the set of elements ofA( resp. K) which are primitive overQ, that is those elements which generateKoverQ. For any primitive polynomialg(x)∈Z[x], we define the integeri(g)by

i(g) =gcd_x∈Zg(x).

Let γ be an algebraic number. When we refer to the minimal polynomial of γ over Z, we mean the unique polynomialg(x) =a_nxⁿ+. . .+a₁x+a₀∈Z[x], irreducible such that a_n>0 andg(γ) =0.The leading coefficienta_nwill be denoted byc(γ). We seti(γ) =i(g).

In [5], Gunji and McQuillan defined the integeri(K)by i(K) =lcm_θ∈Ani(θ).

MacCluer [7] proved that a given prime pdividesi(K)if and only if the number of prime ideals ofAlying over pis at least equal to p. In [5] or [2], it is proved that there exists θ ∈A_nsuch that

i(K) =i(θ).

The smallest positive integerdsuch thatdγ is an algebraic integer is called the denominator ofγ and will be denoted byd(γ).

Date: November 25, 2020.

2010 Mathematics Subject Classification: 11R04, 12Y05.

Key words and phrases: Values of polynomials, Denominators of algebraic numbers, Splitting of prime numbers.

1

Arno et al proved in [1] that the density of the set of the algebraic numbersγ such that c(γ) =d(γ)is equal to 1/ζ(3) =0.8319· · ·. In [3], for a fixed number fieldK, the set

T_p(k) =

t≥1, there existsγ ∈K,ˆ ν_p(d(γ)) =kandν_p(c(γ)) =t is connected to the splitting of the prime pinK.

In this paper, among other results, we study the possible values of ν_p(d) when p| i(γ). After recalling some lemmas in section 2, it is proved in Theorem 3.1 that if θ 6≡0(modpA) and p|i(θ/p^k), for some positive integer k, then the p-adic valuation of the leading coefficient of the minimal polynomial of θ/p^k belongs to the set{k,k+ 1, . . . ,(n−p)k}. Furthermore any element of this set may occur. Section 4 shows that givenθ ∈A_n such thatθ 6≡ 0(mod pA),it is possible to find explicitly the values of k∈N, if any, such that p|i(θ/p^k).Fix θ ∈A_nsuch that θ 6≡0(modpA)and define the setV_p(θ) ={k∈ N; p|i(θ/p^k)}.On the one hand it is shown that|V_p(θ)|forθ ∈A_nis bounded by some constant depending onnand p. On the other hand the values ofkare bounded by a constant depending on pandν_p(N_K/Q(θ)), whereν_p(a)denotes the p-adic valuation ofa. Section 5 deals with this last bound. It is shown that, even if we fix the field K, the values ofkmay be greater than any given positive constant. In this section, we give examples of Galois number fieldsKof degree 4 (resp. 3) for which the set of the values of k, whenθ runs inA_nis{0}( resp. N). Throughout this paper we denote byNthe set of nonnegative integers. The paper ends with remarks and open questions.

2. Indices and denominators of algebraic numbers

Let K be a number field of degree n, A its ring of integers, γ ∈K_n and let g(x) = a_nxⁿ+· · ·+a₁x+a₀∈Z[x] be the unique primitive polynomial of degree n such that a_n>0 andg(γ) =0.This polynomial will be called the minimal polynomial ofγ overZ. The leading coefficienta_nwill be denotedc(γ).

Let

I(γ) ={m∈Z,mγ ∈A},

thenI(γ)is a nonzero ideal ofZ, hence a principal ideal generated by some positive integer denoted byd(γ). The integerd(γ)is called the denominator ofγ. Sincec(γ)∈I(γ), then d(γ)|c(γ). Writeγ = _d(γ)^θ ,whereθ ∈A_n, thenθ is unique and we call it the numerator of γ. Let f(x)be the minimal polynomial ofθ overQ, theng(x) = f(d(γ)x)/cont(f(d(γ)x), where the abbreviationcont(h(x))denotes the content of the polynomialh(x).

From [1] we have the following result:

Lemma 2.1. For any prime p, we have ν_p(d(γ)) =max

0,maxⁿ⁻¹

j=0

νp(a_n)−νp(a_j) n−j

. (1)

From Lemma 2.1 we see that any prime factorpofc(γ)dividesd(γ).

Summarizing the relations betweenc(γ)andd(γ), we have :

Remark 2.1. Let K be a number field of degree n andγ ∈K_n, then d(γ)and c(γ)have the same prime factors and for any prime p, we haveνp(d(γ))≤νp(c(γ))≤nν_p(d(γ)).

For anyθ ∈A_n, letF_θ(x)its minimal polynomial overQ. Following [5], we define the integers

i(θ) =gcd_x∈ZF_θ(x)andi(K) =lcm_θ∈Ani(θ).

(2)

For the integersi(θ)andi(K)we have the following results:

i) C. R. MacCluer proved in [7] that for a given prime number p, pdividesi(K)if and only if the number of prime ideals ofAlying over pis at least equal top.

ii) In [5] and [2], it is proved that there existsθ ∈A_nsuch thati(K) =i(θ),and that i(K) =lcm_θ∈Ai(θ).

We extend the definition ofi(θ)and i(K) to algebraic numbers as follows. Given any γ ∈K_n, we define

i(γ):=gcd_x∈ZF_γ(x)and ˆı(K) =lcm_γ∈Kni(γ).

We quote from [2] the following result.

Lemma 2.2. Let g(x)∈Z[x]. Write g(x)in the form g(x) =b_n

n_x no

+. . .+b₁ n_x

1 o

+b₀, where b₀, . . .b_n∈Z,

nx

0 o

=1andnx

jo

:=x(x−1). . .(x−(j−1)), for j≥1.

Then, we have the identity

i(g) =gcdⁿ_j=0 j!b_j . Corollary 2.1. The integers i(γ), i(K)andı(K)ˆ divide n!.

Remark 2.2. Clearly, from the definitions of i(K)andı(K), we see that i(K)ˆ dividesˆı(K).

3. Study of the denominators of some algebraic numbers We prove the following lemma, which is useful for the rest of this paper:

Lemma 3.1. Let p be a prime number,γ be an algebraic number and d(γ)be its denomi- nator. Write d(γ)in the form d(γ) =p^k·q,wheregcd(p,q) =1and let µ =qγ. Then we have

d(µ) =p^k, ν_p(c(µ)) =ν_p(c(γ)) and ν_p(i(γ)) =ν_p(i(µ)).

Proof : Letnbe the degree ofγ and letg(x)be the minimal polynomial ofγ overZ, then the minimal polynomial ofµ overZis given byh(x) = qⁿ⁻¹g(^x_q)and thatd(µ) = p^k . Suppose that p^u|i(γ) for some positive integeru and let x₀∈Z. Since gcd(p,q) =1, there exists y₀ ∈Z such that the congruence qy₀ ≡ x₀(mod p^u) holds. In particular g(y₀)≡0(modp^u), so that we have

h(x₀)≡qⁿ⁻¹g(y₀)≡0(mod p^u).

Sinceh(qy₀) =qⁿ⁻¹g(y₀)≡0 (mod p^u), it follows thatp^u|h(x₀)and sincex₀was chosen arbitrarily, p^u|i(µ). Conversely, suppose that p^u|i(µ) and let x₀∈Z, then h(qx₀)≡ 0(mod p^u),hence

g(x₀) = 1

qⁿ⁻¹h(qx₀)≡0(mod p^u).

The proof of the equalityν_p(c(µ)) =ν_p(c(γ))is straightforward. 2 We state the main result of this section.

Theorem 3.1. Let K be a number field of degree n overQ,γ ∈K_n, c=c(γ), d=d(γ) and p be a prime number. If k=ν_p(d)≥1, then p<n and k≤ν_p(c)≤(n−p)k.

Proof : Setd=p^kqwithk≥0 andgcd(p,q) =1. First, we supposek=0 and letµ =qγ, thenµ ∈A_nand p|i(µ)by Lemma 3.1, hence p|i(K). Suppose now thatk≥1. By Lemma 3.1, we can assume thatq=1 andd=p^k. Therefore the minimal polynomial ofγ overZ has the form

g(x) =p^t n _x

no

+b_n−1 n x

n−1 o

+. . .+b₁ n_x

1 o

+b₀, withk≤t≤nk.

Since p|n!, then p≤n. If p=n, since p|j!b_j for all j=0, . . . ,n−1, then we conclude thatp|b₀,b₁, . . . ,b_n−1.Thereforeg(x)is reducible inZ[x], which is a contradiction, hence p<n.In this case, since p|b₀,b₁, . . . ,b_p−1, then we may writeg(x)in the form

g(x) =x(x−1)· · ·(x−(p−1))

p^tx^n−p+a˜_n−p−1x^n−p−1+. . .+a˜₁x+a˜₀

+p

c_p−1x^p−1+. . .+c₁x+c₀

where all the coefficients ˜a_i,c_jare integral. Sinceg(x)is irreducible inZ[x], there exists j∈ {0, . . . ,n−p−1} such that p6 |a˜_j.Denote by j_o the greatest of these integers. Let θ ∈A_nbe the unique element such thatγ = ^θ

p^k.Then we have θ(θ−p^k)· · ·(θ−(p−1)p^k)

p^tθ^n−p+p^ka˜_n−p−1θ^n−p−1 +. . .+p^{k(n−p−jo)}a˜_j0θ^j0+. . .+p^k(n−p−1)a˜₁θ+p^k(n−p)a˜₀

+p·p^{k(n−(p−1))}

c_p−1θ^p−1+c_p−2p^kθ^p−2+. . .+c₀p^k(p−1)

=0.

Write this equation in the form:

p^tθⁿ+u_n−1θⁿ⁻¹+. . .+u_pθ^p+. . .+u₁θ+u₀=0.

Sinceθ is integral, it follows in particular that p^t|u_jfor j=p, . . . ,n−1.We can set θ(θ−p^k)· · ·(θ−(p−1)p^k) =θ^p+σ1θ^p−1+. . .+σp−1θ.

Then



















σ₁ = −(p^k+. . .+p^k(p−1)) =p^{k p(p−1)}₂

σ₂ = p^2k

∑

i6=j i,j∈{1,...,p−1}

i j . . .

σ_p−1 = (−1)^p−1p^k(p−1)(p−1)!.

We have

(x^p+σ₁x^p−1+. . .+σ_p−1x)

p^tx^n−p+. . .+p^{k(n−p−j0)}a˜_j0x^j0+. . .+a˜₀p^k(n−p) + pp^{k(n−(p−1))}

c_p−1x^p−1+. . .+c₁x

= p^txⁿ+u_n−1xⁿ⁻¹+. . .+u_px^p+. . .+u₁x+u₀, hence















u_n−1 = p^ka˜_n−p−1+p^tσ₁

u_n−2 = p^2ka˜_n−p−2+p^ka˜_n−p−1σ₁+p^tσ₂ . . .

u_j0+p = a˜_j0p^{k(n−p−j0)}+a˜_j0+1p^{k(n−p−(j0+1)}σ₁+. . .+a˜_j0+mp^{k(n−p−(j0+m)}σ_m + . . .+p^tσ_n−(j0+p)

The first equation implies that p^t|p^ka˜_n−p−1. Then the second implies that p^t|p^2ka˜_n−p−2. Iterating the process, the last equation givesp^t|a˜_j0p^{k(n−p−j0)}. Sincep6 |a˜_j0, then

t≤k(n−p−j₀)≤k(n−p).

Theorem 3.2. Let p be a prime number, n and k be positive integers, p<n.Then for any integer t, such that k≤t≤(n−p)k, there exist infinitely many algebraic numbersγ ∈C of degree n such that p|i(γ),ν_p(c(γ)) =t andν_p(d(γ)) =k.

Proof : Dividingtbyk, we have two possibilities:

t = (n−i)k+α with 0<α <kand p<i≤n−1 (3)

t = (n−i)kwith p≤i≤n−1 (4)

• First case: t= (n−i)k+α with 0<α <kand p<i≤n−1. Then, we have t >(n−i)α+α =α(n−i+1), henceα < t

n−(i+1). On the other hand, choose integersa₀, . . . ,a_nsuch that















gcd(a₀, . . . ,a_n) =1, ν_p(a_j) =t, for j>i

ν_p(a_i) =α,(note thatα6=0) ν_p(a_i−1) =0

νp(a_j) =1, for j<i−1.

(5)

Consider the polynomials

f(x) =

n

∑

a_jnx

jo

=

n j=0

∑

˜ a_jx^j and

g(x) =a˜_nxⁿ+q^en−1a˜_n−1xⁿ⁻¹+. . .+q^e1a˜₁x+qa˜₀:=

n

∑

b_jx^j,

whereqis a prime number such that q≡1 (mod p),q6 |a˜₀ and the exponentse_j are arbitrary fixed positive integers. Clearly g(x) is irreducible in Z[x] by Eisenstein’s Theorem. Letγ be a root of g(x). Since p≤i−1, then by Lemma 2.2, we conclude that p|i(f)and since g(x)≡ f(x) (mod p)for anyx∈Z, then p|i(γ).We look at the p-adic valuations of the ˜a_j. Recall that ˜a_n=c(γ)and ˜a₀=a₀, henceν_p(a˜_n) =t and νp(a˜₀) =1.We claim that







νp(a˜_j)≥t, for j>i νp(a˜_i) =α,

ν_p(a˜_i−1) =0.

(6)

For any j≥1 we have ˜a_j=a_j+

n

∑

l=j+1

a_lc_l, wherec_l∈Zfor anyl.

If j>ithenν_p(a_j) =ν_p(a_j+1) =. . .=ν_p(a_l) =t,henceν_p(a˜_j)≥t.

For j=iwe haveνp(a_j) =α <t andνp(a_l) =t forl=i+1, . . . ,n,henceνp(a˜_i) =α. For j=i−1 we haveνp(a_i−1) =0 andνp(a_l)≥α forl=i, . . . ,n,, henceνp(a˜_i−1) =0.

Thus we obtain the desired claim.

To compute the p-adic valuation of the denominator ofγ, we use Lemma 2.1. For j>i, we have

νp(b_n)−νp(b_j)

n−j = νp(a˜_n)−νp(a˜_j)

n−j =t−t_j

n−j ≤0, becauset_j≥t. For j=i, we have

ν_p(b_n)−ν_p(b_i)

n−i = ν_p(a˜_n)−ν_p(a˜_i)

n−i =t−α n−i =k.

For j<iwe have ν_p(b_n)−ν_p(b_j)

n−j =ν_p(a˜_n)−ν_p(a˜_j)

n−j =t−t_j

n−j≤ t

n−(i−1)≤(n−i)k+α

n−(i−1) <(n−i)k+k n−(i−1) =k, henceνp(d(γ)) =k.

• Second case: t = (n−i)k with p≤ i≤n−1. Choose integers a₀, . . . ,a_n such that gcd(a₀, . . . ,a_n) =1 and











ν_p(a_j)≥t, for j>i, ν_p(a_i) =0,

νp(a_j) =1, for j<i, νp(a_n) =t.

(7)

Consider the polynomials

f(x) =

n

∑

j=0

a_jnx

jo

=

n

∑

j=0

˜ a_jx^j and

g(x) =a˜_nxⁿ+q^en−1a˜_n−1xⁿ⁻¹+. . .+q^e1a˜₁x+qa˜₀:=

n

∑

j=0

b_jx^j,

whereqand the e_j have the same meaning as in the preceding case. Clearly g(x) is irreducible inZ[x]by Eisenstein’s Theorem. Letγ be a root of g(x). Since p≤i, then by Lemma 2.2, we conclude that p|i(f)and sinceg(x)≡ f(x) (mod p)for anyx∈Z, then p|i(γ).We look at thep-adic valuations of the ˜a_j. We have ˜a_n=a_nhenceν_p(a˜_n) =t.For j>i, we have ˜a_j=a_j+

n l=

∑

a_lc_lwherec_l∈Zand we haveνp(a_j)≥tandνp(a_l)≥t forl≥ j+1, henceν_p(a˜_i)≥t.

For j=i, we haveν_p(a_i) =0 andν_p(a_l) =t,l≥ j+1, henceν_p(a˜_i) =0.

For j<i, we haveν_p(a˜_j)≥0.

For j<i, we haveν_p(a˜_j)≥0.

We compute the p-adic valuation of the denominator ofγ by using Lemma 2.1.

For j>i, we have

νp(b_n)−νp(b_j)

n−j = νp(a˜_n)−νp(a˜_j)

n−j =t−t_j

n−j ≤0, becauset_j≥t. For j=i, we have

νp(b_n)−νp(b_i)

n−i = νp(a˜_n)−νp(a˜_i)

n−i = t

n−i=k.

For j<iwe have

νp(b_n)−νp(b_j)

n−j = νp(a˜_n)−νp(a˜_j)

n−j <t−t_j n−i ≤ t

n−i=k, henceν_p(d(γ)) =k.

Since we can chooseqand thee_jin an infinite number of ways, then the number ofγ’s is infinite. 2

4. Upper bounds for the enumeration of the denominators of some algebraic numbers

Proposition 4.1. Letθ ∈A_n,and f(x) =xⁿ+a_n−1xⁿ⁻¹+. . .+a₁x+a₀its minimal polyno- mial overQ. Suppose thatθ 6≡0(modpA).Construct the Newton polygon of f(x)by plot- ting in the(x,y)plan the points A_iwhose coordinates are(i,ν_p(a_i))for all i∈ {0, . . . ,n}

such that a_i6=0. Suppose that there exists k≥1such that p|i(θ/p^k). Then there exists two integers m,M such that1≤m<M≤n−1and the line joining the points A_mand A_M has the following equation

y+kx−u=0, where kM≤u<ν_p(a₀)and u=ν_p(cont(f(p^kx))).

Moreover all the points A_i such that i<m or i>M belong to the domain of all points (x,y)such that y+kx−u>0. If m<i<M, then we haveνp(a_i) +ki−u≥0.

Proof : The minimal polynomial overZofθ/p^kis given by

f(p^kx)/p^u=p^nk−uxⁿ+p^(n−1)k−ua_n−1xⁿ⁻¹+. . .+p^k−ua₁x+p^−ua₀:=g(x), whereu=νp(cont(f(p^kx))). Let

I={i: 1≤i≤n−1,a_i6=0 andik+ν_p(a_i)−u=0}.

Sinceθ/p^kis not integral thennk−u>0. Sinceg(0)≡0(modp), thenν_p(a₀)−u>0.

Adding these two facts to the property thatg(x)is primitive implies thatI6=/0.Furthermore g(1)≡0(modp), hence|I| ≥2.

Let m=inf(I)and M=max(I).Clearly the equation of the line joining the pointsA_m andA_M is given by: y+kx−u=0. Moreover a point(i,νp(a_i))of the Newton polygon belongs to this line if and only ifi∈I. The definition ofmandMimplies the properties of the pointsA_iand ofu. 2

Remark 4.1. Proposition 4.1 shows that−k is the slope of some line joining two points A_mand A_M. Moreover all the others points belong to the same side of the line ( or on the line). Therefore, if we fix a prime p and an algebraic integerθ such thatθ 6=0 (mod pA), it is possible to find explicitly all the values of k such that p|i(θ/p^k).This proposition shows also that the set of such nonnegative integers k is finite (may be empty).

Example 4.1. Let t ≥2be an integer, f(x) =x³+x²+2^tx+2^t+1andθ_t be a root of f(x).

It is seen that f(x)is irreducible overQ: if not, it has a root a/b inQwith a,b∈Zand gcd(a,b) =1. Substitution then yields

a³+a²b+2^tab²+2^t+1b³=0,

implying b|a³. Thus, b=±1, and we then obtain a|2^t+1. Letting a=2ⁱ, we obtain 2³ⁱ+2^t+i=2²ⁱ+2^t+1, implying that t+i≤t+1so i=1which is impossible.

For any nonnegative integer, let γt,k =θt/2^k. We show that V₂(θ_t) ={0,t}. Clearly, 2|i(θ_t), hence0∈V₂(θ_t). The Newton diagram for p=2has the following shape:

A₁ A₀

A₂ A₃

The possible edges of the convex hull which may give rise to values of k∈V₂(θ_t), k≥1, are[A₀A₁]and[A₁A₂]. Their slopes are equal to−t−1and t respectively. Thus, k=t+1 or k=t.

If k=t+1, the minimal polynomial ofγ_t,koverZis given by g(x) =2^t+2x³+2x²+x+1.

This shows that26 |i(g(x)), hence t+16∈V₂(θ_t).

If k=t, the minimal polynomial ofγ_t,koverZis given by h(x) =2x³+x²+x+2. This shows that2|i(h(x)), hence t∈V₂(θ_t). Thus, V₂(θ_t) ={0,t}.

We state now our main result on the upper bounds for the enumeration of the denomina- tors of algebraic numbersγ such that p|i(γ).

Theorem 4.1. Letθ be a root of f(x)∈Z[x], monic irreducible, p a prime number such thatθ 6≡0(modpA)and let a₀= f(0).We set

V_p(θ) =n

k≥0;p|i(θ/p^k)o .

Suppose that V_p(θ)6= /0then we have

|V_p(θ)| ≤ n−1 p−1, (8)

∑

k∈Vp(θ)

k<νp(a₀) p . (9)

For the proof of this theorem, we need the following lemma.

Lemma 4.1. Let p be a prime number and g(x) =a_Mx^M+. . .+a_mx^m, M>m>0such that p6 |a_M.If p|i(g),then M−m≥ p−1.

Proof : Suppose that p|i(g), then clearly p|i(xg₁), whereg₁(x) =a_Mx^M−m+. . .+a_m. Writexg₁in the form

xg₁(x) =a_M

n x

M−m+1 o

+

∑

j<M−m+1

b_j nx

jo ,

then by Lemma 2.2 p|(M−m+1)!a_M, hence p|(M−m+1)!.Therefore p≤M−m+1.

2

Proof of Theorem 4.1. Suppose that the complete list of elements ofV_p(θ)is given by k₁<k₂< . . . <k_z.We havek₁=0 if and only if p|i(θ).Set f(x) =xⁿ+a_n−1xⁿ⁻¹+. . .+ a₁x+a₀.For each j=1, . . . ,z,letg_j(x)be the minimal polynomial ofθ/p^kj overZ. Setg_j(x) =p^tjxⁿ+b⁽_n−1^j) xⁿ⁻¹+. . .+b⁽₁^j)x+b⁽₀^j),we have

g₁(x) =

f(x) ifk₁=0 f(p^k1x)p^−u1 ifk₁≥1 (10)

andg_j+1(x) =g_j(p^kj+1−kjx)p^−uj+1 for j=1, . . . ,z−1 andu₁, . . . ,u_zare positive integers.

For any j=1, . . . ,z, letI_j={i∈ {1, . . . ,n−1},ν_p(b⁽_i^j)) =0}.Sincet_j>0,ν_p(b⁽₀^j))>0 and g_j(x)is irreducible in Z[x], then it follows thatI_j6= /0.Let m_j =inf(I_j) and M_j = sup(I_j). Sinceg_j(1)≡0(modp) then |I_j| ≥2 andm_j <M_j. Clearlym_j≥1 andM_j≤ n−1.

We claim that:

. M_j−m_j≥ p−1 for j=1, . . . ,z.

. u₁≥k₁M₁andu_j≥(k_j−k_j−1)M_jfor j=2, . . . ,z.

. n≥M₁>m₁≥M₂>m₂≥. . .≥M_z>m_z≥1.

The first claim follows from Lemma 4.1. Ifk₁=0, then from (10) we haveu₁=0=k₁M₁. From the definition ofM₁and the definition of the intervalI₁, it follows that

ν_p(b⁽¹⁾_M

1) =0.

Therefore, ifk₁≥1, equation (10) and the above equality imply that 0=ν_p(b⁽¹⁾_M

1) =ν_p(a_M1) +k₁M₁−u₁, henceu₁≥k₁M₁.Similarly for j=2, . . . ,z,we have

0=ν_p(b⁽_M^j)) =ν_p(b⁽_M^j−1)

j ) + (k_j−k_j−1)M_j−u_j,

henceu_j≥(k_j−k_j−1)M_j,which proves the second part of the claim. For the last part of the claim it is sufficient to prove thatm_j≥M_j+1 for j=1, . . . ,z−1. For, suppose that m_j<M_j+1for some j∈ {1, . . . ,z−1}.We have 0=ν_p(b⁽_m^j)_j),hence

νp(b⁽_m^j+1)_j ) =νp(b⁽_m^j)_j.p^{(kj+1−kj)mj}.p^−uj+1) = (k_j+1−k_j)m_j−u_j+1.

We deduce that(k_j+1−k_j)m_j≥u_j+1and then(k_j+1−k_j)M_j+1>u_j+1. It follows that νp(b⁽_M^j+1)

j+1 ) =νp(b⁽_M^j)

j+1) + (k_j+1−k_j)M_j+1−u_j+1>0,

which contradicts the definition ofM_j+1and completes the proof of the claim.

We now come back to the proof of Theorem 4.1.

Completion of proof of Theorem 4.1: We use the first and the third points of the claim. We have

n≥M₁>M₂> . . . >M_z−1>M_z≥p>1.

Using the claim, we obtain

n−p≥M₁−M_z= (M₁−M₂) +. . .+ (M_z−1−M_z)≥(p−1)(z−1)

hence

z≤ n−p

p−1+1= n−1 p−1. Therefore (8) is proved.

We prove the inequality (9) of Theorem 4.1. We haveb⁽¹⁾₀ =a₀p^−u1, b⁽₀^j+1)=b⁽₀^j)p^−uj+1 for j=1, . . . ,z−1 and sinceg_z(0)≡0(modp),thenν_p(b^(z)₀ )>0.Henceu₁+u₂+. . .+ u_z<ν_p(a₀).On the other hand, using the first and the second parts of the claim, we obtain u₁+u₂+. . .+u_z ≥ k₁M₁+ (k₂−k₁)M₂+. . .+ (k_z−k_z−1)M_z

≥ k₁zp+ (k₂−k₁)(z−1)p+. . .+ (k_z−k_z−1)p

= p(k₁z+k₂z−k₁z−k₂+k₁+k₃z−k₂z−2k₃+2k₂+. . .+k_z−k_z−1)

= p((k₁+k₂+. . .+k_z).

Therefore, we have

νp(a₀)>

z

∑

j=1

u_j≥p

z

∑

j=1

k_j, hence

z

∑

k_j<ν_p(a₀)/p.

Remark 4.2. Theorem 4.1, shows that ifν_p(a₀)≤p, then V_p(θ) ={0}or V_p(θ) = /0.

The following result shows that the bound (8) in Theorem 4.1 is the best possible. More precisely, we have

Proposition 4.2. Let p and q be distinct prime numbers such that q≡1 (mod p),θ be a root of

f(x) =xⁿ+

N i=1

∑

a_ix^n−i(p−1)+qp^λ, where N=

(n−1)/(p−1)

,a_i= (−1)ⁱqp(p−1)i(i−1)/2, for i=1, ...,N and λ >2n(N−1)−(p−1) (N−1)²+N−1

2 .

Then

|V_p(θ)|=jn−1 p−1 k

.

Proof : Clearly, by Eisenstein’s criterion, f(x)is irreducible overQ. The coefficient of x^n−(p−1) is coprime top, henceθ 6≡0(modpA).By Theorem 4.1, we have

|V_p(θ)| ≤jn−1 p−1 k

. We show that the integers 0,1, . . . ,j

n−1 p−1

k−1 belong toV_p(θ)and this will complete the proof of Proposition 4.2. Since f(x)≡xⁿ−qx^n−(p−1) (mod pZ[x])andq≡1 (mod p), then f(x)≡x^n−(p−1)(x^p−1) (mod pZ[x]). Thus, p|i(f)and 0∈V_p(θ). Seta₀=1 and fixk∈n

1, . . . ,j

n−1 p−1

k−1o

. We have f(p^kx) =p^nkxⁿ+

N i=1

∑

a_ip^{nk−i(p−1)k}x^n−i(p−1)+qp^λ. We claim, omitting the proofs that

ν_p(a_kp^{nk−k(p−1)k}) =ν_p(a_k+1p^{nk−k(p−1)(k+1)}) = 2nk−(p−1)(k²+k) 2

and

νp(a_ip^{nk−i(p−1)k})> 2nk−(p−1)(k²+k)

2 ifi6=k,k+1.

Moreover, since the functionx7→ψ(x) =2nx−(p−1)(x²+x)is increasing in[0,N−1]

and sinceλ>^{2n(N−1)−(p−1)(N−1)}₂ ^2+N−1, thenλ>2nk−(p−1)(k²+k)

2 .It follows thatcont(f(p^kx)) =

2nk−(p−1)(k²+k)

2 and the minimal polynomial overZofγ_k= ^θ

p^k is given by g_k(x) = f(p^kx)p^{−(2nk−(p−1)k2+k)/2}.

From the above it is seen that

g_k(x)≡(−1)^kx^n−k(p−1)+(−1)^k+1xn−(k+1)(p−1)(modp)≡(−1)^kxn−(k+1)(p−1)(x^p−1−1)(modp), hence p|i(γ_k),thusk∈V_p(θ). 2

Corollary 4.1. Let p be a prime number, andθ be a root of f(x) =xⁿ+a_n−1xⁿ⁻¹+. . .+a₀∈Z[x],

irreducible overQ. Suppose that there exists i∈ {0, . . . ,min(p,n)−1}such thatν_p(a_i) = 0.Then

θ 6≡0(modpA)and V_p(θ) = /0or V_p(θ) ={0}.

Moreover if p>n, then V_p(θ) =/0.

Proof : Letα be an algebraic integer andg(x) =xⁿ+b_n−1xⁿ⁻¹+. . .+b₀be its minimal polynomial overQ. It is easy to prove thatα ≡0(modp)if and only ifν_p(b_i)≥n−ifor i=0, . . . ,n−1.Our assumption then implies thatθ 6≡0(modpA). Suppose thatV_p(θ)6= /0 and letk∈V_p(θ).Assume thatk≥1 and letγ =θ/p^k and u=cont(f(p^kx)).Then the minimal polynomial ofγ overZis given by

g(x) = f(p^kx)p^−u=p^nk−uxⁿ+p^(n−1)k−ua_n−1xⁿ⁻¹+. . .+p^−ua₀.

As in Theorem 4.1, let

I={j∈ {1, . . . ,n−1};νp(a_j) +k j−u=0},m=inf(I),M=sup(I).

Suppose first that m≤i. We have ik−u=ν_p(a_i) +ik−u≥0. Since M−m≥ p−1, thenM>iwhich impliesν_p(a_M) +Mk−u≥Mk−u>ik−u≥0,a contradiction. We deduce thatm>iand thenν_p(a_m)) +km−u≥km−u>ki−u=ν_p(a_i) +ki−u≥0,a contradiction again. ThereforeV_p(θ) ={0}.2

5. A new invariant of number fields and a generalisation of MacCluer’s Theorem Let pbe a fixed prime integer. We have shown that for any algebraic integerθ such that θ 6≡0(modpA), p|i(θ/p^k)for somek≥1, thenk<ν_p(N_Q(θ)/Q(θ))/p. Does there exist some constantc>0 such that ifθ ∈Q,θ 6≡0(modpA)and p|i(θ/p^k)thenk<c?

Even if we fix the degreenofθ and suppose that the constantcdepends onn, the answer is negative as it is shown by the following result.

Proposition 5.1. Let n,N be positive integers and p be a prime number such that p<n.

Then there exists an integer k>N and an algebraic integerθ of degree n such that θ 6≡0(modpA)and p|i(θ/p^k).

Proof : LetF be a number field of degreen−1 such that p|i(F). In particular, we can take F such that pcompletely splits inF, so that p|i(F)by MacCluer’s Theorem. Such a field F exists by Tchebotarev’s theorem [8]. Let α be a primitive element ofF/Q. Suppose thatα is integral andp|i(α). LetF_α(x)be the minimal polynomial ofα overQ. Letqbe a prime number such that

q6=pandq6 |N_F/Q(α).

Let t be an integer such thatt >nN and let g(x) = p^txⁿ+qF_α(x). Then Eisenstein’s criterion shows thatg(x)is irreducible overQ. Obviously g(x)is primitive, hence it is irreducible inZ[x]. Letγ be a root of g(x), then clearly p|i(γ)and d(γ) = p^k for some positive integerksuch thatk≤t≤nk,hencek≥t/n>N. The algebraic integerθ =p^kγ satisfies all the conditions of the proposition and the proof is complete. 2

LetK be a number field of degreenoverQandAbe its ring of integers. We define the integerνp(K)as follows.

Definition 5.1. Let V_p(K) =n

k≥0, there existsθ ∈A_n,θ 6≡0(modpA), and p|i(θ/p^k)o ,

and we define

v_p(K) =







−∞ if V_p(K) = /0,

∞ if V_p(K) is infinite, max(V_p(K)) if V_p(K) is finite.

Remark 5.1. By Theorem 3.1, we have v_p(K) =−∞if and only if p6 |i(K).So there is no need to give examples illustrating this fact. Theorem 3.1 again shows that if the degree of the number field K is a prime p then v_p(K) =0if p|i(K)and v_p(K) =−∞if p6 |i(K).

In the following we compute explicitlyv₂(K)for some number fields of degree 3 or 4 overQ.

Proposition 5.2(Galois field of degree 4). Let K/Qbe a Galois number field of degree4 in which the prime2splits into a product of two prime ideals having their residual degree equal to2. Then we have v₂(K) =0.

Proof : By MacCluer’s theorem, 2|i(K), hence 0∈V_p(K). Letpandp⁰be the conjugate prime ideals of A lying over 2 and having their residual degree equal to 2. Suppose that 2|i(θ/2^k)for somek≥1 andθ ∈A_n such thatθ 6≡0(mod 2A).SinceN_K/Q(θ)≡0 (mod 2), then we may suppose thatp^e||θ andp⁰6 |θ for somee≥1. We suppose that the conjugatesθ1=θ,θ2,θ3,θ4ofθ satisfy the following conditions:

p^e||θ₁,p^e||θ₃,p⁰6 |θ₁θ₃,p^0e||θ₂,p^0e||θ₄,p6 |θ₂θ₄.

Let f(x) =x⁴+a₃x³+a₂x²+a₁x+a₀∈Z[x]be the minimal polynomial ofθ overQ. Let g(x)∈Z[x]be the minimal polynomial ofγ =θ/2^koverZ, then

g(x) = f(2^kx).2^−u=2^4k−ux⁴+2^3k−ua₃x³+2^2k−ua₂x²+2^k−ua₁x+2^−ua₀,

whereuis the content of f(2^kx). Using the elementary symmetric functions of theθ_jand our assumption on theirp-adic andp⁰-adic valuations, we get

ν2(a₀) =2e,ν2(a₁)≥eandν2(a₂) =0.

Ifk≥e, thenν2(2^4k−u)≥4e−u,ν2(2^3k−ua₃)≥3e−u,ν2(2^2k−ua₂) =2k−u≥2e−u, ν₂(2^k−ua₁)≥2e−u, ν₂(2^−ua₀) =2e−u. Since these five 2-adic valuations must be nonnegative, thenu≤2e. Furthermore one (at least) of these valuations must be 0, hence u=2e. In this case, g(0)6≡0(mod 2) which is a contradiction to 2|i(γ). If k<e, then ν₂(2^4k−u) =4k−u, ν₂(2^3k−ua₃)≥3k−u, ν₂(2^2k−ua₂) =2k−u, ν₂(2^k−ua₁)>2k−u, ν₂(2^−ua₀)>2k−u. Using similar arguments as in the preceding case, we obtainu=2k.

We conclude that all the coefficients ofg(x)have their 2-adic valuations positive except the coefficient ofx²which has a 2-adic valuation equal to 0. In this case also we reach a contradiction sinceg(1)6≡0((mod)2).It follows thatV₂(K) ={0}andv₂(K) =0. 2 For the proof of the next proposition, we will need the following lemma.

Lemma 5.1(Engstrom). Let K be a number field, A be its ring of integers and p be a prime integer. Letp₁, . . . ,p_s be distint prime ideals of A lying over p and letΦ₁(x), . . . ,Φ_s(x) be monic irreducible polynomials overFp not necessarily distincts of degree d₁, . . . ,d_s respectively, where d_idivides the residual degree ofp_i. Let h₁, . . . ,h_sbe positive integers.

Then there exists a primitive elementθ ∈A such thatp^h_iⁱ||Φ_i(θ)for i=1, . . . ,s Proof : see [4]. 2

Proposition 5.3(Cubic Galois). Let K/Qbe a Galois number field of degree3in which the prime2splits completely. Then V₂(K) =N.

Proof : Letk and ebe positive integers such thate>k. Letp₁,p₂ andp₃ be the prime ideals ofAlying over 2. By Lemma 5.1 there existsθ ∈A_nsuch that

p^e₁||θ,p^k₂||θ andp₃6 |θ.

Assume that the conjugates ofθ,θ₁=θ,θ₂,θ₃are labelled in order to satisfy the following conditions:

p^e₂||θ₂,p^k₃||θ₂,p₁6 |θ₂, p^e₃||θ₃,p^k₁|||θ₃,p₂6 |θ₃.

Let f(x) =x³+a₂x²+a₁x+a₀∈Z[x]be the minimal polynomial ofθ overQ. Expressing a₀,a₁anda₂in terms ofθ₁,θ₂,θ₃, we get

ν2(a₀) =e+k,ν2(a₁) =kandν2(a₂) =0.

We have

f(2^kx) =2^3kx³+2^2ka₂x²+2^ka₁x+a₀. Set

b₃=2^3k,b₂=2^2ka₂,b₁=2^ka₁,b₀=a₀. Using the 2-adic valuation ofa₀,a₁,a₂we obtain

ν₂(b₁) =ν₂(b₂) =2k,ν₂(b₀) =e+k>2k,ν₂(b₃) =3k>2k.

Thereforecont(f(2^2kx)) =2^2kand the minimal polynomial ofθ/2^kis given by g(x) = f(2^kx).2^−2k.

Clearly we haveg(0)≡g(1)≡0(mod2)hence 2|i(θ/p^k).Since the prime 2 splits com- pletely inK, then 0∈V₂(K). ThereforeV₂(K) =Nandv₂(K) =∞.2

Remark 5.2. Our result in the sequel can be viewed as a generalization of MacCluer’s theorem which establishes a relation between the number of prime ideals of A lying over p and the property of p to be a divisor of i(K).

Fix a prime number p and define, for any primitive elementθ ∈A of K, the integer j_p(θ)as follows.

Definition 5.2. Let F_θ(x)be the minimal polynomial ofθ overQ. Let j_p(θ)be the largest integer y, if it exists,1≤y≤p such that F_θ(1)≡F_θ(2)≡. . .≡F_θ(y)≡0(modp).If not set j_p(θ) =0.We define also j_p(K) =max_θ∈An j_p(θ).

Theorem 5.1. Let r be the number of prime ideals of A lying over p. Then j_p(K) =inf(r,p).

Moreover

p|i(K) ⇐⇒ j_p(K) =p.

Proof : Suppose first thatr≤pand letp₁, . . . ,p_rbe the distinct prime ideals ofAlying over p. By Lemma 5.1 there existsθ ∈A_nsuch thatθ ≡i(modp)fori=1, . . . ,r.It follows that the minimal polynomialF_θ(x)ofθ satisfies the condition

F_θ(x)≡(x−1)(x−2). . .(x−r)g(x)(modp)

hence j_p(θ)≥rwhich implies that j_p(K)≥r.On the other hand, letθ ∈A_nsuch that j_p(K) = j_p(θ):= t,

then

F_θ(x)≡(x−1)(x−2). . .(x−t)g(x)(modp)

hence, by Hensel’s Lemma, we deduce thatF_θ(x)has at leastt irreducible factors overZp, the ring of p-adic integers. Again by Theorem 5.1 of chapt. 2 of [6], we havet ≤r.We conclude that j_p(K) =r=inf(p,r).

Suppose now thatr>p. By Lemma 5.1, letθ∈A_nsuch thatθ ≡i(modp)fori=1, . . . ,p.

Then

F_θ(x)≡(x−1)(x−2). . .(x−p)g(x)(modp).

therefore we have j_p(θ)≥ p which implies j_p(K)≥ p. From the definition we have j_p(K)≤p, hence j_p(K) =p=inf(r,p).

We now prove the last statement of the proposition. We have

p|i(K) ⇐⇒ r≥ p (by MacCluer’s theorem) ⇐⇒ inf(r,p) =p ⇐⇒ j_p(K) =p. 2 6. CONCLUDING REMARKS

LetK be a number field. If[K:Q] =2, then by Corollary 2.1,i(K)and ˆı(K)are equal to 1 or 2. Theorem 3.1 shows that 26 |i(γ)ifγ 6∈A_n, hencei(K) =ı(K)ˆ ∈ {1,2}.

If [K:Q] =3, then i(K)and ˆı(K)∈ {1,2,3,6}. Moreover, Theorem 3.1 shows that 3|ˆı(K)if and only if 3|i(K).

Suppose that there exists γ =θ/2^k withk≥1,k≤t ≤3k and θ 6≡0 (mod p)such that 2|i(γ). Letg(x) =2^tx³+b₂x²+b₁x+b₀ be the minimal polynomial ofγ overZ. Sinceg(0)≡0 (mod 2), thenb₀≡0 (mod 2). Sinceg(1)≡0 (mod 2), thenb₁+b₂≡0 (mod 2), thus b₁ =b₂ (mod 2). Moreover, since g(x) is primitive, then b₁ ≡b₂≡1 (mod 2). By Theorem 3.1,t≤k. Sincek≤t, thenk=t. The minimal polynomial ofθ is then given by

f(x) =x³+b₂x²+b₁2^tx+b₀2^2t.

This shows that 2|i(f)and then 2|i(θ), thus 2|i(K). We conclude thati(K) =ˆı(K).

Let K be a number field of degreenand letγ ∈K_n\A_n. Setγ =θ/d, whered is an integer at least equal to 2 such thatθ 6≡0 (mod p)for any prime divisor pof d. It is proved in Lemma 3.1 that ifd=p^kqwith gcd(p,q) =1 andk≥1, then p|i(γ)if and only if p|i(θ/p^k). We ask that following: Is it true that if p|i(θ/p^k)withk≥1 andθ 6≡0 (mod p), then p|i(K)? Do we have ˆı(K) =i(K)?

Recall that ν_p(K) is the greatest element of the set V_p(K), when this set is finite.

Do we have {0,1, . . . ,ν_p(K)}=V_p(K)? The example given in section 4 shows that V_p(θ_t) ={0,t} 6={0,1, . . . ,t}. We may ask a similar question whenV_p(K)is infinite. Do we haveV_p(K) =N?

Acknowledgments: The authors express their gratitude to the anonymous referee for constructive suggestions which improved the quality of the paper. The third author was supported in part by NSERC.

REFERENCES

[1] S. Arno, M. L. Robinson, and F. S. Wheeler,On denominators of algebraic numbers and integer Polynomials, J. Number theory57(1996), 292-302.

[2] M. Ayad, O. Kihel,Common Divisors of Values of Polynomials and Common Factors of Indices in a Number Field, Inter. J. Number Theory7 (2011), 1173-1194.

[3] M. Ayad, A. Bayad, and O. Kihel, Denominators of Algebraic Numbers in a Number Field, J.

Number Theory149(2015), 1–14.

[4] H. T. Engstrom,On the common index divisors of an algebraic field, Trans. A. M. S32(1930), 223-237.

[5] H. Gunji, D. L. McQuillan,On a class of ideals in an algebraic number field, J. Number Theory2 (1970), 207-222.

[6] G. J. Janusz,Algebraic Number Fields, Gaduate Studies in Math, A. M. S. (1996).

[7] C.R. MacCluer,Common divisors of values of polynomials, J. Number Theory3(1971), 33-34.

[8] O. Neukirch,Algebraic Number Theory, Springer (1999).

MOHAMEDAYAD, LABORATOIRE DEMATH_EMATIQUES´ PURES ETAPPLIQU_EES´ UNIVERSIT_{E DU}´ LITTORAL, F-62228 CALAIS, FRANCE

E-mail address:ayad@lmpa.univ-littoral.fr

ABDELMEJIDBAYAD, D ´EPARTEMENT DE MATHEMATIQUES´ , UNIVERSITE D´ ’EVRYVAL D’ESSONNE, B ˆATIMENTI.B.G.B.I., 3 `EMEETAGE´ , 23 BD.DEFRANCE, 91037 EVRYCEDEX, FRANCE

E-mail address:abayad@maths.univ-evry.fr

OMARKIHEL, DEPARTMENT OFMATHEMATICS, BROCKUNIVERSITY, ONTARIO, CANADAL2S 3A1

E-mail address:okihel@brocku.ca