Proposition 4

Tag 04BR, permanent. Cite it as [Bourbaki, Tag 04BR].

Commutative Algebra, chapter II, § 5. Finitely generated projective modules. Invertible fractional ideals

Let B be a commutative A-algebra and P a projective A-module of rank n. The B-module $P_{(B)} = B \otimes_A P$ is then projective of rank n.

We know that $P_{(B)}$ is projective and finitely generated (Algebra, Chapter II, § 5, no. 1, Corollary to Proposition 4). If q is a prime ideal of B and p its inverse image in A, then

(P_{(B)})_q = (P \otimes_A B) \otimes_B B_q = P @_A B_q = (P \otimes_A A_i) \otimes_A B_q

and, as, by hypothesis, $P \otimes_A A_i$ is a free A,-module of rank $n$ , $(P_{(B)})_q$ is a free $B_q$ -module of rank $n$ .

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

Cited by

Exercise 9, § 5
Proposition 5, § 5
Proposition 7, § 5
Theorem 3, § 5

Cites

*Algebra*, Chapter II, § 5, no. 1, Corollary to Proposition 4