On the Irreducibility of the Generalized Laguerre Polynomials

M. Filaseta^{1} and T.-Y. Lam

May 29, 2001

1The author gratefully acknowledges support from the National Security Agency.

1 Introduction

The generalized Laguerre polynomials are defined by

L_{m}^{(α)}(x) = ∑_{j=0}^{m} \frac{(m + α)(m − 1 + α) ⋯ (j + 1 + α)(−x)^{j}}{(m − j)!j!} ,

where m is a positive integer and α is an arbitrary complex number. In 1929, I. Schur [?] established the irreducibility over the rationals of L_{m}^{(0)}(x), the classical Laguerre polynomials, for every m. In 1931, I. Schur [?] considered L_{m}^{(α)}(x) in general and showed that L_{m}^{(1)}(x) is irreducible over the rationals for every m. The case α ∉ {0, 1} remained open. The purpose of this paper is to establish the following:

Theorem 1. Let α be a rational number which is not a negative integer. Then for all but finitely many positive integers m, the polynomial L_{m}^{(α)}(x) is irreducible over the rationals.

Before going to the proof, it is worth noting that reducible L_{m}^{(α)}(x) do exist even with α = 2. In particular, we give the following examples:

L_{2}^{(2)}(x) = \frac{1}{2}(x − 2)(x − 6)
L_{2}^{(23)}(x) = \frac{1}{2}(x − 20)(x − 30)
L_{4}^{(23)}(x) = \frac{1}{24}(x − 30)(x^{3} − 78x^{2} + 1872x − 14040)
L_{4}^{(12/5)}(x) = \frac{1}{15000}(25x^{2} − 420x + 1224)(25x^{2} − 220x + 264)
L_{5}^{(39/5)}(x) = \frac{−1}{375000}(5x − 84)(625x^{4} − 29500x^{3}
+ 448400x^{2} − 2662080x + 5233536).

It is not difficult to show that in fact there are infinitely many positive integers α for which L_{2}^{(α)}(x) is reducible (a product of two linear polynomials).

Theorem ?? is a direct consequence of the following more general result:

Theorem 2. Let α be a rational number which is not a negative integer. Then for all but finitely many positive integers m, the polynomial

∑_{j=0}^{m} a_{j} \frac{(m + α)(m − 1 + α) ⋯ (j + 1 + α)x^{j}}{(m − j)!j!}

is irreducible over the rationals provided only that a_{j} ∈ ℤ for 0 ≤ j ≤ m and |a_{0}| = |a_{m}| = 1.

I. Schur obtained his irreducibility results for L_{m}^{(0)}(x) and L_{m}^{(1)}(x) through general results similar to the above (except also for all m ≥ 1). Recent work of a similar nature has been done by Filaseta [?, ?] and by Filaseta and Trifonov [?]. We note also that the above results can be made effective so that for any fixed α ∈ ℚ it is possible to determine a finite set S = S(α) of m such that the polynomial in Theorem ?? is irreducible (for a_{j} as stated there) provided m ∉ S.

2 A Proof of Theorem ??

For a prime p and a non-zero integer a, we define ν(a) = ν_{p}(a) = e where p^{e}||a. We set ν(0) = +∞. We define the Newton polygon of a polynomial f(x) = ∑_{j=0}^{n} a_{j}x^{j} with respect to a prime p, where a_{n}a_{0} ≠ 0 as the lower convex hull of the points (j, ν(a_{n−j})). Thus, the slopes of the edges of the Newton polygon of f(x) with respect to p are increasing from left to right. We begin with the following preliminary results.

Lemma 1. Let k and ℓ be integers with k > ℓ ≥ 0. Suppose g(x) = ∑_{j=0}^{n} b_{j}x^{j} ∈ ℤ[x] and p is a prime such that p ∤ b_{n}, p|b_{j} for all j ∈ {0, 1, …, n − ℓ − 1}, and the right-most edge of the Newton polygon for g(x) with respect to p has slope < 1/k. Then for any integers a_{0}, a_{1}, …, a_{n} with |a_{0}| = |a_{n}| = 1, the polynomial f(x) = ∑_{j=0}^{n} a_{j}b_{j}x^{j} cannot have a factor with degree in the interval [ℓ + 1, k].

Lemma 2. Let a, b, c and d be integers with bc − ad ≠ 0. Then the largest prime factor of (am + b)(cm + d) tends to infinity as the integer m tends to infinity.

Lemma ?? is given as Lemma 2 in [?]. Lemma ?? above is a fairly easy consequence of the fact that the Thué equation ux^{3} − vy^{3} = w has finitely many solutions in integers x and y where u, v, and w are fixed integers with w ≠ 0. It also immediately follows from Corollary 1.2 of [?]. We omit the proofs.

Fix α now as in Theorem ??. Throughout the argument we suppose as we may that m is large. Define

c_{j} = \binom{m}{j}(m + α)(m − 1 + α) ⋯ (j + 1 + α) for 0 ≤ j ≤ m.

We want to show that for all but finitely many positive integers m, the polynomial f(x) = ∑_{j=0}^{m} a_{j}c_{j}x^{j} is irreducible over the rationals, where a_{j} are arbitrary integers with |a_{0}| = |a_{n}| = 1. Motivated by Lemma ??, we consider instead g(x) = ∑_{j=0}^{m} c_{j}x^{j}. Let u and v be relatively prime integers with v > 0 such that α = u/v. The condition that α is a non-negative integer implies that for each j ∈ {0, 1, …, m − 1}, m − j + α and, hence, v(m − j) + u cannot be zero. We assume that f(x) has a factor in ℤ[x] of degree k ∈ [1, m/2] and establish the theorem by obtaining a contradiction to Lemma ??. We divide the argument into cases depending on the size of k.

Case 1. k > m/ log^{2} m.

For a and b integers with b > 0, let π(x; b, a) denote the number of primes ≤ x which are ≡ a (mod b). Then the Prime Number Theorem for Arithmetic Progressions implies that if gcd(a, b) = 1, then

π(x; b, a) = \frac{1}{ϕ(b)} ∫_{2}^{x} \frac{dt}{log t} + O (\frac{x}{log^{4} x})

= \frac{1}{ϕ(b)} (\frac{x}{log x} + \frac{x}{log^{2} x} + \frac{2x}{log^{3} x} + O (\frac{x}{log^{4} x})) .

By considering π(x; b, a) − π(x − h; b, a), it follows that for a and b fixed, the interval (x − h, x] contains a prime ≡ a (mod b) if h = x/(2 log^{2} x) and if x is sufficiently large. Taking a = u, b = v, and x = vm + u, we deduce that for some integer j ∈ [0, k), the number v(m − j) + u is prime. Call such a prime p, and observe that p ≥ 2vm/3 (since v is a positive integer and m is large). We deduce that p does not divide v. Observe that

c_{ℓ} = \binom{m}{ℓ}\frac{(vm + u)(v(m − 1) + u) ⋯ (v(ℓ + 1) + u)}{v^{m−ℓ}} for 0 ≤ ℓ ≤ m.

For j ∈ {0, 1, …, k − 1}, the numbers v(m − j) + u appear in the numerator of the fraction on the right-hand side above whenever 0 ≤ ℓ ≤ m − k. Therefore,

ν_{p}(c_{ℓ}) ≥ 1  for 0 ≤ ℓ ≤ m − k.  (1)

Since c_{m} = ±1, ν_{p}(c_{m}) = 0. To obtain a contradiction from Lemma ?? for the case under consideration, we show that ν_{p}(c_{0}) = 1; the contradiction will be achieved since then it will follow that the right-most edge of the Newton polygon of g(x) with respect to p has slope < 1/(m − k) < 1/k. Recall that p ∤ v and that p ≥ 2vm/3. For j ∈ {0, 1, …, m − 1}, we deduce the inequality

2p > vm + u ≥ v(m − j) + u ≥ v + u > −p.

The condition that α is a non-negative integer implies that none of v(m−j)+u can be zero. Hence, p itself is the only multiple of p among the numbers v(m − j) + u with 0 ≤ j ≤ m − 1. Since c_{0} = (vm + u)(v(m − 1) + u) ⋯ (v + u)/v^{m}, we obtain ν_{p}(c_{0}) = 1.

Case 2. k_{0} ≤ k ≤ m/ log^{2} m with k_{0} = k_{0}(u, v) a sufficiently large integer.

Let z = k(log log k)^{1/2}. We first show that there is a prime p > z that divides v(m − j) + u for some j ∈ {0, 1, …, k − 1}. Then (??) follows as before, and we will obtain a contradiction to Lemma ?? by showing that the right-most edge of the Newton polygon of g(x) with respect to p has slope < 1/k.

Let

T = {v(m − j) + u : 0 ≤ j ≤ k − 1}.

Since m is large, we deduce that the elements of T are each ≥ m/2. Also, observe that gcd(u, v) = 1 implies that each element of T is relatively prime to v. For each prime p ≤ z, we consider an element a_{p} = v(m − j) + u ∈ T with ν_{p}(a_{p}) as large as possible. We let

S = T − {a_{p} : p ∤ v, p ≤ z}.

By the Prime Number Theorem,

π(z) ≤ \frac{2k(log log k)^{1/2}}{log k} .

We combine this estimate momentarily with |S| ≥ k−π(z). Since k ≤ m/ log^{2} m, we obtain m ≥ k log^{2} k. Consider a prime p ≤ z with p not dividing v, and let r = ν_{p}(a_{p}). By the definition of a_{p}, if j > r, then there are no multiples of p^{j} in T (and, hence, in S). For 1 ≤ j ≤ r, there are ≤ [k/p^{j}] + 1 multiples of p^{j} in T and, hence, at most [k/p^{j}] multiples of p^{j} in S. Therefore,

ν_{p} (∏_{s∈S} s) ≤ ∑_{j=1}^{r} [\frac{k}{p^{j}} ] ≤ ν_{p}(k!),

and

∏_{s∈S} ∏_{p≤z} p^{ν_{p}(s)} ≤ k! ≤ k^{k}.

On the other hand,

∏_{s∈S} s ≥ (\frac{m}{2})^{|S|} ≥ (\frac{k log^{2} k}{2} )^{k−π(z)} .

Recalling our bound on π(z), we obtain

log ( ∏_{s∈S} s) ≥ (k − π(z))( log k + 2 log log k − log 2)

≥ (k − \frac{2k\sqrt{log log k}}{log k} )( log k + 2 log log k − log 2)
≥ k log k + 2k log log k + O(k\sqrt{log log k}).

Since k ≥ k_{0} where k_{0} is sufficiently large,

log ( ∏_{s∈S} s) > k log k ≥ log ( ∏_{s∈S} ∏_{p≤z} p^{ν_{p}(s)}).

It follows that there is a prime p > z that divides some element of S and, hence, divides some element of T.

Fix a prime p > z that divides an element in T, and let ν = ν_{p}. The right-most edge of the Newton polygon of g(x) with respect to p is

max_{1≤j≤m} \{\frac{ν(c_{0}) − ν(c_{j})}{j}\} .

Fix j ∈ {1, 2, …, m}. To complete the case under consideration, we want to show that the fraction above is < 1/k. Observe that

ν(c_{0}) − ν(c_{j}) ≤ ν ((vj + u)(v(j − 1) + u) ⋯ (v + u))

≤ ν((vj + |u|)!) = ∑_{j=1}^{∞} [\frac{vj + |u|}{p^{j}}]

< ∑_{j=1}^{∞} \frac{vj + |u|}{p^{j}} = \frac{vj + |u|}{p − 1}.

Since p > z = k(log log k)^{1/2} and k ≥ k_{0}, we easily deduce that the right-most edge of the Newton polygon of g(x) with respect to p has slope < 1/k as desired. Hence, as indicated at the beginning of this case, we obtain a contradiction to Lemma ??.

Case 3. 2 ≤ k < k_{0}.

By Lemma ?? (with a = v, b = u, c = v, and d = u − v), the largest prime factor of the product (vm+u)(v(m−1)+u) tends to infinity. Since m is large, we deduce that there is a prime p > (v + |u|)k_{0} that divides (vm + u)(v(m − 1) + u). The argument now follows as in the previous case. In particular,

\frac{ν(c_{0}) − ν(c_{j})}{j} < \frac{vj + |u|}{j(p − 1)} ≤ \frac{v + |u|}{p − 1} ≤ \frac{1}{k_{0}} < \frac{1}{k} for 1 ≤ j ≤ m,

and the right-most edge of the Newton polygon of g(x) with respect to p has slope < 1/k. Hence, in this case, we also obtain a contradiction.

Case 4. k = 1.

¿From Lemma ??, the largest prime factor of m(vm+u) tends to infinity with m. We consider a large prime factor p of this product. In particular, we suppose that p > v + |u|. Note this implies p ∤ v. As in the previous case, we are through if p|(vm + u). So suppose p|m. The binomial coefficient \binom{m}{j} appears in the definition of c_{j}, and this is sufficient to guarantee that ν(c_{j}) ≥ 1 and ν(c_{m−j}) ≥ 1 for 1 ≤ j ≤ p − 1. On the other hand,

c_{j} = \binom{m}{j}\frac{(vm + u)(v(m − 1) + u) ⋯ (v(j + 1) + u)}{v^{m−j}}.

For j ≤ m − p, the numerator of the fraction on the right is a product of ≥ p consecutive terms in the arithmetic progression vt + u with gcd(p, v) = 1; thus, ν(c_{m−j}) ≥ 1 for j ≥ p. This implies that (??) holds with k = 1. It follows in the same manner as before that the slope of the right-most edge is < 1. A contradiction to Lemma ?? is again obtained (and the proof of the theorem is complete).

References

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