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So let $\mathbf{F}_q$ be a finite field with $q$ elements where $q=p^m$,

$p$ a prime number and $m\in\mathbf{Z}_{\geq 1}$. Let $f(x,y)\in \mathbf{F}_q[x,y]$

be a smooth non-constant polynomial and let $A:=\mathbf{F}_q[x,y]/(f)$.

Q: Does there exists an integer $N_0$ (which depends on $A$) such that for $N\geq N_0$ one

may always find a prime ideal $P\subseteq A$ such that $A/P\simeq\mathbf{F}_{q^N}$?

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The answer is "yes", given what we know about the number of points on the corresponding projective curve (or more accurately a smooth model of it, in case of singularities). Since points at infinity and singular points can be bounded terms of the degree of f, they don't affect the asymptotics here. So the question comes down to the existence of points with a given field of definition, of size a large power of q. The calculation that the main term gives the right answer (>0) can be read off the case of the affine line, where it is certainly true.

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    $\begingroup$ Well now that I think it is obvious. If $C$ is the corresponding curve then we have an exact formula for $\#C(\mathbb{F}_{q^n})$. It is $q^n+1+(error term)$. Nous using weak Weil estimates we are done. $\endgroup$
    – Hugo Chapdelaine
    Commented Feb 9, 2011 at 17:44
  • $\begingroup$ and I should say that for a proper divisor $d|n$ $\#C(\mathbb{F}_{q^d})=q^d+1+(error term)$ so weak estimates imply that $C(\mathbb{F}_{q^n})-\bigcup_{d|n,d\neq n}C(\mathbb{F}_{q^d})$ is larger than 1. $\endgroup$
    – Hugo Chapdelaine
    Commented Feb 9, 2011 at 18:23

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