Nonlinear Equations - UFRJ

[SEC. 1.1: BÉZOUT’S THEOREM 5
The derivative of the evaluation function ev : f, x ↦→ f(x) is
ḟ, ẋ ↦→ Df(x)ẋ + ḟ(x).
Assume that f 0 (x 0 ) = 0 with Df 0 (x 0 )ẋ non-degenerate. Then the
derivative of ev with respect to the x variables is an isomorphism. By
the implicit function theorem, there is a neighborhood U ∋ f 0 and a
function x(f) : U → C n so that f(x 0 ) = f 0 and
ev(f(x), x) ≡ 0.
Now, let
{
}
Σ = f : ∃x ∈ P n+1 : f h (1, x) = 0 and (det Df(·)) h (1, x) = 0 .
By elimination theory, Σ is a Zariski closed set. It does not contain
f ini , so its complement is not empty.
Let g be a polynomial system not in Σ and without roots at
infinity. (Fact 3 says that this is true for a generic g). We claim that
g has the same number of roots as f ini .
Since Σ and the set of polynomials with roots at infinity are
Zariski closed, there is a smooth path (or homotopy) between f ini
and g avoiding those sets. Along this path, locally, the root count is
constant. Indeed, let I ⊆ [0, 1] be the maximal interval so that the
implicit function x t for f t (x t ) ≡ 0 can be defined. Let t 0 = sup I.
If 1 ≠ t 0 ∈ I, then (by the implicit function theorem) the implicit
function x t can be extended to some interval (0, t 0 + ɛ) contradicting
that t 0 = sup I. So let’s suppose that t 0 ∉ I. The fact that f t0 has
no root at infinity makes x t convergent when t → t 0 ± ɛ. Hence x t
can be extended to the closed interval [0, t 0 ], another contradiction.
Therefore I = [0, 1].
Thus, f ini and g have the same number of roots.
Until now we counted roots of systems outside Σ. Suppose that
f ∈ Σ has more roots than the Bézout bound. By lower semicontinuity
of the root count, there is a neighborhood of f (in the
usual topology) where there are at least as many roots as in f. However,
this neighborhood is not contained in Σ, contradiction.