# Why is the Grassmannian functor representable by a scheme?

I am having some trouble following a proof that the Grassmannian functor is representable by a scheme. I am following the proof in EGA 9.7.4. It is only a small step that I am stuck on. For reference, let $$\mathcal{E}$$ be a quasicoherent sheaf on a scheme $$S$$ and we define the functor $$F: Sch/S \rightarrow \text{Set}$$ by (just thinking about the n=1 case since that should clarify all I need): $$F(T \stackrel{g}{\rightarrow}S) = \{ \text{invertible sheaves } \mathcal{L} : (g^{*}\mathcal{E} \rightarrow \mathcal{L} \rightarrow 0 )/\sim \}$$ where the equivalence $$\sim$$ is just isomorphisms commuting with the quotient in the obvious way.

It is easy enough to see that $$F$$ is a Zariski sheaf. So consider when $$S = \text{Spec} A$$ is affine. All I need to do is cover $$F$$ by representable open subfunctors, which is where I run into a problem.

From what I understand, Grothendieck's argument is as follows: Let $$\mathcal{E} = \tilde{M}$$ be generated by sections $$\{ m_{i} \}_{i \in I}$$. Then the sections $$\{ g^{*}m_{i} \}_{i \in I}$$ generated $$g^{*} \mathcal{E}$$ on $$T$$ and so correspond to a surjection, $$\bigoplus_{i \in I} \mathcal{O}_{T}^{(i)} \longrightarrow g^{*} \mathcal{E} \longrightarrow 0.$$ He then seems to appeal to the fact that for the $$n=1$$ case, such a surjection must factor through precisely one of the summands. This is where I get lost. It looks like some kind of compact object argument, but any argument I can see would rely on strong finiteness assumptions on the scheme $$T$$, such as $$T$$ being at least quasicompact (and probably quasiseparated). Can anyone explain how the subfunctors are defined, and how they go on to cover $$F$$?

For reference, I found this note which seems to suggest quasicompactness is necessary also.

• This should be asked on mathoverflow. May 2, 2021 at 11:34