Timeline for Calculus over finite fields
Current License: CC BY-SA 3.0
11 events
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| Jul 11, 2013 at 18:50 | comment | added | Ben | Thanks for all the answers and I apologize for the vagueness and my absence yesterday. I do want the polynomial defining $S$ to be irreducible and we assume there are at least $q^{2}$ points on $S$. All polynomials are defined over $F_{q}$, not any extension of it. | |
| Jul 10, 2013 at 1:57 | comment | added | Jason Starr | I agree with Felipe about over-interpreting a vague question. I do agree that it is reasonable to at least assume that $S$ is irreducible (although I wish the OP would have stated that). However, the OP never says anything about varying the field with the polynomials held fixed. I agree that, if one allows the field to grow with the degrees of the polynomials, then Lang-Weil will take over. But, in fact, I don't think you even need something as precise as Lang-Weil. There are infinitely many closed points: I think that is all you need. | |
| Jul 9, 2013 at 23:53 | comment | added | Felipe Voloch | I think it is very frustrating to try to answer vague and ambiguous questions when the OP disappears and it's not worth trying. | |
| Jul 9, 2013 at 23:33 | comment | added | Terry Tao | Fair enough... but then I think if one imposes the condition that $S$ is geometrically irreducible and defined over the field, then one has problem for which the answer is not obviously negative for trivial reasons (and for which I believe, using the ultraproduct argument I sketched earlier, should work; the thing I missed previously was that to use the Lefschetz principle one had to pass from one char 0 field to a larger one and so one needed geometric irreducibility to show that the rational points were Zariski dense). | |
| Jul 9, 2013 at 23:16 | comment | added | Felipe Voloch | @TerryTao Lang-Weil requires absolute irreducibility. An irreducible but not absolutely irreducible (i.e. factoring in an extension) will have very few points and Jason's argument will apply. | |
| Jul 9, 2013 at 23:13 | comment | added | Terry Tao | I assume that the OP wants the polynomial to be defined over the field ${\bf F}_q$ in question, rather than in some extension of that field. In that case, the Lang-Weil inequality ensures plenty of rational points when the field size is large. | |
| Jul 9, 2013 at 23:03 | history | edited | Jason Starr | CC BY-SA 3.0 |
added 960 characters in body
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| Jul 9, 2013 at 22:06 | comment | added | Jason Starr | @Vivek: For a smooth variety, such as mine, reducible is equivalent to connected. If you know precisely what conditions the OP has in mind, I ask that you edit the statement of the question so that the problem is well-posed. | |
| Jul 9, 2013 at 11:05 | comment | added | Vivek Shende | S is worse than reducible, it's not connected; presumably this is meant to be excluded. In particular your objection also works over $\mathbb{C}$... | |
| S Jul 9, 2013 at 8:20 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S Jul 9, 2013 at 8:20 | history | made wiki | Post Made Community Wiki by Jason Starr |