Proposition 9

Let $\alpha$ and $\beta$ be two non-proportional roots.

(i) The set I of integers j such that $\beta + j\alpha$ is a root is an interval $[ -q, p ]$ in $\mathbf{Z}$ containing 0. (ii) Let S be the set of $\beta + j\alpha$ for $j \in \mathrm{I}$. Then,

(iii) $p - q = -n(\beta, \alpha)$.

Clearly, $0 \in \mathrm{I}$. Let $p$ (resp. $-q$) be the largest (resp. smallest) element of I. If not all the integers in $[ -q, p ]$ belong to I, there exist two integers $r, s$ in $[ -q, p ]$ with the following properties: $s > r + 1, s \in \mathrm{I}, r \in \mathrm{I}, r + k \notin \mathrm{I}$ for $1 \leq k \leq s - r - 1$. With the notation of the Cor. of Th. 1, $(\alpha|\beta + s\alpha) \leq 0, (\alpha|\beta + r\alpha) \geq 0$, which is absurd because

This proves (i).

We have $s_{\alpha}(\beta + j\alpha) = \beta - n(\beta, \alpha)\alpha - j\alpha = \beta + j'\alpha$ with $j' = -j - n(\beta, \alpha)$. Thus $s_{\alpha}(S) \subseteq S$ and consequently $s_{\alpha}(S) = S$. Now $j \mapsto -j - n(\beta, \alpha)$ is a decreasing bijection from I to I. It follows that, $j' = -q$ when $j = p$, so that $-q = -p - n(\beta, \alpha)$. This proves (ii) and (iii).

The set S is called the $\alpha$-chain of roots defined by $\beta$, $\beta - q\alpha$ is its origin, $\beta + p\alpha$ is its end, and $p + q$ is its length.