Example

Let M be a K-module; it follows from Proposition 14 of no. 9 that for every S(M)-module E, the mapping D ↦ D | M defines an S(M)-module isomorphism of D_K(S(M), E) onto Hom_K(M, E); on the other hand, since E is an S(M)-module, Hom_K(M, E) is canonically isomorphic to

every K-homomorphism of M into E being uniquely expressible in the form $x \mapsto h(x \otimes 1)$, where

(II, § 5, no. 1). Let D_0 be the K-derivation $S(M) \to M \otimes_K S(M)$ whose restriction to M is the canonical homomorphism $x \mapsto x \otimes 1$; every K-derivation D : S(M) → E can therefore be written uniquely as $h \circ D_0$ with

By the uniqueness of a solution of a universal mapping problem, it is seen that there exists a unique S(M)-module isomorphism

such that $D_0 \circ \omega = d_{S(M)/K}$; in other words, for all $x \in M$, $\omega(x \otimes 1) = dx$.

In particular, if M is a free K-module with basis $(e_\lambda)_{\lambda \in L}$, $\Omega_K(S(M))$ is a free S(M)-module with basis the set of differentials $de_\lambda$. Consider in particular the case where $L = \{1, n\}$, so that S(M) is identified with the polynomial algebra $K[X_1, \ldots, X_n]$ (§ 6, no. 6); for every polynomial $P \in K[X_1, \ldots, X_n]$, we can write uniquely

with $D_i P \in K[X_1, \ldots, X_n]$ and, by virtue of the above, the mappings $P \mapsto D_i P$ are the K-derivations of $K[X_1, \ldots, X_n]$ corresponding to the coordinate forms on $\Omega_K(S(M))$ for the basis $(dX_i)$; we also write $\frac{\partial P}{\partial X_i}$ instead of $D_i P$ and this is called the partial derivative of P with respect to $X_i$.