Hello,

$g$ is a complex semisimple Lie algebra.

There is the result that $U(g)$ is free over $Z(g)$.

There is another result: If $E$ is a finite dimensional representation of $g$, then $Hom(E,U(g)^{ad})$ is a free $Z(g)$-module of rank equals the multiplicity of the zero weight in $E$ (here $U(g)^{ad}$ denotes $U(g)$ as a $g$-module for the adjoint action $v\cdot u = vu-uv$).

My question is: how can one deduce the second result from the first one?

Thanks, Sasha