The Schinzel Hypothesis for Polynomials

TL;DR

This paper extends the Schinzel hypothesis to polynomial rings, proving it for various classes of polynomials over integers and fields, and introduces related conjectures and spectral results.

Contribution

It develops a version of the Hilbert specialization property over rings and proves the Schinzel hypothesis for broad classes of polynomials.

Findings

Schinzel hypothesis proven for polynomials over integers and fields

A polynomial Goldbach conjecture is proposed

Results on spectra of rational functions are established

Abstract

The Schinzel hypothesis is a famous conjectural statement about primes in value sets of polynomials, which generalizes the Dirichlet theorem about primes in an arithmetic progression. We consider the situation that the ring of integers is replaced by a polynomial ring and prove the Schinzel hypothesis for a wide class of them: polynomials in at least one variable over the integers, polynomials in several variables over an arbitrary field, etc. We achieve this goal by developing a version over rings of the Hilbert specialization property. A polynomial Goldbach conjecture is deduced, along with a result on spectra of rational functions.

Equations

$$\left\{\begin{array}[]{c}A(\underline{x})=H(\phi(\underline{x}),\psi(\underline{x}))=\sum_{i=0}^{\delta}h_{i}\hskip 2.0pt\phi(\underline{x})^{i}\psi(\underline{x})^{\delta-i}\hfill\\ B(\underline{x})=H_{0}(\phi(\underline{x}),\psi(\underline{x}))=\sum_{i=0}^{\delta}h_{0i}\hskip 2.0pt\phi(\underline{x})^{i}\psi(\underline{x})^{\delta-i}\hfill\\ BQ_{1}(\underline{x})=H_{1}(\phi(\underline{x}),\psi(\underline{x}))=\sum_{i=0}^{\delta}h_{1i}\hskip 2.0pt\phi(\underline{x})^{i}\psi(\underline{x})^{\delta-i}\hfill\\ \ \vdots\\ BQ_{\ell}(\underline{x})=H_{\ell}(\phi(\underline{x}),\psi(\underline{x}))=\sum_{i=0}^{\delta}h_{\ell i}\hskip 2.0pt\phi(\underline{x})^{i}\psi(\underline{x})^{\delta-i}\\ \end{array}\right.$$

Full text

Acknowledgment. This work was supported in part by the Labex CEMPI (ANR-11-LABX-0007-01) and by the ANR project “LISA” (ANR-17-CE40–0023-01).

1. Introduction

The so-called Schinzel Hypothesis (H), which builds on an earlier conjecture of Bunyakovsky, was stated in [SS58]. Consider a set $\underline{P}=\{P_{1},\ldots,P_{s}\}$ of $s$ polynomials, irreducible in $\mathbb{Z}[y]$, of degree $\geqslant 1$ and such that

(*) there is no prime $p\in\mathbb{Z}$ dividing all values $\prod_{i=1}^{s}P_{i}(m)$, $m\in\mathbb{Z}$. Hypothesis (H) concludes that there are infinitely many $m\in\mathbb{Z}$ such that $P_{1}(m),\ldots,P_{s}(m)$ are prime numbers. If true, the Schinzel hypothesis would solve many classical problems in number theory: the twin prime problem (take ${\underline{P}}=\{y,y+2\}$), the infiniteness of primes of the form $y^{2}+1$ (take $\underline{P}=\{y^{2}+1\}$), the Sophie Germain prime problem ($\underline{P}=\{y,2y+1\}$), etc. However it is wide open except for one polynomial $P_{1}$ of degree one, in which case it is the Dirichlet theorem about primes in an arithmetic progression.

We consider the situation that the ring $\mathbb{Z}$ is replaced by a polynomial ring $R[\underline{x}]$ in $n\geqslant 1$ variables over some ring $R$, and “prime” is understood as “irreducible”. We prove the Schinzel Hypothesis in this situation for a wide class of rings $R$, for example $\mathbb{Z}$, or $k[u]$ with $k$ an arbitrary field. The infiniteness of integers $m$ is replaced by a degree condition.

1.1. Main result

Specifically, let $R$ be a Unique Factorization Domain (UFD) with fraction field $K$. Our assumptions include $K$ being a field with the product formula. Definition is recalled in §4. The basic example is $K=\mathbb{Q}$. The product formula is: $\prod_{p}|a|_{p}\cdot|a|=1$ for every $a\in\mathbb{Q}^{\ast}$, where $p$ ranges over all prime numbers, $|\cdot|_{p}$ is the $p$-adic absolute value and $|\cdot|$ is the standard absolute value. The rational function field $k(u_{1},\ldots,u_{r})$ in $r\geqslant 1$ variables over an arbitrary field $k$ is another example.

Given $n$ indeterminates $x_{1},\ldots,x_{n}$, set $R[\underline{x}]=R[x_{1},\ldots,x_{n}]$ ($n\geqslant 0$)111For $n=0$, we mean $R[\underline{x}]=R$, which is the original context of Schinzel’s hypothesis.. Consider $s\geqslant 1$ polynomials $P_{1},\ldots,P_{s}$, irreducible in $R[\underline{x},y]$, of degree $\geqslant 1$ in $y$. Set $\underline{P}=\{P_{1},\ldots,P_{s}\}$ and let ${\mathcal{I}rr}_{n}(R,\underline{P})$ be the set of polynomials $M\in R[\underline{x}]$ such that $P_{1}(\underline{x},M(\underline{x})),\ldots,P_{s}(\underline{x},M(\underline{x}))$ are irreducible in $R[\underline{x}]$.

For every $n$-tuple $\underline{d}=(d_{1},\ldots,d_{n})$ of integers $d_{i}\geqslant 0$, denote the set of polynomials $M\in R[\underline{x}]$ such that $\deg_{x_{i}}(M)\leqslant d_{i}$, $i=1,\ldots,n$, by ${\mathcal{P}ol}_{R,n,\underline{d}}$. It is an affine space over $R$: the coordinates correspond to the coefficients.

Theorem 1.1**.**

Assume that $n\geqslant 1$ and $R$ is a UFD with fraction field a field $K$ with the product formula, imperfect if $K$ is of characteristic $p>0$ (i.e. $K^{p}\not=K$). For every $\underline{d}\in(\mathbb{N}^{\ast})^{n}$ such that $\displaystyle d_{1}+\cdots+d_{n}\geqslant\max_{1\leqslant i\leqslant s}\deg_{\underline{x}}(P_{i})+2$,

the set ${\mathcal{I}rr}_{n}(R,\underline{P})$ is Zariski-dense in ${\mathcal{P}ol}_{R,n,\underline{d}}$.

In particular, the following Schinzel hypothesis for $R[\underline{x}]$ holds true:

(**) *there exist polynomials $M\in R[\underline{x}]$ with partial degrees any suitably large integers such that $P_{1}(\underline{x},M(\underline{x})),\ldots,P_{s}(\underline{x},M(\underline{x}))$ are irreducible in $R[\underline{x}]$ 222Up to adding $P_{0}=y$ to the set $\underline{P}$, one may also require that $M$ be irreducible in $R[\underline{x}]$.. *

Irreducibility over $R$ is a main point. As a comparison, the Hilbert specialization property provides elements $m\in K$ such that $P_{1}(\underline{x},m),\ldots,P_{s}(\underline{x},m)$ are irreducible over $K$ (provided that all $\deg_{\underline{x}}(P_{i})$ are $\geqslant 1$). Developing a Hilbert property over rings will in fact be the core of our approach; we say more about this in §1.6.

Rings $R$ satisfying the assumptions of Theorem 1.1 include:

(a) the ring $\mathbb{Z}$ of integers, and more generally, every ring ${\mathcal{O}}_{k}$ of integers of a number field $k$ of class number $1$,

(b) every polynomial ring $k[u_{1},\ldots,u_{r}]$ with $r\geqslant 1$ and $k$ an arbitrary field.

(c) fields (so $R=K$) with the product formula, imperfect if of characteristic $p>0$ e.g. $\mathbb{Q}$, $k(u_{1},\ldots,u_{r})$ ($r\geqslant 1$, $k$ arbitrary), their finite extensions.

As to the analog of assumption (*), it is automatically satisfied under our hypotheses (Lemma 2.1). Our approach also allows the situation that the polynomials $P_{i}$ have several variables $y_{1},\ldots,y_{m}$, which leads to a multivariable Schinzel hypothesis for polynomials (Theorem 5.3).

1.2. Examples

Take $R[\underline{x}]$ as above and $P_{i}=b_{i}(\underline{x})\hskip 0.5pty^{\rho_{i}}+a_{i}(\underline{x})$ with $\rho_{i}\in\mathbb{N}^{\ast}$, $a_{i},b_{i}$ relatively prime in $R[\underline{x}]$ (possibly in $R$) and such that $-a_{i}/b_{i}$ satisfies the Capelli condition that makes $b_{i}y^{\rho_{i}}+a_{i}$ irreducible in $K(\underline{x})[y]$, i.e. $-a_{i}/b_{i}\notin K(\underline{x})^{\ell}$ for every prime divisor $\ell$ of $\rho_{i}$ and $-a_{i}/b_{i}\notin-4K(\underline{x})^{4}$ if $4\hskip 2.0pt|\hskip 2.0pt\rho_{i}$. Then

(***) there exist polynomials $M\in R[\underline{x}]$ with partial degrees any suitably large integers such that $b_{1}M^{\rho_{1}}+a_{1},\ldots,b_{s}M^{\rho_{s}}+a_{s}$ are irreducible in $R[\underline{x}]$.

This solves the polynomial analogs of all famous number-theoretic problems mentioned above (twin prime, etc.), and proves the Dirichlet theorem as well.

On the other hand, Schinzel’s hypothesis for $R[\underline{x}]$ obviously fails (hence Theorem 1.1 too) for $n=1$ if $R=K$ is algebraically closed. It also fails for the finite field $R=\mathbb{F}_{2}$ and $\underline{P}=\{y^{8}+x^{3}\}$: from an example of Swan [Swa62, pp.1102-1103], $M(x)^{8}+x^{3}$ is reducible in $\mathbb{F}_{2}[x]$ for every $M\in\mathbb{F}_{2}[x]$. Interestingly enough, results of Kornblum-Landau [KL19] show that it does hold for $\mathbb{F}_{q}[x]$ in the degree one case and for one polynomial, i.e., in the situation of the Dirichlet theorem; see also [Ros02, Theorem 4.7]. The situation that $R=K$ is a finite field, and the related one that $R=K$ is a PAC field333A field $K$ is PAC if every curve over $K$ has infinitely many $K$-rational points. The first examples of PAC fields were ultraproducts of finite fields., and $n=1$, have led to valuable variants; see [BS09], [BS12], [BW05].

1.3. Special rings

The special situation that $R=K$ is a field is easier, and is dealt with in §2. In the addendum to Theorem 1.1 (in §2), $K$ is assumed to be a Hilbertian field, more exactly a totally Hilbertian field (definitions are in §4.1). This provides more fields than those in §1.1(c) for which Theorem 1.1 holds (with $R=K$): every abelian extension of $\mathbb{Q}$, the field $k((u_{1},\ldots,u_{r}))$ of formal power series over a field $k$ in at least two variables, etc.

For $R=k[u]$ with $k$ a field, we have this version of Theorem 1.1 in which the partial degrees of $M$ are prescribed, including the degree in $u$.

Theorem 1.2**.**

With ${\underline{P}}$ as above and $n\geqslant 1$, assume $R=k[u]$ with $k$ an arbitrary field. For every $\underline{d}\in(\mathbb{N}^{\ast})^{n}$ satisfying $\displaystyle d_{1}+\cdots+d_{n}\geqslant\max_{1\leqslant i\leqslant s}\deg_{\underline{x}}(P_{i})+2$, there is an integer $d_{0}\geqslant 1$ such that for every integer $\delta\geqslant d_{0}$, there is a polynomial $M\in{\mathcal{I}rr}_{n}(R,\underline{P})$ satisfying

\left\{\begin{matrix}\deg_{x_{j}}(M)=\hskip 5.69054ptd_{j}\hskip 17.07164ptj=1,\ldots,n\hfill\\ \deg_{u}(M)=\left\{\begin{matrix}\delta\hskip 8.53581pt&\hbox{if {\rm char}(k)=0}\hfill\\ p\delta\hskip 8.53581pt&\hbox{if {\rm char}(k)=p>0}.\hfill\\ \end{matrix}\right.\\ \end{matrix}\right.*

Identifying $k[u][x_{1},\ldots,x_{n}]$ with a polynomial ring in $n+1$ variables, it follows that Schinzel’s hypothesis holds for polynomial rings in at least $2$ variables over a field of characteristic [math]. In characteristic $p>0$, a weak version holds where one degree is allowed to be any suitably large multiple of $p$.

In the degree one case of the Schinzel hypothesis, i.e. in the Dirichlet situation, one can get rid of this last restriction.

Theorem 1.3**.**

Assume that $n\geqslant 2$ and $k$ is an arbitrary field. Let $(A_{1},B_{1}),$ $\ldots,$ $(A_{s},B_{s})$ be $s$ pairs of nonzero relatively prime polynomials in $k[\underline{x}]$. There is an integer $d_{0}\geqslant 1$ with this property: for all integers $d_{1},\ldots,d_{n}$ larger than $d_{0}$, there exists an irreducible polynomial $M\in k[\underline{x}]$ such that $A_{i}+B_{i}M$ is irreducible in $k[\underline{x}]$, $i=1,\ldots,s$, and $\deg_{x_{j}}(M)=d_{j}$, $j=1,\ldots,n$.

To our knowledge, this was unknown, even for $s=1$. When $k$ is infinite, we have a stronger version, not covered either by Theorems 1.1 and 1.2. Let $\overline{k}$ denote an algebraic closure of $k$.

Theorem 1.4**.**

Assume $n\geqslant 2$ and $k$ is an infinite field. Let $A,B\in k[\underline{x}]$ be two nonzero relatively prime polynomials and ${\mathcal{I}rr}_{n}(k,A,B)$ the set of polynomials $M\in k[\underline{x}]$ such that $A+BM$ is irreducible in $k[\underline{x}]$. For every $\underline{d}\in(\mathbb{N}^{\ast})^{n}$, ${\mathcal{I}rr}_{n}(\overline{k},A,B)$ contains a nonempty Zariski open subset of ${\mathcal{P}ol}_{k,n,\underline{d}}(k)$.

1.4. The Goldbach problem

The analog of the Goldbach conjecture for a polynomial ring $R[\underline{x}]$ is that every nonconstant polynomial ${\mathcal{Q}}\in R[\underline{x}]$ is the sum of two irreducible polynomials $F,G\in R[\underline{x}]$ with $\deg(F)\leqslant\deg({\mathcal{Q}})$ (and so $\deg(G)\leqslant\deg({\mathcal{Q}})$ too). Pollack [Pol11] showed it in the $1$-variable case when $R$ is a Noetherian integral domain with infinitely many maximal ideals, or, if $R=S[u]$ with $S$ an integral domain. His method relies on a clever use of the Eisenstein criterion.

Finding Goldbach decompositions for ${\mathcal{Q}}\in R[\underline{x}]$ ($n\geqslant 1$) corresponds to the special situation of the degree $1$ case of the Schinzel hypothesis for which $\underline{P}=\{P_{1},P_{2}\}$ with $P_{1}=-y$ and $P_{2}=y+{\mathcal{Q}}$. We obtain this result.

Corollary 1.5**.**

Let $R$ be a ring as in Theorem 1.1. Every nonconstant polynomial ${\mathcal{Q}}\in R[\underline{x}]$ is the sum of two irreducible polynomials $F,G\in R[\underline{x}]$ with $F=a+b\hskip 1.0ptx_{1}^{d_{1}}\cdots x_{n}^{d_{n}}$ ($a,b\in R$) a binomial of degree $d_{1}+\cdots+d_{n}\leqslant\deg({\mathcal{Q}})$.

One can even take $d_{1}+\cdots+d_{n}=1$ when $R=K$ is a Hilbertian field, or when $n\geqslant 2$ and $R=K$ is an infinite field (the latter was already known from [BDN09, Corollary 4.3(2)]). On the other hand, the Goldbach conjecture fails for $\mathbb{F}_{2}[x]$ and ${\mathcal{Q}}(x)=x^{2}+x$ (note that $x^{2}+x+1$ is the only irreducible polynomial in $\mathbb{F}_{2}[x]$ of degree $2$). From Corollary 1.5 however, it holds true for $\mathbb{F}_{q}[x,y]$ if condition $\deg(F)\leqslant\deg({\mathcal{Q}})$ is replaced by $\deg_{x}(F)\leqslant\deg_{x}({\mathcal{Q}})$.

1.5. Spectra

The following result uses Theorem 1.3 as a main ingredient.

Corollary 1.6**.**

Assume that $n\geqslant 2$ and $k$ is an arbitrary field. Let ${\mathcal{S}}\subset k$ be a finite subset, $a_{0}\in\overline{k}\setminus{\mathcal{S}}$, separable over $k$ and $V\in k[\underline{x}]$ a nonzero polynomial. Then, for all suitably large integers $d_{1},\ldots,d_{n}$ (larger than some $d_{0}$ depending on ${\mathcal{S}}$, $a_{0}$, $V$), there is a polynomial $U\in k[\underline{x}]$ such that:

\begin{matrix}&\hbox{{\rm(a)} U(\underline{x})-aV(\underline{x})$is reducible in$k[\underline{x}]$for every$a\in{\mathcal{S}},}\hfill\\ &\hbox{{\rm(b)} U(\underline{x})-a_{0}V(\underline{x})$is irreducible in$k(a_{0})[\underline{x}]$of degree$\max(\deg(U),\deg(V)),}\hfill\\ &\hbox{{\rm(c)} \deg_{x_{i}}(U)=d_{i}$,$i=1,\ldots,n.}\hfill\\ \end{matrix}**

If ${\mathcal{S}}\not=k$, e.g. if $k$ is infinite, $a_{0}$ can be chosen in $k$ itself.

A more precise version of Corollary 1.6, given in §5.5, shows that one can even prescribe all irreducible factors but one of each polynomial $U(\underline{x})-aV(\underline{x})$, $a\in{\mathcal{S}}$, provided that these factors satisfy some standard condition.

If $k$ is algebraically closed, the irreducibility condition (b) implies that the rational function $U/V$ is indecomposable [Bod08, Theorem 2.2]; “indecomposable” means that $U/V$ cannot be written $h\circ H$ with $h\in k(u)$ and $H\in k(\underline{x})$ with $\deg(h)\geqslant 2$. The set of all $a\in k$ such that $U(\underline{x})-aV(\underline{x})$ is reducible in $k[\underline{x}]$ is called the spectrum of $U/V$ and the indecomposability condition equivalent to the spectrum being finite. Corollary 1.6 rephrases to conclude that given ${\mathcal{S}}$ and $V$ as above, indecomposable rational functions $U/V\in k(\underline{x})$ exist with a spectrum containing ${\mathcal{S}}$ and satisfying (c). See [Naj04] [Naj05] for the special case $V=1$ and [BDN17, §3.1.1] for further results.

1.6. Hilbertian rings

Except for Theorem 1.4 for which we use geometrical tools (§3), we follow a Hilbert like specialization approach.

Given an irreducible polynomial $F(\underline{\lambda},\underline{x})\in R[\underline{\lambda},\underline{x}]$ with $\deg_{\underline{x}}(F)\geqslant 1$, the Hilbert property provides specializations $\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast}\in K$ of the indeterminates from $\underline{\lambda}$ such that $F(\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast},\underline{x})$ is irreducible in $K[\underline{x}]$ (§4.1).

As suggested above and detailed in §2, the challenge for our purpose is to make it work over the ring $R$, i.e., to be able to find $\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast}$ in $R$ such that $F(\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast},\underline{x})$ is irreducible in $R[\underline{x}]$. A problem however is that this is false in general, even with $R=\mathbb{Z}$. Take $F=(\lambda^{2}-\lambda)\hskip 2.0ptx+(\lambda^{2}-\lambda+2)$ in $\mathbb{Z}[\lambda,x]$; for every $\lambda^{\ast}\in\mathbb{Z}$, $F(\lambda^{\ast},\underline{x})$ is divisible by $2$, hence reducible in $\mathbb{Z}[x]$.

To remedy this problem, we develop the notion of Hilbertian ring introduced in [FJ08, §13.4]. The defining property is that, for separable polynomials $F(\underline{\lambda},x)$ in the one variable $x$, tuples $(\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast})$ can be found with coordinates in the ring $R$ and satisfying the specialization property over $K$.

Our approach to reach irreducibility over $R$ can be summarized as follows. It may be of interest for the sole sake of the Hilbertian field theory.

(Hilbert sections 4 and 5) Assume that $K$ is of characteristic [math], or $K$ is of characteristic $p>0$ and imperfect (the imperfectness assumption).

(a) We extend the property of Hilbertian rings to all irreducible polynomials $F(\underline{\lambda},\underline{x})$ (not just the separable ones $F(\underline{\lambda},x)$), and show in fact a stronger version: $\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast}$ can be chosen pairwise relatively prime (Prop.4.2); and for $R=k[u]$, their degrees in $u$ can be prescribed off a finite range (Theorem 4.8).

(b) We show that if $K$ is a field with the product formula, then $R$ is a Hilbertian ring (Theorem 4.6); this improves on [FJ08, Prop.13.4.1] where the assumption is that $R$ is finitely generated over $\mathbb{Z}$, or over $k[u]$ for some field $k$.

(c) For $R$ both a UFD and a Hilbertian ring, we show that our polynomials $F(\underline{\lambda},\underline{x})$, due to their structure, satisfy the specialization property over the ring $R$, and we prove Theorem 1.1 in this situation (§5).

The imperfectness assumption relates to a classical subtlety in positive characteristic. There are two notions of Hilbertian fields, depending on whether the specialization property is requested for all irreducible polynomials or only for the separable ones. We follow [FJ08] and use the name Hilbertian for the weaker (the latter), and we say totally Hilbertian for the stronger (precise definitions are in §4.1). They are equivalent under the imperfectness assumption [Uch80] [FJ08, Proposition 12.4.3].

Final note. The original Schinzel hypothesis has also appeared in Arithmetic Geometry, notably around the question of whether, for appropriate varieties over a number field $k$, the Brauer-Manin obstruction is the only obstruction to the Hasse principle: if rational points exist locally (over all completions of $k$), they should exist globally (over $k$). In 1979, Colliot-Thélène and Sansuc [CTS82] noticed that this is true for a large family of conic bundle surfaces over $\mathbb{P}^{1}_{\mathbb{Q}}$ if one assumes Schinzel’s hypothesis. This conjectural statement has become since a working hypothesis of the area. See for example [HW16] for some last developments. Although the number field environment seems closely tied to the question, it could be interesting to investigate the potential use of our polynomial version of the Schinzel hypothesis to some similar questions over appropriate fields like rational function fields.

The paper is organized as follows. The strategy is detailed in §2. §3 is devoted to the geometric case that $R=k[\underline{x}]$ with $n\geqslant 2$ and $k$ is an infinite field; Theorem 1.4 is proved. §4 is the Hilbert part. The main results from §1 (other than Theorem 1.4) are finally proved in §5.

2. General strategy

Throughout the paper, $R$ is a UFD with fraction field $K$. Recall that a polynomial with coefficients in $R$ is said to be primitive w.r.t. $R$ if its coefficients are relatively prime in $R$.

All indeterminates are algebraically independent over $\overline{K}$.

Let $\underline{x}=(x_{1},\ldots,x_{n})$ ($n\geqslant 1$) and ${\underline{\lambda}}=(\lambda_{0},\lambda_{1},\ldots,\lambda_{\ell})$ ($\ell\geqslant 1$) be two tuples of indeterminates and let ${\underline{Q}}=(Q_{0},Q_{1},\ldots,Q_{\ell}$) with $Q_{0}=1$ be a $(\ell+1)$-tuple of nonzero polynomials in $R[\underline{x}]$, distinct up to multiplicative constants in $K^{\times}$. Set

$M({\underline{\lambda}},\underline{x})=\sum_{i=0}^{\ell}\lambda_{i}\hskip 1.0ptQ_{i}(\underline{x})$.

Consider a set $\underline{P}=\{P_{1},\ldots,P_{s}\}$ of $s$ polynomials

$P_{i}(\underline{x},y)=P_{i\rho_{i}}(\underline{x})\hskip 2.0pty^{\rho_{i}}+\cdots+P_{i1}(\underline{x})\hskip 1.0pty+P_{i0}(\underline{x})$,

irreducible in $R[\underline{x},y]$ and of degree $\rho_{i}\geqslant 1$ in $y$, $i=1,\ldots,s$. Each polynomial $P_{i}(\underline{x},y)$ is irreducible in $K(\underline{x})[y]$ and is primitive w.r.t. $R[\underline{x}]$.

Finally set, for $i=1,\ldots,s$,

$F_{i}({\underline{\lambda}},\underline{x})=P_{i}(\underline{x},M({\underline{\lambda}},\underline{x}))=P_{i}(\underline{x},\sum_{j=0}^{\ell}\lambda_{j}Q_{j}(\underline{x}))$.

In the case $\rho_{i}=1$, i.e., $P_{i}=A_{i}(\underline{x})+B_{i}(\underline{x})\hskip 1.0pty$, the polynomial $F_{i}$ rewrites

$F_{i}({\underline{\lambda}},\underline{x})=A_{i}(\underline{x})+B_{i}(\underline{x})\left(\sum_{j=0}^{\ell}\lambda_{j}Q_{j}(\underline{x}))\right)$.

Lemma 2.1**.**

(a)* Each polynomial $F_{i}({\underline{\lambda}},\underline{x})$ is irreducible in $R[\underline{\lambda},\underline{x}]$ and of degree $\geqslant 1$ in $\underline{x}$. Furthermore, if $\deg_{y}(P_{i})=1$, $F_{i}({\underline{\lambda}},\underline{x})$ is irreducible in $\overline{K}[{\underline{\lambda}},\underline{x}]$.*

(b)* If $R$ is infinite and $\Pi=\prod_{i=1}^{s}P_{i}$, there is no irreducible polynomial $p\in R[\underline{x}]$ dividing all polynomials $\Pi(\underline{x},M(\underline{x}))$ with $M\in R[\underline{x}]$.*

Note that (b) fails if $R$ is finite: with $R=\mathbb{F}_{2}$ and $\underline{P}=\{x,x+1\}$, the polynomial $x$ divides all polynomials $M(x)(M(x)+1)$ ($M\in\mathbb{F}_{2}[x]$).

Proof.

(a) Fix an integer $i\in\{1,\ldots,s\}$. By assumption, the polynomial $P_{i}(\underline{x},\lambda_{0})$ is irreducible in $R[\underline{x},\lambda_{0}]$. It is also irreducible in the bigger ring $R[\underline{x},\underline{\lambda}]$. Consider the ring automorphism $R[\underline{x},\underline{\lambda}]\rightarrow R[\underline{x},\underline{\lambda}]$ that is the identity on $R[\underline{x},\lambda_{1},\ldots,\lambda_{\ell}]$ and maps $\lambda_{0}$ to the polynomial $\lambda_{0}+\sum_{i=1}^{\ell}\lambda_{i}Q_{i}(\underline{x})$. The polynomial $F_{i}({\underline{\lambda}},\underline{x})$ is the image of $P_{i}(\underline{x},\lambda_{0})$ by this isomorphism. Hence it is irreducible in $R[\underline{x},\underline{\lambda}]$.

To see that $\deg_{\underline{x}}(F_{i})\geqslant 1$, write $F_{i}$ as a polynomial in $\lambda_{1}$. The leading coefficient is $P_{i\rho_{i}}(\underline{x})Q_{1}(\underline{x})^{\rho_{i}}$; it is of positive degree in $\underline{x}$ since $Q_{1}$ is by assumption. This proves that $\deg_{\underline{x}}(F_{i})\geqslant 1$.

In the case $\rho_{i}=1$, irreducibility of $F_{i}({\underline{\lambda}},\underline{x})$ in $\overline{K}[\underline{x},\underline{\lambda}]$ follows from the above case, applied with $R$ taken to be $\overline{K}$, and the fact that the polynomial $P_{i}(\underline{x},y)=A_{i}(\underline{x})+B_{i}(\underline{x})\hskip 1.0pty$ is irreducible in $\overline{K}[\underline{x},y]$. Namely $P_{i}(\underline{x},y)$ is of degree $1$ in $y$ and is primitive w.r.t. $\overline{K}[\underline{x}]$. Primitivity follows from the fact that, as $A_{i}$ and $B_{i}$ are relatively prime in $R[\underline{x}]$, then

• they are relatively prime in $K[\underline{x}]$ (an application of Gauss’s lemma), and,

• they are relatively prime in $\overline{K}[\underline{x}]$. For lack of reference for this last point, we provide below a quick argument.

Prove by induction on $n$ that for every field $K$, for every nonzero $A,B\in K[\underline{x}]$, if $A$ and $B$ have a common divisor $D\in\overline{K}[\underline{x}]$ with $\deg(D)>0$, they have a common divisor $C\in K[\underline{x}]$ with $\deg(C)>0$. The case $n=1$ follows from the Bézout theorem. Then, for $n\geqslant 2$, if $D$ is as in the claim, we may assume that $\deg_{(x_{2},\ldots,x_{n})}(D)>0$. Observe then that $D$ divides $A$ and $B$ in $\overline{K(x_{1})}[x_{2},\ldots,x_{n}]$. By induction $A$ and $B$ have a common divisor $C\in K(x_{1})[x_{2},\ldots,x_{n}]$ with $\deg_{(x_{2},\ldots,x_{n})}(C)>0$. Using Gauss’s lemma, one easily constructs a polynomial $C_{0}=c(x_{1})C\in K[x_{1}][x_{2},\ldots,x_{n}]$ (with $c(x_{1})\in K[x_{1}]$) dividing both $A$ and $B$ in $K[x_{1}][x_{2},\ldots,x_{n}]$.

(b) If the claim is false, there is an irreducible polynomial $p\in R[\underline{x}]$ such that $\Pi(\underline{x},M(\underline{x}))=0$ in the quotient ring $R[\underline{x}]/(p(\underline{x}))$ for all $M\in R[\underline{x}]$. But $R[\underline{x}]/(p(\underline{x}))$ is an integral domain, and it is infinite. Indeed, if $p$ is nonconstant, say $d=\deg_{x_{1}}(p)\geqslant 1$, the elements $\sum_{i=0}^{d-1}r_{i}x_{1}^{i}$ with $r_{0},\ldots,r_{d-1}\in R$ are infinitely many different elements in $R[\underline{x}]/(p(\underline{x}))$; and if $p\in R$, then the quotient ring is $R/(p)[\underline{x}]$, which is infinite too. Conclude that the polynomial $\Pi(\underline{x},y)$ which has infinitely many roots in $R[\underline{x}]/(p(\underline{x}))$ is zero in the ring $(R[\underline{x}]/(p(\underline{x}))[y]$. As this ring is an integral domain, there is an index $i\in\{1,\ldots,s\}$ such that $P_{i}(\underline{x},y)$ is zero in $(R[\underline{x}]/(p(\underline{x}))[y]$. This contradicts $P_{i}(\underline{x},y)$ being primitive w.r.t. $R[\underline{x}]$. ∎

Denote the set of polynomials $F_{1},\ldots,F_{s}$ by ${\underline{F}}$ and consider the subset

${H}_{R}({\underline{F}})\subset R^{{\ell}+1}$,

of all $(\ell+1)$-tuples $\underline{\lambda}^{\ast}=(\lambda_{0}^{\ast},\ldots,\lambda_{\ell}^{\ast})$ such that $F_{i}(\underline{\lambda}^{\ast},\underline{x})$ is irreducible in $R[\underline{x}]$, for each $i=1,\ldots,s$. It can be equivalently viewed as the set of all polynomials of the form $m(\underline{x})=\sum_{j=0}^{\ell}m_{j}Q_{j}(\underline{x})$ ($m_{0},\ldots,m_{\ell}\in R$) such that $P_{i}(\underline{x},m(\underline{x}))$ is irreducible in $R[\underline{x}]$, $i=1,\ldots,s$.

Theorems 1.1 – 1.3 will be obtained via the following special case of our situation: for a given $\underline{d}=(d_{1},\ldots,d_{n})\in(\mathbb{N}^{\ast})^{n}$, the polynomials $Q_{i}$ are all the monic monomials $Q_{0},Q_{1},\ldots,Q_{N_{\underline{d}}}$ in ${\mathcal{P}ol}_{R,n,\underline{d}}$. The polynomial

$M_{\underline{d}}({\underline{\lambda}},\underline{x})=\sum_{i=0}^{N_{\underline{d}}}\lambda_{i}\hskip 1.0ptQ_{i}(\underline{x})$

is then the generic polynomial in $n$ variables of $i$-th partial degree $d_{i}$, $i=1,\ldots,n$, and Theorems 1.2 and 1.3 are about the set

${H}_{R}({\underline{F}})={\mathcal{I}rr}_{n}(R,{\underline{P}})\cap{\mathcal{P}ol}_{R,n,{\underline{d}}}$

For example, anticipating on the reminder on Hilbertian fields in §4.1, we can immediately prove this statement, already alluded to in §1.

Addendum to Theorem 1.1. The set ${\mathcal{I}rr}_{n}(R,\underline{P})$ is Zariski-dense in ${\mathcal{P}ol}_{R,n,\underline{d}}$ for every $\underline{d}\in(\mathbb{N}^{\ast})^{n}$, in each of these two situations:

(a)* $R=K$ is a totally Hilbertian field,*

(b)* $R=K$ is a Hilbertian field and $\deg_{y}(P_{1})=\ldots=\deg_{y}(P_{s})=1$. *

Proof.

By definition, ${H}_{K}({\underline{F}})$ is a Hilbert subset. Furthermore, from Remark 5.5, it contains a separable Hilbert subset if $\deg_{y}(P_{1})=\ldots=\deg_{y}(P_{s})=1$. It follows from the definitions that ${H}_{K}({\underline{F}})$ is Zariski-dense in $K^{N_{\underline{d}}+1}={\mathcal{P}ol}_{K,n,{\underline{d}}}$ in both situations. One does not even need to assume that $\displaystyle d_{1}+\cdots+d_{n}\geqslant\max_{1\leqslant i\leqslant s}\deg_{\underline{x}}(P_{i})+2$; the statement holds for example for $d_{1}=\ldots=d_{n}=1$.∎

When $R$ is more generally a ring, we have to further guarantee that:

• the Hilbert subset ${H}_{K}({\underline{F}})$ contains $(\ell+1)$-tuples with coordinates in $R$,

• for some of these $(\ell+1)$-tuples $\underline{\lambda}^{\ast}$, the corresponding polynomials $F_{i}({\underline{\lambda}}^{\ast},\underline{x})$ are primitive w.r.t. $R$, and so irreducible in $R[\underline{x}]$.

For $R=k[u_{1},\ldots,u_{r}]$, polynomials in $R[\underline{x}]$ can be viewed as polynomials in at least two variables over the field $k$. We explain in §3 how geometric specialization techniques can be used, if $k$ is also infinite. For more general rings $R$, more arithmetic specialization tools are needed, which we develop in §4. The specific argument for the primitivity point is given in §5.1; it takes advantage of the special form of the polynomial $F_{i}$ and, as mentioned before, cannot extend to arbitrary polynomials $F\in R[{\underline{\lambda}},\underline{x}]$.

3. The geometric part

Lemma 3.1 is our specialization tool here. Based on results of Bertini, Krull and Noether, it is in the same vein as those from [BDN09], [BDN17]. We prove it in §3.1, then deduce Theorem 1.4 in §3.2.

3.1. The specialization lemma

Notation is as in §2. Consider the special case of the general situation from §2 for which $s=1=\rho_{1}$. One degree $1$ polynomial $P(\underline{x},y)$ is given: $P(\underline{x},y)=A(\underline{x})+B(\underline{x})y$ with $A,B\in R[\underline{x}]$ two nonzero relatively prime polynomials, or $P(\underline{x},y)=y$. We then have:

$\begin{matrix}F({\underline{\lambda}},\underline{x})&=A(\underline{x})+B(\underline{x})\left(\sum_{j=0}^{\ell}\lambda_{j}Q_{j}(\underline{x})\right)\hfill\\ &=A(\underline{x})+\lambda_{0}B(\underline{x})+\lambda_{1}B(\underline{x})Q_{1}(\underline{x})+\cdots+\lambda_{\ell}B(\underline{x})Q_{\ell}(\underline{x})\end{matrix}$

Lemma 3.1**.**

Assume that $n\geqslant 2$, $R=K$ is an algebraically closed field and the following holds (which implies $\ell\geqslant 1$):

(a)* there is an index $i_{0}\in\{1,\ldots,\ell\}$ such that*

- $\deg(Q_{i_{0}})\not\equiv 0$ modulo $p$ if $\mathop{\mathrm{char}}\nolimits(K)=p>0$,

- $\deg(Q_{i_{0}})\not=0$ if $\mathop{\mathrm{char}}\nolimits(K)=0$.

(b)* there is no polynomial $\chi\in K[\underline{x}]$ such that $A,B,Q_{1},\ldots,Q_{\ell}\in K[\chi]$.*

Then the set ${H}_{K}({F})$ of all $(\ell+1)$-tuples $\underline{\lambda}^{\ast}=(\lambda_{0}^{\ast},\ldots,\lambda_{\ell}^{\ast})$ such that $F(\underline{\lambda}^{\ast},\underline{x})$ is irreducible in $K[\underline{x}]$ contains a nonempty Zariski open subset of $K^{\ell+1}$.

Remark 3.2*.*

Assumptions (a) and (b) can probably be improved but the following examples show they cannot be totally removed. In each of them, $F({\underline{\lambda}},\underline{x})$ is reducible in $\overline{K({\underline{\lambda}})}[\underline{x}]$ and every non-trivial factorization yields a Zariski dense subset of ${\underline{\lambda}}^{\ast}\in K^{\ell+1}$ such that $F({\underline{\lambda}}^{\ast},\underline{x})$ is reducible in $K[\underline{x}]$.

$\bullet$ If $A,B,Q_{1},\ldots,Q_{\ell}\in K[\chi]$ for some $\chi\in K[\underline{x}]$, one can write $F({\underline{\lambda}},\underline{x})=h(\chi)$ with $h\in\overline{K({\underline{\lambda}})}[u]$. If $\deg(h)\geqslant 2$, $h$ is reducible and so is $F({\underline{\lambda}},\underline{x})$ in $\overline{K({\underline{\lambda}})}[\underline{x}]$.

$\bullet$ For $A=x_{1}^{2}$, $B=-x_{2}^{2}$, $\ell=1$ and $Q_{0}=Q_{1}=1$, we have

$F({\underline{\lambda}},\underline{x})=x_{1}^{2}-\lambda_{0}x_{2}^{2}-\lambda_{1}x_{2}^{2}=(x_{1}-\sqrt{\lambda_{0}+\lambda_{1}}\hskip 1.5ptx_{2})(x_{1}+\sqrt{\lambda_{0}+\lambda_{1}}\hskip 1.5ptx_{2})$.

$\bullet$ If ${\rm char}(K)=p>0$, for $A=x_{1}^{p}$, $B=x_{2}^{p}$, $\ell=1$, $Q_{0}=1$, $Q_{1}=x_{2}^{p}$, we have

$F({\underline{\lambda}},\underline{x})=x_{1}^{p}+\lambda_{0}x_{2}^{p}+\lambda_{1}x_{2}^{2p}=(x_{1}+\lambda_{0}^{1/p}x_{2}+\lambda_{1}^{1/p}x_{2}^{2})^{p}$.

Proof.

Assume that the conclusion of Lemma 3.1 is false. From the Bertini-Noether theorem [FJ08, Prop. 9.4.3], $F({\underline{\lambda}},\underline{x})$ is reducible in $\overline{K({\underline{\lambda}})}[\underline{x}]$. Clearly then polynomials $F(\underline{x},\underline{\lambda}^{\ast})$ are reducible in $K[\underline{x}]$ for all $\underline{\lambda}^{\ast}\in K^{\ell+1}$ such that $\deg(F(\underline{x},\underline{\lambda}^{\ast}))=\deg_{\underline{x}}(F)$. The Bertini-Krull theorem [Sch00, Theorem 37] then yields that one of the following conditions holds:

1. (1)

$\mathop{\mathrm{char}}\nolimits(K)=p>0$ and $F({\underline{\lambda}},\underline{x})\in K[{\underline{\lambda}},\underline{x}^{p}]$ with $\underline{x}^{p}=(x_{1}^{p},\ldots,x_{n}^{p})$, 2. (2)

there exist $\phi,\psi\in K[\underline{x}]$ with $\deg_{\underline{x}}(F)>\max(\deg(\phi),\deg(\psi))$ satisfying the following: there is an integer $\delta\geqslant 1$ and $\ell+2$ polynomials $H,H_{0},H_{1},\ldots,H_{\ell}\in K[u,v]$ homogeneous of degree $\delta$ such that

[TABLE]

The rest of the proof consists in ruling out both conditions (1) and (2).

For condition (1), this readily follows from the assumption on $\deg(Q_{i_{0}})$: if $\mathop{\mathrm{char}}\nolimits(k)=p>0$, the polynomials $B$ and $BQ_{i_{0}}$ cannot be both in $K[\underline{x}^{p}]$.

Assume condition (2) holds. Note that the polynomials $\phi$ and $\psi$ are relatively prime in $K[\underline{x}]$ as a consequence of $A,B$ being relatively prime in $K[\underline{x}]$. We claim that the two conditions

\left\{\begin{array}[]{c}B(\underline{x})=H_{0}(\phi(\underline{x}),\psi(\underline{x}))\hfill\\ BQ_{i_{0}}(\underline{x})=H_{i_{0}}(\phi(\underline{x}),\psi(\underline{x}))\hfill\\ \end{array}\right.

lead to this conclusion: there is $(\beta,\gamma)\in K^{2}$ such that $\beta\phi(\underline{x})+\gamma\psi(\underline{x})=1$. We show it by induction on the common degree $\delta$ of $H_{0}$ and $H_{i_{0}}$.

For $\delta=1$, write $B=a\phi+b\psi$ and $BQ_{i_{0}}=a^{\prime}\phi+b^{\prime}\psi$ with $a,b,a^{\prime},b^{\prime}\in K$. If $\deg(B)=0$, then $a\phi+b\psi\in K\setminus\{0\}$ and the claim is established. Assume $\deg(B)>0$. If $ab^{\prime}-a^{\prime}b\not=0$, any irreducible factor $\pi$ of $B$ divides $a\phi+b\psi$ and $a^{\prime}\phi+b^{\prime}\psi$, hence divides both $\phi$ and $\psi$ in $K[\underline{x}]$, which contradicts $\phi$ and $\psi$ being relatively prime. As there is at least one such factor $\pi$, we have $(a,b)=\kappa(a^{\prime},b^{\prime})$ for some nonzero $\kappa\in K$. It follows that $B=\kappa BQ_{i_{0}}$ and $\deg(Q_{i_{0}})=0$. This contradicts our assumption. Hence the claim is established for $\delta=1$.

Assume the claim is proved for $\delta\geqslant 1$ and that

$\left\{\begin{matrix}B=\prod_{j=1}^{\delta+1}(a_{j}\phi+b_{j}\psi)\\ BQ_{i_{0}}=\prod_{j=1}^{\delta+1}(a_{j}^{\prime}\phi+b_{j}^{\prime}\psi)\\ \end{matrix}\right.$

for some $(\delta+1)$-tuples $((a_{1},b_{1}),\ldots,(a_{\delta+1},b_{\delta+1}))$ and $((a^{\prime}_{1},b^{\prime}_{1}),\ldots,(a^{\prime}_{\delta+1},b^{\prime}_{\delta+1}))$ with components in $K^{2}$.

If $\deg(B)=0$, all polynomials $a_{j}\phi+b_{j}\psi$, $j=1,\ldots,\delta+1$, are of degree [math]. Hence there exists $(\beta,\gamma)\in K^{2}$ such that $\beta\phi+\gamma\psi=1$. Assume $\deg(B)>0$. As above in the case $\delta=1$, use an irreducible factor of $B$ in $K[\underline{x}]$ to conclude that there exist two indices $j$, $j^{\prime}$ such that this irreducible factor divides both $a_{j}\phi+b_{j}\psi$ and $a_{j^{\prime}}^{\prime}\phi+b_{j^{\prime}}^{\prime}\psi$. We may assume that $j=j^{\prime}=\delta+1$. As above in the case $\delta=1$, it follows from $\phi$, $\psi$ relatively prime in $K[\underline{x}]$ that

$a_{\delta+1}\phi+b_{\delta+1}\psi=\kappa(a_{{\delta+1}}^{\prime}\phi+b_{{\delta+1}}^{\prime}\psi)$

for some nonzero $\kappa\in K$. Consider the polynomial $B_{1}=B/(a_{\delta+1}\phi+b_{\delta+1}\psi)$. It is nonzero and we have

$\left\{\begin{matrix}B_{1}=\prod_{j=1}^{\delta}(a_{j}\phi+b_{j}\psi)\\ \kappa B_{1}Q_{i_{0}}=\prod_{j=1}^{\delta}(a_{j}^{\prime}\phi+b_{j}^{\prime}\psi)\\ \end{matrix}\right.$

From the induction hypothesis, applied to $B_{1}$ and $\kappa B_{1}Q_{i_{0}}$, there is $(\beta,\gamma)\in K^{2}$ such that $\beta\phi+\gamma\psi=1$. This completes the proof of our claim.

Fix $(\beta,\gamma)\in K^{2}$ such that $\beta\phi+\gamma\psi=1$. Pick $(a,b)\in K^{2}$ such that $a\gamma-\beta b\not=0$ and set $\chi=a\phi+b\psi$. We have $\deg(\chi)>0$. Then $K\phi+K\psi=K\chi+K$ and so $A,B,BQ_{1},\ldots,BQ_{\ell}$ are in $K[\chi]$. It follows that $A,B,Q_{1},\ldots,Q_{\ell}$ are in $K[\chi]$ too. Here is an argument. Fix $i\in\{1,\ldots,\ell\}$. Since $B,BQ_{i}\in K[\chi]$, $Q_{i}$ writes $Q_{i}=(p/q)(\chi)$ for some $p,q\in K[t]$ relatively prime. But then there exists $u,v\in K[t]$ such that $u(\chi)p(\chi)+v(\chi)q(\chi)=1$. Since $q(\chi)$ divides $p(\chi)$ in $K[\underline{x}]$, we have $\deg(q)=0$. Hence $Q_{i}\in K[\chi]$. ∎

3.2. Proof of Theorem 1.4

Assume $n\geqslant 2$, fix an infinite field $k$, two nonzero relatively prime polynomials $A$, $B$ in $k[\underline{x}]$ and a $n$-tuple $\underline{d}\in(\mathbb{N}^{\ast})^{n}$. As explained in §2, consider the special case of Lemma 3.1 for which the polynomials $Q_{i}$ are all the monomials $Q_{0},\ldots,Q_{N_{\underline{d}}}$ in ${\mathcal{P}ol}_{k,n,\underline{d}}$ (with $Q_{0}=1$). We then have $F({\underline{\lambda}},\underline{x})=A(\underline{x})+B(\underline{x})M_{\underline{d}}(\underline{\lambda},\underline{x})$ with $M_{\underline{d}}=\sum_{i=0}^{N_{\underline{d}}}\lambda_{i}Q_{i}$ the generic polynomial in $n$ variables of partial degree $d_{i}$ in $x_{i}$, $i=1,\ldots,n$.

Lemma 3.1 concludes that ${H}_{k}({F})={\mathcal{I}rr}_{n}(\overline{k},A,B)\cap{\mathcal{P}ol}_{k,n,{\underline{d}}}(\overline{k})$ contains a nonempty Zariski open subset of ${\mathcal{P}ol}_{k,n,{\underline{d}}}(\overline{k})$. As $k$ is infinite, the set ${\mathcal{I}rr}_{n}(\overline{k},A,B)\cap{\mathcal{P}ol}_{k,n,{\underline{d}}}(k)$ also contains a nonempty Zariski open subset of ${\mathcal{P}ol}_{k,n,{\underline{d}}}(k)$. This proves Theorem 1.4.

Remark 3.3*.*

(a) If $k$ is finite however, non emptyness of ${\mathcal{I}rr}_{n}(k,A,B)$ cannot be guaranteed at this stage: each finite set ${\mathcal{I}rr}_{n}(k,A,B)\cap{\mathcal{P}ol}_{k,n,\underline{d}}(k)$ ($\underline{d}\in(\mathbb{N}^{\ast})^{n}$) could be covered by an hypersurface. For infinite fields, Theorem 1.4 clearly covers Theorem 1.3. We will use a different method, in §4, to prove Theorem 1.3 for finite fields (which will also reprove the infinite case).

(b) Lemma 3.1 can be used in other situations. For example, let $A,B,C\in K[\underline{x}]$ be nonzero polynomials, with $A$, $B$ relatively prime and $C\in K[\underline{x}]$ distinct from $A$, $B$, up to multiplicative constants in $K^{\times}$. Assume hypotheses (a) and (b) of Lemma 3.1 respectively hold for $Q_{i_{0}}=C$ and for $A,B,C$. Lemma 3.1 shows that the set of $(\lambda,\mu)\in K^{2}$ such that $A+B(\lambda C+\mu)$ is irreducible in $K[\underline{x}]$ contains a nonempty Zariski open subset of $\mathbb{A}^{2}_{K}$.

4. The Hilbert side

This section introduces the notion of Hilbertian ring and establishes some corresponding specialization tools, which will be important ingredients of the proofs of the main theorems in §5.

4.1. Basics from the Hilbertian field theory

We recall the basic definitions and refer to chapters 12 and 13 of [FJ08] for more. Other classical references include [Sch82], [Sch00], [Lan83].

Consider a field $K$ and two tuples $\underline{\lambda}=(\lambda_{1},\ldots,\lambda_{r})$ and $\underline{x}=(x_{1},\ldots,x_{n})$ ($r\geqslant 1$, $n\geqslant 1$) of indeterminates. Given $m$ polynomials $f_{1}(\underline{\lambda},\underline{x}),\ldots,f_{m}(\underline{\lambda},\underline{x})$ ($m\geqslant 1$) in $\underline{x}$ with coefficients in $K(\underline{\lambda})$, irreducible in the ring $K(\underline{\lambda})[\underline{x}]$ and a polynomial $g\in K[\underline{\lambda}]$, $g\not=0$, consider the set

$\displaystyle H_{K}(f_{1},\ldots,f_{m};g)=\left\{\underline{\lambda}^{\ast}\in K^{r}\hskip 2.84526pt\left|\begin{matrix}\hskip 2.84526ptf_{i}(\underline{\lambda}^{\ast},\underline{x})\hskip 2.84526pt\hbox{\rm irreducible in }K[\underline{x}]\cr\hskip 2.84526pt\hbox{\rm for each }i=1,\ldots,m,\hfill\cr\hskip 2.84526pt\hbox{\rm and}\ g(\underline{\lambda}^{\ast})\not=0.\hfill\cr\end{matrix}\right.\right\}$

Call $H_{K}(f_{1},\ldots,f_{m};g)$ a Hilbert subset of $K^{r}$. If in addition $n=1$ and each $f_{i}$ is separable in $x$ (i.e., $f_{i}$ has no multiple root in $\overline{K(\underline{\lambda})}$), call $H_{K}(f_{1},\ldots,f_{m};g)$ a separable Hilbert subset of $K^{r}$. The field $K$ is called Hilbertian if every separable Hilbert subset of $K^{r}$ is nonempty and totally Hilbertian if every Hilbert subset of $K^{r}$ is nonempty ($r\geqslant 1$). Equivalently, “nonempty” can be replaced by “Zariski-dense in $K^{r}$” in the definitions. As recalled earlier, a field $K$ is totally Hilbertian if and only if it is Hilbertian and the imperfectness condition holds: $K$ is imperfect if of characteristic $p>0$.

Classical Hilbertian fields include the field $\mathbb{Q}$, the rational function fields $\mathbb{F}_{q}(u)$ (with $u$ some indeterminate) and all of their finitely generated extensions [FJ08, Theorem 13.4.2], every abelian extension of $\mathbb{Q}$ [FJ08, Theorem 16.11.3], fields $k((u_{1},\ldots,u_{r}))$ of formal power series in $r\geqslant 2$ variables over a field $k$ [FJ08, Theorem 15.4.6]; all of them are also totally Hilbertian. Algebraically closed fields, the fields $\mathbb{R}$, $\mathbb{Q}_{p}$ of real, of $p$-adic numbers, more generally Henselian fields are non-Hilbertian. The fraction field of a UFD $R$ need not be Hilbertian (take $R=\mathbb{Z}_{p}$), even if $R$ has infinitely many distinct prime ideals: a counter-example is given in [FJ08, Example 15.5.8].

Fields with the product formula provide other examples of Hilbertian fields. Recall from [FJ08, §15.3] that a nonempty set $S$ of primes ${\mathfrak{p}}$ of $K$, with associated absolute value $|\hskip 5.69054pt|_{\mathfrak{p}}$, is said to satisfy the product formula if for each ${\mathfrak{p}}\in S$, there exists $\beta_{{\mathfrak{p}}}>0$ such that:

(1) For each $a\in K^{\times}$, the set $\{{\mathfrak{p}}\in S\hskip 2.0pt|\hskip 2.0pt|a|_{\mathfrak{p}}\not=1\}$ is finite and $\prod_{{\mathfrak{p}}\in S}|a|_{\mathfrak{p}}^{\beta_{{\mathfrak{p}}}}=1$.

In this case call $K$ a field with the product formula. From a result of Weissauer, such fields are Hilbertian [FJ08, Theorem 15.3.3]. The fields $\mathbb{Q}$, $k(\lambda_{1},\ldots,\lambda_{r})$ with $k$ any field and $r\geqslant 1$, and their finite extensions, are fields with the product formula.

4.2. Hilbertian ring

The following definition is given in [FJ08, §13.4].

Definition 4.1*.*

An integral domain $R$ with fraction field $K$ is said to be a Hilbertian ring if every separable Hilbert subset of $K^{r}$ ($r\geqslant 1$) contains $r$-tuples $\underline{\lambda}^{\ast}=(\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast})$ with coordinates in $R$.

Since Zariski open subsets of Hilbert subsets remain Hilbert subsets, it is equivalent to require that a Zariski dense subset of tuples $\underline{\lambda}^{\ast}$ exist in Definition 4.1. Under the imperfectness assumption, a better property holds for Hilbertian rings, and extends to arbitrary Hilberts sets.

Proposition 4.2**.**

Let $R$ be an integral domain such that the fraction field $K$ is imperfect if of characteristic $p>0$. The following are equivalent.

(i)* $R$ is a Hilbertian ring.*

(ii)* Every separable Hilbert subset of $K$ contains elements $\lambda^{\ast}\in R$.*

(iii)* For every nonzero $\lambda_{0}^{\ast}\in R$ and every $\underline{a}=(a_{1},\ldots,a_{r})\in R^{r}$, every Hilbert subset of $K^{r}$ ($r\geqslant 1$) contains $r$-tuples $\underline{\lambda}^{\ast}=(\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast})$ with nonzero coordinates in $R$ and such that $\lambda_{i}^{\ast}\equiv a_{i}\hskip 2.84526pt[{\rm mod}\hskip 2.84526pt\lambda_{0}^{\ast}\cdots\lambda_{i-1}^{\ast}]$, $i=1,\ldots,r$.*

Clearly, it suffices to prove ${\rm(ii)}\Rightarrow{\rm(iii)}$. This is done in §4.4 by reducing the number of variables to reach the separable situation $r=n=1$ of Definition 4.1. We recall a classical tool.

4.3. The Kronecker substitution

Given an arbitrary field $K$, an irreducible polynomial $f\in K[\underline{\lambda},\underline{y}]$, of degree $\geqslant 1$ in $\underline{y}=(y_{1},\ldots,y_{m})$ and an integer $\displaystyle D>\max_{{1\leqslant i\leqslant m}}\deg_{y_{i}}(f)$, the Kronecker substitution is the map

$S_{D}:{\mathcal{P}ol}_{K(\underline{\lambda}),m,\underline{D}}\rightarrow{\mathcal{P}ol}_{K(\underline{\lambda}),1,D^{m}}$,   with $\underline{D}=(D,\ldots,D)$,

deriving from the substitution of $y^{D^{i-1}}$ for $y_{i}$, $i=1,\ldots,m$, and leaving the coefficients in the field $K(\underline{\lambda})$ unchanged.

Proposition 4.3**.**

There exist a finite set ${\mathcal{S}}(f)$ of irreducible polynomials $g\in K[\underline{\lambda}][y]$ of degree $\geqslant 1$ in $y$ and a nonzero polynomial $\varphi\in K[\underline{\lambda}]$ such that the Hilbert subset ${H}_{K}(f)\subset K^{r}$ contains the Hilbert subset

$H_{K}({\mathcal{S}}(f);\varphi)$

Furthermore, the finite set ${\mathcal{S}}(f)$ can be taken to be the set of irreducible divisors of $S_{D}(f)$ in $K[\underline{\lambda}][y]$.

Proof.

See [FJ08, Lemma 12.1.3]. The statement is only stated for $r=1$ but the proof carries over to the situation $r\geqslant 1$ by merely changing the single variable for an $r$-tuple $\underline{\lambda}=(\lambda_{1},\ldots,\lambda_{r})$ of variables. ∎

We will also use several times the following observation.

Lemma 4.4**.**

Let $R$ be a Hilbertian ring with a fraction field $K$ of characteristic $p>0$ and imperfect. There are infinitely many $a\in R$ that are different modulo $K^{p}$.

Proof.

Let $R$ be a Hilbertian ring. Clearly $K$ is Hilbertian, in particular it is infinite. Assume further that $K$ is of characteristic $p>0$ and imperfect. Then $K\not=K^{p}$ and $K/K^{p}$ is a nonzero vector space over the infinite field $K^{p}$. Thus $K/K^{p}$ is infinite. It follows that if $h\in\mathbb{N}$ is an integer, one can find $h+1$ elements $k_{1},\ldots,k_{h+1}$ of $K$ that are different modulo $K^{p}$. If $\delta\in R$ is a common denominator of $k_{1},\ldots,k_{h+1}$, then $\delta k_{1},\ldots,\delta k_{h+1}$ are elements of $R$ that are distinct modulo $K^{p}$. The conclusion follows. ∎

4.4. Proof of Proposition 4.2

Fix an integral domain $R$ satisfying the imperfectness assumption and assume that condition (ii) holds. Let $\lambda_{0}^{\ast}\in R\setminus\{0\}$, $\underline{a}=(a_{1},\ldots,a_{r})\in R^{r}$ and ${\mathcal{H}}\subset K^{r}$ be a Hilbert subset.

4.4.1. First reductions

Consider the Hilbert subset ${\mathcal{H}}_{\lambda_{0}^{\ast},a_{1}}$ deduced from ${\mathcal{H}}$ by substituting $\lambda_{0}^{\ast}\lambda_{1}+a_{1}$ to $\lambda_{1}$ in the polynomials involved in ${\mathcal{H}}$. This first reduction is used at the end of the proof in §4.4.4.

From the standard reduction Lemma 12.1.1 from [FJ08], the Hilbert subset ${\mathcal{H}}_{\lambda_{0}^{\ast},a_{1}}$ contains a Hilbert subset of the form

$\displaystyle H_{K}(f_{1},\ldots,f_{m};g)=\left\{\underline{\lambda}^{\ast}\in K^{r}\hskip 2.84526pt\left|\begin{matrix}f_{i}(\underline{\lambda}^{\ast},\underline{x})\hskip 2.84526pt\hbox{\rm irreducible in }K[\underline{x}]\cr\hskip 2.84526pt\hbox{for each}\ i=1,\ldots,m,\hfill\cr g(\underline{\lambda}^{\ast})\not=0\cr\end{matrix}\right.\right\}$

with $f_{1},\ldots,f_{m}$ irreducible polynomials in $K[\underline{\lambda},\underline{x}]$,

of degree at least $1$ in $\underline{x}$ and $g\in K[\underline{\lambda}]$, $g\not=0$.

For $i=1,\ldots,m$, view $f_{i}$ as a polynomial in $\underline{y}=(\lambda_{2},\ldots,\lambda_{r},x_{1},\ldots,x_{n})$ with coefficients in $K[\lambda_{1}]$. From Proposition 4.3, there is a finite set ${\mathcal{S}}(f_{i})$ of irreducible polynomials $g\in K[\lambda_{1}][y]$ of degree $\geqslant 1$ in $y$ and a nonzero polynomial $\varphi_{i}\in K[\lambda_{1}]$ such that the Hilbert subset $H_{K}(f_{i})\subset K$ contains the Hilbert subset $H_{K}({\mathcal{S}}(f_{i});\varphi_{i})\subset K$.

Consider the Hilbert subset

${H}_{K}({\mathcal{S}}(f_{1})\cup\cdots\cup{\mathcal{S}}(f_{m});\varphi_{1}\cdots\varphi_{m})\subset K$.

From the standard reduction Lemma 12.1.4 from [FJ08], this Hilbert subset contains a Hilbert subset of the form

$\displaystyle H_{K}(g_{1},\ldots,g_{\nu})=\left\{\lambda_{1}^{\ast}\in K\hskip 2.84526pt\left|\begin{matrix}g_{i}(\lambda_{1}^{\ast},y)\hskip 2.84526pt\hbox{\rm irreducible in }K[y]\cr\hskip 2.84526pt\hbox{\rm for each}\ i=1,\ldots,\nu,\hfill\cr\end{matrix}\right.\right\}$

with $g_{1},\ldots,g_{\nu}$ irreducible polynomials in $K[\lambda_{1},y]$,

monic and of degree at least $2$ in $y$.

4.4.2. 1st case: $g_{1},\ldots,g_{\nu}$ are separable in $y$

From assumption (ii), there is an element $\lambda_{1}^{\ast}\in R\setminus\{-a_{1}/\lambda_{0}^{\ast}\}$ such that, for each $i=1,\ldots,\nu$, $g_{i}(\lambda_{1}^{\ast},y)$ is irreducible in $K[y]$ and $\deg_{\underline{x}}(f_{i}(\lambda_{1}^{\ast},\lambda_{2},\ldots,\lambda_{r},\underline{x}))\geqslant 1$. We refer to §4.4.4 for the end of the proof which is common to 1st and 2nd cases.

4.4.3. 2nd case:

$g_{1},\ldots,g_{\nu}$ are not all separable in $y$.

Necessarily $K$ is of characteristic $p>0$. The following lemma (which we will use a second time) adjusts arguments from [FJ08, Prop. 12.4.3]. For simplicity, set $\lambda=\lambda_{1}$.

Lemma 4.5**.**

Under the 2nd case assumption, for every nonzero $\lambda_{0}^{\ast}\in R$, there is a nonzero $b\in\lambda_{0}^{\ast}R$ with this property: there exist irreducible polynomials $\widetilde{Q}_{1},\ldots,\widetilde{Q}_{\nu}$ in $K[\lambda,y]$, separable, monic of degree $\geqslant 1$ in $y$ such that for all but finitely many $\tau\in H_{K}(\widetilde{Q}_{1},\ldots,\widetilde{Q}_{\nu})$, $\tau^{p}+b$ is in $H_{K}(g_{1},\ldots,g_{\nu})$.

Proof of Lemma 4.5.

Assume $g_{1},\ldots,g_{\ell}$ are not separable in $y$ (with $\ell\geqslant 1$) and $g_{\ell+1},\ldots,g_{\nu}$ are separable in $y$. For each $i=1,\ldots,\ell$, there exists $Q_{i}\in K[\lambda,y]$ irreducible, separable, monic and of degree $\geqslant 1$ in $y$ and $q_{i}$ a power of $p$ different from $1$ such that $g_{i}(\lambda,y)=Q_{i}(\lambda,y^{q_{i}})$. Since $g_{i}(\lambda,y)$ is irreducible in $K[\lambda,y]$, $Q_{i}$ has a coefficient $h_{i}\in K[\lambda]$ which is not a $p$th power. Choose $a_{i}\in R$ with $h_{i}(\lambda+a_{i})\in K^{p}[\lambda]$ if there exists any, otherwise let $a_{i}=0$. Also set $Q_{i}=g_{i}$ for $i=\ell+1,\ldots,\nu$.

Consider the elements $a\in R$ from Lemma 4.4. Among the corresponding elements $a\lambda_{0}^{\ast}\in R$, which are also different modulo $K^{p}$, there is at least one, say $b=a\lambda_{0}^{\ast}$, such that $b\in R\setminus\bigcup_{i=1}^{\ell}(a_{i}+K^{p})$. By [FJ08, Lemma 12.4.2(b)], $h_{i}(\lambda+b)\notin K^{p}[\lambda]$, $i=1,\ldots,\ell$.

Consider the polynomials $\widetilde{Q}_{i}(\lambda,y)=Q_{i}(\lambda^{p}+b,y)$, $i=1,\ldots,\nu$. They are monic and separable in $y$. Furthermore, as detailed in §12.4 from [FJ08] (and [FJ] which clarifies the argument), they are irreducible in $K[\lambda,y]$.

Let $\tau\in H_{K}(\widetilde{Q}_{1},\ldots,\widetilde{Q}_{\nu})$ but not in the set $C$, finite by [FJ08, Lemma 12.4.2(c)], of all elements $c\in R$ with $h_{i}(c^{p}+b)\in K^{p}$ for some $i=1,\ldots,\ell$. For $i=\ell+1,\ldots,\nu$, we have $\widetilde{Q}_{i}(\tau,y)=g_{i}(\tau^{p}+b,y)$ and so $g_{i}(\tau^{p}+b,y)$ is irreducible in $K[y]$. Let $i\in\{1,\ldots,\ell\}$. Since $\tau\notin C$, we have $h_{i}(\tau^{p}+b)\notin K^{p}$. Hence $Q_{i}(\tau^{p}+b,y)=\widetilde{Q}_{i}(\tau,y)\notin K^{p}[y]$. From the choice of $\tau$, this polynomial is irreducible in $K[y]$. By [FJ08, Lemma 12.4.1], we obtain that

$\widetilde{Q}_{i}(\tau,y^{q_{i}})=Q_{i}(\tau^{p}+b,y^{q_{i}})=g_{i}(\tau^{p}+b,y)$

is irreducible in $K[y]$. Whence finally: $\tau^{p}+b\in H_{K}(g_{1},\ldots,g_{\nu})$. ∎

Use then the assumption (ii) of Proposition 4.2 to conclude that for the element $b$ and the polynomials $\widetilde{Q}_{1},\ldots,\widetilde{Q}_{\nu}$ given by Lemma 4.5, the Hilbert subset $H_{K}(\widetilde{Q}_{1},\ldots,\widetilde{Q}_{\nu})$ contains infinitely many elements $\tau\in R$. Fix one off the finite list of exceptions in the final sentence of Lemma 4.5 and such that $\lambda_{1}^{\ast}=\tau^{p}+b$ is different from $-a_{1}/\lambda_{0}^{\ast}$. The element $\lambda_{1}^{\ast}\in R$ is then in $H_{K}(g_{1},\ldots,g_{\nu})$ and $\lambda_{0}^{\ast}\lambda_{1}^{\ast}+a_{1}\not=0$. Up to excluding finitely many more $\tau$ above, we may also assure that $\deg_{\underline{x}}(f_{i}(\lambda_{1}^{\ast},\lambda_{2},\ldots,\lambda_{r},\underline{x}))\geqslant 1$ ($i=1,\ldots,\nu$). (We have only used here that $b\in R$. The possible choice of $b$ in $\lambda_{0}^{\ast}R$ will be used later (§4.6.1)).

4.4.4. End of proof of Proposition 4.2

Applying Prop.4.3 and taking into account the first reduction changing ${\mathcal{H}}$ to ${\mathcal{H}}_{\lambda_{0}^{\ast},a_{1}}$ yields in both cases that

(2) there is $\lambda_{1}^{\ast}\in R\setminus\{0\}$ such that $\lambda_{1}^{\ast}\equiv a_{1}\hskip 2.84526pt[{\rm mod}\hskip 2.84526pt\lambda_{0}^{\ast}]$, $f_{i}(\lambda_{1}^{\ast},\lambda_{2},\ldots,\lambda_{r},\underline{x})$ is irreducible in $K[\lambda_{2},\ldots,\lambda_{r},\underline{x}]$ and is of degree at least $1$ in $\underline{x}$, $i=1,\ldots,m$.

Repeating this argument provides a $r$-tuple $\underline{\lambda}^{\ast}=(\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast})$ in $(R\setminus\{0\})^{r}$ such that $f_{1}(\underline{\lambda}^{\ast},\underline{x}),\ldots,f_{m}(\underline{\lambda}^{\ast},\underline{x})$ are irreducible in $K[\underline{x}]$ (so $\underline{\lambda}^{\ast}$ is in the original Hilbert subset ${\mathcal{H}}$) and such that $\lambda_{i}^{\ast}\equiv a_{i}\hskip 2.84526pt[{\rm mod}\hskip 2.84526pt\lambda_{0}^{\ast}\cdots\lambda_{i-1}^{\ast}]$, $i=1,\ldots,r$.

4.5. UFD with fraction field with the product formula

Theorem 4.6**.**

If $R$ is an integral domain such that the fraction field $K$ has the product formula and is imperfect if of characteristic $p>0$, then $R$ is a Hilbertian ring.

Fix a ring $R$ as in the statement. Theorem 4.6 relies on the following lemma, whose main ingredient is a result for fields with the product formula. Recall a useful tool in a field $K$ with a set $S$ of primes $\mathfrak{p}$ satisfying the product formula. For every $a\in K$, the (logarithmic) height $h(a)$ of $a$ is defined by:

$h(a)=\sum_{{\mathfrak{p}}\in S}\log(\max(1,|a|_{\mathfrak{p}}))$.

Clearly $h(a^{n})=nh(a)$ ($n\in\mathbb{N}$) and $h(1/a)=h(a)$ if $a\not=0$.

Lemma 4.7**.**

Let $f_{1},\ldots,f_{m}$ be $m$ irreducible polynomials in $K(\lambda)[y]$. For all but finitely many $t_{0}\in R$, there is a nonzero element $a\in R$ with the following property: if $b\in R$ is of height $h(b)>0$, the Hilbert subset $H_{K}(f_{1},\ldots,f_{m})$ contains infinitely many elements of $R$ of the form $t_{0}+ab^{\ell}$ ($\ell>0$).

Proof.

[Dèb99, Theorem 3.3] proves the weaker version for which the element $a$ is only asserted to lie in $K$. However the proof can be adjusted so that $a\in R$. Specifically, the same argument there leads to the stronger conclusion provided that, if $K$ is of characteristic $p>0$, infinitely many $a\in R$ can be found that are different modulo $K^{p}$. This is the conclusion of Lemma 4.4. ∎

Proof of Theorem 4.6.

We prove condition (ii) from Proposition 4.2. Let ${\mathcal{H}}\subset K$ be a separable Hilbert subset. From Lemmas 12.1.1 and 12.1.4 of [FJ08], the Hilbert subset ${\mathcal{H}}$ contains a separable Hilbert subset of the form

$\displaystyle H_{K}(f_{1},\ldots,f_{m})=\left\{\lambda^{\ast}\in K\hskip 2.84526pt\left|\begin{matrix}f_{i}(\lambda^{\ast},y)\hskip 2.84526pt\hbox{\rm irreducible in }K[y]\cr\hskip 2.84526pt\hbox{\rm for each}\ i=1,\ldots,m,\hfill\cr\end{matrix}\right.\right\}$

with $f_{1},\ldots,f_{m}$ irreducible polynomials in $K[\lambda,y]$,

monic, separable and of degree at least $2$ in $y$.

Pick an element $t_{0}\in R$ that avoids the finite set of exceptions in Lemma 4.7. Consider an element $a\in R$ associated to this $t_{0}$ in Lemma 4.7. Choose an element $b\in R$ of height $h(b)>0$.

Here is an argument showing that such $b$ exist. Fix a prime ${\mathfrak{p}}\in S$. Recall that by definition, the corresponding absolute value is nontrivial [FJ08, §13.3]: there exists $b\in K$ such that $|b|_{\mathfrak{p}}\not=1$. One may assume that $b\in R$. From the product formula, there is a prime ${\mathfrak{p}}_{0}\in S$ such that $|b|_{\mathfrak{p}_{0}}>1$. We have $h(b)\geqslant\log(\max(1,|b|_{\mathfrak{p}_{0}}))>0$.

From Lemma 4.7, $\lambda_{1}^{\ast}=t_{0}+ab^{\ell}\in R$ is in the Hilbert subset $H_{K}(f_{1},\ldots,f_{m})$, hence in the Hilbert subset ${\mathcal{H}}$, for infinitely many integers $\ell>0$. ∎

4.6. Polynomial rings in one variable

Theorem 4.8**.**

Assume that $R=k[u]$ with $k$ an arbitrary field. Let ${\mathcal{H}}$ be a Hilbert subset of $K^{r}$ ($r\geqslant 1$), $\lambda_{0}^{\ast}\in R$ a nonzero element of $R$ and $d_{1}\geqslant 1$ an integer. Define $\widetilde{p}$ by

\widetilde{p}=\left\{\begin{matrix}1\hskip 14.22636pt\hbox{if {\rm char}(k)=0$or${\mathcal{H}} is a separable Hilbert subset}\hfill\\ p\hskip 14.22636pt\hbox{otherwise}.\hfill\\ \end{matrix}\right.

Denote the subset of ${\mathcal{H}}$ of $r$-tuples $\underline{\lambda}^{\ast}=(\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast})\in R^{r}$ such that $\lambda_{1}^{\ast}$ and $\lambda_{0}^{\ast}\lambda_{2}^{\ast}\cdots\lambda_{r}^{\ast}$ are relatively prime in $R$ and $\max_{1\leqslant i\leqslant r}\deg(\lambda_{i}^{\ast})=\widetilde{p}d_{1}$ by ${\mathcal{H}}_{\lambda_{0}^{\ast},\widetilde{p}d_{1}}$. There is an integer $d_{0}$ such that if $d_{1}\geqslant d_{0}$, the set ${\mathcal{H}}_{\lambda_{0}^{\ast},\widetilde{p}d_{1}}$ is nonempty.

When $R=k[u]$, statement (iii) from Proposition 4.2 also holds for the Hilbert subset ${\mathcal{H}}$: there the congruence conditions are stronger but no control is given on the degree in $u$ of $\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast}$ as in Theorem 4.8.

We divide the proof of Theorem 4.8 into two parts. The situation: one parameter, one variable, is considered in §4.6.1, the general one in §4.6.2.

4.6.1. Proof of Theorem 4.8 – situation $r=n=1$ –

We are given a Hilbert subset ${\mathcal{H}}\subset K=k(u)$, a nonzero element $\lambda_{0}^{\ast}\in k[u]$, an integer $d_{1}\geqslant 1$ and we need to find an element $\lambda_{1}^{\ast}\in k[u]$ such that $\lambda_{1}^{\ast}\in{\mathcal{H}}$, $\lambda_{1}^{\ast}$ and $\lambda_{0}^{\ast}$ are relatively prime and $\deg(\lambda_{1}^{\ast})=\widetilde{p}d_{1}$.

From Lemmas 12.1.1 and 12.1.4 from [FJ08], the Hilbert subset ${\mathcal{H}}$ contains a Hilbert subset of the form

$\displaystyle H_{K}(f_{1},\ldots,f_{m})=\left\{\lambda^{\ast}\in K\hskip 2.84526pt\left|\begin{matrix}f_{i}(\lambda^{\ast},y)\hskip 2.84526pt\hbox{\rm irreducible in }K[y]\cr\hskip 2.84526pt\hbox{\rm for each}\ i=1,\ldots,m.\hfill\cr\end{matrix}\right.\right\}$

with $f_{1},\ldots,f_{m}$ irreducible polynomials in $K[\lambda,y]$,

monic and of degree at least $2$ in $y$.

We distinguish the two cases corresponding to the definition of $\widetilde{p}$.

Separable case: ${\rm char}(k)=0$ or ${\mathcal{H}}$ is a separable Hilbert subset. As $n=1$, the Hilbert subset ${\mathcal{H}}$ is also separable under the assumption ${\rm char}(k)=0$. So we may assume that the polynomials $f_{1},\ldots,f_{m}$ above are separable in $y$. We distinguish two sub-cases.

• 1st sub-case: $k$ is infinite. Use [Lan83, Prop.4.1 p.236] to assert that there exists a nonempty Zariski open subset $V\subset\mathbb{A}_{k}^{2}$ such that for all but finitely many $\gamma\in k$,

$\{\tau+\gamma(u-\beta)^{d_{1}}\in k[u]\hskip 2.0pt|\hskip 2.0pt(\tau,\beta)\in V\}\subset H_{K}(f_{1},\ldots,f_{m})$.

Fix a nonzero $\gamma\in k$ off the finite exceptional list. There are infinitely many different $(\tau,\beta)\in V$ such that no root in $\overline{k}$ of the polynomial $\lambda_{0}^{\ast}\in k[u]$ is a root of $\tau+\gamma(u-\beta)^{d_{1}}$, and so $\tau+\gamma(u-\beta)^{d_{1}}$ and $\lambda_{0}^{\ast}$ are relatively prime. The corresponding elements $\lambda_{1}^{\ast}=\tau+\gamma(u-\beta)^{d_{1}}$ are infinitely many different elements of the set ${\mathcal{H}}_{\lambda_{0}^{\ast},d_{1}}$. In this case, one can take $d_{0}=1$.

• 2nd sub-case: $k$ is finite. Start with another classical reduction, namely [FJ08, Lemma 13.1.2], to conclude that there exist polynomials $Q_{1},\ldots,Q_{\nu}$ in $K[\lambda,y]$, irreducible in $\overline{K}[\lambda,y]$, monic and separable in $y$, of degree $\geqslant 2$ in $y$ and such that the Hilbert subset $H_{K}(f_{1},\ldots,f_{m})$ contains the set

$\displaystyle H^{\prime}_{K}(Q_{1},\ldots,Q_{\nu})=\left\{\lambda^{\ast}\in K\hskip 2.84526pt\left|\hskip 2.84526pt\begin{matrix}Q_{i}(\lambda^{\ast},y)\hskip 2.84526pt\hbox{\rm has no root in }K\cr\hbox{for each}\ i=1,\ldots,\nu\cr\end{matrix}\right.\right\}$

Consider the set $\{{\mathfrak{p}}_{i}\hskip 2.84526pt|\hskip 2.84526pti\in I\}$ of irreducible factors of the given polynomial $\lambda_{0}^{\ast}\in k[u]$; view them as primes of $K$. Apply [FJ08, Lemma 13.3.4] to assert that, for each $j=1,\ldots,\nu$, there are infinitely primes ${\mathfrak{p}}_{j}$ of $K$ such that there is an $a_{{\mathfrak{p}}_{j}}\in R$ with this property: if $a\in R$ satisfies $a\equiv a_{{\mathfrak{p}}_{j}}\ {\rm mod}\ {\mathfrak{p}}_{j}$, then $Q_{j}(a,v)\not=0$ for every $v\in K$. For each $j=1,\ldots,\nu$, pick one such prime ${\mathfrak{p}}_{j}$ that is different from all primes ${\mathfrak{p}}_{i}$ with $i\in I$.

Denote the ideal $(\prod_{j=1}^{\nu}{\mathfrak{p}}_{j})(\prod_{i\in I}{\mathfrak{p}}_{i})\subset R$ by ${\mathcal{I}}$. From the Chinese Remainder Theorem, there exists $a_{0}\in R$ such that every $a\in a_{0}+{\mathcal{I}}$ satisfies

$\displaystyle\left\{\begin{matrix}a\equiv a_{{\mathfrak{p}}_{j}}\ {\rm mod}\ {\mathfrak{p}}_{j}\ {\rm for}\ j=1,\ldots,\nu,\hfill\\ a\equiv 1\ {\rm mod}\ {\mathfrak{p}}_{i}\ {\rm for}\ i\in I.\hfill\\ \end{matrix}\right.$

Consider such an $a$ and rename it $\lambda_{1}^{\ast}$. It follows from the first condition that $\lambda_{1}^{\ast}\in H^{\prime}_{K}(Q_{1},\ldots,Q_{\nu})$ and so $\lambda_{1}^{\ast}\in H_{K}(f_{1},\ldots,f_{m})\subset{\mathcal{H}}$. It follows from the second condition that $\lambda_{1}^{\ast}\not\equiv 0\ {\rm mod}\ {\mathfrak{p}}_{i}$ for every $i\in I$. Hence $\lambda_{1}^{\ast}$ and $\lambda_{0}^{\ast}$ are relatively prime. Finally when $\lambda_{1}^{\ast}=a$ ranges over $a_{0}+{\mathcal{I}}$, $\deg(\lambda_{1}^{\ast})$ assumes all but finitely many values in $\mathbb{N}$. Therefore there is an integer $d_{0}$ such that ${\mathcal{H}}_{\lambda^{\ast}_{0},d_{1}}\not=\emptyset$ for every $d_{1}\geqslant d_{0}$.

2nd case: ${\rm char}(k)=p>0$* and ${\mathcal{H}}$ is not a separable Hilbert subset*. Not all the polynomials $f_{1},\ldots,f_{m}$ are separable in $y$. Proceed as in §4.4.3. From Lemma 4.5, there is a nonzero $b\in\lambda_{0}^{\ast}R$ and some irreducible polynomials $\widetilde{Q}_{1},\ldots,\widetilde{Q}_{m}$ in $K[\lambda,y]$, separable, monic of degree $\geqslant 1$ in $y$ such that for all but finitely many $\tau\in H_{K}(\widetilde{Q}_{1},\ldots,\widetilde{Q}_{m})$, $\tau^{p}+b$ is in $H_{K}(f_{1},\ldots,f_{m})$.

From the separable case of the current proof, there is an integer $d_{0}\geqslant 1$ with the following property: the Hilbert subset $H_{K}(\widetilde{Q}_{1},\ldots,\widetilde{Q}_{\nu})$ contains infinitely many elements $\tau\in R$ such that $\tau$ and $\lambda_{0}^{\ast}$ are relatively prime and $\deg(\tau)=d_{1}$. Fix one off the finite list of exceptions in the final sentence of Lemma 4.5 and set $\lambda_{1}^{\ast}=\tau^{p}+b$. We then have $\lambda_{1}^{\ast}\in H_{K}(f_{1},\ldots,f_{m})$. Furthermore $\lambda_{1}^{\ast}$ and $\lambda_{0}^{\ast}$ are relatively prime in $R$. Finally assuming that $d_{0}$ is also larger than $\deg(b)$, we have $\deg(\lambda_{1}^{\ast})=pd_{1}$ if $d_{1}\geqslant d_{0}$, thus finally proving that $\lambda_{1}^{\ast}\in{\mathcal{H}}_{\lambda_{0}^{\ast},pd_{1}}$.

4.6.2. Proof of Theorem 4.8 – situation $r\geqslant 1$, $n\geqslant 1$ –

As in §4.6.1 we distinguish two cases according to the definition of $\widetilde{p}$.

Separable case: ${\mathcal{H}}$ is a separable Hilbert subset (in particular $n=1$). From Lemma 12.1.1 and Lemma 12.1.4 from [FJ08], the separable Hilbert subset ${\mathcal{H}}\subset K^{r}$ contains a Hilbert subset of the form

$\displaystyle H_{K}(f_{1},\ldots,f_{m})=\left\{\underline{\lambda}^{\ast}\in K^{r}\hskip 2.84526pt\left|\begin{matrix}f_{i}(\underline{\lambda}^{\ast},x)\hskip 2.84526pt\hbox{\rm irreducible in }K[x]\cr\hskip 2.84526pt\hbox{\rm for each}\ i=1,\ldots,m,\hfill\cr\end{matrix}\right.\right\}$

with $f_{1},\ldots,f_{m}$ irreducible polynomials in $K[\underline{\lambda},x]$,

separable, monic and of degree at least $2$ in $x$.

Set ${\mathcal{K}}=K(\lambda_{3},\ldots,\lambda_{r})$ (with ${\mathcal{K}}=K$ if $r=2$) and regard $f_{1},\ldots,f_{m}$ as polynomials in the ring ${\mathcal{K}}(\lambda_{1})[\lambda_{2},x]$. By [FJ08, Proposition 13.2.1], there exists a nonempty Zariski open subset $U\subset\mathbb{A}_{{\mathcal{K}}}^{2}$ such that

$\{a+b\lambda_{1}\hskip 2.0pt|\hskip 2.0pt(a,b)\in U\}\subset H_{{\mathcal{K}}(\lambda_{1})}(f_{1},\ldots,f_{m})$.

Furthermore, up to shrinking $U$, one may require that the polynomials

(4)                       $f_{i}(\lambda_{1},a\lambda_{1}+b,\lambda_{3},\ldots,\lambda_{r},x)$, $i=1,\ldots,m$

are separable and of degree at least $2$ in $x$, and that $b\not=0$. As $R=k[u]\subset{\mathcal{K}}$ is infinite, the open subset $U$ contains elements $(a,b)\in R^{2}$. For such $(a,b)$, the polynomials above in (4) are in $K[\lambda_{1},\lambda_{3},\ldots,\lambda_{r},x]$ and are irreducible in $K(\lambda_{1},\lambda_{3},\ldots,\lambda_{r})[x]$. Repeating this procedure provides an $(r-1)$-tuple $((a_{2},b_{2}),\ldots,(a_{r},b_{r}))\in(R^{2})^{r-1}$ with $b_{2}\cdots b_{r}\not=0$ such that the polynomials

$g_{i}(\lambda_{1},x)=f_{i}(\lambda_{1},a_{2}\lambda_{1}+b_{2},\ldots,a_{r}\lambda_{1}+b_{r},x)$, $i=1,\ldots,m$

are in $K[\lambda_{1},x]$, irreducible in $K(\lambda_{1})[x]$, separable and of degree $\geqslant 2$ in $x$.

From the proof in situation $r=n=1$ and in the separable case (in §4.6.1), there is an integer $\delta_{0}\geqslant 1$ with this property: the Hilbert subset $H_{K}(g_{1},\ldots,g_{m})$ contains an element $\lambda_{1}^{\ast}\in R$ relatively prime to $\lambda_{0}^{\ast}\cdot b_{2}\cdots b_{r}$ and such that $\deg(\lambda_{1}^{\ast})=\delta_{1}$ if $\delta_{1}\geqslant\delta_{0}$. Request further to $\delta_{0}$ to satisfy:

(5)                           $\delta_{0}>\max_{2\leqslant i\leqslant r}\deg(b_{i})$.

Set $d_{0}=\delta_{0}+\max_{2\leqslant i\leqslant r}\deg(a_{i})$ and fix an integer $d_{1}\geqslant d_{0}$. It follows from $d_{1}-\max_{2\leqslant i\leqslant r}\deg(a_{i})\geqslant\delta_{0}$ that the Hilbert subset $H_{K}(g_{1},\ldots,g_{m})$ contains an element $\lambda_{1}^{\ast}\in R$ such that $\deg(\lambda_{1}^{\ast})=d_{1}-\max_{2\leqslant i\leqslant r}\deg(a_{i})$.

Consequently we have the following:

• the $r$-tuple $\underline{\lambda}^{\ast}=(\lambda_{1}^{\ast},a_{2}\lambda_{1}^{\ast}+b_{2},\ldots,a_{r-1}\lambda_{1}^{\ast}+b_{r-1},a_{r}\lambda_{1}^{\ast}+b_{r})\in R^{r}$ is in the original Hilbert subset ${\mathcal{H}}$, and, denoting the $i$-th component of $\underline{\lambda}^{\ast}$ by $\lambda_{i}^{\ast}$,

• $\lambda_{1}^{\ast}$ is relatively prime to $\lambda_{0}^{\ast}\lambda_{2}^{\ast}\cdots\lambda_{r}^{\ast}$,

• the largest degree of $\lambda_{1}^{\ast},\ldots,\lambda_{r}^{\ast}$ is $d_{1}$ (due to condition (5), this largest degree is $\max_{2\leqslant i\leqslant r}\deg(a_{i}\lambda_{1}^{\ast})$).

This proves that $\underline{\lambda}^{\ast}\in{\mathcal{H}}_{\lambda_{0}^{\ast},d_{1}}$.

General case: We will use the Kronecker substitution. The Hilbert subset ${\mathcal{H}}$ contains a Hilbert subset

$\displaystyle H_{K}(f_{1},\ldots,f_{m};g)=\left\{\underline{\lambda}^{\ast}\in K^{r}\hskip 2.84526pt\left|\begin{matrix}\hskip 2.84526ptf_{i}(\underline{\lambda}^{\ast},\underline{x})\hskip 2.84526pt\hbox{\rm irreducible in }K[\underline{x}]\cr\hskip 2.84526pt\hbox{\rm for each}\ i=1,\ldots,m,\hfill\cr g(\underline{\lambda}^{\ast})\not=0\cr\end{matrix}\right.\right\}$

with $f_{1},\ldots,f_{m}$ irreducible polynomials in $K[\underline{\lambda},\underline{x}]$,

of degree at least $1$ in $\underline{x}$ and $g\in K[\underline{\lambda}]$, $g\not=0$.

As in §4.4, Proposition 4.3, followed by [FJ08, Lemma 12.1.4], provides polynomials $g_{1},\ldots,g_{\nu}$, irreducible in $K[\lambda_{1},y]$, monic and of degree $\geqslant 2$ in $y$ with this property. For every $\lambda_{1}^{\ast}\in H_{K}(g_{1},\ldots,g_{\nu})$, each of the polynomials

$f_{i}(\lambda_{1}^{\ast},\lambda_{2},\ldots,\lambda_{r},\underline{x})$, $i=1,\ldots,m$,

is irreducible in $K[\lambda_{2},\ldots,\lambda_{r},\underline{x}]$. From the proof in situation $r=n=1$ (§4.6.1), the Hilbert subset $H_{K}(g_{1},\ldots,g_{\nu})$ contains infinitely many $\lambda_{1}^{\ast}\in R$ relatively prime to $\lambda_{0}^{\ast}$. Repeating this argument $(r-2)$ times provides $\lambda_{1}^{\ast},\ldots,\lambda_{r-1}^{\ast}\in R$ such that $f_{i}(\lambda_{1}^{\ast},\ldots,\lambda_{r-1}^{\ast},\lambda_{r},\underline{x})$ is irreducible in $K[\lambda_{r},\underline{x}]$ ($i=1,\ldots,m$) and $\lambda_{i}^{\ast}$ and $\lambda_{0}^{\ast}\lambda_{1}^{\ast}\cdots\lambda_{i-1}^{\ast}$ are relatively prime ($i=1,\ldots,r-1$).

Repeating the argument once more but applying this time the full conclusion of the case $r=n=1$ of the proof including the degree condition, we obtain that there is an integer $d_{0}$, which we may also choose to be larger than $\max_{1\leqslant i\leqslant r-1}\deg(\lambda_{i}^{\ast})$, with the following property: if $d_{1}\geqslant d_{0}$, there exists an element $\lambda_{r}^{\ast}\in R$ such that

• $f_{i}(\lambda_{1}^{\ast},\ldots,\lambda_{r-1}^{\ast},\lambda_{r}^{\ast},\underline{x})$ is irreducible in $K[\underline{x}]$, $i=1,\ldots,m$,

• $\lambda_{r}^{\ast}$ and $\lambda_{0}^{\ast}\lambda_{1}^{\ast}\cdots\lambda_{r-1}^{\ast}$ are relatively prime,

• $\deg(\lambda_{r}^{\ast})=\widetilde{p}d_{1}$.

Finally the $r$-tuple $\underline{\lambda}^{\ast}$ is in the original Hilbert subset ${\mathcal{H}}$, $\lambda_{i}^{\ast}$ and $\lambda_{0}^{\ast}\lambda_{1}^{\ast}\cdots\lambda_{i-1}^{\ast}$ are relatively prime ($i=1,\ldots,r$), and consequently, $\lambda_{1}^{\ast}$ is relatively prime to $\lambda_{0}^{\ast}\lambda_{2}^{\ast}\cdots\lambda_{r}^{\ast}$, and $\max_{1\leqslant i\leqslant r}\deg(\lambda_{i}^{\ast})=\widetilde{p}d_{1}$. Thus the set ${\mathcal{H}}_{\lambda_{0}^{\ast},d_{1}}$ is nonempty.

5. Proofs of the main results

5.1. Proof of Theorem 1.1 and Theorem 1.2

Recall the notation from §2: $R$ is a UFD with fraction field $K$, $\underline{x}=(x_{1},\ldots,x_{n})$, ${\underline{\lambda}}=(\lambda_{0},\lambda_{1},\ldots,\lambda_{\ell})$ ($n\geqslant 1$, $\ell\geqslant 1$) are two tuples of indeterminates, ${\underline{Q}}=(Q_{0},Q_{1},\ldots,Q_{\ell}$), with $Q_{0}=1$, is a $(\ell+1)$-tuple of nonzero polynomials in $R[\underline{x}]$, distinct up to multiplicative constants in $K^{\times}$, $\underline{P}=\{P_{1},\ldots,P_{s}\}$ is a set of $s$ polynomials

$P_{i}(\underline{x},y)=P_{i\rho_{i}}(\underline{x})\hskip 2.0pty^{\rho_{i}}+\cdots+P_{i1}(\underline{x})\hskip 1.0pty+P_{i0}(\underline{x})$,

irreducible in $R[\underline{x},y]$ and of degree $\rho_{i}\geqslant 1$ in $y$, $i=1,\ldots,s$. We also set

$M({\underline{\lambda}},\underline{x})=\sum_{j=0}^{\ell}\lambda_{j}\hskip 1.0ptQ_{j}(\underline{x})$

and, for $i=1,\ldots,s$,

$F_{i}({\underline{\lambda}},\underline{x})=P_{i}(\underline{x},M({\underline{\lambda}},\underline{x}))=P_{i}(\underline{x},\sum_{j=0}^{\ell}\lambda_{j}Q_{j}(\underline{x}))$

The polynomials $F_{1},\ldots,F_{s}$ are irreducible in $R[\underline{\lambda},\underline{x}]$ (Lemma 2.1). Finally, for $\underline{F}=\{F_{1},\ldots,F_{s}\}$, we introduced the subset

${H}_{R}({\underline{F}})\subset R^{{\ell}+1}$

of all $(\ell+1)$-tuples $\underline{\lambda}^{\ast}$ (or equivalently, of polynomials $\Lambda(\underline{x})=\sum_{j=0}^{\ell}\lambda_{i}^{\ast}\hskip 2.0ptQ_{j}(\underline{x})$) such that $F_{i}(\underline{\lambda}^{\ast},\underline{x})=P_{i}(\underline{x},\Lambda(\underline{x}))$ is irreducible in $R[\underline{x}]$, $i=1,\ldots,s$.

Given a nonzero element $\lambda_{-1}^{\ast}\in R$ and a tuple $\underline{a}=(a_{0},\ldots,a_{\ell})\in R^{\ell+1}$, consider the subset

${H}_{R,\lambda_{-1}^{\ast},\underline{a}}({\underline{F}})\subset{H}_{R}({\underline{F}})$

of those $(\ell+1)$-tuples $\underline{\lambda}^{\ast}=(\lambda_{0}^{\ast},\ldots,\lambda_{\ell}^{\ast})\in{H}_{R}({\underline{F}})$ which further satisfy the congruences $\lambda_{i}^{\ast}\equiv a_{i}\hskip 2.84526pt[{\rm mod}\hskip 2.84526pt\lambda_{-1}^{\ast}\lambda_{0}^{\ast}\cdots\lambda_{i-1}^{\ast}]$, $i=0,\ldots,\ell$.

Make this additional assumption on $Q_{0},\ldots,Q_{\ell}$ (which implies $\ell\geqslant 2$):

(1)   $Q_{0},\ldots,Q_{\ell}$* are monomials with coefficient $1$, $Q_{0}=1$ and*

$\displaystyle\min(\deg(Q_{1}),\deg(Q_{2}))>\max_{1\leqslant i\leqslant s}\deg_{\underline{x}}(P_{i})$.

Theorem 5.1**.**

Let $\lambda_{-1}$ be a nonzero element of $R$ and $\underline{a}=(1,\ldots,1)\in R^{\ell+1}$.

(a)* Assume that $R$ is a UFD and a Hilbertian ring and $K$ is imperfect if it is of characteristic $p>0$. The subset ${H}_{R,\lambda_{-1}^{\ast},\underline{a}}({\underline{F}})$ is Zariski-dense in $R^{\ell+1}$.*

(b)* If $R=k[u]$ with $k$ an arbitrary field and $d_{1}$ is a suitably large integer, then ${H}_{R}({\underline{F}})$ contains a polynomial $\Lambda=\sum_{j=0}^{\ell}\lambda_{i}^{\ast}\hskip 2.0ptQ_{j}(\underline{x})$ with $\underline{\lambda}^{\ast}=(\lambda_{0}^{\ast},\ldots,\lambda_{\ell}^{\ast})\in R^{\ell+1}$ such that $\lambda_{1}^{\ast}$ and $\lambda_{-1}^{\ast}\lambda_{0}^{\ast}\lambda_{2}^{\ast}\cdots\lambda_{\ell}^{\ast}$ are relatively prime and $\deg_{u}(\Lambda)=\widetilde{p}d_{1}$.*

Proof.

The number of monomials $Q_{i}$ is $\ell+1\geqslant 3$. Each $F_{i}$ is of degree $\geqslant 1$ in $\underline{x}$ and is irreducible in $K(\underline{\lambda})[\underline{x}]$, $i=1,\ldots,s$ (Lemma 2.1). Let $g\in K[\underline{\lambda}]$ be a nonzero polynomial and consider the Hilbert subset

${H}_{K}({\underline{F}};g)\subset K^{{\ell}+1}$.

In situation (a), it follows from Proposition 4.2 that the Hilbert subset ${H}_{K}({\underline{F}};g)$ contains an $(\ell+1)$-tuple $\underline{\lambda}^{\ast}=(\lambda_{0}^{\ast},\ldots,\lambda_{\ell}^{\ast})\in R^{\ell+1}$ satisfying the congruences $\lambda_{i}^{\ast}\equiv 1\hskip 2.84526pt[{\rm mod}\hskip 2.84526pt\lambda_{-1}^{\ast}\lambda_{0}^{\ast}\cdots\lambda_{i-1}^{\ast}]$, $i=0,\ldots,\ell$.

In situation (b), from Theorem 4.8, the Hilbert subset ${H}_{K}({\underline{F}};g)$ contains an $(\ell+1)$-tuple $\underline{\lambda}^{\ast}$ such that $\lambda_{1}^{\ast}$ and $\lambda_{-1}^{\ast}\lambda_{0}^{\ast}\lambda_{2}^{\ast}\cdots\lambda_{\ell}^{\ast}$ are relatively prime and $\displaystyle\max_{0\leqslant i\leqslant\ell}\deg(\lambda_{i}^{\ast})=\widetilde{p}d_{1}$, i.e, $\deg_{u}(\Lambda)=\widetilde{p}d_{1}$ for $\Lambda=\sum_{j=0}^{\ell}\lambda_{i}^{\ast}\hskip 2.0ptQ_{j}(\underline{x})$.

Each $F_{i}(\underline{\lambda}^{\ast},\underline{x})$ being irreducible in $K[\underline{x}]$, to finish the proof, it suffices to show that $F_{i}(\underline{\lambda}^{\ast},\underline{x})$ is primitive w.r.t. $R$ ($i=1,\ldots,s$).

Assume otherwise, i.e., for some $i=1,\ldots,s$, there is an irreducible element $\pi\in R$ dividing all the coefficients of $F_{i}(\underline{\lambda}^{\ast},\underline{x})$. The quotient ring $\overline{R}=R/(\pi)$ is an integral domain. Use the notation $\overline{U}$ to denote the class modulo $(\pi)$ of polynomials $U$ with coefficients in $R$. We have:

(2)            $\overline{P}_{i\rho_{i}}(\underline{x})\hskip 2.0pt\overline{M}({\underline{\lambda}}^{\ast},\underline{x})^{\rho_{i}}+\cdots+\overline{P}_{i1}(\underline{x})\hskip 1.0pt\overline{M}({\underline{\lambda}}^{\ast},\underline{x})=-\overline{P}_{i0}(\underline{x})$.

We distinguish two cases.

1st case: $\pi$ divides all polynomials $P_{ij}(\underline{x})$, $j=1,\ldots,\rho_{i}$. From (2), $\pi$ also divides $P_{i0}(\underline{x})$. This contradicts $P_{i}(\underline{x},y)$ being primitive w.r.t. $R$.

2nd case: there is an index $j\in\{1,\ldots,\rho_{i}\}$ such that $\pi$ does not divides $P_{ij}(\underline{x})$. As $\lambda_{1}^{\ast}$ and $\lambda_{2}^{\ast}$ are relatively prime (in both situations (a) and (b)), one of the two is not divisible by $\pi$. Conjoin this with our monomials $Q_{i}$ being of coefficient $1$ to conclude that $\overline{M}({\underline{\lambda}}^{\ast},\underline{x})\not=0$ in $R/(\pi)[\underline{x}]$ and that there is at least one nonzero term $\overline{P}_{ij}(\underline{x})\hskip 2.0pt\overline{M}({\underline{\lambda}}^{\ast},\underline{x})^{j}$ with $j\in\{1,\ldots,\rho_{i}\}$. Furthermore we have:

$\deg(\overline{M}(\underline{\lambda}^{\ast},\underline{x}))\geqslant\min(\deg(Q_{1}),\deg(Q_{2}))$.

Using next the following inequality (coming from assumption (1)):

$\displaystyle\min(\deg(Q_{1}),\deg(Q_{2}))>\max_{\genfrac{}{}{0.0pt}{}{1\leqslant i\leqslant s}{1\leqslant j\leqslant\rho_{i}}}\deg(P_{ij})$,

we obtain that all nonzero terms $\overline{P}_{ij}(\underline{x})\hskip 2.0pt\overline{M}({\underline{\lambda}}^{\ast},\underline{x})^{j}$ with $j\in\{1,\ldots,\rho_{i}\}$ are of different degrees: otherwise, for two integers $j,k\in\{1,\ldots,\rho_{i}\}$ with $k>j$, we would have the following, where $\delta=\deg(\overline{M}({\underline{\lambda}}^{\ast},\underline{x}))$:

$\displaystyle\max_{\genfrac{}{}{0.0pt}{}{1\leqslant i\leqslant s}{1\leqslant h\leqslant\rho_{i}}}\deg({P}_{ih})\geqslant\deg(\overline{P}_{ij})-\deg(\overline{P}_{ik})=(k-j)\delta\geqslant\delta$,

which contradicts the preceding inequalities. It follows that the left-hand side of (2) is of degree $\geqslant\deg(\overline{M}({\underline{\lambda}}^{\ast},\underline{x}))$. But then the following inequality (using again assumption (1)):

$\displaystyle\deg(\overline{M}(\underline{\lambda}^{\ast},\underline{x}))\geqslant\min(\deg(Q_{1}),\deg(Q_{2}))>\max_{1\leqslant i\leqslant s}\deg(P_{i0}(\underline{x}))$.

contradicts identity (2). ∎

Proof of Theorem 1.1 and Theorem 1.2.

From Theorem 4.6, the assumption on $R$ in Theorem 1.1 implies that of Theorem 5.1(a), and $R=k[u]$ in both Theorem 1.2 and Theorem 5.1(b). Theorems 1.1 and 1.2 then correspond to the special case of Theorem 5.1 for which, for a given $\underline{d}\in(\mathbb{N}^{\ast})^{n}$, the $Q_{i}$ are all the monomials $Q_{0},Q_{1},\ldots,Q_{N_{\underline{d}}}$ in ${\mathcal{P}ol}_{k,n,\underline{d}}$ and $Q_{1}$, $Q_{2}$ are monomials of degree $d_{1}+\cdots+d_{n}$ and $d_{1}+\cdots+d_{n}-1$. Assumption on $d_{1},\ldots,d_{n}$ in Theorems 1.1 and 1.2 guarantees assumption (1) of Theorem 5.1. ∎

Remark 5.2*.*

The proof shows that Theorem 1.1 holds under the more general assumption that $R$ is a UFD, a Hilbertian ring and $K$ is imperfect if of characteristic $p>0$. We note that there exist UFD with a Hilbertian fraction field satisfying the imperfectness assumption but not Hilbertian as a ring, e.g. the ring $\mathbb{C}[[u_{1},\ldots,u_{n}]]$ of formal power series with $n\geqslant 2$ [FJ08, Example 15.5]. It is unclear whether Theorem 1.1 holds for these rings.

5.2. The multivariable Schinzel hypothesis

Theorem 5.1 offers more flexibility than Theorem 1.1 and Theorem 1.2. Instead of taking for $Q_{0},\ldots,Q_{\ell}$ all the monomials in ${\mathcal{P}ol}_{k,n,\underline{d}}$, one may want to work with a proper subset of them and construct irreducible polynomials of the form $P_{i}(\underline{x},M(\underline{x}))$ with some of the coefficients in $M(\underline{x})$ equal to [math].

In this manner one can extend Theorems 1.1 and 1.2 to the situation that $P_{1},\ldots,P_{s}$ are polynomials in $m$ variables $y_{1},\ldots,y_{m}$.

Let $R$ be a UFD with fraction field a field $K$ with the product formula, imperfect if $K$ is of characteristic $p>0$. Let $\underline{x}=(x_{1},\ldots,x_{n})$ ($n\geqslant 1$) and $\underline{y}=(y_{1},\ldots,y_{m})$ ($m\geqslant 1$) be two tuples of indeterminates.

Theorem 5.3**.**

Let $\underline{P}=\{P_{1},\ldots,P_{s}\}$ be a set of polynomials, irreducible in $R[\underline{x},\underline{y}]$ and of degree $\geqslant 1$ in $\underline{y}$. Let ${\mathcal{I}rr}_{n}(R,\underline{P})$ be the set of all $m$-tuples $\underline{M}=(M_{1},\ldots,M_{m})\in R[\underline{x}]^{m}$ such that $P_{i}(\underline{x},\underline{M}(\underline{x}))$ is irreducible in $R[\underline{x}]$, $i=1,\ldots,s$. For every $\underline{d}\in(\mathbb{N}^{\ast})^{n}$ such that

$D:=\displaystyle d_{1}+\cdots+d_{n}\geqslant\max_{1\leqslant i\leqslant s}(\deg(P_{i})+2)$,

the set ${\mathcal{I}rr}_{n}(R,\underline{P})$ is Zariski-dense in ${\mathcal{P}ol}_{R,n,\underline{d}}\times\cdots\times{\mathcal{P}ol}_{R,n,D^{m-1}\underline{d}}$.

The proof is an easy induction left to the reader: use Theorem 5.1 to successively specialize in $R[\underline{x}]$ the indeterminates $y_{1},\ldots,y_{m}$.

5.3. Proof of Corollary 1.5 (Goldbach)

Fix an integral domain $R$ as in Theorem 1.1, an integer $n\geqslant 1$ and a nonconstant polynomial ${\mathcal{Q}}\in R[\underline{x}]$.

Let $\underline{P}=\{P_{1},P_{2}\}$ with $P_{1}=-y$ and $P_{2}=y+{\mathcal{Q}}$. We will proceed as in Theorem 5.1 but with only two monomials $Q_{0},Q_{1}$ (so $\ell=1$) and without assuming condition (1) from §5.1.

Assume that we are not in the case $n=1=\deg({\mathcal{Q}})$; this case is dealt with separately. Let $Q_{\infty}$ be a monic nonconstant monomial appearing in ${\mathcal{Q}}$ with a nonzero coefficient. Denote this coefficient by $q_{\infty}$. Let $Q_{1}$ be a nonconstant monomial distinct from $Q_{\infty}$ and of degree $\deg(Q_{1})\leqslant\deg({\mathcal{Q}})$. Denote the coefficient of $Q_{0}=1$ in ${\mathcal{Q}}$ by $q_{0}$ (the constant coefficient).

As in the proof of Theorem 5.1, Proposition 4.2 provides nonzero $\lambda_{0}^{\ast},\lambda_{1}^{\ast}$ in $R$ satisfying the following: for $M=\lambda_{0}^{\ast}+\lambda_{1}^{\ast}Q_{1}$, both $M$ and $M+{\mathcal{Q}}$ are irreducible in $K[\underline{x}]$, $\lambda_{0}^{\ast}\equiv 1-q_{0}\hskip 2.84526pt[{\rm mod}\hskip 2.84526ptq_{\infty}]$ and $\lambda_{1}^{\ast}\equiv 1\hskip 2.84526pt[{\rm mod}\hskip 2.84526pt\lambda_{0}^{\ast}]$ (the elements $q_{\infty},\lambda_{0}^{\ast},\lambda_{1}^{\ast}$ play the respective roles of $\lambda_{0}^{\ast},\lambda_{1}^{\ast},\lambda_{2}^{\ast}$ from Proposition 4.2).

To conclude, it suffices to show that $M$ and $M+{\mathcal{Q}}$ are primitive. As $\lambda_{0}^{\ast}$ and $\lambda_{1}^{\ast}$ are relatively prime, $M$ is primitive. As for $M+{\mathcal{Q}}$, it follows from this: the coefficients of $Q_{\infty}$ and $Q_{0}$ in $M+{\mathcal{Q}}$ are relatively prime. Indeed the former is $q_{\infty}$ and the latter is $\lambda_{0}^{\ast}+q_{0}$ which is congruent to $1$ modulo $q_{\infty}$.

Finally, in the case $n=1=\deg({\mathcal{Q}})$, write ${\mathcal{Q}}=q_{1}x+q_{0}$. We can take:

$\displaystyle\left\{\begin{matrix}&\hbox{\rm if}\ q_{1}\not=1\hfill&{\mathcal{Q}}=[x+(q_{0}-1)]+[(q_{1}-1)x+1]\hfill\cr&\hbox{\rm if}\ q_{1}\not=-1\hfill&{\mathcal{Q}}=[-x+(q_{0}-1)]+[(q_{1}+1)x+1]\hfill\cr&\hbox{\rm if}\ q_{1}=1=-1\hfill&{\mathcal{Q}}=[rx+(rq_{0}+1)]+[(r+1)x+(rq_{0}+q_{0}+1)]\hfill\cr&\hfill&\hskip 5.69054pt\hbox{with}\ r\in R\setminus\{0,1\}.\hfill\cr\end{matrix}\right.$

The more specific conclusion, alluded to in §1.4, that one can further take $\deg(Q_{1})=1$ if $R=K$ is a Hilbertian field, or if $R=K$ is an infinite field and $n\geqslant 2$, can be obtained from similar arguments but using the Addendum to Theorem 1.1 (in §2) and Theorem 1.4 instead of Theorem 5.1.

5.4. Proof of Theorem 1.3

Retain the notation from §5.1 but consider the degree $1$ case. That is, we have, for $i=1,\ldots,s$:

$\left\{\begin{matrix}P_{i}=A_{i}(\underline{x})+B_{i}(\underline{x})\hskip 1.0pty\hfill\\ F_{i}({\underline{\lambda}},\underline{x})=A_{i}(\underline{x})+B_{i}(\underline{x})\left(\sum_{j=0}^{\ell}\lambda_{j}Q_{j}(\underline{x})\right).\\ \end{matrix}\right.$

Assume further that the polynomials $Q_{i}$ are the monomials $Q_{0},Q_{1},\ldots,Q_{N_{\underline{d}}}$ in ${\mathcal{P}ol}_{k,n,\underline{d}}$ for some $\underline{d}\in(\mathbb{N}^{\ast})^{n}$, with as before $Q_{0}=1$ and $Q_{1}$ and $Q_{2}$ monomials of degree $d_{1}+\cdots+d_{n}$ and $d_{1}+\cdots+d_{n}-1$.

Lemma 5.4**.**

If as above $\deg_{y}(P_{1})=\ldots=\deg_{y}(P_{s})=1$, then the Hilbert subset $H_{K}(F_{1},\ldots,F_{s})\subset K^{N_{\underline{d}}+1}$ contains a separable Hilbert subset.

Proof of Lemma 5.4.

Fix $D>\max_{{{1\leqslant j\leqslant n}\atop{1\leqslant i\leqslant s}}}\deg_{x_{j}}(F_{i})$ and consider the Kronecker substitution:

$S_{D}:{\mathcal{P}ol}_{K(\underline{\lambda}),n,\underline{D}}\rightarrow{\mathcal{P}ol}_{K(\underline{\lambda}),1,D^{n}}$,   with $\underline{D}=(D,\ldots,D)$,

mapping $x_{j}$ to $x^{D^{j-1}}$, $j=1,\ldots,n$ (introduced in §4.2). Fix $i\in\{1,\ldots,s\}$. From Proposition 4.3, there exist a finite set ${\mathcal{S}}(F_{i})$ of irreducible polynomials in $K[\underline{\lambda}][x]$ of degree $\geqslant 1$ in $x$ and a nonzero polynomial $\varphi_{i}\in K[\underline{\lambda}]$ such that the Hilbert subset ${H}_{K}(F_{i})\subset K^{N_{\underline{d}}+1}$ contains the Hilbert subset $H_{K}({\mathcal{S}}(F_{i});\varphi_{i})$. Furthermore, one can take for ${\mathcal{S}}(F_{i})$ the set of irreducible divisors in $K[\underline{\lambda}][x]$ of the following polynomial (in which $M_{\underline{d}}=\sum_{h=0}^{N_{{\underline{d}}}}\lambda_{h}\hskip 2.0ptQ_{h}$):

$\displaystyle S_{D}(A_{i}+B_{i}M_{\underline{d}})=S_{D}(A_{i})+S_{D}(B_{i})\sum_{h=0}^{N_{{\underline{d}}}}\lambda_{h}\hskip 2.0ptS_{D}(Q_{h})$.

The polynomials $S_{D}(Q_{h})$ are distinct monomials in $x$ (up to multiplicative constants in $K^{\times}$): this indeed follows from the fact that two different integers between [math] and $D^{n-1}-1$ have different $D$-adic expansions $a_{1}+a_{2}D+\cdots+a_{n-1}D^{n-2}$ with $0\leqslant a_{j}\leqslant D-1$, $j=1,\ldots,n-1$.

Note that $S_{D}(A_{i})$ and $S_{D}(B_{i})$ may not be relatively prime (take for example $A_{i}=x_{2}-1$ and $B_{i}=x_{3}-1$) and so Lemma 2.1 cannot be used directly. Denote the gcd of $S_{D}(A_{i})$ and $S_{D}(B_{i})$ by $\Delta\in K[x]$. Conclude from Lemma 2.1 that the polynomial

$\displaystyle f_{i}:=\frac{S_{D}(A_{i}+B_{i}M_{\underline{d}})}{\Delta}=\frac{S_{D}(A_{i})}{\Delta}+\frac{S_{D}(B_{i})}{\Delta}\sum_{h=0}^{N_{\underline{d}}}\lambda_{h}\hskip 2.0ptS_{D}(Q_{h})$

is irreducible in $\overline{K}[\underline{\lambda},x]$. Since $\Delta\in K[x]$, its irreducible factors $f$ in $K[\underline{\lambda},x]$ are in fact in $K[x]$, and so satisfy $H_{K}(f)=K^{N_{\underline{d}}+1}$. Conclude that one can take ${\mathcal{S}}(F_{i})=\{f_{i}\}$ where $f_{i}$ is the polynomial displayed above.

The polynomial $f_{i}$ has an additional property: it is separable in $x$. Indeed, if $p>0$, not all exponents of $x$ in $f_{i}$ are divisible by $p$ (note that $\sum_{h=0}^{N_{{\underline{d}}}}\lambda_{h}\hskip 2.0ptS_{D}(Q_{h})$ is the generic polynomial in one variable of degree $D^{n}-1$).

We have thus proved that the Hilbert subset $H_{K}(F_{1},\ldots,F_{s})\subset K^{N_{\underline{d}}+1}$ contains the separable Hilbert subset $H_{K}(f_{1},\ldots,f_{s};\varphi_{1}\cdots\varphi_{s})$. ∎

Proof of Theorem 1.3.

The statement is about polynomials in at least $2$ variables that are denoted $x_{1},\ldots,x_{n}$ there. For consistency with the previous notation, we relabel them here $u,x_{1},\ldots,x_{n}$, with $n\geqslant 1$. Set $R=k[u]$ and view $k[u,x_{1},\ldots,x_{n}]$ as $R[\underline{x}]$.

Up to adding it to the given list $(A_{1},B_{1}),\ldots,(A_{s},B_{s})$ of couples of relatively prime polynomials in $R[\underline{x}]$, one may assume that the couple $(1,0)$ is in this list; this will guarantee that the desired polynomial $M$ is itself irreducible in $R[\underline{x}]$ as requested.

With the notation from this subsection, Lemma 5.4 gives that the Hilbert subset $H_{K}(F_{1},\ldots,F_{s})\subset K^{N_{\underline{d}}+1}$ contains a separable Hilbert subset, say $H_{K}(f_{1},\ldots,f_{s};\varphi)$. From the separable case of Theorem 4.8, there is an integer $d_{0}$ such that for every integer $\delta\geqslant d_{0}$, $H_{K}(f_{1},\ldots,f_{s};\varphi)$ contains a tuple $\underline{\lambda}^{\ast}\in R^{N_{\underline{d}}+1}$ such that $\lambda_{1}^{\ast}$ and $\lambda_{2}^{\ast}$ are relatively prime in $R$ and $\deg_{u}(M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x}))=\delta$. We have a fortiori $\underline{\lambda}^{\ast}\in H_{K}(F_{1},\ldots,F_{s})\subset K^{N_{\underline{d}}+1}$:

$F_{i}(\underline{\lambda}^{\ast},\underline{x})=A_{i}(\underline{x})+B_{i}(\underline{x})M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x})$ is irreducible in $K[\underline{x}]$, $i=1,\ldots,s$.

Assume $d_{0}$ large enough so that, if $d_{i}\geqslant d_{0}$, $i=1,\ldots,n$, then

$\displaystyle d_{1}+\cdots+d_{n}-1>\max_{i=1,\ldots,s}\max(\deg(A_{i}),\deg(B_{i}))$.

Irreduciblility of each $A_{i}(\underline{x})+B_{i}(\underline{x})M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x})$ in $R[\underline{x}]$ is deduced by proving it is primitive from $\lambda_{1}^{\ast}$, $\lambda_{2}^{\ast}$ being relatively prime as in the proof of Theorem 5.1.

Finally, up to multiplying $\varphi$ by the coordinate $\lambda_{h}$ corresponding to the monomial $x_{1}^{d_{1}}\cdots x_{n}^{d_{n}}$, one guarantees $\deg_{x_{i}}(M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x}))=d_{i}$, $i=1,\ldots,n$. This completes the proof: $M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x})$ is the requested polynomial. ∎

Remark 5.5*.*

Lemma 5.4 also shows that the degree $1$ case of the Schinzel hypothesis holds when $R$ is a Hilbertian field (totally Hilbertian is not needed), thus completing the proof of the addendum to Theorem 1.1 in situation (b).

5.5. Proof of Corollary 1.6

Assume $n\geqslant 2$, fix an arbitrary field $k$, a subset ${\mathcal{S}}=\{a_{1},\ldots,a_{t}\}\subset k$, $a_{0}\in\overline{k}\setminus{\mathcal{S}}$, separable over $k$, and $V\in k[\underline{x}]$, $V\not=0$. We will show this more precise version of Corollary 1.6.

Corollary 1.6 (explicit form). Let $w_{0},\ldots,w_{t}\in k[\underline{x}]$ be $t+1$ nonzero polynomials with $w_{0}=1$. Assume that $(w_{i})+(w_{j})=k[\underline{x}]$ for $i\not=j$ and each $w_{i}$ is relatively prime to $V$. For all suitably large integers $d_{1},\ldots,d_{n}$ (larger than some $d_{0}$ depending on ${\mathcal{S}},a_{0},V,w_{1},\ldots,w_{t}$), there is a polynomial $U\in k[\underline{x}]$ such that these three conclusions hold:

\begin{matrix}&\hbox{{\rm(a)} U-a_{i}V=w_{i}H_{i}$with$H_{i}\in k[\underline{x}]$irreducible in$k(a_{0})[\underline{x}]}\hfill\\ &\hbox{\hskip 99.58464ptand not dividing w_{i}$,$i=1,\ldots,t, }\hfill\\ &\hbox{{\rm(b)} \deg(U-a_{0}V)=\max(\deg(U),\deg(V)),}\hfill\\ &\hbox{{\rm(c)} \deg_{x_{i}}(U)=d_{i}$,$i=1,\ldots,n.}\hfill\\ \end{matrix}**

In order to obtain the version of Corollary 1.6 from §1, it suffices to choose $w_{1},\ldots,w_{t}$ as in the statement but not in $k$ and

$d_{1}>\max(\deg(V),\deg(w_{1}),\ldots,\deg(w_{t}))$.

It then follows from $\deg(U)\geqslant\deg_{x_{1}}(U)=d_{1}$ (using (c)) that $\deg(U-a_{i}V)=\deg(U)$, and next from (a) that $U-a_{i}V$ is reducible, $i=1,\ldots,t$.

Remark 5.6*.*

The assumption $(w_{i})+(w_{j})=k[\underline{x}]$ is necessary when $V=1$: if we have $U-a_{i}V=w_{i}H_{i}$ and $U-a_{j}V=w_{j}H_{j}$ for two distinct indices $i,j$, then $w_{i}H_{i}-w_{j}H_{j}=(a_{j}-a_{i})V$.

Proof.

As $(w_{i})+(w_{j})=k[\underline{x}]$, $i\not=j$, the Chinese Remainder Theorem may be used to conclude that there is a polynomial $U_{0}\in k[\underline{x}]$ such that

$U_{0}-a_{i}V=w_{i}p_{i}$ with $p_{i}\in k[\underline{x}]$, $i=1,\ldots,t$.

As $w_{0}=1$, we also have $U_{0}-a_{0}V=w_{0}p_{0}$ for some $p_{0}$, but here $p_{0}$ is in $k(a_{0})[\underline{x}]$. Furthermore the polynomials $U\in k(a_{0})[\underline{x}]$ satisfying the same $(t+1)$ conditions are of the form

$\displaystyle U(\underline{x})=U_{0}(\underline{x})+M(\underline{x})\prod_{i=0}^{t}w_{i}(\underline{x})$

for some $M\in k(a_{0})[\underline{x}]$. For such a polynomial $U$, we have

$U-a_{i}V=w_{i}\hskip 2.0pt(p_{i}+M\hskip 1.0pt\prod_{j\not=i}w_{i}(\underline{x})),\hskip 5.69054pti=0,\ldots,t$.

Up to changing $U_{0}$, we may assume that $p_{0},\ldots,p_{t}$ are nonzero.

For each $i=0,\ldots,t$, the polynomials $A_{i}=p_{i}$ and $B_{i}=\prod_{j\not=i}w_{i}(\underline{x})$ are relatively prime in $k(a_{0})[\underline{x}]$. Namely if $\pi\in k(a_{0})[\underline{x}]$ is a common irreducible divisor in $k(a_{0})[\underline{x}]$ of these two polynomials, then $\pi$ divides $p_{i}$ and $\pi$ divides $w_{j}$ for some $j\not=i$ and hence, $\pi$ is a common divisor of $U_{0}-a_{i}V$ and $U_{0}-a_{j}V$. Therefore $\pi$ divides $V$ and $w_{j}$, which contradicts the assumption $(V,w_{j})=1$.

Set $R=k(a_{0})[x_{n}]$, $K=k(a_{0})(x_{n})$, $\underline{x}=(x_{1},\ldots,x_{n-1})$ and, for $\underline{d}\in(\mathbb{N}^{\ast})^{n-1}$ and $i=0,\ldots,t$,

$\left\{\begin{matrix}P_{i}=A_{i}(\underline{x})+B_{i}(\underline{x})\hskip 1.0pty\hfill\\ F_{i}({\underline{\lambda}},\underline{x})=A_{i}(\underline{x})+B_{i}(\underline{x})\left(\sum_{j=0}^{N_{\underline{d}}}\lambda_{j}Q_{j}(\underline{x})\right).\\ \end{matrix}\right.$

As in the proof of Theorem 1.3, the Hilbert subset $H_{K}(F_{0},\ldots,F_{t})$ contains a separable Hilbert subset $H_{K}(f_{0},\ldots,f_{t},\varphi)$ with $f_{0},\ldots,f_{t}\in K[\underline{\lambda},x]$ of degree $\geqslant 1$ in $x$ and $\varphi\in K[\underline{\lambda}]$, $\varphi\not=0$.

The field extension $k(a_{0})/k$ is finite and separable. Setting $R_{0}=k[x_{n}]$ and $K_{0}=k(x_{n})$, so is the extension $K/K_{0}$. From [FJ08, Corollary 12.2.3], $H_{K}(f_{0},\ldots,f_{t},\varphi)$ contains a separable Hilbert subset ${\mathcal{H}}_{K_{0}}$ of $K_{0}^{N_{\underline{d}}+1}$.

Proceed as in the proof of Theorem 1.3 to conclude that there is an integer $d_{0}$ with the following property: if $\delta_{1}$, $\delta_{2},\ldots,\delta_{n}$ are integers $\geqslant d_{0}$, the Hilbert subset ${\mathcal{H}}_{K_{0}}$, and so the Hilbert subset $H_{K}(F_{0},\ldots,F_{t})$ too, contains a tuple $\underline{\lambda}^{\ast}=(\lambda_{0}^{\ast},\ldots,\lambda_{N_{\underline{d}}}^{\ast})\in R_{0}^{N_{\underline{d}}+1}$ such that $\lambda_{1}^{\ast}$ and $\lambda_{2}^{\ast}$ are irreducible in $R_{0}$, and $\deg_{x_{i}}(M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x}))=\delta_{i}$, $i=1,\ldots,n$. Choosing again for $Q_{1},Q_{2}$ monomials of respective degrees $d_{1}+\cdots+d_{n-1}$ and $d_{1}+\cdots+d_{n-1}-1$ and assuming $d_{0}$ suitably large, we obtain as for Theorem 1.3 that each of the polynomials

$F_{i}(\underline{x})=A_{i}(\underline{x})+B_{i}(\underline{x})M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x})$

is irreducible in $k(a_{0})[x_{n}][x_{1},\ldots,x_{n-1}]$, $i=0,\ldots,t$.

Up to increasing $d_{0}$, one can further guarantee that $\delta_{1},\ldots,\delta_{n}$ are large enough so that $\deg(M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x}))>\deg(U_{0})$ and $F_{i}$ does not divide $w_{i}$, $i=1,\ldots,s$. The polynomial

$U(\underline{x})=U_{0}(\underline{x})+M_{\underline{d}}(\underline{\lambda}^{\ast},\underline{x})\prod_{i=0}^{t}w_{i}(\underline{x})$

is in $k[\underline{x}]$ and satisfies the required condition $U-a_{i}V=w_{i}H_{i}$, with $H_{i}=F_{i}$ irreducible in $k(a_{0})[\underline{x}]$, $i=0,\ldots,t$. Up to replacing the Hilbert subset $H_{K}(f_{0},\ldots,f_{t},\varphi)$ by a Zariski open subset of it, one can also request that $\deg(U-a_{0}V)=\max(\deg(U),\deg(V))$. Finally $\deg_{x_{i}}(U)=\delta_{i}+\sum_{j=1}^{t}\deg_{x_{i}}(w_{j})$ can be taken to be any given suitably large integer $d_{i}$, $i=1,\ldots,n$. ∎