Lemma 50.10.5. In the situation above, assume there is a morphism $S \to \mathop{\mathrm{Spec}}(\mathbf{Q})$. Then $\Omega ^\bullet _{X/S} \to \pi _*\Omega ^\bullet _{L/S}$ is a quasi-isomorphism and $H_{dR}^*(X/S) = H_{dR}^*(L/S)$.
Proof. Let $R$ be a $\mathbf{Q}$-algebra. Let $A$ be an $R$-algebra. The affine local statement is that the map
\[ \Omega ^\bullet _{A/R} \longrightarrow \Omega ^\bullet _{A[t]/R} \]
is a quasi-isomorphism of complexes of $R$-modules. In fact it is a homotopy equivalence with homotopy inverse given by the map sending $g \omega + g' \text{d}t \wedge \omega '$ to $g(0)\omega $ for $g, g' \in A[t]$ and $\omega , \omega ' \in \Omega ^\bullet _{A/R}$. The homotopy sends $g \omega + g' \text{d}t \wedge \omega '$ to $(\int g') \omega '$ were $\int g' \in A[t]$ is the polynomial with vanishing constant term whose derivative with respect to $t$ is $g'$. Of course, here we use that $R$ contains $\mathbf{Q}$ as $\int t^ n = (1/n)t^{n + 1}$. $\square$
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