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Let $X$ be a smooth variety over an algebraically closed field (whose characteristic could be positive), $Y\to X$ is a $G_m$-bundle ($G_m=\mathbb{A}\setminus {0}$). Then I want to have a long exact sequence that relates the \'etale cohomology of $Y$ with the one of $X$; this should probably be similar to Theorem 3.57 of http://books.google.ru/books?id=wJJ0fnq0LPMC&pg=PA138&lpg=PA138&dq=long+exact+sequence+circle+bundle+cohomology&source=bl&ots=-brkx3H2yi&sig=HmC1xKeRfXqB4G5CE53uG5PrjtU&hl=ru&sa=X&ei=8N0ZT5PrPMXsOZ-8sJ8L&ved=0CGMQ6AEwCA#v=onepage&q=long%20exact%20sequence%20circle%20bundle%20cohomology&f=false Is such a fact known/true?

flag	asked Jan 26 2012 at 5:10 Mikhail Bondarko 5,936●1●4●21

Angelo · Accepted Answer · 2012-01-26 06:09:31Z UTC

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This is true, if you take étale cohomology with coefficients in a finite abelian group of order not divisible by the characteristic. You can embed $Y$ in the corresponding line bundle $L \to X$. Then by smooth base change the pullback from the cohomology of $X$ to that of $L$ is an isomorphism, and you can apply the Gysin sequence of $Y$ in $L$ (see for example http://www.jmilne.org/math/CourseNotes/LEC.pdf, Chapter 16).

answered Jan 26 2012 at 6:09

Angelo
19.8k●3●47●75

I am sorry for a very silly question: does every $G_m$-bundle necessarily extend to a line bundle? – Mikhail Bondarko Jan 26 2012 at 11:01

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Yes: $\mathbb G_{\rm m}$ acts on $\mathbb A^1$ by multiplication, and the line bundle associated with a $\mathbb G_{\rm m}$-bundle $Y \to X$ is the quotient $(Y \times \mathbb A^1)/\mathbb G_{\rm m}$. – Angelo Jan 27 2012 at 6:42

Thank you!! Do you think that I can say in a paper that this fact is 'well-known', or that some (sketch of) a proof is necessary? – Mikhail Bondarko Feb 5 2012 at 10:23

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A sketch of argument might be in order, if you can't find a reference. – Angelo Feb 5 2012 at 11:15

The cohomology of a $G_m$-bundle

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