Nonsingular indecomposable Kac–Moody algebras are simple

By F1 choose $0\mathrm{\ne }h\in J\cap \mathfrak{h}$. F3 and nonsingularity give $\dim \mathfrak{h}=n$, so the $n$ independent simple roots form a basis of ${\mathfrak{h}}^{\ast }$. Some ${\alpha }_i(h)\mathrm{\ne }0$. The ideal property applied to $[h,e_i]={\alpha }_i(h)e_i$ gives $e_i\in J$, then $[e_i,f_i]=h_i$ gives $h_i\in J$, and $[h_i,f_i]=-2f_i$ gives $f_i\in J$.

The finite graph with edges $a_{ij}\mathrm{\ne }0$ is connected by F4: otherwise its path components supply a zero block partition, using the symmetric zero condition. For an edge from an index already obtained in step 1.1, $[h_i,e_j]=a_{ij}{e}_j$ gives $e_j\in J$, and the same two brackets give $h_j,f_j\in J$. Induction along finite paths reaches every index. The independent $h_i$ span the $n$-dimensional Cartan, so every defining generator lies in $J$, and $J=\mathfrak{g}$. Finally $[e_i,f_i]=h_i\mathrm{\ne }0$ by F2 and Cartan injectivity; hence the algebra is nonabelian.