Sum of all solvable ideals of a Lie algebra and radical

Question

Let $mathfrak{g}$ be a finite dimensional Lie algebra. I know the fact that if the ideals $mathfrak{a}$,$mathfrak{b}$ are solvable, then so is $mathfrak{a+b}$.
Now I want to show the existence of maximal solvable ideal (called "radical") of $mathfrak{g}$ by showing that the (infinite) sum over all solvable ideals in $mathfrak{g}$ is solvable. But why is this infinite sum of solvables again solvable(does the above fact apply immediately?)? Or should I prove the existence of radical in another way?

Tsemo Aristide · Accepted Answer

If $dim{cal G}=n$ is finite. Let $A_1, A_2$ two non zero solvable ideals, suppose that $A_2$ is not contained in $A_1$, $dim(A_1+A_2)>dim(A_1)>2$. Suppose that there exists $A_3$ solvable not contained in $A_1+A_2$, $dim(A_1+A_2+A_3)>dim(A_1+A_2)>3$,...
The sequence needs to stop we cannot construct $A_n$, since it will implies that $dim(A_1+A_2+...+A_n)>n>dim({cal G})$.
This implies that there exists $p$ such that every solvable ideal $B$ is contained in $R=A_1+...+A_p$; $R$ is the radical.

Sum of all solvable ideals of a Lie algebra and radical

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