Proposition 5

Tag 061P, permanent. Cite it as [Bourbaki, Tag 061P].

Algebra, chapter III, § 5. Tensor algebra. Tensors

Let $A, B$ be two commutative rings, $\rho : A \to B$ a ring homomorphism and $M$ an $A$ -module. The canonical extension

\psi : T_B(B \otimes_A M) \to B \otimes_A T_A(M)

of the B-linear mapping $l_B \otimes \phi_M : B \otimes_A M \to B \otimes_A T_A(M)$ is an isomorphism of graded B-algebras.

Consider the two $A$ -algebra homomorphisms: the canonical injection $j : B = T^0(B \otimes_A M) \to T(B \otimes_A M)$ and the homomorphism

h = T(i) : T(M) \to T(B \otimes_A M)

derived (cf. formula (8)) from the canonical $A$ -linear mapping

i : M \to B \otimes_A M.

As $T^0(B \otimes_A M)$ is contained in the centre of $T(B \otimes_A M)$ , Proposition 5 of § 4, no. 2 can be applied and an A-algebra homomorphism

\psi': B \otimes_A T(M) \to T(B \otimes_A M)

is obtained such that, for $\beta \in B, x_i \in M$ for $1 \leq i \leq n$ ,

\psi'(\beta \otimes (x_1 \otimes x_2 \otimes \cdots \otimes x_n)) = \beta((1 \otimes x_1) \otimes (1 \otimes x_2) \otimes \cdots \otimes (1 \otimes x_n)),

which shows immediately that $\psi'$ is also a B-algebra homomorphism. It suffices to prove that $\psi \circ \psi'$ and $\psi' \circ \psi$ are the identity mappings on $B \otimes_A T(M)$ and $T(B \otimes_A M)$ respectively. Now, these two algebras are generated by $B \otimes_A M$ and clearly $\psi \circ \psi'$ and $\psi' \circ \psi$ coincide with the identity mapping on $B \otimes_A M$ , whence the conclusion.

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

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Cites

Proposition 5 of § 4, no. 2