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=8N0ZT5PrPMXsOZ8sJ8L&ved=0CGMQ6AEwCA#v=onepage&q=long%20exact%20sequence%20circle%20bundle%20cohomology&f=false Is such a fact known/true?
1 Answer
$\begingroup$
$\endgroup$
4
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).

$\begingroup$ I am sorry for a very silly question: does every $G_m$bundle necessarily extend to a line bundle? $\endgroup$ Commented Jan 26, 2012 at 11:01

3$\begingroup$ 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 $(Y \times \mathbb A^1)/\mathbb G_{\rm m}$. $\endgroup$– AngeloCommented Jan 27, 2012 at 6:42

$\begingroup$ Thank you!! Do you think that I can say in a paper that this fact is 'wellknown', or that some (sketch of) a proof is necessary? $\endgroup$ Commented Feb 5, 2012 at 10:23

1$\begingroup$ A sketch of argument might be in order, if you can't find a reference. $\endgroup$– AngeloCommented Feb 5, 2012 at 11:15