Nonlinear Equations - UFRJ

[SEC. 2.6: THE NULLSTELLENSATZ 25
of ideals in J. Therefore, there must be an element J ∈ J that is
maximal with respect to the inclusion.
But J is not irreducible itself, so there are J 1 , J 2 with J = J 1 ∩J 2 ,
J ≠ J 1 , J ≠ J 2 .
If J 1 and J 2 are intersections of finitely many irreducible ideals,
then so does J = J 1 ∩ J 2 and hence J ∉ J, contradiction. If however
one of them (say J 1 ) is not the intersection of finitely many irreducible
ideals, then J ⊆ J 1 with J 1 in J. Then J is not maximal with respect
to the inclusion, contradicting the definition.
Thus, J must be empty.
An ideal p in R is primary if and only if, for any x, y ∈ R,
xy ∈ p =⇒ x ∈ p or ∃n ∈ N : y n ∈ p
For instance, (4) ⊂ Z and (x 2 ) ⊂ k[x] are primary ideals, but (12)
is not. Prime ideals are primary but the converse is not always true.
The reader will show a famous theorem:
Theorem 2.24 (Primary Decomposition Theorem). If R is Noetherian,
then every ideal in R is the intersection of finitely many primary
ideals.
Exercise 2.6. Let R be Noetherian. Assume the zero ideal is irreducible.
Show then that the zero ideal (0) = {0} is primary. Hint:
assume that xy = 0 with x ≠ 0. Set J n = {z : zy n = 0}. Using
Noether’s condition, show that there is n such that y n = 0.
Exercise 2.7. Let J be irreducible in R. Show that the zero ideal in
R/J is irreducible.
Exercise 2.8. Let J be and ideal of R, such that R/J is primary.
Show that J is primary. This finishes the proof of Theorem 2.24
2.6 The Nullstellensatz
To each subset X ⊆ k n , we associated the ideal of polynomials vanishing
in X:
I(X) = {f ∈ k[x 1 , . . . , x n ] : ∀x ∈ X, f(x) = 0}.