Bourbaki

Exercise 1, § 1

Commutative Algebra, II, 121

Bourbaki marks this one as harder than the rest.

(a) Show that a group G cannot be the union of two subgroups distinct from G. Show that, for every set I with at least two elements, the commutative group G=F2(I)G = F_2^{(I)} is the union of three subgroups distinct from G.

(b) Let (Hi)i∈I(H_i)_{i \in I} be a finite family of subgroups of a group G such that each of the H_i is a subgroup of G of infinite index. Show that G cannot be the union of a finite number of left cosets of the H_i. (Argue by induction on the number of elements in I; if there exist two distinct indices i, j such that the index (Hi:(H,nHj))(H_i : (H, nH_j)) is finite, H may be suppressed; if on the other hand Hi∩HjH_i \cap H_j is of infinite index in H, for every ordered pair (i, j) of distinct indices, consider an index k such that HkH_k is maximal in the set of HiH_i and show that HkH_k is the union of a finite number of left cosets of the Hk∩HiH_k \cap H_i where i≠ki \neq k.)

(c) Give an example of a commutative ring A and four ideals a,b1,b2,b3a, b_1, b_2, b_3 of A such that a∉bia \notin b_i (i=1,2,3i = 1, 2, 3), but a=⋃ibia = \bigcup_i b_i (use (a)).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.