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The function field of a curve over an algebraically closed field is a C1 field. Consequently its Brauer group is zero. This is the input for cyclic splitting by extension of constants over a finite constant field.

Over an algebraically closed field, a finite-dimensional Lie algebra is nilpotent exactly when every two-dimensional Lie subalgebra is abelian. A nonnilpotent Adjoint representation matrix has a nonzero eigenvalue $\lambda$ with eigenvector $y$, so $[x,y]=\lambda y$ generates a nonabelian two-dimensional subalgebra. Conversely a subalgebra of a nilpotent Lie algebra is nilpotent, while a nonabelian two-dimensional algebra has a basis $u,v$ with $[u,v]=v$ and a nonnilpotent adjoint matrix. Algebraic closure is what guarantees the required eigenvector.

A square matrix $U$ is unipotent when $U-I$ is nilpotent. Over an algebraically closed field, this is equivalent to all eigenvalues being one. If $j$ belongs to a nilpotent ideal of an operator algebra, $I+j$ is a unipotent matrix. This realizes the radical factor in the Levi decomposition of a quiver automorphism group.