# de Rham complex of closed immersion between smooth schemes

Hi,

Let $R$ be a $\mathbb Q$-algebra and let $P$ and $Q$ be (EDIT: smooth) $R$-algebras such that there is a surjective map of $R$-algebras $Q\to P$. The following proof cannot possibly be correct, but I can't find the mistake:

''Theorem'': The natural map $\Omega^\bullet_{Q/R}\to \Omega^\bullet_{P/R}$ is a quasi-isomorphism of $Q$-modules.

''Proof''. Since this assertion is local on $Spec(Q)$, we can assume there is a cartesian diagram of rings

Q ---> P
^      ^
|      |
F ---> G


where $F = R[T_1,\dots,T_{r+n}]$ and $G = F/(T_{r+1},\dots,T_n)$ and both vertical maps are etale. Since etale maps are flat, it is enough to prove the assertion when $Q = F$ and $P = G$. This case is well-known to be true (because $R\supset\mathbb Q$). QED?

Do you see my mistake??

Thanks!

-
The mistake is the fact that the differentials of the de Rham complex are not $Q-$ (resp. $P-$) linear and thus you cannot apply the fact that flat morphisms preserve homology. –  Damian Rössler Apr 9 '13 at 14:52
There is more than one mistake. Since you have imposed no hypotheses on $P$ and $Q$, why should there exist etale morphisms as specified? –  Jason Starr Apr 9 '13 at 14:55
Sorry, I forgot: $P$ and $Q$ are smooth over $R$. –  Nicolás Apr 9 '13 at 15:05