Exercise 13, § 2
Commutative Algebra, VII, 560
Bourbaki marks this one as harder than the rest.
(a) In a Priiferian domain A, let be two finitely generated ideals; show that, if , there exists a finitely generated ideal such that , and (consider the ideal ).
(b) Deduce from (a) that, if is a finitely generated ideal in a Priiferian domain A, then in the ring every element which is not a divisor of zero is invertible. In particular, a prime ideal of A cannot be finitely generated if it is maximal.
(c) Show that in a Priiferian domain A two prime ideals neither of which is contained in the other are relatively prime (consider, for every maximal ideal of A, the ideal ).
(d) In a Priiferian domain A let be a finitely generated ideal. For to be primal (Chapter IV, § 2, Exercise 33), it is necessary and sufficient that be contained only in a single maximal ideal of A (use (b)); is then irreducible and quasi-prime (Chapter IV, § 2, Exercise 34), so that, for finitely generated ideals, these three notions coincide. For to be primary, it is necessary and sufficient that it be contained only in a single prime ideal (necessarily maximal) ; for to be strongly primary (Chapter IV, § 2, Exercise 27), it is necessary and sufficient also that be principal.
Solution
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