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

For an arbitrary finite set S of natural numbers greater 1, we construct an integer-valued polynomial f, whose set of lengths in Int(Z) is S. The set of lengths of f is the set of all natural numbers n, such that f has a factorization as a product of n irreducibles in Int(Z)={g in Q[x] | g(Z) contained in Z}.


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 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 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 as 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.

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][3]. The reader familiar with integer-valued polynomials is encouraged to skip to Sect. 3.
(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, , .
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 In particular, it is easy to deduce that for any f, g ∈ Z[x], Remark 1 (i) Every non-zero polynomial f ∈ Q[x] can be written in a unique way as (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 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.
. 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), where a = a 1 . . . a n and c = c 1 . . . c m are factorizations into primes in Z, There are only finitely many such expressions.
Remark 3 (i) The binomial polynomials 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).

Lemma 4 Let p be a prime, I = ∅ a finite set and for i
Then every factorization of g in Int(Z) is essentially the same as one of the following: Proof Follows from Remark 1 (iii) and the fact that 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. 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 , 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 ∈ ( s k=1 p α k 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 s k=1 p α k 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, where p α i i is the highest power of p i that can divide the fixed divisor of any monic polynomial of degree at most n.

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 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

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 (i, k)) and for 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

then, by Lemma 4, H has a factorization into m + 1 irreducibles
and an essentially different factorization into n + 1 irreducibles and no other essentially different factorization.
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: 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: For 1 ≤ i ≤ n, 1 ≤ j ≤ m i , let T i, j be the set of elements in the (i, j)-th column: Then deg H = N + p; and for each i = 1, . . . , n, H has a factorization into m i + 1 irreducible polynomials in Int(Z): Corollary Every finite subset of N\{1} occurs as the set of lengths of a polynomial f ∈ Int(Z).

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 transfer-homomorphism to a block monoid.  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 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 Then H (x) is irreducible in Int(Z), and where x F(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 x m are irreducible in Int(Z) for m > 0, it suffices to consider , 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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