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Let ${{\bf F}_q}$ be a finite field of order ${q = p^n}$, and let ${C}$ be an absolutely irreducible smooth projective curve defined over ${{\bf F}_q}$ (and hence over the algebraic closure ${k := \overline{{\bf F}_q}}$ of that field). For instance, ${C}$ could be the projective elliptic curve

$\displaystyle C = \{ [x,y,z]: y^2 z = x^3 + ax z^2 + b z^3 \}$

in the projective plane ${{\bf P}^2 = \{ [x,y,z]: (x,y,z) \neq (0,0,0) \}}$, where ${a,b \in {\bf F}_q}$ are coefficients whose discriminant ${-16(4a^3+27b^2)}$ is non-vanishing, which is the projective version of the affine elliptic curve

$\displaystyle \{ (x,y): y^2 = x^3 + ax + b \}.$

To each such curve ${C}$ one can associate a genus ${g}$, which we will define later; for instance, elliptic curves have genus ${1}$. We can also count the cardinality ${|C({\bf F}_q)|}$ of the set ${C({\bf F}_q)}$ of ${{\bf F}_q}$-points of ${C}$. The Hasse-Weil bound relates the two:

Theorem 1 (Hasse-Weil bound) ${||C({\bf F}_q)| - q - 1| \leq 2g\sqrt{q}}$.

The usual proofs of this bound proceed by first establishing a trace formula of the form

$\displaystyle |C({\bf F}_{p^n})| = p^n - \sum_{i=1}^{2g} \alpha_i^n + 1 \ \ \ \ \ (1)$

for some complex numbers ${\alpha_1,\dots,\alpha_{2g}}$ independent of ${n}$; this is in fact a special case of the Lefschetz-Grothendieck trace formula, and can be interpreted as an assertion that the zeta function associated to the curve ${C}$ is rational. The task is then to establish a bound ${|\alpha_i| \leq \sqrt{p}}$ for all ${i=1,\dots,2g}$; this (or more precisely, the slightly stronger assertion ${|\alpha_i| = \sqrt{p}}$) is the Riemann hypothesis for such curves. This can be done either by passing to the Jacobian variety of ${C}$ and using a certain duality available on the cohomology of such varieties, known as Rosati involution; alternatively, one can pass to the product surface ${C \times C}$ and apply the Riemann-Roch theorem for that surface.

In 1969, Stepanov introduced an elementary method (a version of what is now known as the polynomial method) to count (or at least to upper bound) the quantity ${|C({\bf F}_q)|}$. The method was initially restricted to hyperelliptic curves, but was soon extended to general curves. In particular, Bombieri used this method to give a short proof of the following weaker version of the Hasse-Weil bound:

Theorem 2 (Weak Hasse-Weil bound) If ${q}$ is a perfect square, and ${q \geq (g+1)^4}$, then ${|C({\bf F}_q)| \leq q + (2g+1) \sqrt{q} + 1}$.

In fact, the bound on ${|C({\bf F}_q)|}$ can be sharpened a little bit further, as we will soon see.

Theorem 2 is only an upper bound on ${|C({\bf F}_q)|}$, but there is a Galois-theoretic trick to convert (a slight generalisation of) this upper bound to a matching lower bound, and if one then uses the trace formula (1) (and the “tensor power trick” of sending ${n}$ to infinity to control the weights ${\alpha_i}$) one can then recover the full Hasse-Weil bound. We discuss these steps below the fold.

I’ve discussed Bombieri’s proof of Theorem 2 in this previous post (in the special case of hyperelliptic curves), but now wish to present the full proof, with some minor simplifications from Bombieri’s original presentation; it is mostly elementary, with the deepest fact from algebraic geometry needed being Riemann’s inequality (a weak form of the Riemann-Roch theorem).

The first step is to reinterpret ${|C({\bf F}_q)|}$ as the number of points of intersection between two curves ${C_1,C_2}$ in the surface ${C \times C}$. Indeed, if we define the Frobenius endomorphism ${\hbox{Frob}_q}$ on any projective space by

$\displaystyle \hbox{Frob}_q( [x_0,\dots,x_n] ) := [x_0^q, \dots, x_n^q]$

then this map preserves the curve ${C}$, and the fixed points of this map are precisely the ${{\bf F}_q}$ points of ${C}$:

$\displaystyle C({\bf F}_q) = \{ z \in C: \hbox{Frob}_q(z) = z \}.$

Thus one can interpret ${|C({\bf F}_q)|}$ as the number of points of intersection between the diagonal curve

$\displaystyle \{ (z,z): z \in C \}$

and the Frobenius graph

$\displaystyle \{ (z, \hbox{Frob}_q(z)): z \in C \}$

which are copies of ${C}$ inside ${C \times C}$. But we can use the additional hypothesis that ${q}$ is a perfect square to write this more symmetrically, by taking advantage of the fact that the Frobenius map has a square root

$\displaystyle \hbox{Frob}_q = \hbox{Frob}_{\sqrt{q}}^2$

with ${\hbox{Frob}_{\sqrt{q}}}$ also preserving ${C}$. One can then also interpret ${|C({\bf F}_q)|}$ as the number of points of intersection between the curve

$\displaystyle C_1 := \{ (z, \hbox{Frob}_{\sqrt{q}}(z)): z \in C \} \ \ \ \ \ (2)$

and its transpose

$\displaystyle C_2 := \{ (\hbox{Frob}_{\sqrt{q}}(w), w): w \in C \}.$

Let ${k(C \times C)}$ be the field of rational functions on ${C \times C}$ (with coefficients in ${k}$), and define ${k(C_1)}$, ${k(C_2)}$, and ${k(C_1 \cap C_2)}$ analogously )(although ${C_1 \cap C_2}$ is likely to be disconnected, so ${k(C_1 \cap C_2)}$ will just be a ring rather than a field. We then (morally) have the commuting square

$\displaystyle \begin{array}{ccccc} && k(C \times C) && \\ & \swarrow & & \searrow & \\ k(C_1) & & & & k(C_2) \\ & \searrow & & \swarrow & \\ && k(C_1 \cap C_2) && \end{array},$

if we ignore the issue that a rational function on, say, ${C \times C}$, might blow up on all of ${C_1}$ and thus not have a well-defined restriction to ${C_1}$. We use ${\pi_1: k(C \times C) \rightarrow k(C_1)}$ and ${\pi_2: k(C \times C) \rightarrow k(C_2)}$ to denote the restriction maps. Furthermore, we have obvious isomorphisms ${\iota_1: k(C_1) \rightarrow k(C)}$, ${\iota_2: k(C_2) \rightarrow k(C)}$ coming from composing with the graphing maps ${z \mapsto (z, \hbox{Frob}_{\sqrt{q}}(z))}$ and ${w \mapsto (\hbox{Frob}_{\sqrt{q}}(w), w)}$.

The idea now is to find a rational function ${f \in k(C \times C)}$ on the surface ${C \times C}$ of controlled degree which vanishes when restricted to ${C_1}$, but is non-vanishing (and not blowing up) when restricted to ${C_2}$. On ${C_2}$, we thus get a non-zero rational function ${f \downharpoonright_{C_2}}$ of controlled degree which vanishes on ${C_1 \cap C_2}$ – which then lets us bound the cardinality of ${C_1 \cap C_2}$ in terms of the degree of ${f \downharpoonright_{C_2}}$. (In Bombieri’s original argument, one required vanishing to high order on the ${C_1}$ side, but in our presentation, we have factored out a ${\hbox{Frob}_{\sqrt{q}}}$ term which removes this high order vanishing condition.)

To find this ${f}$, we will use linear algebra. Namely, we will locate a finite-dimensional subspace ${V}$ of ${k(C \times C)}$ (consisting of certain “controlled degree” rational functions) which projects injectively to ${k(C_2)}$, but whose projection to ${k(C_1)}$ has strictly smaller dimension than ${V}$ itself. The rank-nullity theorem then forces the existence of a non-zero element ${P}$ of ${V}$ whose projection to ${k(C_1)}$ vanishes, but whose projection to ${k(C_2)}$ is non-zero.

Now we build ${V}$. Pick a ${{\bf F}_q}$ point ${P_\infty}$ of ${C}$, which we will think of as being a point at infinity. (For the purposes of proving Theorem 2, we may clearly assume that ${C({\bf F}_q)}$ is non-empty.) Thus ${P_\infty}$ is fixed by ${\hbox{Frob}_q}$. To simplify the exposition, we will also assume that ${P_\infty}$ is fixed by the square root ${\hbox{Frob}_{\sqrt{q}}}$ of ${\hbox{Frob}_q}$; in the opposite case when ${\hbox{Frob}_{\sqrt{q}}}$ has order two when acting on ${P_\infty}$, the argument is essentially the same, but all references to ${P_\infty}$ in the second factor of ${C \times C}$ need to be replaced by ${\hbox{Frob}_{\sqrt{q}} P_\infty}$ (we leave the details to the interested reader).

For any natural number ${n}$, define ${R_n}$ to be the set of rational functions ${f \in k(C)}$ which are allowed to have a pole of order up to ${n}$ at ${P_\infty}$, but have no other poles on ${C}$; note that as we are assuming ${C}$ to be smooth, it is unambiguous what a pole is (and what order it will have). (In the fancier language of divisors and Cech cohomology, we have ${R_n = H^0( C, {\mathcal O}_C(-n P_\infty) )}$.) The space ${R_n}$ is clearly a vector space over ${k}$; one can view intuitively as the space of “polynomials” on ${C}$ of “degree” at most ${n}$. When ${n=0}$, ${R_0}$ consists just of the constant functions. Indeed, if ${f \in R_0}$, then the image ${f(C)}$ of ${f}$ avoids ${\infty}$ and so lies in the affine line ${k = {\mathbf P}^1 \backslash \{\infty\}}$; but as ${C}$ is projective, the image ${f(C)}$ needs to be compact (hence closed) in ${{\mathbf P}^1}$, and must therefore be a point, giving the claim.

For higher ${n \geq 1}$, we have the easy relations

$\displaystyle \hbox{dim}(R_{n-1}) \leq \hbox{dim}(R_n) \leq \hbox{dim}(R_{n-1})+1. \ \ \ \ \ (3)$

The former inequality just comes from the trivial inclusion ${R_{n-1} \subset R_n}$. For the latter, observe that if two functions ${f, g}$ lie in ${R_n}$, so that they each have a pole of order at most ${n}$ at ${P_\infty}$, then some linear combination of these functions must have a pole of order at most ${n-1}$ at ${P_\infty}$; thus ${R_{n-1}}$ has codimension at most one in ${R_n}$, giving the claim.

From (3) and induction we see that each of the ${R_n}$ are finite dimensional, with the trivial upper bound

$\displaystyle \hbox{dim}(R_n) \leq n+1. \ \ \ \ \ (4)$

Riemann’s inequality complements this with the lower bound

$\displaystyle \hbox{dim}(R_n) \geq n+1-g, \ \ \ \ \ (5)$

thus one has ${\hbox{dim}(R_n) = \hbox{dim}(R_{n-1})+1}$ for all but at most ${g}$ exceptions (in fact, exactly ${g}$ exceptions as it turns out). This is a consequence of the Riemann-Roch theorem; it can be proven from abstract nonsense (the snake lemma) if one defines the genus ${g}$ in a non-standard fashion (as the dimension of the first Cech cohomology ${H^1(C)}$ of the structure sheaf ${{\mathcal O}_C}$ of ${C}$), but to obtain this inequality with a standard definition of ${g}$ (e.g. as the dimension of the zeroth Cech cohomolgy ${H^0(C, \Omega_C^1)}$ of the line bundle of differentials) requires the more non-trivial tool of Serre duality.

At any rate, now that we have these vector spaces ${R_n}$, we will define ${V \subset k(C \times C)}$ to be a tensor product space

$\displaystyle V = R_\ell \otimes R_m$

for some natural numbers ${\ell, m \geq 0}$ which we will optimise in later. That is to say, ${V}$ is spanned by functions of the form ${(z,w) \mapsto f(z) g(w)}$ with ${f \in R_\ell}$ and ${g \in R_m}$. This is clearly a linear subspace of ${k(C \times C)}$ of dimension ${\hbox{dim}(R_\ell) \hbox{dim}(R_m)}$, and hence by Rieman’s inequality we have

$\displaystyle \hbox{dim}(V) \geq (\ell+1-g) (m+1-g) \ \ \ \ \ (6)$

if

$\displaystyle \ell,m \geq g-1. \ \ \ \ \ (7)$

Observe that ${\iota_1 \circ \pi_1}$ maps a tensor product ${(z,w) \mapsto f(z) g(w)}$ to a function ${z \mapsto f(z) g(\hbox{Frob}_{\sqrt{q}} z)}$. If ${f \in R_\ell}$ and ${g \in R_m}$, then we see that the function ${z \mapsto f(z) g(\hbox{Frob}_{\sqrt{q}} z)}$ has a pole of order at most ${\ell+m\sqrt{q}}$ at ${P_\infty}$. We conclude that

$\displaystyle \iota_1 \circ \pi_1( V ) \subset R_{\ell + m\sqrt{q}} \ \ \ \ \ (8)$

and in particular by (4)

$\displaystyle \hbox{dim}(\pi_1(V)) \leq \ell + m \sqrt{q} + 1 \ \ \ \ \ (9)$

and similarly

$\displaystyle \hbox{dim}(\pi_2(V)) \leq \ell \sqrt{q} + m + 1. \ \ \ \ \ (10)$

We will choose ${m}$ to be a bit bigger than ${\ell}$, to make the ${\pi_2}$ image of ${V}$ smaller than that of ${\pi_1}$. From (6), (10) we see that if we have the inequality

$\displaystyle (\ell+1-g) (m+1-g) > \ell \sqrt{q}+m + 1 \ \ \ \ \ (11)$

(together with (7)) then ${\pi_2}$ cannot be injective.

On the other hand, we have the following basic fact:

Lemma 3 (Injectivity) If

$\displaystyle \ell < \sqrt{q}, \ \ \ \ \ (12)$

then ${\pi_1: V \rightarrow \pi_1(V)}$ is injective.

Proof: From (3), we can find a linear basis ${f_1,\dots,f_a}$ of ${R_\ell}$ such that each of the ${f_i}$ has a distinct order ${d_i}$ of pole at ${P_\infty}$ (somewhere between ${0}$ and ${\ell}$ inclusive). Similarly, we may find a linear basis ${g_1,\dots,g_b}$ of ${R_m}$ such that each of the ${g_j}$ has a distinct order ${e_j}$ of pole at ${P_\infty}$ (somewhere between ${0}$ and ${m}$ inclusive). The functions ${z \mapsto f_i(z) g_j(\hbox{Frob}_{\sqrt{q}} z)}$ then span ${\iota_1(\pi_1(V))}$, and the order of pole at ${P_\infty}$ is ${d_i + \sqrt{q} e_j}$. But since ${\ell < \sqrt{q}}$, these orders are all distinct, and so these functions must be linearly independent. The claim follows. $\Box$

This gives us the following bound:

Proposition 4 Let ${\ell,m}$ be natural numbers such that (7), (11), (12) hold. Then ${|C({\bf F}_q)| \leq \ell + m \sqrt{q}}$.

Proof: As ${\pi_2}$ is not injective, we can find ${f \in V}$ with ${\pi_2(f)}$ vanishing. By the above lemma, the function ${\iota_1(\pi_1(f))}$ is then non-zero, but it must also vanish on ${\iota_1(C_1 \cap C_2)}$, which has cardinality ${|C({\bf F}_q)|}$. On the other hand, by (8), ${\iota_1(\pi_1(f))}$ has a pole of order at most ${\ell+m\sqrt{q}}$ at ${P_\infty}$ and no other poles. Since the number of poles and zeroes of a rational function on a projective curve must add up to zero, the claim follows. $\Box$

If ${q \geq (g+1)^4}$, we may make the explicit choice

$\displaystyle m := \sqrt{q}+2g; \quad \ell := \lfloor \frac{g}{g+1} \sqrt{q} \rfloor + g + 1$

and a brief calculation then gives Theorem 2. In some cases one can optimise things a bit further. For instance, in the genus zero case ${g=0}$ (e.g. if ${C}$ is just the projective line ${{\mathbf P}^1}$) one may take ${\ell=1, m = \sqrt{q}}$ and conclude the absolutely sharp bound ${|C({\bf F}_q)| \leq q+1}$ in this case; in the case of the projective line ${{\mathbf P}^1}$, the function ${f}$ is in fact the very concrete function ${f(z,w) := z - w^{\sqrt{q}}}$.

Remark 1 When ${q = p^{2n+1}}$ is not a perfect square, one can try to run the above argument using the factorisation ${\hbox{Frob}_q = \hbox{Frob}_{p^n} \hbox{Frob}_{p^{n+1}}}$ instead of ${\hbox{Frob}_q = \hbox{Frob}_{\sqrt{q}} \hbox{Frob}_{\sqrt{q}}}$. This gives a weaker version of the above bound, of the shape ${|C({\bf F}_q)| \leq q + O( \sqrt{p} \sqrt{q} )}$. In the hyperelliptic case at least, one can erase this loss by working with a variant of the argument in which one requires ${f}$ to vanish to high order at ${C_1}$, rather than just to first order; see this survey article of mine for details.

One of the first non-trivial theorems one encounters in classical algebraic geometry is Bézout’s theorem, which we will phrase as follows:

Theorem 1 (Bézout’s theorem) Let ${k}$ be a field, and let ${P, Q \in k[x,y]}$ be non-zero polynomials in two variables ${x,y}$ with no common factor. Then the two curves ${\{ (x,y) \in k^2: P(x,y) = 0\}}$ and ${\{ (x,y) \in k^2: Q(x,y) = 0\}}$ have no common components, and intersect in at most ${\hbox{deg}(P) \hbox{deg}(Q)}$ points.

This theorem can be proven by a number of means, for instance by using the classical tool of resultants. It has many strengthenings, generalisations, and variants; see for instance this previous blog post on Bézout’s inequality. Bézout’s theorem asserts a fundamental algebraic dichotomy, of importance in combinatorial incidence geometry: any two algebraic curves either share a common component, or else have a bounded finite intersection; there is no intermediate case in which the intersection is unbounded in cardinality, but falls short of a common component. This dichotomy is closely related to the integrality gap in algebraic dimension: an algebraic set can have an integer dimension such as ${0}$ or ${1}$, but cannot attain any intermediate dimensions such as ${1/2}$. This stands in marked contrast to sets of analytic, combinatorial, or probabilistic origin, whose “dimension” is typically not necessarily constrained to be an integer.

Bézout’s inequality tells us, roughly speaking, that the intersection of a curve of degree ${D_1}$ and a curve of degree ${D_2}$ forms a set of at most ${D_1 D_2}$ points. One can consider the converse question: given a set ${S}$ of ${N}$ points in the plane ${k^2}$, can one find two curves of degrees ${D_1,D_2}$ with ${D_1 D_2 = O(N)}$ and no common components, whose intersection contains these points?

A model example that supports the possibility of such a converse is a grid ${S = A \times B}$ that is a Cartesian product of two finite subsets ${A, B}$ of ${k}$ with ${|A| |B| = N}$. In this case, one can take one curve to be the union of ${|A|}$ vertical lines, and the other curve to be the union of ${|B|}$ horizontal lines, to obtain the required decomposition. Thus, if the proposed converse to Bézout’s inequality held, it would assert that any set of ${N}$ points was essentially behaving like a “nonlinear grid” of size ${N}$.

Unfortunately, the naive converse to Bézout’s theorem is false. A counterexample can be given by considering a set ${S = S_1 \cup S_2}$ of ${2N}$ points for some large perfect square ${N}$, where ${P_1}$ is a ${\sqrt{N}}$ by ${\sqrt{N}}$ grid of the form described above, and ${S_2}$ consists of ${N}$ points on an line ${\ell}$ (e.g. a ${1 \times N}$ or ${N \times 1}$ grid). Each of the two component sets ${S_1, S_2}$ can be written as the intersection between two curves whose degrees multiply up to ${N}$; in the case of ${S_1}$, we can take the two families of parallel lines (viewed as reducible curves of degree ${\sqrt{N}}$) as the curves, and in the case of ${S_2}$, one can take ${\ell}$ as one curve, and the graph of a degree ${N}$ polynomial on ${\ell}$ vanishing on ${S_2}$ for the second curve. But, if ${N}$ is large enough, one cannot cover ${S}$ by the intersection of a single pair ${\gamma_1, \gamma_2}$ of curves with no common components whose degrees ${D_1,D_2}$ multiply up to ${D_1 D_2 = O(N)}$. Indeed, if this were the case, then without loss of generality we may assume that ${D_1 \leq D_2}$, so that ${D_1 = O(\sqrt{N})}$. By Bézout’s theorem, ${\gamma_1}$ either contains ${\ell}$, or intersects ${\ell}$ in at most ${O(D_1) = O(\sqrt{N})}$ points. Thus, in order for ${\gamma_1}$ to capture all of ${S}$, it must contain ${\ell}$, which forces ${\gamma_2}$ to not contain ${\ell}$. But ${\gamma_2}$ has to intersect ${\ell}$ in ${N}$ points, so by Bézout’s theorem again we have ${D_2 \geq N}$, thus ${D_1 = O(1)}$. But then (by more applications of Bézout’s theorem) ${\gamma_1}$ can only capture ${O(\sqrt{N})}$ of the ${N}$ points of ${S_1}$, a contradiction.

But the above counterexample suggests that even if an arbitrary set of ${N}$ (or ${2N}$) points cannot be covered by the single intersection of a pair of curves with degree multiplying up to ${O(N)}$, one may be able to cover such a set by a small number of such intersections. The purpose of this post is to record the simple observation that this is, indeed, the case:

Theorem 2 (Partial converse to Bézout’s theorem) Let ${k}$ be a field, and let ${S}$ be a set of ${N}$ points in ${k}$ for some ${N > 1}$. Then one can find ${m = O(\log N)}$ and pairs ${P_i,Q_i \in k[x,y]}$ of coprime non-zero polynomials for ${i=1,\ldots,m}$ such that

$\displaystyle S \subset \bigcup_{i=1}^m \{ (x,y) \in k^2: P_i(x,y) = Q_i(x,y) = 0 \} \ \ \ \ \ (1)$

and

$\displaystyle \sum_{i=1}^m \hbox{deg}(P_i) \hbox{deg}(Q_i) = O( N ). \ \ \ \ \ (2)$

Informally, every finite set in the plane is (a dense subset of) the union of logarithmically many nonlinear grids. The presence of the logarithm is necessary, as can be seen by modifying the ${P_1 \cup P_2}$ example to be the union of logarithmically many Cartesian products of distinct dimensions, rather than just a pair of such products.

Unfortunately I do not know of any application of this converse, but I thought it was cute anyways. The proof is given below the fold.