Exercise 12, § 5
Commutative Algebra, II
Bourbaki marks this one as harder than the rest.
Let A be a ring, I an A-module, B the unique ring whose underlying additive group is A ⊕ I, A being a subring of B, I a square zero ideal, whose A-module structure obtained by restriction of scalars is the given structure (Alg., chap. II, 3rd ed., § 1, exerc. 7).
a) Suppose that every noninvertible element of A belongs to the annihilator of an element ≠ 0 of I. Show that in B every element which is not a divisor of 0 is invertible, in other words that B is equal to its total ring of fractions.
b) Let be a family of maximal ideals of A such that every noninvertible element of A belongs to at least one . Show that, if one takes for I the direct sum of the A-modules , the condition of a) is satisfied.
c) Deduce from b) that, if there exists a projective A-module of rank 1 which is not free, one can determine I so that B is equal to its total ring of fractions, but that there exists a projective B-module of rank 1 which is not free (this module being necessarily noninvertible, since B is the only nondegenerate sub-B-module of B).
d) Suppose that there exists in A a maximal ideal m such that, for every , there exists a maximal ideal of A containing x. If one takes for I the direct sum A-module of the , where runs through the set of maximal ideals of A distinct from m, the corresponding ring is equal to its total ring of fractions; if one puts , which is a maximal ideal of B, show that is isomorphic to ; if A is integral (necessarily different from a field by virtue of the assumption), is therefore an integral ring distinct from its field of fractions. If moreover mAₘ is a principal ideal of the integral ring , show that the ideal of B is a degenerate nonfree projective B-module of rank 1. (One will see in chap. VII that there are Dedekind rings A in which there exist maximal ideals m of which no power is a principal ideal; for such a ring A, all the preceding assumptions are satisfied.)
Solution
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