The Stacks project

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. $\square$