exterior derivation on general forms

We make use of the dual number $\epsilon$ . That means, we consider an algebra $A_\epsilon=A[\epsilon]/(\epsilon^2)$ . We define

$$E= A_\epsilon\otimes_A (\wedge \Omega^1_{B/A}).$$

We assume $\epsilon$ is odd. That means, we equip an $A$ -algebra structure on $E$ by introducing the following commutation relations.

$$\epsilon \omega =-\omega \epsilon \quad (\forall \omega \in \Omega^1_{B/A}.)$$

(An equivalent and (probably) easier way to describe $E$ is given by considering a free $B$ -module $B\epsilon$ . We then define

$$E=\wedge_B (B\epsilon \oplus \Omega^1_{B/A}).$$

This method is also valid when we deal with the interior derivation.)

Let us then define a map

$$\phi_0:B \to E$$

by the following formula.

$$\phi_0(x)=(1+\epsilon d)(x)=x+\epsilon d x.$$

We may easily see that the map $\phi_0$ is an algebra homomorphism. We regard $E$ as a $B$ -algebra via this homomorphism. We then define

$$\phi_1:\Omega^1_{B/A}\to E$$

by

$$\phi_1(\omega)=(1+\epsilon d)\omega.$$

$\phi_1$ is a $B$ -module homomorphism.

So $\phi_0,\phi_1$ together defines an algebra homomorphism

$$\phi:\otimes_A \Omega^1_{B/A} \to E.$$

For any 1-form $\omega \in \Omega^1_{B/A}$ , we have

$$\phi(\omega \otimes \omega ) =(\omega + \epsilon d \omega )\wedge(\omega +\epsilon d \omega ) =0.$$

So $\phi$ factors through the exterior algebra and define an algebra homomorphism

$$\phi_{(\bullet)}:\wedge_A \Omega^1_{B/A} \to E.$$

We decompose the homomorphism above as $\phi_{(\bullet)}=1+\epsilon d$ and obtain the exterior derivation

$$d: \Omega^k_{B/A} \to \Omega^{k+1}_{B/A}$$

for any non negative integer $k$ . It is easy to verify that the exterior derivation $d$ satisfies the rules (EXT1) and (EXT2).

The theory of exterior derivation may of course be generalized to a theory of that on a separated scheme $X$ over a scheme $S$ .