Bourbaki

Proposition 11

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

Algebra, chapter X, § 1. Compléments d’algèbre linéaire

Let B be a ring, F a B-module and P a (B, A)-bimodule. If F is an injective B-module and P a flat A-module, HomB(P,F)\mathrm{Hom}_B(P, F) is an injective A-module.

Let u:M′→Mu : M' \to M be an injective homomorphism of A-modules. We have a commutative diagram

HomA(M,HomB(P,F))→HomA(u,1)HomA(M′,HomB(P,F))β↓β′↓HomB(P⊗AM,F)→Hom(1P⊗u,1F)HomB(P⊗AM′,F) \begin{array}{ccc} \mathrm{Hom}_A(M, \mathrm{Hom}_B(P, F)) & \xrightarrow{\mathrm{Hom}_A(u, 1)} & \mathrm{Hom}_A(M', \mathrm{Hom}_B(P, F)) \\ \beta \downarrow & & \beta' \downarrow \\ \mathrm{Hom}_B(P \otimes_A M, F) & \xrightarrow{\mathrm{Hom}(1_P \otimes u, 1_F)} & \mathrm{Hom}_B(P \otimes_A M', F) \end{array}

where β\beta and β′\beta' are the canonical isomorphisms of II, p. 74. Since P is flat over A, the homomorphism 1P⊗u:P⊗AM′→P⊗AM1_P \otimes u : P \otimes_A M' \to P \otimes_A M is injective. Since F is injective, Hom(1P⊗u,1F)\mathrm{Hom}(1_P \otimes u, 1_F) is surjective, hence also HomA(u,1)\mathrm{Hom}_A(u, 1), which proves that HomF(P,F)\mathrm{Hom}_F(P, F) is an injective A-module (X, p. 16, lemma 3).

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

Cited by

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Cites

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