The Stacks project

Proof. Note that the two assertions are equivalent by Algebra, Lemma 10.30.6. To start the proof let $C$ be an $A$-algebra of finite type and $\mathfrak q$ a prime of $C$ such that $B = C_{\mathfrak q}$. Of course we may assume that $C$ is a domain and that $g \in C$. After replacing $C$ by a localization we see that $\dim (C/\mathfrak m_ AC) = \dim (B/\mathfrak m_ AB) + \text{trdeg}_{\kappa (\mathfrak m_ A)}(\kappa (\mathfrak m_ B))$, see Morphisms, Lemma 29.28.1. Setting $K$ equal to the fraction field of $A$ we see by the same reference that $\dim (C \otimes _ A K) = \text{trdeg}_ A(B)$. Hence assumption (4) means that the generic and closed fibres of the morphism $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ have the same dimension.

Suppose that the lemma is false. Then $(B/gB) \otimes _ A K = 0$. This means that $g \otimes 1$ is invertible in $B \otimes _ A K = C_{\mathfrak q} \otimes _ A K$. As $C_{\mathfrak q}$ is a limit of principal localizations we conclude that $g \otimes 1$ is invertible in $C_ h \otimes _ A K$ for some $h \in C$, $h \not\in \mathfrak q$. Thus after replacing $C$ by $C_ h$ we may assume that $(C/gC) \otimes _ A K = 0$. We do one more replacement of $C$ to make sure that the minimal primes of $C/\mathfrak m_ AC$ correspond one-to-one with the minimal primes of $B/\mathfrak m_ AB$. At this point we apply Lemma 37.52.1 to $X = \mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A) = S$ and the locally closed subscheme $Z = \mathop{\mathrm{Spec}}(C/gC)$. Since $Z_ K = \emptyset $ we see that $Z \otimes \kappa (\mathfrak m_ A)$ has to contain an irreducible component of $X \otimes \kappa (\mathfrak m_ A) = \mathop{\mathrm{Spec}}(C/\mathfrak m_ AC)$. But this contradicts the assumption that $g$ is not contained in any prime minimal over $\mathfrak m_ AB$. The lemma follows. $\square$