As far as vanishing is concerned, the usual motivic cohomology has the following two properties (for a smooth scheme $X$ over a field):

  1. $H^{p,q}(X, \mathbb Z) = 0$, if $p > q + dim(X)$; and

  2. $H^{p,q}(X, \mathbb Z) = 0$, if $q<0$.

Are the analogous properties true for etale (or Lichtenbaum) motivic cohomology? The only statement of this sort that I have heard of is the statement of the motivic Hilbert 90 ($H_L^{n+1,n}(Spec F, \mathbb Z)=0$, for a field $F$). However, it is very hard prove and is a crucial step in the proof of the Bloch-Kato conjecture by Voevodsky.


2 Answers 2


Check out Chapter 10 of Mazza-Voevodsky-Weibel "Lecture notes on motivic cohomology", which discusses étale motivic cohomology. The answers to your questions can be found there:

2) yes: Immediately after Definition 10.1, you find the vanishing $H^{p,q}_L(X,\mathbb{Z})=0$ if $q<0$. This follows directly from the definition of $H^{p,q}_L$ as étale hypercohomology of complexes $\mathbb{Z}(q)$ which are trivial for $q<0$.

1) no: The statement for motivic cohomology is a consequence of the fact that the Nisnevich topology has finite cohomological dimension cohomological dimension (equal to the Krull dimension of the scheme). There is no such thing for étale cohomology, and so you should not actually expect 1) to be true for étale motivic cohomology. Theorem 10.2 gives you an isomorphism $H^{p,q}_L(X,\mathbb{Z}/n)\cong H^{p,q}_{ét}(X,\mu_n^{\otimes q})$ for $q\geq 0$, $p\in\mathbb{Z}$ and $n$ prime to the characteristic of the field $k$ over which $X$ is defined. To get a concrete example of this failure, I guess we can take $X=\operatorname{Spec}\mathbb{R}$, $n=2$. The étale cohomology in this case is cohomology of the Galois group $\operatorname{Gal}(\mathbb{C}/\mathbb{R})$ with coefficients in $\mu_2$. This is 2-periodic (cohomology of a finite group) and $H^1_{ét}(X,\mu_2)=\mathbb{R}^\times/(\mathbb{R}^\times)^2\cong\mathbb{Z}/2$. So there are non-trivial cohomology groups in arbitrarily high dimensions.

  • 1
    $\begingroup$ The correct analogue of 1) would be $H^{p,q}(X,{\mathbb Z})=0$ when $p>q+cd(X)+1$ where $cd(X)$ is the etale cohomological dimension of $X$ (in the Nisnevich case, this agrees with dim(X)) and the +1 has to do with the fact that the sheaves may not be torsion. $\endgroup$
    – Amit H
    Apr 30, 2015 at 15:24

The easiest counterexample is $p=2, q=0$. In this case the usual motivic cohomology vanishes, and if $X$ is normal $$H^2_{et}(X,\mathbf{Z})=H^1_{et}(X,\mathbf{Q}/\mathbf{Z})= \mathrm{Hom}(\pi_1^{et}(X),\mathbf{Q}/\mathbf{Z})$$ which does not vanish even for fields (which are not separably closed).

Another example is $p=3, q=1$. In this case you get the Brauer group which doesn't vanish in general.


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