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The Stacks project

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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