# $H^2(S, f_* \mathbb{G}_m)$ in the fppf versus etale topology for proper $f$

Let $f\colon X \rightarrow S$ be a proper morphism of schemes. Is the cohomology group $H^2(S, f_* \mathbb{G}_m)$ the same regardless of whether it is computed in the etale or the fppf topology? And if so, why?

This is claimed in the middle of p. 203 of "Neron models" by Bosch, Lutkebohmert, and Raynaud, and it is hinted there that Stein factorization is of use for proving the claim (Stein factorization also has a non-Noetherian version, by the way, see http://stacks.math.columbia.edu/tag/03H2). I would be grateful if someone could spell out the argument.

Note that the book mentioned above uses a somewhat strange definition of fppf topology: fppf covers are required to be faithfully flat and of finite presentation (as opposed to faithfully flat and locally of finite presentation); in particular, some Zariski covers are not fppf. It seems to me that this leads to a lot of awkwardness (e.g., on p. 201). For the sake of definiteness though, let us say that for the purposes of this question fppf cohomology is computed when the covers are taken to be faithfully flat and locally of finite presentation, although I would appreciate if someone could comment on what difference this choice makes in comparison to what is used in the book.

• 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 – user74230 Dec 25 '14 at 4:00
• 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 $H^1(S, f_*(\mathbb{G}_m)) = 0$. – Question Mark Dec 25 '14 at 19:59
• 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. – Question Mark Dec 25 '14 at 20:24
• 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 $S$. – user74230 Dec 25 '14 at 21:40
• @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. – Question Mark Dec 25 '14 at 21:42