Polynomials Irreducible by Eisenstein s Criterion

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1 AAECC 14, (2003 Digital Object Identifier (DOI /s Polynomials Irreducible by Eisenstein s Criterion Artūras Dubickas Dept. of Mathematics and Informatics, Vilnius University, Naugarduko 24, 2600 Vilnius, Lithuania ( arturas.dubickas@maf.vu.lt Received: February 21, 2002; revised version: May 5, 2003 Abstract. We compute the probability for a monic univariate integer polynomial to be irreducible by Eisenstein s Criterion. In particular, it follows that less than 1% of the polynomials with at least seven non-zero coefficients are irreducible by Eisenstein. Keywords: Irreducible polynomial, Eisenstein s Criterion. 1 Introduction Let Z and N be the sets of integers and of positive integers, respectively. One of the simplest irreducibility criteria for polynomials with integral coefficients over the field of rational numbers is that named after Eisenstein. The monic polynomial x d + c d 1 x d 1 + +c 1 x + c 0 Z[x] is said to be irreducible (over the field of rational numbers and, equivalently, in Z[x] by Eisenstein s Criterion if there is a prime number p such that the coefficients c i, where i = 0, 1,...,d 1, are all divisible by p and c 0 is not divisible by p 2. Because of its simplicity, often for applications one constructs the polynomials having their irreducibility confirmed either directly by Eisenstein s Criterion or this is done via Eisenstein after a linear change of variable. In their recent paper [1] D.E. Dobbs and L.E. Johnson have raised several probabilistic questions concerning the application of Eisenstein s Criterion. In particular, their main theorem contains some bounds for the probability that a cubic and a quadratic monic polynomial is irreducible by Eisenstein. The purpose of this paper is to compute this probability for polynomials of arbitrary degree. Given k N and some k +1 integers 0 = d 0 <d 1 < <d k 1 <d,what are the chances that a random monic polynomial f(x = x d + k 1 i=0 a ix d i Z[x] is irreducible by Eisenstein s Criterion? More precisely, for N N, there are (2N + 1 k polynomials of the above form, if a i,i= 0, 1,...,d 1, independently take integer values from N to N. Suppose that there are I(k,N

2 128 A. Dubickas irreducible (in Z[x] polynomials among them, E(k,N of which are irreducible by Eisenstein. Following [1], we are interested in E(k,N e(k = lim N (2N + 1 k and e E(k,N (k = lim N I(k,N. We conclude this section with the main result of the paper. Theorem 1. For every k N we have that e(k = e (k = 1 ( 1 1 p + 1, k p k+1 p where the product is taken over every prime number p. 2 Some Computations and Discussion Let µ(j and φ(j be the Möbius and the Euler totient functions, respectively. These are defined on N as µ(1 = φ(1 = 1, φ(j = j p j (1 1/p for j>1, whereas the Möbius function (for j>1 is zero, unless j is square-free, and equals 1 and 1 for j being the product of even and odd number of distinct primes, respectively. Writing 1 p k + p k 1 = 1 φ(pp k 1 and using the multiplicativity of φ(j, one can easily deduce that 1 ( 1 1 p + 1 µ(jφ(j =, k p k+1 j k+1 p giving another formula for the computation of e(k. Evidently, the product formula gives e(1 = e (1 = 1. For k 2, we have the following values for e(k = e (k with 7 true digits. j=2 k e(k

3 Polynomials Irreducible by Eisenstein s Criterion 129 For k = 2 and d {2, 3}, by the main theorem of [1], <e(2 < As it was correctly guessed there, the empirical evidence for d = 3 and N 30 points toward the wrong answer for e(2. (This is because N is too small. Our theorem implies that on average at least one out of four trinomials is irreducible by Eisenstein. From the table we see that there is not too much hope to expect that the irreducibility of a random polynomial with many non-zero coefficients can be confirmed via Eisenstein. There are less than 1% of monic polynomials with at least seven non-zero coefficients which are irreducible by Eisenstein, and less than 0.05% of such polynomials with at least eleven non-zero coefficients. For k = 1, the answer to the question whether the polynomial x d + a 0 is irreducible or not is well-known and is described by the theorem of Capelli (see, for instance, Theorem 16, Ch. VIII in [6] or Theorem 19 in [9]. Reducibility of trinomials, quadrinomials and other lacunary polynomials, where lacunary means that k is small compared to d, was in particular studied by A. Schinzel in more than a dozen of papers starting with [8]. Given a monic polynomial f, let H (f, L(f, L 2 (f, max z =1 f(z, and M(f be the maximum modulus of its coefficients, sum of the absolute values of coefficients, square root of the sum of squares of coefficients, maximum modulus on the unit circle and the absolute value of the product of roots lying outside the unit circle, respectively. Similarly, the probabilities e(k and e (k could be defined not only using this particular height H(f, namely, by the proportion of all polynomials among those with H(f N, but also some other height (for example, H(f N being replaced by M(f N. From the inequalities H (f, M(f L 2 (f max f(z L(f z =1 (d + 1H (f (d + 12 d M(f, it follows that these heights are of approximately the same magnitude. However the constant will apparently be different from that of the theorem. (Here, all inequalities are almost trivial except for M(f L 2 (f which is known as Landau s inequality. See, for instance, the proofs given by E. Landau himself [5] or by M. Mignotte and P. Glesser [7]. Finally, Lemma 1 below (due to H.W. Knobloch implies that almost all polynomials (in the above sense are irreducible. A similar probabilistic question on the proportion of irreducible polynomials, for N and two extreme coefficients being fixed and d tending to infinity, is much more difficult and still open (see, for instance, [4]. 3 Auxiliary Results We will denote by V N the set of (2N + 1 k vectors (a 0,a 1,...,a k 1 Z k, where a i N for i = 0, 1,...,k 1.

4 130 A. Dubickas Lemma 1. Let F(x,y 0,...,y k 1 Z[x,y 0,...,y k 1 ] be an irreducible polynomial, and let R(k, N be the number of k-tuples (a 0,...,a k 1 V N for which F(x,a 0,...,a k 1 is reducible in Z[x]. Then there are some positive constants c 1 and c 2 such that R(k, N < c 1 N k c 2. The lemma was first proved by H.W. Knobloch [3] who generalized an earlier result of K. Dörge [2] given for k = 1. The strongest result in this direction is apparently due to J.-P. Serre (see p. 310 in A. Schinzel s book [8] who showed that R(k, N < c 1 N k 1/2 (log N 1 c 3 with some positive c 1 and 0 <c 3 < 1. In order to state the next lemma we need one more arithmetic function. For j N, let τ(jbe the number of distinct positive divisors of j including 1 and j itself. For every square-free integer l 2, we denote by V (l, N the subset of V N with vectors whose entries a i,i= 0, 1,...,k 1, are all divisible by l and a 0 is not divisible by p 2 for every prime p dividing l. Lemma 2. We have V (l, N (2N + 1k φ(l ( 2N k 1 < (τ(l + k l k+1 l + 1. Proof. Assume without loss of generality that l N. The number of possibilities for each of the coordinates a i,i= 1, 2,...,k 1, is clearly 2[N/l] + 1, where [...] stands for the integral part of a number. It follows that the number of possibilities for (a 1,...,a k 1 is (2[N/l] + 1 k 1. By the inclusion exclusion principle, a 0 can take 2 ( 1 S [ N l p S p ] values. Here, the sum is taken over every subset S of the set of prime divisors of l including the empty set, and S stands for the number of elements of S. The above sum without integral parts (and without the factor 2 is clearly equal to Nφ(l/. Furthermore, 2 ω(l 1 of the elements in the sum are positive and 2 ω(l 1 are negative, where ω(l is the number of distinct prime divisors of l. Since l is square-free, 2 ω(l = τ(l. By estimating each positive error term in this approximation by 1, we see that the number of values taken by a 0 is equal to 2Nφ(l/ + θ(l, Nτ(l, where θ(l, N < 1. It follows that V (l, N = (2[N/l] + 1 k 1 (2Nφ(l/ + θ(l, Nτ(l. Note that V (l, N < (2N/l+1 k 1 (2Nφ(l/ +τ(l and the difference = ( 2N l + 1 k 1 ( 2Nφ(l ( 2N k 1 ( l + 1 τ(l φ(l ( τ(l (2N + 1k φ(l l k+1 ( 2N + 1 k 1 (2N + 1φ(l 2N + l

5 Polynomials Irreducible by Eisenstein s Criterion 131 is positive. On applying the inequality 1 t k 1 = (1 t(1 + t + +t k 2 (1 t(k 1, where 0 <t 1, for t = (2N + 1/(2N + l, we deduce that the above difference is at most (2N/l + 1 k 1 (τ(l + k 1. Similarly, we have that V (l, N > (2N/l 1 k 1 (2Nφ(l/ τ(l and the difference (2N + 1 k φ(l l k+1 ( 2N k 1 ( 2Nφ(l l 1 τ(l ( 2N + 1 k 1 ( ( 2N l k 1 φ(l ( ( ( 2N l k 1 = τ(l N 1 l 2N + 1 2N + 1 is positive. On applying the inequality 1 t k 1 (1 t(k 1 for t = (2N l/(2n + 1, we deduce that the above difference is at most ( 2N + 1 k 1 ( τ(l + φ(l ( 2N(l+ 1(k l 2N + 1 The required bound now easily follows, because, firstly, (2N +1/l < 2N/l+ 1, secondly, 1+2N(l+1(k 1/(2N +1 <(l+1k and, finally, φ(l(l+1 k/ ( 1k/ <k. 4 Proof of the Theorem Note first that the polynomial x d + k 1 i=1 y ix d i + y 0 is linear say in y 0, so it is irreducible in Z[x,y 0,...,y k 1 ]. By applying Lemma 1, we deduce that lim N I(k,N (2N + 1 k = 1 lim N R(k, N (2N + 1 k = 1. It follows that e(k = e (k if say e(k, defined as the limit, exists. In particular, this answers in the affirmative the question raised at the end of Section 2 in [1]. For k = 1, the polynomial x d + a 0 is irreducible by Eisenstein if there is a prime p a 0 such that p 2 does not divide a 0. Let 4 = ν 1 <ν 2 <ν 3 <... be all positive integers such that, given i N, for every prime p which divides ν i we have that p 2 ν i. Then the number 2N + 1 E(1,Nis bounded from above by {i N ν i N}. In order to prove that e(1 = 1 it suffices to show that lim j j/ν j = 0. However this follows by a standard sieve method. Indeed let us fix some η in the range 1/2 <η<1. The series 1 + ν η i = (1 + p 2η + p 3η +...< (1 + 4p 2η, i=1 p p where the products are taken over every prime p, are convergent. By considering the left-hand sum in the range j/2 i j, we deduce that j/ν η j is bounded from above by an absolute constant. Thus j/ν j <cν η 1 j, giving the result.

6 132 A. Dubickas Let now k 2. Assume that 2 = p 1 <p 2 <... are all prime numbers. How many polynomials in E(k,N are irreducible by Eisenstein with respect to the primes p 1,p 2,...,p s? Of course, by the inclusion exclusion principle the are precisely l µ(lv (l, N of them. Here and below, the sum is taken over every l 2 which divides p 1 p 2...p s. The number of polynomials in E(k,N which are irreducible by Eisenstein with respect to some prime p such that p s <p N is at most (2[N/p] + 1 k (3N/p k. Using Lemma 2 we deduce that the modulus of E(k,N + (2N + 1 k µ(lφ(l l k+1 l p 1...p s = (2N + 1 k( E(k,N s (2N (1 p k k i i=1 + p k 1 i is bounded from above by ( 2N k 1 (τ(l + k l + 1 N ( 3N k. + i l p 1...p s i=p s +1 Here, the first sum is at most 2 s (2 s + k(n + 1 k 1, whereas the second sum is at most (3N k ps 1 k /(k 1. Taking say s = [log log N] and letting N, we see that both sums, divided by (2N + 1 k, tend to zero, whence the result. Acknowledgements. I thank to Professor A. Schinzel who pointed out some relevant references. The research was partially supported by the Lithuanian State Science and Studies Foundation. References 1. Dobbs, D.E., Johnson, L.E.: On the probability that Eisenstein s criterion applies to an arbitrary irreducible polynomial. In: Dobbs, D.E. et al. (eds. Advances in commutative ring theory. Proc. of 3rd Intern. Conf., Fez, Morocco. Lect. Notes in Pure Appl. Math. Vol. 205, New York: Marcel Dekker, 1999, pp Dörge, K.: Ein Beitrag zur Theorie der diophantischen Gleichungen mit zwei Unbekannten. Math. Z. 24, ( Knobloch, H.W.: Zum Hilbertschen Irreduzibilitätssatz. Abh. Math. Sem. Univ. Hamburg 19, ( Konyagin, S.V.: On the number of irreducible polynomials with 0,1 coefficients. Acta Arith. 88, ( Landau, E.: Sur quelques thèorémes de M. Petrovi`c relatifs aux zéros des fonctions analytiques. Bull. Math. Soc. France 33, ( Lang, S.: Algebra. Reading, MA: Addison Wesley, Mignotte, M., Glesser, P.: Landau s inequality via Hadamard s. J. Symbolic Comp. 18, ( Schinzel, A.: Reducibility of lacunary polynomials I. Acta Arith. 16, ( Schinzel, A.: Polynomials with special regard to irreducibility. Cambridge: Cambridge University Press 2000