Proposition 6

Let $R_1, . . . ,R_p$ be the irreducible components of R. For $i= 1, . . . , p$, put $\mathfrak{g}_i=\mathfrak{h}_{R_i}+\mathfrak{g}^{R_i}$. Then $\mathfrak{g}_1, . . . ,\mathfrak{g}_p$ are the simple components of $\mathfrak{g}$.

The $\mathfrak{g}_i$ are ideals of $\mathfrak{g}$ (Prop. 1 (ii)). It is clear that $\mathfrak{g}$ is the direct sum of the $\mathfrak{g}_i$, hence the product of the $\mathfrak{g}_i$. Let $\mathfrak{a}$ and $\mathfrak{b}$ be complementary ideals of $\mathfrak{g}$. Then $\mathfrak{a}$ and $\mathfrak{b}$ are semi-simple and stable under ad $\mathfrak{h}$, so there exist subsets $P,Q$ of R such that $\mathfrak{a}=\mathfrak{h}_P+\mathfrak{g}^P,\mathfrak{b}=\mathfrak{h}_Q+\mathfrak{g}^Q$. Then $\mathfrak{h}_P,\mathfrak{h}_Q$ are orthogonal complements of each other in $\mathfrak{h}$ for the Killing form, so P and Q are unions of irreducible components of R. This proves that the $\mathfrak{g}_i$ are minimal ideals of $\mathfrak{g}$.