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Sep 24, 2019 at 15:39 comment added Ken Berner @OliverRoche-Newton, thank you very much.
Sep 24, 2019 at 15:17 comment added Oliver Roche-Newton @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.
Sep 24, 2019 at 15:15 history edited George Shakan CC BY-SA 4.0
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Sep 24, 2019 at 15:10 comment added Ken Berner @OliverRoche-Newton, Thank you. So if we assume |A| < p^{1/2}, then the conjecture is $1-\epsilon$?
Sep 24, 2019 at 11:58 comment added Oliver Roche-Newton 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.
Sep 23, 2019 at 23:24 comment added Ken Berner Thank you for your response. So the conjecture is $1-\epsilon$ like what we have for integers?
Sep 23, 2019 at 22:56 history answered George Shakan CC BY-SA 4.0