If A is a K-algebra, for K a ring, and is a K-derivation, then • If A has a unit 1, then D(1) = D(1^2) = 2D(1), so that D(1) = 0. Thus by K-linearity, D(k) = 0 for all k\in K. • If A is commutative, then D(x^2) = xD(x) + D(x)x = 2xD(x), and D(x^n) = nx^{n-1}D(x), by the Leibniz rule. • More generally, for any x_1, x_2, \ldots, x_n \in A, it follows by
induction that • : D(x_1x_2\cdots x_n) = \sum_i x_1\cdots x_{i-1}D(x_i)x_{i+1}\cdots x_n : which is \textstyle \sum_i D(x_i)\prod_{j\neq i}x_j if for all i, D(x_i) commutes with x_1,x_2,\ldots, x_{i-1}. • For n > 1, D^n is not a derivation, instead satisfying a higher-order Leibniz rule: :: D^n(uv) = \sum_{k=0}^n \binom{n}{k} \cdot D^{n-k}(u)\cdot D^k(v). : Moreover, if M is an A-bimodule, write :: \operatorname{Der}_K(A,M) :for the set of K-derivations from A to M. • \mathrm{Der}_K(A,M) is a
module over K. • \mathrm{Der}_K(A) is a
Lie algebra with Lie bracket defined by the
commutator: :: [D_1,D_2] = D_1\circ D_2 - D_2\circ D_1. : since it is readily verified that the commutator of two derivations is again a derivation. • There is an A-module \Omega_{A/K} (called the
Kähler differentials) with a K-derivation d:A\to\Omega_{A/K} through which any derivation D:A\to M factors. That is, for any derivation D' there is a A-module map \varphi with :: D: A\stackrel{d}{\longrightarrow} \Omega_{A/K}\stackrel{\varphi}{\longrightarrow} M : The correspondence D\leftrightarrow \varphi is an isomorphism of A-modules: :: \operatorname{Der}_K(A,M)\simeq \operatorname{Hom}_{A}(\Omega_{A/K},M) • If k\subset K is a
subring, then A inherits a k-algebra structure, so there is an inclusion :: \operatorname{Der}_K(A,M)\subset \operatorname{Der}_k(A,M) , : since any K-derivation is
a fortiori a k-derivation. == Graded derivations ==