= Diagonalizable adjoint operators force an algebraically closed Lie algebra to be abelian
Let $\mathfrak g$ be a finite-dimensional <Lie algebra> over an <algebraically closed field>. If every operator $\operatorname{ad}(x)$ of the <Adjoint representation of a Lie algebra> is <diagonalizable>, then $\mathfrak g$ is abelian. For if $[x,y]=\lambda y$ with $\lambda\ne0$, then $\operatorname{ad}(y)x=-\lambda y\ne0$ but $(\operatorname{ad}y)^2x=0$. This contradicts $\ker T^2=\ker T$ for a diagonalizable endomorphism $T$. Therefore every adjoint operator has only the eigenvalue zero and is diagonalizable, so every adjoint operator is zero. Hence every <Lie bracket> vanishes.
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