In the following we fix a big étale site $\mathit{Sch}_{\acute{e}tale}$ as constructed in Topologies, Definition 34.4.6. Moreover, a scheme will be an object of this site. Recall that if $x, x'$ are points of a scheme $X$ we say $x$ is a specialization of $x'$ or we write $x' \leadsto x$ if $x \in \overline{\{ x'\} }$. This is true in particular if $x = x'$.
Consider the functor $F : \mathit{Sch}_{\acute{e}tale}\to \textit{Ab}$ defined by the following rules:
Given a scheme $X$ we denote $|X|$ the underlying set of points. An element $a \in F(X)$ will be viewed as a map of sets $|X| \times |X| \to \mathbf{Z}/2\mathbf{Z}$, $(x, x') \mapsto a(x, x')$ which is zero if $x = x'$ or if $x$ is not a specialization of $x'$. Given a morphism of schemes $f : X \to Y$ we define
by the rule that for $b \in F(Y)$ we set
Note that this really does define an element of $F(X)$. We claim that if $f : X \to Y$ and $g : Y \to Z$ are composable morphisms then $F(f) \circ F(g) = F(g \circ f)$. Namely, let $c \in F(Z)$ and let $x' \leadsto x$ be a specialization of points in $X$, then
because $f(x') \leadsto f(x)$. (This also works if $f(x) = f(x')$.)
Let $G$ be the sheafification of $F$ in the étale topology.
I claim that if $X$ is a scheme and $x' \leadsto x$ is a specialization and $x' \not= x$, then $G(X) \not= 0$. Namely, let $a \in F(X)$ be an element such that when we think of $a$ as a function $|X| \times |X| \to \mathbf{Z}/2\mathbf{Z}$ it is nonzero at $(x, x')$. Let $\{ f_ i : U_ i \to X\} $ be an étale covering of $X$. Then we can pick an $i$ and a point $u_ i \in U_ i$ with $f_ i(u_ i) = x$. Since generalizations lift along flat morphisms (see Morphisms, Lemma 29.25.9) we can find a specialization $u'_ i \leadsto u_ i$ with $f_ i(u'_ i) = x'$. By our construction above we see that $F(f_ i)(a) \not= 0$. Hence $a$ determines a nonzero element of $G(X)$.
Note that if $X = \mathop{\mathrm{Spec}}(k)$ where $k$ is a field (or more generally a ring all of whose prime ideals are maximal), then $F(X) = 0$ and for every étale morphism $U \to X$ we have $F(U) = 0$ because there are no specializations between distinct points in fibres of an étale morphism. Hence $G(X) = 0$.
Suppose that $X \subset X'$ is a thickening, see More on Morphisms, Definition 37.2.1. Then the category of schemes étale over $X'$ is equivalent to the category of schemes étale over $X$ by the base change functor $U' \mapsto U = U' \times _{X'} X$, see Étale Cohomology, Theorem 59.45.2. Since it is always the case that $F(U) = F(U')$ in this situation we see that also $G(X) = G(X')$.
As a variant we can consider the presheaf $F_ n$ which associates to a scheme $X$ the collection of maps $a : |X|^{n + 1} \to \mathbf{Z}/2\mathbf{Z}$ where $a(x_0, \ldots , x_ n)$ is nonzero only if $x_ n \leadsto \ldots \leadsto x_0$ is a sequence of specializations and $x_ n \not= x_{n - 1} \not= \ldots \not= x_0$. Let $G_ n$ be the sheaf associated to $F_ n$. In exactly the same way as above one shows that $G_ n$ is nonzero if $\dim (X) \geq n$ and is zero if $\dim (X) < n$.
Lemma 110.56.1. There exists a sheaf of abelian groups $G$ on $\mathit{Sch}_{\acute{e}tale}$ with the following properties
$G(X) = 0$ whenever $\dim (X) < n$,
$G(X)$ is not zero if $\dim (X) \geq n$, and
if $X \subset X'$ is a thickening, then $G(X) = G(X')$.
Proof. See the discussion above. $\square$
Remark 110.56.2. Here are some remarks:
The presheaves $F$ and $F_ n$ are separated presheaves.
It turns out that $F$, $F_ n$ are not sheaves.
One can show that $G$, $G_ n$ is actually a sheaf for the fppf topology.
We will prove these results if we need them.
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