Exercise 10, § 2
Commutative Algebra, VII, 558
Bourbaki marks this one as harder than the rest.
Let be a ring in which every ideal is a product of prime ideals. (a) Show that, for every prime ideal of , is a Dedekind domain (Exercise 9). (b) Show that the set of minimal prime ideals of is finite (decompose as a product of prime ideals). (c) Supposethat is not a field. Show that, if , then, for all , there exists such that . (Consider in the decompositions as products of prime ideals of and and the corresponding decompositions in .) Deduce that, for all non-zero , (consider , decomposing as a product of prime ideals); show finally that and therefore that there exists in an idempotent such that (cf. Chapter II, § 4, Exercise 15). Then is the direct composition of the ring , and the ring , isomorphic to the Dedekind domain . (d) Suppose now that all the are maximal, so that is a direct composition of rings of the form (Chapter II, § 1, no, 2, Proposition 5) and we may therefore confine our attention to the case where is primary or where the unique prime ideal of is nilpotent. Show that in this case the hypothesis implies that is a principal ideal ring (Algebra, Chapter VII, § 1, Exercises 5 and 6). (e) Conclude from (c) and (d) that is the product of a finite number of Dedekind domains and a principal ideal ring.
Solution
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