Bourbaki

Exercise 9, § 2

Algebra, II, 388

Given an A-module E, for every submodule M of E (resp. every submodule M′M' of E∗E^*), let M∘M^\circ (resp. M′∘{M'}^\circ) denote the orthogonal of M in E∗E^* (resp. the orthogonal of M′M' in E). Consider the following four properties: (A) The canonical homomorphism cE:E→E∗∗c_E : E \to E^{**} is bijective. (B) For every submodule M of E, the canonical homomorphism E∗/M∘→M∗E^*/M^\circ \to M^* is bijective. (C) For every submodule M of E, M∘∘=MM^{\circ\circ} = M. (D) For every ordered pair of submodules M, N of E,

(M∩N)∘=M∘+N∘. (M \cap N)^\circ = M^\circ + N^\circ.
(a) Show that condition (B) implies (D) (prove that for x∗∈(M∩N)∘x^* \in (M \cap N)^\circ, there exists y∗∈E∗y^* \in E^* such that ⟨x+y,y∗⟩=⟨x,x∗⟩\langle x + y, y^* \rangle = \langle x, x^* \rangle for x∈Mx \in M and y∈Ny \in N).

(b) Let A be a ring with no divisors of zero, but having no field of left fractions (for example the tensor algebra of a vector space of dimension > 1, cf. III, § 5, Exercise 5). If we take E=AsE = A_s, condition (A) is satisfied, but none of conditions (B), (C), (D) (for an example where (B), (C), (D) hold, but not (A), see § 7, no. 5, Theorem 6).

(c) Take A to be a product ring KIK^I, where K is a commutative field and I an infinite set. Show that the A-module E=AE = A satisfies conditions (A) and (B), but not (C) (observe that the annihilators of the ideals of A are all of the form KJK^J, where J⊂IJ \subset I).

(d) Suppose that for two submodules M, N of E such that N⊂MN \subset M and N≠MN \neq M, the dual (M/N)∗(M/N)^* is non-zero; then show that condition (B) implies (C).

(e) The kernel of cEc_E is the orthogonal (E∗)∘(E^*)^\circ of E∗E^* in E. Give an example where neither E∗E^* nor (E∗)∘(E^*)^\circ is 0 (consider a module containing an element whose annihilator contains an element which is not a divisor of 0).

(f) Let M be a direct factor of E; show that M∘∘=M+(E∗)∘M^{\circ\circ} = M + (E^*)^\circ.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.