The Stacks project

66.11 Generic points

Let $T$ be a topological space. According to the second edition of EGA I, a maximal point of $T$ is a generic point of an irreducible component of $T$. If $T = |X|$ is the topological space associated to an algebraic space $X$, there are at least two notions of maximal points: we can look at maximal points of $T$ viewed as a topological space, or we can look at images of maximal points of $U$ where $U \to X$ is an étale morphism and $U$ is a scheme. The second notion corresponds to the set of points of codimension $0$ (Lemma 66.11.1). The codimension $0$ points are easier to work with for general algebraic spaces; the two notions agree for quasi-separated and more generally decent algebraic spaces (Decent Spaces, Lemma 68.20.1).

Lemma 66.11.1. Let $S$ be a scheme and let $X$ be an algebraic space over $S$. Let $x \in |X|$. Consider étale morphisms $a : U \to X$ where $U$ is a scheme. The following are equivalent

  1. $x$ is a point of codimension $0$ on $X$,

  2. for some $U \to X$ as above and $u \in U$ with $a(u) = x$, the point $u$ is the generic point of an irreducible component of $U$, and

  3. for any $U \to X$ as above and any $u \in U$ mapping to $x$, the point $u$ is the generic point of an irreducible component of $U$.

If $X$ is representable, this is equivalent to $x$ being a generic point of an irreducible component of $|X|$.

Proof. Observe that a point $u$ of a scheme $U$ is a generic point of an irreducible component of $U$ if and only if $\dim (\mathcal{O}_{U, u}) = 0$ (Properties, Lemma 28.10.4). Hence this follows from the definition of the codimension of a point on $X$ (Definition 66.10.2). $\square$

Lemma 66.11.2. Let $S$ be a scheme and let $X$ be an algebraic space over $S$. The set of codimension $0$ points of $X$ is dense in $|X|$.

Proof. If $U$ is a scheme, then the set of generic points of irreducible components is dense in $U$ (holds for any quasi-sober topological space). Thus if $U \to X$ is a surjective étale morphism, then the set of codimension $0$ points of $X$ is the image of a dense subset of $|U|$ (Lemma 66.11.1). Since $|X|$ has the quotient topology for $|U| \to |X|$ we conclude. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BAP. Beware of the difference between the letter 'O' and the digit '0'.