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I am trying to understand the proof of Theorem 4.6.2.1 in the book on algebraic stacks by Laumon and Moret-Bailly. The setting is this: $S$ is a Noetherian scheme, $f\colon X \rightarrow S$ is a projective $S$-scheme such that $f_*\mathscr{O}_X = \mathscr{O}_S$ universally (i.e. also after any base change on $S$), and we are trying to prove that the stack $\mathscr{C}oh_{X/S}$ that parametrizes $U$-flat quasi-coherent locally of finite presentation $\mathscr{O}_{X_{U}}$-modules for variable $S$-schemes $U$ is actually algebraic.

The proof proceeds by arguing that $$ \bigsqcup_{N, n \ge 0} \mathrm{Quot}^{\circ}_{\mathscr{O}_X^N/X/S} \rightarrow \mathscr{C}oh_{X/S},$$ is a smooth atlas, where for a variable $S$-scheme $U$ the $S$-scheme $\mathrm{Quot}^{\circ}_{\mathscr{O}_X^N/X/S}$ parametrizes $S$-flat $\mathscr{O}_{X_{U}}$-module quotients $\mathscr{Q}$: $$(\alpha: \mathscr{O}_{X_U}^N \twoheadrightarrow \mathscr{Q})$$ satisfying

1) $R^p(f_U)_* \mathscr{Q} = 0$ for $p > 0$,

2) $(f_U)_*(\alpha)\colon \mathscr{O}_U^N \rightarrow (f_U)_*(\mathscr{Q})$ is an isomorphism (in particular, $(f_U)_*(\mathscr{Q})$ is locally free of rank $N$).

The atlas map is $\alpha \mapsto \mathscr{Q}(-n)$ (the same $n$ that is part of the index of the disjoint union).

The proof concludes by arguing surjectivity of the atlas map: take any $U \rightarrow \mathscr{C}oh_{X/S}$ with $U$ affine, it corresponds to an $U$-flat $\mathscr{O}_{X_{U}}$-module $\mathscr{M}$. We need to show that etale locally on $U$ the module $\mathscr{M}$ comes from some $\mathscr{Q}$ and $n$. In fact, Zariski locally will suffice: take any large $n$ for which $R^p(f_U)_*(\mathscr{M}(n)) = 0$ for all $p > 0$ and for which $(f_U)^*((f_U)_*(\mathscr{M}(n))) \rightarrow \mathscr{M}(n)$ is surjective (we use projectivity of $f_U$), then subdivide $U$ into $U_{N, n}$ according to the rank $N$ of the locally free $\mathscr{O}_U$-module $(f_U)_*(\mathscr{M}(n))$, and observe that $$U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S} \rightarrow U$$ factors through $U_{N, n}$ and is in fact the $\mathrm{GL}_{N, U_{N, n}}$-torsor $$\mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n))).$$ I agree that the map factors through $U_{N, n}$ but the rest is unclear to me: by definition the displayed fiber product should be parametrizing isomorphisms over $X_{U_{N, n}}$, so I presume given an "element" $\beta$ in this $\mathscr{I}som$ functor one realizes $\mathscr{M}(n)$ as a $\mathscr{Q}$ as above via the composition $$(f_{U_{N, n}})^*(\beta)\colon \mathscr{O}_{X_{U_{N, n}}}^N \xrightarrow{\sim} (f_{U_{N, n}})^*((f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n))) \twoheadrightarrow \mathscr{M}_{U_{N, n}}(n).$$ But to check that this is really a valid $\mathscr{Q}$ one needs to check 2) above, namely, that $(f_{U_{N, n}})_*((f_{U_{N, n}})^*(\beta))$ is an isomorphism. My question is: why is this an isomorphism? This being an isomorphism is like saying that a set of global sections generating the sheaf $\mathscr{M}_{U_{N, n}}(n)$ already generates $\Gamma(X_{U_{N, n}}, \mathscr{M}_{U_{N, n}}(n))$, which seems doubtful. By cohomology and base change, in checking this one may assume that the base is a field, but even this case is unclear to me.

I apologize that the question seems so long. The length came from trying to state it in a self-contained way, but in the end this is a basic question about a key step in the proof of the algebraicity of $\mathscr{C}oh_{X/S}$. I would be very grateful if someone could explain how to check the missing compatibility mentioned above (alternatively, how to modify the argument in the book) -- this would clear up any doubts about the completeness of the proof.

EDIT. Here is a short version of the question summarizing the issue

By definition, $U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S}$ parametrizes a $\mathscr{Q}$ (with its $\alpha$) + an isomorphism $x$ over $X$ between the $\mathscr{Q}$ and the $\mathscr{M}(n)$. On the other hand, $\mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n)))$ parametrizes isomorphisms $u$ over (an open of) $U$. We need to identify the two functors. There are $U$-maps $$U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S} \rightarrow \mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n))), \quad x \mapsto f_*(x\circ \alpha)$$ and $$\mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n))) \rightarrow U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S}, \quad u \mapsto \mathrm{(adjunction\ surjection)} \circ(f^*u),$$ which are candidates for the sought isomorphism. My question is why is the second of these maps well-defined? In other words, why does the surjection $$\mathrm{(adjunction\ surjection)}\circ (f^*u) \colon \mathscr{O}_{X_{U_{N, n}}}^N \twoheadrightarrow \mathscr{M}(n)$$ satisfy the condition 2) above (i.e. why is its pushforward an isomorphism)?

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  • $\begingroup$ I do not understand your concern. By the definition, $(f_{U_N})_*\mathcal{M}_{U_N}(n)$ is locally free of rank $N$. Now you have a flat base change by some $g:I\to U_{N,n}$, where $I$ is the total space of the Isom torsor. Thus, by the usual flat base change result, the natural map $g^*((f_U)_*\mathcal{M}_U(n)) \to (f_I)_*(g_X^*\mathcal{M}_U(n))$ is an isomorphism. If you choose a trivialization $\beta:\mathcal{O}_I^N\xrightarrow{\cong} g^*((f_U)_*\mathcal{M}_U(n))$, then of course the induced homomorphism $\mathcal{O}_I^N \to (f_I)_*(g_X^*\mathcal{M}_U(n))$ is an isomorphism. $\endgroup$ – Jason Starr May 2 '15 at 10:43
  • $\begingroup$ @JasonStarr: Thank you for taking a look into this, I really appreciate it. I have edited the question. Does the edit make my concern clear? I agree with what you say, but, unless I am misunderstanding, it does not resolve my concern. As far as I understand, your suggestion of base changing to $I$ amounts to reducing the required checking to the universal case. $\endgroup$ – O-Ren Ishii May 2 '15 at 18:03
  • $\begingroup$ I am afraid that I still do not understand your concern. $\endgroup$ – Jason Starr May 4 '15 at 6:47
  • $\begingroup$ @JasonStarr: Do you agree that $U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S}$ and $\mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n)))$ parametrize things that are a priori different: the first one some isomorphisms over $X$, the second one some isomorphisms over $U$? $\endgroup$ – O-Ren Ishii May 4 '15 at 17:51
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As I understand it, you have $f: X\to S$ a projective morphism and $\mathcal{F}$ a sheaf on $X$ which is flat and relatively globally generated. Since your question is local we can assume $S$ is affine and $f_*\mathcal{F}$ is free.

You want to find an integer $N$ and a surjective morphism $$\alpha: \mathcal{O}_X^N\to \mathcal{F}$$ such that the pushforward of this map to $S$ is an isomorphism. I suggest taking $N=\text{rank}_S(f_*\mathcal{F})$, and letting $\alpha$ be the evaluation map (or if you'd like, the adjunction map) $$\mathcal{O}_X^N\simeq f^*f_*\mathcal{F}\to \mathcal{F}.$$ (The isomorphism follows as we assumed $f_*\mathcal{F}$ was free. By the relative global generation hypothesis, this map is indeed surjective.

So we have to check that this map is an isomorphism after pushing forward. But this is just the projection formula, where we use the assumption in your question that $f_*\mathcal{O}_X=\mathcal{O}_S$.

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    $\begingroup$ OMG, you're right! I totally overlooked that all that was missing was the projection formula. Thank you so much for taking your time to look into this and for resolving my confusion! $\endgroup$ – O-Ren Ishii May 14 '15 at 2:29
  • $\begingroup$ @O-RenIshii: No problem, glad I was able to help. $\endgroup$ – Daniel Litt May 14 '15 at 4:20

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