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
![\displaystyle \left|V\cap [0,N]^n \cap{\mathbb Z}^n\right|\le c(n,\nu,\epsilon) N^{\mu-1+1/\nu+\epsilon}.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cleft%7CV%5Ccap+%5B0%2CN%5D%5En+%5Ccap%7B%5Cmathbb+Z%7D%5En%5Cright%7C%5Cle+c%28n%2C%5Cnu%2C%5Cepsilon%29+N%5E%7B%5Cmu-1%2B1%2F%5Cnu%2B%5Cepsilon%7D.+&bg=ffffff&fg=000000&s=0&c=20201002)
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:
![\displaystyle \Bigl|\det \bigl[f_i(x_j)\bigr]\Bigr|\le n!\|f_1\|_n\cdots\|f_n\|_n \bigl(\max|x_i-x_j|\bigr)^{n(n-1)/2}.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5CBigl%7C%5Cdet+%5Cbigl%5Bf_i%28x_j%29%5Cbigr%5D%5CBigr%7C%5Cle+n%21%5C%7Cf_1%5C%7C_n%5Ccdots%5C%7Cf_n%5C%7C_n+%5Cbigl%28%5Cmax%7Cx_i-x_j%7C%5Cbigr%29%5E%7Bn%28n-1%29%2F2%7D.+&bg=ffffff&fg=000000&s=0&c=20201002)
The Leibniz identity

shows that

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.)