Bourbaki

Lemma 1

Tag 0CLH, permanent. Cite it as [Bourbaki, Tag 0CLH].

Commutative Algebra, chapter VIII, § 2. Dimension des algèbres

Let ρ:A→B\rho : A \to B be a homomorphism of rings and p a prime ideal of A. The continuous mapping ah:Spec(B⊗Aκ(p))→Spec(B)^a h : \mathrm{Spec}(B \otimes_A \kappa(p)) \to \mathrm{Spec}(B), associated with the canonical homomorphism h:B→B⊗Aκ(p)h : B \to B \otimes_A \kappa(p), induces a homeomorphism of Spec(B⊗Aκ(p))\mathrm{Spec}(B \otimes_A \kappa(p)) onto the subspace (αρ)−1(p)(\alpha\rho)^{-1}(p) of Spec(B)\mathrm{Spec}(B) formed by the prime ideals of BB lying above pp.

The homomorphism hh is composed of the homomorphism of passing to the quotient of BB into B/ρ(p)B/\rho(p) BB and of the canonical homomorphism of B/ρ(p)B/\rho(p) BB into its ring of fractions (ρ(A−p))−1(B/ρ(p) B)(\rho(A - p))^{-1}(B/\rho(p) \ B). According to the remark and the corollary to prop. 13 of II, § 4, no 3, ah^a h therefore induces a homeomorphism of Spec(B⊗Aκ(p))\mathrm{Spec}(B \otimes_A \kappa(p)) onto the subspace of Spec(B)\mathrm{Spec}(B) formed by the prime ideals qq of BB which contain ρ(p)\rho(p) and are disjoint from ρ(A−p)\rho(A - p), that is to say which lie above pp.

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

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Cites

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