## 66.10 Dimension of local rings

The dimension of the local ring of an algebraic space is a well defined concept.

Lemma 66.10.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$ be a point. Let $d \in \{ 0, 1, 2, \ldots , \infty \} $. The following are equivalent

for some scheme $U$ and étale morphism $a : U \to X$ and point $u \in U$ with $a(u) = x$ we have $\dim (\mathcal{O}_{U, u}) = d$,

for any scheme $U$, any étale morphism $a : U \to X$, and any point $u \in U$ with $a(u) = x$ we have $\dim (\mathcal{O}_{U, u}) = d$.

If $X$ is a scheme, this is equivalent to $\dim (\mathcal{O}_{X, x}) = d$.

**Proof.**
Combine Lemma 66.7.4 and Descent, Lemma 35.21.3.
$\square$

Definition 66.10.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$ be a point. The *dimension of the local ring of $X$ at $x$* is the element $d \in \{ 0, 1, 2, \ldots , \infty \} $ satisfying the equivalent conditions of Lemma 66.10.1. In this case we will also say *$x$ is a point of codimension $d$ on $X$*.

Besides the lemma below we also point the reader to Lemmas 66.22.4 and 66.22.5.

Lemma 66.10.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following quantities are equal:

The dimension of $X$.

The supremum of the dimensions of the local rings of $X$.

The supremum of $\dim _ x(X)$ for $x \in |X|$.

**Proof.**
The numbers in (1) and (3) are equal by Definition 66.9.2. Let $U \to X$ be a surjective étale morphism from a scheme $U$. The supremum of $\dim _ x(X)$ for $x \in |X|$ is the same as the supremum of $\dim _ u(U)$ for points $u$ of $U$ by definition. This is the same as the supremum of $\dim (\mathcal{O}_{U, u})$ by Properties, Lemma 28.10.2. This in turn is the same as (2) by definition.
$\square$

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