The Stacks project

Proof. The ideal $J_ i = J \mathfrak p_1 \ldots \hat{\mathfrak p}_ i \ldots \mathfrak p_ r$ is not contained in $\mathfrak p_ i$, see Algebra, Lemma 10.15.1. It follows that every element $\xi $ of $\kappa (\mathfrak p_ i) = \text{Frac}(B/\mathfrak p_ i)$ is of the form $\xi = a/b$ with $a, b \in J_ i$ and $b \not\in \mathfrak p_ i$. Choosing $\xi $ transcendental over $k_ i$ we see that either $a$ or $b$ maps to an element of $\kappa (\mathfrak p_ i)$ transcendental over $k_ i$. We conclude that for every $i = 1, \ldots , r$ we can find an element $x_ i \in J_ i = J \mathfrak p_1 \ldots \hat{\mathfrak p}_ i \ldots \mathfrak p_ r$ which maps to an element of $\kappa (\mathfrak p_ i)$ transcendental over $k_ i$. Then $x = x_1 + \ldots + x_ r$ works. $\square$

Proof. In this paragraph we reduce to the case where $R \to S$ is of finite presentation. Namely, write $S = R[A, B, x_1, \ldots , x_ n]/J$ for some ideal $J \subset R[x_1, \ldots , x_ n]$ where $A$ and $B$ map to $a$ and $b$ in $S$. Then $J$ is the union of its finitely generated ideals $J_\lambda \subset J$. Set $S_\lambda = R[A, B, x_1, \ldots , x_ n]/J_\lambda $ and denote $a_\lambda , b_\lambda \in S_\lambda $ the images of $A$ and $B$. Then for some $\lambda $ the fibres of

have dimension $\leq d$, see Limits, Lemma 32.18.1. Fix such a $\lambda $. If we can find $n_0$ which works for $R \to S_\lambda $, $a_\lambda $, $b_\lambda $, then $n_0$ works for $R \to S$. Namely, the fibres of $f_{a_\lambda ^ n + b_\lambda } : \mathop{\mathrm{Spec}}(S_\lambda ) \to \mathbf{A}^1_ R$ contain the fibres of $f_{a^ n + b} : \mathop{\mathrm{Spec}}(S) \to \mathbf{A}^1_ R$. 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 $\mathbf{Z}$. Choose $\lambda _0 \in \Lambda $ and elements $a_{\lambda _0}, b_{\lambda _0} \in S_{\lambda _0}$ mapping to $a, b \in S$. For $\lambda \geq \lambda _0$ denote $a_\lambda , b_\lambda \in S_\lambda $ the image of $a_{\lambda _0}, b_{\lambda _0}$. Then for $\lambda \geq \lambda _0$ large enough the fibres of

have dimension $\leq d$, see Limits, Lemma 32.18.4. Fix such a $\lambda $. If we can find $n_0$ which works for $R_\lambda \to S_\lambda $, $a_\lambda $, $b_\lambda $, then $n_0$ works for $R \to S$. Namely, any fibre of $f_{a^ n + b} : \mathop{\mathrm{Spec}}(S) \to \mathbf{A}^1_ R$ has the same dimension as a fibre of $f_{a_\lambda ^ n + b_\lambda } : \mathop{\mathrm{Spec}}(S_\lambda ) \to \mathbf{A}^1_{R_\lambda }$ by Morphisms, Lemma 29.28.3. This reduces us the the case discussed in the next paragraph.

Assume $R$ and $S$ are of finite type over $\mathbf{Z}$. In particular the dimension of $R$ is finite, and we may use induction on $\dim (R)$. Thus we may assume the result holds for all situations with $R' \to S'$, $a$, $b$ as in the lemma with $R'$ and $S'$ of finite type over $\mathbf{Z}$ but with $\dim (R') < \dim (R)$.

Since the statement is about the topology of the spectrum of $S$ we may assume $S$ is reduced. Let $S^\nu $ be the normalization of $S$. Then $S \subset S^\nu $ is a finite extension as $S$ is excellent, see Algebra, Proposition 10.162.16 and Morphisms, Lemma 29.54.11. Thus $\mathop{\mathrm{Spec}}(S^\nu ) \to \mathop{\mathrm{Spec}}(S)$ is surjective and finite (Algebra, Lemma 10.36.17). It follows that if the result holds for $R \to S^\nu $ and the images of $a$, $b$ in $S^\nu $, then the result holds for $R \to S$, $a$, $b$. (Small detail omitted.) This reduces us to the case discussed in the next paragraph.

Assume $R$ and $S$ are of finite type over $\mathbf{Z}$ and $S$ normal. Then $S = S_1 \times \ldots \times S_ r$ for some normal domains $S_ i$. If the result holds for each $R \to S_ i$ and the images of $a$, $b$ in $S_ i$, then the result holds for $R \to S$, $a$, $b$. (Small detail omitted.) This reduces us to the case discussed in the next paragraph.

Assume $R$ and $S$ are of finite type over $\mathbf{Z}$ and $S$ a normal domain. We may replace $R$ by the image of $R$ in $S$ (this does not increase the dimension of $R$). This reduces us to the case discussed in the next paragraph.

Assume $R \subset S$ are of finite type over $\mathbf{Z}$ and $S$ a normal domain. Consider the morphism

The assumption tells us that $f_ a$ has fibres of dimension $\leq d$. Hence the fibres of $f : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ have dimension $\leq d + 1$ (Morphisms, Lemma 29.28.2). Consider the morphism of integral schemes

corresponding to the $R$-algebra map $R[x, y] \to S$ sending $x$ to $a$ and $y$ to $b$. There are two cases to consider

1. $\phi $ is dominant, and

2. $\phi $ is not dominant.

We claim that in both cases there exists an integer $n_0$ and a nonempty open $V \subset \mathop{\mathrm{Spec}}(R)$ such that for $n \geq n_0$ the fibres of $f_{a^ n + b}$ at points $q \in \mathbf{A}^1_ V$ have dimension $\leq d$.

Proof of the claim in case (1). We have $f_{a^ n + b} = \pi _ n \circ \phi $ where

is the flat morphism corresponding to the $R$-algebra map $R[x] \to R[x, y]$ sending $x$ to $x^ n + y$. Since $\phi $ is dominant there is a dense open $U \subset \mathop{\mathrm{Spec}}(S)$ such that $\phi |_ U : U \to \mathbf{A}^2_ R$ is flat (this follows for example from generic flatness, see Morphisms, Proposition 29.27.1). Then the composition

is flat as well. Hence the fibres of this morphism have at least codimension $1$ in the fibres of $f|_ U : U \to \mathop{\mathrm{Spec}}(R)$ by Morphisms, Lemma 29.28.2. In other words, the fibres of $f_{a^ n + b}|_ U$ have dimension $\leq d$. On the other hand, since $U$ is dense in $\mathop{\mathrm{Spec}}(S)$, we can find a nonempty open $V \subset \mathop{\mathrm{Spec}}(R)$ such that $U \cap f^{-1}(p) \subset f^{-1}(p)$ is dense for all $p \in V$ (see for example Lemma 37.24.3). Thus $\dim (f^{-1}(p) \setminus U \cap f^{-1}(p)) \leq d$ and we conclude that our claim is true (as any fibres of $f_{a^ n + b} : \mathop{\mathrm{Spec}}(S) \to \mathbf{A}^1_ R$ is contained in a fibre of $f$).

Case (2). In this case we can find a nonzero $g = \sum c_{ij} x^ i y^ j$ in $R[x, y]$ such that $\mathop{\mathrm{Im}}(\phi ) \subset V(g)$. In fact, we may assume $g$ is irreducible over $\text{Frac}(R)$. If $g \in R[x]$, say with leading coefficient $c$, then over $V = D(c) \subset \mathop{\mathrm{Spec}}(R)$ the fibres of $f$ already have dimension $\leq d$ (because the image of $f_ a$ is contained in $V(g) \subset \mathbf{A}^1_ R$ which has finite fibres over $V$). Hence we may assume $g$ is not contained in $R[x]$. Let $s \geq 1$ be the degree of $g$ as a polynomial in $y$ and let $t$ be the degree of $\sum c_{is} x^ i$ as a polynomial in $x$. Then $c_{ts}$ is nonzero and

provided that $n$ is bigger than the degree of $g$ as a polynomial in $x$ (small detail omitted). For such $n$ the polynomial $g(x, -x^ n)$ is a nonzero polynomial in $x$ and maps to a nonzero polynomial in $\kappa (\mathfrak p)[x]$ for all $\mathfrak p \subset R$, $c_{st} \not\in \mathfrak p$. We conclude that our claim is true for $V$ equal to the principal open $D(c_{ts})$ of $\mathop{\mathrm{Spec}}(R)$.

OK, and now we can use induction on $\dim (R)$. Namely, let $I \subset R$ be an ideal such that $V(I) = \mathop{\mathrm{Spec}}(R) \setminus V$. Observe that $\dim (R/I) < \dim (R)$ as $R$ is a domain. Let $n'_0$ be the integer we have by induction on $\dim (R)$ for $R/I \to S/IS$ and the images of $a$ and $b$ in $S/IS$. Then $\max (n_0, n'_0)$ works. $\square$

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. $\square$