Lemma 10.132.1. Suppose that we have ring maps $A \to A'$ and $A \to B$. Set $B' = B \otimes _ A A'$, so that we obtain a diagram as above. Then the canonical map defined above induces an isomorphism $\Omega ^\bullet _{B/A} \otimes _ A A' = \Omega ^\bullet _{B'/A'}$ of complexes.
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
10.132.0.1
\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 elements $b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p$ 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 (10.132.0.1) 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 (10.132.0.1).
Proof. This follows from Lemma 10.131.12 and the fact that taking exterior powers commutes with base change. $\square$
Lemma 10.132.2. 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 whose differential is defined by the rule
\[ \Omega ^0 \to \Omega ^1 \to \Omega ^2 \to \ldots \]
\[ \text{d} : \Omega ^ p \to \Omega ^{p + 1},\quad \text{d}\left(f_0\text{d}f_1 \wedge \ldots \wedge \text{d}f_ p\right) = \text{d}f_0 \wedge \text{d}f_1 \wedge \ldots \wedge \text{d}f_ p \]
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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