# Cohomology of a sheaf with only one stalk

Let $X$ be a proper scheme over a henselian discrete valation ring. I have a Nisnevich sheaf $F$ of which has only one stalk at the generic point of $X$ (and all other stalks vanish).

I believe that this implies that $F$ has no cohomology. This is known, for example, if $X$ itself is the spectrum of a henselian discrete valuation ring (see Milne's arithmetic duality theorems).

Has anybody any idea, or a counter-example?

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Can you give the precise reference in Milne's ADT? And---it is probably a stupid idea---what does the Leray spectral sequence give us in this situation? –  Timo Keller Sep 30 '12 at 9:27
Can Leray really help? After all, won't all higher direct images have only one stalk, which by the mentioned theorem implies that the spectral sequence degenerates immediately? (so higher direct image = cohomology, on the nose?) –  Jacob Bell Sep 30 '12 at 12:28
It seems to me that the case of a henselian DVR is not really significant, since the Nisnevich cohomology of any sheaf on a henselian local rings is 0. –  Angelo Sep 30 '12 at 17:17
@Timo: II Prop. 1.1(a) because $F$ is necessarily of the form $j_!G$ for some $G$ on the generic fiber @Angelo: I mean including $H^0$ –  Thomas Geisser Sep 30 '12 at 19:01
Are there any purity statements for Nisnevich cohomology ? Couldn't this help ? –  Damian Rössler Oct 1 '12 at 6:43