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Let $X$ be a locally noetherian regular scheme and $Y$ be a closed subscheme of codimension $d > 0$ in every point. Why does it "immédiatement" (Grothendieck, Groupe de Brauer III, §6, p. 133 f.) follow that the local cohomology sheaves with supports $\mathcal{H}^i_Y(X,\mathbf{G}_m)$ vanish for $0 \leq i \leq 2$ (if $d \neq 1$ for $i=1$)?

(For $i \neq 0$, it would be clear to me if there were no supports: The stalks are zero since the Picard group of a local ring and the Brauer group of a strictly henselian local ring vanish.)

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These vanishing statements are equivalent to the following statements: put $j:Y\hookrightarrow X$ and $U:=X-Y$. Then (i) $\mathbf{G}_m\rightarrow j_{*}(\mathbf{G}_m|_U)$ is an isomorphism (this will give vanishing for $i=0,1$), and (ii) $\mathrm{R}^1j_{*}(\mathbf{G}_m|_U)=0$ (this will give vanishing for $i=2$). –  Mahdi Majidi-Zolbanin Jan 31 '12 at 20:06
    
I don't get it: $\mathbf{G}_m|_U$ is a sheaf on $U = X \setminus Y$, but the domain of $j$ is $Y$, not $U$. –  Timo Keller Feb 1 '12 at 14:13
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Yes, you are right, $j$ is $U\hookrightarrow X$. –  Mahdi Majidi-Zolbanin Feb 1 '12 at 15:25
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1 Answer 1

Let us assume that $X=Spec A$ for $A$ a strictly henselian ring. Since $Y$ has codimension at least $2$ and $X$ is normal, the restriction $H^0(X,\mathbb{G}_m)\rightarrow H^0(X\setminus Y,\mathbb{G}_m)$ is an isomorphism (it is always an embedding, without any restriction on the codimension of $Y$). This implies the vanishing of $H^0$ and $H^1$ with support in $Y$. To see that $H^2$ vanishes (without condition on codimension), it is enough to know that $H^1(U,\mathbb{G}_m)=0$ (where $U=X\setminus Y$). If $Y$ has codimension $1$, this is clear, since $U$ is then affine (This is wrong. See Moret-Bailly's comment below). Otherwise, the restriction functor from the category of line bundles over $X$ to that over $U$ is an equivalence of categories; see SGA 2, XI.3.

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@Keerthi: are you establishing vanishing of groups $H_Y$ or sheaves $\mathcal{H}_Y$? –  Mahdi Majidi-Zolbanin Jan 31 '12 at 20:15
    
I think the same argument should show that the stalks of the sheaves $\mathcal{H}_Y$ are $0$, though I haven't thought it through. –  Keerthi Madapusi Pera Jan 31 '12 at 23:06
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In the codimension 1 case, the reason is not that "$U$ is affine" (affine schemes may have nontrivial Picard groups) but that $U$ is a point. –  Laurent Moret-Bailly Feb 1 '12 at 7:52
    
Thanks for the correction. –  Keerthi Madapusi Pera Feb 1 '12 at 16:35
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