algebra over an algebraically closed field of characteristic zero; thus, one can choose a basis on with respect to which consists of upper triangular matrices. It follows easily that for each , has diagonal consisting of zeros; i.e., is a strictly upper triangular matrix. This implies that is a nilpotent Lie algebra. Moreover, if the base field is not algebraically closed then solvability and nilpotency of a Lie algebra is unaffected by extending the base field to its algebraic closure. Hence, one concludes the statement (the other implication is obvious):A finite-dimensional Lie algebra over a field of characteristic zero is solvable if and only if the derived algebra is nilpotent.Lie's theorem also establishes one direction in Cartan's criterion for solvability:If V is a finite-dimensional vector space over a field of characteristic zero and a Lie subalgebra, then is solvable if and only if for every and .Indeed, as above, after extending the base field, the implication is seen easily. (The converse is more difficult to prove.) Lie's theorem (for various V) is equivalent to the statement:For a solvable Lie algebra over an algebraically closed field of characteristic zero, each finite-dimensional simple -module (i.e., irreducible as a representation) has dimension one.Indeed, Lie's theorem clearly implies this statement. Conversely, assume the statement is true. Given a finite-dimensional -module V, let be a maximal -submodule (which exists by finiteness of the dimension). Then, by maximality, is simple; thus, is one-dimensional. The induction now finishes the proof. The statement says in particular that a finite-dimensional simple module over an abelian Lie algebra is one-dimensional; this fact remains true over any base field since in this case every vector subspace is a Lie subalgebra. Here is another quite useful application:Let be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero with radical . Then each finite-dimensional simple representation is the tensor product of a simple representation of with a one-dimensional representation of (i.e., a linear functional vanishing on Lie brackets). By Lie's theorem, we can find a linear functional of so that there is the weight space of . By Step 4 of the proof of Lie's theorem, is also a -module; so . In particular, for each , . Extend to a linear functional on that vanishes on ; is then a one-dimensional representation of . Now, . Since coincides with on , we have that is trivial on and thus