Bourbaki

Exercise 10, § 2

Commutative Algebra, VII, 558

Bourbaki marks this one as harder than the rest.

Let AA be a ring in which every ideal is a product of prime ideals. (a) Show that, for every prime ideal pp of AA, A/pA/p is a Dedekind domain (Exercise 9). (b) Show that the set of minimal prime ideals pip_i of AA is finite (decompose OO as a product of prime ideals). (c) Supposethat A/piA/p_i is not a field. Show that, if y∈piy \in p_i, then, for all x∈A−pix \in A - p_i, there exists z∈piz \in p_i such that y=zxy = zx. (Consider in AA the decompositions as products of prime ideals of AxAx and Ax+AyAx + Ay and the corresponding decompositions in A/piA/p_i.) Deduce that, for all non-zero y∈piy \in p_i, Ay=piAy = p_i (consider pi/Ayp_i/Ay, decomposing AyAy as a product of prime ideals); show finally that pi=pi2p_i = p_i^2 and therefore that there exists in pip_i an idempotent eie_i such that pi=Aeip_i = Ae_i (cf. Chapter II, § 4, Exercise 15). Then AA is the direct composition of the ring AeAe, and the ring A(1−ei)A(1 - e_i), isomorphic to the Dedekind domain A/piA/p_i. (d) Suppose now that all the pip_i are maximal, so that AA is a direct composition of rings of the form A/piniA/p_i^{n_i} (Chapter II, § 1, no, 2, Proposition 5) and we may therefore confine our attention to the case where AA is primary or where the unique prime ideal pp of AA is nilpotent. Show that in this case the hypothesis implies that AA is a principal ideal ring (Algebra, Chapter VII, § 1, Exercises 5 and 6). (e) Conclude from (c) and (d) that AA is the product of a finite number of Dedekind domains and a principal ideal ring.

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