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][1].

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**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][1].


  [1]: http://mathoverflow.net/q/239934/82179