Monatsh. Math. 171 (2013) (3-4) 341–350

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

Sophie Frisch^{1}

Abstract. For an arbitrary finite set S of natural numbers greater 1, we construct f ∈ Int(ℤ) = {g ∈ ℚ[x] | g(ℤ) ⊆ 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 Int(ℤ). More generally, we can realize any finite multi-set of natural numbers greater 1 as the multi-set of lengths of the essentially different factorizations of f.

MSC 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 ℤ_{p^{n}}[x], cf. [4], and in the ring of integer-valued polynomials Int(ℤ) = {g ∈ ℚ[x] | g(ℤ) ⊆ 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 Int(ℤ) (Theorem 9 in section 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 ∈ R is called irreducible in R if r is a non-zero non-unit such that r = ab with a, b ∈ R implies that a or b is a unit. A factorization of r in R is an expression r = s_{1} … 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 L(r) of r ∈ 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 ∈ R the elasticity of r is defined

1 Supported by FWF (Austrian Science Fund) grant P23245 as

ρ(r) = sup\{\frac{m}{n} | m, n ∈ L(r)\}

and the elasticity of R is ρ(R) = sup_{r∈R′}(ρ(r)), where R′ 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 ρ(r) for some non-zero non-unit r ∈ R.

Two elements r, s ∈ R are called associated in R if there exists a unit u ∈ R such that r = us. Two factorizations of the same element r = r_{1} ·. . .·r_{m} = s_{1} ·. . .·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 ≤ j ≤ 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 Int(ℤ) and the fixed divisor d(f), to be found in [1], [2] and [3]. The reader familiar with integer-valued polynomials is encouraged to skip to section 3.

Definition. For f ∈ ℤ[x],

(i) the content c(f) is the ideal of ℤ generated by the coefficients of f, (ii) the fixed divisor d(f) is the ideal of ℤ generated by the image f(ℤ).

By abuse of notation we will identify the principal ideals c(f) and d(f) with their non-negative generators. Thus, for f = ∑_{k=0}^{n} a_{k}x^{k} ∈ ℤ[x],

c(f) = gcd (a_{k} | k = 0, …, n) and d(f) = gcd (f(c) | c ∈ ℤ).

A polynomial f ∈ ℤ[x] is called primitive if c(f) = 1.

Recall that a primitive polynomial f ∈ ℤ[x] is irreducible in ℤ[x] if and only if it is irreducible in ℚ[x]. Similarly, f ∈ ℤ[x] with d(f) = 1 is irreducible in ℤ[x] if and only if it is irreducible in Int(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

v_{p}(d(f)) = min_{c∈ℤ}(v_{p}(f(c))).

In particular, it is easy to deduce that for any f, g ∈ ℤ[x],

d(f)d(g) | d(fg).

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

Remark 1.

(i) Every non-zero polynomial f ∈ ℚ[x] can be written in a unique way as

f(x) = \frac{ag(x)}{b} with g ∈ ℤ[x], c(g) = 1, a, b ∈ ℕ, gcd(a, b) = 1.

(ii) When expressed as in (i), f is in Int(ℤ) if and only if b divides d(g). (iii) For non-constant f ∈ Int(ℤ) expressed as in (i) to be irreducible in Int(ℤ) it is

necessary that a = 1 and b = d(g).

Proof. (i) and (ii) are easy. Ad (iii). Note that the only units in Int(ℤ) are ±1. By (ii), b divides d(g). Let d(g) = bc. Then f factors as a · c · (g/bc), where (g/bc) is non-constant and ac is a unit only if a = c = 1. □

Remark 2.

(i) Every non-zero polynomial f ∈ 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

f(x) = \frac{a}{b} ∏_{i∈I} g_{i}(x),

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

(ii) A non-constant polynomial f ∈Int(ℤ) expressed as in (i) is irreducible in Int(ℤ) if and only if a = ±1, b = d(∏_{i∈I} g_{i}), and there do not exist ∅ ≠ J ⫋ I and b_{1}, b_{2} ∈ ℕ with b_{1}b_{2} = b and b_{1} = d(∏_{i∈J} g_{i}), b_{2} = d(∏_{i∈I\J} g_{i}).

(iii) Int(ℤ) is atomic. (iv) Every non-zero non-unit f ∈ Int(ℤ) has only finitely many factorizations into

irreducibles in Int(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 d(∏_{i∈I} g_{i}). Any non-constant irreducible factor would, by Remark 1 (iii), be of the kind (∏_{i∈J} g_{i})/b_{1} with b_{1} = d(∏_{i∈J} g_{i}), and its co-factor would be (∏_{i∈I\J} g_{i})/b_{2} with b_{1}b_{2} = b and b_{2} a divisor of d(∏_{i∈I\J} g_{i}). Also, b_{2} could not be a proper divisor of d(∏_{i∈I\J} g_{i}), because otherwise b_{1}b_{2} = b would be a proper divisor of ∏_{i∈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) = ag(x)/b, g = ∏_{i∈I} g_{i} as in (i), d(g) = cb for some c ∈ ℕ, and f(x)=acg(x)/d(g) with g(x)/d(g)∈Int(ℤ). We can factor ac into irreducibles in ℤ, which are also irreducible in Int(ℤ). Either g(x)/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)/d(g) into irreducibles.

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

f = a_{1} … a_{n}c_{1} … c_{m} ∏_{j=1}^{k} \frac{∏_{i∈I_{j}} g_{i}}{d_{j}} ,

where a = a_{1} … a_{n} and c = c_{1} … c_{m} are factorizations into primes in ℤ, I = I_{1} ∪…∪I_{k} is a partition of I into non-empty sets, d_{1} … d_{k} = bc, d_{j} = d(∏_{i∈I_{j}} g_{i}). There are only finitely many such expressions. □

Remark 3.

(i) The binomial polynomials

\binom{x}{n} = \frac{x(x − 1) … (x − n + 1)}{n!} for n ≥ 0

are a basis of Int(ℤ) as a free ℤ-module. (ii) n!f ∈ ℤ[x] for every f ∈ Int(ℤ) of degree at most n. (iii) Let f ∈ ℤ[x] primitive, deg f = n and p prime. Then

v_{p}(d(f)) ≤ ∑_{k≥1} [\frac{n}{p^{k}} ] = v_{p}(n!).

In particular, if p divides d(f) then p ≤ deg f.

Proof. Ad (i). The binomial polynomials are in Int(ℤ) and they form a ℚ-basis of ℚ[x]. If a polynomial in Int(ℤ) is written as a 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 ∈ Int(ℤ) and d(f)ℤ = (ℤ[x] :_{ℤ} g). Since n! ∈ (ℤ[x] :_{ℤ} g) by (ii), d(f) divides n!. □

3. Useful Lemmata

Lemma 4. Let p be a prime, I ≠ ∅ a finite set and for i ∈ I, f_{i} ∈ ℤ[x] primitive and irreducible in ℤ[x] such that d(∏_{i∈I} f_{i}) = p. Let

g(x) = \frac{∏_{i∈I} f_{i}}{p} .

Then every factorization of g in Int(ℤ) is essentially the same as one of the following:

g(x) = \frac{∏_{j∈J} f_{j}}{p} · ∏_{i∈I\J} f_{i},

where J ⊆ I is minimal such that d(∏_{i∈J} f_{j}) = p.

Proof. Follows from Remark 1 (iii) and the fact that d(f)d(h) divides d(fh) for all f, h ∈ ℤ[x]. □

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

Lemma 5. For every prime p ∈ 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, …, 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} ∈ ℤ[x], i ∈ I, we can construct monic irreducible polynomials F_{i} ∈ ℤ[x], pairwise non-associated in 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 ∈ J (J an arbitrary subset of I) and g_{i} = f_{i} for i ∈ I \ J, then for all K ⊆ I,

d( ∏_{i∈K}g_{i}) = d( ∏_{i∈K}f_{i}).

Proof. Let n = ∑_{i∈I} deg f_{i}. Let p_{1}, …, p_{s} be all the primes with p_{i} ≤ n, and set α_{i} = v_{p_{i}}(n!). Let q > n be a prime. For each i ∈ I, we find by the Chinese Remainder Theorem the coefficients of a polynomial φ_{i} ∈(∏_{k=1}^{s} p_{k}^{α_{k}})ℤ[x] of smaller degree than f_{i}, such that F_{i} = f_{i} + φ_{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 ℚ[x] to any F_{j} of smaller index, we add a suitable non-zero integer divisible by q^{2} ∏_{k=1}^{s} p_{k}^{α_{k}} to F_{i}, to make F_{i} non-associated in ℚ[x] to all F_{j} of smaller index.

The statement about the fixed divisor follows, because for every c ∈ ℤ and every prime p_{i} that could conceivably divide the fixed divisor,

∏_{i∈K}(g_{i}(c)) ≡ ∏_{i∈K}(f_{i}(c)) mod p_{i}^{α_{i}},

where p_{i}^{α_{i}} is the highest power of p_{i} that can divide the fixed divisor of any monic polynomial of degree at most n. □

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 ≥ 0, we can construct H ∈ Int(ℤ) such that H has exactly two essentially different factorizations in Int(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}, …, a_{p} of residues mod p in ℤ that does not contain a complete set of residues mod any prime q < p. Let

f(x) = (x − a_{2})(x − a_{3}) … (x − a_{p}) and g(x) = (x − a_{n+2})(x − a_{n+3}) … (x − a_{p}).

By Lemma 6, we construct monic irreducible polynomials F, G ∈ ℤ[x], not associated in ℚ[x], with deg F = deg f, deg G = deg g, such that any product of a selection of polynomials from (x − a_{1}), …, (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

H(x) = \frac{F (x)(x − a_{1}) … (x − a_{n+1})G(x)}{p}.

By Lemma 4, H factors into two irreducible polynomials in Int(ℤ)

H(x) = F (x) · \frac{(x − a_{1}) … (x − a_{n+1})G(x)}{p}

or into n + 2 irreducible polynomials in Int(ℤ)

H(x) = \frac{F (x)(x − a_{1})}{p} · (x − a_{2})(x − a_{3}) … (x − a_{n+1})G(x).

□ Corollary. (Cahen and Chabert [1]). ρ (Int(ℤ)) = ∞.

Example 8. For 1 ≤ m ≤ n, we can construct a polynomial H ∈ Int(ℤ) that has in Int(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:

R = {r(i, j) | 1 ≤ i ≤ m, 1 ≤ j ≤ n} ∪ {b_{1}, …, b_{s}}.

Let b(x) = ∏_{k=1}^{s}(x − b_{k}). For 1 ≤ i ≤ m let f_{i}(x) = ∏_{k=1}^{n}(x − r(i, k)) and for 1 ≤ j ≤ n let g_{j}(x) = ∏_{k=1}^{m}(x − r(k, j)).

By Lemma 6, we construct monic irreducible polynomials F_{i}, G_{j} ∈ ℤ[x], pairwise non-associated in ℚ[x], such that the product of any selection of the polynomials (x − b_{1}), …, (x − b_{s}), f_{1}, …, f_{m}, g_{1}, …, 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 ≤ i ≤ m and 1 ≤ j ≤ n. Let

H(x) = \frac{1}{p}b(x) ∏_{i=1}^{m} F_{i}(x) ∏_{j=1}^{n} G_{j}(x),

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

H(x) = F_{1}(x) · . . . · F_{m}(x) · \frac{b(x)G_{1}(x) · . . . · G_{n}(x)}{p}

and an essentially different factorization into n + 1 irreducibles

H(x) = \frac{b(x)F_{1}(x) · . . . · F_{m}(x)}{p} · G_{1}(x) · . . . · G_{n}(x)

and no other essentially different factorization. □

Corollary. (Chapman and McClain [3]). Int(ℤ) is fully elastic.

Theorem 9. Given natural numbers 1 ≤ m_{1} ≤ … ≤ m_{n}, we can construct a polynomial H ∈ Int(ℤ) that has exactly n essentially different factorizations into irreducibles in Int(ℤ), the lengths of these factorizations being m_{1} +1, …, m_{n} +1.

Proof. Let N = (∑_{i=1}^{n} m_{i})^{2} − ∑_{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} ∪ {t_{1}, …, t_{s}} with |R_{0}| = N. The elements of R_{0} are indexed as follows:

R_{0} = {r(k, h, i, j) | 1 ≤ k ≤ n, 1 ≤ h ≤ m_{k}, 1 ≤ i ≤ n, 1 ≤ j ≤ m_{i}; i ≠ k},

meaning we arrange the elements of R_{0} in an m×m matrix with m=m_{1}+…+m_{n}, whose rows and columns are partitioned into n blocks of sizes m_{1}, …, 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 ≤ k ≤ n, 1 ≤ h ≤ m_{k}, let S_{k,h} be the set of entries in the (k, h)-th row:

S_{k,h} = {r(k, h, i, j) | 1 ≤ i ≤ n, i ≠ k, 1 ≤ j ≤ m_{i}}.

For 1 ≤ i ≤ n, 1 ≤ j ≤ m_{i}, let T_{i,j} be the set of elements in the (i, j)-th column:

T_{i,j} = {r(k, h, i, j) | 1 ≤ k ≤ n, k ≠ i, 1 ≤ h ≤ m_{k}}.

For 1 ≤ k ≤ n, 1 ≤ h ≤ m_{k}, set

f_{h}^{(k)}(x) = ∏_{r∈S_{k,h}}(x − r) · ∏_{r∈T_{k,h}}(x − r).

Also, let b(x) = ∏_{i=1}^{s}(x − t_{i}).

By Lemma 6, we construct monic irreducible polynomials F_{h}^{(k)}, pairwise nonassociated in ℚ[x], with deg F_{h}^{(k)} = deg f_{h}^{(k)}, such that any product of a selection of polynomials from (x − t_{1}), …, (x − t_{s}) and f_{h}^{(k)} for 1 ≤ k ≤ n, 1 ≤ h ≤ m_{k} has the same fixed divisor as the corresponding product in which the f_{h}^{(k)} have been replaced by the F_{h}^{(k)}. Let

H(x) = \frac{1}{p}b(x) ∏_{k=1}^{n} ∏_{h=1}^{m_{k}} F_{h}^{(k)}(x).

Then deg H = N + p; and for each i = 1, …, n, H has a factorization into m_{i} + 1 irreducible polynomials in Int(ℤ):

H(x) = F_{1}^{(i)}(x) · . . . · F_{m_{i}}^{(i)}(x) · \frac{b(x) ∏_{k≠i} ∏_{h=1}^{m_{k}} F_{h}^{(k)}(x)}{p}

These factorizations are essentially different, since the F_{j}^{(i)} are pairwise nonassociated in ℚ[x] and hence in Int(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 ≤ i ≤ s and F_{h}^{(k)} for 1 ≤ k ≤ n, 1 ≤ h ≤ m_{k} must consist of all the linear factors (x − t_{i}) together with a minimal selection of F_{h}^{(k)} such that all r ∈ R_{0} occur as roots in the product of the corresponding f_{h}^{(k)}. For all linear factors (x − r) with r ∈ R_{0} to occur in a set of polynomials f_{h}^{(k)}, it must contain for all but one k all f_{h}^{(k)}, h = 1, … m_{k}. If, for i ≠ k, f_{h}^{(k)} and f_{j}^{(i)} are missing, then r(k, h, i, j) and r(i, j, k, h) do not occur among the roots of the polynomials f_{h}^{(k)}. A set consisting of all f_{h}^{(k)} for n − 1 different values of k, however, has the property that all linear factors (x − r) for r ∈ R_{0} occur. □

Corollary. Every finite subset of ℕ\{1} occurs as the set of lengths of a polynomial f ∈ Int(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.

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

Proposition 10. For every n ≥ 1 there exist irreducible elements H, G_{1}, …, G_{n+1} in Int(ℤ) such that xH(x) = G_{1}(x) … G_{n+1}(x).

Proof. Let p_{1} < p_{2} < … < p_{n} be n distinct odd primes, P = {p_{1}, p_{2}, …, p_{n}}, and Q the set of all primes q ≤ p_{n} + n. By the Chinese remainder theorem construct a_{1}, …, a_{n} with a_{i} ≡ 0 mod p_{i} and a_{i} ≡ 1 mod q for all q ∈ Q with q ≠ p_{i}. Similarly, construct b_{1}, … b_{p_{n}} such that, firstly, for all p ∈ P, b_{k} ≡ k mod p if k ≤ p and b_{k} ≡ 1 mod p if k > p and, secondly, b_{k} ≡ 1 mod q for all q ∈ Q \ P. So, for each p_{i} ∈ P, a complete set of residues mod p_{i} is given by b_{1}, … 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 \ P.

Set f(x) = (x−b_{1}) … (x−b_{p_{n}}) and let F (x) be a monic irreducible polynomial in 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}), …, (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

H(x) = \frac{F (x)(x − a_{1}) … (x − a_{n})}{p_{1} … p_{n}}.

Then H(x) is irreducible in Int(ℤ), and

xH(x) = \frac{xF (x)}{p_{1} … p_{n}} · (x − a_{1}) · . . . · (x − a_{n}),

where xF (x)/(p_{1} … p_{n}) and, of course, (x − a_{1}), …, (x − a_{n}), are irreducible in Int(ℤ). □

Remark 11. Thanks to Roger Wiegand for suggesting an easier proof of Proposition 10: Using the well-known fact that the binomial polynomials \binom{x}{m} are irreducible in Int(ℤ) for all m > 0, it suffices to consider

x\binom{x − 1}{m − 1} = m\binom{x}{m}

with m chosen to have exactly n prime factors in ℤ.

Remark 12. Thanks to Alfred Geroldinger for pointing this out: Proposition 10 implies that there does not exist a transfer-homomorphism from the multiplicative monoid (Int(ℤ) \ {0}, ·) 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.

References

[1] P.-J. Cahen and J.-L. Chabert, Elasticity for integral-valued polynomials, J. Pure Appl. Algebra 103 (1995), 303–311.

[2] P.-J. Cahen and J.-L. Chabert, Integer-valued polynomials, vol. 48 of Mathematical Surveys and Monographs, Amer. Math. Soc., 1997.

[3] S. T. Chapman and B. A. McClain, Irreducible polynomials and full elasticity in rings of integer-valued polynomials, J. Algebra 293 (2005), 595–610.

[4] Ch. Frei and S. Frisch, Non-unique factorization of polynomials over residue class rings of the integers, Comm. Algebra 39 (2011), 1482–1490.

[5] A. Geroldinger and F. Halter-Koch, Non-unique factorizations, vol. 278 of Pure and Appl. Math., Chapman & Hall/CRC, Boca Raton, FL, 2006.

[6] F. Kainrath, Factorization in Krull monoids with infinite class group, Colloq. Math. 80 (1999), 23–30.

Institut für Mathematik A Technische Universität Graz Steyrergasse 30 A-8010 Graz, Austria

frisch@tugraz.at