Let be a Lie algebra over a field of characteristic . Then, is semisimple if and only if there exists unique and simple ideals such that
Proof We prove the theorem by induction on , where the base case is clear. Suppose the statement is true for all semisimple Lie algebras with . If is simple then it is its own decomposition and we are done. So suppose otherwise and let be a proper ideal. Then is also an ideal, where To show this note that for and we havesince . So, as required. By Theorem 9.4 we know that is non-degenerate, so . This implies We now show that , which will imply that . Let , thenHenceSo is solvable by Cartan’s criterion, however is semisimple, so as desired. Thus , and from the proof we also obtain that . We now show thatWe know thatHowever since we obtain . Thus as required. The same argument shows that this is also true for . Lastly we show that is semisimple, observe thatThen note that there exists such that if and only if , so by the non-degeneracy of no such exists. So, , meaning is semisimple, while a similar argument shows that is also semisimple.
By our induction hypothesis we know since are semisimple that there exist simple ideals and such that Further we know that each , so we concludeSo is the direct sum of simple ideals as required. This shows that such a decomposition exists.
To show that our decomposition is unique, we consider an arbitrary decomposition (and our existence proof above gives us one such decomposition): , a direct sum of simple ideals. We will show that if an ideal has , then for some . Note that this is sufficient because we can then compare two decompositions and show that they are equivalent, as each component simple ideal is equal to one component simple ideal in the other decomposition.
By the Jacobi identity, . Note that this is equivalent to saying that and since ideal and closed under Lie bracket.
Now we have two cases, since is a simple ideal, and or . The first implies that commutes with and so , but is a solvable ideal as it is abelian, and solvable ideals are ( 0 ) in semisimple g . Hence .
Thence, . By our argument earlier, and so by simple, or , and if it is the latter, then it is also precisely .
Hence there exists unique such that and our uniqueness proof is done.