# Voevodsky's 'split standard triple' argument: an explanation; does it work with $Z/nZ$-coefficients?

For varieties over a perfect field (of some characteristic $p$ that could be 0) Voevodsky defines the notion of a 'standard triple' (see http://books.google.ru/books?id=TzUmk87bN9cC&pg=PA85&lpg=PA85&dq=Voevodsky+standard+triple&source=bl&ots=lqFZojfUU-&sig=Jtt57xtmlQwX7XvShXHPyaKVP68&hl=ru&sa=X&ei=SaeuUIzEGZH24QSEwIG4DA&redir_esc=y#v=onepage&q=Voevodsky%20standard%20triple&f=false). One says that a triple is split over $U$ if a certain line bundle is trivial (see Definition 11.11); this has certain consequences for cohomology of varieties with coefficients in a homotopy invariant presheaf with transfers $F$ (see Proposition 11.15).

My question is: if $nF=0$, is it sufficient to consider triviality modulo $n$ instead, i.e. could one replace all the Picard groups considered in this section by their $\mathbb{Z}/n\mathbb{Z}$-analogues? I looked at the proofs, and it seems that the answer is positive; yet possibly I miss something.

Alternatively, one can find Voevodsky's (Mazza's-Weibel's) book here http://www.claymath.org/library/monographs/cmim02c.pdf an earlier exposition of this argument can be found in section 4 of http://www.math.illinois.edu/K-theory/0368/s3.pdf

Upd. Possibly, a more clear reference to Voevodsky's argument is http://www.math.uiuc.edu/K-theory/0832/motvo.pdf, section 5.1.1; yet I would be deeply grateful for any 'explanation' of this reasoning.

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It seems that the answer is "yes".:) –  Mikhail Bondarko Nov 28 '12 at 19:31