Bourbaki

Corollary 4

Tag 046E, permanent. Cite it as [Bourbaki, Tag 046E].

Commutative Algebra, chapter II, § 2. Rings and modules of fractions

Let A,BA, B be two rings and ρ\rho a homomorphism from AA to BB. (i) Suppose that there exists a BB-module EE such that ρ∗(E)\rho_*(E) is a faithfully flat AA-module. Then, for every prime ideal pp of AA, there exists a prime ideal p′p' of BB such that ρˉ−1(p′)=p\bar{\rho}^{-1}(p') = p. (ii) Conversely, suppose that BB is a flat AA-module. Then, if, for every prime ideal pp of AA, there exists an ideal p′p' of BB such that ρˉ−1(p′)=p\bar{\rho}^{-1}(p') = p, BB is a faithfully flat AA-module.

(i) The hypothesis implies that, for every ideal aa of AA, ρˉ−1(Bρ(a))=a\bar{\rho}^{-1}(B_{\rho}(a)) = a (Chapter I, § 3, no. 5, Proposition 8 (ii)) and it is sufficient to apply Corollary 3. (ii) It is sufficient to show that, for every maximal ideal mm of AA, there exists a maximal ideal m′m' of BB such that ρˉ−1(m′)=m\bar{\rho}^{-1}(m') = m (Chapter I, § 2, no. 5, Proposition 9 (e)). Now there exists by hypothesis an ideal qq of BB such that ρˉ−1(q)=m\bar{\rho}^{-1}(q) = m; as q≠Bq \neq B, there exists a maximal ideal m′m' of BB containing qq and consequently ρˉ−1(m′)⊇m\bar{\rho}^{-1}(m') \supseteq m; but, as ρˉ−1(m′)\bar{\rho}^{-1}(m') cannot contain 1, ρˉ−1(m′)=m\bar{\rho}^{-1}(m') = m.

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

Cited by

Cites

Transcribed from the printed volume. Not translated.