Bourbaki

Exercise 3, § 3

Commutative Algebra, II, 136

Bourbaki marks this one as harder than the rest.

Let AA be a local ring (not necessarily commutative, cf. Exercise 1) and mm its Jacobson radical.

(a) Let LL be a free left AA-module and x≠0x \neq 0 an element of LL. Let (ei)i∈I(e_i)_{i \in I} be a basis of LL for which the number of components ≠0\neq 0 of xx is the least possible; if x=∑i∈Iξieix = \sum_{i \in I} \xi_i e_i, let JJ be the set of i∈Ii \in I such that ξi≠0\xi_i \neq 0. Show that none of the ξi\xi_i (i∈Ji \in J) belongs to the right ideal of AA generated by the others.

(b) With the hypotheses and notation of (a), let P, Q be two complementary submodules of L such that x∈Px \in P; for all t∈It \in I, let et=yt+zte_t = y_t + z_t, where yt∈Py_t \in P, zt∈Qz_t \in Q. Show that the family of yty_t for t∈Jt \in J and ete_t for t∉Jt \notin J is a basis of L. (Prove, using (a), that, if zk=∑t∈Iζktetz_k = \sum_{t \in I} \zeta_{kt} e_t, the components ζkt\zeta_{kt} such that k∈Jk \in J and t∈Jt \in J necessarily belong to m and use Corollary 2 to Proposition 4.) Deduce that there exists a free submodule of P which contains x and is a direct factor of P.

(c) Show that every projective A-module P is free (using Kaplansky's Theorem (Algebra, Chapter 11, § 2, Exercise 4), reduce it to the case where P is generated by a countable family of elements and apply (b) to this case).

(d) Give an example of a local ring A and a flat A-module M which is not free and satisfies mM=MmM = M (take A=Z(p)A = \mathbf{Z}_{(p)}). (In Chapter III, § 3, we shall give examples of faithfully flat A-modules which are not free over a Noetherian local ring A.)

(e) Let A be a local ring and M a finitely generated flat A-module. Show that M is a free A-module (use Exercise 23 (e) of Chapter I, § 2).

Solution

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