The Stacks project

Proof. Let $C$ be an $A$-algebra of finite type and $\mathfrak q$ a prime of $C$ such that $B = C_{\mathfrak q}$. We may assume $C$ is a domain. We have $\dim _{\mathfrak q}(C/\mathfrak m_ AC) = \dim (B/\mathfrak m_ AB) + \text{trdeg}_{\kappa (\mathfrak m_ A)}(\kappa (\mathfrak m_ B))$, see Morphisms, Lemma 29.28.1. Setting $K$ equal to the fraction field of $A$ we see by the same reference that $\dim (C \otimes _ A K) = \text{trdeg}_ A(B)$. Thus we are really trying to prove that $\dim _{\mathfrak q}(C/\mathfrak m_ AC) = \dim (C \otimes _ A K)$. Choose a valuation ring $A'$ in $K$ dominating $A$, see Algebra, Lemma 10.50.2. Set $C' = C \otimes _ A A'$. Choose a prime $\mathfrak q'$ of $C'$ lying over $\mathfrak q$; such a prime exists because

which proves that $C/\mathfrak m_ AC \to C'/\mathfrak m_{A'}C'$ is faithfully flat. This also proves that $\dim _{\mathfrak q}(C/\mathfrak m_ AC) = \dim _{\mathfrak q'}(C'/\mathfrak m_{A'}C')$, see Algebra, Lemma 10.116.6. Note that $B' = C'_{\mathfrak q'}$ is a localization of $B \otimes _ A A'$. Hence $B'$ is flat over $A'$. The generic fibre $B' \otimes _{A'} K$ is a localization of $B \otimes _ A K$. Hence $B'$ is a domain. If we prove the lemma for $A' \subset B'$, then we get the equality $\dim _{\mathfrak q'}(C'/\mathfrak m_{A'}C') = \dim (C' \otimes _{A'} K)$ which implies the desired equality $\dim _{\mathfrak q}(C/\mathfrak m_ AC) = \dim (C \otimes _ A K)$ by what was said above. This reduces the lemma to the case where $A$ is a valuation ring.

Let $A \subset B$ be as in the lemma with $A$ a valuation ring. As before write $B = C_{\mathfrak q}$ for some domain $C$ of finite type over $A$. By Algebra, Lemma 10.125.9 we obtain $\dim (C/\mathfrak m_ AC) = \dim (C \otimes _ A K)$ and we win. $\square$