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Let be a finite field of order , and let be an absolutely irreducible smooth projective curve defined over (and hence over the algebraic closure of that field). For instance, could be the projective elliptic curve

in the projective plane , where are coefficients whose discriminant is non-vanishing, which is the projective version of the affine elliptic curve

To each such curve one can associate a genus , which we will define later; for instance, elliptic curves have genus . We can also count the cardinality of the set of -points of . The *Hasse-Weil bound* relates the two:

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

for some complex numbers independent of ; 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 is rational. The task is then to establish a bound for all ; this (or more precisely, the slightly stronger assertion ) is the Riemann hypothesis for such curves. This can be done either by passing to the Jacobian variety of and using a certain duality available on the cohomology of such varieties, known as Rosati involution; alternatively, one can pass to the product surface 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 . 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 is a perfect square, and , then .

In fact, the bound on can be sharpened a little bit further, as we will soon see.

Theorem 2 is only an upper bound on , 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 to infinity to control the weights ) 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 as the number of points of intersection between two curves in the surface . Indeed, if we define the Frobenius endomorphism on any projective space by

then this map preserves the curve , and the fixed points of this map are precisely the points of :

Thus one can interpret as the number of points of intersection between the diagonal curve

and the Frobenius graph

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

with also preserving . One can then also interpret as the number of points of intersection between the curve

Let be the field of rational functions on (with coefficients in ), and define , , and analogously )(although is likely to be disconnected, so will just be a ring rather than a field. We then (morally) have the commuting square

if we ignore the issue that a rational function on, say, , might blow up on all of and thus not have a well-defined restriction to . We use and to denote the restriction maps. Furthermore, we have obvious isomorphisms , coming from composing with the graphing maps and .

The idea now is to find a rational function on the surface of controlled degree which vanishes when restricted to , but is non-vanishing (and not blowing up) when restricted to . On , we thus get a non-zero rational function of controlled degree which vanishes on – which then lets us bound the cardinality of in terms of the degree of . (In Bombieri’s original argument, one required vanishing to high order on the side, but in our presentation, we have factored out a term which removes this high order vanishing condition.)

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

Now we build . Pick a point of , which we will think of as being a point at infinity. (For the purposes of proving Theorem 2, we may clearly assume that is non-empty.) Thus is fixed by . To simplify the exposition, we will also assume that is fixed by the square root of ; in the opposite case when has order two when acting on , the argument is essentially the same, but all references to in the second factor of need to be replaced by (we leave the details to the interested reader).

For any natural number , define to be the set of rational functions which are allowed to have a pole of order up to at , but have no other poles on ; note that as we are assuming 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 .) The space is clearly a vector space over ; one can view intuitively as the space of “polynomials” on of “degree” at most . When , consists just of the constant functions. Indeed, if , then the image of avoids and so lies in the affine line ; but as is projective, the image needs to be compact (hence closed) in , and must therefore be a point, giving the claim.

For higher , we have the easy relations

The former inequality just comes from the trivial inclusion . For the latter, observe that if two functions lie in , so that they each have a pole of order at most at , then some linear combination of these functions must have a pole of order at most at ; thus has codimension at most one in , giving the claim.

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

*Riemann’s inequality* complements this with the lower bound

thus one has for all but at most exceptions (in fact, exactly 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 in a non-standard fashion (as the dimension of the first Cech cohomology of the structure sheaf of ), but to obtain this inequality with a standard definition of (e.g. as the dimension of the zeroth Cech cohomolgy 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 , we will define to be a tensor product space

for some natural numbers which we will optimise in later. That is to say, is spanned by functions of the form with and . This is clearly a linear subspace of of dimension , and hence by Rieman’s inequality we have

Observe that maps a tensor product to a function . If and , then we see that the function has a pole of order at most at . We conclude that

and in particular by (4)

We will choose to be a bit bigger than , to make the image of smaller than that of . From (6), (10) we see that if we have the inequality

(together with (7)) then cannot be injective.

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

*Proof:* From (3), we can find a linear basis of such that each of the has a distinct order of pole at (somewhere between and inclusive). Similarly, we may find a linear basis of such that each of the has a distinct order of pole at (somewhere between and inclusive). The functions then span , and the order of pole at is . But since , these orders are all distinct, and so these functions must be linearly independent. The claim follows.

This gives us the following bound:

Proposition 4Let be natural numbers such that (7), (11), (12) hold. Then .

*Proof:* As is not injective, we can find with vanishing. By the above lemma, the function is then non-zero, but it must also vanish on , which has cardinality . On the other hand, by (8), has a pole of order at most at 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.

If , we may make the explicit choice

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 (e.g. if is just the projective line ) one may take and conclude the absolutely sharp bound in this case; in the case of the projective line , the function is in fact the very concrete function .

Remark 1When is not a perfect square, one can try to run the above argument using the factorisation instead of . This gives a weaker version of the above bound, of the shape . In the hyperelliptic case at least, one can erase this loss by working with a variant of the argument in which one requires to vanish to high order at , 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 be a field, and let be non-zero polynomials in two variables with no common factor. Then the two curves and have no common components, and intersect in at most 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 or , but cannot attain any intermediate dimensions such as . 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 and a curve of degree forms a set of at most points. One can consider the converse question: given a set of points in the plane , can one find two curves of degrees with and no common components, whose intersection contains these points?

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

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

But the above counterexample suggests that even if an arbitrary set of (or ) points cannot be covered by the single intersection of a pair of curves with degree multiplying up to , 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 be a field, and let be a set of points in for some . Then one can find and pairs of coprime non-zero polynomials for such that

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 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.

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