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There is a paper by Bourgain, Katz and Tao

Bourgain, Jean; Katz, N.; Tao, Terence C., A sum-product estimate in finite fields, and applications, Geom. Funct. Anal. 14, No. 1, 27-57 (2004). ZBL1145.11306.

where they proved:

$$ \begin{array}{l}{\text { Theorem } 1.1 \text { (Sum-product estimate). } \text {Let } F :=\mathbf{Z} / q \mathbf{Z} \text { for some prime } q} \\ {\text { and let } A \text { be a subset of } F \text { such that }} \\ {\qquad|F|^{\delta}<|A|<|F|^{1-\delta}} \\ {\text { for some } \delta>0 \text { . Then one has a bound of the form }} \\ {\max (|A+A|,|A \cdot A|) \geq c(\delta)|A|^{1+\varepsilon}} \\ {\text { for some } \varepsilon=\varepsilon(\delta)>0}\end{array} $$

What is conjectured to be the value of $\epsilon$?

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First off, we can remove the condition that $|A| \geq |F|^{\delta}$. One expects to be able to take any $\epsilon < 1$, as long as $|A| \leq |F|^{1/2}$ (not quite, see Oliver's comment below). The most recent progress is contained in this joint work with Rudnev and Shkredov. It is known you cannot take $\epsilon =1$, say by work on the multiplication table problem or a slightly different construction in the original paper of Erdos and Szemeredi on the sum-product conjecture.

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  • $\begingroup$ Thank you for your response. So the conjecture is $1-\epsilon$ like what we have for integers? $\endgroup$
    – Ken Berner
    Commented Sep 23, 2019 at 23:24
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    $\begingroup$ Hi George. You wrote that "One expects to be able to take any $\epsilon<1$, as long as $|A|≤|F|1/2$". This is not strictly true, as it is possible that $A$ has size $\sqrt p$ and $\max \{ |A+A|, |AA| \} \ll p^{3/4}$. See this paper of Garaev. Probably the correct conjecture is that $$ \max \{ |A+A|, |AA| \} \gg \min \{|A|^{2-\epsilon}, \sqrt{p|A|} \}.$$ M.Z. Garaev, The sum-product estimate for large subsets of prime fields. Proc. Amer. Math. Soc. 136 (2008), no. 8, 2735–2739. $\endgroup$
    – Oliver Roche-Newton
    Commented Sep 24, 2019 at 11:58
  • $\begingroup$ @OliverRoche-Newton, Thank you. So if we assume |A| < p^{1/2}, then the conjecture is $1-\epsilon$? $\endgroup$
    – Ken Berner
    Commented Sep 24, 2019 at 15:10
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    $\begingroup$ @KenBerner the construction of Garaev continues to give counterexamples to your statement when $A$ is smaller than $\sqrt p$. But most people expect that $$ \max \{|A+A|,|AA| \} \gg |A|^{2-\epsilon}$$ provided that $A$ is sufficiently small. And it may be enough to assume that $|A| <p^{1/3}$ to reach this conclusion. $\endgroup$
    – Oliver Roche-Newton
    Commented Sep 24, 2019 at 15:17
  • $\begingroup$ @OliverRoche-Newton, thank you very much. $\endgroup$
    – Ken Berner
    Commented Sep 24, 2019 at 15:39

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