Show that $B^{(1)} = 0$ and $B^{(2)}$ have basis $\{[x_i,x_j]; i>j\}$

Question

Definition: A polynomial $f \in K \langle X \rangle$ is called a proper polynomial, if it is a linear combination of products of commutators:

$$f(x_1,x_2,...x_n) = \sum \alpha_{i,...,j}[x_{i_1},...,x_{i_p}]...[x_{j_1},...,x_{j_q}], \alpha_{i,...,j} \in K$$

(Assume that $1$ is a product of an emply ser of commutators). We denote by $B$ the set of all proper polynomial in $K \langle X \rangle$.

Definition: The commutator of $y$ and $z$ is $[y,z] = yz - zy$, $[y,z,w] = [[y,z],w]$, ...

Let $B^{(n)}$ be the homogeneous component of degree $n$ of B. Show that $B^{(1)} = 0$ and $B^{(2)}$ have basis $\{[x_i,x_j]; i>j\}$. (Remark: $B^{(1)}$ is the polynomial proper of degree $1$, $B^{(2)}$ the polynomial proper of degree $2$)

Answer 1

The degree of each commutator is either 0 (for constants) or at least two. The degree of a product of commutators is the product of the degrees. It follows that you cannot have a product of commutators of degree 1, and the only possibility to get a product of commutators of degree 2, is the commutator $[x_i,x_j]$.