Lemma 10.132.1. Let $A \to B$ be a ring map. Let $\pi : \Omega _{B/A} \to \Omega $ be a surjective $B$-module map. Denote $\text{d} : B \to \Omega $ the composition of $\pi $ with the universal derivation $\text{d}_{B/A} : B \to \Omega _{B/A}$. Set $\Omega ^ i = \wedge _ B^ i(\Omega )$. Assume that the kernel of $\pi $ is generated, as a $B$-module, by elements $\omega \in \Omega _{B/A}$ such that $\text{d}_{B/A}(\omega ) \in \Omega _{B/A}^2$ maps to zero in $\Omega ^2$. Then there is a de Rham complex

\[ \Omega ^0 \to \Omega ^1 \to \Omega ^2 \to \ldots \]

whose differential is defined by the rule

\[ \text{d} : \Omega ^ p \to \Omega ^{p + 1},\quad \text{d}\left(f_0\text{d}f_1 \wedge \ldots \wedge \text{d}f_ p\right) = \text{d}f_0 \wedge \text{d}f_1 \wedge \ldots \wedge \text{d}f_ p \]

**Proof.**
We will show that there exists a commutative diagram

\[ \xymatrix{ \Omega _{B/A}^0 \ar[d] \ar[r]_{\text{d}_{B/A}} & \Omega _{B/A}^1 \ar[d]_\pi \ar[r]_{\text{d}_{B/A}} & \Omega _{B/A}^2 \ar[d]_{\wedge ^2\pi } \ar[r]_{\text{d}_{B/A}} & \ldots \\ \Omega ^0 \ar[r]^{\text{d}} & \Omega ^1 \ar[r]^{\text{d}} & \Omega ^2 \ar[r]^{\text{d}} & \ldots } \]

the description of the map $\text{d}$ will follow from the construction of the differentials $\text{d}_{B/A} : \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A}$ of the de Rham complex of $B$ over $A$ given above. Since the left most vertical arrow is an isomorphism we have the first square. Because $\pi $ is surjective, to get the second square it suffices to show that $\text{d}_{B/A}$ maps the kernel of $\pi $ into the kernel of $\wedge ^2\pi $. We are given that any element of the kernel of $\pi $ is of the form $\sum b_ i\omega _ i$ with $\pi (\omega _ i) = 0$ and $\wedge ^2\pi (\text{d}_{B/A}(\omega _ i)) = 0$. By the Leibniz rule for $\text{d}_{B/A}$ we have $\text{d}_{B/A}(\sum b_ i\omega _ i) = \sum b_ i \text{d}_{B/A}(\omega _ i) + \sum \text{d}_{B/A}(b_ i) \wedge \omega _ i$. Hence this maps to zero under $\wedge ^2\pi $.

For $i > 1$ we note that $\wedge ^ i \pi $ is surjective with kernel the image of $\mathop{\mathrm{Ker}}(\pi ) \wedge \Omega ^{i - 1}_{B/A} \to \Omega _{B/A}^ i$. For $\omega _1 \in \mathop{\mathrm{Ker}}(\pi )$ and $\omega _2 \in \Omega ^{i - 1}_{B/A}$ we have

\[ \text{d}_{B/A}(\omega _1 \wedge \omega _2) = \text{d}_{B/A}(\omega _1) \wedge \omega _2 - \omega _1 \wedge \text{d}_{B/A}(\omega _2) \]

which is in the kernel of $\wedge ^{i + 1}\pi $ by what we just proved above. Hence we get the $(i + 1)$st square in the diagram above. This concludes the proof.
$\square$

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