Absolute irreducibility of the binomial polynomials
AABSOLUTE IRREDUCIBILITY OF THE BINOMIALPOLYNOMIALS
ROSWITHA RISSNER AND DANIEL WINDISCH
Abstract.
In this paper we investigate the factorization behaviour ofthe binomial polynomials (cid:0) xn (cid:1) = x ( x − ··· ( x − n +1) n ! and their powers in thering of integer-valued polynomials Int( Z ). While it is well-known thatthe binomial polynomials are irreducible elements in Int( Z ), the factor-ization behaviour of their powers has not yet been fully understood.We fill this gap and show that the binomial polynomials are absolutelyirreducible in Int( Z ), that is, (cid:0) xn (cid:1) m factors uniquely into irreducible ele-ments in Int( Z ) for all m ∈ N . By reformulating the problem in termsof linear algebra and number theory, we show that the question can bereduced to determining the rank of, what we call, the valuation matrixof 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, . . . , n − ( k −
1) are composite integers, then there exists a primenumber p > k that divides one of these integers. Introduction
In this work, our main objects of interest are the so-called binomial poly-nomials xn ! = x ( x − · · · ( x − n + 1) n !for integers n ≥ 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 (cid:0) xn (cid:1) indeed is an integer-valued polynomial for all n ≥
1. More-over, the binomial polynomials are known to be irreducible in Int( Z ), cf. thesurvey of Cahen and Chabert [3]. 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 factoriza-tion behaviour of the powers a m . An irreducible element a is said to be ab-solutely irreducible (or a strong atom ) if a m factors uniquely into irreducibleelements for all integers m ≥
1. There are plenty of non-absolutely irre-ducible elements in Int( Z ), as it was shown lately by Nakato [12]. Moreover,the recent work of Frisch and Nakato [6] gives a criterion for the absoluteirreducibility of integer-valued polynomials with square-free denominatorsover Z . In particular, it follows from their Theorem 2 that, for all prime R. Rissner is supported by the Austrian Science Fund (FWF): P 28466.D. Windisch is supported by the Austrian Science Fund (FWF): P 30934. a r X i v : . [ m a t h . A C ] S e p ROSWITHA RISSNER AND DANIEL WINDISCH numbers p , the binomial polynomial (cid:0) xp (cid:1) is absolutely irreducible (see [6,Example 2.6]).In the present paper we show that the binomial polynomial (cid:0) xn (cid:1) ∈ Int( Z )is absolutely-irreducible for all n ≥
1. Our approach also covers the casethat n = p is a prime number, and hence serves in this special case also asan alternative proof to the one in [6].The binomial polynomials play a central role in the study of Int( Z ) asthey form a so-called regular Z -module basis, that is, a basis which containsexactly one polynomial of each degree. Implicitly, this fact was already ap-plied by Newton who used integer-valued polynomials to interpolate integer-valued functions on Z .Our viewpoint on the binomial polynomials in this paper relates to thewhole area of investigating non-unique factorizations. When mathemati-cians first explored that factorizations into irreducible elements do not haveto be unique in general, they considered this behaviour as pathological andpassed over to unique factorizations of ideals into prime ideals in Dedekinddomains. It is a movement of the last few decades that non-unique factor-izations are viewed in their own right. Since then, the machinery for theirinvestigation has been developed mostly in the direction of Krull domainsand monoids, including for instance the concept of the divisor class groupof a Krull domain or monoid. For an introduction to this topic, we refer tothe textbook of Geroldinger and Halter-Koch [8].However, almost nothing is known in the area of non-unique factorizationsin non-Noetherian Prüfer domains. In this context, general rings of integer-valued polynomials Int( D ) = { f ∈ K [ x ] | f ( D ) ⊆ D } where D is an integral domain with quotient field K are of interest. Theintegral domain Int( D ) has been studied very intensively during the lastdecades and is a standard source for examples and counterexamples. Forinstance, it is well-known that Int( D ) is a non-Noetherian Prüfer domainprovided that D is a Dedekind domain with finite residue fields. A pro-found introduction to the theory of integer-valued polynomial is given inthe textbook of Cahen and Chabert [2], and a more recent survey is theirwork [3].Coming back to factorization theory, Frisch [4] proved that in Int( Z ) ar-bitrary sets of lengths can be realized. These results were generalized byFrisch, Nakato and Rissner [7] to rings of integer-valued polynomials onDedekind domains with infinitely many maximal ideal which are all of finiteindex. 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 gen-eral far from being trivial. A partial answer for Int( Z ) using an algorithmicapproach can be found in [1]. On the other hand, the behaviour of prod-ucts of general irreducible elements in rings of integer-valued polynomials isnot at all understood. Our result on the absolute irreducibility of the bi-nomial polynomials leads to a better understanding of one important classof irreducible elements in Int( Z ). It also broadens the knowledge of a dif-ferent aspect of rings of integer-valued polynomials. It has been already BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 3 mentioned that Int( D ) is a non-Noetherian Prüfer domain and is thereforenon-Krull, when D is a Dedekind domain with finite residue fields. Never-theless, Reinhart [13] showed that if D is any factorial domain, then Int( D )is monadically Krull , i.e., the so-called monadic submonoid [[ f ]] = { g ∈ Int( D ) | g divides f n for some n ∈ N } is a Krull monoid for each f ∈ Int( D ). Frisch [5] extended this result tointeger-valued polynomial rings on Krull domains. Moreover, Reinhart [14]showed that for a factorial domain D the divisor class group of a monadicsubmonoid [[ f ]] of Int( D ) is free Abelian for every f ∈ D [ x ] \ { } . Forgeneral polynomials f ∈ Int( D ), the structure of the divisor class groupof the monadic submonoid generated by f is not known, but it can beeasily 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 divisorclass groups of monadic submonoids of Int( Z ).This paper is structured as follows. In Section 2 we present our two mainresults. Theorem 1 states that the binomial polynomials are absolutelyirreducible in Int( Z ). A second result, Theorem 2, is a number-theoreticresult which we need in the proof of Theorem 1 but is interesting in its ownright. It states that given a sequence of k consecutive composite integerssuch that the largest integer is greater than 10, then one of the numbers ofthis sequence has a prime divisor p > k .The remaining paper is dedicated to the proofs of the main results. InSection 3 we explain the strategy for the proof of Theorem 1 and introducethe necessary notation. The main idea is to rephrase the question of absoluteirreducibility of the binomial polynomial (cid:0) xn (cid:1) in terms of linear algebra andnumber theory. For this purpose, we introduce, what we call, the valuationmatrix A n (Definition 3.10) and show that (cid:0) xn (cid:1) is absolutely irreducible inInt( Z ) if rank( A n ) = n − A n ) = n − n ∈ N which is the final piece for proving theabsolute irreducibility of the binomial polynomial (cid:0) xn (cid:1) .2. Results
In this section, we present the main results of this paper. The subsequentsections are dedicated to their proofs.
Theorem 1.
The binomial polynomial (cid:0) xn (cid:1) is absolutely irreducible in Int( Z ) for all n ∈ N . The proof of Theorem 1 is a consequence of the results of Sections 3and 5. To be more specific, the theorem is the summarized statements ofCorollaries 5.12 and 5.15 and Remark 3.3.
Corollary 2.1.
The monadic submonoid [[ (cid:0) xn (cid:1) ]] of Int( Z ) is factorial for all n ≥ . In particular, it has trivial divisor class group. ROSWITHA RISSNER AND DANIEL WINDISCH
The theory developed in Section 5 heavily depends on some number-theoretic results which are built up in Section 4. Next to well-known factswhich are collected there, we also prove the following theorem in this section.
Theorem 2.
Let n > be an integer and P the maximal prime numberwith P ≤ n .If ≤ k ≤ n − P then there exists a prime number p > k which dividesone of the numbers n , n − , . . . , n − ( k − . Absolute irreducibility of binomial polynomials and thevaluation matrix
In this section we introduce the valuation matrix A n which is associatedto the binomial polynomial (cid:0) xn (cid:1) and explain how the question of its absoluteirreducibility can be answered by determining the rank of A n . First, wediscuss the notion of absolute irreducibility. Definition 3.1.
Let D be an integral domain and b ∈ D be an irreducibleelement. We say b is absolutely irreducible if b m factors uniquely into irre-ducible elements for every m ∈ N . Remark 3.2.
Let D be an integral domain and b ∈ D be an irreducibleelement. A straight forward verification shows that the following assertionsare equivalent:(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 followsimmediately that (cid:0) x (cid: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. Ourgoal is to show that for every positive integer m and f , g ∈ Int( Z ) thefollowing property holds: xn ! m = f · g = ⇒ f = ± xn ! k and g = ± xn ! ‘ with k, ‘ ∈ N . Once this is shown, it follows by Remark 3.2 that (cid:0) xn (cid:1) ∈ Int( Z ) is absolutelyirreducible.Our first step is to give a precise description of f and g exploiting thefact that Q [ x ] is a UFD. Proposition 3.5.
Let n , m ≥ and f , g ∈ Int( Z ) with (cid:0) xn (cid:1) m = f · g .Then there exist k i and ‘ i ∈ N with k i + ‘ i = m for ≤ i ≤ n − suchthat f = ± n − Y i =0 (cid:18) x − in − i (cid:19) k i and g = ± n − Y i =0 (cid:18) x − in − i (cid:19) ‘ i holds. BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 5
Proof.
Since both sides of the equality (cid:0) xn (cid:1) m = f · g factor uniquely in Q [ x ]into irreducible elements, it follows that(1) f = q n − Y i =0 ( x − i ) k i and g = q n − Y i =0 ( x − i ) ‘ i for q , q ∈ Q and non-negative integers k , . . . , k n − and ‘ , . . . , ‘ n − with k i + ‘ i = m for all i ∈ { , . . . , n − } .Evaluating (cid:0) xn (cid:1) m = f · g at x = n implies f ( n ) · g ( n ) = 1. Moreover, as f , g ∈ Int( Z ), it follows that f ( n ) = ± g ( n ) = ±
1. This observationtogether with Equations (1) imply that1 q = ± n − Y i =0 ( n − i ) k i and 1 q = ± n − Y i =0 ( n − i ) ‘ i . The assertion follows. (cid:3)
Remark 3.6.
Let k , . . . , k n − and ‘ , . . . , ‘ n − ∈ N be the exponents ofthe factors f and g of (cid:0) xn (cid:1) m , cf. Proposition 3.5. If k = k = · · · = k n − (and hence ‘ = ‘ = · · · = ‘ n − ), then f = (cid:0) xn (cid:1) k and g = (cid:0) xm (cid:1) ‘ .Having Remark 3.4 in mind, we aim to show that, for all possible factors f and g , the corresponding exponents satisfy k = k = · · · = k n − and ‘ = ‘ = · · · = ‘ n − in order to prove that the binomial polynomial (cid:0) xn (cid:1) isabsolutely irreducible.We reformulate the task at hand into a homogeneous system of linearequations for which the vectors ( k , . . . , k n − ) t and ( ‘ , . . . , ‘ n − ) t are so-lutions. Given the form of possible factors f and g of (cid:0) xn (cid:1) m in Int( Z ) (seeProposition 3.5), it is natural to rephrase the integer-valued condition interms of p -adic valuations. Notation 3.7.
For w ∈ Q and a prime number p , we denote by v p ( w ) the p -adic valuation of w . Remark 3.8.
Let k , . . . , k n − and ‘ , . . . , ‘ n − be the exponents of thefactors f and g of (cid:0) xn (cid:1) m , cf. Proposition 3.5. Since f and g are integer-valuedpolynomials, it follows that v p ( f ( s )) = n − X j =0 ( v p ( s − j ) − v p ( n − j )) k j ≥ v p ( g ( s )) = n − X j =0 ( v p ( s − j ) − v p ( n − j )) ‘ j ≥ s ∈ Z and all p ∈ P .For our purposes, it turns out to be sufficient to consider p -adic valuationsfor prime numbers p ≤ n and integers of the form s = n + r for r ∈{ , . . . , p − r n,p − } where r n,p is the uniquely determined integer with0 ≤ r n,p ≤ p − n ≡ r n,p mod p . However, there are two cases forwhich this range of r is not sufficient, that is, when n = 2 s with s > p = 2 we also need r = 2 and in case n = 9 and p = 3 we also need r = 3and r = 4. This motivates the following notation which we use throughoutthe remainder of this paper. ROSWITHA RISSNER AND DANIEL WINDISCH
Notation 3.9.
For n ∈ N , we use the following notation.(1) P n = { p | < 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 .(3) For p ∈ P n , we set R n,p = { , } if n = 2 s with s > p = 2 { , , , } if n = 9 and p = 3 { r | ≤ r ≤ p − r n,p − } elseAs mentioned above, our goal is to reformulate the question whether thebinomial polynomial (cid:0) xn (cid:1) is absolutely irreducible as a homogeneous equationsystem. Definition 3.10.
For n ≥
2, we define the valuation matrix A n of n by A n = ( v p ( n + r − j ) − v p ( n − j )) p ∈P n ,r ∈R n,p ≤ j ≤ n − cf. Notation 3.9. Remark 3.11.
Let k = ( k , . . . , k n − ) t and l = ( ‘ , . . . , ‘ n − ) t ∈ N n be (thevectors of) the exponents of the factors f and g of (cid:0) xn (cid:1) m , cf. Proposition 3.5.The condition that f and g are integer-valued immediately implies that A n k ≥ A n l ≥ , cf. Remark 3.8.Our next step is to show that the exponent vectors k = ( k , . . . , k n − ) t and l = ( ‘ , . . . , ‘ n − ) t ∈ N n of possible integer-valued factors f and g of (cid:0) xn (cid:1) m are actually solutions to the homogeneous equation system A n x = 0,cf. Remark 3.11. Before we prove this in Proposition 3.15, we need thefollowing 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 n − X j =0 ( v p ( n + r − j ) − v p ( n − j )) = 0 Proof.
Let p ∈ P n and q and r n,p be the uniquely determined integers suchthat n = qp + r n,p and 0 ≤ r n,p ≤ p −
1. First, we assume that 0 ≤ r ≤ p − r n,p − n is a power of 2 and r = p = 2 and the one where r ∈ { , } , p = 3 and n = 9). Then for all0 ≤ j ≤ n − n + r − j and n − j is divisible by p . Since2 = n + 1 − ( n − ≤ n + r − j ≤ n + ( p − r n,p −
1) = qp + ( p − , it follows that n + r − j = kp if and only if k ∈ { , . . . , q } and hence n − X j =0 v p ( n + r − j ) = q X k =1 v p ( kp ) . BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 7
Similarly, since 1 ≤ n − j ≤ n = qp + r n,p , it follows that n − X j =0 v p ( n − j ) = q X k =1 v p ( kp )which proves the assertion in this case.Next, we discuss the case where n = 2 s is a power of 2 with s > r = p = 2. We want to show that P n − j =0 ( v ( n + 2 − j ) − v ( n − j )) = 0.Let v j = v ( n + 2 − j ) − v ( n − j ). As n is even, v j = 0 whenever j is odd.Moreover, for j ∈ Z , v ( n + 2 − j ) = v ( n − − j ) = 1 and v ( n − j ) > j ∈ Z , v j + v j +2 = v ( n + 2 − j ) − v ( n − j ) + v ( n − j ) − v ( n − − j ) = 0which implies that n − X j =0 v j = n − X j =0 j ∈ Z v j = n − X j =0 j ∈ Z v j + v j +2 = 0 . Finally, it remains to discuss the case where n = 9, p = 3 and r = 3, 4.The two rows are (cid:18) − − − − (cid:19) . An easy computation verifies the claim and the assertion follows. (cid:3)
Lemma 3.12 is key to show that the exponent vectors k = ( k , . . . , k n − ) t and l = ( ‘ , . . . , ‘ n − ) t of potential factors f and g of (cid:0) xn (cid:1) m can be viewed asthe solutions of the homogeneous equation system A n x = 0, or equivalently,as elements of ker( A n ). Proposition 3.13.
Let n , m ≥ be integers and k = ( k , . . . , k n − ) t and l = ( ‘ , . . . , ‘ n − ) t ∈ N n such that n − Y i =0 (cid:18) x − in − i (cid:19) k i , n − Y i =0 (cid:18) x − in − i (cid:19) ‘ i ∈ Int( Z ) and k j + ‘ j = m for all ≤ j ≤ n − .Then k , l ∈ ker( A n ) .Proof. The integer-valued condition in the hypothesis implies that A n k ≥ A n l ≥
0, cf. Remark 3.11.Moreover, by Lemma 3.12,0 = m · n − X j =0 ( v p ( n + r − j ) − v p ( n − j ))= n − X j =0 ( v p ( n + r − j ) − v p ( n − j )) · ( k j + l j )= A n ( k + l ) = A n k + A n l holds and therefore A n k = A n l = 0. (cid:3) ROSWITHA RISSNER AND DANIEL WINDISCH
Remark 3.14.
It follows from Lemma 3.12 that ( k ) ≤ j ≤ n − are elements ofker( A n ) for all k ∈ Q . In particular, this implies that (1) ≤ j ≤ n − ∈ ker( A n ),or equivalently, rank( A n ) < n .Moreover, note that the exponent vector of (cid:0) xn (cid:1) k (written in the form ofProposition 3.5) is a scalar multiple of the element (1) ≤ j ≤ n − .This brings us to the main result of this section which states that, in orderto show that (cid:0) xn (cid:1) is absolutely irreducible, it suffices to prove that the rankof the valuation matrix A n of n is exactly n − Proposition 3.15.
Let n ≥ be an integer and A n the valuation matrix of n . If rank( A n ) = n − , then (cid:0) xn (cid:1) is absolutely irreducible.Proof. Let m ≥ f , g ∈ Int( Z ) such that (cid:0) xn (cid:1) m = f · g . According to Propositions 3.5 and 3.13 it follows that f = ± n − Y i =0 (cid:18) x − in − i (cid:19) k i and g = ± n − Y i =0 (cid:18) x − in − i (cid:19) ‘ i where ( k , . . . , k n − ) t , ( ‘ , . . . , ‘ n − ) t ∈ ker( A n ) ∩ N n .Since dim ker( A n ) = n − rank( A n ) = 1 and (1) ≤ j ≤ n − ∈ ker( A n ) byLemma 3.12, it follows that ker( A n ) = span Q { (1) ≤ j ≤ n } . Therefore, k = k = · · · = k n − and ‘ = ‘ = · · · = ‘ n − which implies that f = (cid:0) xn (cid:1) k and g = (cid:0) xn (cid:1) ‘ .It follows that (cid:0) xn (cid:1) is absolutely irreducible by Remark 3.4. (cid:3) To be able to prove that rank( A n ) = n − n in Section 5, we firstneed to show Theorem 2 which is part of our number-theoretic toolbox.4. Number-theoretic toolbox
The goal of this section is to prove Theorem 2. Given an integer n > P be the maximal prime number with P ≤ n . We show that for every2 ≤ k ≤ n − P there exists a prime number p > k which divides one ofthe numbers n , n −
1, . . . , n − k + 1. Note that the condition k ≤ n − P isequivalent 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 proofinto cases and present partial results on their own. We start with the casefor large n . Proposition 4.1.
Let ≤ k < n be positive integers with n ≥ , , .If n , n − ,. . . , n − k + 1 are composite numbers, then one of them has aprime divisor p > k . For the proof we use the following facts from the literature.
Fact 4.2 (Bertrand’s postulate) . For all integers n ≥ , there exists a primenumber p with n < p < n . Fact 4.3 ([15, Theorem 12]) . Let m be an integer with m ≥ , , .Then there exists a prime number p such that m < p < (1 + ) m . BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 9
Fact 4.4 ([9, Theorem 1]) . Let m and k ≥ be positive integers such that m > max { k + 13 , k } .Then the product m ( m + 1) · · · ( m + k − has a prime factor greater than k .Proof. Let P be the largest prime number with P ≤ n . First, we prove thefollowing Claim. P + 1 > max { ( n − P ) + 13 , ( 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 n − k + 1 ≥ P + 1 > max { ( n − P ) + 13 , n − P ) } ≥ max { k + 13 , k } and the assertion follows from Fact 4.4 with m = n − k + 1.It remains to prove the claim. Let Q be the smallest prime number with Q > n . According to Fact 4.3 it follows that
P > n ≥ , ,
760 and
Q < (1 + ) P .Therefore, 2 P − n > P − Q > (1 − ) P >
12 which implies that P + 1 > ( n − P ) + 13. Moreover, since 2(1 + ) < it follows that n − < n < Q < (1 + ) P and hence P + 1 > ( n − P ). Thiscompletes the proof of the claim. (cid:3) It remains to discuss the case where n < , , Fact 4.5. [10, Theorem 1]
Let k < n such that n − k + 1 ≤ . · .If n , n − , . . . , n − k + 1 are composite numbers then there exist pairwisedistinct prime numbers p , p , . . . , p k − with p i | n − i for all ≤ i ≤ k − . Remark 4.6.
Let p k denote the k -th prime number. It is easily seen that p k > k for k ≥
5. Therefore, under the hypothesis of Fact 4.5, there existsa prime number p > k 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 thefollowing facts. Fact 4.7 (Catalan’s conjecture, [11]) . Mihăilescu proved Catalan’s conjec-ture which states that the only consecutive positive integers which are non-prime prime powers are and . Fact 4.8 (Pillai’s conjecture for difference 2, [16]) . The only non-primeprime powers p x and q y less than with p x − q y = 2 are and . Finally, we restate the desired theorem and give a complete proof.
Theorem 2.
Let n > be an integer and P the maximal prime numberwith P ≤ n .If ≤ k ≤ n − P then there exists a prime number p > k which dividesone of the numbers n , n − , . . . , n − ( k − . Remark 4.9.
It is immediately clear that n has a prime divisor p > 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 divisorsin case n = 2 x , cf. Definition 3.10. Proof.
For n ≥ , , n < , ,
520 and k ≥
5, the assertion follows from Fact 4.5, cf. Re-mark 4.6.We treat the cases k = 2, 3, 4 separately. Case k = 2 : Since n and n − n and n − n ( n −
1) has three distinct prime divisors, one of which is necessarilyat least 5 ≥ · Case k = 3 : We use a similar argument as above. If n = 27 then p = 13 > · n − n , n − n − | n .Moreover, either both n and n − n − n , n − n − n n − n − Table 1.
Distribution of 2 and 3 as prime factors of n , n − n − > · Case k = 4 : If 27 ∈ { n, n − } , then p = 13 > · n − n − n , n − n − n − n and n − n is even.At most two of the numbers are divisible by 3. Moreover, we assume herethat at most one of the numbers n , n − n − n − n and n − n , n − n − n − > · (cid:3) BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 11 n n − n − n − Table 2.
Distribution of 2 and 3 as prime factors of n , n − n − n − A n ) and the p -blocks In this section we show that rank( A n ) = n − n ≥
2. Once this isshown, it follows from Proposition 3.15 that (cid:0) xn (cid:1) is absolutely irreducible.Our strategy is to show that the columns j ∈ { , , . . . , n − } \ { n − P } are linearly independent where P = max P n is the maximal prime numberwhich is less than or equal n . To do so, we split these n − A n into two groups, namely the “outer” 2( n − P ) columns indexed with { , . . . , n − P − } ∪ { P, P + 1 , . . . , n − } and the “inner” 2 P − n − { n − P + 1 , . . . , P − } . According to the next lemma, n − P < P − n − Lemma 5.1.
Let n ≥ be an integer and P = max P n be the largest primenumber which is less than or equal n (cf. Notation 3.9).Then P − > n − P .Proof. According to Bertrand’s postulate (see Fact 4.2), there is a primenumber q with P < q < P . Since P = max P n is the maximal primenumber at most n , it follows that n < q < P . Therefore, n < P − n − P < P − (cid:3) Next, we introduce the Q -spans of groups of columns of a matrix thatare of interest for our investigation. This is done slightly more general asthe two column groups of A n we mentioned above. On the one hand, wewant to switch between the whole matrix A n and certain submatrices whichwe introduce below (the p -blocks). On the other hand, to show that theoutermost 2( n − P ) columns or A n are linearly independent, we use aninductive argument for which the definition below is convenient. Definition 5.2.
Let n ≥ ‘ be positive integers and C ∈ Q ‘ × n withcolumns c , . . . , c n − . Further, let P = max P n be the largest prime numberless than or equal to n , cf. Notation 3.9.(1) We denote by I ( C ) = span { c n − P +1 , c n − P +1 , . . . , c P − } be the spanof the “inner” 2 P − n − C .(2) For 0 ≤ s < t ≤ n − P −
1, we denote by O [ s,t ] ( C ) = span { c j | s ≤ j ≤ t and n − ( t + 1) ≤ j ≤ n − ( s + 1) } 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 the columns n − ( t + 1) and n − ( s + 1) inthe left 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. ∗ ∗ ∗ ∗ s t − s + 1 n − P − t n − P − t t − s + 1 sj = s j = t j = n − ( s + 1) j = n − ( t − j = n − P I ( C ) j = n − P + 1 j = P − Figure 1. I ( C ) is spanned by the columns in the greyzigzag-area; O [ s,t ] ( C ) is spanned by the columns in the greenrectangles Remark 5.3.
Let n ∈ N and P = max P n , cf. Notation 3.9.Our goal is to show that(i) dim O [0 ,n − P − ( A n ) = 2( n − P ),(ii) dim I ( A n ) = 2 P − n − I ( A n ) ∩ O [0 ,n − P − ( A n ) = .Together this then implies that the sum I ( A n ) + O [0 ,n − P − ( A n ) is directanddim (cid:16) I ( A n ) ⊕ O [0 ,n − P − ( A n ) (cid:17) = 2( n − P ) + (2 P − n −
1) = n − . Since n > rank( A n ) by Remark 3.14, it further follows that n > rank( A n ) ≥ dim (cid:16) I ( A n ) ⊕ O [0 ,n − P − ( A n ) (cid:17) = n − A n ) = n − . Then, by Proposition 3.15, (cid:0) xn (cid:1) 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 valuationmatrix A n into blocks of rows, each block corresponding to a prime number p ∈ P n .5.1. 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 B n,p = ( v p ( n + r − j ) − v p ( n − j )) r ∈R n,p ≤ j ≤ n − For our purposes, we focus on the distribution of zero and non-zero entriesof leftmost p and rightmost p − p -blocks. In this sense B n,p has a “structure” which is described in Propositions 5.6 and 5.7. BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 13
Remark 5.5.
Note that in the exceptional cases where R n.p = { r | ≤ r ≤ p − r n,p − } , Propositions 5.6 and 5.7 only give information about the first p − r n,p − p -blocks. However, in all situationswhere apply the two propositions below, we can rule out these exceptionalcases. 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 ≤ j ≤ p − and ≤ r ≤ p − r n,p − the following holds: b r,j = − v p ( n − r n,p ) j = r n,p v p ( n − r n,p ) j = r + r n,p else . where, as usual, ≤ r n,p < p with n ≡ r n,p mod p . A visualization of the statement in Proposition 5.6 can be found in Fig-ure 2. − v v ∗ − v v p − r n,p − r n,p j = r n,p j = p − Figure 2.
Leftmost p columns of (the first p − r n,p − B n,p where v = v p ( n − r n,p ), cf. Proposition 5.6 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 that1 + r n,p ≤ r + r n,p ≤ p − v p ( n + r − j ) = 0 if and only if j = r n,p + r .Therefore, all entries of first r n,p columns of B n,p are zero and the entriesof 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 − r = j − r n,p whichamounts to v p ( n − r n,p ). The assertion follows. (cid:3) 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 − 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 − ≤ j ≤ n − and r ∈ R n,p the following holds: b r,j = ( j = n − p + r else . A visualization of the statement in Proposition 5.7 can be found in Fig-ure 3. 1 ∗ p − r n,p − j = n − ( p − j = n − ( r n,p + 1) r n,p Figure 3.
Rightmost p − p − r n,p − B n,p , cf. Proposition 5.7 Proof.
For n − ( p − ≤ j ≤ n −
1, it follows that 1 ≤ n − j ≤ p − 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 ifand only if n − j + r = p in which case v p ( n + r − j ) = 1. The assertionfollows. (cid:3) The inner columns of A n . In this section we prove the results con-cerning 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, thisleads to the special case of Theorem 1 where n = P is a prime number, seeCorollary 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 ≥ P = max P n , cf. Notation 3.9. If n = 2 s with s ≥
2, then
P > n = 9, then P = 7 >
3. Therefore, R n,P = { r | ≤ r ≤ P − r n − P − } .Hence, the P -block B n,P always consists of P − r n,P − P -block, cf. Remark 5.5. Proposition 5.9.
Let n ≥ be an integer and P = max P n be the largestprime number which is less than or equal to n .Then(1) dim I ( B n,P ) = 2 P − n − (2) O [0 ,n − P − ( B n,P ) = Proof.
By Lemma 5.1, n − P < P − r n,P = n − P . Moreover, R n,P = { , , . . . , P − n − } , cf. Remark 5.8. ByPropositions 5.6 and 5.7 the P -block B n,P is of the following form, see alsoFigure 4: BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 15 (a) All columns indexed with j ∈ { , . . . , n − P − } ∪ { P, . . . , P + 1 } arezero columns.(b) Each entry of the column j = r n,P = n − P is equal − n − P + 1 ≤ j ≤ P − r = j + P − n is equal 1. − − P − n − n − P j = n − P j = P − Figure 4. B n,P It immediately follows from (a), that O [0 ,n − P − ( B n,P ) = . Moreover,as I ( B n,P ) is spanned by 2 P − n − P − n −
1. By (c), the submatrix of B n,P consisting of columns n − P + 1 ≤ j ≤ P − P − − ( n − P −
1) + 1 =2 P − n −
1. As all these columns are elements of I ( B n,P ) it follows thatdim I ( B n,P ) = 2 P − n − (cid:3) Corollary 5.10.
Let n ≥ be an integer and P = max P n be the largestprime number less than or equal to n , cf. Notation 3.9.Then(i) dim I ( A n ) = 2 P − n − and(ii) O [0 ,n − P − ( A n ) ∩ I ( A n ) = . Note that the assertions in Corollary 5.10 are exactly the Goals (ii)and (iii) of Remark 5.3.
Proof.
Since I ( A n ) is spanned by 2 P − n − I ( A n ) ≤ P − n − π : I ( A n ) → I ( B n,P ) is an epimorphism,it follows that dim I ( A n ) = 2 P − n − v ∈ O [0 ,n − P − ( A n ) ∩ I ( A n ). Let a n − P +1 ,. . . , a P − denote the columns of A n which span I ( A n ). Then there exist λ n − P +1 , . . . , λ P − ∈ Q such that(2) v = P − X j = n − P +1 λ j a j ∈ O [0 ,n − P − ( A n ) ∩ I ( A n ) . Then(3) π ( v ) = P − X j = n − P +1 λ j b ( P ) j ∈ O [0 ,n − P − ( B n,P ) ∩ I ( B n,P )where b ( P ) j denotes the j -th column of B n,P for n − P + 1 ≤ j ≤ P − O [0 ,n − P − ( B n,P ) = and the columns b ( P ) j are linearly independent. Hence π ( v ) = and Equation (7) implies that λ j = 0 for n − P + 1 ≤ j ≤ P −
1. Therefore, by plugging into Equation (2), it followsthat v = 0 which completes the proof. (cid:3) It follows from Corollary 5.10 that the “inner” 2 P − n − A n , that is, the columns which span I ( A n ) are linearly independent. Thisimmediately implies the next corollary. Corollary 5.11.
Let n ≥ be an integer and A n its valuation matrix.Then rank( A n ) ≥ P − n − . As shown below, the results so far imply the absolute irreducibility of (cid:0) xn (cid:1) in the special case where n = P is a prime number. Note that this hasalready been shown by Frisch and Nakato [6, Example 2.6] and our proofonly serves as an alternative. Corollary 5.12.
Let P be a prime number, then (cid:0) xP (cid:1) is absolutely irre-ducible.Proof. If n = P , then rank( A P ) ≥ P − P − P − A P ) < P by Remark 3.14, it follows rank( A P ) = P −
1. Accord-ing to Proposition 3.15, (cid:0) xP (cid:1) is absolutely irreducible. (cid:3) The outer columns of A n . For a composite number n ≥
2, the spanof the “outer” columns of A n is not trivial. The goal of this section is toshow that dim O [0 ,n − P − ( A n ) = 2( n − P ), see Proposition 5.14. This is thefinal ingredient to prove that rank( A n ) = n − (cid:0) xn (cid:1) is absolutely irreducible, cf. Remark 5.3.As above, we have to exploit the structure of certain p -blocks to reach thisgoal. In contrast to the arguments in the previous subsection, we need to findmore than one suitable p -block. Which choices to make is explained in detailin the proof of Proposition 5.14. The next proposition gives informationabout the dimension of O [ r n,p ,k − ( B n,p ) for certain choices of k . Proposition 5.13.
Let n ≥ be a composite number, p ∈ P n (cf. Nota-tion 3.9) be a prime number and B n,p the corresponding p -block.If r n,p + 1 ≤ k ≤ min { n − P, p + r n,p − } , then(1) dim O [ r n,p ,k − ( B n,p ) = 2( k − r n,p ) and(2) O [0 ,r n,p − ( B n,p ) = .Proof. We first treat the special case n = 9 and p = 3. Figure 5 displays B , . Since r , = 0, the second assertion of the proposition follows triv-ially. Moreover, since k ∈ { , } , the first assertion can be verified by directcomputation.For the remainder of the proof, we assume that n = 9 or p = 3. Itfollows from r n,p + 1 ≤ k ≤ min { n − P, p + r n,p − } that r n,p ≤ p − k ≤ p + r n,p −
1. In particular, p > p -block B n,p has p − r n,p − ≥ k − r n,p ) rows, cf. Remark 5.5.Moreover, since k ≤ p + r n,p − ≤ p −
1, we can apply Propositions 5.6and 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: BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 17 B , = − − − − I ( B , ) . Figure 5.
The 3-block B , of 9.(a) The columns indexed with j ∈ { , . . . , r n,p − } ∪ { n − r n,p , . . . , n − } arezero columns (these are the r n,p left outermost and r n,p right outermostcolumns).(b) The entries of column j = r n,p are all equal − v p ( n − r n,p ) = 0.(c) The columns r n,p + 1 ≤ j ≤ k − v p ( n − r n,p ), in row r = j − r n,p , that is, the first k − ( r n,p + 1) rows ofthese columns form a diagonal matrix with v ( n − r n,p ) on the diagonaland all entries below are zero.(d) The columns n − k ≤ j ≤ n − ( r n,p + 1) contain exactly one non-zeroentry, namely 1, in row r = j − ( n − p ), that is, the lower k − r n,p rowsof these columns form the identity matrix and all entries above are zero.(e) Since k < p + r n,p 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 themiddle, namely there is at least one row indexed with k − r n,p ≤ r ≤ p − k − − v v ∗ ∗ − v v − v ∗ ∗ − v − v ∗ ∗ − v k − ( r n,p + 1) p + r n,p − kk − r n,p r n,p k − ( r n,p + 1) n − P − k n − P − k k − r n,p r n,p j = r n,p I ( B n,p ) Figure 6. p -block B n,p with p + r n,p > k It immediately follows, that O [0 ,r n,p − ( B n,p ) = since it is spanned onlyby zero columns. It remains to show that dim O [ r n,p ,k − ( B n,p ) = 2( k − r n,p ). 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 − } ∪ { n − k, . . . , n − ( r n,p + 1) } such that(4) k − X j = r n,p λ j b j + n − ( r n,p +1) X j = n − k λ j b j = 0Since b r n,p is the only column which has a non-zero entry in the rowsindexed with k − r n,p ≤ r ≤ p − k −
1, Equation (4) implies that λ r n,p = 0.Now for each of the remaining rows, there is exactly one column with anon-zero entry in this (and no other) row. With the same reasoning we canconclude that λ j = 0 for r n,p + 1 ≤ j ≤ k − n − k ≤ j ≤ n − ( r n,p + 1).Therefore, these 2( k − r n,p ) columns of B n,p are linearly independent. (cid:3) Proposition 5.14.
Let n ≥ be a composite integer and A n its valuationmatrix.Then dim O [0 ,n − P − ( A n ) = 2( n − P ) .Proof. We treat the cases n = 9 and n = 10 first as we want to excludethem below. For n = 9, observe that P = 7 and hence n − P = 2. Itfollows from Proposition 5.13, that dim O [0 , ( B , ) = 4, see also Figure 5for direct verification. Since the projection O [0 , ( A ) → O [0 , ( B , ) is anepimorphism, it follows that dim O [0 , ( A ) = 4.The valuation matrix A is displayed in Figure 7. A direct computationverifies that rank( O [0 , ( A )) = 6. A = − − − − − − − − − I ( A ) B , B , B , B , . Figure 7. O [0 , ( A ) has dimension 6.From now on, assume that n ≥ n = 9 and n = 10. We prove by induction on 1 ≤ k ≤ n − P thatdim O [0 ,k − ( A n ) = 2 k holds. Base case k = 1. Since the projection O [0 , ( A n ) → O [0 , ( B n,p ) is anepimorphims, it follows that 2 ≥ dim O [0 , ( A n ) ≥ dim O [0 , ( B n,p ) for all p ∈ P n . Therefore, it suffices to show that there always exists a primenumber p with dim O [0 , ( B n,p ) = 2.If n = 2 s is a proper power of 2, then R s , = { , } by definition which im-plies that B s , has two rows. Moreover, the outermost columns of B s , aredisplayed in Figure 8. As v ( n ) = s ≥
2, it follows that dim O [0 , ( B , ) = 2. BSOLUTE IRREDUCIBILITY OF THE BINOMIAL POLYNOMIALS 19 B s , = v ( n ) 11 − v ( n ) 0 I ( B s , ) . Figure 8. B s , 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 − } . Therefore,we can apply Proposition 5.13 and conclude that dim O [0 , ( B n,p ) = 2. Thiscompletes the proof of the base case.Observe that the induction base case also covers the cases n = 4, n = 6and n = 8 because in each of these cases n − P = 1. Therefore, we alreadyhandled all cases where n ≤ induction step and prove the following claim. Claim . There exists p ∈ P n with r n,p ≤ k − O [ r n,p ,k − ( A n ) = 2( k − r n,p ) and(2) O [0 ,r n,p − ( A n ) ∩ O [ r n,p ,k − ( A n ) = Assume for a moment that this claim holds. Since r n,p < k we can applythe induction hypothesis, that is, dim O [0 ,r n,p − ( A n ) = 2 r n,p and hencedim O [0 ,k − ( A n ) = dim (cid:16) O [0 ,r n,p − ( A n ) ⊕ O [ r n,p ,k − ( A n ) (cid:17) = dim O [0 ,r n,p − ( A n ) + dim O [ r n,p ,k − ( A n )= 2 r n,p + 2( k − r n,p )= 2 k and the assertion follows.Next, we prove the claim. By Theorem 2 (for which we needed to assumethat n > p ∈ P n such that p > k and r n,p ≤ k − r n,p + 1 ≤ k ≤ min { n − P, p − } ≤ min { n − P, p + r n,p − } andhence, by Proposition 5.13, dim O [ r n,p ,k − ( B n,p ) = 2( k − r n,p ) holds.Let π : O [ r n,p ,k − ( A n ) → O [ r n,p ,k − ( B n,p ) be the projection. As π isan epimorphism, it follows that dim O [ r n,p ,k − ( A n ) ≥ dim O [ r n,p ,k − ( B n,p ).Hence (1) of the claim above follows.To prove (2) of the claim, assume that v ∈ O [0 ,r n,p − ( A n ) ∩O [ r n,p ,k − ( A n )and let J = { j | ≤ j ≤ r n,p − n − r n,p ≤ j ≤ n − } and J = { j | r n,p ≤ j ≤ k − n − k ≤ j ≤ n − ( r n,p + 1) } be the indices of the columnsof A n which span O [0 ,r n,p − ( A n ) and O [ r n,p ,k − ( A n ), respectively.For 0 ≤ j ≤ n −
1, let a j with denote the columns of A n and λ j ∈ Q suchthat(5) v = X j ∈J λ j a j = X j ∈J λ j a j ∈ O [0 ,r n,p − ( A n ) ∩ O [ r n,p ,k − ( A n ) . Then(6) π ( v ) = X j ∈J λ j b ( p ) j = X j ∈J λ j b ( p ) j ∈ O [0 ,r n,p − ( B n,p ) ∩ O [ r n,p ,k − ( B n,p ) where b ( p ) j denotes the j -th column of B n,p . However, by Proposition 5.13, O [0 ,r n,p − ( B n,P ) = and hence Equation (6) reduces to(7) 0 = π ( v ) = X j ∈J λ j b ( p ) j . As the columns b ( p ) j with j ∈ J of B n,p are linearly independent by Propo-sition 5.13, it follows that λ j = 0 for all j ∈ J . Therefore, plugging intoEquation (5), it follows that v = 0 which completes the proof of the claim. (cid:3) Proposition 5.14 together with Corollary 5.10 imply that rank( A n ) = n − n ≥
2, cf. Remark 5.3. Using Proposition 3.15yields the corollary below.
Corollary 5.15.
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Institut für Mathematik, Alpen-Adria-Universität Kla-genfurt, Universitätsstraße 65-67, 9020 Klagenfurt am Wörthersee, Austria
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