Nearly sharp Lang–Weil bounds for a hypersurface

Abstract

We improve to nearly optimal the known asymptotic and explicit bounds for the number of $\mathbb {F}_q$ -rational points on a geometrically irreducible hypersurface over a (large) finite field. The proof involves a Bertini-type probabilistic combinatorial technique. Namely, we slice the given hypersurface with a random plane.

1 Introduction

Let $n\geq 2$ and $d\geq 1$ , and let ${\mathbb {F}}_q$ be a finite field. Let $X\subset \mathbb {P}^n$ be a geometrically irreducible hypersurface of degree $d$ over ${\mathbb {F}}_q$ . Lang and Weil [4] have established the bound

(1.1) $$ \begin{align} |\#X({\mathbb{F}}_q)-\#\mathbb{P}^{n-1}({\mathbb{F}}_q)|\leq (d-1)(d-2)q^{n-3/2}+ O_{n,d}(q^{n-2}), \end{align} $$

where the implicit constant can depend only on $d$ and $n$ (but not on $q$ or $X$ ). We prove that, in fact, the implicit constant can be taken to be an absolute constant—independent of $n$ and $d$ altogether—in the regime of interest $q\gg _d 1$ .

Theorem 1.1 Let $X\subset \mathbb {P}^n_{{\mathbb {F}}_q}$ be a geometrically irreducible hypersurface of degree $d$ . Then

$$ \begin{align*} |X({\mathbb{F}}_q)|&\geq q^{n-1}-(d-1)(d-2)q^{n-3/2}-O_d(q^{n-5/2}) \qquad\text{and}\\ |X({\mathbb{F}}_q)|&\leq q^{n-1}+(d-1)(d-2)q^{n-3/2} +(1+\pi^2/6)q^{n-2}+O_d(q^{n-5/2}). \end{align*} $$

Example 1.2 (Cone over a maximal curve)

Let $(d,q_0)$ be such that there exists a (nonsingular) maximal curve $C=\{f=0\}$ in $\mathbb {P}^2$ over ${\mathbb {F}}_{q_0}$ of degree $d$ . Let $q$ be a power of $q_0$ , and let $X=\{f=0\}\subset \mathbb {P}^n_{{\mathbb {F}}_q}$ be a projective cone over $C$ . Then

$$\begin{align*}\#X({\mathbb{F}}_q)=q^{n-1}\pm (d-1)(d-2)q^{n-3/2}+q^{n-2}+q^{n-3}+\cdots+1,\end{align*}$$

with $\pm $ depending on whether $q$ is an odd or an even power of $q_0$ . Thus, the constant $1+\pi ^2/6$ in the upper bound exhibited in Theorem 1.1 cannot possibly be improved by more than $\pi ^2/6$ , and the constant $0$ in the lower bound in Theorem 1.1 cannot be improved by more than $1$ .

In most of this article, we work in affine space. For a geometrically irreducible hypersurface $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ of degree $d$ , [4] states that

(1.2) $$ \begin{align} |\#X({\mathbb{F}}_q)-q^{n-1}|\leq (d-1)(d-2)q^{n-3/2}+ C_d q^{n-2}, \end{align} $$

where $C_d$ can depend only on $d$ and $n$ . Our notation highlights the more important dependence of $C_d$ on $d$ and suppresses the dependence on $n$ (usually one thinks of $n$ as being fixed from the beginning).

The problem of giving explicit versions of (1.2) and of improving the dependence of $C_d$ on $d$ has a long history, which we now briefly summarize. See [2] for a more detailed account.

• • Schmidt has shown that in the case of the lower bound, one can take $C_d=6d^2$ for $q\gg _{n,d} 1$ (see [5]) and in the case of the upper bound, one can take $C_d=4d^2 k^{2^k}$ , where $k=\binom {d+1}{2}$ (see Theorem 4C on page 208 and Theorem 5A on page 210 in [6]).

• • Ghorpade and Lachaud [3] use $\ell $ -adic étale cohomology techniques to prove that one can take $C_d$ to be a polynomial in $d$ (of degree that depends on $n$ ) in the case of the upper bound as well. Explicitly, one can take $C_d=12(d+3)^{n+1}$ in (1.2).

• • Cafure and Matera [2] prove that one can take $C_d=5d^{13/3}$ in (1.2); moreover, if $q>15d^{13/3}$ , one can take $C_d=5d^2+d+1$ (this is a polynomial whose degree does not grow with $n$ ).

• • The author [7] has established the lower bound (for any $\varepsilon>0$ )

  $$\begin{align*}|X({\mathbb{F}}_q)|\geq q^{n-1}-(d-1)(d-2)q^{n-3/2}-(d+2+\varepsilon)q^{n-2}\end{align*}$$
  for $q\gg _\varepsilon 1$ .

• • The author’s Theorem 8 in the preprint [8] implies that for every $\varepsilon>0$ and $\varepsilon '>0$ , we have

  $$\begin{align*}|X({\mathbb{F}}_q)|\leq q^{n-1}+(d-1)(d-2)q^{n-3/2}+((2+\varepsilon)d+1+\varepsilon')q^{n-2}\end{align*}$$
  as long as $q\gg _{\varepsilon ,\varepsilon '} 1$ .

• • When $\dim X=1$ (equivalently, $n=2$ ), Aubry and Perret have proved (apply Corollary 2.5 in [1] to the closure of $X$ in $\mathbb {P}^2$ ) that one can take $C_d=d-1$ in the case of the lower bound and $C_d=1$ in the case of the upper bound:

  (1.3) $$ \begin{align} q-(d-1)(d-2)\sqrt{q}-d+1\leq |X({\mathbb{F}}_q)| \leq q+(d-1)(d-2)\sqrt{q}+1. \end{align} $$

1.1 Upper bounds

The affine version of the asymptotic upper bound in Theorem 1.1 reads as follows.

Theorem 1.3 Let $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a geometrically irreducible hypersurface of degree $d$ . Then

(1.4) $$ \begin{align} |X({\mathbb{F}}_q)|\leq q^{n-1}+(d-1)(d-2)q^{n-3/2}+\left(1+\pi^2/6\right)q^{n-2}+O_{d}(q^{n-5/2}), \end{align} $$

where the implied constant depends only on $d$ and can be computed effectively.

We can give an explicit bound, as in the following theorem.

Theorem 1.4 Let $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a geometrically irreducible hypersurface of degree $d$ . Suppose that $q>15d^{13/3}$ . Then

(1.5) $$ \begin{align} |X({\mathbb{F}}_q)|\leq q^{n-1}+(d-1)(d-2)q^{n-3/2}+5q^{n-2}. \end{align} $$

Example 1.5 (Cylinder over a maximal curve)

Let $d\geq 3$ be such that $d-1$ is a prime power. Let $q$ be an odd power of $(d-1)^2$ . Consider the curve $C=\{y^{d-1}+y=x^d\}$ in $\mathbb {A}^2_{{\mathbb {F}}_q}$ . It is known (see, for example, [9]) that $\#C({\mathbb {F}}_q)=q+(d-1)(d-2)\sqrt {q}$ . Then the number of ${\mathbb {F}}_q$ -points on $C\times \mathbb {A}^{n-2}$ is $q^{n-1}+(d-1)(d-2)q^{n-3/2}$ . Thus, the constant $5$ in (1.4) cannot possibly be improved by more than $5$ .

Remark 1.6 While the cylinder $C\times \mathbb {A}^{n-2}$ in Example 1.5 is nonsingular, its Zariski closure in $\mathbb {P}^n$ has a large (in fact, $(n-3)$ -dimensional) singular locus. In general, let $X\subset \mathbb {A}^n$ be a geometrically irreducible hypersurface such that $\#X({\mathbb {F}}_q)\geq q^{n-1}+ (d-1)(d-2)q^{n-3/2}-O_d(q^{n-2})$ for large $q$ . Theorem 6.1 in [3] implies that the Zariski closure of $X$ in $\mathbb {P}^n$ must have singular locus of dimension $n-3$ or $n-2$ .

We exhibit a forbidden interval for $|X({\mathbb {F}}_q)|$ that improves Theorem 4 in [7]. The statement below does not require $X$ to be geometrically irreducible.

Theorem 1.7 Let $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a hypersurface of degree $d$ . If

(1.6) $$ \begin{align} |X({\mathbb{F}}_q)|\leq \frac{3}{2}q^{n-1}-(d-1)(d-2)q^{n-3/2}- (d^2+d+1)q^{n-2}, \end{align} $$

then in fact

(1.7) $$ \begin{align} |X({\mathbb{F}}_q)|\leq q^{n-1}+(d-1)(d-2)q^{n-3/2}+12q^{n-2}. \end{align} $$

Remark 1.8 Let us write $g(d)+\cdots $ for an effectively computable $g(d)+g_1(d)$ , where $g_1(d)=o(g(d))$ for $d\to \infty $ . Theorem 1.7 has content when the right-hand side of (1.6) exceeds the right-hand side of (1.7), which takes place for $q>16d^4+\cdots $ . Thus, in the presence of Theorem 1.4, Theorem 1.7 addresses the range $16d^4+\cdots <q<15d^{13/3}$ . Notice that in the Lang–Weil bound (1.2), the approximation term $q^{n-1}$ dominates the error precisely when $q>d^4+\cdots $ . This is why it is reasonable to frame the entire discussion of the Lang–Weil bound in the range $q>d^4+\cdots $ . For example, any lower Lang–Weil bound is trivial for $q$ below this threshold.

1.2 Lower bounds

The proof of Theorem 4 in [7] actually gives a lower bound which is tighter for $q\gg 1$ than the one stated in [7].

Theorem 1.9 Let $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a geometrically irreducible hypersurface of degree $d$ . Then

(1.8) $$ \begin{align} |X({\mathbb{F}}_q)|\geq q^{n-1}-(d-1)(d-2)q^{n-3/2}-dq^{n-2}-O_d(q^{n-5/2}), \end{align} $$

where the implied constant depends only on $d$ and can be computed explicitly.

We give a version with an explicit lower bound as well.

Theorem 1.10 Let $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a geometrically irreducible hypersurface of degree $d$ . Suppose that $q>15d^{13/3}$ . Then

(1.9) $$ \begin{align} |X({\mathbb{F}}_q)|\geq q^{n-1}-(d-1)(d-2)q^{n-3/2}-(d+0.6)q^{n-2}. \end{align} $$

Example 1.11 As in Example 1.5, let $d\geq 3$ be such that is a prime power. The curve $\{y^{d-1}z+yz^{d-1}=x^d\}$ in $\mathbb {P}^2$ over ${\mathbb {F}}_{q_0}$ intersects the line $x=0$ at $d$ distinct points defined over an extension ${\mathbb {F}}_{q_1}$ of ${\mathbb {F}}_{q_0}$ . Let $q$ be an even power of $q_1$ . Then the affine curve in $\mathbb {A}^2_{{\mathbb {F}}_q}$ satisfies $\#C({\mathbb {F}}_q)= q-(d-1)(d-2)\sqrt {q}-d+1$ . Consequently, the number of ${\mathbb {F}}_q$ -points on the hypersurface $C\times \mathbb {A}^{n-2}$ in $\mathbb {A}^n$ is $q^{n-1}-(d-1)(d-2)q^{n-3/2}-(d-1)q^{n-2}$ . Therefore, the constant $d+0.6$ in (1.9) cannot possibly be improved by more than $1.6$ .

We can elaborate on (1.8) by pushing the implied constant further down.

Corollary 1.12 Let $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a geometrically irreducible hypersurface of degree $d$ . Then

(1.10) $$ \begin{align} |X({\mathbb{F}}_q)|&\geq q^{n-1}-(d-1)(d-2)q^{n-3/2}-dq^{n-2} -2(d-1)(d-2)q^{n-5/2}\nonumber\\ &-(2(d-1)^2(d-2)^2+d^2/2+d+2+\pi^2/6)q^{n-3}-O_d(q^{n-7/2}). \end{align} $$

A lower Lang–Weil bound can be useful in proving that a geometrically irreducible hypersurface $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ has an ${\mathbb {F}}_q$ -rational point. It is known (see Theorem 5.4 in [2] and its proof) that if $q>1.5d^4+\cdots $ , then $X({\mathbb {F}}_q)\neq \emptyset $ . Notice that the approximation term $q^{n-1}$ in (1.10) dominates the remaining explicit terms already for $q>d^4+\cdots $ . Based on this heuristic, we state the following conjecture.

Conjecture 1.13 There exists an effectively computable function $g_1(d)=O(d^{7/2})$ as $d\to \infty $ with the following property. Let $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a geometrically irreducible hypersurface of degree $d$ . Then $X({\mathbb {F}}_q)\neq \emptyset $ as long as $q>d^4+g_1(d)$ .

Remark 1.14 In contrast to the upper bounds, all lower bounds in the affine cases above (including (1.3) and Example 1.11) contain a $d$ in the coefficient of $q^{n-2}$ . This is an artifact of affine space; the discrepancy disappears in projective space (Theorem 1.1).

1.3 Outline

This paper builds upon the author’s earlier work [7] and is inspired by Tao’s discussion [10] of the Lang–Weil bound through random sampling and the idea of Cafure–Matera [2] to slice $X$ with planes (a plane is a two-dimensional affine linear subvariety of $\mathbb {A}^n_{{\mathbb {F}}_q}$ ). If $H\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ is any plane, then $\#(X\cap H)({\mathbb {F}}_q)$ is either $q^2$ , $0$ , or $\approx kq$ , where $k$ is the number of geometrically irreducible ${\mathbb {F}}_q$ -irreducible components of $X\cap H$ . For $0\leq k\leq d$ , we exhibit a small interval $I_k=[a_k,b_k]$ containing $kq$ so that if we also define $I_\infty =\{q^2\}$ , then each $\#(X\cap H)({\mathbb {F}}_q)$ belongs to $\bigcup I_k$ .

The problem when it comes to the upper bound is that when $k$ is large, planes $H$ with $\#(X\cap H)({\mathbb {F}}_q)\in I_k$ contribute significantly toward the count $\#X({\mathbb {F}}_q)$ . However, it turns out that the number of such $H$ ’s decreases quickly as $k$ grows.

2 A collection of small intervals

Lemma 2.1 [5, Lemma 5]

Let $C\subset \mathbb {A}^2_{{\mathbb {F}}_q}$ be a curve of degree $d$ . Let $k$ be the number of geometrically irreducible ${\mathbb {F}}_q$ -irreducible components of $C$ . Then

$$\begin{align*}|\#C({\mathbb{F}}_q)-kq|\leq (d-1)(d-2)\sqrt{q}+d^2+d+1.\end{align*}$$

It will be crucial to give a refined upper bound when $k=1$ .

Lemma 2.2 Let $C\subset \mathbb {A}^2_{{\mathbb {F}}_q}$ be a curve of degree $d$ . Suppose that $C$ has exactly one geometrically irreducible ${\mathbb {F}}_q$ -irreducible component. Then

$$\begin{align*}|C({\mathbb{F}}_q)|\leq q+(d-1)(d-2)\sqrt{q}+1.\end{align*}$$

Proof Let $C_1,\dots ,C_s$ be the ${\mathbb {F}}_q$ -irreducible components of $C$ . Suppose that $C_1$ is geometrically irreducible, but $C_i$ is not for $i\geq 2$ . Let $e=\deg (C_1)$ . Note that $(d,e)\neq (2,1)$ .

Using the Aubry–Perret bound (1.3) for $C_1$ and Lemma 2.3 in [2] for each $C_i$ with $i\geq 2$ , we estimate

$$ \begin{align*} |C({\mathbb{F}}_q)| &\leq |C_1({\mathbb{F}}_q)|+\sum_{i=2}^s |C_i({\mathbb{F}}_q)|\\&\leq q+(e-1)(e-2)\sqrt{q}+1+\sum_{i=2}^s (\deg C_i)^2/4\\&\leq q+(e-1)(e-2)\sqrt{q}+1+(d-e)^2/4\\[3pt]&\leq q+(d-1)(d-2)\sqrt{q}+1; \end{align*} $$

to justify the last inequality in the chain, note that it is equivalent to

$$\begin{align*}(d-e)\left((d+e-3)\sqrt{q}-\frac{d-e}{4}\right)\geq 0\end{align*}$$

and holds true because either $e=d$ , or else $d-e>0$ , and we can write

$$\begin{align*}(d+e-3)\sqrt{q}-\frac{d-e}{4} \geq (d+e-3)\sqrt{2}-\frac{d-e}{4} \geq \frac{(4\sqrt{2}-1)d+(4\sqrt{2}+1)e-12\sqrt{2}}{4}>0 \end{align*}$$

(using that $e\geq 1$ and $d\geq 3$ on the last step).

Let $a_0=0$ , $b_0=d^2/4$ , $a_1=q-(d-1)(d-2)\sqrt {q}-d+1$ , and $b_1=q+(d-1) (d-2)\sqrt {q}+1$ . For $2\leq k\leq d$ , set $a_k=kq-(d-1)(d-2)\sqrt {q}-d^2-d-1$ and $b_k=kq+(d-1)(d-2)\sqrt {q}+d^2+d+1$ . Finally, set $a_\infty =b_\infty =q^2$ . Define for $k\in \{0,\dots ,d\}\cup \{\infty \}$ .

Lemma 2.3 Let $X\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a hypersurface of degree $d$ . Let $H\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ be a plane. Then $\#(X\cap H)({\mathbb {F}}_q)\in I_k$ for some $k\in \{0,\dots ,d\}\cup \{\infty \}$ .

Proof If $X\cap H=\emptyset $ , then $\#(X\cap H)({\mathbb {F}}_q)=0\in I_0$ . If $H\subset X$ , then $X\cap H=H$ and $\#(X\cap H)({\mathbb {F}}_q)=q^2\in I_\infty $ . Suppose that $X\cap H\neq \emptyset $ and $H\not \subset X$ . Let $k$ be the number of geometrically irreducible ${\mathbb {F}}_q$ -irreducible components of the degree $d$ plane curve $X\cap H\subset H\simeq \mathbb {A}^2_{{\mathbb {F}}_q}$ . Then $0\leq k\leq d$ . If $k=0$ , the proof of Lemma 11 in [7] gives $\#(X\cap H)({\mathbb {F}}_q)\leq d^2/4$ . If $k=1$ , we use Lemma 2.2 and the lower bound from (1.3) applied to a geometrically irreducible ${\mathbb {F}}_q$ -irreducible component (necessarily of degree $\leq d$ ) of $X$ . For $2\leq k\leq d$ , use Lemma 2.1.

Alternatively, one could take $b_d=dq$ by the Schwartz–Zippel lemma.

When it comes to giving an upper bound for $|X({\mathbb {F}}_q)|$ , it will be more convenient to work with and for $i\in \{2,\dots ,d\}\cup \{\infty \}$ .

3 Probability estimates

We spell out in detail the proof of Theorem 1.3; the proofs of the remaining results will then require only slight modifications. The implied constant in each $O$ -notation is allowed to depend only on $d$ (a priori, possibly also on $n$ ), but not on $q$ or $X$ .

Proof of Theorem 1.3

Set . For a plane $H\subset \mathbb {A}^n_{{\mathbb {F}}_q}$ chosen uniformly at random, consider $\#(X\cap H)({\mathbb {F}}_q)$ as a random variable. Let $\mu $ and $\sigma ^2$ denote its mean and variance. Lemma 10 in [7] and (1.2) imply

(3.1) $$ \begin{align} \mu=\frac{N}{q^{n-2}}\quad\text{and}\quad\sigma^2\leq \frac{N}{q^{n-2}}\leq q+O(\sqrt{q}). \end{align} $$

Write

(3.2) $$ \begin{align} \frac{N}{q^{n-2}}=\mu\leq \sum_{k\in\{1,\dots,d\}\cup\{\infty\}}\operatorname{\mathrm{Prob}}\Big(\#(X\cap H)({\mathbb{F}}_q)\in J_k\Big)b_k. \end{align} $$

For $k\in \{1,\dots ,d\}\cup \{\infty \}$ , denote

We can assume that $q$ is large enough so that the intervals $J_1,\dots ,J_d$ are pairwise disjoint.

Let $k\in \{2,\dots ,d\}$ . If $H$ is a plane such that $\#(X\cap H)({\mathbb {F}}_q)\in J_k\cup \dots \cup J_d$ , then

(3.3) $$ \begin{align} |\#(X\cap H)({\mathbb{F}}_q)-\mu|\geq a_k-\frac{N}{q^{n-2}}\geq (k-1)q-O(\sqrt{q}). \end{align} $$

Define $t$ via $(k-1)q-O(\sqrt {q})=t\sigma $ ; then Chebyshev’s inequality and the variance bound (3.1) imply

(3.4) $$ \begin{align} p_k+\cdots+p_d =\operatorname{\mathrm{Prob}}\Big(\#(X\cap H)({\mathbb{F}}_q)\in J_k\cup\dots\cup J_d\Big) &\leq\frac{1}{t^2}\notag \\ &=\frac{\sigma^2}{((k-1)q-O(\sqrt{q}))^2}\notag \\ &\leq\frac{q+O(\sqrt{q})}{((k-1)q-O(\sqrt{q}))^2}\notag \\ &=\frac{1}{(k-1)^2 q}+O(q^{-3/2}). \end{align} $$

If $H$ is a plane such that $\#(X\cap H)({\mathbb {F}}_q)=q^2$ , then

$$\begin{align*}|\#(X\cap H)({\mathbb{F}}_q)-\mu|= q^2-\frac{N}{q^{n-2}}\geq q^2-O(q).\end{align*}$$

Define $t$ via $q^2-O(q)=t\sigma $ ; then

$$\begin{align*}p_\infty\leq \frac{1}{t^2}=\frac{\sigma^2}{(q^2-O(q))^2}\leq \frac{q+O(\sqrt{q})}{(q^2-O(q))^2}= q^{-3}+O(q^{-7/2}),\!\quad\text{and hence}\quad\! p_\infty b_\infty=O(q^{-1}). \end{align*}$$

Note that $b_k-b_{k-1}=q+O(1)$ for $2\leq k\leq d$ . We now go back to (3.2) and apply the Abel summation formula:

$$ \begin{align*} \frac{N}{q^{n-2}}=\mu&\leq (p_1+\cdots+p_d)b_1+(p_2+\cdots+p_d)(b_2-b_1)+\cdots+p_d(b_d-b_{d-1})+p_\infty b_\infty\\ &\leq b_1+\frac{1}{1^2}+\cdots+\frac{1}{(d-1)^2}+O(q^{-1/2})\\ &\leq q+(d-1)(d-2)\sqrt{q}+1+\pi^2/6+O(q^{-1/2}). \end{align*} $$

Multiply both sides by $q^{n-2}$ to arrive at (1.4).

Going through all the explicit inequalities with an $O$ -term, one can compute explicitly a possible value of the constant implicit in (1.4). In fact, since the Cafure–Matera bound gives a choice of $C_d$ in the Lang–Weil bound that depends only on $d$ and not on $n$ , a second look at all the inequalities written down in the proof above reveals that the implied constant in (1.4) can likewise be chosen not to depend on $n$ .

For the rest of the paper, we follow the notation and proof of Theorem 1.3.

Proof of Theorem 1.9

Say that a plane $H$ is “bad” if $\#(X\cap H)({\mathbb {F}}_q)\in I_0$ and “good” otherwise. If $H\subset \mathbb {A}^2_{{\mathbb {F}}_q}$ is a bad plane, then

$$\begin{align*}|\#(X\cap H)({\mathbb{F}}_q)-\mu|\geq \frac{N}{q^{n-2}}-\frac{d^2}{4}\geq q-O(\sqrt{q}).\end{align*}$$

By computations similar to the ones in the proof of Theorem 1.3, the probability that a plane is bad is at most $q^{-1}+O(q^{-3/2})$ . Every good plane contributes at least $a_1$ to the mean. Therefore,

$$\begin{align*}\frac{N}{q^{n-2}}=\mu \geq (1-q^{-1}-O(q^{-3/2}))(q-(d-1)(d-2)\sqrt{q}-d+1), \end{align*}$$

giving (1.8).

Proof of Corollary 1.12

In fact, the proofs of Theorems 1.3 and 1.9 give an algorithm that takes as input a half-integer $r\geq 0$ and constants ¹ $C_d^{(j)}$ and $D_d^{(j)}$ for each half-integer $1/2\leq j\leq r$ such that

$$\begin{align*}|X({\mathbb{F}}_q)|\leq q^{n-1}+\sum_{j=1/2}^r C_d^{(j)} q^{n-1-j}+O_d(q^{n-r-3/2}) \!\!\qquad\text{(summation over half-integers)}\end{align*}$$

and

$$\begin{align*}|X({\mathbb{F}}_q)|\geq q^{n-1}-\sum_{j=1/2}^r D_d^{(j)} q^{n-1-j}-O_d(q^{n-r-3/2}) \!\!\!\qquad\text{(summation over half-integers)},\end{align*}$$

and returns as output four additional $C_d^{(r+1/2)}$ , $C_d^{(r+1)}$ , $D_d^{(r+1/2)}$ , and $D_d^{(r+1)}$ such that

$$\begin{align*}|X({\mathbb{F}}_q)|\leq q^{n-1}+\sum_{j=1/2}^{r+1} C_d^{(j)} q^{n-1-j}+O_d(q^{n-r-5/2}) \!\!\qquad\text{(summation over half-integers)}\end{align*}$$

and

$$\begin{align*}|X({\mathbb{F}}_q)|\geq q^{n-1}-\sum_{j=1/2}^{r+1} D_d^{(j)} q^{n-1-j}-O_d(q^{n-r-5/2}) \!\!\!\qquad\text{(summation over half-integers)}.\end{align*}$$

Initiating the algorithm with $r=0$ and the rather weak version

$$\begin{align*}q^{n-1}-O_d(q^{n-3/2})\leq |X({\mathbb{F}}_q)|\leq q^{n-1}+O_d(q^{n-3/2})\end{align*}$$

of (1.2), we obtained (1.4) and (1.8). In turn, taking the upper bound for $N$ from (1.4) and the lower bound for $N$ from (1.8) as input, we obtain (1.10).

Proof of Theorem 1.1

We now slice with a random plane $H\subset \mathbb {P}^n_{{\mathbb {F}}_q}$ . The mean $\mu $ of $\#(X\cap H)({\mathbb {F}}_q)$ is $N\rho _1$ , where $N=|X({\mathbb {F}}_q)|$ and $\rho _1=(q^3-1)/(q^{n+1}-1)$ is the probability that a plane passes through a given point. Let $\rho _2$ be the probability that a plane passes through two distinct given points. Explicitly (in terms of $q$ -binomial coefficients), $\rho _2=\binom {n-1}{1}_q/\binom {n+1}{3}_q$ . One verifies directly that $\rho _2\leq \rho _1^2$ and expresses $\sigma ^2$ as in [10]:

$$\begin{align*}N^2\rho_1^2+\sigma^2=\mu^2+\sigma^2=\mu+N(N-1)\rho_2\leq\mu+N^2\rho_2\end{align*}$$

to deduce $\sigma ^2\leq \mu $ .

We can still take $I_0=[0,d^2/4]$ . Use the projective version of (1.3) (Corollary 2.5 in [1]). Adapt $I_1$ with $a_1=q-(d-1)(d-2)\sqrt {q}+1$ . Use $I_\infty =\{q^2+q+1\}$ . Up to a summand $d$ to account for points at infinity, the remaining $a_k$ and $b_k$ are unchanged.

Proceed as in the proof of Theorems 1.3 and 1.9. On the very last step in proving either bound, multiply by $1/\rho _1$ rather than by $q^{n-2}$ and use that $1/\rho _1=q^{n-2}+ O(q^{n-5})$ .

4 Explicit versions

Proof of Theorem 1.4

The statement clearly holds for $d=1$ , so assume that $d\geq 2$ . We will use the explicit Cafure–Matera bound for $N$ . Replace the variance bound (3.1) by

$$\begin{align*}\sigma^2\leq\frac{N}{q^{n-2}}\leq q+(d-1)(d-2)\sqrt{q}+5d^2+d+1\leq (8.44/7.44)q;\end{align*}$$

to verify the last inequality above, we argue as follows. For any $c_1>0$ and $c_2>0$ , the function $q\mapsto q/(c_1\sqrt {q}+c_2)$ is increasing. Therefore,

$$\begin{align*}\frac{q}{(d-1)(d-2)\sqrt{q}+5d^2+d+1}>\frac{15d^{13/3}}{(d-1)(d-2)\sqrt{15}d^{13/6}+5d^2+d+1}.\end{align*}$$

It remains to check that the function $g(d)$ on the right-hand side above satisfies $g(d)>7.44$ for any integer $d\geq 2$ . On the one hand, $g$ grows like $d^{1/6}$ , so one easily exhibits a $d_0$ such that $g(d)>7.44$ for $d>d_0$ . Then a simple computer calculation checks that $g(d)>7.44$ for integers $d\in \{2,\dots ,d_0\}$ as well.

In the same way, one readily checks that the intervals $J_1,\dots ,J_d$ are pairwise disjoint.

For $k\in \{2,\dots ,d\}$ , replace (3.3) by

$$\begin{align*}a_k-\frac{N}{q^{n-2}}\geq (k-1)q-2(d-1)(d-2)\sqrt{q}-2(3d^2+d+1)\geq (5.45/7.45)(k-1)q;\end{align*}$$

to check the last inequality, one has to consider only $k=2$ and to argue as above.

For $k\in \{2,\dots ,d\}$ , (3.4) is now replaced by

$$\begin{align*}p_k+\cdots+p_d\leq \frac{(8.44/7.44)q}{\left((5.45/7.45)(k-1)q\right)^2}<\frac{2.12}{(k-1)^2q}.\end{align*}$$

To bound $p_\infty b_\infty $ , note that $q>15d^{13/3}>15\times 2^{13/3}>302$ , so

$$\begin{align*}p_\infty b_\infty\leq\frac{(8.44/7.44)q}{(q^2-(8.44/7.44)q)^2}q^2 =\frac{8.44\times 7.44q}{(7.44q-8.44)^2}<0.01.\end{align*}$$

Since $b_k-b_{k-1}=q$ for $3\leq k\leq d$ , but $b_2-b_1=q+d^2+d$ , we have to estimate $(d^2+d)/q<(d^2+d)/15d^{13/3}<0.02$ . The Abel summation argument now gives

$$\begin{align*}\frac{N}{q^{n-2}}\leq q+(d-1)(d-2)\sqrt{q}+1+2.12(\pi^2/6+0.02)+0.01<q+(d-1)(d-2)\sqrt{q}+5.\end{align*}$$

Proof of Theorem 1.7

Again, assume $d\geq 2$ . We can assume that the right-hand side of (1.7) is less than the right-hand side of (1.6); i.e.,

$$\begin{align*}4(d-1)(d-2)\sqrt{q}+2(d^2+d+13)<q. \end{align*}$$

This inequality implies in particular that the intervals $J_1,\dots ,J_d$ are pairwise disjoint. Note that it is equivalent to $q>r(d)^2$ , where $r(d)$ is the positive root of the quadratic equation $x^2-4(d-1)(d-2)x-2(d^2+d+13)=0$ .

Due to (1.6), now we can use the variance bound $\sigma ^2\leq N/q^{n-2}\leq (3/2)q$ . Furthermore, (1.6) gives

$$\begin{align*}a_k-\frac{N}{q^{n-2}}= kq-(d-1)(d-2)\sqrt{q}-(d^2+d+1)-\frac{N}{q^{n-2}}\geq \frac{k-1}{2}q\end{align*}$$

for $2\leq k\leq d$ . Therefore, $p_k+\cdots +p_d$ is now bounded by $6/((k-1)^2q)$ .

We bound $(d^2+d)/q$ by $(d^2+d)/(r(d))^2<0.16$ for $d\geq 2$ . Finally, note that $q>r(2)^2=38$ , so $q\geq 41$ , and we can bound $p_\infty b_\infty $ by $6q/(2q-3)^2<0.04$ . Therefore,

$$\begin{align*}\frac{N}{q^{n-2}}\leq q+(d-1)(d-2)\sqrt{q}+1+6(\pi^2/6+0.16)+0.04<q+(d-1)(d-2)\sqrt{q}+12.\end{align*}$$

Proof of Theorem 1.10

As above, assume that $d\geq 2$ . We bound the variance as

$$\begin{align*}\sigma^2\leq\frac{N}{q^{n-2}}\leq q+(d-1)(d-2)\sqrt{q}+5d^2+d+1\leq (8.44/7.44)q.\end{align*}$$

Moreover,

$$\begin{align*}\frac{N}{q^{n-2}}-\frac{d^2}{4}\geq q-(d-1)(d-2)\sqrt{q}-21d^2/4-d-1\geq (6.44/7.44)q.\end{align*}$$

From here, we bound the probability that a plane is bad by $1.6/q$ . Thus,

$$\begin{align*}\frac{N}{q^{n-2}}\geq \left(1-\frac{1.6}{q}\right)(q-(d-1)(d-2)\sqrt{q}-d+1)\geq q-(d-1)(d-2)\sqrt{q}-(d+0.6).\end{align*}$$