## 64.10 Dimension of local rings

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

Lemma 64.10.1. 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

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

2. 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 64.7.4 and Descent, Lemma 35.18.3. $\square$

Definition 64.10.2. 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 64.10.1. 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 64.22.4 and 64.22.5.

Lemma 64.10.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following quantities are equal:

1. The dimension of $X$.

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

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

Proof. The numbers in (1) and (3) are equal by Definition 64.9.2. 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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