## 50.2 The de Rham complex

Let $p : X \to S$ be a morphism of schemes. There is a complex

$\Omega ^\bullet _{X/S} = \mathcal{O}_{X/S} \to \Omega ^1_{X/S} \to \Omega ^2_{X/S} \to \ldots$

of $p^{-1}\mathcal{O}_ S$-modules with $\Omega ^ i_{X/S} = \wedge ^ i(\Omega _{X/S})$ placed in degree $i$ and differential determined by the rule $\text{d}(g_0 \text{d}g_1 \wedge \ldots \wedge \text{d}g_ p) = \text{d}g_0 \wedge \text{d}g_1 \wedge \ldots \wedge \text{d}g_ p$ on local sections. See Modules, Section 17.29.

Given a commutative diagram

$\xymatrix{ X' \ar[r]_ f \ar[d] & X \ar[d] \\ S' \ar[r] & S }$

of schemes, there are canonical maps of complexes $f^{-1}\Omega _{X/S}^\bullet \to \Omega ^\bullet _{X'/S'}$ and $\Omega _{X/S}^\bullet \to f_*\Omega ^\bullet _{X'/S'}$. See Modules, Section 17.29. Linearizing, for every $p$ we obtain a linear map $f^*\Omega ^ p_{X/S} \to \Omega ^ p_{X'/S'}$.

In particular, if $f : Y \to X$ be a morphism of schemes over a base scheme $S$, then there is a map of complexes

$\Omega ^\bullet _{X/S} \longrightarrow f_*\Omega ^\bullet _{Y/S}$

Linearizing, we see that for every $p \geq 0$ we obtain a canonical map

$\Omega ^ p_{X/S} \otimes _{\mathcal{O}_ X} f_*\mathcal{O}_ Y \longrightarrow f_*\Omega ^ p_{Y/S}$

$\xymatrix{ X' \ar[r]_ f \ar[d] & X \ar[d] \\ S' \ar[r] & S }$

be a cartesian diagram of schemes. Then the maps discussed above induce isomorphisms $f^*\Omega ^ p_{X/S} \to \Omega ^ p_{X'/S'}$.

Proof. Combine Morphisms, Lemma 29.32.10 with the fact that formation of exterior power commutes with base change. $\square$

Lemma 50.2.2. Consider a commutative diagram of schemes

$\xymatrix{ X' \ar[r]_ f \ar[d] & X \ar[d] \\ S' \ar[r] & S }$

If $X' \to X$ and $S' \to S$ are étale, then the maps discussed above induce isomorphisms $f^*\Omega ^ p_{X/S} \to \Omega ^ p_{X'/S'}$.

Proof. We have $\Omega _{S'/S} = 0$ and $\Omega _{X'/X} = 0$, see for example Morphisms, Lemma 29.36.15. Then by the short exact sequences of Morphisms, Lemmas 29.32.9 and 29.34.16 we see that $\Omega _{X'/S'} = \Omega _{X'/S} = f^*\Omega _{X/S}$. Taking exterior powers we conclude. $\square$

## Comments (2)

Comment #5373 by Lor Gro on

Ctrl+F "schemss"

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