Bourbaki

Proposition 1

Tag 05PP, permanent. Cite it as [Bourbaki, Tag 05PP].

Algebra, chapter II, § 4. Relations between tensor products and homomorphism modules

(a) Let E be a (B, A)-bimodule, F a left A-module and G a left B-module. For every mapping g∈HomB(E⊗AF,G)g \in \mathrm{Hom}_B(E \otimes_A F, G), let g′g' be the mapping of F into HomB(E,G)\mathrm{Hom}_B(E, G) defined by (g′(y))(x)=g(x⊗y)(g'(y))(x) = g(x \otimes y) for x∈E,y∈Fx \in E, y \in F. The mapping g↦g′g \mapsto g' is an isomorphism

(5)

β:HomB(E⊗AF,G)→HomA(F,HomB(E,G)). \beta : \mathrm{Hom}_B(E \otimes_A F, G) \to \mathrm{Hom}_A(F, \mathrm{Hom}_B(E, G)).

(b) Let E be a right A-module, F an (A, C)-bimodule and G a right C-module. For every mapping h∈HomC(E⊗AF,G)h \in \mathrm{Hom}_C(E \otimes_A F, G), let h′h' be the mapping of E into HomC(F,G)\mathrm{Hom}_C(F, G) defined by (h′(x))(y)=h(x⊗y)(h'(x))(y) = h(x \otimes y) for x∈E,y∈Fx \in E, y \in F. The mapping h↦h′h \mapsto h' is an isomorphism

(6)

γ:HomC(E⊗AF,G)→HomA(E,HomC(F,G)). \gamma : \mathrm{Hom}_C(E \otimes_A F, G) \to \mathrm{Hom}_A(E, \mathrm{Hom}_C(F, G)).

In particular B and C may be taken to be a subring Γ\Gamma of the centre of the ring A; then for every Γ\Gamma-module G, the three Γ\Gamma-modules

HomΓ(E⊗AF,G),HomA(E,HomΓ(F,G)),HomA(F,HomΓ(E,G)) \mathrm{Hom}_\Gamma(E \otimes_A F, G), \quad \mathrm{Hom}_A(E, \mathrm{Hom}_\Gamma(F, G)), \quad \mathrm{Hom}_A(F, \mathrm{Hom}_\Gamma(E, G))

are canonically isomorphic to the Γ\Gamma-module of Γ\Gamma-*bilinear* mappings of E×FE \times F into G which satisfy (1). More particularly:

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

Cited by

Cites

Nothing.

Transcribed from the printed edition, 1998, Springer. Not translated.