I am trying to use a well known result of Grothendieck to show that if $S$ is a scheme, and $\mathcal{B}$ is a quasi coherent sheaf of $\mathcal{O}_{S}$-algebras, then there is a relative affine spectrum which is a scheme over $S$.

Denote by $\text{Sch}_{/S}$ the category of schemes over $S$. I am trying to use the result which states that for a functor $$ F: \text{Sch}_{/S} \longrightarrow \text{Set} $$ to be representable, it is necessary and sufficient that it be a Zariski sheaf which is covered by representable open subfunctors.

Grothendieck proves a corollary of this (EGA, which states that if $\{S_{i} \}$ is an affine (or indeed any open) cover of $S$ and $F$ is a Zariski sheaf, then we need only show that the functors $$ F_{i} = F \times_{h_{S}} h_{S_{i}} $$ are representable.

My problem is that I have no idea what this pullback diagram even is. The functor $h_{S}$ is (presumably) the functor which sends any $S$-scheme $(f: X \rightarrow S)$ to the set with one element, namely $\{ f \}$. But then the $h_{S_{i}}$ certainly are not going to be subfunctors of $h_{S}$, and by extension the $F_{i}$ will not be subfunctors of $F$, let alone open subfunctors.

What am I missing here? What is $h_{S}$ actually supposed to be? If I assume that it's an abuse of notation and that $h_{S}$ is really the functor that sends a scheme (not an $S$-scheme) $X$ to the set $\text{Hom}(X, S)$, then what is the morphism of functors $F \rightarrow h_{S}$ supposed to be?

  • $\begingroup$ Luke, was there something else you needed clarified about this problem? $\endgroup$ Commented Apr 7, 2019 at 20:04

1 Answer 1


I think what you're looking for is the following. Let $F$ be a sheaf on $X$ and let $\{U_i\}$ be an open cover of $X$. Then, we know by the definition of a sheaf that the diagram

$$0\to F\to \prod_i (\iota_i)_\ast(F\mid_{U_i})\to \prod_{ij}(\iota_{i,j})_\ast(F\mid_{U_i\cap U_j})\qquad (1)$$

is exact where $\iota_i:U_i\hookrightarrow X$ and $\iota_{ij}:U_i\cap U_j\hookrightarrow X$.

Suppose now that $F\mid_{U_i}\cong h_{S_i}$ for some $U_i$-schemes $S_i$. The isomorphisms $(F\mid_{U_i})\mid_{U_j}\cong (F\mid_{U_j})\mid_{U_i}$ give rise to isomorphisms $\varphi_{ij}:(S_i)_{U_i\cap U_j}\xrightarrow{\approx} (S_j)\mid_{U_i\cap U_j}$. We then get a $U$-scheme $S$ (e.g. see Tag01JA). Then, check that $h_S$ sits in the same exact sequence as in $(1)$ (e.g. see the description of the functor of points as in Lemma 25.14.1 of loc. cit.).

I guess technically I only addressed the question on the small Zariski site of $X$, but you can get the claim on the big Zariski site by just noting that if you have any map $f:X'\to X$ you can pull the arguments back to the small Zariski site of $X'$ using the open cover $f^{-1}(U_i)$ and proceeding in the same way.

  • $\begingroup$ Thank you for the response. I'm still not entirely sure how that would give a natural transformation $F \rightarrow h_{S}$. I should also clarify that $h_{S}$ is really an abuse of notation for $h_{(S, \text{id}_{S})}$, right? $\endgroup$
    – Luke
    Commented Apr 7, 2019 at 2:19
  • $\begingroup$ @Luke To show that they have the same description of points functorially? Also, what does $h_{(S,\id)}$ mean? Be careful, I just realized that I switched notation from your post. In my language it's the fact that we've constructed an $X$-scheme $S$, let's say with structure morphism $f:S\to X$, and thus $h_S$ means $h_{(S,f)}$ if I'm interpreting your question correctly. $\endgroup$ Commented Apr 7, 2019 at 2:22
  • $\begingroup$ @Luke chat.stackexchange.com/rooms/92097/chat $\endgroup$ Commented Apr 7, 2019 at 2:24

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