Proposition 9

Suppose that the field $\mathbf{A}$ is maximal ordered, and that $\Phi$ is a positive form. For every integer $n > 1$, the number of elements $\theta$ of the group $\mathfrak{A}_0$ of angles of lines (resp. $\mathfrak{A}$ of angles of half-lines) such that $n\theta = 0$ is equal to $n$.

Since $\mathfrak{A}_0, S^+/H, \mathfrak{A}$ and $O^+$ are isomorphic (Prop. 3), it is enough to give the proof for $O^+$, that is to say, to show that there are exactly $n$ rotations $\varphi$ such that $\varphi^n = 1$. Now, since $\mathbf{A}$ is a maximal ordered field and $\mathbf{A}(\Phi)$ is an extension field of degree 2 of $\mathbf{A}$ (Prop. 1 a)), $\mathbf{A}(\Phi)$ is an algebraically closed field (Chapter VI, § 2, No. 6, Theorem 3). Hence the $n$-th roots of unity in $\mathbf{A}(\Phi)$ form a cyclic group of order $n$ (Chapter V, § 11, No. 1, Theorem 1). Since one has $N(u) = u \overline{u} \geqslant 0$ for every $u \in \mathbf{A}(\Phi)$, the relation $u^n = 1$ implies that one has $N(u) = 1$, hence that $u$ is a rotation (No. 1, Remark 1). This proves our assertion.