Proposition 13

Tag 068F, permanent. Cite it as [Bourbaki, Tag 068F].

Algebra, chapter III, § 11. Cogebras, products of multilinear forms, inner products and duality

With the hypotheses and notation of Proposition 11, for every endomorphism g of the A-module M.

(79)

(\det g)\phi = \Lambda^{(tg)} \circ \phi \circ \Lambda(g).

Clearly $\Lambda^{(tg)} = \theta_{\Lambda}^{-1} \circ (t\Lambda(g)) \circ \theta_{\Lambda}$ ; since $\Lambda(g)$ is an endomorphism of the algebra $\Lambda(M)$ and by definition, for all

z \in \Lambda(M), \quad \theta_{\Lambda}(\Lambda(g)(z) \perp e^*) = \theta_{\Lambda}(e^*) \perp \theta_{\Lambda}(\Lambda(g)(z)),

we deduce from formula (42) of no. 6 that

((\theta_{\Lambda}^{-1} \circ (t\Lambda(g)) \circ \theta_{\Lambda}) \circ \phi \circ \Lambda(g))(z) = \theta_{\Lambda}^{-1}(t\Lambda(g)(\theta_{\Lambda}(e^*)) \perp z) = z \perp (\Lambda^{(tg)}(e^*)) = (\det g)(z \perp e^*) = (\det g)\phi(z)

taking account of § 8, no. 4, Proposition 8.

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

Proposition 13

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