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

TL;DR

This paper establishes precise conditions on positive integers to identify when prime ideals exist such that a polynomial's reduction has roots but its k-th power polynomial's reduction does not, refining previous results in field arithmetic.

Contribution

It provides exact criteria for the existence of primes where polynomial reductions behave differently for P(T) and P(T^k), improving earlier general results.

Findings

Identifies conditions on k for prime ideals with specific reduction properties.

Refines previous results by providing precise sufficient conditions.

Enhances understanding of prime divisors of polynomial values in number fields.

Abstract

Let $F$ be a number field, $O_F$ the integral closure of $\mathbb{Z}$ in $F$ and $P(T) \in O_F[T]$ a monic separable polynomial such that $P(0) \not=0$ and $P(1) \not=0$. We give precise sufficient conditions on a given positive integer $k$ for the following condition to hold: there exist infinitely many non-zero prime ideals $\mathcal{P}$ of $O_F$ such that the reduction modulo $\mathcal{P}$ of $P(T)$ has a root in the residue field $O_F/\mathcal{P}$, but the reduction modulo $\mathcal{P}$ of $P(T^k)$ has no root in $O_F/\mathcal{P}$. This makes a result from a previous paper (motivated by a problem in field arithmetic) asserting that there exist (infinitely many) such integers $k$ more precise.

Equations

$$t_1,\dots ,t_{r_1},t_{r_1+1},\dots ,t_{r_1+r_2},t_{r_1+r_2+1},\dots ,t_{r_1+r_2+r_3}=t_r$$

$$d_1=\frac{|{\bigcup }_{j=1}^r\mathrm{Gal}(L_1/F(t_j))|}{|\mathrm{Gal}(L_1/F)|}>0.$$

$$A>\frac{\log ([F:\mathbb{Q}])+\log (r_1)-\log (d_1)}{\log (p_{\min })}.$$

$$a_j=\gcd (|a_{j,1}|,\dots ,|a_{j,v}|)\in \mathbb{N}\setminus \{0\}$$

$$f(k)=\frac{|{\bigcup }_{j=1}^r{\bigcup }_{l=0}^{k-1}\mathrm{Gal}(L_k/F(e^{2i\pi l/k}\sqrt[k]{t_j}))|}{|{\bigcup }_{j=1}^r\mathrm{Gal}(L_k/F(t_j))|}\le 1,$$

$$f_{1}(k)=\left\{\begin{array}[]{ccc}\displaystyle{\frac{|\bigcup_{j=1}^{r_{1}}\bigcup_{l=0}^{k-1}{\rm{Gal}}(L_{k}/F(e^{2i\pi l/k}\sqrt[k]{t_{j}}))|}{|\bigcup_{j=1}^{r}{\rm{Gal}}(L_{k}/F(t_{j}))|}}&{\rm{if}}&r_{1}>0\\ 0&{\rm{if}}&r_{1}=0,\\ \end{array}\right.$$

$$f_{2}(k)=\left\{\begin{array}[]{ccc}\displaystyle{\frac{|\bigcup_{j=r_{1}+1}^{r}\bigcup_{l=0}^{k-1}{\rm{Gal}}(L_{k}/F(e^{2i\pi l/k}\sqrt[k]{t_{j}}))|}{|\bigcup_{j=1}^{r}{\rm{Gal}}(L_{k}/F(t_{j}))|}}&{\rm{if}}&r_{2}+r_{3}>0\\ 0&{\rm{if}}&r_{2}+r_{3}=0.\\ \end{array}\right.$$

$$A>\frac{\log ([F:\mathbb{Q}])+\log (r_1)-\log (d_1)-\log (ϵ)}{\log (p_{\min })}.$$

$$A>\frac{\log ([F:\mathbb{Q}])+\log (r_1)-\log (d_1)-\log (1-f_2(p_0))}{\log (p_{\min })}.$$

$$f(k)\le f_1(k)+f_2(k)<1-f_2(p_0)+f_2(p_0)=1.$$

$$h_k(n)=\prod _{\begin{array}{c}p|k\\ p|n\end{array}}{p}^{v_p(k)}.$$

$$\bigcup _{l=0}^{k-1}\mathrm{Gal}(L_k/F(e^{2i\pi l/k}\sqrt[k]{t_j}))=\mathrm{Gal}(L_k/F(e^{2i\pi /(h_k(n_j)\cdot n_j)})).$$

$$F(e^{2i\pi l/k}\sqrt[k]{t_j})=F(e^{2i\pi (l{n}_j+m_j)/(k\cdot n_j)})$$

$$=F(e^{2i\pi \cdot \gcd (l{n}_j+m_j,k)/(k\cdot n_j)}).$$

$$F(e^{2i\pi /(h_k(n_j)\cdot n_j)})=F(e^{2i\pi (k/h_k(n_j))/(k\cdot n_j)})\subseteq F(e^{2i\pi \cdot \gcd (l{n}_j+m_j,k)/(k\cdot n_j)}).$$

$$\bigcup _{l=0}^{k-1}\mathrm{Gal}(L_k/F(e^{2i\pi l/k}\sqrt[k]{t_j}))\subseteq \mathrm{Gal}(L_k/F(e^{2i\pi /(h_k(n_j)\cdot n_j)})).$$

$$f_1(k)\le \sum _{j=1}^{r_1}\frac{[F:\mathbb{Q}]}{d_1\cdot h_k(n_j)}.$$

$$f_1(k)$$

$$\le \sum _{j=1}^{r_1}\frac{|\mathrm{Gal}(L_k/F(e^{2i\pi /(h_k(n_j)\cdot n_j)}))|}{d_1\cdot |\mathrm{Gal}(L_k/F)|}$$

$$=\sum _{j=1}^{r_1}\frac{1}{d_1\cdot [F(e^{2i\pi /(h_k(n_j)\cdot n_j)}):F]}.$$

$$[F(e^{2i\pi /n}):F]\ge \frac{[\mathbb{Q}(e^{2i\pi /n}):\mathbb{Q}]}{[F:\mathbb{Q}]}=\frac{\phi (n)}{[F:\mathbb{Q}]}.$$

$$f_1(k)\le \sum _{j=1}^{r_1}\frac{[F:\mathbb{Q}]}{d_1\cdot \phi (h_k(n_j)\cdot n_j)}.$$

$$f_1(k)\le \sum _{j=1}^{r_1}\frac{[F:\mathbb{Q}]}{d_1\cdot p_{\min }^A}=\frac{[F:\mathbb{Q}]\cdot r_1}{d_1\cdot p_{\min }^A}.$$

$$A>\frac{\log ([F:\mathbb{Q}])+\log (r_1)-\log (d_1)-\log (ϵ)}{\log (p_{\min })}$$

$$g_2(k):=\frac{|{\bigcup }_{j=r_1+1}^r\mathrm{Gal}(L_k/L_1(e^{2i\pi /k},\sqrt[k]{t_j}))|}{|\mathrm{Gal}(L_k/L_1(e^{2i\pi /k}))|}<1.$$

$$\bigcup _{j=r_1+1}^r\mathrm{Gal}(L_k/F(t_j))\setminus \bigcup _{j=r_1+1}^r\bigcup _{l=0}^{k-1}\mathrm{Gal}(L_k/F(e^{2i\pi l/k}\sqrt[k]{t_j})).$$

$$g_2(k)\le \frac{r_2+r_3}{k}.$$

$$g_2(k)$$

$$=\sum _{j=r_1+1}^r\frac{1}{[L_1(e^{2i\pi /k},\sqrt[k]{t_j}):L_1(e^{2i\pi /k})]}.$$

$$g_2(k)\le \sum _{j=r_1+1}^r\frac{1}{k}=\frac{r-r_1}{k},$$

$$x={\zeta }^{\prime }\cdot u_1^{w_1}\cdots u_v^{w_v},$$

$${\zeta }_j\cdot u_1^{a_{j,1}}\cdots u_v^{a_{j,v}}=t_j=x^k={\zeta }^{\prime k}\cdot u_1^{k\cdot w_1}\cdots u_v^{k\cdot w_v}.$$

Full text

1. Introduction

Given a number field $F$, recall that a (non-zero) prime ideal $\mathcal{P}$ of the integral closure $O_{F}$ of $\mathbb{Z}$ in $F$ is a prime divisor of a monic separable polynomial $P(T)\in O_{F}[T]$ if the reduction of $P(T)$ modulo $\mathcal{P}$ has a root in the residue field $O_{F}/\mathcal{P}$. 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 $P(T)\in O_{F}[T]$ has infinitely many prime divisors and (2) if $P(T)$ is irreducible over $F$, then all but finitely many prime ideals of $O_{F}$ are prime divisors of $P(T)$ if and only if $P(T)$ 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 $F$ of the branch points of a finite Galois extension $E/F(T)$ with $F$ algebraically closed in $E$ corresponds to the set of all prime ideals of $F$ that ramify in at least one specialization of $E/F(T)$; see [Bec91] and [Leg16b]. Another application deals with the following classical question in diophantine geometry: does a given superelliptic curve $C:y^{n}=P(t)$ over $F$ has twists without non-trivial rational points? Under the abc-conjecture, Granville [Gra07] proved that, in the case of hyperelliptic curves over $\mathbb{Q}$, only a few (quadratic) twists of $C$ have non-trivial rational points, provided that $C$ has genus at least 3. Partial unconditional results were then given by Sadek [Sad14] (in the case of hyperelliptic curves over $\mathbb{Q}$) and the author [Leg16c], asserting that, to guarantee the existence of twists of $C$ without non-trivial rational points, it suffices that $n$ divides the degree of $P(T)$ and there exist infinitely many prime ideals of $O_{F}$ which are not prime divisors of $P(T)$.

A more recent application in field arithmetic consists in producing examples of “non-parametric extensions” over number fields [Leg16b], i.e., finite Galois extensions $E/F(T)$ with Galois group $G$ and $F$ algebraically closed in $E$ which do not provide all the Galois extensions of $F$ with Galois group $G$ by specializing the indeterminate $T$. For example, the main result of [Leg16a] asserts that such an extension of $F(T)$ with given non-trivial Galois group $G$ exists, unless $G$ is the Galois group of no Galois extension $E/F(T)$ with $F$ algebraically closed in $E$. 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 $P(0)\not=0$ and $P(1)\not=0$. Then there exist infinitely many positive integers $k$ such that the following condition holds:

($*$/$k$)* there exist infinitely many prime ideals of $O_{F}$ each of which is a prime divisor of $P(T)$ but not of $P(T^{k})$.*

Note that the result fails trivially if either $P(0)=0$ or $P(1)=0$.

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 $k$, which depend on the nature of the roots of the polynomial $P(T)$, for condition ($*$/$k$) to hold. In particular, we provide upper bounds on the number of distinct prime factors of such an integer $k$ and the multiplicity of each of them. This is motivated by the trivial fact that condition ($*$/$k$) holds for every multiple $k\geq 1$ of any given integer $k_{0}\geq 1$, as soon as condition ($*$/$k_{0}$) holds. Proposition 2 below, which is Theorem 2.2 in the sequel, gives an idea of the effectiveness of our method.

Proposition 2. Assume $P(0)\not=0$ and no root of unity is a root of $P(T)$. Then there exists an effective positive constant $c$ such that condition ($*$/$k$) of Proposition 1 holds for any positive integer $k$ which has at least one prime factor $p\geq c$.

See §2.4 for the definition of $c$, as well as Theorems 2.1 and 2.3 for the case where each root of $P(T)$ 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 $L$, $O_{L}$ is the integral closure of $\mathbb{Z}$ in $L$ and, given a non-zero prime ideal $\mathcal{P}$ of $O_{L}$, $v_{\mathcal{P}}$ is the associated valuation over $L$.

Given a number field $F$, let $P(T)\in O_{F}[T]$ be a monic separable polynomial such that $P(0)\not=0$ and $P(1)\not=0$. Let

• $L_{1}$ be the splitting field of $P(T)$ over $F$,

• $r_{1}$ the number (possibly zero) of roots of $P(T)$ that are roots of unity,

• $r_{2}$ the number (possibly zero) of roots of $P(T)$ that are units of $O_{L_{1}}$, but not roots of unity,

• $r_{3}$ the number (possibly zero) of roots of $P(T)$ that are not units of $O_{L_{1}}$.

Set $r=r_{1}+r_{2}+r_{3}>0.$ We denote the (distinct) roots of $P(T)$ by

[TABLE]

and assume that

• $t_{1},\dots,t_{r_{1}}$ are roots of unity (if $r_{1}>0$),

• $t_{r_{1}+1},\dots,t_{r_{1}+r_{2}}$ are units of $O_{L_{1}}$, but not roots of unity (if $r_{2}>0$),

• $t_{r_{1}+r_{2}+1},\dots,t_{r}$ are not units of $O_{L_{1}}$ (if $r_{3}>0$).

Pick a positive integer $k_{0}$ such that, for each prime number $p\geq k_{0}$, the fields $\mathbb{Q}(e^{2i\pi/p})$ and $L_{1}$ are linearly disjoint over $\mathbb{Q}$ 111Since the extension $\mathbb{Q}(e^{2i\pi/p})/\mathbb{Q}$ totally ramifies at $p$, the fields $\mathbb{Q}(e^{2i\pi/p})$ and $L_{1}$ are linearly disjoint over $\mathbb{Q}$ as soon as $p$ does not ramify in the extension $L_{1}/\mathbb{Q}$.. Finally, set

[TABLE]

2.2. Data associated with $t_{1},\dots,t_{r_{1}}$

Assume $r_{1}>0$. For each $j\in\{1,\dots,r_{1}\}$, $t_{j}$ is a root of unity and $t_{j}\not=1$ (as $P(1)\not=0$). Then there exist two coprime integers $m_{j}$ and $n_{j}$ such that $1\leq m_{j}<n_{j}$ and $t_{j}=e^{2i\pi m_{j}/n_{j}}.$ Denote the set of all prime factors of $n_{j}$ by $\mathcal{S}_{j}$, the smallest element of $\mathcal{S}_{1}\cup\cdots\cup\mathcal{S}_{r_{1}}$ by $p_{\rm{min}}$ and, by $A_{0},$ we mean the smallest positive integer $A$ that satisfies

[TABLE]

2.3. Data associated with $t_{r_{1}+1},\dots,t_{r_{1}+r_{2}}$

Assume $r_{2}>0$. Let $\{u_{1},\dots,u_{v}\}$ be a system of fundamental units of $O_{L_{1}}$, i.e., $u_{1},\dots,u_{v}$ are units of $O_{L_{1}}$ such that each unit $u$ of $O_{L_{1}}$ can be uniquely written as $u=\zeta\cdot u_{1}^{a_{1}}\cdots u_{v}^{a_{v}},$ where $\zeta$ is a root of unity and $a_{1},\dots,a_{v}$ are integers.

For each $j\in\{r_{1}+1,\dots,r_{1}+r_{2}\}$, set $t_{j}=\zeta_{j}\cdot u_{1}^{a_{j,1}}\cdots u_{v}^{a_{j,v}},$ where $\zeta_{j}$ is a root of unity and ${a_{j,1}},\dots,{a_{j,v}}$ are integers. As $t_{j}$ is not a root of unity, one has $|{a_{j,l_{j}}}|\geq 1$ for some $l_{j}\in\{1,\dots,v\}$. Finally, set

[TABLE]

and $a_{0}={\rm{lcm}}(a_{r_{1}+1},\dots,a_{r_{1}+r_{2}}).$

2.4. Data associated with $t_{r_{1}+r_{2}+1},\dots,t_{r}$

Assume $r_{3}>0$. Given $j\in\{r_{1}+r_{2}+1,\dots,r\}$, one has $t_{j}\not=0$ (as 0 is not a root of $P(T)$) and $t_{j}$ is an element of $O_{L_{1}}$ which is not a unit. Pick a non-zero prime ideal $\mathcal{P}_{j}$ of $O_{L_{1}}$ such that $v_{\mathcal{P}_{j}}(t_{j})$ is a positive integer. Finally, set $v_{0}={\rm{lcm}}(v_{\mathcal{P}_{r_{1}+r_{2}+1}}(t_{r_{1}+r_{2}+1}),\dots,v_{\mathcal{P}_{r}}(t_{r}))$ and, in the case where $r_{2}+r_{3}>0$, we define the integer $c$ as follows:

• $c={\rm{max}}\,(r_{2}+r_{3}+1,k_{0},{\rm{lcm}}(a_{0},v_{0})+1)$ if $r_{2}>0$ and $r_{3}>0$,

• $c={\rm{max}}\,(r_{3}+1,v_{0}+1)$ if $r_{2}=0$,

• $c={\rm{max}}\,(r_{2}+1,k_{0},a_{0}+1)$ if $r_{3}=0$.

2.5. Statements of Theorems 2.1, 2.2 and 2.3

First, we consider the case where each root of $P(T)$ is a root of unity.

Theorem 2.1**.**

Assume $P(1)\not=0$ and each root of $P(T)$ is a root of unity. Then condition ($*/{\rm{lcm}}(p_{1}^{A_{0}},\dots,p_{r_{1}}^{A_{0}})$) of Proposition 1 holds for every $r_{1}$-tuple $(p_{1},\dots,p_{r_{1}})$ of prime numbers in $\mathcal{S}_{1}\times\cdots\times\mathcal{S}_{r_{1}}$.

Next, we handle the case where no root of $P(T)$ is a root of unity.

Theorem 2.2**.**

Assume $P(0)\not=0$ and no root of unity is a root of $P(T)$. Then condition ($*/p$) of Proposition 1 holds for every prime number $p\geq c$.

Finally, we deal with the mixed case.

Theorem 2.3**.**

Assume $P(0)\not=0$, $P(1)\not=0$, $P(T)$ has a root that is a root of unity and $P(T)$ has a root that is not a root of unity. Then, given a prime number $p_{0}\geq c$, there exists an integer $A_{0}(p_{0})\geq 1$ 222See §3.3.3 for the precise definition of $A_{0}(p_{0})$. such that condition ($*/{\rm{lcm}}(p_{0},p_{1}^{A_{0}(p_{0})},\dots,p_{r_{1}}^{A_{0}(p_{0})})$) of Proposition 1 holds for every $(p_{1},\dots,p_{r_{1}})\in\mathcal{S}_{1}\times\cdots\times\mathcal{S}_{r_{1}}$.

3. Proofs of Theorems 2.1, 2.2 and 2.3

3.1. Notation

Given a positive integer $k$, denote the splitting field of $P(T^{k})$ over $F$ by $L_{k}$. For each $j\in\{1,\dots,r\}$, fix a $k$-th root $\sqrt[k]{t_{j}}$ of $t_{j}$. If $r_{1}>0$ and $j\in\{1,\dots,r_{1}\}$, we choose $\sqrt[k]{t_{j}}=e^{2i\pi m_{j}/(k\cdot n_{j})}$. Finally, set

[TABLE]

[TABLE]

[TABLE]

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 $k$ be a positive integer.

(1)* [Leg16a, Lemma 2.2] Condition ($*$/$k$) holds if and only if $f(k)<1$.*

(2)* Assume $r_{1}>0$. Let $(p_{1},\dots,p_{r_{1}})\in\mathcal{S}_{1}\times\cdots\times\mathcal{S}_{r_{1}}$, $\epsilon$ be a positive real number and $A$ a positive integer such that*

[TABLE]

Then one has $f_{1}(k)<\epsilon$ if $k$ is a multiple of ${\rm{lcm}}(p_{1}^{A},\dots,p_{r_{1}}^{A})$.

(3)* Assume $r_{2}+r_{3}>0$. Then one has $f_{2}(k)<1$ if $k$ is a prime $\geq c$.*

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

3.3.1. Proof of Theorem 2.1

Given $(p_{1},\dots,p_{r_{1}})\in\mathcal{S}_{1}\times\cdots\times\mathcal{S}_{r_{1}}$, set $k={\rm{lcm}}(p_{1}^{A_{0}},\dots,p_{r_{1}}^{A_{0}})$. By the definition of $A_{0}$, we may apply part (2) of Theorem 3.1 with $\epsilon=1$ and $A=A_{0}$ to get $f_{1}(k)<1$. As each root of $P(T)$ is a root of unity, one has $f_{1}(k)=f(k)$. Hence $f(k)<1$. It then remains to apply part (1) of Theorem 3.1 to get Theorem 2.1.

3.3.2. Proof of Theorem 2.2

Assume that $p$ is a prime number $\geq c$. Then we may apply part (3) of Theorem 3.1 to get $f_{2}(p)<1$. As no root of unity is a root of $P(T)$, one has $f_{2}(p)=f(p)$. This gives $f(p)<1$. It then remains to apply part (1) of Theorem 3.1 to conclude.

3.3.3. Proof of Theorem 2.3

Let $p_{0}$ be a prime number satisfying $p_{0}\geq c$. We may then apply part (3) of Theorem 3.1 to get $f_{2}(p_{0})<1$. Let $A_{0}(p_{0})$ be the smallest positive integer $A$ that satisfies

[TABLE]

Given $(p_{1},\dots,p_{r_{1}})\in\mathcal{S}_{1}\times\cdots\times\mathcal{S}_{r_{1}}$, set $k={\rm{lcm}}(p_{0},p_{1}^{A_{0}(p_{0})},\dots,p_{r_{1}}^{A_{0}(p_{0})})$. By the definition of $A_{0}(p_{0})$, we may apply part (2) of Theorem 3.1 with $\epsilon=1-f_{2}(p_{0})$ and $A=A_{0}(p_{0})$ to get $f_{1}(k)<1-f_{2}(p_{0})$. As $p_{0}$ divides $k$, one has $f_{2}(k)\leq f_{2}(p_{0})$. Hence

[TABLE]

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 $\varphi$. From now on, we assume that $r_{1}$ is positive.

Given an integer $n\geq 2$, set

[TABLE]

The easy lemma below will be used in the sequel.

Lemma 3.2**.**

One has $\varphi(h_{k}(n)\cdot n)=h_{k}(n)\cdot\varphi(n)$ for each $n\geq 2$.

Lemma 3.3**.**

Given $j\in\{1,\dots,r_{1}\}$, one has

[TABLE]

Proof.

Let $j\in\{1,\dots,r_{1}\}$ and $l\in\{0,\dots,k-1\}$. From our choice of $\sqrt[k]{t_{j}}$ and since ${\rm{gcd}}(m_{j},n_{j})=1$, one has

[TABLE]

Obviously, ${\rm{gcd}}(ln_{j}+m_{j},k)$ divides $k=h_{k}(n_{j})\cdot({k}/{h_{k}(n_{j})}).$ As $m_{j}$ and $n_{j}$ are coprime, the same is true for ${\rm{gcd}}(ln_{j}+m_{j},k)$ and $h_{k}(n_{j})$. Hence ${\rm{gcd}}(ln_{j}+m_{j},k)$ divides $k/h_{k}(n_{j})$. Then

[TABLE]

Hence $F(e^{2i\pi/(h_{k}(n_{j})\cdot n_{j})})\subseteq F(e^{2i\pi l/k}\sqrt[k]{t_{j}}).$ This provides

[TABLE]

For the converse, it suffices to find $l_{0}\in\{0,\dots,k-1\}$ such that $k/h_{k}(n_{j})$ divides ${\rm{gcd}}(l_{0}n_{j}+m_{j},k)$. First, assume that $k$ has a prime factor not dividing $n_{j}$. By the Chinese Remainder Theorem, there exists $l_{0}\in\{0,\dots,k-1\}$ such that $l_{0}\equiv-m_{j}/n_{j}\,\,{\rm{mod}}\,\,p^{v_{p}(k)}$ for each prime factor $p$ of $k$ not dividing $n_{j}$. Hence $k/h_{k}(n_{j})$ divides $l_{0}n_{j}+m_{j}$, as needed. Now, assume that each prime factor of $k$ divides $n_{j}$. One then has $k/h_{k}(n_{j})=1$, thus ending the proof. ∎

Lemma 3.4**.**

One has

[TABLE]

Proof.

By Lemma 3.3 and the definitions of $f_{1}(k)$ and $d_{1}$, one has

[TABLE]

For every positive integer $n$, one has

[TABLE]

We then get

[TABLE]

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

Let $(p_{1},\dots,p_{r_{1}})\in\mathcal{S}_{1}\times\cdots\times\mathcal{S}_{r_{1}}$ and $\epsilon>0$. Given an integer $A\geq 1$, assume ${\rm{lcm}}(p_{1}^{A},\dots,p_{r_{1}}^{A})|k$. One then has $h_{k}(n_{j})\geq p_{j}^{v_{p_{j}}(k)}\geq p_{j}^{A}\geq p_{\rm{min}}^{A}$ for each $j\in\{1,\dots,r_{1}\}$. Then apply Lemma 3.4 to get

[TABLE]

It then suffices to take

[TABLE]

to get $f_{1}(k)<\epsilon$, thus ending the proof of part (2) of Theorem 3.1.

3.5. Proof of part (3) of Theorem 3.1

Assume $r_{2}+r_{3}>0$.

3.5.1. Refining the condition $f_{2}(k)<1$

Lemma 3.5**.**

One has $f_{2}(k)<1$ if

[TABLE]

Proof.

Assume that there exists some $\sigma$ in ${\rm{Gal}}(L_{k}/L_{1}(e^{2i\pi/k}))$ which is not in $\bigcup_{j=r_{1}+1}^{r}{\rm{Gal}}(L_{k}/L_{1}(e^{2i\pi/k},\sqrt[k]{t_{j}}))$. Then such an element $\sigma$ lies in

[TABLE]

This provides $f_{2}(k)<1$, as needed for the lemma. ∎

Next, we need the following conditional bound.

Lemma 3.6**.**

Assume that the polynomials $T^{k}-t_{r_{1}+1},\dots,T^{k}-t_{r_{1}+r_{2}},T^{k}-t_{r_{1}+r_{2}+1},\dots,T^{k}-t_{r}$ all are irreducible over $L_{1}(e^{2i\pi/k})$. Then one has

[TABLE]

Proof.

By the definition of $g_{2}(k)$, one has

[TABLE]

For each $j\in\{r_{1}+1,\dots,r\}$, one has $[L_{1}(e^{2i\pi/k},\sqrt[k]{t_{j}}):L_{1}(e^{2i\pi/k})]=k$ as $T^{k}-t_{j}$ is irreducible over $L_{1}(e^{2i\pi/k})$. We then get

[TABLE]

thus ending the proof. ∎

3.5.2. On the irreducibility of the polynomials $T^{k}-t_{r_{1}+1},\dots,T^{k}-t_{r_{1}+r_{2}},T^{k}-t_{r_{1}+r_{2}+1},\dots,$ $T^{k}-t_{r}$.

We start with the case where $t_{j}$ is a unit of $O_{L_{1}}$.

Lemma 3.7**.**

Assume $r_{2}>0$ and let $j\in\{r_{1}+1,\dots,r_{1}+r_{2}\}$. The polynomial $T^{k}-t_{j}$ is irreducible over $L_{1}(e^{2i\pi/k})$ if $k$ is a prime number $\geq k_{0}$ not dividing $a_{j}$.

Proof.

First, assume that $T^{k}-t_{j}$ is reducible over $L_{1}$. By the Capelli lemma [Lan02, Chapter VI, §9, Theorem 9.1] and as $k$ is a prime number, there exists $x\in O_{L_{1}}$ such that $t_{j}=x^{k}.$ As $t_{j}$ is a unit of $O_{L_{1}}$, the same is true for $x$. Set

[TABLE]

where $\zeta^{\prime}$ is a root of unity and ${w_{1}},\dots,w_{v}$ are integers. We then get

[TABLE]

In particular, we get $a_{j,l}=k\cdot w_{l}$ for each $l\in\{1,\dots,v\}$. Then $k$ divides $a_{j}$, which cannot happen. Hence $T^{k}-t_{j}$ is irreducible over $L_{1}$.

Now, we show that $T^{k}-t_{j}$ is irreducible over $L_{1}(e^{2i\pi/k})$. By the definition of $k_{0}$ and as $k$ is a prime number $\geq k_{0}$, the fields $L_{1}$ and $\mathbb{Q}(e^{2i\pi/k})$ are linearly disjoint over $\mathbb{Q}$, i.e., one has

[TABLE]

(since $k$ is a prime number). By the above, one has $[L_{1}(\sqrt[k]{t_{j}}):L_{1}]=k.$ Since $k$ and $k-1$ are coprime, the fields $L_{1}(\sqrt[k]{t_{j}})$ and $L_{1}(e^{2i\pi/k})$ are linearly disjoint over $L_{1}$. Hence we get

[TABLE]

as needed for the lemma. ∎

Now, we consider the case where $t_{j}$ is not a unit of $O_{L_{1}}$.

Lemma 3.8**.**

Suppose $r_{3}>0$ and let $j\in\{r_{1}+r_{2}+1,\dots,r\}$. Then $T^{k}-t_{j}$ is irreducible over $L_{1}(e^{2i\pi/k})$ if $k$ is a prime not dividing $v_{\mathcal{P}_{j}}(t_{j})$.

Proof.

Assume that $T^{k}-t_{j}$ is reducible over $L_{1}(e^{2i\pi/k})$. By the Capelli lemma and as $k$ is a prime, there exists $x\in O_{L_{1}(e^{2i\pi/k})}$ such that $t_{j}=x^{k}.$ Let $\mathcal{Q}_{j}$ be a non-zero prime ideal of $O_{L_{1}(e^{2i\pi/k})}$ lying over $\mathcal{P}_{j}$. One has

[TABLE]

This gives $e_{j}\cdot v_{\mathcal{P}_{j}}(t_{j})=k\cdot v_{\mathcal{Q}_{j}}(x)$, where $e_{j}$ is the ramification index of $\mathcal{P}_{j}$ in $L_{1}(e^{2i\pi/k})/L_{1}$. As $v_{{\mathcal{P}_{j}}}(t_{j})$ is a positive integer, this is also true for $v_{\mathcal{Q}_{j}}(x)$. Hence the prime $k$ divides $e_{j}$ or $v_{{\mathcal{P}_{j}}}(t_{j})$. As $e_{j}\leq k-1$, the prime $k$ has to divide $v_{{\mathcal{P}_{j}}}(t_{j})$, which cannot happen. ∎

3.5.3. Conclusion

For simplicity, assume $r_{2}>0$ and $r_{3}>0$ (the other two cases are similar). Suppose $k$ is a prime number $\geq c$. As $k$ satisfies $k\geq k_{0}$ and $k{\not|}\,{\rm{lcm}}(a_{0},v_{0})$, one may apply Lemmas 3.7 and 3.8 to get that the polynomials $T^{k}-t_{r_{1}+1},\dots,T^{k}-t_{r_{1}+r_{2}},T^{k}-t_{r_{1}+r_{2}+1},\dots,T^{k}-t_{r}$ all are irreducible over $L_{1}(e^{2i\pi/k})$. Then, by Lemma 3.6 and since $k>r_{2}+r_{3}$, we get $g_{2}(k)<1$. 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 $g_{2}(k)<1$ (and then the condition $f_{2}(k)<1$ too) holds if $k$ is a prime such that

• $k\geq{\rm{max}}\,(r_{2}+r_{3}+1,k_{0})$ and $k{\not|}\,{\rm{lcm}}(a_{0},v_{0})$ if $r_{2}>0$ and $r_{3}>0$,

• $k\geq r_{3}+1$ and $k{\not|}\,v_{0}$ if $r_{2}=0$,

• $k\geq{\rm{max}}\,(r_{2}+1,k_{0})$ and $k{\not|}\,a_{0}$ if $r_{3}=0$.

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 $P(T)$ has a root that is a root of unity. This suggests that our strategy to handle the roots of $P(T)$ 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:

${\rm{(H)}}\bigcup_{j=r_{1}+1}^{r}\hbox{\rm Gal}(L_{1}/F(t_{j}))\subset\bigcup_{j=1}^{r_{1}}\hbox{\rm Gal}(L_{1}/F(t_{j})).$**

Then $r_{1}>0$ and condition ($*$/$k$) of Proposition 1 fails for each positive integer $k$ which is coprime to ${\rm{lcm}}(n_{1},\dots,n_{r_{1}})$. 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 $P(T)$ is a root of unity,

(2) $P(T)$ has a root that is a root of unity and that is in $F$.

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 ${\rm{Gal}}(L_{1}/F)$.

Proof.

Assume that condition ${\rm{(H)}}$ holds. Then one has $r_{1}>0$ as the polynomial $P(T)$ is not constant. Let $k\geq 1$ be an integer such that $k$ and ${\rm{lcm}}(n_{1},\dots,n_{r_{1}})$ are coprime. By Lemma 3.3, one has

[TABLE]

By the assumption on $k$, one has

[TABLE]

This gives

[TABLE]

Moreover, as condition ${\rm{(H)}}$ holds, one has

[TABLE]

We then get

[TABLE]

(as condition (H) holds). Hence $f(k)=1$. It then remains to apply part (1) of Theorem 3.1 to finish the proof. ∎

In Proposition 4.2 below, we give a more precise conclusion in the previous case (2).

Proposition 4.2**.**

Assume that $P(T)$ has a root that is a root of unity and that is in $F$. Denote the number of such roots by $r^{\prime}_{1}$ and assume that $t_{1},\dots,t_{r^{\prime}_{1}}$ are in $F$. Then condition ($*$/$k$) of Proposition 1 fails for each integer $k\geq 1$ that is coprime to $n_{j}$ for some $j\in\{1,\dots,r^{\prime}_{1}\}$.

In particular, if $P(T)$ has at least two roots that are roots of unity, that are in $F$ and that have coprime orders, then condition ($*$/$k$) of Proposition 1 fails for all prime numbers $k$.

Proof.

Let $k$ be a positive integer. Assume that $k$ is coprime to $n_{j_{0}}$ for some $j_{0}\in\{1,\dots,r^{\prime}_{1}\}$. By Lemma 3.3, one has

[TABLE]

As $k$ and $n_{j_{0}}$ are coprime, one has $h_{k}(n_{j_{0}})=1$. This gives

[TABLE]

(as $t_{j_{0}}$ is in $F$). Hence

[TABLE]

Moroever, as $t_{j_{0}}$ is in $F$, one has $\bigcup_{j=1}^{r}{\rm{Gal}}(L_{k}/F(t_{j}))={\rm{Gal}}(L_{k}/F).$ This provides $f(k)=1$, thus ending the proof. ∎