Proof of Cartan’s criterion for solvability

In class February 26 I presented a proof of Theorem 0.1 (Cartan). Suppose V is a finite-dimensional vector space over a field k of characteristic zero, and g ⊂ gl(V) is a Lie subalgebra. Write B for the trace form B(X, Y) = def tr(XY) (X, Y ∈ g ⊂ gl(V)). Then g is solvable if and only if the radical of B| g contains the commutator subalgebra [g, g]. In other words, g is solvable if and only if tr(XY) = 0 The proof was complicated and not so clearly presented in class, and the result is not proved in the text; so these notes attempt to reproduce the argument from class, with some details and explanations added. Proof. First, the easy parts of the argument. Let K be any extension field of k. The homework for March 3 introduces extension of scalars for vector spaces, algebras, linear maps, and bilinear forms. Extension of scalars preserves dimension of vector spaces, and in particular vanishing or non-vanishing of vector spaces. It's very easy to check that some obvious natural maps define isomorphisms [g (p) ] K [g K ] (p). Consequently g is solvable if and only if g K is solvable. If T : V → W is a linear map, then ker(T) K = ker(T K). A bilinear form B on V is the same thing as a linear map from V to Hom(V, k); the radical of B is the kernel of the map. From these two facts, we deduce that [Rad(B)] K = Rad(B K).