论文标题
关于有限估值环的总和估计值的注释
A note on sum-product estimates over finite valuation rings
论文作者
论文摘要
令$ \ Mathcal r $为订单$ q^r $的有限估值环，$ q $ a os os od prime数字，而$ \ mathcal a $是$ \ mathcal r $的设置。在本文中，由于Yazici（2018），我们改善了最新结果，该结果对总产品类型问题。更确切地说，我们将证明 1. If $|\mathcal A|\gg q^{r-\frac{1}{3}}$, then $$\max\left\lbrace |\mathcal A+\mathcal A|, |\mathcal A^2+\mathcal A^2|\right\rbrace \gg q^{\ frac {r} {2}} | \ Mathcal a |^{\ frac {\ frac {1} {2}}。 $ \ max \ left \ lbrace | \ Mathcal a+\ Mathcal a |，| \ Mathcal A^2+\ Mathcal A^2 | \ right \ rbrace \ rbrace \ gg \ gg \ gg \ frac {| \ Mathcal a |^2} a+\ \ \ \ mathcal a || \ mathcal a |^2 \ gg q^{3R-1} $和$ 2Q^{r-1} \ le | \ le | \ Mathcal a | \ ll Q^{r- \ frac {3} | \ Mathcal A^2+\ Mathcal A^2 | \ right \ rbrace \ gg q^{r/3} | \ Mathcal a |^{2/3}。$$
Let $\mathcal R$ be a finite valuation ring of order $q^r$ with $q$ a power of an odd prime number, and $\mathcal A$ be a set in $\mathcal R$. In this paper, we improve a recent result due to Yazici (2018) on a sum-product type problem. More precisely, we will prove that 1. If $|\mathcal A|\gg q^{r-\frac{1}{3}}$, then $$\max\left\lbrace |\mathcal A+\mathcal A|, |\mathcal A^2+\mathcal A^2|\right\rbrace \gg q^{\frac{r}{2}}|\mathcal A|^{\frac{1}{2}}.$$ 2. If $q^{r-\frac{3}{8}}\ll |\mathcal A|\ll q^{r-\frac{1}{3}}$, then $$\max\left\lbrace |\mathcal A+\mathcal A|, |\mathcal A^2+\mathcal A^2|\right\rbrace \gg \frac{|\mathcal A|^2}{q^{\frac{2r-1}{2}}}.$$ 3. If $|\mathcal A+\mathcal A||\mathcal A|^2\gg q^{3r-1}$ and $2q^{r-1}\le |\mathcal A|\ll q^{r-\frac{3}{8}}$, then $$\max\left\lbrace |\mathcal A+\mathcal A|, |\mathcal A^2+\mathcal A^2|\right\rbrace \gg q^{r/3}|\mathcal A|^{2/3}.$$
