Bourbaki

Exercise 35, § 7

Algebra, II, 407

Bourbaki marks this one as harder than the rest.

(a) Let KK be a field and EE a (K,K)(K, K)-bimodule. Suppose that the dimensions of EE, as a left and right vector space over KK, are equal to the same finite number nn. Show that there exists a family (ei)(e_i) of nn elements of EE, which is a basis for each of the two vector space structures on EE. (Note that if (bj)1≤j≤m(b_j)_{1 \leq j \leq m} is a family of m<nm < n elements of EE, which is free for each of the two vector space structures on EE and VV is the left vector subspace and WW the right vector subspace generated by (bj)(b_j), either V+W≠EV + W \neq E, or V∩CWV \cap CW and W∩CVW \cap CV are non-empty; in the latter case, if y∈V∩CW,z∈W∩CVy \in V \cap CW, z \in W \cap CV, y+zy + z forms with the bjb_j a family of m+1m + 1 elements which is free for each of the two vector space structures on EE.)

(b) Let FF be a subbimodule of EE, whose two vector space structures over KK have the same dimension p<np < n. Show that there exists a family (ei)1≤i≤n(e_i)_{1 \leq i \leq n} of elements of EE which is a basis of EE for each of its vector space structures and such that (ei)1≤i≤p(e_i)_{1 \leq i \leq p} is a basis of FF for each of its two vector space structures (same method).

(c) Let (bj)1≤j≤n−1(b_j)_{1 \leq j \leq n-1} be a family of n−1n - 1 elements of EE, which is free for each of the two vector space structures on EE. Let VV (resp. WW) be the hyperplane generated by the (bj)(b_j) in EE considered as a left (resp. right) vector space.

Show that if V⊂WV \subset W (resp. W⊂VW \subset V), then necessarily V=WV = W (note that if a∉Wa \notin W, the set of λ∈K\lambda \in K such that λa∈W\lambda a \in W is a left ideal).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.