5.20 Dimension functions
It scarcely makes sense to consider dimension functions unless the space considered is sober (Definition 5.8.6). Thus the definition below can be improved by considering the sober topological space associated to $X$. Since the underlying topological space of a scheme is sober we do not bother with this improvement.
Definition 5.20.1. Let $X$ be a topological space.
Let $x, y \in X$, $x \not= y$. Suppose $x \leadsto y$, that is $y$ is a specialization of $x$. We say $y$ is an immediate specialization of $x$ if there is no $z \in X \setminus \{ x, y\} $ with $x \leadsto z$ and $z \leadsto y$.
A map $\delta : X \to \mathbf{Z}$ is called a dimension function1 if
whenever $x \leadsto y$ and $x \not= y$ we have $\delta (x) > \delta (y)$, and
for every immediate specialization $x \leadsto y$ in $X$ we have $\delta (x) = \delta (y) + 1$.
It is clear that if $\delta $ is a dimension function, then so is $\delta + t$ for any $t \in \mathbf{Z}$. Here is a fun lemma.
Lemma 5.20.2. Let $X$ be a topological space. If $X$ is sober and has a dimension function, then $X$ is catenary. Moreover, for any $x \leadsto y$ we have
\[ \delta (x) - \delta (y) = \text{codim}\left(\overline{\{ y\} }, \ \overline{\{ x\} }\right). \]
Proof.
Suppose $Y \subset Y' \subset X$ are irreducible closed subsets. Let $\xi \in Y$, $\xi ' \in Y'$ be their generic points. Then we see immediately from the definitions that $\text{codim}(Y, Y') \leq \delta (\xi ) - \delta (\xi ') < \infty $. In fact the first inequality is an equality. Namely, suppose
\[ Y = Y_0 \subset Y_1 \subset \ldots \subset Y_ e = Y' \]
is any maximal chain of irreducible closed subsets. Let $\xi _ i \in Y_ i$ denote the generic point. Then we see that $\xi _ i \leadsto \xi _{i + 1}$ is an immediate specialization. Hence we see that $e = \delta (\xi ) - \delta (\xi ')$ as desired. This also proves the last statement of the lemma.
$\square$
Lemma 5.20.3. Let $X$ be a topological space. Let $\delta $, $\delta '$ be two dimension functions on $X$. If $X$ is locally Noetherian and sober then $\delta - \delta '$ is locally constant on $X$.
Proof.
Let $x \in X$ be a point. We will show that $\delta - \delta '$ is constant in a neighbourhood of $x$. We may replace $X$ by an open neighbourhood of $x$ in $X$ which is Noetherian. Hence we may assume $X$ is Noetherian and sober. Let $Z_1, \ldots , Z_ r$ be the irreducible components of $X$ passing through $x$. (There are finitely many as $X$ is Noetherian, see Lemma 5.9.2.) Let $\xi _ i \in Z_ i$ be the generic point. Note $Z_1 \cup \ldots \cup Z_ r$ is a neighbourhood of $x$ in $X$ (not necessarily closed). We claim that $\delta - \delta '$ is constant on $Z_1 \cup \ldots \cup Z_ r$. Namely, if $y \in Z_ i$, then
\[ \delta (x) - \delta (y) = \delta (x) - \delta (\xi _ i) + \delta (\xi _ i) - \delta (y) = - \text{codim}(\overline{\{ x\} }, Z_ i) + \text{codim}(\overline{\{ y\} }, Z_ i) \]
by Lemma 5.20.2. Similarly for $\delta '$. Whence the result.
$\square$
Lemma 5.20.4. Let $X$ be locally Noetherian, sober and catenary. Then any point has an open neighbourhood $U \subset X$ which has a dimension function.
Proof.
We will use repeatedly that an open subspace of a catenary space is catenary, see Lemma 5.11.5 and that a Noetherian topological space has finitely many irreducible components, see Lemma 5.9.2. In the proof of Lemma 5.20.3 we saw how to construct such a function. Namely, we first replace $X$ by a Noetherian open neighbourhood of $x$. Next, we let $Z_1, \ldots , Z_ r \subset X$ be the irreducible components of $X$. Let
\[ Z_ i \cap Z_ j = \bigcup Z_{ijk} \]
be the decomposition into irreducible components. We replace $X$ by
\[ X \setminus \left( \bigcup \nolimits _{x \not\in Z_ i} Z_ i \cup \bigcup \nolimits _{x \not\in Z_{ijk}} Z_{ijk} \right) \]
so that we may assume $x \in Z_ i$ for all $i$ and $x \in Z_{ijk}$ for all $i, j, k$. For $y \in X$ choose any $i$ such that $y \in Z_ i$ and set
\[ \delta (y) = - \text{codim}(\overline{\{ x\} }, Z_ i) + \text{codim}(\overline{\{ y\} }, Z_ i). \]
We claim this is a dimension function. First we show that it is well defined, i.e., independent of the choice of $i$. Namely, suppose that $y \in Z_{ijk}$ for some $i, j, k$. Then we have (using Lemma 5.11.6)
\begin{align*} \delta (y) & = - \text{codim}(\overline{\{ x\} }, Z_ i) + \text{codim}(\overline{\{ y\} }, Z_ i) \\ & = - \text{codim}(\overline{\{ x\} }, Z_{ijk}) - \text{codim}(Z_{ijk}, Z_ i) + \text{codim}(\overline{\{ y\} }, Z_{ijk}) + \text{codim}(Z_{ijk}, Z_ i) \\ & = - \text{codim}(\overline{\{ x\} }, Z_{ijk}) + \text{codim}(\overline{\{ y\} }, Z_{ijk}) \end{align*}
which is symmetric in $i$ and $j$. We omit the proof that it is a dimension function.
$\square$
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