A note on prime divisors of polynomials P(T^{k}),k\geq 1

# A note on prime divisors of polynomials P(Tk),k≥1

## Abstract.

Let be a number field, the integral closure of in and a monic separable polynomial such that and . We give precise sufficient conditions on a given positive integer for the following condition to hold: there exist infinitely many non-zero prime ideals of such that the reduction modulo of has a root in the residue field , but the reduction modulo of has no root in . This makes a result from a previous paper (motivated by a problem in field arithmetic) asserting that there exist (infinitely many) such integers more precise.

## 1. Introduction

Given a number field , recall that a (non-zero) prime ideal of the integral closure of in is a prime divisor of a monic separable polynomial if the reduction of modulo has a root in the residue field . Examples of standard results on this classical number theoretic notion, which may be obtained as applications of the Chebotarev density theorem, are (1) any monic separable polynomial has infinitely many prime divisors and (2) if is irreducible over , then all but finitely many prime ideals of are prime divisors of if and only if has degree 1. See e.g. [Nag69], [Sch69] and [GB71] for more classical results.

Prime divisors of polynomials are known to appear in several areas of interest. For instance, in field arithmetic, the set of prime divisors of the product of the irreducible polynomials over of the branch points of a finite Galois extension with algebraically closed in corresponds to the set of all prime ideals of that ramify in at least one specialization of ; see [Bec91] and [Leg16b]. Another application deals with the following classical question in diophantine geometry: does a given superelliptic curve over has twists without non-trivial rational points? Under the abc-conjecture, Granville [Gra07] proved that, in the case of hyperelliptic curves over , only a few (quadratic) twists of have non-trivial rational points, provided that has genus at least 3. Partial unconditional results were then given by Sadek [Sad14] (in the case of hyperelliptic curves over ) and the author [Leg16c], asserting that, to guarantee the existence of twists of without non-trivial rational points, it suffices that divides the degree of and there exist infinitely many prime ideals of which are not prime divisors of .

A more recent application in field arithmetic consists in producing examples of “non-parametric extensions” over number fields [Leg16b], i.e., finite Galois extensions with Galois group and algebraically closed in which do not provide all the Galois extensions of with Galois group by specializing the indeterminate . For example, the main result of [Leg16a] asserts that such an extension of with given non-trivial Galois group exists, unless is the Galois group of no Galois extension with algebraically closed in . See also [Dèb16] for other results about non-parametric extensions.

The proof of the main result of [Leg16a] requires the following statement on prime divisors of polynomials, whose an ineffective proof may be found in [Leg16a, §4].

Proposition 1. [Leg16a, Proposition 3.5] Assume and . Then there exist infinitely many positive integers such that the following condition holds:

(/) there exist infinitely many prime ideals of each of which is a prime divisor of but not of .

Note that the result fails trivially if either or .

The aim of the present note is to make Proposition 1 effective by elementary techniques, thus allowing to make previous applications partially effective as well (in particular, that to superelliptic curves). We give precise sufficient conditions on a given positive integer , which depend on the nature of the roots of the polynomial , for condition (/) to hold. In particular, we provide upper bounds on the number of distinct prime factors of such an integer and the multiplicity of each of them. This is motivated by the trivial fact that condition (/) holds for every multiple of any given integer , as soon as condition (/) holds. Proposition 2 below, which is Theorem 2.2 in the sequel, gives an idea of the effectiveness of our method.

Proposition 2. Assume and no root of unity is a root of . Then there exists an effective positive constant such that condition (/) of Proposition 1 holds for any positive integer which has at least one prime factor .

See §2.4 for the definition of , as well as Theorems 2.1 and 2.3 for the case where each root of is a root of unity and the “mixed” case. See also §4 where we show that the conclusion of Proposition 2 fails in general for polynomials with roots of unity among their roots.

Acknowledgments. This work is partially supported by the Israel Science Foundation (grants No. 696/13, No. 40/14 and No. 577/15).

## 2. Statements of Theorems 2.1, 2.2 and 2.3

This section is organized as follows. In §2.1-4, we state the needed notation for Theorems 2.1, 2.2 and 2.3, which are then given in §2.5.

### 2.1. General notation

For any number field , is the integral closure of in and, given a non-zero prime ideal of , is the associated valuation over .

Given a number field , let be a monic separable polynomial such that and . Let

- be the splitting field of over ,

- the number (possibly zero) of roots of that are roots of unity,

- the number (possibly zero) of roots of that are units of , but not roots of unity,

- the number (possibly zero) of roots of that are not units of .

Set We denote the (distinct) roots of by

 t1,…,tr1,tr1+1,…,tr1+r2,tr1+r2+1,…,tr1+r2+r3=tr

and assume that

- are roots of unity (if ),

- are units of , but not roots of unity (if ),

- are not units of (if ).

Pick a positive integer such that, for each prime number , the fields and are linearly disjoint over 1. Finally, set

 d1=|⋃rj=1\rm Gal(L1/F(tj))||\rm Gal(L1/F)|>0.

### 2.2. Data associated with t1,…,tr1

Assume . For each , is a root of unity and (as ). Then there exist two coprime integers and such that and Denote the set of all prime factors of by , the smallest element of by and, by we mean the smallest positive integer that satisfies

 A>log([F:Q])+log(r1)−log(d1)log(pmin).

### 2.3. Data associated with tr1+1,…,tr1+r2

Assume . Let be a system of fundamental units of , i.e., are units of such that each unit of can be uniquely written as where is a root of unity and are integers.

For each , set where is a root of unity and are integers. As is not a root of unity, one has for some . Finally, set

 aj=gcd(|aj,1|,…,|aj,v|)∈N∖{0}

and

### 2.4. Data associated with tr1+r2+1,…,tr

Assume . Given , one has (as 0 is not a root of ) and is an element of which is not a unit. Pick a non-zero prime ideal of such that is a positive integer. Finally, set and, in the case where , we define the integer as follows:

- if and ,

- if ,

- if .

### 2.5. Statements of Theorems 2.1, 2.2 and 2.3

First, we consider the case where each root of is a root of unity.

###### Theorem 2.1.

Assume and each root of is a root of unity. Then condition () of Proposition 1 holds for every -tuple of prime numbers in .

Next, we handle the case where no root of is a root of unity.

###### Theorem 2.2.

Assume and no root of unity is a root of . Then condition () of Proposition 1 holds for every prime number .

Finally, we deal with the mixed case.

###### Theorem 2.3.

Assume , , has a root that is a root of unity and has a root that is not a root of unity. Then, given a prime number , there exists an integer 2 such that condition () of Proposition 1 holds for every .

## 3. Proofs of Theorems 2.1, 2.2 and 2.3

### 3.1. Notation

Given a positive integer , denote the splitting field of over by . For each , fix a -th root of . If and , we choose . Finally, set

 f(k)=|⋃rj=1⋃k−1l=0Gal(Lk/F(e2iπl/kk√tj))||⋃rj=1Gal(Lk/F(tj))|≤1,
 f1(k)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩|⋃r1j=1⋃k−1l=0Gal(Lk/F(e2iπl/kk√tj))||⋃rj=1Gal(Lk/F(tj))|ifr1>00ifr1=0,
 f2(k)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩|⋃rj=r1+1⋃k−1l=0Gal(Lk/F(e2iπl/kk√tj))||⋃rj=1Gal(Lk/F(tj))|ifr2+r3>00ifr2+r3=0.

### 3.2. Statement of an auxiliary result

Theorems 2.1, 2.2 and 2.3 rest essentially on Theorem 3.1 below.

###### Theorem 3.1.

Let be a positive integer.

(1) [Leg16a, Lemma 2.2] Condition (/) holds if and only if .

(2) Assume . Let , be a positive real number and a positive integer such that

 A>log([F:Q])+log(r1)−log(d1)−log(ϵ)log(pmin).

Then one has if is a multiple of .

(3) Assume . Then one has if is a prime .

Theorem 3.1 is proved in §3.4-5.

### 3.3. Proofs of Theorems 2.1, 2.2 and 2.3 under Theorem 3.1

#### Proof of Theorem 2.1

Given , set . By the definition of , we may apply part (2) of Theorem 3.1 with and to get . As each root of is a root of unity, one has . Hence . It then remains to apply part (1) of Theorem 3.1 to get Theorem 2.1.

#### Proof of Theorem 2.2

Assume that is a prime number . Then we may apply part (3) of Theorem 3.1 to get . As no root of unity is a root of , one has . This gives . It then remains to apply part (1) of Theorem 3.1 to conclude.

#### Proof of Theorem 2.3

Let be a prime number satisfying . We may then apply part (3) of Theorem 3.1 to get . Let be the smallest positive integer that satisfies

 A>log([F:Q])+log(r1)−log(d1)−log(1−f2(p0))log(pmin).

Given , set . By the definition of , we may apply part (2) of Theorem 3.1 with and to get . As divides , one has . Hence

 f(k)≤f1(k)+f2(k)<1−f2(p0)+f2(p0)=1.

It then remains to apply part (1) of Theorem 3.1 to conclude.

### 3.4. Proof of part (2) of Theorem 3.1

Denote the Euler function by . From now on, we assume that is positive.

Given an integer , set

 hk(n)=∏p|kp|npvp(k).

The easy lemma below will be used in the sequel.

###### Lemma 3.2.

One has for each .

###### Lemma 3.3.

Given , one has

 k−1⋃l=0Gal(Lk/F(e2iπl/kk√tj))=Gal(Lk/F(e2iπ/(hk(nj)⋅nj))).
###### Proof.

Let and . From our choice of and since , one has

 F(e2iπl/kk√tj)=F(e2iπ(lnj+mj)/(k⋅nj)) =F(e2iπ⋅gcd(lnj+mj,k⋅nj)/(k⋅nj)) =F(e2iπ⋅gcd(lnj+mj,k)/(k⋅nj)).

Obviously, divides As and are coprime, the same is true for and . Hence divides . Then

 F(e2iπ/(hk(nj)⋅nj))=F(e2iπ(k/hk(nj))/(k⋅nj))⊆F(e2iπ⋅gcd(lnj+mj,k)/(k⋅nj)).

Hence This provides

 k−1⋃l=0Gal(Lk/F(e2iπl/kk√tj))⊆Gal(Lk/F(e2iπ/(hk(nj)⋅nj))).

For the converse, it suffices to find such that divides . First, assume that has a prime factor not dividing . By the Chinese Remainder Theorem, there exists such that for each prime factor of not dividing . Hence divides , as needed. Now, assume that each prime factor of divides . One then has , thus ending the proof. ∎

###### Lemma 3.4.

One has

 f1(k)≤r1∑j=1[F:Q]d1⋅hk(nj).
###### Proof.

By Lemma 3.3 and the definitions of and , one has

 f1(k) =|⋃r1j=1Gal(Lk/F(e2iπ/(hk(nj)⋅nj)))|d1⋅|\rm Gal(Lk/F)| ≤r1∑j=1|Gal(Lk/F(e2iπ/(hk(nj)⋅nj)))|d1⋅|Gal(Lk/F)| =r1∑j=11d1⋅[F(e2iπ/(hk(nj)⋅nj)):F].

For every positive integer , one has

 [F(e2iπ/n):F]≥[Q(e2iπ/n):Q][F:Q]=φ(n)[F:Q].

We then get

 f1(k)≤r1∑j=1[F:Q]d1⋅φ(hk(nj)⋅nj).

It then remains to apply Lemma 3.2 to finish the proof. ∎

Let and . Given an integer , assume . One then has for each . Then apply Lemma 3.4 to get

 f1(k)≤r1∑j=1[F:Q]d1⋅pAmin=[F:Q]⋅r1d1⋅pAmin.

It then suffices to take

 A>log([F:Q])+log(r1)−log(d1)−log(ϵ)log(pmin)

to get , thus ending the proof of part (2) of Theorem 3.1.

### 3.5. Proof of part (3) of Theorem 3.1

Assume .

#### Refining the condition f2(k)<1

###### Lemma 3.5.

One has if

 g2(k):=|⋃rj=r1+1Gal(Lk/L1(e2iπ/k,k√tj))||Gal(Lk/L1(e2iπ/k))|<1.
###### Proof.

Assume that there exists some in which is not in . Then such an element lies in

 r⋃j=r1+1Gal(Lk/F(tj))∖r⋃j=r1+1k−1⋃l=0Gal(Lk/F(e2iπl/kk√tj)).

This provides , as needed for the lemma. ∎

Next, we need the following conditional bound.

###### Lemma 3.6.

Assume that the polynomials all are irreducible over . Then one has

 g2(k)≤r2+r3k.
###### Proof.

By the definition of , one has

 g2(k) ≤r∑j=r1+1|Gal(Lk/L1(e2iπ/k,k√tj))||Gal(Lk/L1(e2iπ/k))| =r∑j=r1+11[L1(e2iπ/k,k√tj):L1(e2iπ/k)].

For each , one has as is irreducible over . We then get

 g2(k)≤r∑j=r1+11k=r−r1k,

thus ending the proof. ∎

#### On the irreducibility of the polynomials Tk−tr1+1,…,Tk−tr1+r2,Tk−tr1+r2+1,…,Tk−tr.

###### Lemma 3.7.

Assume and let . The polynomial is irreducible over if is a prime number not dividing .

###### Proof.

First, assume that is reducible over . By the Capelli lemma [Lan02, Chapter VI, §9, Theorem 9.1] and as is a prime number, there exists such that As is a unit of , the same is true for . Set

 x=ζ′⋅uw11⋯uwvv,

where is a root of unity and are integers. We then get

 ζj⋅uaj,11⋯uaj,vv=tj=xk=ζ′k⋅uk⋅w11⋯uk⋅wvv.

In particular, we get for each . Then divides , which cannot happen. Hence is irreducible over .

Now, we show that is irreducible over . By the definition of and as is a prime number , the fields and are linearly disjoint over , i.e., one has

 [L1(e2iπ/k):L1]=[Q(e2iπ/k):Q]=k−1

(since is a prime number). By the above, one has Since and are coprime, the fields and are linearly disjoint over . Hence we get

 [L1(e2iπ/k,k√tj):L1(e2iπ/k)]=[L1(k√tj):L1]=k,

as needed for the lemma. ∎

Now, we consider the case where is not a unit of .

###### Lemma 3.8.

Suppose and let . Then is irreducible over if is a prime not dividing .

###### Proof.

Assume that is reducible over . By the Capelli lemma and as is a prime, there exists such that Let be a non-zero prime ideal of lying over . One has

 vQj(tj)=k⋅vQj(x).

This gives , where is the ramification index of in . As is a positive integer, this is also true for . Hence the prime divides or . As , the prime has to divide , which cannot happen. ∎

#### Conclusion

For simplicity, assume and (the other two cases are similar). Suppose is a prime number . As satisfies and , one may apply Lemmas 3.7 and 3.8 to get that the polynomials all are irreducible over . Then, by Lemma 3.6 and since , we get . It then remains to apply Lemma 3.5 to finish the proof of part (3) of Theorem 3.1.

###### Remark 3.9.

More generally, the proof shows that the condition (and then the condition too) holds if is a prime such that

- and if and ,

- and if ,

- and if .

## 4. On the converse of Theorem 2.2

In Propositions 4.1 and 4.2 below, we show that the conclusion of Theorem 2.2 does not hold in general if has a root that is a root of unity. This suggests that our strategy to handle the roots of that are not roots of unity, which leads to a better conclusion in Theorem 2.2, cannot be extended to the case of roots of unity.

###### Proposition 4.1.

Assume that the following condition holds:

Then and condition (/) of Proposition 1 fails for each positive integer which is coprime to . In particular, the conclusion of Theorem 2.2 does not hold.

For example, condition (H) holds in each of the following situations:

(1) each root of is a root of unity,

(2) has a root that is a root of unity and that is in .

Indeed, in case (1), the left-hand side in condition (H) is empty while, in case (2), the right-hand side is equal to the whole group .

###### Proof.

Assume that condition holds. Then one has as the polynomial is not constant. Let be an integer such that and are coprime. By Lemma 3.3, one has

 r1⋃j=1k−1⋃l=0Gal(Lk/F(e2iπl/kk√tj))=r1⋃j=1Gal(Lk/F(e2iπ/(h