Exercise 1, § 10

If A and B are K-algebras and $\alpha$ and $\beta$ two K-homomorphisms of A into B, an $(\alpha, \beta)$-derivation of A into B is a K-linear mapping $d$ of A into B satisfying the relation

in other words, a derivation in the sense of Definition 1 (no. 2) where $A = A' = A''$, $B = B' = B''$, $d = d' = d''$, $\mu : A \times A \to A$ is multiplication, $\lambda_1 : B \times A \to B$ is the K-bilinear mapping $(z, x) \mapsto z\alpha(x)$ and $\lambda_2 : A \times B \to B$ the K-bilinear mapping $(x, z) \mapsto \beta(x)z$.

Suppose henceforth that A and B are commutative fields and that $\alpha \neq \beta$. Show that there then exists an element $b \in B$ such that $d$ is the $(\alpha, \beta)$-derivation $x \mapsto b\alpha(x) - \beta(x)b$. (It can be assumed that $d \neq 0$. Show first that the kernel N of $d$ is the set of $x \in A$ such that $\alpha(x) = \beta(x)$ and then show that for $x \notin N$ the element $b = d(x)/(\alpha(x) - \beta(x))$ is independent of the choice of $x$.)