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This is a lemma on pg 10 of Kac's Infinite Dimensional Lie Algebras. Here $A=(a_{ij})$ is an $n\times n$ matrix, $g(A)$ its Kac-Moody algebra.

Let $I_1,I_2\subset\{1,\dots,n\}$ be disjoint such that $a_{ij}=a_{ji}=0$ when $i\in I_1,j\in I_2$. Let $\beta_s=\sum_{i\in I_s} k_i^{(s)}\alpha_i$ for $s=1,2$. Suppose $\alpha=\beta_1+\beta_2$ is a root of $g(A)$. Then either $\beta_1$ or $\beta_2$ is zero.

Pf: Let $i\in I_1,j\in I_2$. It's clear that $[\alpha_i^\vee,e_j]=0$, $[\alpha_j^\vee,e_i]=0$, $[e_i,f_j]=0$, $[e_j,f_i]=0$, and one checks $[e_i,e_j]=0$, $[f_i,f_j]=0$. Let $g^{(s)}$ denote the subalgebra of $g(A)$ generated by the $e_i,f_i$ for $i\in I_s$ ($s=1,2$). So $g^{(1)}$ and $g^{(2)}$ commute. Since $g_\alpha$ is in the subalgebra generated by $g^{(1)}$ and $g^{(2)}$, we deduce $g_\alpha$ lies in either $g^{(1)}$ or in $g^{(2)}$.

The one thing I don't get, is why does $g_\alpha$ have to be in either $g^{(1)}$ or $g^{(2)}$? Since they commute, I think the generated Lie subalgebra is just $g^{(1)}\oplus g^{(2)}$, but I can't see why that forces $g_\alpha$ to only be in one factor.

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The roots of $\mathfrak{g}^{(1)}\oplus \mathfrak{g}^{(2)}$ are $\Phi^{(1)}\sqcup\Phi^{(2)}$, where $\Phi^{(s)}$ is the root system for $\mathfrak{g}^{(s)}$. In particular, since $\alpha$ is a root of $\mathfrak{g}^{(1)}\oplus \mathfrak{g}^{(2)}$, $\alpha\in\Phi^{(1)}$ or $\alpha\in\Phi^{(2)}$. In turn, this forces $\mathfrak{g}_\alpha$ to belong to either $\mathfrak{g}^{(1)}$ or to $\mathfrak{g}^{(2)}$.

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