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10.132 The de Rham complex

Let $A \to B$ be a ring map. Denote $\text{d} : B \to \Omega _{B/A}$ the module of differentials with its universal $A$-derivation constructed in Section 10.131. Let $\Omega _{B/A}^ i = \wedge ^ i_ B(\Omega _{B/A})$ for $i \geq 0$ be the $i$th exterior power as in Section 10.13. The de Rham complex of $B$ over $A$ is the complex

\[ \Omega _{B/A}^0 \to \Omega _{B/A}^1 \to \Omega _{B/A}^2 \to \ldots \]

with $A$-linear differentials constructed and described below.

The map $\text{d} : \Omega ^0_{B/A} \to \Omega ^1_{B/A}$ is the universal derivation $\text{d} : B \to \Omega _{B/A}$. Observe that this is indeed $A$-linear.

For $p \geq 1$ we claim there is a unique $A$-linear map $\text{d} : \Omega _{B/A}^ p \to \Omega _{B/A}^{p + 1}$ such that
\begin{equation} \label{algebra-equation-rule} \text{d}\left(b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p\right) = \text{d}b_0 \wedge \text{d}b_1 \wedge \ldots \wedge \text{d}b_ p \end{equation}

Recall that $\Omega _{B/A}$ is generated as a $B$-module by the elements $\text{d}b$. Thus $\Omega ^ p_{B/A}$ is generated as an $A$-module by the element $b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p$ and it follows that the map $\text{d} : \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A}$ if it exists is unique.

Construction of $\text{d} : \Omega _{B/A}^1 \to \Omega _{B/A}^2$. By Definition 10.131.2 the elements $\text{d}b$ freely generate $\Omega _{B/A}$ as a $B$-module subject to the relations $\text{d}a = 0$ for $a \in A$ and $\text{d}(b' + b'') = \text{d}b' + \text{d}b''$ and $\text{d}(b'b'') = b'\text{d}b'' + b''\text{d}b'$ for $b', b'' \in B$. Hence to show that the rule

\[ \sum b'_ i \text{d}b_ i \longmapsto \sum \text{d}b'_ i \wedge \text{d}b_ i \]

is well defined we have to show that the elements

\[ b\text{d}a, \quad \text{and}\quad b\text{d}(b' + b'') - b\text{d}b' - b\text{d}b'' \quad \text{and}\quad b\text{d}(b'b'') - bb'\text{d}b'' - bb''\text{d}b' \]

for $a \in A$ and $b, b', b'' \in B$ are mapped to zero. This is clear by direct computation using the Leibniz rule for $\text{d}$.

Observe that the composition $\Omega ^0_{B/A} \to \Omega ^1_{B/A} \to \Omega ^2_{B/A}$ is zero as $\text{d}(\text{d}(b)) = \text{d}(1 \text{d}b) = \text{d}(1) \wedge \text{d}(b) = 0 \wedge \text{d}b = 0$. Here $\text{d}(1) = 0$ as $1 \in B$ is in the image of $A \to B$. We will use this below.

Construction of $\text{d} : \Omega _{B/A}^ p \to \Omega _{B/A}^{p + 1}$ for $p \geq 2$. We will show the $A$-linear map

\[ \gamma : \Omega ^1_{B/A} \otimes _ A \ldots \otimes _ A \Omega ^1_{B/A} \longrightarrow \Omega _{B/A}^{p + 1} \]

defined by the formula

\[ \omega _1 \otimes \ldots \otimes \omega _ p \longmapsto \sum (-1)^{i + 1} \omega _1 \wedge \ldots \wedge \text{d}(\omega _ i) \wedge \ldots \wedge \omega _ p \]

factors over the natural surjection $\Omega ^1_{B/A} \otimes _ A \ldots \otimes _ A \Omega ^1_{B/A} \to \Omega ^ p_{B/A}$ to give the desired map $\text{d} : \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A}$. According to Lemma 10.13.4 the kernel of $\Omega ^1_{B/A} \otimes _ A \ldots \otimes _ A \Omega ^1_{B/A} \to \Omega ^ p_{B/A}$ is generated as an $A$-module by the elements $\omega _1 \otimes \ldots \otimes \omega _ p$ with $\omega _ i = \omega _ j$ for some $i \not= j$ and $\omega _1 \otimes \ldots \otimes f\omega _ i \otimes \ldots \otimes \omega _ p - \omega _1 \otimes \ldots \otimes f\omega _ j \otimes \ldots \otimes \omega _ p$ for some $f \in B$. A direct computation shows the first type of element is mapped to $0$ by $\gamma $, in other words, $\gamma $ is alternating. To finish we have to show that

\[ \gamma ( \omega _1 \otimes \ldots \otimes f\omega _ i \otimes \ldots \otimes \omega _ p) = \gamma ( \omega _1 \otimes \ldots \otimes f\omega _ j \otimes \ldots \otimes \omega _ p) \]

for $f \in B$. By $A$-linearity and the alternating property, it is enough to show this for $p = 2$, $i = 1$, $j = 2$, $\omega _1 = b \text{d}b'$ and $\omega _2 = c \text{d} c'$ for $b, b', c, c' \in B$. Thus we need to show that

\begin{align*} & \text{d}(fb) \wedge \text{d}b' \wedge c \text{d}c' - fb \text{d}b' \wedge \text{d}c \wedge \text{d}c' \\ & = \text{d}b \wedge \text{d}b' \wedge fc\text{d}c' - b \text{d}b' \wedge \text{d}(fc) \wedge \text{d}c' \end{align*}

in other words that

\[ (c \text{d}(fb) + fb \text{d}c - fc \text{d}b - b \text{d}(fc)) \wedge \text{d}b' \wedge \text{d}c' = 0. \]

This follows from the Leibniz rule. Observe that the value of $\gamma $ on the element $b_0\text{d}b_1 \otimes \text{d}b_2 \otimes \ldots \otimes \text{d}b_ p$ is $\text{d}b_0 \wedge \text{d}b_1 \wedge \ldots \wedge \text{d}b_ p$ and hence ( will be satisfied for the map $\text{d} : \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A}$ so obtained.

Finally, since $\Omega ^ p_{B/A}$ is additively generated by the elements $b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p$ and since $\text{d}(b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p) = \text{d}b_0 \wedge \ldots \wedge \text{d}b_ p$ we see in exactly the same manner that the composition $ \Omega ^ p_{B/A} \to \Omega ^{p + 1}_{B/A} \to \Omega ^{p + 2}_{B/A} $ is zero for $p \geq 1$. Thus the de Rham complex is indeed a complex.

Given just a ring $R$ we set $\Omega _ R = \Omega _{R/\mathbf{Z}}$. This is sometimes called the absolute module of differentials of $R$; this makes sense: if $\Omega _ R$ is the module of differentials where we only assume the Leibniz rule and not the vanishing of $\text{d}1$, then the Leibniz rule gives $\text{d}1 = \text{d}(1 \cdot 1) = 1 \text{d}1 + 1 \text{d}1 = 2 \text{d}1$ and hence $\text{d}1 = 0$ in $\Omega _ R$. In this case the absolute de Rham complex of $R$ is the corresponding complex

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

where we set $\Omega ^ i_ R = \Omega ^ i_{R/\mathbf{Z}}$ and so on.

Suppose we have a commutative diagram of rings

\[ \xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] } \]

There is a natural map of de Rham complexes

\[ \Omega ^\bullet _{B/A} \longrightarrow \Omega ^\bullet _{B'/A'} \]

Namely, in degree $0$ this is the map $B \to B'$, in degree $1$ this is the map $\Omega _{B/A} \to \Omega _{B'/A'}$ constructed in Section 10.131, and for $p \geq 2$ it is the induced map $\Omega ^ p_{B/A} = \wedge ^ p_ B(\Omega _{B/A}) \to \wedge ^ p_{B'}(\Omega _{B'/A'}) = \Omega ^ p_{B'/A'}$. The compatibility with differentials follows from the characterization of the differentials by the formula (

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. $\square$

Comments (6)

Comment #4933 by Théo de Oliveira Santos on

Very minor typo: "The map is the univeral derivation ".

Comment #7458 by ? on

?The most difficult part is omitted?

Comment #7459 by on

The omitted part is something about exterior powers and has nothing to do with differentials. It just says that given a ring , an -module , and an integer the map , is the universal map which is -linear in each entry and alternating. This belongs in Section 10.13.

Comment #8617 by Brian Nugent on

Minor typo, in the paragraph below equation, I think you mean "generated by elements of the form" and not "generated by the element"

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