The Stacks project

Proof. The corresponding algebra statement is Algebra, Lemma 10.113.1. $\square$

Proof. Let $X_1, \ldots , X_ n$ be the irreducible components of $X$ containing $x$ endowed with their reduced induced scheme structure. These correspond to the minimal primes $\mathfrak q_ i$ of $\mathcal{O}_{X, x}$ and hence there are finitely many of them (Schemes, Lemma 26.13.2 and Algebra, Lemma 10.31.6). Then $\dim (\mathcal{O}_{X, x}) = \max \dim (\mathcal{O}_{X, x}/\mathfrak q_ i) = \max \dim (\mathcal{O}_{X_ i, x})$. The $\xi $'s occurring in the definition of $E$ are exactly the generic points $\xi _ i \in X_ i$. Let $Z_ i = \overline{\{ f(\xi _ i)\} } \subset S$ endowed with the reduced induced scheme structure. The composition $X_ i \to X \to S$ factors through $Z_ i$ (Schemes, Lemma 26.12.7). Thus we may apply the dimension formula (Lemma 29.52.1) to see that $\dim (\mathcal{O}_{X_ i, x}) \leq \dim (\mathcal{O}_{Z_ i, x}) + \text{trdeg}_{\kappa (f(\xi ))}(\kappa (\xi )) - \text{trdeg}_{\kappa (s)} \kappa (x)$. Putting everything together we obtain the lemma. $\square$

Proof. Let $f : X \to S$ be locally of finite type. Let $x \leadsto y$, $x \not= y$ be a specialization in $X$. We have to show that $\delta _{X/S}(x) > \delta _{X/S}(y)$ and that $\delta _{X/S}(x) = \delta _{X/S}(y) + 1$ if $y$ is an immediate specialization of $x$.

Choose an affine open $V \subset S$ containing the image of $y$ and choose an affine open $U \subset X$ mapping into $V$ and containing $y$. We may clearly replace $X$ by $U$ and $S$ by $V$. Thus we may assume that $X = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$ and that $f$ is given by a ring map $R \to A$. The ring $R$ is universally catenary (Lemma 29.17.2) and the map $R \to A$ is of finite type (Lemma 29.15.2).

Let $\mathfrak q \subset A$ be the prime ideal corresponding to the point $x$ and let $\mathfrak p \subset R$ be the prime ideal corresponding to $f(x)$. The restriction $\delta '$ of $\delta $ to $S' = \mathop{\mathrm{Spec}}(R/\mathfrak p) \subset S$ is a dimension function. The ring $R/\mathfrak p$ is universally catenary. The restriction of $\delta _{X/S}$ to $X' = \mathop{\mathrm{Spec}}(A/\mathfrak q)$ is clearly equal to the function $\delta _{X'/S'}$ constructed using the dimension function $\delta '$. Hence we may assume in addition to the above that $R \subset A$ are domains, in other words that $X$ and $S$ are integral schemes, and that $x$ is the generic point of $X$ and $f(x)$ is the generic point of $S$.

Note that $\mathcal{O}_{X, x} = R(X)$ and that since $x \leadsto y$, $x \not= y$, the spectrum of $\mathcal{O}_{X, y}$ has at least two points (Schemes, Lemma 26.13.2) hence $\dim (\mathcal{O}_{X, y}) > 0$ . If $y$ is an immediate specialization of $x$, then $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, y}) = \{ x, y\} $ and $\dim (\mathcal{O}_{X, y}) = 1$.

Write $s = f(x)$ and $t = f(y)$. We compute

where we use equality in (29.52.1.1) in the last step. Since $\delta $ is a dimension function on the scheme $S$ and $s \in S$ is the generic point, the difference $\delta (s) - \delta (t)$ is equal to $\text{codim}(\overline{\{ t\} }, S)$ by Topology, Lemma 5.20.2. This is equal to $\dim (\mathcal{O}_{S, t})$ by Properties, Lemma 28.10.3. Hence we conclude that

and the lemma follows from what we said above about $\dim (\mathcal{O}_{X, y})$. $\square$

Proof. Let $f : X \to Y$ be as in the lemma. Let $\xi _0 \leadsto \xi _1 \leadsto \ldots \leadsto \xi _ e$ be a sequence of specializations in $X$. Set $x = \xi _ e$ and $y = f(x)$. Observe that $e \leq \dim (\mathcal{O}_{X, x})$ as the given specializations occur in the spectrum of $\mathcal{O}_{X, x}$, see Schemes, Lemma 26.13.2. By the dimension formula, Lemma 29.52.1, we see that

Hence we conclude that $e \leq \dim (Y) + \text{trdeg}_{R(Y)} R(X)$ as desired.

Next, assume $f$ is also closed. Say $\overline{\xi }_0 \leadsto \overline{\xi }_1 \leadsto \ldots \leadsto \overline{\xi }_ d$ is a sequence of specializations in $Y$. We want to show that $\dim (X) \geq d + r$. We may assume that $\overline{\xi }_0 = \eta $ is the generic point of $Y$. The generic fibre $X_\eta $ is a scheme locally of finite type over $\kappa (\eta ) = R(Y)$. It is nonempty as $f$ is dominant. Hence by Lemma 29.16.10 it is a Jacobson scheme. Thus by Lemma 29.16.8 we can find a closed point $\xi _0 \in X_\eta $ and the extension $\kappa (\eta ) \subset \kappa (\xi _0)$ is a finite extension. Note that $\mathcal{O}_{X, \xi _0} = \mathcal{O}_{X_\eta , \xi _0}$ because $\eta $ is the generic point of $Y$. Hence we see that $\dim (\mathcal{O}_{X, \xi _0}) = r$ by Lemma 29.52.1 applied to the scheme $X_\eta $ over the universally catenary scheme $\mathop{\mathrm{Spec}}(\kappa (\eta ))$ (see Lemma 29.17.5) and the point $\xi _0$. This means that we can find $\xi _{-r} \leadsto \ldots \leadsto \xi _{-1} \leadsto \xi _0$ in $X$. On the other hand, as $f$ is closed specializations lift along $f$, see Topology, Lemma 5.19.7. Thus, as $\xi _0$ lies over $\eta = \overline{\xi }_0$ we can find specializations $\xi _0 \leadsto \xi _1 \leadsto \ldots \leadsto \xi _ d$ lying over $\overline{\xi }_0 \leadsto \overline{\xi }_1 \leadsto \ldots \leadsto \overline{\xi }_ d$. In other words we have

which means that $\dim (X) \geq d + r$ as desired. $\square$

Proof. Immediate consequence of Lemma 29.52.2 and Properties, Lemma 28.10.2. $\square$