Can you use Chevalley‒Warning to prove existence of a solution?

Question

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.

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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.

Answer 1

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, 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.

Answer 2

The answer by js21 is a great answer, and it gives an explicit choice of $B$. However, you have not asked for any restrictions on $B$ other than it depends only on $n$, $r$, and $d_1,\dots,d_r$ (I assume you want it to be independent of $q$ and of the coefficients of $f_1,\dots,f_r$). There is, in fact, such a $B$ for every choice of $n$, $r$, and $d_1,\dots,d_r$, even if $r>n$, and this only uses the basic techniques of existence of the Hilbert scheme. Over $\text{Spec}(\mathbb{Z})$, you can form the following parameter space as a product of projective spaces $$\Pi = \mathbb{P}H^0(\mathbb{P}^n,\mathcal{O}(d_1))\times \dots \times \mathbb{P}H^0(\mathbb{P}^n,\mathcal{O}(d_r)).$$ Inside the product $\Pi\times \mathbb{P}^n$, you can form the universal closed subscheme $Z$ that is the simultaneous zero locus of an $r$-tuple of homogeneous polynomials of degrees $d_1,\dots,d_r$.

The scheme $\Pi$ is finite type over $\text{Spec}(\mathbb{Z})$, and the projection morphism $\pi:Z\to \Pi$ is projective. Thus, only a finite number of Hilbert polynomials may occur for geometric fibers of $\pi$. More precisely, for every scheme-theoretic fiber of $\pi$ over a point, and for every nonempty open subset of the fiber, we may consider the scheme-theoretic closure of that open subset as a closed subscheme of the fiber. There are only finitely many Hilbert polynomials that can occur for all of these schemes. This follows quicky from the existence of a flattening stratification of $\pi$.

So now let $\mathcal{E}=(e_1(t),\dots,e_N(t))$ be a finite set of numerical polynomials. We say that $\mathcal{E}$ is "saturated" if for every field $F$, both (a) for every $e(t)$ in $\mathcal{E}$, the first difference $e(t) - e(t-1)$ is also in $\mathcal{E}$, and (b) for every closed subscheme $Y\subset \mathbb{P}^n_F$ whose Hilbert polynomial is in $\mathcal{E}$, for every nonempty open subset of $Y$ with its reduced structure, the scheme-theoretic closure of that open set also has Hilbert polynomial in $\mathcal{E}$. By the existence of the Hilbert scheme with fixed Hilbert polynomial as a finite type scheme, by the existence of the flattening stratification, etc., every finite $\mathcal{E}$ is contained in a finite $\mathcal{E}$ that is saturated with respect to (b). It is straightforward to saturate with respect to first differences without altering $\mathcal{E}_m$. Thus, by induction on $m$, every $\mathcal{E}$ is contained in a finite $\mathcal{E}$ that is saturated.

We can partition $\mathcal{E}$ as $\mathcal{E}_0\sqcup \dots \sqcup \mathcal{E}_m$, where $\mathcal{E}_k \subset \mathcal{E}$ is the subset of numerical polynomials of degree $k$. Fix a hyperplane $H=\mathbb{P}^{n-1}\subset \mathbb{P}^n$, i.e., a "hyperplane at infinity" whose complement is an affine space $\mathbb{A}^n$. The claim is that there exists an integer $d=d(\mathcal{E})$ such that for every field $F$ (e.g., a finite field), for every closed subscheme $Y\subset \mathbb{P}^n_F$ whose Hilbert polynomial is in $\mathcal{E}$ and such that $Y$ is not contained in $\mathbb{P}^{n-1}_F$, then there exists a field extension $F'/F$ of degree $\leq d$ such that $Y(F')$ contains an element that is not in $\mathbb{P}^{n-1}(F')$. Up to replacing $Y$ by one irreducible component $Y_j$ that is not contained in $\mathbb{P}^{n-1}_F$ (this is allowed since $\mathcal{E}$ is saturated), we may assume that $Y$ is integral and $Y\cap \mathbb{P}^{n-1}_F$ is the zero scheme of a regular section of $\mathcal{O}_Y(1)$. In particular, the Hilbert polynomial of this zero scheme is the first difference of the Hilbert polynomial of $Y$, thus it is also in $\mathcal{E}$ since $\mathcal{E}$ is saturated.

The claim is proved by induction on $m$. The base case is when $m$ equals $0$. In this case, every $e_i(t)$ is a constant polynomial, and we let $d$ be the maximum of these integers. Since the Hilbert polynomial of $Y$ is constant, $Y$ is zero-dimensional. Then $Y_j$ has total degree $\leq d$. Thus, the associated reduced scheme of $Y_j$ has total degree $\leq d$. This reduced scheme is a singleton set of a closed point that is not contained in $\mathbb{P}^{n-1}_F$. Thus, the residue field of this closed point gives $F'/F$ of degree $\leq d$. This proves the claim when $m$ equals $0$.

By way of induction, assume that $m>0$ and assume the result has been proved for smaller $m$. By the previous paragraph, there exists an integer $d$ that works for $\mathcal{E}_0$. Thus, without loss of generality, assume that the Hilbert polynomial of $Y$ has degree $>0$, i.e., $Y$ has dimension $>0$. Since the Hilbert polynomial of $Y\cap \mathbb{P}^{n-1}_F$ is in $\mathcal{E}$, and since $\mathcal{E}$ is saturated, every irreducible component $D_j$ of $Y\cap \mathbb{P}^{n-1}_F$ has Hilbert polynomial in $\mathcal{E}$. The number $\ell$ of such components is bounded by the degree of $Y$, which in turn is bounded by the maximum $d_0$ of the (normalized) leading coefficients of the finitely many elements of $\mathcal{E}$.

By Gotzmann's Regularity Theorem, there exists an integer $\rho=\rho(\mathcal{E})$ such that the Castelnuovo-Mumford regularity of $Y$ and every $D_j$ is $\leq \rho$ (actually the existence of such $\rho$ is older, presumably due to Mumford, but Gotzmann gave an explicit upper bound on $\rho$). Thus for the ideal sheaf $\mathcal{I}_Y$ of $Y$ in $\mathbb{P}^n$, $\mathcal{I}_{Y}(\rho)$ is globally generated, and $H^0(\mathbb{P}^n_F,\mathcal{O}(\rho))\to H^0(Y_j,\mathcal{O}_{Y}(\rho))$ is surjective, and the same is true for $D_j$. In particular, for distinct components $D_j$ and $D_k$ of $D$, the image of the restriction map $$H^0(\mathbb{P}^n_F,\mathcal{I}_{D_j}(\rho))\to H^0(D_k,\mathcal{I}_{D_j}\cdot\mathcal{O}_{D_k}(\rho))$$ has nonzero image. This can be checked after base change to an algebraic closure of $F$, and it follows there from the fact that $\mathcal{I}_{D_j}(\rho)$ is globally generated. Thus, there exists $f_{j,k}\in H^0(\mathbb{P}^n_F,\mathcal{O}(\rho)$ such that $f_{j,k}$ vanishes on $D_j$ but not on $D_k$. The element $f_k = \prod_{j\neq k} f_{j,k}$ is an element of $H^0(\mathbb{P}^n_F,\mathcal{O}((\ell-1)\rho))$ that vanishes on every $D_j$ with $j\neq k$, yet does not vanish on $D_k$. Thus, $f=\sum_k f_k$ is an element of $H^0(\mathbb{P}^n_F,\mathcal{O}((\ell - 1)\rho))$ that does not vanish identically on any $D_k$. The same is true for the power $f^s$. Thus, there exists a section of $H^0(\mathbb{P}^n_k,\mathcal{O}((d_0-1)!\rho)))$ whose zero scheme $Z$ on $Y$ does not contain any $D_j$. In particular, $Z$ is not contained in $\mathbb{P}^{n-1}_F$ (there may have been a faster way to arrange that).

By the same quasi-compactness arguments as above, the Hilbert polynomials of $Z$'s are bounded and have degree $<m$. Thus, letting $\mathcal{F}$ be the union of $\mathcal{E}_k$ for $k<m$ and the Hilbert polynomials of $Z$s as above, $\mathcal{F}$ is a finite set of Hilbert polynomials with smaller $m$. Thus, by the induction hypothesis, there exists an integer $d$ such that $Z(F')\setminus \mathbb{P}^{n-1}(F')$ is nonempty for some $F'/F$ of degree $<d$. Thus, this same $d$ works for $\mathcal{E}$, and the claim is proved by induction on $m$.