# How do I complete the sketch of proof in 'FGA explained', 5.5.8?

It seems to me that the one part that is difficult to transfer to general coherent sheaves is given only one sentence: "Because we have such a common $m$, we get as before an injective morphism from the functor $\mathfrak{Quot}^{\Phi,L}_{E/X/S}$ into the Grassmanian functor $\mathfrak{Grass}(\pi_\ast E(r), \Phi(r))$." But I see at least two serious hardships with this proof: First, why is $R^1 \pi_{T,\ast} \left( \mathcal{G}_t (r) \right) = 0$? Second, even if that is true, the values on $T$ are quotients of $\pi_{T,\ast} E_T$, not of $(\pi_\ast E)_T$, where the latter is the pullback of the pushforward. These two sheaves are the same for some $r$ by Theorem 5.4 (3.1 in the standalone paper), but that $r$ might depend on $T$.

Edit: The statement of the theorem that has a more or less thorough proof in the book "Fundamental algebraic geometry: Grothendieck's FGA explained.", in part 5, by Nitin Nitsure (it exists as a separate paper [Nitsure, Nitin (2005), "Construction of Hilbert and Quot schemes", Fundamental algebraic geometry, Math. Surveys Monogr. 123, Providence, R.I.: Amer. Math. Soc., pp. 105–137, arXiv:math/0504590, MR 2223407]), is as follows: (theorem 5.2 in the paper, 5,15 in the book)

Let $S$ be a Noetherian scheme, and $X$ a closed subscheme in $\mathbb{P}(V)$ for some vector bundle $V$, let $\pi$ denote the stuctral morphism $X \to S$. Let $E$ be a coherent factor-sheaf of $\pi^\ast(W)(\nu)$ where $W$ is an $S$-vector bundle and $\nu$ is an integer. Then for any integer-valued polynomial $\Phi$, the functor $\mathfrak{Quot}_{E/X/S}^{\Phi,\mathcal{O}(1)}$ is representable by a closed subscheme of the Grassmanian $Gr(W \otimes Sym^r(V),\Phi(r))$ for large enough r.

It differs from the original Grothendieck's statement in that in general, not all projective schemes embed into projectivisations of vector bundles (if there is an ample bundle on $S$, they do) Also, the bundle $E$ is just any coherent sheaf in Grothendieck's version of the theorem.

Because I don't know what the author meant when he wrote that the part that I pointed out generalises easily, I guess I will just write down that part of the proof (it is implicit in the text, I had to guess the specifics myself) and note where I see difficulty.

First of all, $X$ is obviously replaced with $\mathbb{P}(V)$. Then the factor-sheaf of $\pi^\ast (W) (\nu)$ is replaced by $\pi^\ast(W)$ - this induces a closed embedding on the $\mathfrak{Quot}$ sheaves. Here I don't see how to generalise - to express an arbitrary coherent sheaf as a quotient of a vector bundle. So, over an $S$-scheme $T$, on $\mathbb{P}(H) \times_S T$ there is an exact sequence $0 \to \mathcal{G} \to E_T \to \mathcal{F} \to 0.$ And what seems crucial is that all these three sheaves are flat over $T$. On fibers $s$, $\mathcal{G}_s$ is a subsheaf of a free sheaf $E_s$, and we know the Hilbert polynome of $\mathcal{G}$ so Mumford's theorem (5.3 in the book, 2.3 in the paper) is used to show that there exists an $m$ which does not depend on $s$ or the particular $\mathcal{F}$ such that all three sheaves in the exact sequence on the fiber are (Castelnuovo-Mumford) $m$-regular. (perhaps for general $E$ one can emulate the proof of Mumford's theorem, uniformly bounding the dimensions of global sections of $E$ on subspaces in fibers) Then after twisting by $r$ for $r \geq m$, the three sheaves have no higher cohomology on fibers. Then by the Flatness and Base Change Theorem (strongly using that the sheaves are flat over $S$), the higher cohomology vanish. They are also relatively globally generated (maybe after a twist by n), I hae thought of the following argument: they become "relatively 0-regular" after thet twist, so they are relatively globally generated. Since $R^1\pi_{T,\ast}(\mathcal{G(r)})=0$, we get an exact sequence $0 \to \pi_{T,\ast}(\mathcal{G}(r)) \to \pi_{T,\ast}(E(r)) \to \pi_{T,\ast}(\mathcal{F}(r)) \to 0$, and thus a morphism to the Grassmanian. The global generation (of $\mathcal{G}(r)$) is used to show that the morphism is injective and to find its image.

The statement of Mumford's theorem is as follows: There exists a polynome with integer coefficients $F_{d,n}$ in $n+1$ variables such that for any $\mathcal{F}$ a coherent subsheaf of $\oplus^d \mathcal{O}_{\mathbb{P}^n_k}$, $\mathcal{F}$ is $F_{n,d}(a_0, \cdots a_n)$-regular where $a_i$ are the coordinates of the hilbert polynome of $\mathcal{F}$ in the binomial basis.

• You're far more likely to get an answer to your question if you make it self-contained. Maybe someone knows some alg geom but doesn't have FGA explained (e.g. maybe they read FGA!) and they're not going to be answering your question as it stands. – znt Jun 2 '16 at 21:52
• As @znt says, you should make your question self-contained. As a minimum, you should include the statement of the result that you are asking about, with a proper reference to the document in which it appears, preferably with a link. – Neil Strickland Jun 3 '16 at 6:13
• I hope the text that I wrote is helpful. It is really hard to make the question self-contained because the statements in the text are interconnected and many of them I have not met before. Maybe I should have just posted the link to the paper, which contains the text available for anyone. – Alexei Tsybyshev Jun 3 '16 at 14:20