most recent 30 from http://mathoverflow.net 2013-05-22T10:39:30Z http://mathoverflow.net/feeds/question/48594 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48594/motivic-cohomology-with-finite-coefficients-for-singular-varieties Leonid Positselski 2010-12-07T22:29:21Z 2012-09-25T08:40:58Z

<p>Let $X$ be a smooth variety over a field $K$ whose characteristic does not divide a positive integer $m$. Then the motivic cohomology of $X$ with coefficients in $\mathbb Z/m(j)$ can be computed in terms of the etale and Zariski topologies by the familiar rule $$ H_M^i(X,\mathbb Z/m(j)) = H_{Zar}^i(X,\tau_{\le j}R\pi_*\mu_m^{\otimes j}), $$ where $\pi:Et\to Zar$ is the natural map between the (big or small) etale and Zariski sites, $\mu_m$ is the etale sheaf of $m$-roots of unity, and $\tau$ denotes the canonical truncation of complexes of Zariski sheaves. In fact, this result is a combination of two: the Beilinson-Lichtenbaum etale descent <code>$$ \mathbb Z(j) = \tau_{\le j}R\pi_*\pi^*\mathbb Z(j) = \tau_{\le j+1}R\pi_*\pi^*\mathbb Z(j) $$</code> and (a version of) the Suslin rigidity theorem <code>$$ \pi^*\mathbb Z/m(j) = \mu_m^{\otimes j}. $$</code> These results also remain true if one replaces the Zariski topology with the Nisnevich one.</p> <p>For singular varieties, the above formula for motivic cohomology no longer holds. It suffices to consider the example when $X$ is the affine line with two points glued together. Then one has $H^1_M(X,\mathbb Z/m(0))=\mathbb Z/m$, though the formula would give $0$ as the answer for this motivic cohomology group. It appears that in this particular case ($X$ as above and $j=0$) the rigidity assertion still holds, so it must be the etale descent that breaks down.</p> <p>Is there any way to make any or all of the above assertions true for singular varieties by changing the topologies one works with? E.g., replace the etale topology with the h topology and the Zariski/Nisnevich with cdh? If there is more than one way to do this, I would be greatly interested to hear about each and every of them.</p>

http://mathoverflow.net/questions/48594/motivic-cohomology-with-finite-coefficients-for-singular-varieties/108030#108030 Thomas Geisser 2012-09-25T08:40:58Z 2012-09-25T08:40:58Z

<p>The problem is the definition of the motivic complex: it is a priori only defined for smooth schemes. Voevodsky defines motivic cohomology of non-smooth schemes by pulling the complex back to the category of all (separated, finite type) schemes and taking cdh-cohomology (etale cohomology has cdh-descent by the proper base-change theorem). If you do this, your counterexample disappears , and I think the conjecture holds for non-smooth schemes as well.</p>