Proof. We have seen that there exists p(x) ∈ R such that D(g(x)) =p(x) d/dx (g(x)) for all g(x) ∈ R. If f(x) ∈ R
with f(x) ∈ R1 and D(f(x)) = 0,then 0 = D(f(x)) = p(x) d/dx(f(x)).
Since R is a domain, either p(x) = 0 or d/dx (f(x)) = 0. But if f ∈ R1, then d/dx (f(x)) = 0, so it follows that p(x) = 0.
This implies that D = 0.
Note. This is not true in general. As examples of nontrivial derivations with
kernels of dimension more than one,take the inner derivations of the associative
algebra of square matrices.
Another set of known facts about derivations follows.