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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 $$ \mathbb Z(j) = \tau_{\le j}R\pi_*\pi^*\mathbb Z(j) = \tau_{\le j+1}R\pi_*\pi^*\mathbb Z(j) $$ and (a version of) the Suslin rigidity theorem $$ \pi^*\mathbb Z/m(j) = \mu_m^{\otimes j}. $$ These results also remain true if one replaces the Zariski topology with the Nisnevich one.

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.

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.

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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.

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