Absolute irreducibility of the binomial polynomials

In this paper we investigate the factorization behaviour of the binomial polynomials $\binom{x}{n} = \frac{x(x-1)\cdots (x-n+1)}{n!}$ and their powers in the ring of integer-valued polynomials $\operatorname{Int}(\mathbb{Z})$. While it is well-known that the binomial polynomials are irreducible elements in $\operatorname{Int}(\mathbb{Z})$, the factorization behaviour of their powers has not yet been fully understood. We fill this gap and show that the binomial polynomials are absolutely irreducible in $\operatorname{Int}(\mathbb{Z})$, that is, $\binom{x}{n}^m$ factors uniquely into irreducible elements in $\operatorname{Int}(\mathbb{Z})$ for all $m\in \mathbb{N}$. By reformulating the problem in terms of linear algebra and number theory, we show that the question can be reduced to determining the rank of, what we call, the valuation matrix of $n$. A main ingredient in computing this rank is the following number-theoretical result for which we also provide a proof: If $n>10$ and $n$, $n-1$, \ldots, $n-(k-1)$ are composite integers, then there exists a prime number $p>2k$ that divides one of these integers.


Introduction
In this work, our main objects of interest are the so-called binomial polynomials x(x − 1) · · · (x − n + 1) n! for integers n ≥ 1 as elements of Int(Z) = {f ∈ Q[x] | f (Z) ⊆ Z}, where Z denotes the ring of integers and Q is the field of rational numbers. It is well-known that n! divides the product of any n consecutive integers, and therefore x n indeed is an integer-valued polynomial for all n ≥ 1. Moreover, the binomial polynomials are known to be irreducible in Int(Z), cf. the survey of Cahen and Chabert [5]. A non-zero non-unit a in a (commutative) integral domain D is said to be irreducible if a = bc implies that either b or c is a unit for all b, c ∈ D.
However, the irreducibility of a does not tell anything about the factorization behaviour of the powers a m . An irreducible element a is said to be absolutely irreducible (or a strong atom) if a m factors uniquely into irreducible elements for all integers m ≥ 1. There are plenty of non-absolutely irreducible elements in Int(Z), as it was shown lately by Nakato [16]. Moreover, the recent work of Frisch and Nakato [9] gives a criterion for the absolute irreducibility of integer-valued polynomials with square-free denominators over Z. In particular, it follows from their Theorem 2 that, for all prime numbers p, the binomial polynomial x p is absolutely irreducible (see [9,Example 2.6]). That the latter holds was shown before by McClain [14].
In the present paper we show that the binomial polynomial x n ∈ Int(Z) is absolutely irreducible for all n ≥ 1. Our approach also covers the case that n = p is a prime number, and hence serves in this special case also as an alternative proof to the one in [9].
The binomial polynomials play a central role in the study of Int(Z) as they form a so-called regular Z-module basis, that is, a basis which contains exactly one polynomial of each degree. Implicitly, this fact was already applied by Newton who used integer-valued polynomials to interpolate integervalued functions on Z.
Our viewpoint on the binomial polynomials in this paper relates to the whole area of investigating non-unique factorizations. When mathematicians first explored that factorizations into irreducible elements do not have to be unique in general, they considered this behaviour as pathological and passed over to unique factorizations of ideals into prime ideals in Dedekind domains. It is a movement of the last few decades that non-unique factorizations are viewed in their own right. Since then, the machinery for their investigation has been developed mostly in the direction of Krull domains and monoids, including for instance the concept of the divisor class group of a Krull domain or monoid. For an introduction to this topic, we refer to the textbook of Geroldinger and Halter-Koch [11].
However, almost nothing is known in the area of non-unique factorizations in non-Noetherian Prüfer domains. In this context, general rings of integervalued polynomials where D is an integral domain with quotient field K are of interest. The integral domain Int(D) has been studied very intensively during the last decades and is a standard source for examples and counterexamples. For instance, it is well-known that Int(D) is a non-Noetherian Prüfer domain provided that D is a Dedekind domain with finite residue fields. A profound introduction to the theory of integer-valued polynomial is given in the textbook of Cahen and Chabert [4], and a more recent survey is their work [5].
Coming back to factorization theory, Cahen and Chabert [3] studied elasticity in rings of integer-valued polynomials and proved that Int(D) has infinite elasticity for broad classes of domains D. In a joint publication, Anderson, Cahen, Chapman, Scott and Smith [1] expanded this investigation and further studied factorization properties such as atomicity and elasticity in rings of integer-valued polynomials. Strengthening previous results on infinite elasticity, Chapman and McClain [6] proved that every rational number r ≥ 1 is the elasticity of a polynomial f that is integer-valued on an infinite subset S of a unique factorization domain D, that is, r = L where L and are the maximum and mininum lengths, respectively, of factorizations of f . Frisch [7] proved that in Int(Z) arbitrary sets of lengths can be realized. These results were generalized by Frisch, Nakato and Rissner [10] to rings of integer-valued polynomials on Dedekind domains with infinitely many maximal ideals which are all of finite index. Both papers use a very specific type of irreducible elements of Int(D) to realize the aimed sets of lengths.
On the one hand, recognizing irreducible elements in Int(D) is in general far from being trivial. A partial answer for Int(Z) using an algorithmic approach can be found in [2]. On the other hand, the behavior of products of general irreducible elements in rings of integer-valued polynomials is not at all understood. Our result on the absolute irreducibility of the binomial polynomials leads to a better understanding of one important class of irreducible elements in Int(Z). It also broadens the knowledge of a different aspect of rings of integer-valued polynomials. It has been already mentioned that Int(D) is a non-Noetherian Prüfer domain and is therefore non-Krull, when D is a Dedekind domain with finite residue fields. Nevertheless, Reinhart [18] showed that if D is any factorial domain, then Int(D) is monadically Krull, i.e., the so-called monadic submonoid is a Krull monoid for each f ∈ Int(D). Frisch [8] extended this result to integer-valued polynomial rings on Krull domains. Moreover, Reinhart [19] showed that for a factorial domain D the divisor class group of a monadic submonoid For general polynomials f ∈ Int(D), the structure of the divisor class group of the monadic submonoid generated by f is not known, but it can be easily seen that the following two assertions are equivalent for an irreducible f ∈ Int(D): • The polynomial f ∈ Int(D) is absolutely irreducible.
• The monoid [[f ]] is factorial, i.e., its divisor class group is trivial. Therefore our main result gives a description of a whole new family of divisor class groups of monadic submonoids of Int(Z). This paper is structured as follows. In Section 2 we present our two main results. Theorem 1 states that the binomial polynomials are absolutely irreducible in Int(Z). A second result, Theorem 2, is a number-theoretic result which we need in the proof of Theorem 1 but is interesting in its own right. It states that given a sequence of k consecutive composite integers such that the largest integer is greater than 10, then one of the numbers of this sequence has a prime divisor p > 2k.
The remaining paper is dedicated to the proofs of the main results. In Section 3 we explain the strategy for the proof of Theorem 1 and introduce the necessary notation. The main idea is to rephrase the question of absolute irreducibility of the binomial polynomial x n in terms of linear algebra and number theory. For this purpose, we introduce, what we call, the valuation matrix A n (Definition 3.10) and show that x n is absolutely irreducible in Int(Z) if rank(A n ) = n − 1 (Proposition 3.15). This point of view also motivates the content of Section 4, where we present our number-theoretic toolbox, including a proof of Theorem 2. Finally, in Section 5 we prove that rank(A n ) = n − 1 holds for all n ∈ N which is the final piece for proving the absolute irreducibility of the binomial polynomial x n .

Results
In this section, we present the main results of this paper. The subsequent sections are dedicated to their proofs. Theorem 1. The binomial polynomial x n is absolutely irreducible in Int(Z) for all n ∈ N.
The proof of Theorem 1 is a consequence of the results of Sections 3 and 5. To be more specific, the theorem is the summarized statements of Corollaries 5.12 and 5.15 and Remark 3.3. The theory developed in Section 5 heavily depends on some numbertheoretic results which are built up in Section 4. Next to well-known facts which are collected there, we also prove the following theorem in this section.

Theorem 2.
Let n > 10 be an integer and P the maximal prime number with P ≤ n.

Absolute irreducibility of binomial polynomials and the valuation matrix
In this section we introduce the valuation matrix A n which is associated to the binomial polynomial x n and explain how the question of its absolute irreducibility can be answered by determining the rank of A n . First, we discuss the notion of absolute irreducibility.  (1) b is absolutely irreducible.
(2) For every non-negative integer m and for all f , g ∈ D with b m = f ·g, there exist non-negative integers k, and units u, v ∈ D such that f = ub k and g = vb .

Remark 3.3.
Since x m factors uniquely in Int(Z) for all m ≥ 1, it follows immediately that x 1 = x is absolutely irreducible. This covers the case n = 1.

Remark 3.4.
For the rest of this work, fix a positive integer n ≥ 2. Our goal is to show that for every positive integer m and f , g ∈ Int(Z) the following property holds: Once this is shown, it follows by Remark 3.2 that x n ∈ Int(Z) is absolutely irreducible.
Our first step is to give a precise description of f and g exploiting the fact that Q[x] is a UFD. Then there exist k i and i ∈ N 0 with k i + i = m for 0 ≤ i ≤ n − 1 such that Proof. Since both sides of the equality x n m = f · g factor uniquely in Q[x] into irreducible elements, it follows that for q 1 , q 2 ∈ Q and non-negative integers k 0 , . . . , k n−1 and 0 , . . . , n−1 with , it follows that f (n) = ±1 and g(n) = ±1. This observation together with Equations (1) imply that The assertion follows.
We reformulate the task at hand into a homogeneous system of linear equations for which the vectors (k 0 , . . . , k n−1 ) t and ( 0 , . . . , n−1 ) t are solutions. Given the form of possible factors f and g of x n m in Int(Z) (see Proposition 3.5), it is natural to rephrase the integer-valued condition in terms of p-adic valuations.
for all s ∈ Z and all prime numbers p.
For our purposes, it turns out to be sufficient to consider p-adic valuations for prime numbers p ≤ n and integers of the form s = n + r for r ∈ {1, . . . , p − r n,p − 1} where r n,p is the uniquely determined integer with 0 ≤ r n,p ≤ p − 1 and n ≡ r n,p mod p. However, there are two cases for which this range of r is not sufficient, that is, when n = 2 s with s > 1 and p = 2 we also need r = 2 and in case n = 9 and p = 3 we also need r = 3 and r = 4. This motivates the following notation which we use throughout the remainder of this paper. Notation 3.9. For n ∈ N, we use the following notation.
(1) P n = {p | 0 < p ≤ n prime number} (2) For p ∈ P n , let 0 ≤ r n,p < p be the uniquely determined integer with n ≡ r n,p mod p.
As mentioned above, our goal is to reformulate the question whether the binomial polynomial x n is absolutely irreducible as a homogeneous linear equation system. Definition 3.10. For n ≥ 2, we define the valuation matrix A n of n by The condition that f and g are integer-valued immediately implies that A n k ≥ 0 and A n l ≥ 0, cf. Remark 3.8.
Our next step is to show that the exponent vectors k = (k 0 , . . . , k n−1 ) t and l = ( 0 , . . . , n−1 ) t ∈ N n 0 of possible integer-valued factors f and g of x n m are actually solutions to the homogeneous linear equation system A n x = 0, cf. Remark 3.11. Before we prove this in Proposition 3.15, we need the following lemma which states that the row sums of A n equal zero.

Lemma 3.12.
Let n ∈ N, p ∈ P n and r ∈ R n,p , cf. Notation 3.9. Then Proof. Let p ∈ P n and q and r n,p be the uniquely determined integers such that n = qp + r n,p and 0 ≤ r n,p ≤ p − 1. First, we assume that 1 ≤ r ≤ p − r n,p − 1 (this covers all cases except the one where n is a power of 2 and r = p = 2 and the one where r ∈ {3, 4}, p = 3 and n = 9). Then for all 0 ≤ j ≤ n − 1 at most one of the numbers n + r − j and n − j is divisible by p. Since it follows that n + r − j = kp if and only if k ∈ {1, . . . , q} and hence Similarly, since 1 ≤ n − j ≤ n = qp + r n,p , it follows that which proves the assertion in this case. Next, we discuss the case where n = 2 s is a power of 2 with s > 1 and Finally, it remains to discuss the case where n = 9, p = 3 and r = 3, 4. The two rows are An easy computation verifies the claim and the assertion follows.
Then k, l ∈ ker(A n ).
Proof. The integer-valued condition in the hypothesis implies that A n k ≥ 0 and A n l ≥ 0, cf. Remark 3.11. Moreover, by Lemma 3.12, holds and therefore A n k = A n l = 0.
Remark 3.14. It follows from Lemma 3.12 that (k) 0≤j≤n−1 are elements of ker(A n ) for all k ∈ Q. In particular, this implies that (1) 0≤j≤n−1 ∈ ker(A n ) and rank(A n ) < n. Moreover, note that the exponent vector of x n k (written in the form of This brings us to the main result of this section which states that, in order to show that x n is absolutely irreducible, it suffices to prove that the rank of the valuation matrix A n of n is exactly n − 1. If rank(A n ) = n − 1, then x n is absolutely irreducible. Proof. Let m ≥ 2 be an integer and assume that f , g ∈ Int(Z) such that x n m = f · g. According to Propositions 3.5 and 3.13 it follows that It follows that x n is absolutely irreducible by Remark 3.4. To be able to prove that rank(A n ) = n − 1 for all integers n ≥ 2 in Section 5, we first need to show Theorem 2 which is part of our numbertheoretic toolbox.

Number-theoretic toolbox
The goal of this section is to prove Theorem 2. Given an integer n > 10, let P be the maximal prime number with P ≤ n. We show that for every 2 ≤ k ≤ n − P there exists a prime number p > 2k which divides one of the numbers n, n − 1, . . . , n − k + 1. Note that the condition k ≤ n − P is equivalent to n, n − 1, . . . , n − k + 1 being composite numbers.
The literature provides us a collection of number-theoretic facts which, putting the pieces together, give a proof of Theorem 2. We split the proof into cases and present partial results on their own. We start with the case for large n.    Proof of Proposition 4.1. Let P be the largest prime number with P ≤ n. First, we prove the following Claim. P + 1 > max{(n − P ) + 13, 279 262 (n − P )} Assume for a moment that the claim holds. By hypothesis, the numbers n, n−1, . . . , n−k+1 are composite numbers which implies that n−k+1 > P or, equivalently, n − P ≥ k. Therefore It remains to discuss the case where n < 4, 021, 520. Here, we can use a result of Laishram and Shorey [13]. They proved Grimm's conjecture for sequences of consecutive positive integers whose minimum does not exceed a certain bound.  4.6. Let p k denote the k-th prime number. It is easily seen that p k > 2k for k ≥ 5. Therefore, under the hypothesis of Fact 4.5, there exists a prime number p > 2k which divides one of the numbers n, n − 1, . . . , n − k + 1.
We treat the cases k = 2, 3, 4 in the proof below where we use the following facts.  Finally, we restate the desired theorem and give a complete proof.

Theorem 2.
Let n > 10 be an integer and P the maximal prime number with P ≤ n.

Remark 4.9.
It is immediately clear that n has a prime divisor p > 2 (k = 1) if and only if n is not a power of 2.
It turns out that we need to work around the lack of other prime divisors in case n = 2 x , cf. Definition 3.10.
We treat the cases k = 2, 3, 4 separately. Case k = 2: Note that n and n − 1 are composite numbers which have no common prime factors. Moreover, not both of them are prime powers by Fact 4.7 which implies that n(n − 1) has three pairwise distinct prime factors, one of which is necessarily at least 5 > 2 · 2.
Case k = 3: We use a similar argument as above. If n = 27 then p = 13 > 2 · 3 divides n − 1 = 26. By Fact 4.8, we can therefore assume that at most one of the numbers n, n−1 and n−2 is a prime power and the other two are composite numbers with at least two distinct prime divisors each. Since n, n − 1, and n − 2 can have at most the prime factor 2 in common, it follows that their product n(n − 1)(n − 2) has at least four pairwise distinct prime divisors. In particular, one of these prime divisors is at least 7 > 2 · 3.
Case k = 4: If 27 ∈ {n, n − 1}, then p = 13 > 2 · 4 divides n − 1 or n − 2. We can therefore assume that either at most one of the numbers n, n − 1, n − 2, n − 3 is a prime power or n and n − 3 are the only prime powers among them (cf. Fact 4.8). First, we assume that at most one of the numbers n, n − 1, n − 2 and n − 3 is a prime power and the other three are composite numbers with at least two distinct prime divisors each. Since n, n − 1, n − 2 and n − 3 can have at most the prime factors 2 (occurring twice) and 3 (occurring at most twice) in common, it follows that their product n(n − 1)(n − 2)(n − 3) has at least five pairwise distinct prime divisors. In particular, one of these prime divisors is at least 11 > 2 · 4. Finally, we assume that both n and n − 3 are prime powers and n − 1 and n − 2 are composite numbers with at least two distinct prime divisors each. Note that n and n − 3 are prime powers of two distinct primes and both are not divisible by 3. Therefore, n, n−1, n−2 and n−3 can have at most the prime factor 2 (occurring twice) in common. Again, it follows that their product n(n − 1)(n − 2)(n − 3) has at least five pairwise distinct prime divisors, one of them being at least 11.

rank(A n ) and the p-blocks
In this section we show that rank(A n ) = n − 1 for all n ≥ 2. Once this is shown, it follows from Proposition 3.15 that x n is absolutely irreducible. Our strategy is to show that the columns j ∈ {0, 1, . . . , n − 1} \ {n − P } are linearly independent where P = max P n is the maximal prime number which is less than or equal to n. To do so, we split these n − 1 columns of A n into two groups, namely the "outer" 2(n − P ) columns indexed with {0, . . . , n − P − 1} ∪ {P, P + 1, . . . , n − 1} and the "inner" 2P − n − 1 columns indexed with {n − P + 1, . . . , P − 1}. According to the next lemma, n − P < P − 1 which implies that we can always partition the n − 1 columns in this way.

Lemma 5.1. Let n ≥ 2 be an integer and P = max P n be the largest prime number which is less than or equal to n (cf. Notation 3.9).
Then P − 1 > n − P .
Proof. According to Bertrand's postulate (see Fact 4.2), there is a prime number q with P < q < 2P . Since P = max P n is the maximal prime number at most n, it follows that n < q < 2P . Therefore, n < 2P − 1 which is equivalent to n − P < P − 1.
Next, we introduce the Q-spans of groups of columns of a matrix that are of interest for our investigation. This is done slightly more general as the two column groups of A n we mentioned above. On the one hand, we want to switch between the whole matrix A n and certain submatrices which we introduce below (the p-blocks). On the other hand, to show that the outermost 2(n − P ) columns of A n are linearly independent, we use an inductive argument for which the definition below is convenient.

Definition 5.2.
Let n ≥ 2 and be positive integers and C ∈ Q ×n with columns c 0 , . . . , c n−1 . Further, let P = max P n be the largest prime number less than or equal to n, cf. Notation 3.9.
be the Q-span of the 2(t − s + 1) columns in the range between s and t in the left half of C and in the range between n − (t + 1) and n − (s + 1) in the right half of C (the same range of columns on the "left" and the "right" side of C). For a visualization of the generating columns of these spans, see Figure 1 Then, by Proposition 3.15, x n is absolutely irreducible. Note that Assertion (i) is shown in Proposition 5.14 and Assertions (ii) and (iii) are proven in Corollary 5.10 below.
For our further investigation, it turns out to be useful to split the valuation matrix A n into blocks of rows, each block corresponding to a prime number p ∈ P n .

The structure of a p-block.
Definition 5.4. For n ∈ N and p ∈ P n , we define the p-block B n,p as the |R n,p | × n integer matrix defined by For our purposes, we focus on the distribution of zero and non-zero entries of leftmost p and rightmost p − 1 columns of the p-blocks. In this sense B n,p has a "structure" which is described in Propositions 5.6 and 5.7.

Remark 5.5.
Note that in the exceptional cases where R n.p = {r | 1 ≤ r ≤ p − r n,p − 1}, Propositions 5.6 and 5.7 only give information about the first p − r n,p − 1 rows of the corresponding p-blocks. However, in all situations where we apply the two propositions below, we can rule out these exceptional cases.
Proposition 5.6. Let n ∈ N, p ∈ P n (cf. Notation 3.9) and B n,p = (b r,j ) the corresponding p-block.
For 0 ≤ j ≤ p − 1 and 1 ≤ r ≤ p − r n,p − 1 the following holds: where, as usual, 0 ≤ r n,p < p with n ≡ r n,p mod p.
A visualization of the statement in Proposition 5.6 can be found in Figure  Proof. Let 0 ≤ j ≤ p − 1. In this range of j it follows from the definition of r n,p that v p (n − j) = 0 ⇐⇒ j = r n,p holds. Moreover, v p (n+r−j) = 0 if and only if p | r n,p +r−j, or equivalently, j ≡ r n,p + r mod p. However, since 1 ≤ r ≤ p − r n,p − 1, it follows that 1 + r n,p ≤ r + r n,p ≤ p − 1 and therefore v p (n + r − j) = 0 if and only if j = r n,p + r. Therefore, all entries of the first r n,p columns of B n,p are zero and the entries of the r n,p + 1-st column equal −v p (n − r n,p ) = 0 and the remaining columns r n,p + 1 ≤ j ≤ p − 1 are zero except for the entry in row r = j − r n,p which amounts to v p (n − r n,p ). The assertion follows. Proposition 5.6 describes the zero and non-zero entries of the leftmost p columns of the p-block B n,p . Next, we have a closer look at the rightmost p − 1 columns. Proposition 5.7. Let n ∈ N, p ∈ P n (cf. Notation 3.9) and B n,p = (b r,j ) the corresponding p-block.
For n − (p − 1) ≤ j ≤ n − 1 and 1 ≤ r ∈ p − r n,p − 1 the following holds: A visualization of the statement in Proposition 5.7 can be found in Figure 3. Proof. For n − (p − 1) ≤ j ≤ n − 1, it follows that 1 ≤ n − j ≤ p − 1 and therefore v p (n − j) = 0. Moreover, v p (n + r − j) = 0 is equivalent to p | n − j + r. However, given the ranges of n − j and r, this is the case if and only if n − j + r = p in which case v p (n + r − j) = 1. The assertion follows.

The inner columns of A n .
In this section we prove the results concerning the "inner" columns of A n , namely we show that the Goals (ii) and (iii) formulated in Remark 5.3 hold, see Corollary 5.10. Moreover, this leads to the special case of Theorem 1 where n = P is a prime number, see Corollary 5.12 below.
In this section we exploit the structure of the P -block B n,P where P = max P n is the maximal prime number which is less than or equal to n. Remark 5.8. Let n ≥ 2 and P = max P n , cf. Notation 3.9. If n = 2 s with s ≥ 2, then P > 2 and if n = 9, then P = 7 > 3. Therefore, R n,P = {r | 1 ≤ r ≤ P − r n,P − 1}.
Hence, the P -block B n,P always consists of P − r n,P − 1 rows and Propositions 5.6 and 5.7 refer to the whole P -block, cf. Remark 5.5. Proposition 5.9. Let n ≥ 2 be an integer and P = max P n be the largest prime number which is less than or equal to n.
Then (c) Each column indexed with n − P + 1 ≤ j ≤ P − 1 contains exactly one non-zero entry, namely, the entry in row r = j + P − n is equal 1.  For the second assertion, let v ∈ O [0,n−P −1] (A n ) ∩ I(A n ). Let a n−P +1 , . . . , a P −1 denote the columns of A n which span I(A n ). Then there exist λ n−P +1 , . . . , λ P −1 ∈ Q such that independent. Hence π(v) = 0 and Equation (3) implies that λ j = 0 for n − P + 1 ≤ j ≤ P − 1. Therefore, by plugging into Equation (2), it follows that v = 0 which completes the proof.
It follows from Corollary 5.10 that the "inner" 2P − n − 1 columns of A n , that is, the columns which span I(A n ) are linearly independent. This immediately implies the next corollary.
Corollary 5.11. Let n ≥ 2 be an integer and A n its valuation matrix.
As shown below, the results so far imply the absolute irreducibility of x n in the special case where n = P is a prime number. Note that this has already been shown by Frisch and Nakato [9, Example 2.6] and our proof only serves as an alternative.
Corollary 5.12. Let P be a prime number, then x P is absolutely irreducible.
Proof. If n = P , then rank(A P ) ≥ 2P − P − 1 = P − 1 by Corollary 5.11. Since rank(A P ) < P by Remark 3.14, it follows rank(A P ) = P − 1. According to Proposition 3.15, x P is absolutely irreducible. 5.3. The outer columns of A n . For a composite number n ≥ 2, the span of the "outer" columns of A n is not trivial. The goal of this section is to show that dim O [0,n−P −1] (A n ) = 2(n − P ), where (for the remainder of this work) P = max P n is the largest prime number which is less than or equal to n (cf. Notation 3.9), see Proposition 5.14. This is the final ingredient to prove that rank(A n ) = n−1 which then further implies that x n is absolutely irreducible, cf. Remark 5.3.
As above, we have to exploit the structure of certain p-blocks to reach this goal. In contrast to the arguments in the previous subsection, we need to find more than one suitable p-block. Which choices to make is explained in detail in the proof of Proposition 5.14. The next proposition gives information about the dimension of O [rn,p,k−1] (B n,p ) for certain choices of k. Proposition 5.13. Let n ≥ 2 be a composite number, p ∈ P n (cf. Notation 3.9) be a prime number and B n,p the corresponding p-block.
For the remainder of the proof, we assume that n = 9 or p = 3. It follows from r n,p + 1 ≤ k ≤ min{n − P, p+rn,p− 1 2 } that r n,p ≤ p − 3 and 2k ≤ p + r n,p − 1. In particular, p > 2 holds and it follows that the p-block B n,p has p − r n,p − 1 ≥ 2(k − r n,p ) rows, cf. Remark 5.5.
Moreover, since k ≤ p+rn,p−1 2 ≤ p − 1, we can apply Propositions 5.6 and 5.7 to describe the outermost left k and right k columns of the (whole) p-block B n,p , depicted in Figure 6. They are of the following form: (c) The columns r n,p + 1 ≤ j ≤ k − 1 have exactly one non-zero entry, v p (n − r n,p ), in row r = j − r n,p , that is, the first k − (r n,p + 1) rows of these columns form a diagonal matrix with v p (n − r n,p ) on the diagonal and all entries below are zero. (d) The columns n − k ≤ j ≤ n − (r n,p + 1) contain exactly one non-zero entry, namely 1, in row r = j − (n − p), that is, the lower k − r n,p rows of these columns form the identity matrix and all entries above are zero. (e) Since k < p+rn,p 2 by assumption, the first k − (r n,p + 1) rows and the last k − r n,p rows are disjoint and there is at least one row in the part in the middle, namely there is at least one row indexed with k − r n,p ≤ r ≤ p − k − 1. Let b j denote the j-th column of B n,p and assume that there are λ j ∈ Q with j ∈ {r n,p , . . . , k − 1} ∪ {n − k, . . . , n − (r n,p + 1)} such that Since b rn,p is the only column which has a non-zero entry in the rows indexed with k − r n,p ≤ r ≤ p − k − 1, Equation (4) implies that λ rn,p = 0. Now for each of the remaining rows, there is exactly one column with a non-zero entry in this (and no other) row. With the same reasoning we can conclude that λ j = 0 for r n,p + 1 ≤ j ≤ k − 1 and n − k ≤ j ≤ n − (r n,p + 1). Therefore, these 2(k − r n,p ) columns of B n,p are linearly independent.
Proposition 5.14. Let n ≥ 2 be a composite integer and A n its valuation matrix.
Proof. We treat the cases n = 9 and n = 10 first as we want to exclude them below. For n = 9, observe that P = 7 and hence n − P = 2. It follows from a direct verification that the outer four columns of B 9,3 (displayed in Figure 5) are linearly independent and hence dim The valuation matrix A 10 is displayed in Figure 7. A direct computation verifies that rank(O [0,2] (A 10 )) = 6.  From now on, assume that n ≥ 2 is a composite integer with n = 9 and n = 10. We prove by induction on 1 ≤ k ≤ n − P that If n = 2 s is a proper power of 2, then R 2 s ,2 = {1, 2} by definition which implies that B 2 s ,2 has two rows. The outermost columns of B 2 s ,2 are displayed in Figure 8. As v 2 (n) = s ≥ 2, it follows that dim O [0,0] (B 2 s ,2 ) = 2. If n is not a power of 2, then n has a prime divisor p ≥ 3. It follows that r n,p = 0 and hence 1 = r n,p + 1 ≤ k = 1 ≤ min{n − P, p−1 2 }. Therefore, we can apply Proposition 5.13 and conclude that dim O [0,0] (B n,p ) = 2. This completes the proof of the base case.
Observe that the induction base case also covers the cases n = 4, n = 6 and n = 8 because in each of these cases n − P = 1. Therefore, we already handled all cases where n ≤ 10.
We proceed with the induction step and prove the following claim. Claim. There exists p ∈ P n with r n,p ≤ k − 1 such that Assume for a moment that this claim holds. Since r n,p < k we can apply the induction hypothesis, that is, dim O = 2r n,p + 2(k − r n,p ) = 2k and the assertion follows.
Next, we prove the claim. By Theorem 2 (for which we needed to assume that n > 10), there exists p ∈ P n such that p > 2k and r n,p ≤ k − 1.
For 0 ≤ j ≤ n − 1, let a j denote the j-th column of A n and λ j ∈ Q such that As the columns b (p) j with j ∈ J 1 of B n,p are linearly independent by Proposition 5.13, it follows that λ j = 0 for all j ∈ J 1 . Therefore, plugging into Equation (5), it follows that v = 0 which completes the proof of the claim. Proposition 5.14 together with Corollary 5.10 imply that rank(A n ) = n−1 for all composite numbers n ≥ 2, cf. Remark 5.3. Using Proposition 3.15 yields the corollary below.