For algebraically closed fields of characteristic p>0 Lie's theorem holds provided the dimension of the representation is less than p, but can fail for representations of dimension p. An example is given by the 3-dimensional nilpotent Lie algebra spanned by 1, x, and d/dx acting on the p-dimensional vector space k/, which has no eigenvectors. Taking the semidirect product of this 3-dimensional Lie algebra by the p-dimensional representation gives a solvable Lie algebra whose derived algebra is not nilpotent.
Proof
The proof is by induction on the dimension of and consists of several steps. The basic case is trivial and we assume the dimension of is positive. We also assume V is not zero. For simplicity, we write. Step 1: Observe that the theorem is equivalent to the statement: Step 2: Find an ideal of codimension one in. Step 3: There exists some linear functional in such that is nonzero. Step 4: is a -module. Step 5: Finish up the proof by finding a common eigenvector.
Consequences
The theorem applies in particular to the adjoint representation of a solvable Lie algebra ; 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 nilpotent matrix. By Engel's theorem, this implies that is a nilpotent Lie algebra; the converse is obviously true as well. Moreover, whether a linear transformation is nilpotent or not can be determined after extending the base field to its algebraic closure. Hence, one concludes the statement: Lie's theorem also establishes one direction in Cartan's criterion for solvability: if V is a finite-dimensional vector 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. Lie's theorem is equivalent to the statement: 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. 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 without the assumption that the base field has characteristic zero. Here is another quite useful application: 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 is the restriction of a representation of.