Lemma 32.20.3. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme.

1. The topological space of $X$ is catenary (Topology, Definition 5.11.4).

2. For $x \in X$ closed, we have $\dim _ x(X) = \dim (\mathcal{O}_{X, x})$.

3. For $X$ irreducible we have $\dim (X) = \dim (U)$ for any nonempty open $U \subset X$ and $\dim (X) = \dim _ x(X)$ for any $x \in X$.

4. For $X$ irreducible any chain of irreducible closed subsets can be extended to a maximal chain and all maximal chains of irreducible closed subsets have length equal to $\dim (X)$.

5. For $x \in X$ we have $\dim _ x(X) = \max \dim (Z) = \min \dim (\mathcal{O}_{X, x'})$ where the maximum is over irreducible components $Z \subset X$ containing $x$ and the minimum is over specializations $x \leadsto x'$ with $x'$ closed in $X$.

6. If $X$ is irreducible with generic point $x$, then $\dim (X) = \text{trdeg}_ k(\kappa (x))$.

7. If $x \leadsto x'$ is an immediate specialization of points of $X$, then we have $\text{trdeg}_ k(\kappa (x)) = \text{trdeg}_ k(\kappa (x')) + 1$.

8. The dimension of $X$ is the supremum of the numbers $\text{trdeg}_ k(\kappa (x))$ where $x$ runs over the generic points of the irreducible components of $X$.

9. If $x \leadsto x'$ is a nontrivial specialization of points of $X$, then

1. $\dim _ x(X) \leq \dim _{x'}(X)$,

2. $\dim (\mathcal{O}_{X, x}) < \dim (\mathcal{O}_{X, x'})$,

3. $\text{trdeg}_ k(\kappa (x)) > \text{trdeg}_ k(\kappa (x'))$, and

4. any maximal chain of nontrivial specializations $x = x_0 \leadsto x_1 \leadsto \ldots \leadsto x_ n = x$ has length $n = \text{trdeg}_ k(\kappa (x)) - \text{trdeg}_ k(\kappa (x'))$.

10. For $x \in X$ we have $\dim _ x(X) = \text{trdeg}_ k(\kappa (x)) + \dim (\mathcal{O}_{X, x})$.

11. If $x \leadsto x'$ is an immediate specialization of points of $X$ and $X$ is irreducible or equidimensional, then $\dim (\mathcal{O}_{X, x'}) = \dim (\mathcal{O}_{X, x}) + 1$.

Proof. Instead on relying on the more general results proved earlier we will reduce the statements to the corresponding statements for finite type $k$-algebras and cite results from the chapter on commutative algebra.

Proof of (1). This is local on $X$ by Topology, Lemma 5.11.5. Thus we may assume $X = \mathop{\mathrm{Spec}}(A)$ where $A$ is a finite type $k$-algebra. We have to show that $A$ is catenary (Algebra, Lemma 10.104.2). We can reduce to $k[x_1, \ldots , x_ n]$ using Algebra, Lemma 10.104.7 and then apply Algebra, Lemma 10.113.3. Alternatively, this holds because $k$ is Cohen-Macaulay (trivially) and Cohen-Macaulay rings are universally catenary (Algebra, Lemma 10.104.9).

Proof of (2). Choose an affine neighbourhood $U = \mathop{\mathrm{Spec}}(A)$ of $x$. Then $\dim _ x(X) = \dim _ x(U)$. Hence we reduce to the affine case, which is Algebra, Lemma 10.113.6.

Proof of (3). It suffices to show that any two nonempty affine opens $U, U' \subset X$ have the same dimension (any finite chain of irreducible subsets meets an affine open). Pick a closed point $x$ of $X$ with $x \in U \cap U'$. This is possible because $X$ is irreducible, hence $U \cap U'$ is nonempty, hence there is such a closed point because $X$ is Jacobson by Lemma 32.14.1. Then $\dim (U) = \dim (\mathcal{O}_{X, x}) = \dim (U')$ by Algebra, Lemma 10.113.4 (strictly speaking you have to replace $X$ by its reduction before applying the lemma).

Proof of (4). Given a chain of irreducible closed subsets we can find an affine open $U \subset X$ which meets the smallest one. Thus the statement follows from Algebra, Lemma 10.113.4 and $\dim (U) = \dim (X)$ which we have seen in (3).

Proof of (5). Choose an affine neighbourhood $U = \mathop{\mathrm{Spec}}(A)$ of $x$. Then $\dim _ x(X) = \dim _ x(U)$. The rule $Z \mapsto Z \cap U$ is a bijection between irreducible components of $X$ passing through $x$ and irreducible components of $U$ passing through $x$. Also, $\dim (Z \cap U) = \dim (Z)$ for such $Z$ by (3). Hence the statement follows from Algebra, Lemma 10.113.5.

Proof of (6). By (3) this reduces to the case where $X = \mathop{\mathrm{Spec}}(A)$ is affine. In this case it follows from Algebra, Lemma 10.115.1 applied to $A_{red}$.

Proof of (7). Let $Z = \overline{\{ x\} } \supset Z' = \overline{\{ x'\} }$. Then it follows from (4) that $Z \supset Z'$ is the start of a maximal chain of irreducible closed subschemes in $Z$ and consequently $\dim (Z) = \dim (Z') + 1$. We conclude by (6).

Proof of (8). A simple topological argument shows that $\dim (X) = \sup \dim (Z)$ where the supremum is over the irreducible components of $X$ (hint: use Topology, Lemma 5.8.3). Thus this follows from (6).

Proof of (9). Part (a) follows from the fact that any open $U \subset X$ containing $x'$ also contains $x$. Part (b) follows because $\mathcal{O}_{X, x}$ is a localization of $\mathcal{O}_{X, x'}$ hence any chain of primes in $\mathcal{O}_{X, x}$ corresponds to a chain of primes in $\mathcal{O}_{X, x'}$ which can be extended by adding $\mathfrak m_{x'}$ at the end. Both (c) and (d) follow formally from (7).

Proof of (10). Choose an affine neighbourhood $U = \mathop{\mathrm{Spec}}(A)$ of $x$. Then $\dim _ x(X) = \dim _ x(U)$. Hence we reduce to the affine case, which is Algebra, Lemma 10.115.3.

Proof of (11). If $X$ is equidimensional (Topology, Definition 5.10.5) then $\dim (X)$ is equal to the dimension of every irreducible component of $X$, whence $\dim _ x(X) = \dim (X) = \dim _{x'}(X)$ by (5). Thus this follows from (7). $\square$

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