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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