I am trying to show that a complex Lie algebra $\mathfrak g$ is the direct sum of simple ideals iff it is semisimple.
In fact, I have already proved that it is sufficient. I am now trying to prove that $\mathfrak g$ being a direct sum of simple ideals implies that it is semisimple.
Currently, I would like to prove that if $J$ is an ideal of the direct sum of simple ideals, then $J$ is itself a direct sum of some subset of these simple ideals.
So considering $J$ to be a direct sum of $L_i$, I am able to show that if $J$ is not equal to the direct sum of all the $L_i$, then there exists $i$ such that $[L_i,J]=0$.
From this how can I conclude that $J$ is contained in the direct sum of the $L_j$s with $j\neq i$?
Best Answer
Let $L=L_1 \oplus \cdots \oplus L_n$ be the decomposition of a semisimple Lie algebra into simple ideals. Let $J$ be an ideal of $L$. For each $1\leq i\leq n$, $[JL_i]$ is an ideal of $L_i$, so either $[JL_i]=0$ or $[JL_i]=L_i$. Let $I_0$ be the subset of $\{1,...,n\}$ for which $[JL_i]=0$ and $I_1$ be the subset of $\{1,...,n\}$ for which $[JL_i]=L_i$.
We claim $J=\oplus _{i\in I_1} L_i$. Let $x\in J$. Then $x=x_1+\cdots +x_n$ with $x_i\in L_i$, $1\leq i\leq n$. We show $x_i=0$ for each $i\in I_0$. Suppose $x_i\neq0$ for some $i\in I_0$. Then $x_i\notin Z(L_i)$, because $Z(L_i)=0$ as a solvable ideal of $L$. Hence, there exists $y\in L_i$ such that $[x_i,y]\neq0$. Then $0=[x,y]=[x_1,y]+\cdots +[x_n,y]=[x_i,y]$, using $[JL_i]=0$ and $[L_jL_k]=0$ for $j\neq k$. This is contradictory, so we must have $x_i=0$ for all $i\in I_0$, which gives $x=\sum_{i\in I_1}x_i$. Therefore, $J\subseteq \oplus_{i\in I_1}L_i$. But since $[JL_i]=L_i$ implies $L_i\subseteq J$, we get $J= \oplus_{i\in I_1}L_i$.