Bourbaki

Example

Tag 060K, permanent. Cite it as [Bourbaki, Tag 060K].

Algebra, chapter III, § 4. Tensor products of algebras

(1) Let B be an A-algebra; the ring B has canonically a (B, B)-bimodule structure (II, § 1, no. 14, Example 1) and, if e is the unit element of B, then (αe)x=x(αe)=αx(\alpha e)x = x(\alpha e) = \alpha x for all x∈Bx \in B and all α∈A\alpha \in A; B can therefore be considered as a left bimodule over the algebras B and BoB^o (opposite to B); there is therefore associated with the (B, B)-bimodule structure on B a (B ⊗ABo\otimes_A B^o)-module structure such that, for b,xb, x and b′b' in B,

(b⊗b′).x=bxb′ (b \otimes b').x = bx b'

the right-hand side being the product in the ring B.

(2) Let E and F be two A-algebras, e,e′e, e' their respective unit elements, M an E-module and N an F-module; these module structures define on M a bimodule structure over the rings A and E and on N a bimodule structure over the rings A and F; from these is therefore derived a bimodule structure over the rings E and F on the tensor product M⊗ANM \otimes_A N, defined by

x.(m⊗n)=(x.m)⊗n,y.(m⊗n)=m⊗(y.n) x.(m \otimes n) = (x.m) \otimes n, \quad y.(m \otimes n) = m \otimes (y.n)

for x∈E,y∈F,m∈M,n∈Nx \in E, y \in F, m \in M, n \in N (II, § 3, no. 4); it is also seen that conditions (8) hold and hence the above bimodule structure is associated with an (E⊗AF)(E \otimes_A F)-module structure on M⊗ANM \otimes_A N, such that

(x⊗y).(m⊗n)=(x.m)⊗(y.n) (x \otimes y).(m \otimes n) = (x.m) \otimes (y.n)

for x∈E,y∈F,m∈M,n∈Nx \in E, y \in F, m \in M, n \in N.

In particular, taking M=Es,Es⊗ANM = E_s, E_s \otimes_A N has canonically an (E⊗AF)(E \otimes_A F)-module structure; on the other hand, E⊗ANE \otimes_A N is canonically identified with E⊗A(Fd⊗FN)=(E⊗AF)⊗FNE \otimes_A (F_d \otimes_F N) = (E \otimes_A F) \otimes_F N, where E⊗AFE \otimes_A F is considered as having its right F-module structure defined by the canonical homomorphism v:F→E⊗AFv : F \to E \otimes_A F; for x,x′x, x' in E, y∈F,n∈Ny \in F, n \in N, x′⊗nx' \otimes n is thus identified with (x′⊗e′)⊗n(x' \otimes e') \otimes n and (x⊗y).(x′⊗n′)=(xx′)⊗(y.n)(x \otimes y).(x' \otimes n') = (xx') \otimes (y.n) with ((xx′)⊗y)⊗n((xx') \otimes y) \otimes n. The (E⊗AF)(E \otimes_A F)-module Es⊗ANE_s \otimes_A N is thus identified with the (E⊗AF)(E \otimes_A F)-module derived from N by extending the scalars to E⊗AFE \otimes_A F by means of the homomorphism vv (II, § 5, no. 1). The canonical mapping n↦e⊗nn \mapsto e \otimes n of N into Es⊗ANE_s \otimes_A N is identified with the canonical mapping n↦(e⊗e′)⊗nn \mapsto (e \otimes e') \otimes n of N into (E⊗AF)⊗FN(E \otimes_A F) \otimes_F N; this is known to be an F-homomorphism.

With the same notation, let P be a right (E \otimes_A F)-module; then there is a canonical Z-module isomorphism

(11)

P⊗E⊗AF(Es⊗AN)→P⊗FN P \otimes_{E \otimes_A F} (E_s \otimes_A N) \to P \otimes_F N

where on the right-hand side P is considered as a right F-module by means of the canonical homomorphism vv. For P is canonically identified with P⊗E⊗AF(E⊗AF)P \otimes_{E \otimes_A F} (E \otimes_A F) and (E⊗AF)⊗FN(E \otimes_A F) \otimes_F N with E⊗A(F⊗FN)E \otimes_A (F \otimes_F N) and hence with E⊗ANE \otimes_A N, which establishes the stated isomorphism (II, § 3, no. 8, Proposition 8 and II, § 3, no. 4, proposition 4).

All the above extends to multimodules (II, § 1, no. 14).

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

Cited by

Nothing.

Cites

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