The Stacks project

Lemma 66.12.5. Let $S$ be a scheme. Dimension as defined in Properties of Spaces, Section 64.9 behaves well on decent algebraic spaces $X$ over $S$.

  1. If $x \in |X|$, then $\dim _ x(|X|) = \dim _ x(X)$, and

  2. $\dim (|X|) = \dim (X)$.

Proof. Proof of (1). Choose a scheme $U$ with a point $u \in U$ and an ├ętale morphism $h : U \to X$ mapping $u$ to $x$. By definition the dimension of $X$ at $x$ is $\dim _ u(|U|)$. Thus we may pick $U$ such that $\dim _ x(X) = \dim (|U|)$. Let $d$ be an integer. If $\dim (U) \geq d$, then there exists a sequence of nontrivial specializations $u_ d \leadsto \ldots \leadsto u_0$ in $U$. Taking the image we find a corresponding sequence $h(u_ d) \leadsto \ldots \leadsto h(u_0)$ each of which is nontrivial by Lemma 66.12.1. Hence we see that the image of $|U|$ in $|X|$ has dimension at least $d$. Conversely, suppose that $x_ d \leadsto \ldots \leadsto x_0$ is a sequence of specializations in $|X|$ with $x_0$ in the image of $|U| \to |X|$. Then we can lift this to a sequence of specializations in $U$ by Lemma 66.12.2.

Part (2) is an immediate consequence of part (1), Topology, Lemma 5.10.2, and Properties of Spaces, Section 64.9. $\square$


Comments (2)

Comment #1429 by Matthew Emerton on

You state that part (2) is an immediate consequence of the definitions (given part (1)), but is it? The dimension of is defined as a sup, but not the dimension of . E.g. for schemes, the fact that the dimension of can be computed as the sup over the dimensions at the various points is proved as a particular fact about schemes, not as a statement in general topology. So maybe something more has to be said? (Sorry if I'm just confused, or missing something obvious!)

Comment #1443 by on

OK, I think it is just a generality for topological spaces. I added a lemma to the chapter on topology (will be online soon), so you can criticize it if it is wrong. See here for the latex code. Thanks!


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