Bourbaki

Exercise 17, § 2

Algebra, II, 391

Bourbaki marks this one as harder than the rest.

(a) Let M, N, M', N' be four submodules of a module E such that M ∩ N = M' ∩ N' = P. Show that

M=(M+(N∩M′))∩(M+(N∩N′)) M = (M + (N \cap M')) \cap (M + (N \cap N'))

Deduce that if M is irreducible in E, then necessarily P = M' ∩ N or P = N' ∩ N.

(b) Let (N_i)_{1 ≤ i ≤ n} be a finite family of submodules of E irreducible in E and M its intersection. Show that if (P_j)_{1 ≤ j ≤ m} is a finite family of submodules of E such that M = ⋂jPj\bigcap_j P_j, then for every index i such that 1 ≤ i ≤ n there exists an index φ(i) such that 1 ≤ φ(i) ≤ m and such that, if we write Ni′=⋂k≠iNkN'_i = \bigcap_{k \neq i} N_k, then M = P_{φ(i)} ∩ N'_i (use (a)). Deduce that if none of the P_j contains the intersection of the P_k of index ≠ j, then m ≤ n (successively replace the N_i by suitable P_j, using the above result). Conclude that two reduced irreducible decompositions of M in E, one of which is finite, necessarily have the same number of terms (cf. Exercise 23).

Solution

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Transcribed from the printed volume. Not translated.