The Stacks project

Lemma 37.48.4. 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 10.50.6) 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 37.48.1 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. $\square$


Comments (2)

Comment #1612 by Martin Bright on

I've noticed a couple of typos in this proof. Halfway through there is a rogue which should be a . Also, in the statement "... we may assume that for each ", I think should be .

By the way, the version of this statement when is the spectrum of a DVR seems already interesting: do you know if it's in the literature anywhere?

Comment #1669 by on

Thanks for the corrections. I changed them here. Most of these statements are in the paper by Osserman and Payne mentioned at the beginning of the sections.


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 053U. Beware of the difference between the letter 'O' and the digit '0'.