Proposition 1

— Suppose that the family $x = (x_i)_{i \in I}$, generates the K-algebra A. Let $(m_i)_{i \in I}$, be a family of elements of M and $(f_\lambda)_\lambda$, a family of polynomials generating the ideal $a$ of all polynomials $f \in K[X_i]_{i \in I}$ such that $f(x) = 0$. For a derivation D of A into M to exist such that $D(c \cdot 1) = \Delta(c)$ for all $c \in K$ and $D(x_i) = m_i$ for all $i \in I$ it is necessary and sufficient that

If this is so, the derivation D is unique and satisfies

Put $E = K[X_i]_{i \in I}$ and write $\varphi$ for the K-homomorphism of E onto A which maps $X_i$ to $x_i$ for all $i \in I$; we thus have $\varphi(f) = f(x)$ for all $f \in E$. We consider M as an E-module by means of the homomorphism $\varphi : E \to A$ and define a mapping $D'$ of E into M by $D'(f) = f^\Delta(x) + \sum_{i \in I} D_i f(x) \cdot m_i$ (note that the family $(D_i f)_{i \in I}$ has finite support for each $f \in E$). It is clear that $D'$ is the unique derivation of E into M extending A and mapping $X_i$ to $m_i$ for each $i \in I$.

Let D be a derivation of A into M such that $D(c \cdot 1) = \Delta(c)$ for all $c \in K$ and $D(x_i) = m_i$ for all $i \in I$. Then $D \circ \varphi$ is a derivation of E into M extending A and mapping $X_i$ to $m_i$ for all $i \in I$. We thus have $D \circ \varphi = D'$, that is, the relation (2) holds. This proves the uniqueness of D ; moreover (1) is a consequence of $f, (x) = 0$ and (2).

Conversely assume that (1) holds; in other words, we have $D'(f_\lambda) = 0$ for all $\lambda \in A$. Let $f \in a$; there exists a family with finite support $(q_\lambda)_{\lambda \in \Lambda}$ in $E$ such that $f = \sum_{\lambda \in \Lambda} q_\lambda \cdot f_\lambda$. We have

whence $D'(f) = 0$ because $f_\lambda(x)$ and $D'(f_\lambda)$ are zero for all $\lambda \in A$. Since $D'$ vanishes on $a$, there exists a $Z$-linear mapping $D$ of $A$ into $M$ such that $D' = D \circ \varphi$; clearly $D$ is the required derivation of $A$ into $M$.