I am looking for an example of a contravariant functor from $Sch \to Set$ which is a Zariski sheaf, but which is not representable.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
You just have to take the constant sheaf defined by an infinite set.
answered Jan 15, 2016 at 3:49
-
$\begingroup$ Even the constant sheaf with two elements would do. $\endgroup$– Zhen LinJan 15, 2016 at 8:20
-
1$\begingroup$ I'm confused - I'm pretty convinced that the infinite disjoint union of $\operatorname{spec} \mathbb{Z}$ works (or at least the finite disjoint union for ZhenLins example), but maybe I'm missing something... why do you think this sheaf isn't representable? $\endgroup$ Jan 15, 2016 at 19:58
-
$\begingroup$ @ZhenLin Why not take $X$ to be the disjoint union of two copies of $\operatorname{Spec} \mathbb{Z}$? There is a natural transformation between the functor represented by $X$ and the constant sheaf with two elements functor, given by collapsing each $\operatorname{Spec} \mathbb{Z}$ to a point. The inverse is given by taking a $f : Z \to \{0,1\}$, to the map $Z \to X$ glued from the unique maps $f^{-1}(0) \to \operatorname{Spec} \mathbb{Z} \to_0 X$ and $f^{-1}(1) \to \operatorname{Spec} \mathbb{Z} \to_1 X$. Is this wrong? $\endgroup$ Jan 16, 2016 at 22:07
-
$\begingroup$ Yes, sorry. I was thinking of the constant presheaf, which is different. $\endgroup$– Zhen LinJan 16, 2016 at 22:11