Bourbaki

Exercise 7, § 1

Commutative Algebra, IV, 287

Bourbaki marks this one as harder than the rest.

Let A be a Noetherian ring, M a finitely generated A-module and a=Ann⁡(M)a = \operatorname{Ann}(M).

(a) Show that, if a is prime, a is the least element of Ass⁡(M)\operatorname{Ass}(M). (Note that a=⋂p∈Ass⁡(M)pa = \bigcap_{p \in \operatorname{Ass}(M)} p.)

(b) Show that every prime ideal associated with A/aA/a is associated with M. (Note that if p is a prime ideal of A the annihilator of the class in A/aA/a of an element a∈Aa \in A, then p=Ann⁡(aM)p = \operatorname{Ann}(aM) and use (a).)

(c) Let p, q be two distinct prime ideals of A such that p⊂qp \subset q. Show that, if M=(A/p)⊕(A/q)M = (A/p) \oplus (A/q), then Ass⁡(A/a)≠Ass⁡(M)\operatorname{Ass}(A/a) \neq \operatorname{Ass}(M).

Solution

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Transcribed from the printed volume. Not translated.