65.22 Stalks of the structure sheaf

This section is the analogue of Étale Cohomology, Section 59.33.

Lemma 65.22.1. 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 65.19.9 part (4). The second equality is explained in Étale Cohomology, Lemma 59.33.1. $\square$

Definition 65.22.2. 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 $x \in |X|$.

1. 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: $\mathcal{O}_{X, \overline{x}}$.

2. 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 65.19.11.

Lemma 65.22.3. 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 65.22.1 and Theorem 65.19.12. 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 65.22.4. 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 65.10.2) 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 65.19.11. Using Lemma 65.22.1 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 65.22.5. 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 65.9.1 for (1). Part (3) is an immediate consequence of (1), see Definition 65.9.2. $\square$

Lemma 65.22.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$ be a point. The following are equivalent

1. the local ring of $X$ at $x$ is reduced (Remark 65.7.6),

2. $\mathcal{O}_{X, \overline{x}}$ is reduced for some geometric point $\overline{x}$ lying over $x$, and

3. $\mathcal{O}_{X, \overline{x}}$ 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 65.19.11. Using Lemma 65.22.1 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 on

The reference in the first paragraph goes wrong, I guess you want to refer to something 59.33, for which you'll have to use the filename too in the ref.

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