When does a variety have a point over a finite field for sufficiently large primes p?

Question

Let $X$ be an algebraic variety over the rational numbers. Suppose that $X$ has positive dimension. I would like to say that $X(\mathbb{F}_p)$ is non-empty for sufficiently large primes $p$. One idea is to use the Weil conjectures, but that seems like overkill. So my two questions are:

1. When can one deduce that a system of polynomial equations has a solution over $\mathbb{F}_p$ for sufficiently large primes $p$? I might have guessed that $\dim(X) > 0 $ is sufficient, but it seems like the affine variety $X^2+Y^2 = 0$ minus the point $(0,0)$ (which has no solutions for $p = -1 \mod 4$) suggests that extra conditions are necessary.

2. Is there an elementary proof of the (correct) version of part 1?

Answer 1

If $k$ is the field of rational numbers and $A$ is a finitely generated $k$-algebra with $X:=Spec(A)$, it follows $X(\mathbb{F}_p)$ is empty for any prime number $p>0$. If $\mathfrak{m}\subseteq A$ is a maximal ideal it follows $k \subseteq \kappa(\mathfrak{m})$ is a finite field extension of $k$, but $k$ has characteristic zero and hence this is impossible.

If $B:=\mathbb{Z}[x_1,..,x_n]/I$ where $\mathbb{Z}$ is the ring of integers and $Y:=Spec(B)$, we get a canonical map $\pi: Y \rightarrow S:=Spec(\mathbb{Z})$. If you accept Zorns's Lemma/the "Axiom of Choice", it follows there is a maximal ideal $\mathfrak{m} \subseteq B$. The residue field $\kappa(\mathfrak{m})\cong \mathbb{F}_{q^r}$ is finite with characteristic $q>0$. Hence $\mathfrak{m}\cap \mathbb{Z}=(q)$, and it follows the fiber $\pi^{-1}(q)\subseteq Y$ is non-empty. Hence Zorn's Lemma proves that $Y(\mathbb{F}_{q^r})$ is always non-empty for some $q,r$. The fact that the residue field $\kappa(\mathfrak{m})$ of a maximal ideal in $B$ is finite, is proved in the exercises in Atiyah-Macdonald's book "Commutative Algebra", hence this argument is in some sense "elementary". I have also seen simple online proofs that the residue field $\kappa(\mathfrak{m})$ is finite as well.

Answer 2

Using the Lang-Weil bound and maybe also the Chebotarev density theorem or something like that, I believe the correct statement is that if $\dim X \ge 1$ then $X(\mathbb{F}_p)$ is non-empty for all sufficiently large primes $p$ if and only if the action of the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on the set of irreducible components of $X_{\overline{\mathbb{Q}}}$ fixes at least one component of positive dimension. This condition doesn't hold in your example; there are two irreducible components over $\mathbb{Q}(i)$, given by the lines $X + iY = 0$ and $X - iY = 0$ (minus the origin), and the action of complex conjugation switches them.

I don't see any way to use a weaker result than the Lang-Weil bound.

I also want to point out that strictly speaking if $X$ is a variety over the rational numbers then it doesn't make sense to talk about $X(\mathbb{F}_p)$ because $\text{Spec } \mathbb{F}_p$ is not a variety (or scheme or anything) over the rational numbers; when people write expressions like this they are implicitly choosing a lift of $X$ to a scheme defined over some localization of $\mathbb{Z}$ defined by the denominators of some collection of polynomials cutting out $X$, and the behavior of $X(\mathbb{F}_p)$ for any fixed prime $p$ could a priori depend on this choice. I think it's true that the question of whether $X(\mathbb{F}_p)$ is non-empty for sufficiently large $p$ does not depend on this choice but I can't quote a theorem off the top of my head that guarantees this. In the Lang-Weil argument above let's say we're fixing a particular lift to keep things safe.