Bourbaki

Exercise 17, § 11

Algebra, III, 650

Let K be a commutative field and E a vector space over K.

(a) Let zz be a pp-vector ≠0\neq 0 over E and let VzV_z be the vector subspace of E consisting of the vectors xx such that z∧x=0z \wedge x = 0. Then dim⁡(Vz)≤p\dim(V_z) \leq p; if (xi)1≤i≤q(x_i)_{1 \leq i \leq q} is a free system of vectors of VzV_z, there exists a (p−q)(p - q)-vector vv such that z=v∧x1∧⋯∧xqz = v \wedge x_1 \wedge \cdots \wedge x_q. If E is finite-dimensional, then Vz=ϕ‾(Mz0)V_z = \overline{\phi}(M_z^0), where Mz0M_z^0 is the orthogonal in E∗E^* of the subspace MzM_z associated with zz. For zz to be a pure pp-vector, it is necessary and sufficient that dim⁡(Vz)=p\dim(V_z) = p and then Vz=MzV_z = M_z.

(b) Let vv be a pure pp-vector, ww a pure qq-vector and V and W the subspaces of E associated respectively with vv and ww. In order that V⊂WV \subset W, it is necessary and sufficient that q≥pq \geq p and that there exist a (q−p)(q - p)-vector uu such that w=u∧vw = u \wedge v. In order that V∩W={0}V \cap W = \{0\}, it is necessary and sufficient that v ∧ w ≠ 0; the subspace V + W is then associated with the pure (p + q)-vector v ∧ w.

(c) Let u and v be two pure p-vectors and U and V the subspaces associated respectively with u and v. For u + v to be pure, it is necessary and sufficient that dim(U ∩ V) ≥ p − 1.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.