Subjects:
Links: Lie Algebras, Module and Algebra, Graded Ring, Rings and Fields

If $A$ is an $R$-algebra over a commutative ring $R$, a derivation of $A$ is a $R$-linear map $D:A\to A$ such that $$D(ab) = (Da) b + a(Db)$$
The set of all derivations of $A$ is closed under addition and scalar multiplication and thus forms a vector space, denoted by $\text{Der}(A)$. We don't have that $\text{Der}(A)$ is an algebra since the composition of derivations isn't necessarily a derivation.

Let $D_1$ and $D_2$ be two derivations, then $D_1{D}_2-D_2{D}_1$ is a derivation, denoted as $[D_1,D_2]$, and it is called the commutator of $D_1$ and $D_2$.

Prop: For any algebra $A$, the set $\text{Der}(A)$ along with the commutator $[,]$ is a Lie algebra.

Obs: One way to remember the Jacobi identity is to notice that is equivalent to the operador $[D,\cdot ]:\text{Der}(A)\to \text{Der}(A)$ is a derivation of the algebra $\text{Der}(A)$, for each $D\in \text{Der}(A)$.

Graded Derivations

Let $A=\underset{k=0}{\overset{\mathrm{\infty }}{⨁}}{A}^k$ be a graded algebra over a field $K$. The antiderivation of a graded algebra $A$ is a $K$-linear map $D:A\to A$ such that $a\in A^k$ and $b\in A^{\ell }$, $$D(ab) = (Da) b + (-1)^k a (Db)$$If there is an integer $m$ such an antiderivation $D$ sends $A^k$ to $A^{k+m}$ for all $k$, then we say that is an antiderivation of degree $m$. By defining $A_k=0$ for $k<0$, we can extend the grading of a graded algebra $A$ to negative integers. With this extension in mind, the degree $m$ of an antiderivation can be negative.