Timeline for $H^2(S, f_* \mathbb{G}_m)$ in the fppf versus etale topology for proper $f$
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| Dec 30, 2014 at 4:03 | history | edited | Question Mark |
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| Dec 26, 2014 at 1:57 | comment | added | user74230 | Question Mark: I recommend to take the definition to be fplpf cohomology and not dwell on the comparison since scheme cohomology for a topology not refining Zariski seems pathological. | |
| Dec 26, 2014 at 1:29 | comment | added | Question Mark | @user74230: The link with geometric components of the fiber works equally well in the non-Noetherian case, see stacks.math.columbia.edu/tag/03H2 and use that the $S$-fibers there are the disjoint unions of the corresponding (nonempty, since connected) $S'$-fibers (also that the fibers of an integral map are totally disconnected: going up/down). I understand that fplpf is what is usually meant by the "fppf topology," but this is not the case for this book, since they explicitly specify otherwise; hence my question: are the fppf cohomology groups in this book the same as the usual ones? | |
| Dec 26, 2014 at 0:50 | comment | added | user74230 | Question Mark: The link with geometric components of the fiber over the closed point rests on limit arguments from the noetherian case (the latter is where Stein factorization is developed), so I don't see how you avoid limit methods. As for merging fppf and Zariski, certainly one has to use "fplpf", which is what is always meant by "fppf topology". By openness of fplpf maps, an fplpf cover of an qcqs base has a genuinely fppf refinement, so no confusion occurs; the fpqc analogue has this "qc refinement" aspect inserted manually, but fpqc cohomology (rather than sheaf aspects) is irrelevant. | |
| Dec 25, 2014 at 22:04 | comment | added | Question Mark | @user74230: By the way, could you elaborate on your comment about merging Zariski and fppf? I agree that the authors do explain what they mean by an fppf sheaf/sheafification (i.e., that one has to take into account Zariski covers, too), but what site (or topos--the one where sheaves are both wrt Zariski and fppf?) do they use to define and talk about fppf cohomology and derived pushforwards? Are the latter two the same as the ones computed using the usual definition: fppf = faithfully flat and locally of finite presentation? | |
| Dec 25, 2014 at 21:42 | comment | added | Question Mark | @user74230: Because the maximal ideals of $T$ correspond to the geometric connected components of the fiber of $f$ over the closed point of $S$ and there is a finite number of such components since $f$ is proper. A similar argument is used in the first half of the proof of 8.1/3. | |
| Dec 25, 2014 at 21:40 | comment | added | user74230 | Question Mark: If you don't pass to the noetherian case then why would the Stein factorization $T$ be semi-local (i.e., only finitely many maximal ideals)? That is why I suggested to use limit methods to pass to noetherian $S$. | |
| Dec 25, 2014 at 20:24 | comment | added | Question Mark | As for your "exercise," a combination of the proof of Thm. A.1 of Conrad, B. "Deligne's notes on ..." with Cor. 6.6 of Rydh "Noetherian approximation ..." proves that for a proper morphism $X \rightarrow S$ with $S$ qcqs, there is a closed immersion $X \hookrightarrow \overline{X}$ (over $S$) with $\overline{X} \rightarrow S$ proper and of finite presentation. | |
| Dec 25, 2014 at 19:59 | comment | added | Question Mark | Thank you, this is very helpful. The Grothendieck reference indeed handles the claim and, I think, no limit arguments are needed: since we've reduced to the strictly local case, letting $T \rightarrow S$ be the semi-local Stein factorization of $X \rightarrow S$, we have (due to the semi-locality of $T$) that every line bundle on $T$ is trivial, in particular, that every line bundle on $T$ that is trivialized over the pullback of an fppf cover of $S$ is trivial, i.e., that $H^1(S, f_*(\mathbb{G}_m)) = 0$. | |
| Dec 25, 2014 at 4:00 | comment | added | user74230 | The bottom of p.201 and just above Prop. 1 on p.200 merge Zariski and fppf. Their Grothendieck reference answers your question; did you try? By Lemma 11.1 ($n=1$, $G_{\rm{pl}}=f_{\ast}(\mathbf{G}_m)$), showing H$^1(S_{\rm{fppf}},f_{\ast}(\mathbf{G}_m))=1$ for strictly henselian $S$ suffices. But $X$ is closed in a finitely presented proper $S$-scheme (hard exercise), so by limit methods WLOG $S$ is noetherian. Then H$^1$ is the limit of H$^1(S'/S,\cdot)$'s for finite flat $S'\to S$ by EGA IV$_4$ 17.16.2 since $S$ is henselian. This is Pic of the semi-local Stein factorization. QED | |
| Dec 24, 2014 at 23:51 | history | edited | Question Mark | CC BY-SA 3.0 |
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| Dec 24, 2014 at 23:41 | history | asked | Question Mark | CC BY-SA 3.0 |