Bourbaki

Exercise 12, § 4

Commutative Algebra, X, p. 164

Bourbaki marks this one as harder than the rest.

Let AA be a regular local ring, MM a finitely generated AA-module, such that the AA-module EndA(M)\mathrm{End}_A(M) is free. We propose to prove that the following conditions are equivalent: (i) the AA-module MM is free ; (ii) the AA-module MM is reflexive ; (iii) ExtA1(M,M)=0\mathrm{Ext}_A^1(M, M) = 0. a) Prove the implication (iii) ⇒\Rightarrow (ii) (use Exercise 2 of VII, §4§ 4).

b) Prove the opposite implication (use Exercise 4 if dim(A) ≤2\leq 2, then reason by induction on dim(A) with the aid of Exercise 18 of § 1).

c) Suppose that dim(A) ≤3\leq 3. Prove the implication (ii) ⇒\Rightarrow (i) (use Exercise 4 if dim(A) ≤2\leq 2; if dim(A) = 3, observe that (ii) implies dp_A(M) ≤1\leq 1, 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 mA−mA2m_A - m_A^2; denote by B the ring A/xAA/xA, by N the B-module M/xMM/xM and by N∗N^* 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 ≥4\geq 4; 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)) ≥1\geq 1, then that Ext^1_A(M, A), which is of finite length by the induction hypothesis, is zero ; conclude that the B-module M∗/xM∗M^*/xM^* 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, (x1,…,xn)(x_1, \ldots, x_n) a family of elements of A, generating an ideal I. A family (x1,…,xn)(x_1, \ldots, x_n) is said to be A-independent if the classes of the xix_i in I/I2I/I^2 form a basis of this A/I-module.

a) Every completely transversal family for A is A-independent.

b) Let (x1,…,xn)(x_1, \ldots, x_n) be an A-independent family, with x1=x1′x1′′x_1 = x'_1 x''_1. Prove that the families (x1′,…,xn)(x'_1, \ldots, x_n) and (x1′′,…,xn)(x''_1, \ldots, x_n) are A-independent, and that one has

lg⁡A(A/(x1,…,xn))=lg⁡A(A/(x1′,…,xn))+lg⁡A(A/(x1′′,…,xn)) \lg_A(A/(x_1, \ldots, x_n)) = \lg_A(A/(x'_1, \ldots, x_n)) + \lg_A(A/(x''_1, \ldots, x_n))

(observe that the quotient (x1′,…,xn)/(x1,…,xn)(x'_1, \ldots, x_n)/(x_1, \ldots, x_n) is isomorphic to A/(x1′′,…,xn)A/(x''_1, \ldots, x_n)).

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.