The Stacks project

37.52 Closed points in fibres

Some of the material in this section is taken from the preprint [Osserman-Payne].

Lemma 37.52.1. Let $f : X \to S$ be a morphism of schemes. Let $Z \subset X$ be a closed subscheme. Let $s \in S$. Assume

  1. $S$ is irreducible with generic point $\eta $,

  2. $X$ is irreducible,

  3. $f$ is dominant,

  4. $f$ is locally of finite type,

  5. $\dim (X_ s) \leq \dim (X_\eta )$,

  6. $Z$ is locally principal in $X$, and

  7. $Z_\eta = \emptyset $.

Then the fibre $Z_ s$ is (set theoretically) a union of irreducible components of $X_ s$.

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

\[ \xi \leadsto \xi _1 \leadsto \xi '. \]

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

\[ 0 \to I \to A[X_1, \ldots , X_ n] \to B \to 0 \]

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

\[ B_0 \otimes _{A_0} K_0 \otimes _{K_0} K = B \otimes _ A K \quad \text{and}\quad B_0 \otimes _{A_0} \kappa (\mathfrak p_0) \otimes _{\kappa (\mathfrak p_0)} \kappa (\mathfrak p) = B \otimes _ A \kappa (\mathfrak p). \]

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. $\square$

Here is an algebraic application of the lemma above. The fourth assumption of the lemma holds if $A \to B$ is flat, see Lemma 37.52.3.

Lemma 37.52.2. Let $A \to B$ be a local homomorphism of local rings, and $g \in \mathfrak m_ B$. Assume

  1. $A$ and $B$ are domains and $A \subset B$,

  2. $B$ is essentially of finite type over $A$,

  3. $g$ is not contained in any minimal prime over $\mathfrak m_ AB$, and

  4. $\dim (B/\mathfrak m_ AB) + \text{trdeg}_{\kappa (\mathfrak m_ A)}(\kappa (\mathfrak m_ B)) = \text{trdeg}_ A(B)$.

Then $A \subset B/gB$, i.e., the generic point of $\mathop{\mathrm{Spec}}(A)$ is in the image of the morphism $\mathop{\mathrm{Spec}}(B/gB) \to \mathop{\mathrm{Spec}}(A)$.

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. $\square$

Lemma 37.52.3. Let $A \to B$ be a local homomorphism of local rings. Assume

  1. $A$ and $B$ are domains and $A \subset B$,

  2. $B$ is essentially of finite type over $A$, and

  3. $B$ is flat over $A$.

Then we have

\[ \dim (B/\mathfrak m_ AB) + \text{trdeg}_{\kappa (\mathfrak m_ A)}(\kappa (\mathfrak m_ B)) = \text{trdeg}_ A(B). \]

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

\[ C'/\mathfrak m_{A'}C' = C/\mathfrak m_ AC \otimes _{\kappa (\mathfrak m_ A)} \kappa (\mathfrak m_{A'}) \]

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. $\square$

Lemma 37.52.4. Let $f : X \to S$ be a morphism of schemes. Let $x \leadsto x'$ be a specialization of points in $X$. Set $s = f(x)$ and $s' = f(x')$. Assume

  1. $x'$ is a closed point of $X_{s'}$, and

  2. $f$ is locally of finite type.

Then the set

\[ \{ x_1 \in X \text{ such that } f(x_1) = s \text{ and } x_1\text{ is closed in }X_ s \text{ and } x \leadsto x_1 \leadsto x' \} \]

is dense in the closure of $x$ in $X_ s$.

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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[rd] \ar[r] & X_ A \ar[d] \ar[r] & X \ar[d] \\ & \mathop{\mathrm{Spec}}(A) \ar[r] & S } \]

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. $\square$

Remark 37.52.5. The proof of Lemma 37.52.4 actually shows that there exists a sequence of specializations

\[ x \leadsto x_1 \leadsto x_2 \leadsto \ldots \leadsto x_ d \leadsto x' \]

where all $x_ i$ are in the fibre $X_ s$, each specialization is immediate, and $x_ d$ is a closed point of $X_ s$. The integer $d = \text{trdeg}_{\kappa (s)}(\kappa (x)) = \dim (\overline{\{ x\} })$ where the closure is taken in $X_ s$. Moreover, the points $x_ i$ can be chosen to avoid any closed subset of $X_ s$ which does not contain the point $x$.

Examples, Section 110.39 shows that the following lemma is false if $A$ is not assumed Noetherian.

Lemma 37.52.6. Let $\varphi : A \to B$ be a local ring map of local rings. Let $V \subset \mathop{\mathrm{Spec}}(B)$ be an open subscheme which contains at least one prime not lying over $\mathfrak m_ A$. Assume $A$ is Noetherian, $\varphi $ essentially of finite type, and $A/\mathfrak m_ A \subset B/\mathfrak m_ B$ is finite. Then there exists a $\mathfrak q \in V$, $\mathfrak m_ A \not= \mathfrak q \cap A$ such that $A \to B/\mathfrak q$ is the localization of a quasi-finite ring map.

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

\[ V \setminus \{ \mathfrak q \subset B \mid \mathfrak m_ A = \mathfrak q \cap A\} . \]

(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. $\square$

Lemma 37.52.7. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ with image $s \in S$. Let $U \subset X$ be an open subscheme. Assume $f$ locally of finite type, $S$ locally Noetherian, $x$ a closed point of $X_ s$, and assume there exists a point $x' \in U$ with $x' \leadsto x$ and $f(x') \not= s$. Then there exists a closed subscheme $Z \subset X$ such that (a) $x \in Z$, (b) $f|_ Z : Z \to S$ is quasi-finite at $x$, and (c) there exists a $z \in Z$, $z \in U$, $z \leadsto x$ and $f(z) \not= s$.

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 $x$. $\square$

Remark 37.52.8. We can use Lemma 37.52.6 or its variant Lemma 37.52.7 to give an alternative proof of Lemma 37.52.4 in case $S$ is locally Noetherian. Here is a rough sketch. Namely, first replace $S$ by the spectrum of the local ring at $s'$. Then we may use induction on $\dim (S)$. The case $\dim (S) = 0$ is trivial because then $s' = s$. Replace $X$ by the reduced induced scheme structure on $\overline{\{ x\} }$. Apply Lemma 37.52.7 to $X \to S$ and $x' \mapsto s'$ and any nonempty open $U \subset X$ containing $x$. This gives us a closed subscheme $x' \in Z \subset X$ a point $z \in Z$ such that $Z \to S$ is quasi-finite at $x'$ and such that $f(z) \not= s'$. Then $z$ is a closed point of $X_{f(z)}$, and $z \leadsto x'$. As $f(z) \not= s'$ we see $\dim (\mathcal{O}_{S, f(z)}) < \dim (S)$. Since $x$ is the generic point of $X$ we see $x \leadsto z$, hence $s = f(x) \leadsto f(z)$. Apply the induction hypothesis to $s \leadsto f(z)$ and $z \mapsto f(z)$ to win.

Lemma 37.52.9. Suppose that $f : X \to S$ is locally of finite type, $S$ locally Noetherian, $x \in X$ a closed point of its fibre $X_ s$, and $U \subset X$ an open subscheme such that $U \cap X_ s = \emptyset $ and $x \in \overline{U}$, then the conclusions of Lemma 37.52.7 hold.

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 053Q. Beware of the difference between the letter 'O' and the digit '0'.