Bourbaki

Remark

Tag 05P1, permanent. Cite it as [Bourbaki, Tag 05P1].

Algebra, chapter II, § 3. Tensor products

(1) Let A, B be two commutative rings, ρ:B→A\rho : B \to A a ring homomorphism and E and F two A-modules; then the canonical mapping (6) of no. 3 is a B-linear mapping

ρ∗(E)⊗ρ∗(F)→ρ∗(E⊗AF). \rho_*(E) \otimes \rho_*(F) \to \rho_*(E \otimes_A F).

(2) What has been said in this no. can be generalized to the following case: let E be a right A-module, F a left A-module, C a commutative ring and ρ:C→A\rho : C \to A a homomorphism of C into A such that ρ(C)\rho(C) is contained in the centre of A (cf. III, § 1, no. 3). We may then consider the C-modules ρ∗(E)\rho_*(E) and ρ∗(F)\rho_*(F) and the hypothesis on ρ\rho implies that these C-module structures are compatible respectively with the A-module structures on E and F (§ 1, no. 14). The tensor product E⊗AFE \otimes_A F is thus (by virtue of no. 4) given two C-module structures such that γ(x⊗y)=(xρ(γ))⊗y\gamma(x \otimes y) = (x \rho(\gamma)) \otimes y and

γ(x⊗y)=x⊗(ρ(γ)y) \gamma(x \otimes y) = x \otimes (\rho(\gamma)y)

respectively for γ∈C\gamma \in C, x∈Ex \in E, y∈Fy \in F and Definition 1 (no. 1) shows also that these two structures are identical. If E' (resp. F') is a right (resp. left) A-module and u:E→E′u : E \to E', v:F→F′v : F \to F' are two A-linear mappings, then u⊗v:E⊗AF→E′⊗AF′u \otimes v : E \otimes_A F \to E' \otimes_A F' is C-linear for the C-module structures just defined; the mapping (u,v)↦u⊗v(u, v) \mapsto u \otimes v:

HomA(E,E′)×HomA(F,F′)→HomC(E⊗AF,E′⊗AF′) \mathrm{Hom}_A(E, E') \times \mathrm{Hom}_A(F, F') \to \mathrm{Hom}_C(E \otimes_A F, E' \otimes_A F')

is C-bilinear (for the C-module structures on HomA(E,E′)\mathrm{Hom}_A(E, E') and HomA(F,F′)\mathrm{Hom}_A(F, F')) defined in § 1, no. 14, Remark 1), whence we also derive a C-linear mapping, called canonical

(13) Hom⁡A(E,E′)⊗CHom⁡A(F,F′)→Hom⁡C(E⊗AF,E′⊗AF′)\operatorname{Hom}_A(E, E') \otimes_C \operatorname{Hom}_A(F, F') \to \operatorname{Hom}_C(E \otimes_A F, E' \otimes_A F').

(3) Let A be an integral domain and K its field of fractions. If E and F are two vector K-spaces, the canonical mapping

(E[A])⊗A(F[A])→E⊗KF (E_{[A]}) \otimes_A (F_{[A]}) \to E \otimes_K F

(no. 3 and § 1, no. 13) is bijective. It suffices (no. 4) to prove that if ff is an A-bilinear mapping of E×FE \times F into a vector K-space G, ff is also K-bilinear. Now, for all α≠0\alpha \neq 0 in A.

αf(α−1x,y)=f(x,y)=αf(x,α−1y) \alpha f(\alpha^{-1}x, y) = f(x, y) = \alpha f(x, \alpha^{-1}y)
whence
f(α−1x,y)=f(x,α−1y)=α−1f(x,y) f(\alpha^{-1}x, y) = f(x, \alpha^{-1}y) = \alpha^{-1}f(x, y)
since G is a vector K-space.

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

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