MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbf{Q}_p$?

share|cite|improve this question
4  
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 – Kevin Buzzard Apr 18 '11 at 19:03
    
OK, thank you for the explanation. – Timo Keller Apr 18 '11 at 19:06
2  
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). – Donu Arapura Apr 18 '11 at 19:34
    
That sounds reasonable. – Timo Keller Apr 18 '11 at 19:40
2  
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. – Mikhail Bondarko Apr 18 '11 at 21:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.