Lemma 2

Let ρ : A → B be a homomorphism of rings such that, for every prime ideal p of A, the κ(p)-algebra B ⊗ₓₐ κ(p) is integral (V, § 1, No. 1, def. 2). Let q and q’ be two prime ideals of B such that q ⊂ q’ and q ≠ q’. Then ρ⁻¹(q) ≠ ρ⁻¹(q’).

Indeed, if q and q’ are over the same prime ideal p of A, one has dim(B ⊗ₓₐ κ(p)) ≥ 1 by Lemma 1 of No. 1, which contradicts the fact that dim(B ⊗ₓₐ κ(p)) ≤ 0 (§ 1, No. 3, example 6).