Breaking the 6/5 threshold for sums and products modulo a prime

Ilya Shkredov and I just released our preprint, “Breaking the 6/5 threshold for sums and products modulo a prime.”

Since this paper of Roche–Newton, Rudnev, and Shkredov, the best bound for the sum–product problem in {\mathbb{F}_p} was given by the following.

Theorem (Sum–Product): Let {A \subset \mathbb{F}_p} of size at most {p^{5/8}}. Then

\displaystyle |A+A| + |AA| \gg |A|^{6/5}. \ \ \spadesuit

The major breakthrough was a point–plane incidence theorem of Rudnev. Let us recall their proof, which has been simplified since their original.

We consider the set of points and lines

\displaystyle P = (A+A) \times (AA), \ \ \ \mathcal{L} = \{(u,v) \in \mathbb{F}_p^2 : v = a(u - c)\}_{a , c \in A}.

Note that each line contains the points {(b + c , ab)} and so there are at least {|A|^3} incidences. On the other hand, one may apply a point–line incidence, for instance the cartesian version found in Theorem 4 of this paper of Stevens and Zeeuw to obtain an upper bound for the number of incidences between {P} and {\mathcal{L}}. Combining these two bounds gives Theorem 1.

We are able to make the following improvement.

Theorem 1: Let {A \subset \mathbb{F}_p} of size at most {p^{3/5}}. Then

\displaystyle |A+A| + |AA| \gg |A|^{6/5 + c}, \ \ \ c = 4/305. \ \ \spadesuit

We can improve this application of a Szemerédi–Trotter type bound in the specific case of the sum–product problem. The basic idea is to evaluate the above proof and obtain a bound for the {\tau}–rich lines, instead of incidences. We then obtain the bound for the higher order energy:

\displaystyle d_4^+(A) \lesssim |AA|^2 |A|^{-2} , \ \ d_4^+(A) := E_4^+(A,B) |A|^{-1} |B|^{-3}.\ \ \ \ \ (1)

A simple application of Cauchy–Schwarz gives

\displaystyle |A+A|^{4/3} d_4^+(A)^{-1/3} \leq |A+A|. \ \ \ \ \ (2)

Combining (1) and (2) recovers Theorem 1. We are able to make an improvement to (2) and in turn to the sum–product problem.

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