The algebraic de Rham complex

Question

Let $A$ be a commutative $R$-algebra (or more generally a morphism of ringed spaces). Then there is an "algebraic de Rham complex" of $R$-linear maps $A=\Omega^0_{A/R} \xrightarrow{d^0} \Omega^1_{A/R} \xrightarrow{d^1} \Omega^2_{A/R} \to \dotsc$. The construction is given as an exercise in Lang's Algebra (XIX, Theorem 3.2). Somehow this is not completely formal since $\Omega^i_{A/R}$ only has a universal property as an $A$-module, not as an $R$-module, right?

We depand that $d^{p+q}(\omega \wedge \eta) = d^p(\omega) \wedge \eta + (-1)^p \omega \wedge d^q(\eta)$ for $\omega \in \Omega^p, \eta \in \Omega^q$. This implies $d^1(a * d^0(b)) = d^0(a) \wedge d^0(b)$, and more generally $d^1$ maps $\sum_i a_i * d^0(b_i) \mapsto \sum_i d^0(a_i) \wedge d^0(b_i)$. Conversely, one could try to define $d^1$ this way. But then it is not clear to me how to show well-definedness. I will sketch what I have done to remedy this, but probably it's way too complicated, and I wonder if there is a more direct approach.

There is an isomorphism $\Omega^1_{A/R} \to I/I^2$ mapping $a * d(b) \mapsto [ab \otimes 1-a \otimes b]$, where $I$ is the kernel of the multiplication map $A \otimes_R A \to A$. It fits into a long exact sequence $\dotsc A^{\otimes {n+1}} \to A^{\otimes n} \to \dotsc$ where the differentials are alternating sums, for example $A^{\otimes 4} \to A^{\otimes 3}$ maps $a \otimes b \otimes c \otimes d$ to $ab \otimes c \otimes d - a \otimes bc \otimes d + a \otimes b \otimes cd$. Exactness implies that $I$ is isomorphic to the cokernel of this differential. Now define $A^{\otimes 3} \to \Omega^2_{A/R}$ by $a \otimes b \otimes c \mapsto d^0(ac) \wedge d^0(b)$. This is a well-defined $R$-linear map. One checks that it vanishes on the image of $A^{\otimes 4} \to A^{\otimes 3}$, hence extends to $I$. One then checks that it also vanishes on $I^2$, so that it extends to $I/I^2 \cong \Omega^1_{A/R}$. The resulting map $d^1$ has the desired description.

Once one has $d^1$, one might define the other differentials inductively by $d^{p+1}(\omega \wedge \eta) = d^p(\omega) \wedge \eta + (-1)^p \omega \wedge d^1(\eta)$. But then again well-definedness seems to be not so clear.

Of course there are no such problems when $\mathrm{Spec}(A)$ is a vector bundle over $\mathrm{Spec}(R)$, which also explains that they don't occur for the de Rham complex on a smooth manifold.

Answer 1

I have found a more simple construction. In fact, there is a universal property of $\Omega^1_{A/R}$ as an $R$-module: It is the cokernel of the $A$-linear (and therefore $R$-linear) map $$A \otimes_R A \otimes_R A \to A \otimes_R A,\quad a \otimes b \otimes c \mapsto ab \otimes c - b \otimes ac - a \otimes bc.$$ Here, the $A$-module structure comes from the last factor in each case.

This alternative construction seems to be quite unknown(?), see also here. The differential is given by $d(a)=[a \otimes 1]$. We exactly mod out the Leibniz rule $d(a \cdot b)=a \cdot d(b)+b \cdot d(a)$.

Now consider the $R$-linear map $A \otimes_R A \to \Omega^2_{A/R},~ a \otimes b \mapsto d(b) \wedge d(a)$. It extends to $\Omega^1_{A/R}$ since $$\begin{align*} & d(c) \wedge d(ab) - d(ac) \wedge d(b) - d(bc) \wedge d(a) \\ &=a \cdot d(c) \wedge d(b) +b \cdot d(c) \wedge d(a) \\ & -a \cdot d(c) \wedge d(b) - c \cdot d(a) \wedge d(b) \\ & -b \cdot d(c) \wedge d(a)-c \cdot d(b) \wedge d(a)\\ &=0 \end{align*}$$

More generally, $\Omega^p_{A/R}$ is an explicit quotient of $A^{\otimes 2p}$, and $A^{\otimes 2p} \to \Omega^{p+1}_{A/R}$ defined by $a_1 \otimes b_1 \otimes \dotsc \otimes a_p \otimes b_p \mapsto d(b_1 \cdot \dotsc \cdot b_p) \wedge d(a_1) \wedge \dotsc \wedge d(a_p)$ preserves the relations (easy calculation as above), so that it extends to $d^p : \Omega^p_{A/R} \to \Omega^{p+1}_{A/R}$.

This construction even generalizes to arbitrary commutative algebra objects in semi-additive cocomplete symmetric monoidal categories. In the abelian case we can define de Rham cohomology. Details appear in my thesis in the section about derivations.

Answer 2

I think there is an easier proof.

Write $A$ as a quotient $B/J$ where $B$ is a polynomial ring over $R$ (not necessarily finite type). Then $\Omega^1_{B/R}$ is free over $B$, and you pointed out, the construction is easy in this case. We have the exact sequence $$ J\stackrel{\delta}{\to} \Omega_{B/R}^1\otimes A \to \Omega_{A/R}^1\to 0$$ which implies that $\Omega_{A/R}^i$ is canonically the quotient of $\Omega_{B/R}^i$ by $J\Omega_{B/R}^i$ and the submodule generated by $d(J)\wedge \Omega_{B/R}^{i-1}$. To define $d^i : \Omega_{A/R}^i \to \Omega_{A/R}^{i+1}$, it is enough to check that $d^i : \Omega_{B/A}\to \Omega_{B/R}^{i+1}$ maps $J\Omega_{B/R}^{i}$ and $d(J)\wedge \Omega_{B/R}^{i-1}$ to the submodule generated by $d(J)\wedge \Omega_{B/R}^{i}$ and $J\Omega_{B/R}^{i+1}$, but this is almost obvious.

It remains to check the condition on $d^{p+q}(\eta\wedge \omega)$. This trivially follows from the same condition on the $\Omega_{B/R}^i$'s.

Answer 3

Maybe a bit late, but how about this construction?
By definition, $\Omega_{A/R} = I/I^2$, where $I= \ker[A\otimes_RA\rightarrow A, a\otimes b\mapsto ab]$. We have a direct sum decomposition $(A\otimes_RA)/I^2\cong A \oplus \Omega_{A/R}$ induced by $a\otimes b\mapsto (ab, a\cdot db)$ (where $d\colon A\rightarrow \Omega_{A/R}$, $a\mapsto (1\otimes a - a\otimes 1)\bmod I^2$ is the universal derivation). Now, to get the morphism $d^1\colon \Omega_{A/R}^1\rightarrow \Omega_{A/R}^2$, consider $$f\colon A\otimes_RA\rightarrow \Omega_{A/R}^2,\quad a\otimes b\mapsto da\wedge db$$ Claim: $I^2\subseteq \ker f$. To see this, let $\sum_ia_i\otimes b_i, \sum_jx_j\otimes y_j\in I$, i. e. $\sum_ia_ib_i = \sum_jx_jy_j=0$. Since $d$ is a derivation, we have $$\begin{align} \sum_ia_i\cdot db_i = d\left(\sum_ia_ib_i\right) - \sum_ib_i\cdot da_i = -\sum_ib_i\cdot da_i. \tag{1} \end{align}$$ It follows that $$\begin{align*} f\left(\sum_{i,j}(a_ix_j)\otimes (b_iy_j)\right) &= \sum_{i,j} d(a_ix_j)\wedge d(b_iy_j)\\ &=\sum_{i,j}(a_i\cdot dx_j\wedge b_i\cdot dy_j + a_i\cdot dx_j \wedge y_j\cdot db_i) +\\ &\quad+ \sum_{i,j}(x_j\cdot da_i\wedge b_i\cdot dy_j + x_j\cdot da_i \wedge y_j\cdot db_i)\\ &= \sum_{i,j}(y_j\cdot dx_j \wedge a_i\cdot db_i + b_i\cdot da_i\wedge x_j\cdot dy_j)\\ &\stackrel{(1)}{=}\sum_{i,j}(x_j\cdot dy_j\wedge b_i\cdot da_i + b_i\cdot da_i \wedge x_j\cdot dy_j)\\ &= 0, \end{align*}$$ where we have used $\sum_{i,j}a_i\cdot dx_j\wedge b_i\cdot dy_j = \left(\sum_{i}a_ib_i\right) \cdot \left(\sum_{j}dx_j\wedge dy_j\right)=0$ and similarly $\sum_{i,j}x_j\cdot da_i\wedge y_j\cdot b_i = 0$. This shows the claim.

Now, define $d^1\colon \Omega_{A/R}^1\rightarrow \Omega^2_{A/R}$ as the restriction of $\overline{f}\colon (A\otimes_RA)/I^2\rightarrow \Omega_{A/R}^2$ to $\Omega_{A/R}$. This shows that the naive definition $d^1\colon \Omega_{A/R}^1\rightarrow \Omega_{A/R}^2$ given by $a\cdot db\mapsto da\wedge db$ is well-defined.

Edit: To get the $R$-linear map $d^n\colon \Omega_{A/R}^n\rightarrow \Omega_{A/R}^{n+1}$ (where $\Omega_{A/R}^n = \bigwedge^n \Omega_{A/R}^1$ is the $n$-th $A$-linear exterior power of $\Omega_{A/R}^1$), we define $$ d^n(x_1\wedge\dotsb\wedge x_n) := \sum_{i=1}^n (-1)^{i-1} x_1\wedge \dotsb\wedge d^1(x_i)\wedge\dotsb\wedge x_n. $$ For well-definedness, we need to check two things:
(i) For all $x_1,\dotsc,x_n\in \Omega_{A/R}^1$ with $x_i=x_j$ for some $i\neq j$ we have $d^n(x_1\wedge \dotsb \wedge x_n) = 0$.\

(ii) $d^n(x_1\wedge\dotsb \wedge (ax_{i_0}) \wedge\dotsb \wedge x_n)$ does not depend on $i_0$ for all $x_1,\dotsc,x_n\in \Omega_{A/R}^1$ and $a\in A$.

Ad (i): Let $x_1,\dotsc,x_n\in \Omega_{A/R}^1$ with $x_i= x_j$ for some $i<j$. Then \begin{align*} d^n(x_1\wedge\dotsb\wedge x_n) & = \sum_{r=1}^n (-1)^{r-1} x_1\wedge\dotsb \wedge d^1(x_r)\wedge\dotsb\wedge x_n\\ &= (-1)^{i-1} x_1\wedge\dotsb\wedge d^1(x_i)\wedge\dotsb \wedge x_n + (-1)^{j-1} x_1\wedge\dotsb\wedge d^1(x_j)\wedge\dotsb\wedge x_n\\ &= 0, \end{align*} where for the last equality, we have used that
$$ x_{i+1}\wedge\dotsb\wedge x_j = (-1)^{j-i-1} x_j\wedge x_{i+1}\wedge\dotsb\wedge x_{j-1} $$
and that $d^1(x_i)$ is of degree 2 (so that moving around $d^1(x_i)$ does not produce signs).

Ad (ii): Given $a,x_1,\dotsc,x_n,y_1,\dotsc,y_n\in A$, we compute \begin{align*} & d^n(x_1 dy_1\wedge\dotsb\wedge (ax_{i_0} dy_{i_0})\wedge\dotsb \wedge x_ndy_n)\\ =& \sum_{\substack{i=1\\ i\neq i_0}}^n (-1)^{i-1} x_1dy_1\wedge\dotsb\wedge d^1(x_idy_i)\wedge\dotsb\wedge x_ndy_n + \\ &\qquad + (-1)^{i_0-1} x_1dy_1\wedge\dotsb\wedge d(ax_{i_0})\wedge dy_{i_0}\wedge\dotsb \wedge x_ndy_n\\ =& a\cdot \sum_{i=1}^n (-1)^{i-1} x_1dy_1\wedge\dotsb\wedge d^1(x_idy_i) \wedge\dotsb\wedge x_ndy_n + \\ &\qquad + (-1)^{i_0-1} x_1dy_1\wedge\dotsb\wedge x_{i_0-1}dy_{i_0-1} \wedge da\wedge x_{i_0}dy_{i_0} \wedge \dotsb\wedge x_ndy_n\\ =& a\cdot d^n(x_1dy_1\wedge\dotsb\wedge x_ndy_n) + d(a)\wedge x_1dy_1\wedge \dotsb\wedge x_ndy_n, \end{align*} and this is independent of $i_0$.

Therefore, $d^n$ is a well-defined $R$-linear map on $\Omega_{A/R}^n$.