# Example of sheaf on $\mathrm{Ring}$ that does not come from $\mathrm{Sch}$.

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 $$G:\mathbf{Aff}^{op}\to\mathbf{Set}$$ 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 $$U\mapsto G(U)$$. 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 $$F:\mathbf{Aff}^{op}\to\mathbf{Ring}$$ between the category of affine schmes and the category of rings, that also defines an equivalence $$\mathbf{Hom(Aff^{op},Set)}\cong\mathbf{Hom(Ring,Set)}$$. Hece we can define a sheaf on the category of rings as a (covariant) functor $$\mathbf{Ring}\to\mathbf{Set}$$ 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 $$h_X:Y\mapsto\mathrm{Hom}(Y,X)$$ is a sheaf on $$\mathbf{Sch}$$, and since $$h:X\mapsto h_X$$ is fully faithful, we can identify the category of schemes with a subcategory of the sheaves on $$\mathbf{Ring}$$ by the previous equivalences.

• In the title, did you mean 'does not come from Sch' instead of 'does not come from a sheaf on Sch'?
– Marc
Jan 4, 2019 at 19:57
• What topology are you using on these categories? Also I can't find the remark in the first edition of EGA I, are you using the second?
– jgon
Jan 4, 2019 at 23:02
• @jgon For these particular purposes no topology is used on these categories, since we are saying that a functor is a sheaf when the map $U\mapsto G(U)$ is a sheaf for each $X$. I'm not using that edition, the one I'm using is published by Springer, so maybe it is the second, I don't know.
– Javi
Jan 4, 2019 at 23:18
• @MarcPaul Yes, I wasn't understading the text well.
– Javi
Jan 4, 2019 at 23:25
• Oh, yeah, sorry I missed that
– jgon
Jan 5, 2019 at 0:02

Define $$\tilde{G}(X)=\{f^2:f\in\mathrm{Hom}(X,\mathbb{A}^1)\}$$. 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 $$f:X\to\mathbb{A}^1$$ such that there exists an open cover $$\{U_i\}$$ of $$X$$ and functions $$g_i:U_i\to\mathbb{A}^1$$ 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 $$\mathrm{Spec}\,L\to\mathrm{Spec}\,K$$ is an étale covering, so there should be an equalizer diagram $$G(\mathrm{Spec}\,K)\to G(\mathrm{Spec}\,L)\rightrightarrows G(\mathrm{Spec}\,L \times_{\mathrm{Spec}\, K}\mathrm{Spec}\, L)=G(\mathrm{Spec}\, L\otimes_K L).$$ Now $$L\otimes_K L$$ is a product of copies of $$L$$ indexed by $$\mathrm{Gal}(L/K)$$, so $$G(\mathrm{Spec}\,L\otimes_K L)=\prod_{g\in\mathrm{Gal}(L/K)} G(\mathrm{Spec}(L))$$ and the two maps $$G(\mathrm{Spec}(L))\to\prod_{g\in\mathrm{Gal}(L/K)} G(\mathrm{Spec}(L))$$ 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 $$G(\mathrm{Spec}\,K)= G(\mathrm{Spec}\,L)^{\mathrm{Gal}(L/K)}$$.