Bourbaki

Proposition 3

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

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

Let ρ:A→B\rho : A \to B be a homomorphism of rings. Put n=sup⁡p∈Spec(A)dim⁡(B⊗Aκ(p))n = \sup_{p \in \mathrm{Spec}(A)} \dim(B \otimes_A \kappa(p)). One has the inequality

dim⁡(B)+1≤(dim⁡(A)+1).(n+1). \dim(B) + 1 \leq (\dim(A) + 1).(n + 1) .

We may suppose dim⁡(A)≠−∞\dim(A) \neq -\infty and n<+∞n < +\infty. Let q0⊂...⊂qmq_0 \subset ... \subset q_m be a chain of prime ideals of B; put pi=ρ−1(qi)p_i = \rho^{-1}(q_i). The sequence of the pip_i is increasing, hence the set of its values has cardinal ≤dim⁡(A)+1\leq \dim(A) + 1. For each p∈Spec(A)p \in \mathrm{Spec}(A), the set of the qjq_j such that pj=pp_j = p is a chain of the subset aρ−1(p)^a\rho^{-1}(p) of Spec(B)\mathrm{Spec}(B), hence has cardinal less than dim⁡(B⊗Aκ(p))+1\dim(B \otimes_A \kappa(p)) + 1 (no 1, Lemma 1), and consequently less than or equal to (n+1)(n + 1). It follows that m+1≤(dim⁡(A)+1)(n+1)m + 1 \leq (\dim(A) + 1)(n + 1), whence the proposition.

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

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