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.

  • $\begingroup$ In the title, did you mean 'does not come from Sch' instead of 'does not come from a sheaf on Sch'? $\endgroup$
    – Marc
    Jan 4, 2019 at 19:57
  • $\begingroup$ 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? $\endgroup$
    – jgon
    Jan 4, 2019 at 23:02
  • $\begingroup$ @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. $\endgroup$
    – Javi
    Jan 4, 2019 at 23:18
  • $\begingroup$ @MarcPaul Yes, I wasn't understading the text well. $\endgroup$
    – Javi
    Jan 4, 2019 at 23:25
  • $\begingroup$ Oh, yeah, sorry I missed that $\endgroup$
    – jgon
    Jan 5, 2019 at 0:02

1 Answer 1


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 $G(\mathrm{Spec}\,K)= G(\mathrm{Spec}\,L)^{\mathrm{Gal}(L/K)}$.

  • $\begingroup$ Could you ellaborate why it is not a sheaf for the étale topology? I know what a sheaf for this topology is, but I don't see where this one fails to be a sheaf. Is it related to complex conjugation not being an étale morphism? $\endgroup$
    – Javi
    Jan 5, 2019 at 12:00
  • 1
    $\begingroup$ I added an explanation of the étale sheaf condition for a finite Galois extension. $\endgroup$ Jan 5, 2019 at 17:05

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