Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with finite-fields
Search options not deleted
user 4008
A finite field is a field with a finite number of elements. For each prime power $q^k$, there is a unique (up to isomorphism) finite field with $q^k$ elements. Up to isomorphism, these are the only finite fields.
1
vote
A family of hypersurfaces with many points
This is very far from a solution but just some ideas that could be possibly be
used as a start.
First a simple observation on an Artin-Schreier type extension $X\rightarrow Y$,
where $X:=\mathrm{Spec …
- 22.6k
25
votes
Accepted
How does Tate verify his own conjecture for the Fermat hypersurface?
I don't know how Tate did it but here is one way. Let $\zeta$ be such that
$\zeta^{q+1}=-1$ and put $a_j=(0\colon\cdots\colon1\colon\zeta:\cdots\colon0)$,
$j=0,\ldots,i$ with the $1$ in coordinate $2 …
- 22.6k