The Stacks project

Proof. Let $X$ be a locally algebraic $k$-scheme of dimension $0$. Let $U = \mathop{\mathrm{Spec}}(A) \subset X$ be an affine open subscheme. Since $\dim (X) = 0$ we see that $\dim (A) = 0$. By Noether normalization, see Algebra, Lemma 10.115.4 we see that there exists a finite injection $k \to A$, i.e., $\dim _ k(A) < \infty $. Hence $A$ is Artinian, see Algebra, Lemma 10.53.2. This implies that $A = A_1 \times \ldots \times A_ r$ is a product of finitely many Artinian local rings, see Algebra, Lemma 10.53.6. Of course $\dim _ k(A_ i) < \infty $ for each $i$ as the sum of these dimensions equals $\dim _ k(A)$.

The arguments above show that $X$ has an open covering whose members are finite discrete topological spaces. Hence $X$ is a discrete topological space. It follows that $X$ is isomorphic to the disjoint union of its connected components each of which is a singleton. Since a singleton scheme is affine we conclude (by the results of the paragraph above) that each of these singletons is the spectrum of a local Artinian $k$-algebra $A$ with $\dim _ k(A) < \infty $.

Finally, if $X$ is an algebraic $k$-scheme of dimension $0$, then $X$ is quasi-compact hence is a finite disjoint union $X = \mathop{\mathrm{Spec}}(A_1) \amalg \ldots \amalg \mathop{\mathrm{Spec}}(A_ r)$ hence affine (see Schemes, Lemma 26.6.8) and we have seen the finiteness of $X \to \mathop{\mathrm{Spec}}(k)$ in the first paragraph of the proof. $\square$

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 10.105.2). We can reduce to $k[x_1, \ldots , x_ n]$ using Algebra, Lemma 10.105.7 and then apply Algebra, Lemma 10.114.3. Alternatively, this holds because $k$ is Cohen-Macaulay (trivially) and Cohen-Macaulay rings are universally catenary (Algebra, Lemma 10.105.9).

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 10.114.6.

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 33.14.1. Then $\dim (U) = \dim (\mathcal{O}_{X, x}) = \dim (U')$ by Algebra, Lemma 10.114.4 (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 10.114.4 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 10.114.5.

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 10.116.1 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 10.116.3.

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). $\square$

Proof. The morphism $f$ is locally of finite type by Morphisms, Lemma 29.15.8. Hence the base change $X_ y \to \mathop{\mathrm{Spec}}(\kappa (y))$ is locally of finite type. This proves (1). In the rest of the proof we will freely use the results of Lemma 33.20.3 for $X$, $Y$, and the fibres of $f$.

Proof of (2). Let $\eta \in X$ be the generic point and set $\xi = f(\eta )$. Then $Z = \overline{\{ \xi \} }$. Hence

Thus parts (2)(a) and (2)(b) follow from Morphisms, Lemma 29.28.4. Parts (2)(c) and (2)(d) follow from Lemmas 33.19.3 and 33.19.1.

Proof of (3). Let $x \in X$. Let $X' \subset X$ be a irreducible component of $X$ passing through $x$ of dimension $\dim _ x(X)$. Then (2) implies that $\dim _ x(X_ y) \geq \dim (X') - \dim (Z')$ where $Z' \subset Y$ is the closure of the image of $X'$. This proves (3). $\square$

Proof. Proof of (1). Consider the factorization

of the structure morphism. The first morphism $p : X \times Y \to Y$ is flat as a base change of the flat morphism $X \to \mathop{\mathrm{Spec}}(k)$ by Morphisms, Lemma 29.25.8. Moreover, we have $\dim _ z(p^{-1}(y)) = \dim _ x(X)$ by Morphisms, Lemma 29.28.3. Hence $\dim _ z(X \times Y) = \dim _ x(X) + \dim _ y(Y)$ by Morphisms, Lemma 29.28.2. Part (2) is a direct consequence of (1). $\square$