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Mar 30, 2020 at 3:31 vote accept user1092847
Jan 11, 2020 at 22:06 answer added dorebell timeline score: 9
Jan 10, 2020 at 18:05 comment added user1092847 @MarcHoyois Thanks, now I see that what I was objecting to really is the failure of $\mathbb A^1$ invariance.
Jan 10, 2020 at 7:51 comment added Marc Hoyois This claim about crystalline cohomology is only for smooth proper varieties. Berthelot's rigid cohomolgy extends crystalline cohomology to more general varieties, and it is $\mathbb A^1$-invariant, but it has rational coefficients. A reasonable $\mathbb A^1$-invariant theory with mod $p$ coefficients for $\mathbb F_p$-varieties is logarithmic de Rham cohomology (= motivic cohomology).
Jan 9, 2020 at 23:02 answer added SashaP timeline score: 9
Jan 9, 2020 at 18:45 history edited YCor CC BY-SA 4.0
fixed spelling of de Rham
Jan 9, 2020 at 17:56 history edited user1092847 CC BY-SA 4.0
added 6 characters in body; edited tags
Jan 9, 2020 at 17:39 history asked user1092847 CC BY-SA 4.0