ANTS-XII August 30, 2016
Sparsity-Dependant Bounds f(x) = c 0 + c 1 x a 1 + c 2 x a 2 + + c t x at R[x]. f is sparse if t deg f. Descartes Rule of Signs: The number of positive, real roots of f is bounded by the number of sign alternations in the sequence (c 0, c 1,..., c t ). Thus, f has no more than 2t real roots. Question: Do similar sparsity-dependent bounds exists for other non-algebraically closed fields?
Sparsity-Dependant Bound over F q f(x) = c 0 + c 1 x a 1 + c 2 x a 2 + + c t x at F q [x]. Theorem (Canetti, Friedlander, Konyagin, Larsen, Lieman, Shparlinski - 2002) ( #roots(f) 2(q 1) 1 1/t D 1/t + O (q 1) 1 2/t D 2/t), where D = min max gcd(a i a j, q 1). i j i For ϑ F p, the associated Diffie-Hellman distribution is the set of triples (ϑ x, ϑ y, ϑ xy ) with x, y {1, 2,..., p}. Application: Diffie-Hellman distributions are nearly uniform in [0, p) 3 when p is large for ϑ of high order.
Improved Bound f(x) = c 0 + c 1 x a 1 + c 2 x a 2 + + c t x at F q [x]. Theorem (ZK - 2016) #roots(f) 2(q 1) 1 1/t C 1/t, where C = max{#h : H F q and f ah 0 for some a F q}. Proposition C(f) {k (q 1) : a i, a j i with a i a j mod k} C(f) D(f) = min i max j i gcd(a i a j, q 1). C(f) Q(f) = gcd i lcm j i gcd(a i a j, q 1).
Sketch of Proof f(x) = c 0 + c 1 x a 1 + c 2 x a 2 + + c t x at F q [x]. Suppose gcd(e, q 1) = 1: then, the map x x e is a bijection which simply shuffles the elements of F q. Let g(x) = f(x e ) = c 0 + c 1 x ea 1 + c 2 x ea 2 + + c t x eat. Let h(x) = c 0 + c 1 x ea 1 mod (q 1) + + c t x eat mod (q 1). We have #roots(f) = #roots(g) = #roots(h) degree(h). Idea: find e so that all of the exponents of h are small. If k = gcd(e, q 1) > 1, then we still have #roots(f) = 1 k k #roots(f(σ i x e )) degree(h), i=0 unless f(σ i x e ) is identically zero for some i. Thus we are safe to choose e {1, 2,..., (q 1)/C(f) 1}.
A Short Vector mod q 1 by Volume-Packing Lemma Let a 1, a 2,..., a t, N N. If 1 < n N, there is an e {1, 2,..., n 1} and a v NZ t so that proof: max ea i + v i N/n 1/t. 1 i t Consider the vectors l i = i(a 1,..., a t ) = (ia 1,..., ia t ) (R/NZ) t for each i {1, 2,... n}. Define l N = min v NZ t l + v. We need only to find two nearby vectors l i and l j, since we can set e = j i and l e = l j i = l j l i.
A Short Vector mod q 1 by Volume-Packing Let d = min 1 i<j n l j l i N. Each of the vector l i sits in its own personal box B i = {x (R/NZ) t : x l i N < d/2}. By representing these sets in the fundamental domain [0, N) t, we get the volume constraint nd t N t = d N/n 1/t.
How good is the bound? Let f(x) = t i=1 c ix a i F q [x]. Let R(f) = #{x F q : f(x) = 0}. Recall that R(f) 2(q 1) 1 1/(t 1) C(f) 1/(t 1). When t q 1, f(x) = (x q 1 1)/(x (q 1)/t 1) is a t-nomial with C(f) = (q 1)/t and R(f) = (1 1/t)(q 1). When q is an odd square, f(x) = x q1/2 + x 2 has C(f) = 1 and R(f) = q 1/2. Cheng, Gao, Rojas, and Wan provide a family of t-nomials with C(f) t/2 and R(f) q 1 2/t. Observation: all known examples of sparse polynomials which attain a large number of roots do so by vanishing on entire cosets of subgroups or on entire translations of subspaces.
Computer Data for Prime Fields F(p) = {f F p [x] : deg f < p 1}. F(p, t) = {f F(p) : f has t terms}. F 1 (p) = {f F(p) : C(f) = 1}. F 1 (p, t) = {f F(p, t) : C(f) = 1}. Let R p,t = max{r(f) : f F 1 (p, t)}. R p,3 < 1.8 log p for p 139571. R p,4 < 2.5 log p for p 907. R p,5 < 2.9 log p for p 101. (Compare to the current bound R p,t = O(p 1 1/(t 1) )).
A Possible Explanation F(p) = {f F p [x] : deg f < p 1}. F(p, t) = {f F(p) : f has t terms}. F 1 (p) = {f F(p) : C(f) = 1}. F 1 (p, t) = {f F(p, t) : C(f) = 1}. Fact: #{f F(p) : R(f) = r} #F(p) 1 r!. Heuristic: R(f) and t(f) are statistically independent properties of a random f F 1 (p). Conjecture There exists a constant γ > 0 such that #{f F 1 (p, t) : R(f) = r} #F 1 (p, t) for all p prime, t N, and r N. ( ) 1 γ r!
A Possible Explanation Conjecture There exists a constant γ > 0 such that #{f F 1 (p, t) : R(f) = r} #F 1 (p, t) for all p prime, t N, and r N. ( ) 1 γ r! We have checked by computer that this inequality holds with γ = 1/2 in the following cases. - t = 3, p 30977 - t = 4, p 907 - t = 5, p 101 This inequality is true if we restrict to trinomials of low degree (by the function field Chebotarev density theorem). If this conjecture is true, we have R p,t = O(t log p).
ANTS-XII August 30, 2016