Is there some kind of universal coefficient theorem for motivic cohomology? In particular, suppose we have a ring morphism $R\to S$, then I would like to know when $$ H^{\star\star}(,S)\simeq H^{\star\star}(,R)\otimes_{R}S\; ?$$ Does this for example hold when $R$ is a field? In particular, does it hold for $R=\mathbb{Q}$? Or do we need additional assumption on $S$ as well? E.g. that $S$ is a semisimple or Noetherian $R$algebra?
Yes, there is a universal coefficient theorem: the corresponding object of the derived category (of $S$modules) could be obtained by tensoring by $S$. This is easy, since motivic cohomology is defined as the cohomology of a complex of free modules (over $R$ and $S$, respectively).

1$\begingroup$ By the way, do you know of a place that gives a quick and dirty definition of motivic cohomology like you just gave above (albeit with a bit more background)? $\endgroup$ – Harry Gindi Feb 19 '11 at 11:15

1$\begingroup$ For motivic cohomology of smooth varieties both the Bloch complex and the Suslin complex are 'quick'. I also believe that one can reduce the cohomology of motives to the one of smooth varieties (though possibly here some work is needed). $\endgroup$ – Mikhail Bondarko Feb 19 '11 at 14:53

$\begingroup$ To WesleyT: if you don't want to bother with derived categories. you will have to assume that $S$ is flat over $R$. This is certainly true if $R$ is a field. $\endgroup$ – Mikhail Bondarko Feb 19 '11 at 19:04

$\begingroup$ OK, thank you. I will go over it and maybe get back at you. $\endgroup$ – WesleyT Feb 19 '11 at 20:45