Bourbaki

Proposition 5

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

Algebra, chapter II, § 5. Extension of the ring of scalars

Let E be a left A-module and F a left B-module; the composite mappings

ρ∗(E)→ρ∗(ϕE)ρ∗(ρ∗(ρ∗(E)))→ψρ∗(E)ρ∗(E) \rho^*(E) \xrightarrow{\rho^*(\phi_E)} \rho^*(\rho_*(\rho^*(E))) \xrightarrow{\psi_{\rho^*(E)}} \rho^*(E)
ρ∗(E)→ϕρ∗(F)ρ∗(ρ∗(ρ∗(F)))→ρ∗(ψF)ρ∗(F) \rho_*(E) \xrightarrow{\phi_{\rho_*(F)}} \rho_*(\rho^*(\rho_*(F))) \xrightarrow{\rho_*(\psi_F)} \rho_*(F)

are respectively equal to the identity mappings of ρ∗(E)\rho^*(E) and ρ∗(F)\rho_*(F).

We give the proof, for example, for (12); for all x∈Ex \in E, the mapping ρ∗(ϕE)\rho^*(\phi_E) maps 1⊗x1 \otimes x to the element 1⊗(1⊗x)1 \otimes (1 \otimes x) and the mapping ψρ∗(E)\psi_{\rho^*(E)} maps 1⊗(1⊗x)1 \otimes (1 \otimes x) to the element 1⊗x1 \otimes x; the conclusion follows from the fact that the elements of the form 1⊗x1 \otimes x generate the B-module ρ∗(E)\rho^*(E); the proof is even simpler for (13).

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

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Cites

Nothing.

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