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1.2. prime ideals.
Definition 1.12. A prime ideal is a proper ideal whose complement is closed under multiplication.
This is equivalent to saying: ab โ p โโ a โ p or b โ p
Proposition 1.13. An ideal a is prime iff A/a is an integral domain (ring in which D = 0). In particular, maximal ideals are prime.
Corollary 1.14. For any ring homomorphism f : A โ B (which by definition sends 1 to 1 ), f โ^1 (p) is prime for any prime p โ B.
Definition 1.15. The radical of an ideal is defined to be
r(a) := {x โ A | x n^ โ a for some n โฅ 1 }
Thus, r(0) = nilrad A.
Exercise 1.16 (A-M 1.13). Prove the following
(1) r(r(a)) = r(a) (2) r(ab) = r(a โฉ b) = r(a) โฉ r(b). Recall that a product of ideals ab is defined to be the ideal generated by all products ab where a โ a, b โ b. (3) If p is prime then r(p n^ ) = p.
Theorem 1.17 (A-M 1.8). The nilradical is the intersection of all prime ideals of A:
r(0) = nilrad A =
p
Corollary 1.18 (A-M 1.14). The radical of an ideal a is equal to the intersection all primes containing a:
r(a) =
pโa
p
The proof of this well-known theorem uses properties of unions of prime ideals.
1.2.1. unions of prime ideals. The question is: Which sets are unions of prime ideals?
Definition 1.19. A multiplicative set in A is a subset S โ A which contains 1, does not contain 0 and is closed under multiplication.
For example, the complement of a prime ideal is a multiplicative set.
Lemma 1.20. Let S be a multiplicative set and let a be an ideal which is disjoint from S and maximal with this property. Then a is prime.
Theorem 1.21. Let X be a nonempty proper subset of A. Then X is a union of prime ideals if and only if it satisfies the following condition.
(1.1) ab โ X โโ either a โ X or b โ X.
Furthermore, any maximal a โ X is prime.
Example 1.22 (A-M exercise 1.14). The set D of zero divisors of A is a union of prime ideals.
Letโs prove the theorem first, assuming the lemma.
Proof of Theorem 1.21. It is clear that any union of primes satisfies condition (1.1). Conversely, suppose that X satisfies (1.1). Then in particular, the complement of X is a multiplicative set. Take any x โ X. Then (x) โ X. Let (x) โ a โ X where a is maximal with this property. Then a is prime by the lemma. So, X is the union of the prime ideals contained in X.!
The proof of the lemma uses the notation:
(a : B) := {x โ A | xB โ a}
for any ideal a and subset B. This is an ideal which contains a.
Exercise 1.23 (I-M 1.12(iii)). Show that
((a : b) : c) = (a : bc) = ((a : c) : b)
Proof of Lemma 1.20. Suppose that a โ A โ S is maximal among the ideals disjoint from S. Suppose that bc โ a and b /โ a. Then we want to show that c โ a. Certainly, c โ (a : b) and a โ (a : b). So, it suffices to prove the following: Claim: (a : b) is disjoint from S. By maximality of a this would imply that c โ (a : b) = a. To prove this claim, take any x โ (a : b). Then (a : x)! a since it contains b /โ a. So, by maximality of a, S โฉ (a : x) is nonempty. Say s โ S โฉ (a : x). Then sx โ a is disjoint from S. So, x /โ S as claimed.!
Proof of Theorem 1.17. It is clear that any nilpotent element of A is contained in every prime ideal. Conversely, suppose that f โ A is not nilpotent. Then S = { 1 , f, f 2 , ยท ยท ยท } is a multiplicative set and 0 is an ideal disjoint from S. Take a ideal a disjoint from S and maximal with this property. Then a is prime by the lemma. So, f is not in
p.!
