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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 semi-simple or Noetherian $R$-algebra?

flag	asked Feb 19 2011 at 3:26 WesleyT 23●2

Mikhail Bondarko · Accepted Answer · 2011-02-19 11:06:56Z UTC

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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).

answered Feb 19 2011 at 11:06

Mikhail Bondarko
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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)? – Harry Gindi Feb 19 2011 at 11:15

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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). – Mikhail Bondarko Feb 19 2011 at 14:53

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. – Mikhail Bondarko Feb 19 2011 at 19:04

OK, thank you. I will go over it and maybe get back at you. – WesleyT Feb 19 2011 at 20:45

Is there a universal coefficient theorem for motivic cohomology?

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