Start with $I_0=I$ and set $I_{k+1}=[L,I_k]$. Every term lies in $I$, and $I_k\subseteq L_{k+1}$, so the sequence eventually vanishes. Choose its last nonzero term $I_j$. Then $[L,I_j]=0$, hence $0\ne I_j\subseteq I\cap Z(L)$. Therefore $\boxed{I\cap Z(L)\ne\{0\}}$.


Solution

Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 / A / 1 / b / Solution

 Solution (source code)