Timeline for Question on calculating character sums
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2021 at 19:00 | comment | added | matt stokes | @WillSawin By calculate I mean finding an exact value like you can do for quadratic Gauss sums. I realize this might be hoping for too much. My real concern is what kind of $p$ make this sum zero mod $p$. An example of what I have in mind is $\sum_{\substack{a=1 \\ p \nmid a}}^{np} \chi(a)(-1)^a = (1 - \chi(p))\sum_{a = 1}^{n-1}\chi(a)(-1)^a$. So one condition I get is $\sum_{\substack{a=1 \\ p \nmid a}}^{np} \chi(a)(-1)^a =0 $ if $p$ splits in the field associated to $\chi$. I would like to do something similar here, so I don't need to calculate this sum exactly (but it would be nice). | |
| Jan 10, 2021 at 17:29 | comment | added | Will Sawin | What do you mean by "calculate" here? If you want to bound the sum you can do a Polya-Vinogradov method. I don't see any evidence that there is a formula for the sum simpler than what you have here. | |
| Jan 7, 2021 at 17:46 | comment | added | Henri Cohen | sorry, indeed since you assume $n$ odd. | |
| Jan 7, 2021 at 17:36 | comment | added | matt stokes | @HenriCohen I think the alternating part messes that up since $(-1)^{np-a} = (-1)(-1)^a$, so summing up to $np-1$ gives you zero. | |
| Jan 7, 2021 at 10:16 | comment | added | Henri Cohen | Since both $\chi$ and $\omega$ are odd characters, the sum to $np-1$ is twice that up to $(np-1)/2$, or do I misunderstand ? | |
| Jan 6, 2021 at 22:50 | comment | added | matt stokes | @HenriCohen Doesn't summing up to $(np-1)/2$ mess up the Chinese remainder theorem since it is an isomorphism from $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/np\mathbb{Z}$? | |
| Jan 6, 2021 at 17:33 | comment | added | Henri Cohen | Using the Chinese remainder theorem, doesn't your sum factor into a product of two independent Gauss sums ? | |
| Jan 6, 2021 at 16:34 | history | edited | matt stokes | CC BY-SA 4.0 |
edited body
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| Jan 5, 2021 at 22:08 | history | edited | LSpice | CC BY-SA 4.0 |
Name of article; deleted “thanks”
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| Jan 5, 2021 at 21:36 | history | asked | matt stokes | CC BY-SA 4.0 |