Bourbaki

Exercise 13, § 2

Commutative Algebra, VII, 560

Bourbaki marks this one as harder than the rest.

(a) In a Priiferian domain A, let a,ba, b be two finitely generated ideals; show that, if b∉ab \notin a, there exists a finitely generated ideal cc such that a⊂ca \subset c, a≠ca \neq c and bc⊂abc \subset a (consider the ideal a+ba + b).

(b) Deduce from (a) that, if aa is a finitely generated ideal in a Priiferian domain A, then in the ring A/aA/a 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 p,p′p, p' neither of which is contained in the other are relatively prime (consider, for every maximal ideal mm of A, the ideal pAm+p′AmpA_m + p'A_m).

(d) In a Priiferian domain A let aa be a finitely generated ideal. For aa to be primal (Chapter IV, § 2, Exercise 33), it is necessary and sufficient that aa be contained only in a single maximal ideal of A (use (b)); aa is then irreducible and quasi-prime (Chapter IV, § 2, Exercise 34), so that, for finitely generated ideals, these three notions coincide. For aa to be primary, it is necessary and sufficient that it be contained only in a single prime ideal (necessarily maximal) mm; for aa to be strongly primary (Chapter IV, § 2, Exercise 27), it is necessary and sufficient also that mAmmA_m be principal.

Solution

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Transcribed from the printed volume. Not translated.