A construction of integer-valued polynomials with prescribed sets of lengths of factorizations

Abstract

For an arbitrary finite non-empty set $S$ of natural numbers greater $1$, we construct $f\in \text{ Int }(\mathbb{Z })=\{g\in \mathbb{Q }[x]\mid g(\mathbb{Z })\subseteq \mathbb{Z }\}$ such that $S$ is the set of lengths of $f$, i.e., the set of all $n$ such that $f$ has a factorization as a product of $n$ irreducibles in $\text{ Int }(\mathbb{Z })$. More generally, we can realize any finite non-empty multi-set of natural numbers greater 1 as the multi-set of lengths of the essentially different factorizations of $f$.

1 Introduction

Non-unique factorization has long been studied in rings of integers of number fields, see the monograph of Geroldinger and Halter-Koch [5]. More recently, non-unique factorization in rings of polynomials has attracted attention, for instance in $\mathbb{Z }_{p^n}[x]$, cf. [4], and in the ring of integer-valued polynomials $\text{ Int }(\mathbb{Z })=\{g\in \mathbb{Q }[x]\mid g(\mathbb{Z })\subseteq \mathbb{Z }\}$ (and its generalizations) [1, 3].

We show that every finite set of natural numbers greater $1$ occurs as the set of lengths of factorizations of an element of $\text{ Int }(\mathbb{Z })$ (Theorem 9 in Sect. 4).

Our proof is constructive, and allows multiplicities of lengths of factorizations to be specified. For example, given the multiset {2,2,2,5,5}, we construct a polynomial that has three different factorizations into 2 irreducibles and two different factorizations into 5 irreducibles, and no other factorizations. Perhaps a quick review of the vocabulary of factorizations is in order:

Notation and Conventions

$R$ denotes a commutative ring with identity. An element $r\in R$ is called irreducible in $R$ if $r$ is a non-zero non-unit such that $r=ab$ with $a,b\in R$ implies that $a$ or $b$ is a unit. A factorization of $r$ in $R$ is an expression $r=s_1\ldots s_n$ of $r$ as a product of irreducible elements in $R$. The number $n$ of irreducible factors is called the length of the factorization. The set of lengths $\mathcal{L }(r)$ of $r\in R$ is the set of all natural numbers $n$ such that $r$ has a factorization of length $n$ in $R$.

$R$ is called atomic if every non-zero non-unit of $R$ has a factorization in $R$.

If $R$ is atomic, then for every non-zero non-unit $r\in R$ the elasticity of $r$ is defined as

$$\begin{aligned} \rho (r)=\sup \left\{ {m\over n}\mid m,n\in \mathcal{L }(r)\right\} \end{aligned}$$

and the elasticity of $R$ is $\rho (R)=\sup _{r\in R^{\prime }}(\rho (r))$, where $R^{\prime }$ is the set of non-zero non-units of $R$. An atomic domain $R$ is called fully elastic if every rational number greater than $1$ occurs as $\rho (r)$ for some non-zero non-unit $r\in R$.

Two elements $r,s\in R$ are called associated in $R$ if there exists a unit $u\in R$ such that $r=us$. Two factorizations of the same element $r=r_1\cdot \ldots \cdot r_m=s_1\cdot \ldots \cdot s_n$ are called essentially the same if $m=n$ and, after re-indexing the $s_i, r_j$ is associated to $s_j$ for $1\le j\le m$. Otherwise, the factorizations are called essentially different.

2 Review of factorization of integer-valued polynomials

In this section we recall some elementary properties of $\text{ Int }(\mathbb{Z })$ and the fixed divisor $\mathrm{d}(f)$, to be found in [1–3]. The reader familiar with integer-valued polynomials is encouraged to skip to Sect. 3.

Definition

For $f\in \mathbb{Z }[x]$,

• (i) the content $\mathrm{c}(f)$ is the ideal of $\mathbb{Z }$ generated by the coefficients of $f$,

• (ii) the fixed divisor $\mathrm{d}(f)$ is the ideal of $\mathbb{Z }$ generated by the image $f(\mathbb{Z })$.

By abuse of notation we will identify the principal ideals $\mathrm{c}(f)$ and $\mathrm{d}(f)$ with their non-negative generators. Thus, for $f=\sum _{k=0}^n a_kx^k\in \mathbb{Z }[x]$,

$$\begin{aligned} \mathrm{c}(f)=\gcd {(a_k\mid k=0,\ldots , n)} \quad \text{ and } \quad \mathrm{d}(f)=\gcd {(f(c)\mid c\in \mathbb{Z })}. \end{aligned}$$

A polynomial $f\in \mathbb{Z }[x]$ is called primitive if $\mathrm{c}(f)=1$.

Recall that a primitive polynomial $f\in \mathbb{Z }[x]$ is irreducible in $\mathbb{Z }[x]$ if and only if it is irreducible in $\mathbb{Q }[x]$. Similarly, $f\in \mathbb{Z }[x]$ with $\mathrm{d}(f)=1$ is irreducible in $\mathbb{Z }[x]$ if and only if it is irreducible in $\text{ Int }(\mathbb{Z })$.

We denote $p$-adic valuation by $v_p$. Almost everything that we need to know about the fixed divisor follows immediately from the fact that

$$\begin{aligned} v_p(\mathrm{d}(f))=\min _{c\in \mathbb{Z }}(v_p(f(c))). \end{aligned}$$

In particular, it is easy to deduce that for any $f,g\in \mathbb{Z }[x]$,

$$\begin{aligned} \mathrm{d}(f)\mathrm{d}(g)\,\big \vert \,\mathrm{d}(fg). \end{aligned}$$

Unlike $\mathrm{c}(f)$, which satisfies $\mathrm{c}(f)\mathrm{c}(g)=\mathrm{c}(fg)$, $\mathrm{d}(f)$ is not multiplicative: $\mathrm{d}(f)\mathrm{d}(g)$ is in general a proper divisor of $\mathrm{d}(fg)$.

Remark 1

• (i) Every non-zero polynomial $f\in \mathbb{Q }[x]$ can be written in a unique way as

  $$\begin{aligned} f(x)={{a g(x)}\over b} \quad \text{ with }\quad g\in \mathbb{Z }[x],\; c(g)=1,\quad a,b\in \mathbb{N },\; \gcd (a,b)=1. \end{aligned}$$

• (ii) When expressed as in (i), $f$ is in $\text{ Int }(\mathbb{Z })$ if and only if $b$ divides $\mathrm{d}(g)$.

• (iii) For non-constant $f\in \text{ Int }(\mathbb{Z })$ expressed as in (i) to be irreducible in $\text{ Int }(\mathbb{Z })$ it is necessary that $a=1$ and $b=\mathrm{d}(g)$.

Proof

(i) and (ii) are easy. Ad (iii). Note that the only units in $\text{ Int }(\mathbb{Z })$ are $\pm 1$. By (ii), $b$ divides $\mathrm{d}(g)$. Let $\mathrm{d}(g)=bc$. Then $f$ factors as $a\cdot c\cdot (g/bc)$, where $(g/bc)$ is non-constant and $ac$ is a unit only if $a=c=1$. $\square$

Remark 2

• (i) Every non-zero polynomial $f\in \mathbb{Q }[x]$ can be written in a unique way up to the sign of $a$ and the signs and indexing of the $g_i$ as

  $$\begin{aligned} f(x)={a\over b}\prod _{i\in I} g_i(x), \end{aligned}$$

  with $g_i$ primitive and irreducible in $\mathbb{Z }[x]$ for $i\in I$ (a finite set) and $a\in \mathbb{Z }$, $b\in \mathbb{N }$ with $\gcd (a,b)=1$.

• (ii) A non-constant polynomial $f\in \text{ Int }(\mathbb{Z })$ expressed as in (i) is irreducible in $\text{ Int }(\mathbb{Z })$ if and only if $a=\pm 1$, $b=\mathrm{d}(\prod _{i\in I} g_i)$, and there do not exist $\emptyset \ne J\subsetneq I$ and $b_1,b_2\in \mathbb{N }$ with $b_1b_2=b$ and $b_1=\mathrm{d}(\prod _{i\in J} g_i)$, $b_2=\mathrm{d}(\prod _{i\in I\setminus J} g_i)$.

• (iii) $\text{ Int }(\mathbb{Z })$ is atomic.

• (iv) Every non-zero non-unit $f\in \text{ Int }(\mathbb{Z })$ has only finitely many factorizations into irreducibles in $\text{ Int }(\mathbb{Z })$.

Proof

Ad (ii). If $f$ is irreducible, the conditions on $f$ follow from Remark 1 (ii) and (iii). Conversely, if the conditions hold, what chance does $f$ have to be reducible? By Remark 1 (ii), we cannot factor out a non-unit constant, because no proper multiple of $b$ divides $\mathrm{d}(\prod _{i\in I} g_i)$. Any non-constant irreducible factor would, by Remark 1 (iii), be of the kind $(\prod _{i\in J} g_i)/b_1$ with $b_1=\mathrm{d}(\prod _{i\in J} g_i)$, and its co-factor would be $(\prod _{i\in I\setminus J} g_i)/b_2$ with $b_1b_2=b$ and $b_2$ a divisor of $\mathrm{d}(\prod _{i\in I\setminus J} g_i)$. Also, $b_2$ could not be a proper divisor of $\mathrm{d}(\prod _{i\in I\setminus J} g_i)$, because otherwise $b_1b_2=b$ would be a proper divisor of $\prod _{i\in I} g_i$. So, the existence of a non-constant irreducible factor would imply the existence of $J$ and $b_1,b_2$ of the kind we have excluded.

Ad (iii). With $f(x)={{a g(x)}/ b}$, $g=\prod _{i\in I}g_i$ as in (i), $\mathrm{d}(g)=cb$ for some $c\in \mathbb{N }$, and $f(x)={{ac g(x)}/\mathrm{d}(g)}$ with $g(x)/\mathrm{d}(g)\in \text{ Int }(\mathbb{Z })$. We can factor $ac$ into irreducibles in $\mathbb{Z }$, which are also irreducible in $\text{ Int }(\mathbb{Z })$. Either $g(x)/\mathrm{d}(g)$ is irreducible, or (ii) gives an expression as a product of two non-constant factors of smaller degree. By iteration we arrive at a factorization of $g(x)/\mathrm{d}(g)$ into irreducibles.

Ad (iv). Let $f\in \text{ Int }(\mathbb{Z })=(ag(x)/b)$ with $g=\prod _{i\in I}g_i$ as in (i). Then all factorizations of $f$ are of the form, for some $c\in \mathbb{N }$ such that $bc$ divides $\mathrm{d}(g)$,

$$\begin{aligned} f= a_1\ldots a_n c_1\ldots c_m \prod _{j=1}^k {{\prod _{i\in I_j}g_i}\over {d_j}}, \end{aligned}$$

where $a=a_1\ldots a_n$ and $c=c_1\ldots c_m$ are factorizations into primes in $\mathbb{Z }, I=I_1\cup \ldots \cup I_k$ is a partition of $I$ into non-empty sets, $d_1\ldots d_k= bc$, $d_j=\mathrm{d}(\prod _{i\in I_j}g_i)$. There are only finitely many such expressions. $\square$

Remark 3

• (i) The binomial polynomials

  $$\begin{aligned} \left( \begin{array}{c} {x}\\ {n} \end{array}\right) ={{x(x-1)\ldots (x-n+1)}\over {n!}} \quad \text{ for }\quad n\ge 0 \end{aligned}$$

  are a basis of $\text{ Int }(\mathbb{Z })$ as a free $\mathbb{Z }$-module.

• (ii) $n!f\in \mathbb{Z }[x]$ for every $f\in \text{ Int }(\mathbb{Z })$ of degree at most $n$.

• (iii) Let $f\in \mathbb{Z }[x]$ primitive, $\deg f=n$ and $p$ prime. Then

  $$\begin{aligned} v_p(\mathrm{d}(f))\le \sum _{k\ge 1}\left[ {{n}\over {p^k}}\right] =v_p(n!). \end{aligned}$$

  In particular, if $p$ divides $\mathrm{d}(f)$ then $p\le \deg f$.

Proof

Ad (i). The binomial polynomials are in $\text{ Int }(\mathbb{Z })$ and they form a $\mathbb{Q }$-basis of $\mathbb{Q }[x]$. If a polynomial in $\text{ Int }(\mathbb{Z })$ is written as a $\mathbb{Q }$-linear combination of binomial polynomials then an easy induction shows that the coefficients must be integers. (ii) follows from (i).

Ad (iii). Let $g=f/d(f)$. Then $g\in \text{ Int }(\mathbb{Z })$ and $\mathrm{d}(f)\mathbb{Z }=(\mathbb{Z }[x]:_{\mathbb{Z }} g)$. Since $n!\in (\mathbb{Z }[x]:_{\mathbb{Z }} g)$ by (ii), $\mathrm{d}(f)$ divides $n!$ $\square$

3 Useful Lemmata

Lemma 4

Let $p$ be a prime, $I\ne \emptyset$ a finite set and for $i\in I$, $f_i\in \mathbb{Z }[x]$ primitive and irreducible in $\mathbb{Z }[x]$ such that $\mathrm{d}(\prod _{i\in I} f_i)=p$. Let

$$\begin{aligned} g(x)={{\prod _{i\in I} f_i}\over p}. \end{aligned}$$

Then every factorization of $g$ in $\mathrm{Int}(\mathbb{Z })$ is essentially the same as one of the following:

$$\begin{aligned} g(x)={{\prod _{j\in J} f_j}\over p}\cdot \prod _{i\in I\setminus J}f_i, \end{aligned}$$

where $J\subseteq I$ is minimal such that $\mathrm{d}(\prod _{i\in J} f_j)=p$.

Proof

Follows from Remark 1 (iii) and the fact that $\mathrm{d}(f)\mathrm{d}(h)$ divides $\mathrm{d}(fh)$ for all $f,h\in \mathbb{Z }[x]$. $\square$

The following two easy lemmata are constructive, since the Euclidean algorithm makes the Chinese Remainder Theorem in $\mathbb{Z }$ effective.

Lemma 5

For every prime $p\in \mathbb{Z }$, we can construct a complete system of residues mod $p$ that does not contain a complete system of residues modulo any other prime.

Proof

By the Chinese Remainder Theorem we solve, for each $k=1,\ldots , p$ the system of congruences $s_k= k$ mod $p$ and $s_k=1$ mod $q$ for every prime $q<p$.

Lemma 6

Given finitely many non-constant monic polynomials $f_i\in \mathbb{Z }[x]$, $i\in I$, we can construct monic irreducible polynomials $F_i\in \mathbb{Z }[x]$, pairwise non-associated in $\mathbb{Q }[x]$, with $\deg {F_i}=\deg {f_i}$, and with the following property:

Whenever we replace some of the $f_i$ by the corresponding $F_i$, setting $g_i=F_i$ for $i\in J$ ($J$ an arbitrary subset of $I$) and $g_i=f_i$ for $i\in I{\setminus } J$, then for all $K\subseteq I$,

$$\begin{aligned} \mathrm{d}\left( \prod _{i\in K} g_i\right) = \mathrm{d}\left( \prod _{i\in K} f_i\right) . \end{aligned}$$

Proof

Let $n=\sum _{i\in I}\deg f_i$. Let $p_1,\ldots , p_s$ be all the primes with $p_i\le n$, and set $\alpha _i=v_{p_i}(n!)$. Let $q>n$ be a prime. For each $i\in I$, we find by the Chinese Remainder Theorem the coefficients of a polynomial $\varphi _i\in (\prod _{k=1}^s p_k^{\alpha _k})\mathbb{Z }[x]$ of smaller degree than $f_i$, such that $F_i=f_i+\varphi _i$ satisfies Eisenstein’s irreducibility criterion with respect to the prime $q$. Then, with respect to some linear ordering of $I$, if $F_i$ happens to be associated in $\mathbb{Q }[x]$ to any $F_j$ of smaller index, we add a suitable non-zero integer divisible by $q^2\prod _{k=1}^s p_k^{\alpha _k}$ to $F_i$, to make $F_i$ non-associated in $\mathbb{Q }[x]$ to all $F_j$ of smaller index.

The statement about the fixed divisor follows, because for every $c\in \mathbb{Z }$ and every prime $p_i$ that could conceivably divide the fixed divisor,

$$\begin{aligned} \prod _{i\in K}(g_i(c))\equiv \prod _{i\in K}(f_i(c)) \quad \text{ mod }\; p_i^{\alpha _i}, \end{aligned}$$

where $p_i^{\alpha _i}$ is the highest power of $p_i$ that can divide the fixed divisor of any monic polynomial of degree at most $n$. $\square$

4 Constructing polynomials with prescribed sets of lengths

We precede the general construction by two illustrative examples of special cases, corresponding to previous results by Cahen, Chabert, Chapman and McClain.

Example 7

For every $n\ge 0$, we can construct $H\in \text{ Int }(\mathbb{Z })$ such that $H$ has exactly two essentially different factorizations in $\text{ Int }(\mathbb{Z })$, one of length 2 and one of length $n+2$.

Proof

Let $p>n+1, p$ prime. By Lemma 5 we construct a complete set $a_1,\ldots ,a_p$ of residues mod $p$ in $\mathbb{Z }$ that does not contain a complete set of residues mod any prime $q<p$. Let

$$\begin{aligned} f(x)\!=\!(x\!-\!a_2)(x\!-\!a_3)\ldots (x\!-\!a_{p})\quad \text{ and }\quad g(x)\!=\!(x-a_{n+2})(x\!-\!a_{n+3})\ldots (x\!-\!a_{p}). \end{aligned}$$

By Lemma 6, we construct monic irreducible polynomials $F,G\in \mathbb{Z }[x]$, not associated in $\mathbb{Q }[x]$, with $\deg F=\deg f$, $\deg G=\deg g$, such that any product of a selection of polynomials from $(x-a_{1}),\ldots ,(x-a_{n+1}), f(x), g(x)$ has the same fixed divisor as the corresponding product with $f$ replaced by $F$ and $g$ by $G$.

Let

$$\begin{aligned} H(x)={{F(x)(x-a_{1})\ldots (x-a_{n+1})G(x)}\over p}. \end{aligned}$$

By Lemma 4, $H$ factors into two irreducible polynomials in $\text{ Int }(\mathbb{Z })$

$$\begin{aligned} H(x)=F(x)\cdot {{(x-a_{1})\ldots (x-a_{n+1})G(x)}\over p} \end{aligned}$$

or into $n+2$ irreducible polynomials in $\text{ Int }(\mathbb{Z })$

$$\begin{aligned} H(x)= {{F(x)(x-a_{1})}\over p}\cdot (x-a_2)(x-a_3)\ldots (x-a_{n+1})G(x). \end{aligned}$$

$\square$

Corollary

(Cahen and Chabert [1]) $\rho \,(\mathrm{Int}(\mathbb{Z }))=\infty$.

Example 8

For $1\le m\le n$, we can construct a polynomial $H\in \text{ Int }(\mathbb{Z })$ that has in $\text{ Int }(\mathbb{Z })$ a factorization into $m+1$ irreducibles and an essentially different factorization into $n+1$ irreducibles, and no other essentially different factorization.

Proof

Let $p>mn$ be prime, $s=p-mn$. By Lemma 5 we construct a complete system of residues $R$ mod $p$ that does not contain a complete system of residues for any prime $q<p$. We index $R$ as follows:

$$\begin{aligned} R=\{r(i,j)\mid 1\le i\le m,\> 1\le j\le n\}\cup \{b_1,\ldots , b_s\}. \end{aligned}$$

Let $b(x)=\prod _{k=1}^s (x-b_k)$. For $1\le i\le m$ let $f_i(x)=\prod _{k=1}^n (x-r(i,k))$ and for $1\le j\le n$ let $g_j(x)=\prod _{k=1}^m (x-r(k,j))$.

By Lemma 6, we construct monic irreducible polynomials $F_i, G_j\in \mathbb{Z }[x]$, pairwise non-associated in $\mathbb{Q }[x]$, such that the product of any selection of the polynomials $(x-b_1),\ldots ,(x-b_s), f_1,\ldots , f_m, g_1,\ldots , g_n$ has the same fixed divisor as the corresponding product in which $f_i$ has been replaced by $F_i$ and $g_j$ by $G_j$ for $1\le i\le m$ and $1\le j\le n$. Let

$$\begin{aligned} H(x)={1\over p}b(x)\prod _{i=1}^m F_i(x)\prod _{j=1}^n G_j(x), \end{aligned}$$

then, by Lemma 4, $H$ has a factorization into $m+1$ irreducibles

$$\begin{aligned} H(x)=F_1(x)\cdot \ldots \cdot F_m(x)\cdot {{b(x)G_1(x)\cdot \ldots \cdot G_n(x)}\over p} \end{aligned}$$

and an essentially different factorization into $n+1$ irreducibles

$$\begin{aligned} H(x)={{b(x)F_1(x)\cdot \ldots \cdot F_m(x)}\over p}\cdot G_1(x)\cdot \ldots \cdot G_n(x) \end{aligned}$$

and no other essentially different factorization. $\square$

Corollary

(Chapman and McClain [3]) $\mathrm{Int}(\mathbb{Z })$ is fully elastic.

Theorem 9

Given natural numbers $1\le m_1\le \cdots \le m_n$, we can construct a polynomial $H\in \mathrm{Int}(\mathbb{Z })$ that has exactly $n$ essentially different factorizations into irreducibles in $\mathrm{Int}(\mathbb{Z })$, the lengths of these factorizations being $m_1+1,\ldots , m_n+1$.

Proof

Let $N=(\sum _{i=1}^n m_i)^2-\sum _{i=1}^n m_i^2$, and $p>N$ prime, $s=p-N$. By Lemma 5, we construct a complete system of residues $R$ mod $p$ that does not contain a complete system of residues for any prime $q<p$. We partition $R$ into disjoint sets $R=R_0\cup \{t_1,\ldots , t_s\}$ with $\left| R_0\right| = N$. The elements of $R_0$ are indexed as follows:

$$\begin{aligned} R_0=\{r(k, h, i, j)\mid 1\le k\le n,\> 1\le h\le m_k,\> 1\le i\le n,\> 1\le j\le m_i;\> i\ne k\}, \end{aligned}$$

meaning we arrange the elements of $R_0$ in an $m\times m$ matrix with $m=m_1+\cdots +m_n$, whose rows and columns are partitioned into $n$ blocks of sizes $m_1,\ldots , m_n$. Now $r(k, h, i, j)$ designates the entry in the $h$-th row of the $k$-th block of rows and the $j$-th column of the $i$-th block of columns. Positions in the matrix whose row and column are each in block $i$ are left empty: there are no elements $r(k, h, i, j)$ with $i=k$.

For $1\le k\le n, 1\le h\le m_k$, let $S_{k,h}$ be the set of entries in the $(k,h)$-th row:

$$\begin{aligned} S_{k,h}=\{r(k,h,i,j)\mid 1\le i\le n,\> i\ne k,\> 1\le j\le m_i\}. \end{aligned}$$

For $1\le i\le n, 1\le j\le m_i$, let $T_{i,j}$ be the set of elements in the $(i,j)$-th column:

$$\begin{aligned} T_{i,j}=\{r(k,h,i,j)\mid 1\le k\le n,\> k\ne i,\> 1\le h\le m_k\}. \end{aligned}$$

For $1\le k\le n, 1\le h\le m_k$, set

$$\begin{aligned} f^{(k)}_h(x)=\prod _{r\in S_{k,h}}(x-r)\cdot \prod _{r\in T_{k,h}}(x-r). \end{aligned}$$

Also, let $b(x)=\prod _{i=1}^s(x-t_i)$.

By Lemma 6, we construct monic irreducible polynomials $F^{(k)}_h$, pairwise non-associated in $\mathbb{Q }[x]$, with $\deg F^{(k)}_h =\deg f^{(k)}_h$, such that any product of a selection of polynomials from $(x-t_1),\ldots ,(x-t_s)$ and $f^{(k)}_h$ for $1\le k\le n$, $1\le h\le m_k$ has the same fixed divisor as the corresponding product in which the $f^{(k)}_h$ have been replaced by the $F^{(k)}_h$. Let

$$\begin{aligned} H(x)={1\over p}b(x)\prod _{k=1}^n\prod _{h=1}^{m_k} F^{(k)}_h(x). \end{aligned}$$

Then $\deg H = N+p$; and for each $i=1,\ldots , n$, $H$ has a factorization into $m_i+1$ irreducible polynomials in $\text{ Int }(\mathbb{Z })$:

$$\begin{aligned} H(x)=F^{(i)}_1(x)\cdot \ldots \cdot F^{(i)}_{m_i}(x)\cdot {{b(x)\prod _{k\ne i}\prod _{h=1}^{m_k}F^{(k)}_h(x)}\over p} \end{aligned}$$

These factorizations are essentially different, since the $F^{(i)}_j$ are pairwise non-associated in $\mathbb{Q }[x]$ and hence in $\text{ Int }(\mathbb{Z })$.

By Lemma 4, $H$ has no further essentially different factorizations. This is so because a minimal subset with fixed divisor $p$ of the polynomials $(x-t_i)$ for $1\le i\le s$ and $F^{(k)}_h$ for $1\le k\le n$, $1\le h\le m_k$ must consist of all the linear factors $(x-t_i)$ together with a minimal selection of $F^{(k)}_h$ such that all $r\in R_0$ occur as roots in the product of the corresponding $f^{(k)}_h$. For all linear factors $(x-r)$ with $r\in R_0$ to occur in a set of polynomials $f^{(k)}_{h}$, it must contain for all but one $k$ all $f^{(k)}_{h}$, $h=1,\ldots m_k$. If, for $i\ne k$, $f^{(k)}_{h}$ and $f^{(i)}_j$ are missing, then $r(k,h,i,j)$ and $r(i,j,k,h)$ do not occur among the roots of the polynomials $f^{(k)}_h$. A set consisting of all $f^{(k)}_{h}$ for $n-1$ different values of $k$, however, has the property that all linear factors $(x-r)$ for $r\in R_0$ occur. $\square$

Corollary

Every finite subset of $\mathbb{N }{\setminus } \{1\}$ occurs as the set of lengths of a polynomial $f\in \mathrm{Int}(\mathbb{Z })$.

5 No transfer homomorphism to a block-monoid

For some monoids, results like the above Corollary have been shown by means of transfer-homomorphisms to block monoids. For instance, by Kainrath [6], in the case of a Krull monoid with infinite class group such that every divisor class contains a prime divisor.

$\text{ Int }(\mathbb{Z })$, however, doesn’t admit this method: We will show a property of the multiplicative monoid of $\text{ Int }(\mathbb{Z })\setminus \{0\}$ that excludes the existence of a transfer-homomorphism to a block monoid.

Proposition 10

For every $n\ge 1$ there exist irreducible elements $H,G_1,\ldots ,G_{n+1}$ in $\mathrm{Int}(\mathbb{Z })$ such that $xH(x)=G_1(x)\ldots G_{n+1}(x)$.

Proof

Let $p_1<p_2<\cdots <p_n$ be $n$ distinct odd primes, $P=\{p_1,p_2,\ldots ,p_n\}$, and $Q$ the set of all primes $q\le p_n +n$. By the Chinese remainder theorem construct $a_1,\ldots , a_n$ with $a_i\equiv 0$ mod $p_i$ and $a_i\equiv 1$ mod $q$ for all $q\in Q$ with $q\ne p_i$. Similarly, construct $b_1,\ldots b_{p_n}$ such that, firstly, for all $p\in P$, $b_k\equiv k$ mod $p$ if $k\le p$ and $b_k\equiv 1$ mod $p$ if $k>p$ and, secondly, $b_k\equiv 1$ mod $q$ for all $q\in Q{\setminus } P$. So, for each $p_i\in P$, a complete set of residues mod $p_i$ is given by $b_1,\ldots b_{p_i},a_i$, while all remaining $a_j$ and $b_k$ are congruent to $1$ mod $p_i$. Also, all $a_j$ and $b_k$ are congruent to $1$ for all primes in $Q{\setminus } P$.

Set $f(x)=(x-b_1)\ldots (x-b_{p_n})$ and let $F(x)$ be a monic irreducible polynomial in $\mathbb{Z }[x]$ with $\deg F=\deg f$ such that the fixed divisor of any product of a selection of polynomials from $f(x),(x-a_1), \ldots , (x-a_n)$ is the same as the fixed divisor of the corresponding set of polynomials in which $f$ has been replaced by $F$. Such an $F$ exists by Lemma 6. Let

$$\begin{aligned} H(x)={{F(x)(x-a_1)\ldots (x-a_n)}\over {p_1\ldots p_n}}. \end{aligned}$$

Then $H(x)$ is irreducible in $\text{ Int }(\mathbb{Z })$, and

$$\begin{aligned} xH(x)={{xF(x)}\over {p_1\ldots p_n}}\cdot (x-a_1)\cdot \ldots \cdot (x-a_n), \end{aligned}$$

where $xF(x)/(p_1\ldots p_n)$ and, of course, $(x-a_1)$, $\ldots$, $(x-a_n)$, are irreducible in $\text{ Int }(\mathbb{Z })$. $\square$

Remark

Thanks to Roger Wiegand for suggesting an easier proof of Proposition 10: Using the well-known fact that the binomial polynomials ${x\atopwithdelims ()m}$ are irreducible in $\mathrm{Int}(\mathbb Z )$ for m > 0, it suffices to consider

$$x{{x-1}\atopwithdelims (){m-1}}=m{x\atopwithdelims ()m}$$

with m chosen to have exactly n prime factors in $\mathbb Z$

Remark

Thanks to Alfred Geroldinger for pointing this out: Proposition 10 implies that there does not exist a transfer-homomorphism from the multiplicative monoid $(\text{ Int }(\mathbb{Z }){\setminus } \{0\}, \cdot )$ to a block-monoid. (For the definition of block-monoid and transfer-homomorphism see [5], Def. 2.5.5 and Def. 3.2.1], respectively.)

This is so because, in a block-monoid, the length of factorizations of elements of the form $cd$ with $c$, $d$ irreducible, $c$ fixed, is bounded by a constant depending only on $c$, cf.  [5, Lemma 6.4.4]. More generally, applying [5, Lemma 3.2.2], one sees that every monoid that admits a transfer-homomorphism to a block-monoid has this property, in marked contrast to Proposition 10.