There is a counterexample of Serre showing that there is no Weil cohomology theory with coefficients in $\mathbf{Q}, \mathbf{Q}_p, \mathbf{R}$ over $\mathbf{F}_{p^2}$ (a supersingular elliptic curve). So what happens if we tensor motivic cohomology with $\mathbf{Q}$ or crystalline cohomology with $\mathbf{Q}_p$?
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4$\begingroup$ I know nothing about motivic cohomology, but the crystalline cohomology question isn't hard to answer: the coefficients for crystalline cohomology aren't $\mathbf{Z}$ or $\mathbf{Z}_p$ so no contradiction---indeed, this is explicitly mentioned on the wikipedia article on crystalline cohomology, motivated precisely by Serre's example $\endgroup$– Kevin BuzzardApr 18, 2011 at 19:03
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3$\begingroup$ I don't claim to know much about motivic cohomology either, but if you what you propose, you'll see things like $CH^1(X)_Q=Pic(X)\otimes \mathbb{Q}$ lurking about. So it's much bigger than what would expect from a Weil cohomology (if that was your question). $\endgroup$– Donu ArapuraApr 18, 2011 at 19:34
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$\begingroup$ That sounds reasonable. $\endgroup$– user19475Apr 18, 2011 at 19:40
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2$\begingroup$ Yes, Donu is quite right; the cohomology theory that people usually call motivic cohomology (this is probably not what you think it is) is far from being a Weil cohomology theory. $\endgroup$– Mikhail BondarkoApr 18, 2011 at 21:16
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