On the irreducibility of a olynomial associated with the Strong Factorial Conecture Michael Filaseta Mathematics Deartment University of South Carolina Columbia, SC 29208 USA E-mail: filaseta@math.sc.edu Brady Rocks Mathematics Deartment University of South Carolina Columbia, SC 29208 USA E-mail: ROCKS@math.sc.edu Abstract Asymtotically, more than 2/3 of the olynomials from a sequence of olynomials in Z[x], arising from an examle associated with the Strong Factorial Conecture, are shown to be irreducible in Z[x]. 1 Introduction The Strong Factorial Conecture of E. Edo and A. van den Essen [3] is concerned with the linear functional L on the sace of comlex olynomials defined by sending a monomial generator z a 1 1 zn an to (a 1! (a n!. The conecture asserts that for a non-zero multi-variable comlex olynomial 2010 Mathematics Subect Classification: Primary 11R09; Secondary 11C08, 12E05. Key words and hrases: Strong Factorial Conecture, irreducible, Newton olygon. 1
2 M. Filaseta and B. Rocks F, the maximum number of consecutive zeroes that may aear in the sequence {L(F n : n 1} is N(F 1, where N(F is the number of monomials aearing in F with nonzero coefficient. In the second author s dissertation [12], he considered the irreducibility in Z[x] of the olynomials f n,m (x = n =0 ( n (m! x in connection with his studies on the Strong Factorial Conecture, secifically in the case F = 1 + λz m where λ C. Among other results, f n,m (x was established in [12] to be irreducible when n = r where is a rime > m and r is a ositive integer. In this aer, we rove the following. Theorem 1.1. Fix a ositive integer m. Then lim inf X {n X : f n,m (x is irreducible} X log 2. As log 2 = 0.693147..., we deduce that more than 2/3 of the olynomials f n,m (x are irreducible in Z[x] for a fixed ositive integer m. We do not know of an instance where f n,m (x is reducible, so resumably a much stronger result than Theorem 1.1 holds. 2 Preliminaries on Newton olygons Let f(x = n =0 a x Z[x] with a 0 a n 0. Let be a rime. For an integer m 0, we denote by ν (m the exonent in the largest ower of dividing m. We define ν (0 = +. Let S be the set of lattice oints (, ν (a n, for 0 n, in the extended lane. We consider the lower edges along the convex hull of these oints. The left-most edge has an endoint ( 0, ν (a n and the right-most edge has ( n, ν (a 0 as an endoint. The olygonal ath along the lower edges of the convex hull from ( 0, ν (a n to ( n, ν (a 0 is called the Newton olygon of f(x with resect to the rime. The endoints of every edge belong to the set S, and each edge has a distinct sloe that increases as we move along the Newton olygon from left to right. The following imortant theorem due to G. Dumas [2] connects the Newton olygon of f(x with resect to a rime with the Newton olygon of its factors with resect to the same rime.
Irreducibility of a olynomial for the Strong Factorial Conecture 3 Theorem 2.1. Let g(x and h(x be in Z[x] with g(0h(0 0, and let be a rime. Let k be a non-negative integer such that k divides the leading coefficient of g(xh(x but k+1 does not. Then the edges of the Newton olygon for g(xh(x with resect to can be formed by constructing a olygonal ath beginning at (0, k and using translates of the edges in the Newton olygons for g(x and h(x with resect to the rime, using exactly one translate for each edge of the Newton olygons for g(x and h(x. Necessarily, the translated edges are translated in such a way as to form a olygonal ath with the sloes of the edges increasing. As a articular consequence of Theorem 2.1, we have the following. Let f(x Z[x] with f(0 0. Let (x 0, y 0, (x 1, y 1,..., (x r, y r, with 0 = x 0 < x 1 < < x r = deg f, denote the lattice oints along the edges of the Newton olygon of an f(x with resect to a rime. Set d = x x 1 for 1 r. Then the set {1, 2,..., r} can be written as a disoint union of sets S 1, S 2,..., S t where t is the number of irreducible factors of f(x (counted with multilicities and the t numbers u S d u, for 1 t, are the degrees of the irreducible factors of f(x. Note that it is imortant here to consider all lattice oints along the edges of the Newton olygon of f(x with resect to and not ust lattice oints of the form (, ν (a n used in the construction of the Newton olygon. Before alying Theorem 2.1 to obtain information about the factorization of f n,m (x, we first obtain information on Newton olygons of f n,m (x. We begin with a classical result on the largest ower of a rime dividing a binomial coefficient that we use to comute ν (a where a = ( n (m! is the coefficient of x in f n,m (x. Lemma 2.2. Let n and be nonnegative integers with n > 0, and let be a rime. If b is the number of borrows needed when is subtracted from n in base, then (( n ν = b. Lemma 2.2 is due to E. E. Kummer [8] but originally stated in the form of carries when adding and n in base. Kummer uses another classical result connecting the largest ower of dividing n! with the sum of the base digits of n due to A. M. Legendre [9]. The next lemma can be found in [12]. The roof given here is based on a somewhat different analysis.
4 M. Filaseta and B. Rocks Lemma 2.3. Let k, m and r be ositive integers, and let q be a rime > mk. Let n = kq r. Then the Newton olygon of f n,m (x with resect to q consists of a single edge which has sloe m(q r 1/ ( q r (q 1. Proof. For 0 n, we set a = ( n (m! so that fn,m (x = n =0 a x. In articular, ν q (a 0 = ν q (1 = 0. Since q > mk, we have ( mn ν q (a n = ν q (mn! = = r mkq r = r mkq r = mk(qr 1. q 1 We deduce that the line through ( 0, ν q (a n and ( n, ν q (a 0 has sloe equal to m(q r 1/ ( q r (q 1 and equation y = m(qr 1 q r (q 1 x + mk(qr 1. q 1 We want to rove that, for 0 < < n, the oint ( n, ν q (a is above this line, that is ν q (a m(qr 1 q r (q 1 (n + mk(qr 1 q 1 = m(qr 1 q r (q 1. Note that n in base q consists of the single digit mk followed by r zeroes. Fix (0, n, and let t = ν q (. Then < n imlies t [0, r] and in base q ends with exactly t digits that are zero. It follows that when is subtracted from n in base q, exactly r t borrows are required. Hence, (( n ν q = r t. Using that q t, we now deduce that (( (( n n ( ν q (a ν q (m! = ν q + ν q (m! m t m = r t + = r t + + = r t + t m + r u=t+1 m r u=t+1 m
Irreducibility of a olynomial for the Strong Factorial Conecture 5 r t + = r The lemma follows. t m + r u=t+1 m = m(qr 1 q r (q 1. ( m 1 Lemma 2.4. Let k and m be ositive integers, and let q be a rime number (m + 1 2 /(km. Let be a rime in the interval (kqm/(m + 1, kq], and let n = kq. Then the Newton olygon of f n,m (x with resect to has an edge with sloe m/. Comment: Though not needed for this aer, the statement of Lemma 2.4 seemingly holds for a larger range of rimes. Proof. Again, we set f n,m (x = n =0 a x where a = ( n (m! for 0 n. Observe that 2 > 2kqm kq n, m + 1 so ν (n! = 1. One checks that (( { n 1 if n < < (2.1 ν = 0 otherwise. If the exression (m! is divisible by, then /m. On the other hand, the condition > kqm/(m + 1 is equivalent to /m > n. Thus, (( n ν (m! = 0 for 0 n. The inequality q (m + 1 2 /(km imlies 2 > From ( kqm/(m + 1, kq ], we have ( 2 mn mn. m + 1 m mn < m + 1. Hence, ( ( mn ν an = ν (mn! = + mn mn + = = m. 2
6 M. Filaseta and B. Rocks We ustify that the Newton olygon of f n,m (x with resect to consists of the segment s from (0, m to (, 0 together with the segment from (, 0 to (n, 0. What is left to establish is that the oints ( n, ν (a, for n < < n, lie on or above the segment s. Since the line through (0, m and (, 0 has equation y = ( m/ x + m, we want to rove (2.2 ν ( a m(n for n < < n. As n, we have + m m(n + m = mn + m + m m + m + m = m. Thus, for (n, n, it suffices to show that either (2.2 holds or (2.3 ν ( a m. For n < <, using (2.1, we see that (( ( n ( m ν a = ν (m! = 1 + ν (m! 1 + so that (2.3 holds for such. For < n, we have ( ( m m ν a = ν (m! = m, imlying (2.2 for these. The lemma follows. > m, 3 Proof of Theorem 1.1 H. Cramér [1] showed that if the Riemann Hyothesis holds and n is the nth rime number, then n+1 n = O ( n log n. According to C. J. Moreno [10], P. Erdős osed the related roblem of establishing that, for every ε > 0, almost all numbers n are a distance n (1/2+ε from a rime. More secifically, Erdős asked whether there is a constant c < 1 such that ( n+1 n x c. n+1 n>x (1/2+ε n+1 x Moreno establishes this asymtotic in a weaker form with x c relaced nevertheless by a function which tends to 0 as x tends to infinity. D. Wolke [13] resolved the roblem of Erdős in the affirmative, and a number of other
Irreducibility of a olynomial for the Strong Factorial Conecture 7 authors (cf., [5, 6, 7, 11] have since imroved on the value of c in the asymtotic. In articular, K. Matomäki s work [7] imlies that (3.1 n+1 n> n n x ( n+1 n x 2/3. For our uroses, the weaker result of Moreno would suffice, but we use (3.1. Fix a ositive integer m. Let M = (m + 1 2 /m. Note that M 4. Let A be the set of ositive integers n that have a rime factor q > Mn. Let B be the set of ositive integers n for which there exists a rime satisfying n n < n. Set C = A B. We obtain next the asymtotic densities of the sets A and B in the set of integers, that is the values of lim x {n x : n A} x and lim x {n x : n B}. x The asymtotic density of A is connected to the distribution of smooth numbers (numbers with only small rime factors and is easily exlained. Using the notation π(x for the number of rimes x and to reresent a rime, observe that {x < n 2x : n A} = = Mx< 2x ( 2x ( x + O ( x + O ( π(2x ( + O Mx< 2x Mx< 2Mx x Mx< 2Mx ( 2x. x Using Merten s estimate for the sum of the recirocals of the rimes (cf. Theorem 427 in [4] and a Chebyshev estimate (cf. Theorem 7 in [4], we can deduce from the above that {n x : n A} (3.2 lim = log 2. x x For the asymtotic density of B, we consider first the asymtotic density of the comlement of B in the set of ositive integers. Fix a ositive integer n in the comlement of B. Let and be the consecutive rimes for which n <. Since n B, we have n n. Thus, > n ( n n = n.
8 M. Filaseta and B. Rocks Therefore, such n lie in an interval [, where and are consecutive rimes for which >. By (3.1, the n in the comlement of B have asymtotic density 0. Therefore, {n x : n B} (3.3 lim = 1. x x Combining (3.2 and (3.3, we deduce that {n x : n C} lim x x = log 2. Thus, to establish Theorem 1.1, it suffices to show that if n is a sufficiently large element of C, then f n,m (x is irreducible. Consider such an n. Then n A imlies that we can write n = kq where k is a ositive integer and q is a rime satisfying q > Mn = Mkq = q > Mk > mk. By Lemma 2.3, we deduce that the Newton olygon of f n,m (x with resect to the rime q consists of a single edge with sloe m/q. Since q is a rime > m, the fraction m/q is reduced. As a consequence of Theorem 2.1, we can deduce that each irreducible factor of f n,m (x has degree divisible by q (as noted in [12]. Next, we aly Lemma 2.4. Since q > Mk where M = (m + 1 2 /m, we see that q > (m + 12 k (m + 12 m km. We set to be the largest rime n. To aly Lemma 2.4, we want to show that > nm m + 1. Since n is sufficiently large and m is fixed, this inequality is an easy consequence of the Prime Number Theorem (i.e., that there is a rime number in the interval ( (1 εn, n ], where ε = 1/(m + 1. Lemma 2.4 imlies that the Newton olygon of f n,m (x with resect to the rime has an edge with sloe m/. Theorem 2.1 now imlies that f n,m (x has an irreducible factor of degree. To establish that f n,m (x is irreducible, it is sufficient now to show that the smallest multile of q that is is n = kq. This is equivalent to establishing that n q <. Since q > Mn > n, we need only show that n n <. The latter inequality follows from n B, comleting the roof of Theorem 1.1.
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