SETS OF LENGTHS OF FACTORIZATIONS OF INTEGER-VALUED POLYNOMIALS ON DEDEKIND DOMAINS WITH FINITE RESIDUE FIELDS

LetD be a Dedekind domain with infinitely many maximal ideals, all of finite index, and K its quotient field. Let Int(D) = {f ∈ K[x] | f(D) ⊆ D} be the ring of integer-valued polynomials on D. Given any finite multiset {k1, . . . , kn} of integers greater than 1, we construct a polynomial in Int(D) which has exactly n essentially different factorizations into irreducibles in Int(D), the lengths of these factorizations being k1, . . . , kn. We also show that there is no transfer homomorphism from the multiplicative monoid of Int(D) to a block monoid.


Introduction
By factorization we mean an expression of an element of a ring as a product of irreducible elements. Until not so long ago, the fact that such a factorization, if it exists, need not be unique, was seen as a pathology. When mathematicians were shocked to find that uniqueness of factorization does not hold in rings of integers in number fields, they did not immediately study the details of this non-uniqueness, but moved on to unique factorization of ideals into prime ideals. Non-uniqueness of factorization was avoided, whenever possible.
Only in the last few decades, some mathematicians, notably Geroldinger and Halter-Koch [9], came around to the view that the precise details of non-uniqueness of factorization actually are a fascinating topic: the underlying phenomena give a lot of information about the arithmetic of a ring.
One important object of study is the set of lengths of factorizations of a fixed element, cf. [8]. The length of a factorization is the number of irreducible factors, and the set of lengths of an element is the set of all natural numbers that occur as lengths of factorizations of the element. Geroldinger and Halter-Koch [9] found that the sets of lengths of algebraic integers exhibit a certain structure.
In stark contrast to this, we show in Section 4 that every finite set of natural numbers not containing 1 occurs as the set of lengths of a polynomial in the ring of integer-valued polynomials on D, where D is a Dedekind domain with infinitely many maximal ideals, all of them of finite index, and K denotes the quotient field of D. The special case of D = Z has been shown by Frisch [6].
The study of non-uniqueness of factorization has mostly concentrated on Krull monoids so far. Krull monoids are characterized by having a "divisor theory". The multiplicative monoid D \ {0} of an integral domain D is Krull exactly if D is a Krull ring, cf. [9].
The rings Int(D) for which we study non-uniqueness of factorization are not Krull, but Prüfer, cf. [4,14]. All factorizations of a single polynomial in Int(D), however, take place in a Krull monoid, namely, in the divisor-closed submonoid of Int(D) generated by f . Following Reinhart [16], we call this monoid, consisting of all divisors in Int(D) of all powers of f , the monadic submonoid generated by f . That all monadic submonoids of Int(D) are Krull was shown by Reinhart [16] in the case where D is a unique factorization domain, and, by a different method, by Frisch [7] in the case where D is a Krull ring. Thus, our Theorem 1, concerning nonunique factorization in the Prüfer ring Int(D), also serves to show that quite wild factorization behavior is possible in Krull monoids.
Among Krull monoids, the best studied ones are multiplicative monoids of rings of algebraic integers. We should keep in mind, however, that the multiplicative monoids of rings of algebraic integers are very special, in that unique factorization of ideals always lurks in the background. In technical terms this means that there is a transfer homomorphism to a block monoid.
In Section 5, we show that there is no transfer homomorphism to a block monoid from the multiplicative monoid of Int(D). This is relevant for two reasons: Firstly, because the rings of whose multiplicative monoid it is known that it does not admit such a transfer homomorphism are few and far between, see [5,10,11]; and secondly, because most, if not all, results so far concerning arbitrary finite sets occurring as sets of lengths have been obtained using transfer homomorphisms to block monoids [13].
Our main results are in Sections 4 and 5; in Section 2 we introduce the necessary notation and Section 3 contains some useful lemmas.

Preliminaries
We start with a short review of some elementary facts on factorizations, Dedekind domains and integer-valued polynomials, and introduce some notation.
Factorizations. We define here only the notions that we need throughout this paper, and refer to the monograph by Geroldinger and Halter-Koch [9] for a systematic introduction to non-unique factorizations.
Let R be a commutative ring with identity and r, s ∈ R.
(i) If r is a non-zero non-unit, we say r is irreducible in R if it cannot be written as the product of two non-units of R. (ii) A factorization of r in R is an expression r = a 1 · · · a n (1) where n ≥ 1 and a i is irreducible in R for 1 ≤ i ≤ n. (iii) The number n of irreducible factors is called the length of the factorization in (1). (iv) The set of lengths of r is the set of all natural numbers n such that r has a factorization of length n. (v) We say r and s are associated in R if there exists a unit u ∈ R such that r = us. We denote this by r ∼ s. (vi) Two factorizations of the same element, are called essentially the same if n = m and, after reindexing, a j ∼ b j for 1 ≤ j ≤ m. If this is not the case, the factorizations in (2) are called essentially different.
Dedekind domains. Recall that an integral domain D is a Dedekind domain if and only if every non-zero ideal is a product of prime ideals. This is equivalent to every non-zero ideal being invertible. It is also equivalent to D being a Noetherian domain such that the localization at every non-zero maximal ideal is a discrete valuation domain. And it is further equivalent to the following list of properties: From now on, we only consider Dedekind domains that are not fields. For a Dedekind domain D with quotient field K, let max-spec(D) denote the set of maximal ideals of D. Every prime ideal P ∈ max-spec(D) defines a discrete valuation v P by v P (a) = max{n ∈ Z | a ∈ P n } for a ∈ K \ {0}. v P is called the P -adic valuation on K.
For a non-zero ideal I of D, let v P (I) = min{v P (a) | a ∈ I}. This is compatible with the definition of v P (a) for a ∈ K \ {0}, in the sense that v P (aD) = v P (a). With this notation, the factorization of I into prime ideals is Note that v P (I) > 0 is equivalent to I ⊆ P . There are only finitely many prime overideals of I in D and hence the product in Equation (3) is finite.
For two ideals I and J of D, I ⊆ J is equivalent to v P (J) ≤ v P (I) for all P ∈ max-spec(D). Note that I ⊆ J is equivalent to the fact that there exists an ideal L of D such that JL = I, in which case we say that J divides I and write J | I. This last equivalence is often summarized as "to contain is to divide." For a thorough introduction to Dedekind domains, we refer to Bourbaki [1, Ch. VII, § 2].
Dedekind domains with finite residue fields. Let D be a Dedekind domain. For a maximal ideal P with finite residue field we write P for |D/P | and call this number the index of P . In what follows we will only consider Dedekind rings with infinitely many maximal ideals, all of whose residue fields are finite. We will frequently use the fact that there are only finitely many maximal ideals of each individual finite index. This holds in every Noetherian domain, as Samuel [17] has shown; see also Gilmer [12]. We include a short proof by F. Halter-Koch for the special case of Dedekind domains.
Proposition 2.1 (Samuel [17], Gilmer [12]). Let D be a Dedekind domain. Then for each given q ∈ N, there are at most finitely many maximal ideals P of D with P = q.
Proof (Halter-Koch, personal communication). Suppose that for some q ≥ 2 there exist infinitely many prime ideals of index q, and let 0 = a ∈ D. Then there exist infinitely many prime ideals P of D such that P = q and a / ∈ P . For each such prime ideal P we obtain a q−1 ≡ 1 mod P , hence a q−1 − 1 ∈ P and thus a q−1 = 1. So, every non-zero element of D is a (q − 1)-st root of unity. Impossible! Integer-valued polynomials. If D is a domain with quotient field K, the ring of integer-valued polynomials on D is defined as Every non-zero f ∈ K[x] can be written as a quotient f = g b where g ∈ D[x] and b ∈ D \ {0}. Clearly, f = g b is an element of Int(D) if and only if b | g(a) for all a ∈ D. Definition 2.2. Let D be a domain and g ∈ Int(D). The fixed divisor of g is the ideal d(g) of D generated by the elements g(a) with a ∈ D: We say that g is image primitive if d(g) = D. By abuse of notation, this is also denoted d(g) = 1.
, then every factorization of g in Int(D) as a product of two (not necessarily irreducible) elements is of the form c g c with c ∈ D and d(g) ⊆ cD.
For a general introduction to integer-valued polynomials we refer to the monograph by Cahen and Chabert [2] and to their more recent survey paper [3].

Auxiliary results
In this section we develop tools to construct, first, split polynomials in D[x] with prescribed fixed divisor (Lemma 3.2), then, irreducible polynomials in D[x] with prescribed fixed divisor (Lemma 3.3), and, finally, polynomials of a special form whose essentially different factorizations in Int(D) we have complete control over (Lemma 3.7).
Remark 3.1. In the following, we want to consider the multiplicity of roots of polynomials. For this purpose, we introduce some notation for multisets. Let m S (a) denote the multiplicity of an element a in a multiset S (with m S (a) = 0 if a / ∈ S). For multisets S and T , let S T denote the collection of elements a in the union of the sets underlying S and T with multiplicities m S T (a) = m S (a) + m T (a) (the disjoint union of S and T ). Note that |S T | = |S| + |T |.
is a complete system of residues modulo Q and the respective representatives of the same residue class in each Proof. If T does not contain a complete system of residues modulo Q, then there exists an element Conversely, if T contains a complete system of residues modulo Q then, for all a ∈ D, there exists r ∈ T such that a ≡ r mod Q.
For 1 ≤ i ≤ e, let a i ∈ T i with a i ≡ z mod Q. Since the elements a i are in the same residue class modulo Q 2 , there exists d ∈ D in the same residue class modulo Q as z and all the a i , but in a different residue class modulo Q 2 from all the a i .
For such a d,  Then, there exist monic polynomials Proof. Let P 1 , . . . , P n be all maximal ideals P of D with P ≤ deg i∈I f i . Suppose the prime factorization of the fixed divisor of the product of the f i is Let Q ∈ max-spec(D) \ {P 1 , . . . , P n }. Using the Chinese Remainder Theorem, we add elements to the coefficients of the f i such that the resulting polynomials can be seen to be irreducible according to Eisenstein's irreducibility criterion with respect to Q, while retaining all relevant properties with respect to sufficiently high powers of the P i . Let Since the g ik satisfying the above conditions are only determined modulo Q 2 n i=1 P ei+1 i , there are infinitely many choices for each g ik . We use this flexibility to implement that g i0 +f i0 = g j0 +f j0 for i = j. Then, for i ∈ I, we set As the resulting F i are monic and distinct, they are pairwise non-associated in K[x].
According to Eisenstein's irreducibility criterion, the polynomials By construction, for all i ∈ I. Now, if g(x) is the product of any selection of the polynomials f i , and G(x) the modified product in which some of the f i have been replaced by F i , then g(x) is congruent to G(x) Hence, for all a ∈ D, g(a) ≡ G(a) modulo n j=1 P ej +1 j and, therefore, where c is a non-unit of D and for each i ∈ I, Let P ⊆ max-spec(D) be the finite set of prime ideal divisors of cD. If f = g 1 · · · g m is a factorization of f into (not necessarily irreducible) non-units in Int(D) then each g j is of the form where ∅ = I j ⊆ I and a j ∈ K, such that I 1 . . . I m = I, a 1 · · · a m = c −1 and (i) v P (a j ) ≤ 0 for all P ∈ max-spec(D) and all 1 ≤ j ≤ m; and (ii) v P (a j ) = 0 for all P ∈ max-spec(D) \ P and all 1 ≤ j ≤ m.
Proof. Let f = g 1 · · · g m be a factorization of f into (not necessarily irreducible) non-units in Int(D).
Since d(f ) = 1, no g i is a constant, by Remark 2.3.(iv). Each factor g j is, therefore, of the form where I j is a non-empty subset of I and a j ∈ K, such that I 1 . . . I m = I and a 1 · · · a m = c −1 .
Note that for all P ∈ max-spec(D) Suppose v P (a t ) > 0 for some maximal ideal P and some 1 ≤ t ≤ m. Then j =t v P (a j ) < −v P (c).

Remark 2.3.(iii) and the fact that
a contradiction. We have established that v P (a j ) ≤ 0 for all P ∈ max-spec(D) and all 1 ≤ j ≤ m. Now Equation (6) and the fact that v P (c) = 0 for all P / ∈ P imply v P (a j ) = 0 for all P / ∈ P and all 1 ≤ j ≤ m. Definition 3.5. Let D be a Dedekind domain, P ∈ max-spec(D) and f i ∈ D[x] for a finite set I = ∅. We say f k is indispensable for P (among the polynomials f i with i ∈ I) if there exists an element z ∈ D such that v P (f k (z)) > 0 and v P (f i (z)) = 0 for all i = k. Remark 3.6. Note that f k indispensable for P (among the polynomials f i with i ∈ I) implies: Lemma 3.7. Let D be a Dedekind domain with quotient field K and f ∈ Int(D) of the following form: where c is a non-unit of D and for each i ∈ I, . Let P ⊆ max-spec(D) be the finite set of prime ideal divisors of cD. Suppose, for each P ∈ P, Λ P is a subset of I such that f i is indispensable for P for each i ∈ Λ P . Let Λ = P ∈P Λ P .
If P ∈P Λ P = ∅ then all essentially different factorizations of f into irreducibles in Int(D) are given by: f j (each f j with j ∈ J 2 counted as an individual factor), where I = Λ J 1 J 2 such that J 1 is minimal with d i∈Λ∪J1 f i = cD.
Proof. Let f = g 1 · · · g m be a factorization of f into (not necessarily irreducible) non-units in Int(D). As in Lemma 3.4, where I j is a non-empty subset of I and a j ∈ K, such that I 1 . . . I m = I and a 1 · · · a m = c −1 . Furthermore, v P (a j ) ≤ 0 for all P ∈ max-spec(D) and all 1 ≤ j ≤ m and v P (a j ) = 0 for all P / ∈ P and all 1 ≤ j ≤ m.
We know there exists a polynomial f i0 that is indispensable for all P ∈ P. We may assume that i 0 ∈ I 1 . By the definition of indispensable polynomial, v P d i∈Ij f i = 0, for 2 ≤ j ≤ m and all P ∈ P. From this and the fact that g j = a j i∈Ij f i is in Int(D), we infer that v P (a j ) = 0 for all 2 ≤ j ≤ m and all P ∈ P. We have shown that a 2 , . . . , a m are units of D.
Now u = a 2 · · · a m is a unit of D such that a 1 u = c −1 . Since g 1 ∈ Int(D), we must have for all P ∈ P and, by Remark 3.6, Λ ⊆ I 1 .
So far we have shown that every factorization f = g 1 · · · g m of f into (not necessarily irreducible) non-units of Int(D) is -up to reordering of factors and multiplication of factors by units in Dthe same as one of the following: where I = I 1 · · · I m and I 1 = Λ J 1 . It remains to characterize, among the factorizations of the above form, those in which all factors are irreducible in Int(D).
Since (v). The criterion for each factor g j = i∈Ij f i with j ≥ 2 to be irreducible is, therefore, |I j | = 1 for all j ≥ 2.
Now, concerning the irreducibility of g 1 , the same arguments that lead to Equation (8), applied to g 1 = c −1 i∈Λ∪J1 f i instead of f , show that g 1 is irreducible in Int(D) if and only if we cannot split off any factors f i with i ∈ J 1 . This is equivalent to d i∈Λ∪J f i = cD for every proper subset J J 1 , in other words, to J 1 being minimal such that d i∈Λ∪J1 f i = cD. In this case we set J 2 = m j=2 I j and the assertion follows.
Remark 3.8. When |P| > 1, the hypothesis P ∈P Λ P = ∅ in Lemma 3.7 can be replaced by a weaker condition: Consider the prime ideals P ∈ P as vertices of an undirected graph G and let (P, Q) be an edge of G if and only if there exists a polynomial f t which is indispensable for both P and Q. If G is a connected graph, then the conclusion of Lemma 3.7 holds. The proof of Lemma 3.7 generalizes readily.

Construction of polynomials with prescribed sets of lengths
We are now ready to prove the main result of this paper. Proof. If n = 1, then H(x) = x m1+1 ∈ Int(D) is a polynomial which has exactly one factorization, and this factorization has length m 1 + 1. From now on, assume n ≥ 2.
First, we construct H(x).
and P a prime ideal of D with P > N + 1. Let c ∈ D such that v P (c) = 1 and c is not contained in any maximal ideal of index 2.
Say the prime factorization of cD is cD = P Q e1 1 · · · Q et t . Let τ = ( P −N ) and σ the maximum of the following numbers: τ , and e i Q i for 1 ≤ i ≤ t.
We now choose two subsets of D: a set R of order N , and S = {s 0 , . . . , s σ−1 }. Using the Chinese Remainder Theorem, we arrange that R and S have the following properties: (i) s 0 ≡ 0 mod P , and {s 0 , . . . , s τ −1 } ∪ R is a complete system of residues modulo P . (ii) s i ≡ 0 mod P for all i ≥ τ . (iii) For each Q i , S contains e i disjoint complete systems of residues, in which the respective representatives of the same residue class in different systems are congruent modulo Q 2 i . (iv) For each Q i , no more than e i elements of S are congruent to 1 modulo Q i .
(vi) R ∪ S does not contain a complete system of residues for any prime ideal Q of D other than P and Q 1 , . . . , Q t .
We now assign indices to the elements of R as follows This allows us to visualize the elements of R as entries in a square matrix B with m = n i=1 m i rows and columns, in which the positions in the blocks of a block-diagonal matrix with block sizes m 1 , . . . , m n are left empty, see Fig. 1.
The rows and columns of B are divided into n blocks each, such that the k-th block of rows consists of m k rows, and similarly for columns. Now r (k,i,h,j) designates the entry in row (k, i), that is, in the i-th row of the k-th block of rows, and in column (h, j), that is, in the j-th column of the h-th block of columns. Since no element of R has row and column index in the same block, the positions of a block-diagonal matrix with blocks of sizes m 1 , . . . , m n are left empty. Then is the set of all possible row indices, or, equivalently, column indices. For (k, i) ∈ I k , let B[k, i] be the set of all elements r ∈ R which are either in row or in column (k, i) of B, that is, In order to construct H ∈ Int(D), we set s(x) = Then, let S(x) ∈ D[x], and, for each (k, i) ∈ I, F  Second, we show that d(G(x)) = cD, which implies H(x) ∈ Int(D) and d(H(x)) = 1. Note that (10) By construction, the multiset R R S contains a complete system of residues modulo P , and the residue class modulo P of s 1 ∈ S occurs only once among the elements of R R S. Equation (10) and Lemma 3.2, applied to Q = P and T = R R S, e = 1, and z = s 1 , together imply that One can argue similarly for Q i , 1 ≤ i ≤ t: The multiset R R S contains e i disjoint complete systems of residues modulo Q i in which the respective representatives of the same residue class in different systems are congruent modulo Q 2 i . No more than e i elements of R R S are congruent 1 modulo Q i , and these e i elements are all in the same residue class modulo Q 2 i . By Lemma 3.2, applied to Q = Q i , T = R R S, e = e i and z = 1, and Equation (10), it follows that for 1 ≤ i ≤ t. Since R R S does not contain a complete system of residues modulo any prime ideal other than P , Q 1 , . . . , Q t , we conclude (by Lemma 3.2) that This shows H(x) ∈ Int(D) and d(H(x)) = 1.
Third, we prove that the essentially different factorizations of H(x) into irreducibles in Int(D) are given by: where 1 ≤ h ≤ n. It follows from the properties of R and S that the polynomial S(x) is indispensable for the prime ideals P and Q 1 , . . . , Q t (among the polynomials S(x) and F (k) i for (k, i) ∈ I). Thus, by Lemma 3.7, the essentially different factorizations of H(x) into irreducibles in Int(D) are given by: where and it follows from Lemma 3.2 that the fixed divisor in Equation (14)     We now demonstrate for the irreducible element c = x in Int(D) that the lengths of factorizations of elements of the form c · d with d irreducible in Int(D) are not bounded. We infer from this that there does not exist a transfer homomorphism from the multiplicative monoid Int(D) \ {0} to a block monoid.
Theorem 2. Let D be a Dedekind domain with infinitely many maximal ideals, all of them of finite index.
Then for every n ≥ 1 there exist irreducible elements H, G 1 , . . . , G n+1 in Int(D) such that Proof. Let P 1 , . . . , P n be distinct maximal ideals of D, none of them of index 2. By v i we denote the discrete valuation associated to P i . Let c ∈ D such that v i (c) = 1 for i = 1, . . . , n, and c is not contained in any maximal ideal of D of index 2. Say the prime factorization of cD is cD = P 1 · . . . · P n · Q e1 1 · . . . · Q em m , and define Let R be a subset of D of order N with the following properties (which can be realized by the Chinese Remainder Theorem): (i) R contains an element r 0 ∈ ( No element of R other than r 0 is in any P i ∈ P. (iii) For each P i ∈ P, R contains a complete system of residues modulo P i . (iv) For each Q i ∈ P 1 , R contains e i disjoint complete systems of residues, in which the respective representatives of the same residue class in different systems are congruent modulo Q 2 i ; (v) No more than e i elements of R are in Q i . (vi) For all Q ∈ P 2 , all elements of R are contained in Q. We set B = R \ {r 0 }. Also, let a 1 , . . . , a n ∈ D with the following properties (which, again, can be realized by the Chinese Remainder Theorem): (i) For all i = 1, . . . , n, a i ≡ 0 mod P i .
(ii) For all i = 1, . . . , n, a i ≡ 1 mod P j for all j = i.
(iii) For all Q ∈ P 1 , a n ≡ 0 mod Q 2 and a i ≡ 1 mod Q for all 1 ≤ i < n, (iv) For all Q ∈ P 2 and all 1 ≤ i ≤ n, a i ≡ 0 mod Q. Let f (x) = b∈B (x − b) and let F (x) ∈ D[x] be monic and irreducible in K[x] such that for every selection of polynomials from the set {x, f } ∪ {(x − a i ) | 1 ≤ i ≤ n} the product of the polynomials has the same fixed divisor as the modified product in which f has been replaced by F , as in Lemma 3.3. Lemmas 3.3 and 3.2, applied to T = B ∪ {a 1 , . . . , a n } and each of the prime ideals in P ∪ P 1 , imply that Similarly, Lemmas 3.3 and 3.2, applied to T = B ∪ {0} and each of the prime ideals in P ∪ P 1 , imply that d (xF (x)) = d (xf (x)) = cD. We set Then G(x) and H(x) are elements of Int(D) with d(G(x)) = d(H(x)) = 1 such that xH(x) = G(x)(x − a 1 ) · · · (x − a n ).
It remains to show that H(x) and G(x) are irreducible in Int(D). Observe that x is indispensable for all P ∈ P, and F (x) is indispensable for all P ∈ P and all Q ∈ P 1 simultaneously (among the polynomials F (x) and x). Hence G(x) is irreducible in Int(D) by Lemma 3.7.
Finally, again by Lemma 3.7, H(x) is irreducible in Int(D), since among the polynomials F (x) and x − a j with 1 ≤ j ≤ n.
As discussed at the beginning of this section, we may conclude: