Bourbaki

Remark 10

Tag 0DJX, permanent. Cite it as [Bourbaki, Tag 0DJX].

Algebra, chapter IX, § 1. Formes sesquilinéaires

Let AA be a field, E,FE, F two left vector spaces over AA, Φ\Phi a non-degenerate sesquilinear form on E×FE \times F for an antiautomorphism JJ of AA.

a) Let MM be a subspace of EE, NN a subspace of FF such that N⊃M0N \supset M^0 and M⊃N0M \supset N^0. Show that if one of the spaces N/M0,M/N0N/M^0, M/N^0 is of finite dimension, the same is true of the other, and the dimensions of these two spaces are equal.

b) Let M,M′M, M' be two subspaces of EE such that M00=MM^{00} = M and that M′M' is of finite dimension; show that one has (M∩M′)0=M0+M′0(M \cap M')^0 = M^0 + {M'}^0 and (M+M′)00=M+M′(M + M')^{00} = M + M'. (Applying aa) to the subspaces M′M' and M0+M′0M^0 + {M'}^0, show that dim⁡(M∩M′)=codim⁡(M0+M′0)\dim(M \cap M') = \operatorname{codim}(M^0 + {M'}^0); applying aa to the subspaces M+M′M + M' and M0M^0, show that

dim⁡((M+M′)00/M)=dim⁡((M+M′)/M)). \dim((M + M')^{00}/M) = \dim((M + M')/M)).

c) If E=FE = F and if MM is a subspace of EE such that E=M0+M00E = M^0 + M^{00}, show that EE is direct sum of M0M^0 and M00M^{00}.

d) Let EE be a vector space over a commutative field AA admitting an infinite countable basis (en)n⩾0(e_n)_{n \geqslant 0}, and let Φ\Phi be the symmetric bilinear form on EE such that Φ(en,en)=1\Phi(e_n, e_n) = 1 for all nn, Φ(ei,ej)=0\Phi(e_i, e_j) = 0 for i⩾1,j⩾1i \geqslant 1, j \geqslant 1 and i≠ji \neq j, and Φ(e0,en)=1\Phi(e_0, e_n) = 1 for all n⩾1n \geqslant 1. Now prove that Φ\Phi is nondegenerate. Let MM (resp. NN) be the subspace of EE generated by the e2ke_{2k} (resp. e2k−1e_{2k-1}) for k⩾1k \geqslant 1, and let H=M+NH = M + N, which is a hyperplane in EE. Now prove that one has M0=N,N0=M,H00=E≠H,(M∩N)0≠M0+N0M^0 = N, N^0 = M, H^{00} = E \neq H, (M \cap N)^0 \neq M^0 + N^0 and (M+N)00≠M+N(M + N)^{00} \neq M + N, though M00=M,N00=NM^{00} = M, N^{00} = N; if LL is the subspace of finite dimension 2 generated by e0e_0 and e1e_1, one has (L∩H)0≠L0+H0(L \cap H)^0 \neq L^0 + H^0.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Translated with machine help, not yet reviewed by a person.