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Hi,

Let $R$ be a $\mathbb Q$-algebra and let $A$ be a smooth $R$-algebra. If $A$ is a polynomial algebra $A = R[T_1,\dots,T_n]$, then it is easy to see that the natural map $R \to \Omega^\bullet_{A/R}$ is a quasi-isomorphism of $R$-modules.

Question: is the same true for $A$ a smooth $R$-algebra if there is an etale map $F\to A$ with $F = R[T_1,\dots,T_n]$ (this is always the case locally on $A$)?

Thanks!

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This is not so. Let $R={\bf Q}$. In the situation you describe, the cohomology of the de Rham complex $\otimes{\bf C}$ is then actually the singular cohomology of the space ${\rm Spec\ A}({\bf C})$. This is a theorem of Atiyah-Hodge-Grothendieck. See "On the de Rham cohomology..." by A. Grothendieck, Th. 1. In general, you may find an $A$ so that the singular cohomology of the space ${\rm Spec A}({\bf C})$ does not vanish. In the case of ${\bf C}[T_1,\dots, T_n]$, it holds because affine space is contractible. –  Damian Rössler Apr 9 '13 at 11:02
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