q)$?

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11 events

when toggle format	what		by	license	comment
Mar 9, 2018 at 15:51	vote	accept	Johnny T.
Feb 27, 2018 at 21:07	answer	added	Will Sawin		timeline score: 2
Feb 27, 2018 at 19:34	answer	added	LSpice		timeline score: 3
Feb 24, 2018 at 22:21	comment	added	LSpice		@alpoge, maybe it's worth promoting that long string of comments to an answer?
Feb 24, 2018 at 15:22	comment	added	Johnny T.		@alpoge I will try going through your comments thoroughly. Thank you very much!
Feb 24, 2018 at 6:35	comment	added	alpoge		Otherwise, $x$ has $O(1)$ choices, $m=0$, and the sum over $n$ is $p^{\lfloor \frac{t-2}{2}\rfloor}$, so it seems like you get a bound of $\ll p^{\lfloor \frac{t-2}{2}\rfloor}$ when $t > 2$. Forgive me if I've made a mistake! Just lemme know and I'll try to fix it, I admit I'm just writing what comes to mind immediately so it could be total nonsense. (For example, it seems the exponent of $p$ should have a ceiling or something, since for $t=3$ one shouldn't get a bound of $O(1)$.)
Feb 24, 2018 at 6:35	comment	added	alpoge		Otherwise, when $t\geq 3$ and the linear term does vanish, instead write $k =: m + p^{\lceil \frac{t-2}{2}\rceil} n$ with $0\leq m < p^{\lceil \frac{t-2}{2}\rceil}$ and $n\in \mathbb{Z}/p^{\lfloor \frac{t-2}{2}\rfloor}$, and observe that the sum becomes linear in $n$, with coefficient depending on $m$. Unless that coefficient vanishes, which I think only happens when $m=0$ (or the $a_i$ and $b_i$ are in remarkable position, but let's ignore that possibility) [forgive me if I'm being sloppy --- I figured I'd give a quick answer to this just to help a bit], the sum over $n$ is $0$.
Feb 24, 2018 at 6:34	comment	added	alpoge		Now by writing $k =: m + p^{t-2} n$ with $0\leq m < p^{t-2}$ and $n\in \mathbb{Z}/p$ and examining the sum over $n$, one sees that unless $A_\chi \sum_i (x-a_i)^{-1} - (x-b_i)^{-1}\equiv C\pmod{p^{t-1}}$ (which has $O(1)$ many solutions for $x\in \mathbb{F}_p$) the inner sum is zero. If $t=2$ we're then done, and we get a bound of $O(1)\cdot p = O(p)$, since there are $O(1)$ $x$ remaining and for those $x$ the sum over $k$ is $p$.
Feb 24, 2018 at 6:33	comment	added	alpoge		Otherwise write $h =: x + pk$ with $0\leq x < p$ and $k\in \mathbb{Z}/p^{t-1}$ to obtain $\sum_x e_q(x) \prod_i \chi(x-a_i)\bar{\chi}(x-b_i) \sum_k e_q(Cpk - A_\chi \sum_{1\leq j\ll t} \frac{(-1)^{j-1} p^j k^j}{j}\sum_i (x-a_i)^{-j} - (x-b_i)^{-j})$, where $A_\chi\in (\mathbb{Z}/p^{t-1})^\times$. [Here I've written, via the $p$-adic log (i.e. $(\mathbb{Z}/q)^\times\simeq \mathbb{F}_p^\times\times \mathbb{Z}/p^{t-1}$), $\chi(1 - pt) = e_q(-A_\chi\sum_{1\leq i\ll t} p^i t^i/i)$, i.e.\ as an additive character on $\mathbb{Z}/p^{t-1}$.]
Feb 24, 2018 at 6:33	comment	added	alpoge		Hey so let's say $p > 5$ or something. When $t=1$ this is a standard mixed character sum and can be bounded using the usual techniques (e.g.\ see Katz's webpage and click any paper with "mixed char. sums" in the title). Also unless $\chi$ has conductor $q$ the sum is $0$ (split the sum into fibres mod $p^{t-1}$ and you're summing a nontrivial additive character).
Feb 23, 2018 at 19:04	history	asked	Johnny T.	CC BY-SA 3.0