The Stacks project

Proof. Suppose that $R \subset S' \subset S$ is a finitely generated $R$-subalgebra of $S$. In this case set $\mathfrak q' = S' \cap \mathfrak q$. The lemma for the ring maps $R \to S'$ and $S' \to S$ implies the lemma for $R \to S$ by additivity of transcendence degree in towers of fields (Fields, Lemma 9.26.5). Hence we can use induction on the number of generators of $S$ over $R$ and reduce to the case where $S$ is generated by one element over $R$.

Case I: $S = R[x]$ is a polynomial algebra over $R$. In this case we have $\text{trdeg}_ R(S) = 1$. Also $R \to S$ is flat and hence

see Lemma 10.112.7. Let $\mathfrak r = \mathfrak pS$. Then $\text{trdeg}_{\kappa (\mathfrak p)} \kappa (\mathfrak q) = 1$ is equivalent to $\mathfrak q = \mathfrak r$, and implies that $\dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$. In the same vein $\text{trdeg}_{\kappa (\mathfrak p)} \kappa (\mathfrak q) = 0$ is equivalent to having a strict inclusion $\mathfrak r \subset \mathfrak q$, which implies that $\dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 1$. Thus we are done with case I with equality in every instance.

Case II: $S = R[x]/\mathfrak n$ with $\mathfrak n \not= 0$. In this case we have $\text{trdeg}_ R(S) = 0$. Denote $\mathfrak q' \subset R[x]$ the prime corresponding to $\mathfrak q$. Thus we have

By the previous case we have $\dim ((R[x])_{\mathfrak q'}) = \dim (R_{\mathfrak p}) + 1 - \text{trdeg}_{\kappa (\mathfrak p)} \kappa (\mathfrak q)$. Since $\mathfrak n \not= 0$ we see that the dimension of $S_{\mathfrak q}$ decreases by at least one, see Lemma 10.60.13, which proves the inequality of the lemma. To see the equality in case $R$ is universally catenary note that $\mathfrak n \subset R[x]$ is a height one prime as it corresponds to a nonzero prime in $F[x]$ where $F$ is the fraction field of $R$. Hence any maximal chain of primes in $S_\mathfrak q = R[x]_{\mathfrak q'}/\mathfrak nR[x]_{\mathfrak q'}$ corresponds to a maximal chain of primes with length 1 greater between $\mathfrak q'$ and $(0)$ in $R[x]$. If $R$ is universally catenary these all have the same length equal to the height of $\mathfrak q'$. This proves that $\dim (S_\mathfrak q) = \dim (R[x]_{\mathfrak q'}) - 1$ and this implies equality holds as desired. $\square$

Proof. By the dimension formula (Lemma 10.113.1) for any prime $\mathfrak q$ lying over $\mathfrak p$ we have

As the domain $B_\mathfrak q$ has at least $2$ prime ideals we see that $\dim (B_{\mathfrak q}) \geq 1$. We conclude that $\dim (B_{\mathfrak q}) = 1$ and that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is algebraic. Hence $\mathfrak q$ defines a closed point of its fibre $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$, see Lemma 10.35.9. Since $B \otimes _ A \kappa (\mathfrak p)$ is a Noetherian ring the fibre $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$ is a Noetherian topological space, see Lemma 10.31.5. A Noetherian topological space consisting of closed points is finite, see for example Topology, Lemma 5.9.2. $\square$