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Let $U \subset \mathbb P^1$ be an open subset of projective line (over $\mathbb C$) after removing $r$ points and $j: U\hookrightarrow \mathbb P^1$ an open immersion. How do I compute $R^1j_*\mathbb G_m$ ?

It should vanish, shouldn't it? In this case, it would be enough to show that its stalks vanish. Then, if $p \in \mathbb P^1$ we have

$(R^1j_*\mathbb G_m)_p = \varprojlim H^1(j^{-1}V,\mathbb G_m) =H^1(\varprojlim j^{-1}V,\mathbb G_m)$

where the limit is taken over etale (fppf?) neighborhoods $V$ of point $p$. I could not proceed from this point, please help.

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    $\begingroup$ For a dense open immersion $j:U \hookrightarrow X$ into a Dedekind scheme $X$, ${\rm{R}}^i j_{\ast}|_U$ vanishes for $i > 0$ since $j_{\ast}|_U$ is the identity functor. Hence, these higher direct images are skyscraper with support at the points of $X-U$. The formation of this higher direct image commutes with strict henselization at such points, so we're reduced to the case $X = {\rm{Spec}}(R)$ for a strictly henselian dvr $R$ with fraction field $K$. There it is ${\rm{H}}^1(K, \mathbf{G}_m)$, which vanishes by Hilbert 90. $\endgroup$
    – user27920
    Commented Sep 19, 2014 at 5:57
  • $\begingroup$ Thank you for your answer! Three more questions about this argument: 1) How did we use fact that $X$ - Dedekind to conclude that $j_*|_U$ is identity? 2) Surely, we can do the base change $\mathcal O^{sh}_p \to \mathcal O_p$ and compute group $H^1(j^{-1}(\mathcal O^{sh}_p), \mathbb G_m) = 0$. But why for any $V$ containing point $p$ we always have a map $\mathcal O^{sh}_p \to V$ (to conclude that whatever $H^1(J^{-1}(V), \mathbb G_m)$ is, its elements vanish in inverse limit) ? 3) Why $j^{-1}(\mathcal O^{sh}_p) \cong K$? $\endgroup$
    – lime
    Commented Sep 19, 2014 at 18:31
  • $\begingroup$ (1) We don't, it is relevant in the later parts. (2) There are general commutation theorems for etale cohomology and limits; learn them. (3) This is where we use the Dedekind property (to be sure the strict henselization is a local domain). $\endgroup$
    – user27920
    Commented Sep 19, 2014 at 20:13

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As $R^1 j_*\mathbb{G}_m$ is the sheafification of $(V\to \mathbb{P}^1 \text{ etale})\mapsto H^1(V_U, \mathbb{G}_m)$ (here $V_U = V\times_{\mathbb{P}^1} U$), it suffices to prove that given an etale $V\to \mathbb{P}^1$, an element $\zeta\in H^1(V_U, \mathbb{G}_m)$, and a geometric point $\bar x$ of $V$, there exists an etale neighborhood $V'$ of $\bar x$ in $V$ such that the image of $\zeta$ in $H^1(V'_U, \mathbb{G}_m)$ is zero.

Note that for a scheme $X$, we have $H^1(X, \mathbb{G}_m) = Pic(X)$ (this is still true in the etale topology by descent). So we have to prove that given a line bundle $L$ on $V_U$ and a geometric point $\bar x$ of $V$, there is an etale neighborhood $V'$ of $\bar x$ in $V$ such that the restriction of $L$ to $V'_U$ is trivial. If $x\in V_U$, this is clear, as $L$ is locally trivial. In any case, $L$ extends to a line bundle on $V$ (write $L=\mathcal{O}_{V_U}(D)$ for some divisor $D$, then the extension can be taken as $\tilde L = \mathcal{O}_V(D)$), hence is trivial in a neighborhood of $\bar x$.

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  • $\begingroup$ Piotr, why do we need to do this exactly for geometric points $\bar{p}$? Isn't that enough to show that for any $V$ there exists $V'$ such that corresponding cohomology vanishes? Apparently I am missing some obvious conception. $\endgroup$
    – lime
    Commented Sep 19, 2014 at 18:49
  • $\begingroup$ To show that some presheaf $F$ sheafifies to zero is the same as showing that given a section $s\in F(V)$, we can cover $V$ by $V'_i$ such that $s$ maps to zero in $F(V'_i)$. In other words, for every $x\in V$ there is a $V'_x$... $\endgroup$
    – Piotr Achinger
    Commented Sep 19, 2014 at 19:37

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