Proposition 13

Assume that R is irreducible, non-reduced and of rank $\geq 2$.

(i) The set $R_0$ of indivisible roots is a root system in V; this system is irreducible and reduced; and $W(R_0) = W(R)$.

(ii) Let A be the set of roots $\alpha$ for which $(\alpha|\alpha)$ takes the smallest value $\lambda$. Then any two non-proportional elements of A are orthogonal.

(iii) Let B be the set of $\beta \in R$ such that $(\beta|\beta) = 2\lambda$. Then $B \neq \varnothing$, $R_0 = A \cup B$, $R = A \cup B \cup 2A$.

If $\alpha \in R - R_0$, then $\frac{1}{2} \alpha \in R$, but $\frac{1}{2} (\frac{1}{2} \alpha) \notin R$ (Prop. 8), so $\frac{1}{2} \alpha \in R_0$. This proves that $R_0$ satisfies (RS$_1$). It is clear that, for all $\alpha \in R$, $s_{\alpha, \alpha^-}(R_0) = R_0$, so $R_0$ satisfies (RS$_{II}$) and (RS$_{III}$). Since $\alpha \in R - R_0$ implies that $\frac{1}{2} \alpha \in R_0$ and since $s_\alpha = s_{\alpha/2}$, we have $W(R) = W(R_0)$. Thus $R_0$ is irreducible (Cor. of Prop. 5), and it is evidently reduced.

Since $R$ is not reduced, there exists $\alpha \in R_0$ such that $2\alpha \in R$. Since $R_0$ is irreducible and $\dim V \geq 2$, $\alpha$ cannot be proportional or orthogonal to every root. Let $\beta \in R_0$ be such that $n(\beta, \alpha) \neq 0$ and $\beta$ is not proportional to $\alpha$. Changing $\beta$ to $-\beta$ if necessary, we can assume that $n(\beta, \alpha) > 0$. Now $\frac{1}{2} n(\beta, \alpha) = n(\beta, 2\alpha) \in \mathbf{Z}$, so $n(\beta, \alpha) \in 2\mathbf{Z}$. From the list in no. 3, $n(\beta, \alpha) = 2$, $(\beta|\beta) = 2(\alpha|\alpha)$. Since $R_0$ is reduced, Prop. 12 shows that, for all $\gamma \in R_0$, either $(\gamma|\gamma) = (\alpha|\alpha)$ or $(\gamma|\gamma) = 2(\alpha|\alpha)$. Also, the above shows that, for all $\gamma \in R - R_0$, the vector $\frac{1}{2} \gamma$ is an element of $R_0$ such that $(\frac{1}{2} \gamma|\frac{1}{2} \gamma) = (\alpha|\alpha)$. Thus, $\lambda = (\alpha|\alpha)$, $B \neq \varnothing$, $R_0 = A \cup B$, and $R \subseteq A \cup B \cup 2A$; on the other hand, if $\gamma \in A$, there exists $g \in W(R)$ such that $\gamma = g(\alpha)$ (Prop. 11), so $2\gamma = g(2\alpha) \in R$; thus $2A \subseteq R$ and $R = A \cup B \cup 2A$. Finally, let $\gamma, \gamma'$ be two non-proportional elements of $A$. Then

since $\gamma$ and $\gamma'$ have the same length, so $n(\gamma, \gamma') = 0$ and $(\gamma|\gamma') = 0$.