Exercise 12, § 4
Commutative Algebra, X, p. 164
Bourbaki marks this one as harder than the rest.
Let be a regular local ring, a finitely generated -module, such that the -module is free. We propose to prove that the following conditions are equivalent: (i) the -module is free ; (ii) the -module is reflexive ; (iii) . a) Prove the implication (iii) (ii) (use Exercise 2 of VII, ).
b) Prove the opposite implication (use Exercise 4 if dim(A) , then reason by induction on dim(A) with the aid of Exercise 18 of § 1).
c) Suppose that dim(A) . Prove the implication (ii) (i) (use Exercise 4 if dim(A) ; if dim(A) = 3, observe that (ii) implies dp_A(M) , then that (iii) and Remark 2 of No. 3 imply that M is projective).
d) Suppose that M is reflexive. Let x be an element of ; denote by B the ring , by N the B-module and by its dual. Prove with the aid of b) that the B-module End_B(N) is free, then that End_B(N*) is free (deduce from VII, § 4, No. 2 that End_B(N*) identifies with the bidual of End_B(N)).
e) Prove that (ii) implies (i) (reason by induction on the dimension of A, assumed ; with the notations of d), the induction hypothesis implies that the B-module N* is free ; deduce with the aid of Exercise 18 of § 1 and Proposition 7 of No. 4 that one has prof_A(Ext^1_A(M, B)) , then that Ext^1_A(M, A), which is of finite length by the induction hypothesis, is zero ; conclude that the B-module identifies with N*, hence is free, then that the A-module M* is free and finally that M is free.)
f) Deduce from these results another proof of the fact that the ring A is factorial.
12) Let A be a ring, a family of elements of A, generating an ideal I. A family is said to be A-independent if the classes of the in form a basis of this A/I-module.
a) Every completely transversal family for A is A-independent.
b) Let be an A-independent family, with . Prove that the families and are A-independent, and that one has
(observe that the quotient is isomorphic to ).
Solution
No solution has been written yet.