Nonlinear Equations - UFRJ

[SEC. 2.6: THE NULLSTELLENSATZ 29
Corollary 2.29. If m is a maximal ideal then Z(m) is a point.
Proof. Let m be a maximal ideal. Would Z(m) be empty, J would
contain 1, contradiction. So Z(m) contains at least one point a.
Assume now that it contains a second point b ≠ a. They differ in
at least one coordinate, say a 1 ≠ b 1 . Let J be the ideal generated
by the elements of m and by x 1 − a 1 . Then a ∈ Z(J) but b ≠ Z(J).
Hence m J R.
Thus, I induces a bijection between points of k n and maximal
ideals of k[x 1 , . . . , x n ].
Corollary 2.30. Every non-empty Zariski-closed set can be written
as a finite union of irreducible Zariski-closed sets.
Proof. Let X be Zariski closed. By Theorem 2.24, I(X) is a finite
intersection of primary ideals:
I(X) = J 1 ∩ · · · ∩ J r .
Let X i = Z(J i ), for i = 1, . . . , r. By the Nullstellensatz, I(X i ) =
√
Ji . An ideal that is radical and primary is prime. Hence (Proposition
2.22) X i is irreducible.
An irreducible Zariski-closed set X is called an (affine) algebraic
variety.i Its dimension r is the transcendence degree of A = k[x 1 , . . . ,
x n ] over the prime ideal Z(X). Its degree is the degree of A as an
extension of k[x 1 , . . . , x r ].
We restate an important consequence of Lemma 2.15 in the new
language.
Lemma 2.31. Let X be a variety of dimension r and degree d. Then,
the number of isolated intersections of X with an affine hyperplane
of codimension r is at most d. This number is attained for a generic
choice of the hyperplane.
Exercise 2.9. Let J be an ideal. Show that √ J is an ideal.
Exercise 2.10. Prove that m is a maximal ideal in k[x 1 , . . . , x n ] if and
only if, A = k[x 1 , . . . , x n ]/m is a field.