Monatshefte für Mathematik

, Volume 171, Issue 3–4, pp 341–350

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

• Sophie Frisch
Open Access
Article

## 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 $$f$$.

## Mathematics Subject Classification (2000)

Primary 13A05 Secondary 13B25 13F20 20M13 11C08

## 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, 2, 3]. The reader familiar with integer-valued polynomials is encouraged to skip to Sect. 3.

### Definition

For $$f\in \mathbb{Z }[x]$$,
1. (i)

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

2. (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

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}

2. (ii)

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

3. (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

1. (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 $$\gcd (a,b)=1$$.

2. (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 $$b_1=\mathrm{d}(\prod _{i\in J} g_i)$$, $$b_2=\mathrm{d}(\prod _{i\in I\setminus J} g_i)$$.

3. (iii)

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

4. (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

1. (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.

2. (ii)

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

3. (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.

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