The Stacks project

Proof. In this paragraph we reduce to the case where $R \to S$ is of finite presentation. Namely, write $S = R[x_1, \ldots , x_ n]/J$ for some ideal $J \subset R[x_1, \ldots , x_ n]$. Then $J$ is the union of its finitely generated ideals $J_\lambda \subset J$. Set $S_\lambda = R[x_1, \ldots , x_ n]/J_\lambda $. Then for some $\lambda $ the fibres of $\mathop{\mathrm{Spec}}(S_\lambda ) \to \mathop{\mathrm{Spec}}(R)$ have dimension $\leq d$, see Limits, Lemma 32.18.1. Fix such a $\lambda $. If we can find a quasi-finite $R[t_1, \ldots , t_ d] \to S_\lambda $, then of course the composition $R[t_1, \ldots , t_ d] \to S$ is quasi-finite. This reduces us to the case discussed in the next paragraph.

Assume $R \to S$ is of finite presentation. In this paragraph we reduce to the case where $R$ is of finite type over $\mathbf{Z}$. By Algebra, Lemma 10.127.18 we can find a directed set $\Lambda $ and a system of ring maps $R_\lambda \to S_\lambda $ over $\Lambda $ whose colimit is $R \to S$ such that $S_\mu = S_\lambda \otimes _{R_\lambda } R_\mu $ for $\mu \geq \lambda $ and such that each $R_\lambda $ and $S_\lambda $ is of finite type over $\mathbf{Z}$. Then for $\lambda $ large enough the fibres of $\mathop{\mathrm{Spec}}(S_\lambda ) \to \mathop{\mathrm{Spec}}(R_\lambda )$ have dimension $\leq d$, see Limits, Lemma 32.18.4. Fix such a $\lambda $. If we can find a quasi-finite ring map $R_\lambda [t_1, \ldots , t_ d] \to S_\lambda $, then the base change $R[t_1, \ldots , t_ d] \to S$ is quasi-finite too (Algebra, Lemma 10.122.8). This reduces us the the case discussed in the next paragraph.

Assume $R$ and $S$ are of finite type over $\mathbf{Z}$. If $d = 0$, then the ring map is quasi-finite and we are done. Assume $d > 0$. We will find an element $a \in S$ such that the fibres of the $R$-algebra map $R[x] \to S$, $x \mapsto a$ have dimension $< d$. This will finish the proof by induction.

We will prove the existence of $a$ by induction on $\dim (R)$.

Let $\mathfrak q_1, \ldots , \mathfrak q_ r \subset S$ be those among the minimal primes of $S$ such that $\dim _{\mathfrak q_ i}(S/R) = d$. For notation, see Algebra, Definition 10.125.1. Say $\mathfrak q_ i$ lies over the prime $\mathfrak p_ i \subset R$. We have $\text{trdeg}_{\kappa (\mathfrak p_ i)}(\kappa (\mathfrak q_ i)) = d$ as $\mathfrak q_ i$ is a generic point of its fibre; for example apply Algebra, Lemma 10.116.3 to $S \otimes _ R \kappa (\mathfrak p_ i)$. Hence by Lemma 37.31.1 we can find an element $a \in S$ such that the image of $a$ in $\kappa (\mathfrak q_ i)$ is transcendental over $\kappa (\mathfrak p_ i)$ for $i = 1, \ldots , r$. Consider the morphism

corresponding the $R$-algebra homomorphism $R[x] \to S$ to mapping $x$ to $a$. Let $U \subset \mathop{\mathrm{Spec}}(S)$ be the open subset where the fibres have dimension $\leq d - 1$, see Morphisms, Lemma 29.28.4. By construction $U$ contains all the generic points of $\mathop{\mathrm{Spec}}(S)$. In particular we see that $U$ contains all generic points of all the generic fibres of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ as such points are necessarily generic points of $\mathop{\mathrm{Spec}}(S)$. Set $T = \mathop{\mathrm{Spec}}(S) \setminus U$ viewed as a reduced closed subscheme of $\mathop{\mathrm{Spec}}(S)$. It follows from what we just said and the assumption that $\dim (S/R) \leq d$ that the generic fibres of $T \to \mathop{\mathrm{Spec}}(R)$ have dimension $\leq d - 1$. Hence by Lemma 37.30.1, applied several times to produce open neighbourhoods of the generic points of $\mathop{\mathrm{Spec}}(R)$, we can find a dense open $V \subset \mathop{\mathrm{Spec}}(R)$ such that $T_ V \to V$ has fibres of dimension $\leq d - 1$. We conclude that for $q \in \mathbf{A}^1_ V$ the fibre of $f_ a$ over $q$ has dimension $\leq d - 1$ (as we have bounded the dimension of the fibre of $f_ a|_ U$ and of the fibre of $f_ a|_ T$).

By prime avoidance, we may assume that $V = D(f)$ for some $f \in R$. Then we see that the ring map $R_ f[x] \to S_ f$, $x \mapsto a$ has fibres of dimension $\leq d - 1$. We may replace $a$ by $fa$ and assume $a \in (f)$. By induction on $\dim (R)$ we can find an element $\overline{b} \in S/fS$ such that the fibres of $\mathop{\mathrm{Spec}}(S/fS) \to \mathop{\mathrm{Spec}}(R/fR[x])$, $x \mapsto \overline{b}$ have dimension $\leq d - 1$. Let $b \in S$ be a lift of $\overline{b}$. By Lemma 37.31.2 there exists an $n > 0$ such that $a^ n + b$ still works for $R_ f \to S_ f$. On the other hand, the image of $a^ n + b$ in $S/fS$ is $\overline{b}$ and the proof is complete. $\square$