Ytukyg

At the end of Remarque 2.3.6 (p. 221-222) of EGA I, the author says that there are functors in $mathbf{Fais}|_{mathbf{Ann}}$ (sheaf on the category of Rings) that are not isomorphic to sheaves that come from schemes. I would like to know one such example or if such example is constructed later on the book.

I'm adding the definition and context of each concept below:

A functor $G:mathbf{Aff}^{op}tomathbf{Set}$ from the opposite category of affine schemes to the category of sets is called a presheaf. Given an affine scheme $X$, for any open subscheme $U$, one can consider the map $Umapsto G(U)$. We say that $G$ is a sheaf when this map is always a sheaf in the usual sense.

Since there exist an equivalence of categories $F:mathbf{Aff}^{op}tomathbf{Ring}$ between the category of affine schmes and the category of rings, that also defines an equivalence $mathbf{Hom(Aff^{op},Set)}congmathbf{Hom(Ring,Set)}$. Hece we can define a sheaf on the category of rings as a (covariant) functor $mathbf{Ring}tomathbf{Set}$ whose image under the previous equivalence is a sheaf in the sense defined earlier.

Similarily we can define a sheaf on the category of schemes $mathbf{Sch}$, but it turns out that the category of such sheaves is equivalent to that of sheaves on affine schemes. One can prove that, given an scheme $X$, the functor $h_X:Ymapstomathrm{Hom}(Y,X)$ is a sheaf on $mathbf{Sch}$, and since $h:Xmapsto h_X$ is fully faithful, we can identify the category of schemes with a subcategory of the sheaves on $mathbf{Ring}$ by the previous equivalences.

Define $tilde{G}(X)={f^2:finmathrm{Hom}(X,mathbb{A}^1)}$. Then $tilde{G}$ is a presheaf, and we take $G$ to be the sheaffification of $tilde{G}$. Concretely, $G(X)$ is the set of functions $f:Xtomathbb{A}^1$ such that there exists an open cover ${U_i}$ of $X$ and functions $g_i:U_itomathbb{A}^1$ such that $f|_{U_i}=g_i^2$. I claim that $G$ is not representable by a scheme. To see this, observe that $G(mathrm{Spec},mathbb{R})=mathbb{R}_{geq 0}$, $G(mathrm{Spec},mathbb{C})=mathbb{C}$, and the action of complex conjugation on $mathrm{Spec},mathbb{C}$ induces complex conjugation on $G(mathrm{Spec},mathbb{C})=mathbb{C}$. If $X$ is any scheme, then $mathrm{Hom}(mathrm{Spec},mathbb{R},X)$ is always the subset of $mathrm{Hom}(mathrm{Spec},mathbb{C},X)$ of elements fixed by complex conjugation.

The point here is that although $G$ is a sheaf for the Zariski topology (meaning $G$ gives an ordinary sheaf on each affine scheme), $G$ is not a sheaf for the etale topology. In general, let $G$ be an arbitrary étale sheaf, and let $L/K$ be a finite Galois extension. Then $mathrm{Spec},Ltomathrm{Spec},K$ is an étale covering, so there should be an equalizer diagram
$$
G(mathrm{Spec},K)to G(mathrm{Spec},L)rightrightarrows G(mathrm{Spec},L times_{mathrm{Spec}, K}mathrm{Spec}, L)=G(mathrm{Spec}, Lotimes_K L).
$$
Now $Lotimes_K L$ is a product of copies of $L$ indexed by $mathrm{Gal}(L/K)$, so
$$
G(mathrm{Spec},Lotimes_K L)=prod_{ginmathrm{Gal}(L/K)} G(mathrm{Spec}(L))
$$
and the two maps $ G(mathrm{Spec}(L))toprod_{ginmathrm{Gal}(L/K)} G(mathrm{Spec}(L))$ are the diagonal, and the map whose $g$-th component is induced by $g$. The equalizer in question is then the set of $mathrm{Gal}(L/K)$-invariants, so $G(mathrm{Spec},K)= G(mathrm{Spec},L)^{mathrm{Gal}(L/K)}$.