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Mar 17, 2010 at 13:43 comment added BCnrd @Pete: Correct. That's why I spoke of "fixed" d and "most" k; I was avoiding the consideration of things which are specific to "small" finite fields.
Mar 17, 2010 at 13:28 comment added Pete L. Clark @Brian: not to belabor the point, but you're not suggesting that this is the case even for $n = d \gg \# k$, right? I ask because (i) this is relevant to the original question and (ii) I am interested myself.
Mar 17, 2010 at 5:40 comment added BCnrd Yes Pete, I was only considering d=n as in the question, and so I was only pointing out that if we want to make constructions that include most finite fields and use a fixed d then blah-blah-blah... –
Mar 17, 2010 at 3:45 comment added Pete L. Clark @Brian: inevitable only if $\# k \gg d$, you mean, right? Certainly there are smooth counterexamples when $d$ is large compared to $\# k$ and $n$.
Mar 17, 2010 at 2:02 comment added BCnrd You're asking that the projective hypersurface $(f=0)$ of degree $d$ in projective $(d-1)$-space has no $k$-rational points. If $k$ has sufficiently large size purely in terms of $d$ then the RH bound provides rational points if this hypersurface is smooth and geometrically connected. So if we restrict attention to irreducible $f$ over $k$ then it seems inevitable for it to be a $k^{\times}$-multiple of a norm of a lower-degree homogeneous form over a nontrivial finite extension of degree dividing $d$, or that $(f=0)$ has singularities.
Mar 17, 2010 at 1:20 comment added Wanderer Thanks for the reply. Yes, I know that it is optimal for every $d > 0$, and norms are the obvious examples to show this - I should have mentioned that in the question. But I'm rather interested in "other types" of examples, whatever that may mean.
Mar 17, 2010 at 0:57 history answered BCnrd CC BY-SA 2.5