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


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.


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 \(b=\mathrm{d}(g)\).



(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 \(\text{ Int }(\mathbb{Z })\).



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 \(p\le \deg f\).


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 \(\mathrm{d}(\prod _{i\in J} f_j)=p\).


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.


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


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 \(n+2\).


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


(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.


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. \(\square \)


(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 \(m_1+1,\ldots , m_n+1\).


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. \(\square \)


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 \(xH(x)=G_1(x)\ldots G_{n+1}(x)\).


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 \(\text{ Int }(\mathbb{Z })\). \(\square \)


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 \(\mathbb Z \)


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.


  1. 1.
    Cahen, P.-J., Chabert, J.-L.: Elasticity for integral-valued polynomials. J. Pure Appl. Algebra 103, 303–311 (1995)MathSciNetzbMATHCrossRefGoogle Scholar
  2. 2.
    Cahen, P.-J., Chabert, J.-L.: Integer-valued polynomials. Mathematical Surveys and Monographs, vol. 48. American Mathematical Society, Providence (1997)Google Scholar
  3. 3.
    Chapman, S.T., McClain, B.A.: Irreducible polynomials and full elasticity in rings of integer-valued polynomials. J. Algebra 293, 595–610 (2005)MathSciNetzbMATHCrossRefGoogle Scholar
  4. 4.
    Frei, Ch., Frisch, S.: Non-unique factorization of polynomials over residue class rings of the integers. Commun. Algebra 39, 1482–1490 (2011)MathSciNetzbMATHCrossRefGoogle Scholar
  5. 5.
    Geroldinger, A., Halter-Koch, F.: Non-unique factorizations. Pure and Appl. Mathematics, vol. 278. Chapman & Hall/CRC, Boca Raton (2006)Google Scholar
  6. 6.
    Kainrath, F.: Factorization in Krull monoids with infinite class group. Colloq. Math. 80, 23–30 (1999)MathSciNetzbMATHGoogle Scholar

Copyright information

© The Author(s) 2013

Open AccessThis article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.

Authors and Affiliations

  1. 1.Institut für Mathematik ATechnische Universität GrazGrazAustria

Personalised recommendations