We have shown in [4] that a universal axiom is satisfied by a generic poly-
nomial gdof degree das soon as d≥nand that an inductive axiom is satisfied
as soon as d≥n(nr +n+r+ 1). More precisely, we have obtained the following
result.
Theorem 1 Let φ(x1,...,x
n,y
1,...,y
n, z)be a conjunction of polynomial
equations with coefficients in Q. Given a tuple zof parameters we denote by Vz
the algebraic subset of C2ndefined by φ.
Fix a tuple zwhich satisfies the associated condition θ(z)from (3). Let rbe
the transcendence degree of Q(z)over Q. For any k≥0,ifd≥n(nr+n+r+1)
there exists x∈Cnsuch that (x, gd(x)) is a generic point of Vz(i.e., a point of
transcendence degree nover Q(z)).
2.2 Axiomatization `a la Wilkie
Given I⊆{1,...,n},sayI={i1,...,i
k}(in increasing order), and a n-
tuple x=(x1,...,x
n) of elements or variables, we denote by xIthe k-tuple
(xi1,...,x
ik).
Consider npolynomials fi(x1,...,x
n,y
1,...,y
n,z
1,...,z
r) with integer co-
efficients. For each value of the parameter zwe obtain a polynomial map
Fz:Kn×Kn→Kn. Recall that a zero (a, b)ofFzis said to be regular if the
Jacobian matrix of Fzat (a, b)hasrankn,orinotherwordsif
det ∂Fz
∂(xI, yJ)(a, b)=0 (4)
for some I,J ⊆{1,...,n}such that |I|+|J|=n.LetV(Fz) be the set of zeros
of Fz. A regular zero lies on a unique irreducible component of V(Fz)andthis
component is of dimension n. Following Wilkie, we say that (a, b) is a balanced
zero1of Fzif a1,...,a
nare all non-zero and pairwise distinct, and if one can
choose Iand Jso that I∪J={1,...,n}(or equivalently, so that I∩J=∅).
In a geometric language, condition (4) means that the tangent space to
V(Fz)at(a, b) has dimension n. If this tangent space is not included in a
subspace of the form xi=cfor some i∈{,...,n}and some constant c∈K
and if additionally (a, b) is a balanced zero of Fz,wesaythat(a, b)isawell
balanced zero.
In this axiomatization we keep the axioms 1,2 and 3 from section 2.1 but
we replace the inductive axioms by:
4’. Let θ(z) be a formula of the language of fields which expresses that Fz
has a well balanced zero. We add the following axiom.
∀z1,...,z
r∃x1,...,x
nθ(z)→F(x, H(x), z)=0.(5)
The following lemma is standard.
1Wilkie uses the terminology “balanced, non-singular zero”. We just write “balanced zero”
for short.
3