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Let $S=\mathrm{Spec}(R)$, $s=\mathrm{Spec}(k)$ and $\eta=\mathrm{Spec}(K)$, where $R$ is a d.v.r. with fraction field $K$. Let $j:\eta\rightarrow S$

Now how to compute the sheaf $R^1j_*(\mathbb{G}_{m,\eta})$ in the fppf topology?

The case for etale topology is zero by considering the stalks and use the Hilbert 90.

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  • $\begingroup$ is there any general way to compute such staff in flat topology? $\endgroup$
    – Heer
    Jul 16, 2012 at 22:07

2 Answers 2

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It seems to me that (equation) 5.2 and Appendice 11.7 of "Le Groupe de Brauer, III" imply that the cohomology for the \'etale topology equals the cohomology for the fppf topology.

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  • $\begingroup$ @Jason Starr: thanks a lot for the reference. The condition (L) is satisfied obviously for $j_*\mathbb{G}_m$. the condition (R) is less obvious, in particular what should be the open subfunctor? Is it $\mathbb{G}_m$. $\endgroup$
    – Heer
    Jul 18, 2012 at 23:06
  • $\begingroup$ @Jason Starr: what is the definition of an open subfunctor? $\endgroup$
    – Heer
    Jul 18, 2012 at 23:07
  • $\begingroup$ I am not suggesting to apply Theorem 11.7 to $j_*\mathbb{G}_m$, but rather to $\mathbb{G}_m$, where condition (R) is tautological (the open subfunctor is the entire functor). Grothendieck proves that for any scheme $X$, for every integer $p$, the natural map $H^p(X_{\text{\'et}},\mathbb{G}_m) \to H^p(X_{\text{fppf}},\mathbb{G}_m)$ is an isomorphism. I claim that this implies that for every morphism $f:X\to Y$, also the natural map `$R^p f_*\mathbb{G}_{m,\text{\'et}} \to R^p f_*\mathbb{G}_{m,\text{fppf}}$ is an isomorphism, i.e., the derived etale and fppf pushforwards agree ... $\endgroup$
    – Jason Starr
    Jul 19, 2012 at 13:23
  • $\begingroup$ ... To go from Grothendieck's theorem to the claim, use Prop. 5.1, Exp. V, SGA 4. $\endgroup$
    – Jason Starr
    Jul 19, 2012 at 14:26
  • $\begingroup$ @Jason Starr: I see what you meant. But the natural map should be $R^p f_*\mathbb{G}_{m,\text{et}} \to u_{Y*}R^p f_*\mathbb{G}_{m,\text{fppf}}$, where $u_Y:Y_{\text{fppf}}\rightarrow Y_{\text{et}}$. Then $u_{Y*}R^p f_*\mathbb{G}_{m,\text{fppf}}=0$ provided $R^p f_*\mathbb{G}_{m,\text{et}}$, But this doesn't imply that $R^p f_*\mathbb{G}_{m,\text{fppf}}=0$ $\endgroup$
    – Heer
    Jul 19, 2012 at 20:41
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Here's another way of seeing this. Since $R$ is a dvr, any fppf covering can be refined to a connected cover $p:U\to Spec(R)$ where $U=Spec(A)$ and $A$ is a faithfully flat $R$ algebra of finite type. Also, for our purposes, we might as well replace $R$ with its strict henselization. Now taking generic hyperplane sections of $A$ through a point over $m_R$ refines the covering $p:Spec(A)→Spec(R)$ to a finite, faithfully flat cover $\overline p:Spec(A/(f_1,...,f_n))\to Spec(R)$ which we may assume is connected. But there are no non-trivial line bundles on a connected algebra that is finite over a field.

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