The Stacks project

66.20 Generic points

This section is a continuation of Properties of Spaces, Section 64.11.

Lemma 66.20.1. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x \in |X|$. The following are equivalent

  1. $x$ is a generic point of an irreducible component of $|X|$,

  2. for any étale morphism $(Y, y) \to (X, x)$ of pointed algebraic spaces, $y$ is a generic point of an irreducible component of $|Y|$,

  3. for some étale morphism $(Y, y) \to (X, x)$ of pointed algebraic spaces, $y$ is a generic point of an irreducible component of $|Y|$,

  4. the dimension of the local ring of $X$ at $x$ is zero, and

  5. $x$ is a point of codimension $0$ on $X$

Proof. Conditions (4) and (5) are equivalent for any algebraic space by definition, see Properties of Spaces, Definition 64.10.2. Observe that any $Y$ as in (2) and (3) is decent by Lemma 66.6.6. Thus it suffices to prove the equivalence of (1) and (4) as then the equivalence with (2) and (3) follows since the dimension of the local ring of $Y$ at $y$ is equal to the dimension of the local ring of $X$ at $x$. Let $f : U \to X$ be an étale morphism from an affine scheme and let $u \in U$ be a point mapping to $x$.

Assume (1). Let $u' \leadsto u$ be a specialization in $U$. Then $f(u') = f(u) = x$. By Lemma 66.12.1 we see that $u' = u$. Hence $u$ is a generic point of an irreducible component of $U$. Thus $\dim (\mathcal{O}_{U, u}) = 0$ and we see that (4) holds.

Assume (4). The point $x$ is contained in an irreducible component $T \subset |X|$. Since $|X|$ is sober (Proposition 66.12.4) we $T$ has a generic point $x'$. Of course $x' \leadsto x$. Then we can lift this specialization to $u' \leadsto u$ in $U$ (Lemma 66.12.2). This contradicts the assumption that $\dim (\mathcal{O}_{U, u}) = 0$ unless $u' = u$, i.e., $x' = x$. $\square$

Lemma 66.20.2. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $T \subset |X|$ be an irreducible closed subset. Let $\xi \in T$ be the generic point (Proposition 66.12.4). Then $\text{codim}(T, |X|)$ (Topology, Definition 5.11.1) is the dimension of the local ring of $X$ at $\xi $ (Properties of Spaces, Definition 64.10.2).

Proof. Choose a scheme $U$, a point $u \in U$, and an étale morphism $U \to X$ sending $u$ to $\xi $. Then any sequence of nontrivial specializations $\xi _ e \leadsto \ldots \leadsto \xi _0 = \xi $ can be lifted to a sequence $u_ e \leadsto \ldots \leadsto u_0 = u$ in $U$ by Lemma 66.12.2. Conversely, any sequence of nontrivial specializations $u_ e \leadsto \ldots \leadsto u_0 = u$ in $U$ maps to a sequence of nontrivial specializations $\xi _ e \leadsto \ldots \leadsto \xi _0 = \xi $ by Lemma 66.12.1. Because $|X|$ and $U$ are sober topological spaces we conclude that the codimension of $T$ in $|X|$ and of $\overline{\{ u\} }$ in $U$ are the same. In this way the lemma reduces to the schemes case which is Properties, Lemma 28.10.3. $\square$

Lemma 66.20.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Assume

  1. every quasi-compact scheme étale over $X$ has finitely many irreducible components, and

  2. every $x \in |X|$ of codimension $0$ on $X$ can be represented by a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$.

Then $X$ is a reasonable algebraic space.

Proof. Let $U$ be an affine scheme and let $a : U \to X$ be an étale morphism. We have to show that the fibres of $a$ are universally bounded. By assumption (1) the scheme $U$ has finitely many irreducible components. Let $u_1, \ldots , u_ n \in U$ be the generic points of these irreducible components. Let $\{ x_1, \ldots , x_ m\} \subset |X|$ be the image of $\{ u_1, \ldots , u_ n\} $. Each $x_ j$ is a point of codimension $0$. By assumption (2) we may choose a monomorphism $\mathop{\mathrm{Spec}}(k_ j) \to X$ representing $x_ j$. Then

\[ U \times _ X \mathop{\mathrm{Spec}}(k_ j) = \coprod \nolimits _{a(u_ i) = x_ j} \mathop{\mathrm{Spec}}(\kappa (u_ i)) \]

is finite over $\mathop{\mathrm{Spec}}(k_ j)$ of degree $d_ j = \sum _{a(u_ i) = x_ j} [\kappa (u_ i) : k_ j]$. Set $n = \max d_ j$.

Observe that $a$ is separated (Properties of Spaces, Lemma 64.6.4). Consider the stratification

\[ X = X_0 \supset X_1 \supset X_2 \supset \ldots \]

associated to $U \to X$ in Lemma 66.8.2. By our choice of $n$ above we conclude that $X_{n + 1}$ is empty. Namely, if not, then $a^{-1}(X_{n + 1})$ is a nonempty open of $U$ and hence would contain one of the $x_ i$. This would mean that $X_{n + 1}$ contains $x_ j = a(u_ i)$ which is impossible. Hence we see that the fibres of $U \to X$ are universally bounded (in fact by the integer $n$). $\square$

Lemma 66.20.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent

  1. $X$ is decent and $|X|$ has finitely many irreducible components,

  2. every quasi-compact scheme étale over $X$ has finitely many irreducible components, there are finitely many $x \in |X|$ of codimension $0$ on $X$, and each of these can be represented by a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$,

  3. there exists a dense open $X' \subset X$ which is a scheme, $X'$ has finitely many irreducible components with generic points $\{ x'_1, \ldots , x'_ m\} $, and the morphism $x'_ j \to X$ is quasi-compact for $j = 1, \ldots , m$.

Moreover, if these conditions hold, then $X$ is reasonable and the points $x'_ j \in |X|$ are the generic points of the irreducible components of $|X|$.

Proof. In the proof we use Properties of Spaces, Lemma 64.11.1 without further mention. Assume (1). Then $X$ has a dense open subscheme $X'$ by Theorem 66.10.2. Since the closure of an irreducible component of $|X'|$ is an irreducible component of $|X|$, we see that $|X'|$ has finitely many irreducible components. Thus (3) holds.

Assume $X' \subset X$ is as in (3). Let $\{ x'_1, \ldots , x'_ m\} $ be the generic points of the irreducible components of $X'$. Let $a : U \to X$ be an étale morphism with $U$ a quasi-compact scheme. It suffices to show that $U$ has finitely many irreducible components whose generic points lie over $\{ x'_1, \ldots , x'_ m\} $. It suffices to prove this for the members of a finite affine open cover of $U$, hence we may and do assume $U$ is affine. Note that $U' = a^{-1}(X') \subset U$ is a dense open. The generic points of irreducible components of $U'$ are the points lying over $\{ x'_1, \ldots , x'_ m\} $ and since $x'_ j \to X$ is quasi-compact there are finitely many points of $U$ lying over $x'_ j$ (Lemma 66.4.5). Hence $U'$ has finitely many irreducible components, which implies that the closures of these irreducible components are the irreducible components of $U$. Thus (2) holds.

Assume (2). This implies (1) and the final statement by Lemma 66.20.3. (We also use that a reasonable algebraic space is decent, see discussion following Definition 66.6.1.) $\square$


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