Letρ:A→Bbe a homomorphism of rings. Putn=supp∈Spec(A)dim(B⊗Aκ(p)). One has the inequality
dim(B)+1≤(dim(A)+1).(n+1).
We may suppose dim(A)=−∞ and n<+∞. Let q0⊂...⊂qm be a chain of prime ideals of B; put pi=ρ−1(qi). The sequence of the pi is increasing, hence the set of its values has cardinal ≤dim(A)+1. For each p∈Spec(A), the set of the qj such that pj=p is a chain of the subset aρ−1(p) of Spec(B), hence has cardinal less than dim(B⊗Aκ(p))+1 (no 1, Lemma 1), and consequently less than or equal to (n+1). It follows that m+1≤(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.