Lemma 66.22.4. 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. the dimension of the local ring of $X$ at $x$ (Definition 66.10.2) is $d$,

2. $\dim (\mathcal{O}_{X, \overline{x}}) = d$ for some geometric point $\overline{x}$ lying over $x$, and

3. $\dim (\mathcal{O}_{X, \overline{x}}) = d$ for any geometric point $\overline{x}$ lying over $x$.

Proof. The equivalence of (2) and (3) follows from the fact that the isomorphism type of $\mathcal{O}_{X, \overline{x}}$ only depends on $x \in |X|$, see Remark 66.19.11. Using Lemma 66.22.1 the equivalence of (1) and (2)$+$(3) comes down to the following statement: Given any local ring $R$ we have $\dim (R) = \dim (R^{sh})$. This is More on Algebra, Lemma 15.45.7. $\square$

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