Recall the Chevalley‒Warning theorem:



> 
> **Theorem.** Let $f\_1, \ldots, f\_r \in \mathbb F\_q[x\_1,\ldots,x\_n]$ be polynomials of degrees $d\_1, \ldots, d\_r$. If
>  $$d\_1 + \ldots + d\_r < n,$$
>  then the number of common zeroes of $f\_1, \ldots, f\_r$ is $0$ modulo $p$.
> 
> 
> 


In particular, if there exists a zero (e.g. if all the $f\_i$ have no constant coefficient), then there exists another one.


On the other hand, it can easily happen that the $f\_i$ have no common zeroes at all, e.g. if $n = 3$ and $f\_1 = x\_1$, $f\_2 = x\_1 + 1$. The problem is that the scheme
$$\operatorname{Spec} \mathbb F\_q [x\_1,x\_2,x\_3]/(x\_1, x\_1 + 1)$$
is empty, so you're never going to find any solutions.


A more interesting example is when $n = 3$ and $f\_1 = x\_1^2 - a$ for $a \in \mathbb F\_q$ not a square. Then
$$\operatorname{Spec} \mathbb F\_q[x\_1,x\_2,x\_3]/(x\_1^2 -a ) \neq \varnothing,$$
but there are no rational solutions.



> 
> **Question.** Let $f\_i$ as in the Chevalley‒Warning theorem. Assume that
>  $$X = \operatorname{Spec} \mathbb F\_q [x\_1,\ldots,x\_n]/(f\_1, \ldots, f\_r) \neq \varnothing.$$
>  Is there a bound $B$ in terms of the $d\_i$ and $n$ such that $X(\mathbb F\_{q^k}) \neq \varnothing$ for some $k \leq B$?
> 
> 
> 


In particular, I was hoping that there is some trick to reduce this to Chevalley‒Warning (e.g. introduce an extra variable, cf. the example below). However, other methods for attacking this question are welcome as well.


This question is motivated by, but not identical to, the finite field case of [this question](https://mathoverflow.net/q/239934/82179).




---


**Example.** An example where one can use such a phenomenon is with the reduced norm on central simple algebras over $\mathbb F\_q$: 


Let $A$ be a division algebra over $\mathbb F\_{q}$ of dimension $n$. Then the reduced norm is a homogeneous polynomial $P$ of degree $n$ in $n$ variables. For any $a \in \mathbb F\_{q}$, consider the homogeneous polynomial
$$P(x) - a z^n$$
of degree $n$ in $n + 1$ variables. It has a solution $(0,\ldots,0)$ since it is homogeneous, so by Chevalley‒Warning it has to have a nontrivial solution. But since $A$ is a division algebra, we cannot have $P(x) = 0$ for $x \neq 0$, so the solution has to satisfy $z \neq 0$. This proves that the reduced norm is surjective. (This basically proves that the Brauer group of a finite field (more generally a $C\_1$ field) is trivial.)


**Remark.** A standard trick for this type of question is to view $\mathbb F\_{q^k}$ as a $k$-dimensional vector space over $\mathbb F\_q$. If $\alpha$ is a primitive element, we can introduce new variables $x\_{1,1}, \ldots, x\_{n,k}$ corresponding to
$$x\_i = x\_{i,1} + \alpha x\_{i,2} + \ldots + \alpha^{k-1} x\_{i,k}.$$
Then solutions of the $f\_i$ over $\mathbb F\_{q^k}$ correspond to solutions of suitable polynomials $\tilde{f}\_i$ of the *same degree* $d\_i$ over $\mathbb F\_q$. This allows us to get rid of the degree assumption in my question: take $k$ such that $d\_1 + \ldots + d\_{r} < k n$, and do this substitution to reduce to the question I posed above.


I guess that this means that my question is equivalent to the finite field case of [the question I linked earlier](https://mathoverflow.net/q/239934/82179).



Can you use Chevalley‒Warning to prove existence of a solution?
Let $f\_i \in \mathbb{F}\_q[X\_1,\dots,X\_n]$ have degree at most $d$ for each $i \in [|1,r|]$, and assume that the affine scheme
$$
X = \operatorname{Spec} \mathbb F\_q [x\_1,\ldots,x\_n]/(f\_1, \ldots, f\_r)
$$
is nonempty. Take a prime number $\ell \neq p$. Consider the Frobenius $F$ as an endomorphism of the vector space
$$
V = \bigoplus\_{i \in \mathbb{Z}} H\_c^i(X\_{\bar{\mathbb{F}\_q}}, \mathbb{Q\_{\ell}})
$$
By Theorem 1 of [this paper by Katz](https://web.math.princeton.edu/~nmk/BettiSum14.pdf), the dimension of $V$ is bounded above by an explicit constant $B = B(n,r,d)$. Hence one can find elements $\lambda\_1,\dots,\lambda\_B \in \mathbb{Q}\_{\ell}$ such that one has
$$
F^{B+1} = \lambda\_1 F + \lambda\_2 F^2 + \dots + \lambda\_B F^B
$$
on $V$. It follows that for any $k > B$, the integer
$$
|X(\mathbb{F}\_{q^k})| = \sum\_{i \in \mathbb{Z}} (-1)^i \mathrm{Tr}\left( F^k | H\_c^i(X\_{\bar{\mathbb{F}\_q}}, \mathbb{Q\_{\ell}})\right)
$$
is a linear combination of $|X(\mathbb{F}\_{q})|,|X(\mathbb{F}\_{q^2})|,\cdots,|X(\mathbb{F}\_{q^B})|$. Since $|X(\mathbb{F}\_{q^k})|$ is nonzero for some $k$, it must be nonzero for some $k \leq B$. Hence this constant $B$ yields a positive answer to your question.


