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Sep 15, 2011 at 13:02 comment added user2035 Actually, I was hoping that one could get a clearer picture by separating the local question of how the sheaf kernel of $\mathcal O_X\to\Omega_{X/Y}$ compares to $f^{-1}\mathcal O_{Y_\mathrm{et}}$ ("locally constant functions") from the global question whether $\Gamma(Y,\mathcal O_Y)\to\Gamma(X,f^{-1}\mathcal O_{Y_\mathrm{et}})$ is an isomorphism.
Sep 14, 2011 at 19:26 vote accept Allen Knutson
Sep 14, 2011 at 11:11 comment added Allen Knutson a-fortiori, I do point out that it's false! Basically your worry is that the fiber of Spec B -> Spec A may not be connected. In Qing Liu's very nice sufficient condition, the generic fiber is geometrically integral, which is how he addresses your issue.
Sep 14, 2011 at 8:29 comment added Donu Arapura Qing Liu, thanks. Nice answer, by the way. A-fortiori, certainly, although I don't think he meant it literally.
Sep 14, 2011 at 8:02 history edited user2035
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Sep 14, 2011 at 7:45 comment added user2035 Shouldn't that be: "if the derivative is zero, the function is locally constant"?
Sep 14, 2011 at 0:20 comment added Qing Liu @Donu: I like (and use) your idea of passing to Frac($A$). But over a field you just have to be carefull with finite extensions $B/A$ for which $H^0_{DR}(B)=B$.
Sep 14, 2011 at 0:16 answer added Qing Liu timeline score: 30
Sep 13, 2011 at 18:59 comment added Donu Arapura After teaching 3 hrs, my brain is shot. Here some thoughts which may only be quasicoherent. If $A$ is a field of char $p>0$, there is not much hope. If $A$ is a field of char $0$, and $B$ has a a domain of finite type, then you should be OK by the algebraic de Rham theorem which implies $H^0_{DR}(B)=A$. Perhaps in your situation, you can pass to the fraction field?
Sep 13, 2011 at 16:29 history asked Allen Knutson CC BY-SA 3.0