Proposition 13

There exists a unique k-antiderivation $d : \Omega_{A/k} \to \Omega_{A/k}$ of degree 1, square zero, which extends the derivation $d^0 : A \to \Omega^1_{A/k}$.

Let us prove the uniqueness of the antiderivation d. Since $d \circ d = 0$, one has for $y, x_1, \ldots, x_p \in A$:

The A-module $\Omega^p_{A/k}$ being generated by the elements $dx_1 \wedge \ldots \wedge dx_p$, this proves the uniqueness of d.

To prove existence, it suffices to construct a $k$-homomorphism $d^1 : \Omega_{A/k}^1 \to \Omega_{A/k}^2$ such that $d^1 \circ d^0 = 0$ and

Indeed, it then follows from III, p. 128, prop. 14 (taking into account III, p. 118, remark 2) that there exists an antiderivation $d : \Omega_{A/k} \to \Omega_{A/k}$ which coincides with $d^0$ in degree 0 and with $d^1$ in degree 1. Since $d^0$ is zero on $A$, the antiderivation $d$ is $k$-linear; since $d^1 \circ d^0 = 0$, one has $d \circ d = 0$ since $\Omega_{A/k}$ is generated as an $A$-algebra by the elements $d^0 a$ for $a \in A$.

To define $d^1$, let us recall (III, p. 133) that $\Omega_{A/k}^1$ is equal to the $A$-module $\mathfrak{J}/\mathfrak{J}^2$, where $\mathfrak{J}$ is the kernel of the multiplication $m : A \otimes_k A \to A$. Consider the $k$-linear mapping $u : A \otimes_k A \to \Omega_{A/k}^2$ defined by $u(x \otimes y) = d^0(y) \wedge d^0(x)$. We have

for $x, y$ and $a$ in $A$, whence Since $\mathfrak{J}$ is generated as a left $A$-module by the elements $(a \otimes 1 - 1 \otimes a)$ for $a \in A$, we deduce that $u(\mathfrak{J}^2) = 0$; consequently $u$ defines by restriction to $\mathfrak{J}$ and passing to the quotient a $k$-linear mapping $d^1 : \mathfrak{J}/\mathfrak{J}^2 \to \Omega_{A/k}^2$.

By setting $\xi = b \otimes 1$ in (12), with $b \in A$, we obtain $d^1(b d^0(a)) = d^0(b) \wedge d^0(a)$; it follows that $d^1 \circ d^0 = 0$ and that $d^1(c \omega) = d^0(c) \wedge \omega + c d^1(\omega)$ for $c \in A$ and $\omega = b d^0(a)$. Since $\Omega_{A/k}^1$ is generated as a $k$-module by the elements $b d^0(a)$, for $a$ and $b$ in $A$, formula (11) is satisfied for all $\omega \in \Omega_{A/k}$, which completes the proof of the proposition.

One sometimes says that the elements $\omega \in \Omega_{A/k}^p$ are the exterior differential forms of degree $p$ of $A$ over $k$, and that the antiderivation $d$ is the exterior differential of $\Omega_{A/k}$; the complex $(\Omega_{A/k}, d)$ is called the de Rham complex of A over k, and its homology is the de Rham cohomology of A over k.