On Quadratic Fields Generated by Discriminants of Irreducible Trinomials

A. Mukhopadhyay, M. R. Murty and K. Srinivas (http://arxiv.org/abs/0808.0418) have recently studied various arithmetic properties of the discriminant $\Delta_n(a,b)$ of the trinomial $f_{n,a,b}(t) = t^n + at + b$, where $n \ge 5$ is a fixed integer. In particular, it is shown that, under the $abc$-conjecture, for every $n \equiv 1 \pmod 4$, the quadratic fields $\Q(\sqrt{\Delta_n(a,b)})$ are pairwise distinct for a positive proportion of such discriminants with integers $a$ and $b$ such that $f_{n,a,b}$ is irreducible over $\Q$ and $|\Delta_n(a,b)|\le X$, as $X\to \infty$. We use the square-sieve and bounds of character sums to obtain a weaker but unconditional version of this result.


Introduction
For a fixed integer n ≥ 2, we use ∆ n (a, b) to denote the discriminant of the trinomial f n,a,b (t) = t n + at + b.
A. Mukhopadhyay, M. R. Murty and K. Srinivas [9] have recently studied the arithmetic structure of ∆ n (a, b). In particular, it is shown in [9], under the abc-conjecture, that if n ≡ 1 (mod 4) then for a sufficiently large positive A and B such that B ≥ A 1+δ with some fixed δ > 0, there are at least γAB integers a, b with and such that f n,a,b is irreducible and ∆ n (a, b) is square-free, where γ > 0 depends only on n and δ.
Then this result is used to derive (still under the abc-conjecture) that the quadratic fields Q ∆ n (a, b) are pairwise distinct for a positive proportion of such discriminants with integers a and b such that f n,a,b is irreducible over Q.
More precisely, for a real X ≥ 1, let Q n (X) be the number of distinct fields Q ∆ n (a, b) taken for all of pairs of integers a, b such that f n,a,b is irreducible over Q and |∆ n (a, b)| ≤ X.
Throughout the paper, we use U = O(V ), U ≪ V , and V ≫ U as equivalents of the inequality |U| ≤ cV with some constant c > 0, which may depend only on n.
It is shown in [9] that for a fixed n ≡ 1 (mod 4), where κ n = 1 n + 1 n − 1 and c 0 > 0 is a constant depending only on n.
It is also noted in [9] that the Galois groups of irreducible trinomials f n,a,b have some interesting properties, see also [2,5,10]. We remark that, since ∆ n (a, b) = (n − 1) n−1 a n + n n b n−1 (2) for n ≡ 1 (mod 4), there are O X 1 n + 1 n−1 integers a and b with |∆ n (a, b)| ≤ X and thus indeed (1) means that We use the square-sieve and bounds of character sums to obtain a weaker but unconditional version of this result. We note that, without the irreducibility of f n,a,b condition, the problem of estimating Q(X) can be viewed as a bivariate analogue of the question, considered in [8], on the number of distinct quadratic fields of the form Q F (n) for n = M + 1, . . . , M + N, . Accordingly we use similar ideas, however we also exploit the specific shape of the polynomial ∆ n (a, b) given by (2).

Main result
In fact as in [8] we consider a more general quantity than Q(X). Namely, for real positive A, B, C and D and a square-free integer s, we denote by such that ∆ n (a, b) = sr 2 for some integer r.
We write log x for the maximum of the natural logarithm of x and 1, thus we always have log x ≥ 1.
. Now, for real A, B, C and D we denote by S n (A, B, C, D) the number of distinct quadratic fields Q ∆ n (a, b) taken for all pairs of integers such that f n,a,b is irreducible over Q. Using that the bound of Theorem 1 is uniform in s, we derive . The results of Theorems 1 and 2 are nontrivial in a very wide range of parameters A, B, C and D and apply to very short intervals. In particular, AB could be logarithmically small compared to CD. Furthermore, taking which, although is weaker than (1), does not depend on any unproven conjectures.

Character Sums with the Discriminant
Our proofs rest on some bounds for character sums. For an odd integer m we use (w/m) to denote, as usual, the Jacobi symbol of w modulo m. We also put e m (w) = exp(2πiw/m).
Given an odd integer m ≥ 3 and arbitrary integers λ, µ, we consider the double character sums We need bounds of these sums in the case of m = ℓ 1 ℓ 2 being a product of two primes ℓ 1 > ℓ 2 ≥ n. However, using the multiplicative property of character sums (see [6,Equation (12.21)] for single sums, double sums behaves exactly the same way) we see that it is enough to estimate S n (ℓ; λ, µ) for primes ℓ.
We start with evaluating these sums in the special case of λ = µ = 0 where we define S n (ℓ) = S n (ℓ; 0, 0).

Lemma 3.
For n ≡ 1 (mod 4) and a prime ℓ, we have Proof. We can certainly assume that ℓ ≥ n as otherwise the bound is trivial.
Recalling (2), we derive Substituting uv instead of u, we obtain since n − 1 is even. We now rewrite it in a slightly more convenient form as As gcd(ℓ, n − 1) = 1, making the change of variables (n − 1) n−1 u n v + n n = w, we note that for every u = 1, . . . , ℓ − 1 if v = 1, . . . , ℓ, then w runs through the complete residue system modulo ℓ. Hence, which concludes the proof.

⊓ ⊔
The following result can be derived from [3, Theorem 1.1], however we give a self-contained and more elementary proof.
Proof. As in the proof of Lemma 3, we can certainly assume that ℓ ≥ n as otherwise the bound is trivial. Also as in the proof of Lemma 3, we obtain As gcd(ℓ, n − 1) = 1, making the change of variables (n − 1) n−1 u n v + n n = w, we obtain The sums over w is the Gauss sum, thus for some complex ϑ ℓ with |ϑ ℓ | = 1 (which depends only on the residue class of ℓ modulo 4), we refer to [6,7] for details. Since n ≡ 1 (mod 4) we have Thus, combining the above identities we obtain Since gcd(λ, µ, ℓ) = 1, the Weil bound (see [6,Bound (12.23)]) applies and implies that the sum over u is O(ℓ 1/2 ) which concludes the proof.
Finally, using the standard reduction between complete and incomplete sums (see [6, Section 12.2]), we derive from Lemma 5 Lemma 6. For n ≡ 1 (mod 4), an integer m = ℓ 1 ℓ 2 which is a product of two distinct primes ℓ 1 = ℓ 2 and real positive A, B, C and D, we have

Irreducibility
As in [9] we recall a very special case of a results of S. D. Cohen [1] about the distribution of irreducible polynomials over a finite field F q of q elements.

Proof of Theorem 1
For a real number z ≥ 1 we let L z be the set of primes ℓ ∈ [z, 2z]. For a positive integer k we write ω(k) for the number of prime factors of k.
We note that if k ≥ 1 is a perfect square, then for z ≥ 3, For each pair (a, b), counted in T n (A, B, C, D; s), we see t hat s∆ n (a, b) is a perfect square and that s | ∆ n (a, b). Hence, ω(s∆ n (a, b)) = ω (∆ n (a, b)). a, b)).
In particular, by the Cauchy inequality, We note that Squaring out and changing the order of summation, we obtain We now estimate the double sum over a and b trivially as O(AB) on the "diagonal" ℓ 1 = ℓ 2 and use Lemma 6 otherwise, getting By the prime number theorem we have #L z ≫ z/ log z so we derive from (3) that Clear the first term always dominates the second one, so the second term can be simply dropped. Thus taking z = (AB) 1/3 to balance the terms ABz −1 log z and z 2 log z, we obtain the desired estimate.

Proof of Theorem 2
Let p 0 be smallest prime for which there exists an irreducible trinomial (p 0 exists by Lemma 7).
We now define the sets of integers Clearly #A ≫ A and #B ≫ B and every trinomials t n + at + b with a ∈ A, b ∈ B is irreducible over Z.
Using Theorem 1 to estimate the number of pairs (a, b) ∈ A ×B for which Q ∆ n (a, b) is a given quadratic field, we obtain the desired result.

Remarks
Similar results can be obtained for more general trinomials t n + at m + b with fixed integers n > m ≥ 1. Some properties of the Galois group of these trinomials have been studied in [2,5,10] where one can also find an explicit formula for their discriminant (which generalises (2)). In the case of a = b = 1 it becomes (−1) n(n−1)/2 (n n − (−1) n m m (n − m) n−m ). Studying arithmetic properties of this expression, for example, its square-free part, when n and m vary in the region N ≥ n > m ≥ 1 for a sufficiently large N, is a very challenging question.