Exercise 6, § 1
Commutative Algebra, II, 122
Bourbaki marks this one as harder than the rest.
(a) Let A be a ring and a, b two ideals of A such that b is finitely generated; show that, if the quotient rings A/a and A/b are Noetherian, so is A/ab (observe that b/ab is a finitely generated (A/a)-module).
(b) Show that, if a ring A is such that every prime ideal of A is finitely generated, A is Noetherian. (Argue by reductio ad absurdurn by showing that in the set of ideals of A which are not finitely generated there would be a maximal element c which is not prime by hypothesis; there would therefore be two ideals a ⊃ c, b ⊃ c distinct from c and satisfying ab ⊂ c; then use (a).)
Solution
No solution has been written yet.