Definition 33.20.1. Let $k$ be a field. An algebraic $k$-scheme is a scheme $X$ over $k$ such that the structure morphism $X \to \mathop{\mathrm{Spec}}(k)$ is of finite type. A locally algebraic $k$-scheme is a scheme $X$ over $k$ such that the structure morphism $X \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type.
33.20 Algebraic schemes
The following definition is taken from [I Definition 6.4.1, EGA].
Note that every (locally) algebraic $k$-scheme is (locally) Noetherian, see Morphisms, Lemma 29.15.6. The category of algebraic $k$-schemes has all products and fibre products (unlike the category of varieties over $k$). Similarly for the category of locally algebraic $k$-schemes.
Lemma 33.20.2. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme of dimension $0$. Then $X$ is a disjoint union of spectra of local Artinian $k$-algebras $A$ with $\dim _ k(A) < \infty $. If $X$ is an algebraic $k$-scheme of dimension $0$, then in addition $X$ is affine and the morphism $X \to \mathop{\mathrm{Spec}}(k)$ is finite.
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. $\square$
The following lemma collects some statements on dimension theory for locally algebraic schemes.
Lemma 33.20.3. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme.
The topological space of $X$ is catenary (Topology, Definition 5.11.4).
For $x \in X$ closed, we have $\dim _ x(X) = \dim (\mathcal{O}_{X, x})$.
For $X$ irreducible we have $\dim (X) = \dim (U)$ for any nonempty open $U \subset X$ and $\dim (X) = \dim _ x(X)$ for any $x \in X$.
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 $\dim (X)$.
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 $X$.
If $X$ is irreducible with generic point $x$, then $\dim (X) = \text{trdeg}_ k(\kappa (x))$.
If $x \leadsto x'$ is an immediate specialization of points of $X$, then we have $\text{trdeg}_ k(\kappa (x)) = \text{trdeg}_ k(\kappa (x')) + 1$.
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 $X$.
If $x \leadsto x'$ is a nontrivial specialization of points of $X$, then
$\dim _ x(X) \leq \dim _{x'}(X)$,
$\dim (\mathcal{O}_{X, x}) < \dim (\mathcal{O}_{X, x'})$,
$\text{trdeg}_ k(\kappa (x)) > \text{trdeg}_ k(\kappa (x'))$, and
any maximal chain of nontrivial specializations $x = x_0 \leadsto x_1 \leadsto \ldots \leadsto x_ n = x$ has length $n = \text{trdeg}_ k(\kappa (x)) - \text{trdeg}_ k(\kappa (x'))$.
For $x \in X$ we have $\dim _ x(X) = \text{trdeg}_ k(\kappa (x)) + \dim (\mathcal{O}_{X, x})$.
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 10.105.2). 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). $\square$
Lemma 33.20.4. Let $k$ be a field. Let $f : X \to Y$ be a morphism of locally algebraic $k$-schemes.
For $y \in Y$, the fibre $X_ y$ is a locally algebraic scheme over $\kappa (y)$ hence all the results of Lemma 33.20.3 apply.
Assume $X$ is irreducible. Set $Z = \overline{f(X)}$ and $d = \dim (X) - \dim (Z)$. Then
$\dim _ x(X_{f(x)}) \geq d$ for all $x \in X$,
the set of $x \in X$ with $\dim _ x(X_{f(x)}) = d$ is dense open,
if $\dim (\mathcal{O}_{Z, f(x)}) \geq 1$, then $\dim _ x(X_{f(x)}) \leq d + \dim (\mathcal{O}_{Z, f(x)}) - 1$,
if $\dim (\mathcal{O}_{Z, f(x)}) = 1$, then $\dim _ x(X_{f(x)}) = d$,
For $x \in X$ with $y = f(x)$ we have $\dim _ x(X_ y) \geq \dim _ x(X) - \dim _ y(Y)$.
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
\[ d = \dim (X) - \dim (Z) = \text{trdeg}_ k \kappa (\eta ) - \text{trdeg}_ k \kappa (\xi ) = \text{trdeg}_{\kappa (\xi )} \kappa (\eta ) = \dim _\eta (X_\xi ) \]
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). $\square$
Lemma 33.20.5.slogan Let $k$ be a field. Let $X$, $Y$ be locally algebraic $k$-schemes.
For $z \in X \times Y$ lying over $(x, y)$ we have $\dim _ z(X \times Y) = \dim _ x(X) + \dim _ y(Y)$.
We have $\dim (X \times Y) = \dim (X) + \dim (Y)$.
Proof. Proof of (1). Consider the factorization
\[ X \times Y \longrightarrow Y \longrightarrow \mathop{\mathrm{Spec}}(k) \]
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). $\square$
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