Nonlinear Equations - UFRJ

[SEC. 2.6: THE NULLSTELLENSATZ 27
In general, the radical of an ideal J is defined as
√
J = {f ∈ k[x1 , . . . , x n ] : ∃r ∈ N, f r ∈ J}.
The reader shall check as an exercise that √ J is an ideal.
Theorem 2.26 (Hilbert Nullstellensatz). Let k be an algebraically
closed field. Then, for all ideal J in k[x 1 , . . . , x n ],
I(Z(J)) = √ J.
We will derive this theorem from a weaker version.
Theorem 2.27 (weak Nullstellensatz). Assume that f 1 , . . . , f s ∈
k[x 1 , . . . , x n ] have no common root. Then, there are g 1 , . . . , g s ∈
k[x 1 , . . . , x n ] such that
f 1 g 1 + · · · + f s g s ≡ 1.
Proof. Let J = (f 1 , · · · , f s ) and assume that 1 ∉ J. In that case, the
algebra
A = k[x 1 , . . . , x n ]/J
is not the zero algebra. By Lemma 2.15, there is a surjective projection
from the zero-set of J onto some r-dimensional subspace of k n ,
r ≥ 0. Thus the f i have a common root.
Proof of Theorem 2.26(Hilbert Nullstellensatz).
The inclusion I(Z(J)) ⊇ √ J is easy, so let h ∈ I(Z(J)).
Let (f 1 , . . . , f s ) be a basis of J (Theorem 2.6). Assume that
(f 1 , . . . , f s ) ∌ 1 (or else h ∈ J ⊆ √ J and we are done).
Consider now the ideal K = (f 1 , . . . , f s , (1 − x n+1 h)) ∈ k[x 1 , . . . ,
x n+1 ]. The set Z(K) is empty. Otherwise, there would be (x 1 , . . . ,
x n+1 ) ∈ k n+1 so that f i (x 1 , . . . , x n ) would vanish for all i. But then
by hypothesis h(x 1 , . . . , x n ) = 0 and 1 − x n+1 h ≠ 0.
By the weak Nullstellensatz (Theorem 2.27), 1 ∈ K. Thus, there
are polynomials G 1 , . . . , G n+1 with
1 = f 1 G 1 + · · · + f n G n + (1 − x n+1 h)G n+1