I am looking for a proof (or references) for the following result:



> 
> If $P\in \mathbb{Z}[x]$ then there exists a nonzero polynomial $Q\in \mathbb{Z}[x]$
> such that
> $$H(PQ)\leq M(P)$$
> where
> $H(R)=\max\{|a\_0|,|a\_1|,\dots,|a\_n|\}$ and $
> M(R)=|a\_n|\prod\_{k=1}^n\max\{1,|\alpha\_k|\}$ are respectively the height and the Mahler measure of a polynomial $R(x)=\sum\_{k=o}^na\_kx^k=a\_n\prod\_{k=1}^n (x-\alpha\_k).$
> 
> 
> 


Comments are also welcome.



Given $P\in \mathbb{Z}[x]$ there is a nonzero $Q\in \mathbb{Z}[x]$ such that $H(PQ)\leq M(P)$
In Chapter 3 of Jonas Jankauskas's dissertation thesis, entitled ["Heights of Polynomials"](https://epublications.vu.lt/object/elaba:1917005/1917005.pdf), we learn that the inequality


$$\min\_{Q \in \mathbb{Z}[x] \setminus \{0\}} H(PQ) \le \lfloor M(P)\rfloor$$


where $P \in \mathbb{Z}[x]$ and $\lfloor r \rfloor$ denotes the integral part of $r \in \mathbb{R}$, follows from [Siegel's lemma](https://en.wikipedia.org/wiki/Siegel%27s_lemma) [3].


The references given by Jonas Jankauskas are


* [[2]](https://zbmath.org/0405.12021), written in French, which gives a simplified proof of an enhanced [1, Théorème], also written in French,
* [3], written in English.


I can confirm that the original theorem of Martine Pathiaux [1, Théorème] yields the desired result.


Taking more time to parse [3], I might be able to update this answer with the precise statement of [3] which proves the above inequality.


**Key words:** *Reduced height, Siegel's lemma*.




---


**Addendum.** After reading further [3], but still superficially, I didn't find any direct means to derive the desired result from Siegel's lemma. (If $P$ is monic, we can indeed consider a linear homogenous system of equations with coefficients in $\mathbb{Z}$. As I haven't been able to conclude from there, I'll ask around if such an approach can be successful).


The papers [1] and [2] both establish the result for $P$ an **irreducible** polynomial. In her final remark [1, page 13-08], Martin Pathiaux asserts that the reducible case can be addressed by simply "adding more inequations" to her system. This is accurate, but it requires a good understanding of her proof.


For the sake of seeing **a complete proof of the fact referred to by the OP**, I have written below a proof along the lines of [2].
This proof doesn't add anything original. It consists only of the details which quickly convince me that the reducible case holds also true.


**Generalizing [1, Théorème 1] to reducible polynomials.**


Let $P(x) = a\_0 + a\_1 x + \cdots + a\_d x^d \in \mathbb{C}[x]$. The *height* of $P$ is the real number $H(P) = \max\_{i} \vert a\_i \vert$.
The *Mahler measure* of $P$ is the real number
$M(P) = \vert a\_d \cdot \lambda\_1 \cdots \lambda\_r \vert$ where $\lambda\_1, \dots, \lambda\_r \in \mathbb{C}$ are the roots of $P$ in $\mathbb{C}$ whose complex modulus is at least $1$.


We set ourselves to prove:



> 
> **Claim.** [1, Final remark]
> Let $P \in \mathbb{Z}[x]$. Then there is $Q \in \mathbb{Z}[x] \setminus \{0\}$ such that $H(PQ) \le M(P)$.
> 
> 
> 


We first mention the two results which have allowed Maurice Mignotte to significantly simplify the original proof of [1, Théorème 1] by Martine Pathiaux.



> 
> **Proposition.** [2, Corollaire 1]
> Let $a\_{i,j} \in \mathbb{C}$, with
> $i \in \{1, \dots, m\}, j \in \{1, \dots, n\}$ and
> $L\_{i}(\mathbf{x}) = \sum\_{i = 1}^n a\_{i,j} x\_j$ for
> $\mathbf{x} = (x\_1, \dots, x\_n) \in \mathbb{C}$. Let $A\_i = \max\_j \vert a\_{i,j} \vert$.
> Let $M \in \mathbb{N}\_{> 0}$.
> Then we can find $\mathbf{x} \in \mathbb{Z}^n \setminus \{0\}$ such that $\vert x\_j \vert \le M$ for every $j \in \{1, \dots, n\}$
> and
> $$\vert L\_i(\mathbf{x}) \vert < \sqrt{2n \log(18m)} A\_i \frac{M + 1}{(M + 1)^{n/m} - 2}$$
> for every $i \in \{1, \dots, m\}$, provided that $(M + 1)^{n/m} > 2$.
> 
> 
> 


The following lemma is a quantitative version of the [fundamental theorem for symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#Fundamental_theorem_of_symmetric_polynomials).



> 
> **Lemma.** [2, Section III, see remark at the bottom of page 3] Let $R$ be a commutative ring with identity. Let $Q \in R[x\_1, \dots, x\_d]$ by a symmetric polynomial.
> Then there is a polynomial $\tilde{Q} \in R[x\_1, \dots, x\_d]$, with total degree $\deg(\tilde{Q}) \le \max\_i \deg\_{x\_i}(Q)$ and such that
> $$Q(x\_1, \dots, x\_d) = \tilde{Q}(e\_{1, d}, e\_{2, d}, \dots, e\_{d,d})$$ where
> $e\_{1, d} = \sum\_{i = 1}^d x\_i, e\_{2, d} = \sum\_{1 \le i < j \le d} x\_i x\_j, \dots, e\_{d, d} = x\_1 \cdots x\_d$
> are [the elementary symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial) on $x\_1, \dots, x\_d$.
> 
> 
> 



> 
> *Proof of the lemma.* Revisit one of the classical proofs of the [fundamental theorem for symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#Fundamental_theorem_of_symmetric_polynomials) and
> check in the induction step that $\deg(\tilde{Q}) \le \max\_i\deg\_{x\_i}(Q)$ additionally holds for every $i$.
> 
> 
> 


We are now in position to prove the claim.



> 
> *Proof of the claim.* We can assume that $P$ is not constant, since otherwise the result is trivial. (The result is also obvious if $\deg(P) = 1$, but we won't need this fact).
> We can also assume, without loss of generality, that $P$ is primitive, i.e., the coefficients of $P$ are coprime integers,
> and that the leading coefficient of $P$ is positive.
> As $\mathbb{Z}[x]$ is a [unique factorization domain](https://en.wikipedia.org/wiki/Unique_factorization_domain#:%7E:text=7%20References-,Definition,n%20with%20n%20%E2%89%A5%200), we can write
> $P(x) = P\_1^{\mu\_1}(x) \cdots P\_k^{\mu\_k}(x)$ where $\mu\_l \in \mathbb{N}\_{> 0}$ and
> $P\_i \in \mathbb{Z}[x]$ is an irreducible primitive polynomial of degree $d\_i > 0$, with positive leading coefficient, for
> every $i \in \{1, \dots, k\}$.
> For $i \in \{1, \dots, k\}$, we denote by $\theta\_{i, 1}, \dots, \theta\_{i, d\_i}$ the roots of $P\_i$ in $\mathbb{C}$.
> We shall prove that there exists, for all $n \in \mathbb{N}\_{> 0}$ sufficiently large, a polynomial
> $D(x) = d\_0 + d\_1x + \cdots + d\_{n - 1}x^{n - 1} \in \mathbb{Z}[x]$
> which satisfies the inequality $\vert d\_j \vert \le M(P)$ for every $j \in \{0, \dots, n - 1\}$ together with the equation
> $D^{(\mu)}(\theta\_{i, j}) = 0$ for every
> $i \in \{1, \dots, k\}, j \in \{1, \dots, d\_i\}$ and $\mu \in \{0, \dots, \mu\_i - 1\}$
> where $D^{(\mu)}$ denotes the $\mu$-th derivative of $D$.
> Let us observe that such a polynomial $D$ obviously satisfies $H(D) \le M(P)$. In addition, this polynomial $D$ is divisible by $P$ in $\mathbb{Q}[x]$.
> The fact that $P$ divides $D$ in $\mathbb{Z}[x]$ follows immediately from [Gauss's lemma](https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomials)).
> The proof of the existence of $D$ is a straightforward application of the method developed in [2].
> We proceed by showing, for all $n$ sufficiently large, the existence of a polynomial
> $D \in \mathbb{Z}[x] \setminus \{0\}$ satisfying the inequations
> $\vert D^{(\mu)}(\theta\_{i, j}) \vert < \epsilon\_{i, j, \mu}(n)$ for some upper bound $\epsilon\_{i, j, \mu}(n)$
> which can be rendered sufficiently small to enforce $D^{(\mu)}(\theta\_{i, j}) = 0$
> for every $i, j$ and $\mu$.
> To this end, we set $M = \lfloor M(P) \rfloor$ and apply the above proposition to the linear forms
> $L\_{i, j, \mu}(\mathbf{x}) = \sum\_{l = 0}^n a\_{i, j, \mu, l} x\_l$ with
> $a\_{i, j, \mu, l} = \max(l(l - 1) \cdots (l - \mu + 1), 0) \cdot \theta\_{i, j}^{l - \mu}$ and $n > 0$. (There are $m = \deg(P)$ such forms.)
> Since we have
> $A\_{i, j, \mu} := \max\_l \vert a\_{i, j, \mu, l} \vert \le \mu\_i ! \max(\vert\theta\_{i, j}\vert, 1)^{n - 1}$, the above
> proposition provides us with some
> $\mathbf{d} = (d\_0, \dots, d\_{n - 1}) \in \mathbb{Z}^n \setminus \{0\}$ such that $\vert L\_{i, j, \mu}(\mathbf{d}) \vert < \epsilon\_{i, j, \mu}(n)$ for every $i, j$ and $\mu$, where  
> 
> $$\epsilon\_{i, j, \mu}(n) = \sqrt{2n \log(18 \deg(P))}\mu\_i ! \frac{\max(\vert\theta\_{i, j}\vert^{n - 1}, 1)(M + 1)}{(M + 1)^{n/\deg(P)} - 2}.$$
> Let us set $D(x) := d\_0 + d\_1x + \cdots + d\_{n - 1}x^{n - 1}$, so that $L\_{i, j, \mu}(\mathbf{d}) = D^{(\mu)}(\theta\_{i, j})$ by the definitions of $D$ and $L\_{i, j, \mu}$.
> Let $r\_{i, \mu}(n) = \vert D^{(\mu)}(\theta\_{i, 1}) \cdots D^{(\mu)}(\theta\_{i, d\_i}) \vert$ and let $q\_i > 0$ be the leading coefficient of $P\_i$.
> We have thus
> $$q\_i^{n - 1} r\_{i, \mu}(n) < q\_i^{n - 1} \epsilon\_{i, 1, \mu}(n) \cdots \epsilon\_{i, d\_i, \mu}(n) = O\left(\frac{n^{d\_i/2} M(P\_i)^{n - 1}}{(M + 1)^{nd\_i/\deg(P)}}\right).$$
> It follows that
> $$r\_i (n):= q\_i^{n - 1} r\_{i, 0}(n) \cdots q\_i^{n - 1} r\_{i, \mu\_i - 1}(n) = O\left(\frac{n^{\mu\_i d\_i/2} M(P\_i^{\mu\_i})^{n - 1}}{(M + 1)^{n \mu\_i d\_i/\deg(P)}}\right)$$
> for every $i \in \{1, \dots, k\}$ and hence
> $$r\_1(n) \cdots r\_k(n) = O\left( n^{\deg(P)/2} (\frac{M}{M + 1})^n\right) = o(1).$$
> Since $\theta\_{i, 1}, \dots, \theta\_{i, d\_i}$ are the roots of $P\_i$, we have  
> 
> $e\_{j, d\_i}(\theta\_{i, 1}, \dots, \theta\_{i, d\_i}) \in \frac{1}{q\_i} \mathbb{Z}$ for every elementary symmetric polynomial $e\_{j, d\_i}$.
> By the above lemma, we know that $q\_i^{n - 1} r\_{i, \mu}(n) \in \mathbb{N}$ for every $i \in \{1, \dots, d\_i\}$ and every $\mu \in \{0, \dots, \mu\_i - 1\}$.
> Thus we have $r\_1(n) \cdots r\_k(n) \in \mathbb{N}$ and $r\_1(n) \cdots r\_k(n) = o(1)$,
> which entails that $r\_i(n) = r\_{i, \mu}(n) = D^{(\mu)}(\theta\_{i, j}) = 0$ for all $n$ large enough and for every $i, j$ and $\mu$, as desired.
> 
> 
> 




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[1] M. Pathiaux, "Sur les multiples de polynômes irréductibles associés à certains nombres algébriques", 1972.  

[2] M. Mignotte, "Sur les multiples des polynômes irréductibles", 1975.  

[3] E. Bombieri and J. Vaaler, "On Siegel's Lemma", 1984.


