Lemma 29.50.3. Let $S$ be a locally Noetherian and universally catenary scheme. Let $\delta : S \to \mathbf{Z}$ be a dimension function. Let $f : X \to S$ be a morphism of schemes. Assume $f$ locally of finite type. Then the map

\begin{align*} \delta = \delta _{X/S} : X & \longrightarrow \mathbf{Z} \\ x & \longmapsto \delta (f(x)) + \text{trdeg}_{\kappa (f(x))} \kappa (x) \end{align*}

is a dimension function on $X$.

**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 29.16.2) and the map $R \to A$ is of finite type (Lemma 29.14.2).

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

\begin{align*} \delta _{X/S}(x) - \delta _{X/S}(y) & = \delta (s) + \text{trdeg}_{\kappa (s)} \kappa (x) - \delta (t) - \text{trdeg}_{\kappa (t)} \kappa (y) \\ & = \delta (s) - \delta (t) + \text{trdeg}_{R(S)} R(X) - \text{trdeg}_{\kappa (t)} \kappa (y) \\ & = \delta (s) - \delta (t) + \dim (\mathcal{O}_{X, y}) - \dim (\mathcal{O}_{S, t}) \end{align*}

where we use equality in (29.50.1.1) 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

\[ \delta _{X/S}(x) - \delta _{X/S}(y) = \dim (\mathcal{O}_{X, y}) \]

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

## Comments (0)

There are also: