In the previous post we discussed the following result of Bombieri and Pila (1989): let be a compact segment of a real analytic transcendental plane curve; then for any positive integer
Bombieri and Pila remark that a similar (and even stronger) statement holds if is algebraic but does not admit polynomial parametrization.
The obvious disadvantage of this result is dependence of the constant in . It turns out that a slightly weaker inequality holds uniformly for all algebraic curves of given degree. Precisely:
Theorem 1 let be segment of an irreducible plane algebraic curve of degree contained in the unit square . Then for
Equivalently, if is a segment of an irreducible algebraic curve of degree contained in the square , then
One cannot do better than the exponent , as the example shows.
Due to the uniformity, this result easily extends to arbitrary dimension by projection and slicing (Pila 1995): let be an irreducible affine variety of dimension and degree , contained in the affine space ; then
The proof of Theorem 1 goes along similar lines as that in the transcendence case, but is more involved, because now we have to keep track of the dependence of all parameters in the function . For a function we define the -norm by
where is the sup-norm. With this definition we can make inequality (1) of the previous post totally explicit:
The Leibniz identity
In particular, if then
In the sequel by a -curve we mean a plane algebraic curve of degree less than in and less then in . Arguing as in the previous post, we prove the following: let be a sufficiently smooth function, defined on a compact interval . Let be points on the curve defined by , not lying on on a -curve. Assume that the coordinates and are rational numbers with denominator ; then
with . (Here unnumbered constants may vary from equation to equation, while numbered constants , , etc. are individual.)
From now on we assume that . If the set does not lie on a single -curve, then the right-hand side of (1) does not exceed the length of . Splitting into subintervals shorter than this right-hand side, it follows immediately that the set lies on at most
-curves. (We have to write to take into account the case when the set lies on a single -curve.)
Now assume that is a -algebraic function; that is, it satisfies a polynomial equation , where is an irreducible real polynomial of -degree and -degree . In this case the curve , defined by has at most intersections with any -curve. Hence
Our next goal is get rid of the norm of . To do this, Bombieri and Pila show that the norm is “large” only on short intervals. Precisely:
Proposition 2 Let be a -algebraic function defined on an interval . Let be a positive integer. Then for every there is at most intervals of length at most each such that outside these intervals.
We postpone the proof of this proposition until one of my subsequent posts, and show now how it allows one to replace by in (2). Precisely:
Proposition 3 For as in Proposition 2 and any we have
In particular, if takes values in then
The proof is by induction in . The definition of under which the induction works will appear in the course of the proof.
Applying Proposition 2 with and , we split our interval into intervals of length (call them “short”) where is “large”, and the remaining part where . This remaining part itself splits into at most intervals (call them “long”). Applying to every “long” interval estimate (2), we see that the curve has at most
points with denominator above these “long” intervals.
Now let be a “short” interval, and let be its length. We may assume that with , so that is an integral multiple of :
If then has only point over . If then the number of points with denominator on over is the same as the number of points with denominator on the curve over the interval . By induction, the latter quantity is bounded by
Since , the latter quantity does not exceed
Hence there is at most
points over “short” intervals.
Thus, the total number of points with denominator on is bounded by the sum of (5) and (6). Now let be a so large that for the quantity (6) does not exceed . Then
Now if then (3) holds trivially for , which gives the base of our induction. And if then the right-hand side of (7) does not exceed that of (3), which gives the induction step. This proves Proposition 3.
Now we easily complete the proof of Theorem 1. Let be the irreducible equation of our curve, so that . We may assume that . Indeed, one may find a unimodular integral matrix with entries non-exceeding in absolute value such that the polynomial is of both -degree and -degree . The image of the unit square under the linear transformation defined by the inverse of this matrix is contained in the square , and it remains to split the latter into unit squares and translate the variables in each.
Further, a compact segment of an irreducible algebraic curve of -degree and -degree splits into compact segments of the type or , where is a -algebraic function. Applying to each of the latter estimate (4), we obtain the result.
(I thank all participants of the Groupe de travail “Géométrie diophantienne” in Bordeaux for their precious comments.)