Proposition 3

Assume that W is irreducible. Let $u_1, \ldots, u_l$ be homogeneous elements of the symmetric algebra $S = S(V)$, algebraically independent over $\mathbf{R}$ and generating the algebra of elements of S invariant under W (§ 5, no. 3, Th. 3). If $p_j$ is the degree of $u_j$, the exponents of W are $p_1 - 1, \ldots, p_l - 1$.

Put $V' = V \otimes_{\mathbf{R}} \mathbf{C}$, $S' = S(V') = S \otimes_{\mathbf{R}} \mathbf{C}$, and extend the scalar product on V to a hermitian form on $V'$. If $c$ is a Coxeter transformation of W, there exists an orthonormal basis $(X_i)_{1 \leq i \leq l}$ of $V'$ consisting of eigenvectors of $c \otimes 1$ (Algebra, Chap. IX, § 7, no. 3, Prop. 4); moreover, we can assume that, for $1 \leq j \leq l$, $X_j$ corresponds to the eigenvalue $\exp \frac{2i\pi m_j}{h}$ of $c \otimes 1$. It is clear that $S'$ can be identified with the algebra $\mathbf{C}[X_1, \ldots, X_l]$, and that we can write $u_j \otimes 1 = f_j(X_1, \ldots, X_l)$, where $f_j$ is a homogeneous polynomial of degree $p_j$ in $\mathbf{C}[X_1, \ldots, X_l]$. Put $D_j = \frac{\partial}{\partial X_j}$ and $J = \det(D_k f_j)$. Recall (§ 5, no. 4, Prop. 5) that $J(X_1, \ldots, X_l)$ is proportional to the product in $S'$ of $\mathrm{Card}(\mathfrak{H})$ vectors $y_k$ of V each of which is orthogonal to a hyperplane of $\mathfrak{H}$. Since we can assume that $X_1 \notin H \otimes \mathbf{C}$ for all $H \in \mathfrak{H}$ (Remark), the $X_1$ component of each of the vectors $y_k$ is non-zero, so $J(1, 0, \ldots, 0) \neq 0$. The rule for expanding determinants now proves the existence of a permutation $\sigma$ of $\{1, 2, \ldots, l\}$ such that $(D_{\sigma(j)} f_j)(1, 0, \ldots, 0) \neq 0$ for all $j$. Since $D_{\sigma(j)} f_j$ is homogeneous of degree $p_j - 1$, the coefficient of $X_1^{p_j - 1} X_{\sigma(j)}$ in $f_j(X_1, \ldots, X_l)$ is non-zero. Now $f_j(X_1, \ldots, X_l)$ is invariant under $c \otimes 1$, and

This proves that $p_j - 1 + m_{\sigma(j)} \equiv 0$ (mod. $h$). Now $h - m_{\sigma(j)}$ is an exponent (formula (2)). Permuting the $u_j$ if necessary, we can assume that $p_j - 1 \equiv m_j$ (mod. $h$) for all $j$. Since $p_j - 1 \geq 0$ and $m_j < h$, we have $p_j - 1 = m_j + \mu_j h$ with $\mu_j$ an integer $\geq 0$. By § 5, Prop. 3, we see that

Taking into account formula (3) and Th. 1 (ii), we obtain $h \sum_{j=1}^{l} \mu_j = 0$, so $\mu_j = 0$ for all $j$, and finally $p_j - 1 = m_j$ for all $j$.