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Monatshefte für Mathematik

, Volume 171, Issue 3–4, pp 341–350 | Cite as

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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Authors and Affiliations

  1. 1.Institut für Mathematik ATechnische Universität GrazGrazAustria

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