Proposition 12

Assume that R is irreducible and reduced.

(i) The ratio $\frac{(\beta|\beta)}{(\alpha|\alpha)}$ for $\alpha \in R, \beta \in R$ must take one of the values 1, 2, $\frac{1}{2}$, 3, $\frac{1}{3}$.

(ii) The set of the $(\alpha|\alpha)$ for $\alpha \in R$ has at most two elements.

Since R is irreducible, the transforms of a root by $W(R)$ generate V (no. 2, Cor. of Prop. 5). Hence, for any roots $\alpha, \beta$, there exists a root $\beta'$ such that $(\alpha|\beta') \neq 0$ and $(\beta'|\beta') = (\beta|\beta)$. By formula (9) of no. 1 and the list of no. 3, $\frac{(\beta'|\beta')}{(\alpha|\alpha)}$ takes one of the values 1, 2, $\frac{1}{2}$, 3, $\frac{1}{3}$ (recall that the system is assumed to be reduced). By multiplying $(x|y)$ by a suitable scalar, we can assume that $(\alpha|\alpha) = 1$ for certain roots and that the other possible values of $(\beta|\beta)$ for $\beta \in R$ are 2 and 3. The values 2 and 3 cannot both be attained, since in that case there would exist $\beta \in R, \gamma \in R$ such that $\frac{(\gamma|\gamma)}{(\beta|\beta)} = \frac{3}{2}$, contrary to what we have seen above.