The Stacks project

Proof. Let $X_{red}$ denote the reduction of $X$. Then $Z \cap X_{red}$ is a locally principal closed subscheme of $X_{red}$, see Divisors, Lemma 31.13.11. Hence we may assume that $X$ is reduced. In other words $X$ is integral, see Properties, Lemma 28.3.4. In this case the morphism $X \to S$ factors through $S_{red}$, see Schemes, Lemma 26.12.7. Thus we may replace $S$ by $S_{red}$ and assume that $S$ is integral too.

The assertion that $f$ is dominant signifies that the generic point of $X$ is mapped to $\eta $, see Morphisms, Lemma 29.8.6. Moreover, the scheme $X_\eta $ is an integral scheme which is locally of finite type over the field $\kappa (\eta )$. Hence $d = \dim (X_\eta ) \geq 0$ is equal to $\dim _\xi (X_\eta )$ for every point $\xi $ of $X_\eta $, see Algebra, Lemmas 10.114.4 and 10.114.5. In view of Morphisms, Lemma 29.28.4 and condition (5) we conclude that $\dim _ x(X_ s) = d$ for every $x \in X_ s$.

In the Noetherian case the assertion can be proved as follows. If the lemma does not holds there exists $x \in Z_ s$ which is a generic point of an irreducible component of $Z_ s$ but not a generic point of any irreducible component of $X_ s$. Then we see that $\dim _ x(Z_ s) \leq d - 1$, because $\dim _ x(X_ s) = d$ and in a neighbourhood of $x$ in $X_ s$ the closed subscheme $Z_ s$ does not contain any of the irreducible components of $X_ s$. Hence after replacing $X$ by an open neighbourhood of $x$ we may assume that $\dim _ z(Z_{f(z)}) \leq d - 1$ for all $z \in Z$, see Morphisms, Lemma 29.28.4. Let $\xi ' \in Z$ be a generic point of an irreducible component of $Z$ and set $s' = f(\xi )$. As $Z \not= X$ is locally principal we see that $\dim (\mathcal{O}_{X, \xi }) = 1$, see Algebra, Lemma 10.60.11 (this is where we use $X$ is Noetherian). Let $\xi \in X$ be the generic point of $X$ and let $\xi _1$ be a generic point of any irreducible component of $X_{s'}$ which contains $\xi '$. Then we see that we have the specializations

As $\dim (\mathcal{O}_{X, \xi }) = 1$ one of the two specializations has to be an equality. By assumption $s' \not= \eta $, hence the first specialization is not an equality. Hence $\xi ' = \xi _1$ is a generic point of an irreducible component of $X_{s'}$. Applying Morphisms, Lemma 29.28.4 one more time this implies $\dim _{\xi '}(Z_{s'}) = \dim _{\xi '}(X_{s'}) \geq \dim (X_\eta ) = d$ which gives the desired contradiction.

In the general case we reduce to the Noetherian case as follows. If the lemma is false then there exists a point $x \in X$ lying over $s$ such that $x$ is a generic point of an irreducible component of $Z_ s$, but not a generic point of any of the irreducible components of $X_ s$. Let $U \subset S$ be an affine neighbourhood of $s$ and let $V \subset X$ be an affine neighbourhood of $x$ with $f(V) \subset U$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$ so that $f|_ V$ is given by a ring map $A \to B$. Let $\mathfrak q \subset B$, resp. $\mathfrak p \subset A$ be the prime corresponding to $x$, resp. $s$. After possibly shrinking $V$ we may assume $Z \cap V$ is cut out by some element $g \in B$. Denote $K$ the fraction field of $A$. What we know at this point is the following:

1. $A \subset B$ is a finitely generated extension of domains,

2. the element $g \otimes 1$ is invertible in $B \otimes _ A K$,

3. $d = \dim (B \otimes _ A K) = \dim (B \otimes _ A \kappa (\mathfrak p))$,

4. $g \otimes 1$ is not a unit of $B \otimes _ A \kappa (\mathfrak p)$, and

5. $g \otimes 1$ is not in any of the minimal primes of $B \otimes _ A \kappa (\mathfrak p)$.

We are seeking a contradiction.

Pick elements $x_1, \ldots , x_ n \in B$ which generate $B$ over $A$. For a finitely generated $\mathbf{Z}$-algebra $A_0 \subset A$ let $B_0 \subset B$ be the $A_0$-subalgebra generated by $x_1, \ldots , x_ n$, denote $K_0$ the fraction field of $A_0$, and set $\mathfrak p_0 = A_0 \cap \mathfrak p$. We claim that when $A_0$ is large enough then (1) – (5) also hold for the system $(A_0 \subset B_0, g, \mathfrak p_0)$.

We prove each of the conditions in turn. Part (1) holds by construction. For part (2) write $(g \otimes 1) h = 1$ for some $h \otimes 1/a \in B \otimes _ A K$. Write $g = \sum a_ I x^ I$, $h = \sum a'_ I x^ I$ (multi-index notation) for some coefficients $a_ I, a'_ I \in A$. As soon as $A_0$ contains $a$ and the $a_ I, a'_ I$ then (2) holds because $B_0 \otimes _{A_0} K_0 \subset B \otimes _ A K$ (as localizations of the injective map $B_0 \to B$). To achieve (3) consider the exact sequence

which defines $I$ where the second map sends $X_ i$ to $x_ i$. Since $\otimes $ is right exact we see that $I \otimes _ A K$, respectively $I \otimes _ A \kappa (\mathfrak p)$ is the kernel of the surjection $K[X_1, \ldots , X_ n] \to B \otimes _ A K$, respectively $\kappa (\mathfrak p)[X_1, \ldots , X_ n] \to B \otimes _ A \kappa (\mathfrak p)$. As a polynomial ring over a field is Noetherian there exist finitely many elements $h_ j \in I$, $j = 1, \ldots , m$ which generate $I \otimes _ A K$ and $I \otimes _ A \kappa (\mathfrak p)$. Write $h_ j = \sum a_{j, I}X^ I$. As soon as $A_0$ contains all $a_{j, I}$ we get to the situation where

By either Morphisms, Lemma 29.28.3 or Algebra, Lemma 10.116.5 we see that the dimension equalities of (3) are satisfied. Part (4) is immediate. As $B_0 \otimes _{A_0} \kappa (\mathfrak p_0) \subset B \otimes _ A \kappa (\mathfrak p)$ each minimal prime of $B_0 \otimes _{A_0} \kappa (\mathfrak p_0)$ lies under a minimal prime of $B \otimes _ A \kappa (\mathfrak p)$ by Algebra, Lemma 10.30.6. This implies that (5) holds. In this way we reduce the problem to the Noetherian case which we have dealt with above. $\square$

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. $\square$

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. $\square$

Proof. We apply Schemes, Lemma 26.20.4 to the specialization $x \leadsto x'$. This produces a morphism $\varphi : \mathop{\mathrm{Spec}}(B) \to X$ where $B$ is a valuation ring such that $\varphi $ maps the generic point to $x$ and the closed point to $x'$. We may also assume that $\kappa (x)$ is the fraction field of $B$. Let $A = B \cap \kappa (s)$. Note that this is a valuation ring (see Algebra, Lemma 10.50.7) which dominates the image of $\mathcal{O}_{S, s'} \to \kappa (s)$. Consider the commutative diagram

The generic (resp. closed) point of $B$ maps to a point $x_ A$ (resp. $x'_ A$) of $X_ A$ lying over the generic (resp. closed) point of $\mathop{\mathrm{Spec}}(A)$. Note that $x'_ A$ is a closed point of the special fibre of $X_ A$ by Morphisms, Lemma 29.20.4. Note that the generic fibre of $X_ A \to \mathop{\mathrm{Spec}}(A)$ is isomorphic to $X_ s$. Thus we have reduced the lemma to the case where $S$ is the spectrum of a valuation ring, $s = \eta \in S$ is the generic point, and $s' \in S$ is the closed point.

We will prove the lemma by induction on $\dim _ x(X_\eta )$. If $\dim _ x(X_\eta ) = 0$, then there are no other points of $X_\eta $ specializing to $x$ and $x$ is closed in its fibre, see Morphisms, Lemma 29.20.6, and the result holds. Assume $\dim _ x(X_\eta ) > 0$.

Let $X' \subset X$ be the reduced induced scheme structure on the irreducible closed subscheme $\overline{\{ x\} }$ of $X$, see Schemes, Definition 26.12.5. To prove the lemma we may replace $X$ by $X'$ as this only decreases $\dim _ x(X_\eta )$. Hence we may also assume that $X$ is an integral scheme and that $x$ is its generic point. In addition, we may replace $X$ by an affine neighbourhood of $x'$. Thus we have $X = \mathop{\mathrm{Spec}}(B)$ where $A \subset B$ is a finite type extension of domains. Note that in this case $\dim _ x(X_\eta ) = \dim (X_\eta ) = \dim (X_{s'})$, and that in fact $X_{s'}$ is equidimensional, see Algebra, Lemma 10.125.9.

Let $W \subset X_\eta $ be a proper closed subset (this is the subset we want to “avoid”). As $X_ s$ is of finite type over a field we see that $W$ has finitely many irreducible components $W = W_1 \cup \ldots \cup W_ n$. Let $\mathfrak q_ j \subset B$, $j = 1, \ldots , r$ be the corresponding prime ideals. Let $\mathfrak q \subset B$ be the maximal ideal corresponding to the point $x'$. Let $\mathfrak p_1, \ldots , \mathfrak p_ s \subset B$ be the minimal primes lying over $\mathfrak m_ AB$. There are finitely many as these correspond to the irreducible components of the Noetherian scheme $X_{s'}$. Moreover, each of these irreducible components has dimension $> 0$ (see above) hence we see that $\mathfrak p_ i \not= \mathfrak q$ for all $i$. Now, pick an element $g \in \mathfrak q$ such that $g \not\in \mathfrak q_ j$ for all $j$ and $g \not\in \mathfrak p_ i$ for all $i$, see Algebra, Lemma 10.15.2. Denote $Z \subset X$ the locally principal closed subscheme defined by $g$. Let $Z_\eta = Z_{1, \eta } \cup \ldots \cup Z_{n, \eta }$, $n \geq 0$ be the decomposition of the generic fibre of $Z$ into irreducible components (finitely many as the generic fibre is Noetherian). Denote $Z_ i \subset X$ the closure of $Z_{i, \eta }$. After replacing $X$ by a smaller affine neighbourhood we may assume that $x' \in Z_ i$ for each $i = 1, \ldots , n$. By construction $Z \cap X_{s'}$ does not contain any irreducible component of $X_{s'}$. Hence by Lemma 37.52.1 we conclude that $Z_\eta \not= \emptyset $! In other words $n \geq 1$. Letting $x_1 \in Z_1$ be the generic point we see that $x_1 \leadsto x'$ and $f(x_1) = \eta $. Also, by construction $Z_{1, \eta } \cap W_ j \subset W_ j$ is a proper closed subset. Hence every irreducible component of $Z_{1, \eta } \cap W_ j$ has codimension $\geq 2$ in $X_\eta $ whereas $\text{codim}(Z_{1, \eta }, X_\eta ) = 1$ by Algebra, Lemma 10.60.11. Thus $W \cap Z_{1, \eta }$ is a proper closed subset. At this point we see that the induction hypothesis applies to $Z_1 \to S$ and the specialization $x_1 \leadsto x'$. This produces a closed point $x_2$ of $Z_{1, \eta }$ not contained in $W$ which specializes to $x'$. Thus we obtain $x \leadsto x_2 \leadsto x'$, the point $x_2$ is closed in $X_\eta $, and $x_2 \not\in W$ as desired. $\square$

Proof. Since $A$ is Noetherian and $A \to B$ is essentially of finite type, we know that $B$ is Noetherian too. By Properties, Lemma 28.6.4 the topological space $\mathop{\mathrm{Spec}}(B) \setminus \{ \mathfrak m_ B\} $ is Jacobson. Hence we can choose a closed point $\mathfrak q$ which is contained in the nonempty open

(Nonempty by assumption, open because $\{ \mathfrak m_ A\} $ is a closed subset of $\mathop{\mathrm{Spec}}(A)$.) Then $\mathop{\mathrm{Spec}}(B/\mathfrak q)$ has two points, namely $\mathfrak m_ B$ and $\mathfrak q$ and $\mathfrak q$ does not lie over $\mathfrak m_ A$. Write $B/\mathfrak q = C_{\mathfrak m}$ for some finite type $A$-algebra $C$ and prime ideal $\mathfrak m$. Then $A \to C$ is quasi-finite at $\mathfrak m$ by Algebra, Lemma 10.122.2 (2). Hence by Algebra, Lemma 10.123.13 we see that after replacing $C$ by a principal localization the ring map $A \to C$ is quasi-finite. $\square$

Proof. This is a reformulation of Lemma 37.52.6. Namely, set $A = \mathcal{O}_{S, s}$ and $B = \mathcal{O}_{X, x}$. Denote $V \subset \mathop{\mathrm{Spec}}(B)$ the inverse image of $U$. The ring map $f^\sharp : A \to B$ is essentially of finite type. By assumption there exists at least one point of $V$ which does not map to the closed point of $\mathop{\mathrm{Spec}}(A)$. Hence all the assumptions of Lemma 37.52.6 hold and we obtain a prime $\mathfrak q \subset B$ which does not lie over $\mathfrak m_ A$ and such that $A \to B/\mathfrak q$ is the localization of a quasi-finite ring map. Let $z \in X$ be the image of the point $\mathfrak q$ under the canonical morphism $\mathop{\mathrm{Spec}}(B) \to X$. Set $Z = \overline{\{ z\} }$ with the induced reduced scheme structure. As $z \leadsto x$ we see that $x \in Z$ and $\mathcal{O}_{Z, x} = B/\mathfrak q$. By construction $Z \to S$ is quasi-finite at $x$. $\square$

Proof. Namely, we can reduce this to the cited lemma as follows: First we replace $X$ and $S$ by affine neighbourhoods of $x$ and $s$. Then $X$ is Noetherian, in particular $U$ is quasi-compact (see Morphisms, Lemma 29.15.6 and Topology, Lemmas 5.9.2 and 5.12.13). Hence there exists a specialization $x' \leadsto x$ with $x' \in U$ (see Morphisms, Lemma 29.6.5). Note that $f(x') \not= s$. Thus we see all hypotheses of the lemma are satisfied and we win. $\square$