 Finite-dimensional division algebra over an algebraically closed field

Let $D$ be a finite-dimensional division algebra over an algebraically closed field $k$. Left multiplication by $a\in D$ has an eigenvalue $\lambda \in k$ and a nonzero eigenvector $v$. Thus $(a-\lambda )v=0$; multiplying by $v^{-1}$ gives $a=\lambda$. Every element is scalar, so $D=k$.