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In these notes, page $10$, bullet $(5)$, it is stated that if $X$ is a scheme of finite type over a field $k$, then the motivic cohomology $\mathrm{H}^{p,q}(X,R)$ of $X$ over $k$, where $R$ is a ring, vanishes if $q<0$. Is there a reference available online (for free) that proves the above-mentioned fact?

Thoughts

I think it has something to do with the Chow-groups-definition $\mathrm{H}^{p,q}(X,\mathbb{Z}) = \mathrm{CH}^{q}(X,2q-p)$, since if $q<0$, then $2q-p<0$, and hence $\mathrm{CH}^{q}(X,2q-p)=0$. However, I don't understand why it would hold for a general ring $R$.

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  • $\begingroup$ This is actually by definition. Motivic cohomology is calculated via an explicit complex $R(q)$ where only $q \geq 0$ is defined. See the Mazza-Weibel-Voevodsky book for details $\endgroup$ Jan 2, 2016 at 15:57
  • $\begingroup$ @EldenElmanto So there's no notion of negative cohomology groups? $\endgroup$
    – user62675
    Jan 2, 2016 at 16:40
  • $\begingroup$ Not for negative $q$. For negative $p$, it's related to the beilinson-soule conjectures $\endgroup$ Jan 2, 2016 at 18:57

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I know very little, but my impression from looking at these lecture notes is that it vanishes for non-negative p simply by using bullet (4). Bullet (4) follows from Theorem 2.9 on the previous page which says how to relate these K-groups to algebraic K-groups for smooth varieties. Using Theorem 2.9 again when both p and q are negative, shows that the cohomology vanishes as the negative algebraic K-groups are trivial when X is smooth per Bass' work, cf. Introduction to K-theory by E. M. Friedlander, section 5.4:

http://users.ictp.it/~pub_off/lectures/lns023/Friedlander/Friedlander.pdf

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