66.22 Stalks of the structure sheaf
This section is the analogue of Étale Cohomology, Section 59.33.
Lemma 66.22.1. Let S be a scheme. Let X be an algebraic space over S. Let \overline{x} be a geometric point of X. Let (U, \overline{u}) be an étale neighbourhood of \overline{x} where U is a scheme. Then we have
\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{U, \overline{u}} = \mathcal{O}_{U, u}^{sh}
where the left hand side is the stalk of the structure sheaf of X, and the right hand side is the strict henselization of the local ring of U at the point u at which \overline{u} is centered.
Proof.
We know that the structure sheaf \mathcal{O}_ U on U_{\acute{e}tale} is the restriction of the structure sheaf of X. Hence the first equality follows from Lemma 66.19.9 part (4). The second equality is explained in Étale Cohomology, Lemma 59.33.1.
\square
Definition 66.22.2. Let S be a scheme. Let X be an algebraic space over S. Let \overline{x} be a geometric point of X lying over the point x \in |X|.
The étale local ring of X at \overline{x} is the stalk of the structure sheaf \mathcal{O}_ X on X_{\acute{e}tale} at \overline{x}. Notation: \mathcal{O}_{X, \overline{x}}.
The strict henselization of X at \overline{x} is the scheme \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \overline{x}}).
The isomorphism type of the strict henselization of X at \overline{x} (as a scheme over X) depends only on the point x \in |X| and not on the choice of the geometric point lying over x, see Remark 66.19.11.
Lemma 66.22.3. Let S be a scheme. Let X be an algebraic space over S. The small étale site X_{\acute{e}tale} endowed with its structure sheaf \mathcal{O}_ X is a locally ringed site, see Modules on Sites, Definition 18.40.4.
Proof.
This follows because the stalks \mathcal{O}_{X, \overline{x}} are local, and because S_{\acute{e}tale} has enough points, see Lemmas 66.22.1 and Theorem 66.19.12. See Modules on Sites, Lemma 18.40.2 and 18.40.3 for the fact that this implies the small étale site is locally ringed.
\square
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
the dimension of the local ring of X at x (Definition 66.10.2) is d,
\dim (\mathcal{O}_{X, \overline{x}}) = d for some geometric point \overline{x} lying over x, and
\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
Lemma 66.22.5. Let S be a scheme. Let f : X \to Y be an étale morphism of algebraic spaces over S. Let x \in X. Then (1) \dim _ x(X) = \dim _{f(x)}(Y) and (2) the dimension of the local ring of X at x equals the dimension of the local ring of Y at f(x). If f is surjective, then (3) \dim (X) = \dim (Y).
Proof.
Choose a scheme U and a point u \in U and an étale morphism U \to X which maps u to x. Then the composition U \to Y is also étale and maps u to f(x). Thus the statements (1) and (2) follow as the relevant integers are defined in terms of the behaviour of the scheme U at u. See Definition 66.9.1 for (1). Part (3) is an immediate consequence of (1), see Definition 66.9.2.
\square
Lemma 66.22.6. Let S be a scheme. Let X be an algebraic space over S. Let x \in |X| be a point. The following are equivalent
the local ring of X at x is reduced (Remark 66.7.6),
\mathcal{O}_{X, \overline{x}} is reduced for some geometric point \overline{x} lying over x, and
\mathcal{O}_{X, \overline{x}} is reduced 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: a local ring is reduced if and only if its strict henselization is reduced. This is More on Algebra, Lemma 15.45.4.
\square
Comments (2)
Comment #1372 by Pieter Belmans on
Comment #1385 by Johan on