Corollary 1

For all $\lambda \in \mathfrak{h}^*$, let $\chi_{\lambda}$ be the homomorphism $z \rightarrow (\varphi (z))(\lambda )$ from Z to $k$.

(i) If $k$ is algebraically closed, every homomorphism from Z to $k$ is of the form $\chi_{\lambda}$ for some $\lambda \in \mathfrak{h}^*$.

(ii) Let $\lambda , \mu\in \mathfrak{h}^*$. Then $\chi_{\lambda}=\chi_\mu$ if and only if $\mu+\rho \in W(\lambda +\rho )$.

If $k$ is algebraically closed, every homomorphism from $\mathbf{S}(\mathfrak{h})^W$ to $k$ extends to a homomorphism from $\mathbf{S}(\mathfrak{h})$ to $k($Commutative Algebra, Chap. V, §1, no. 9, Prop. 22, and §2, no. 1, Cor. 4 of Th. 1), and every homomorphism from $\mathbf{S}(\mathfrak{h})$ to $k$ is of the form $f \rightarrow f(\lambda )$ for some $\lambda \in \mathfrak{h}^*$ (Chap. VII, App. I, Prop. 1). Hence, if $\chi$ is a homomorphism from Z to $k$, there exists (Th. 2) a $\mu\in \mathfrak{h}^*$ such that, for all $z\in Z$,

hence (i).

Let $\lambda , \mu\in \mathfrak{h}^*$ and assume that $\chi_{\lambda}=\chi_\mu$. Then, for all $z\in Z$,

in other words, the homomorphisms from $\mathbf{S}(\mathfrak{h})$ to $k$ defined by $\lambda +\rho$ and $\mu+\rho$ coincide on $\mathbf{S}(\mathfrak{h})^W$; thus, assertion (ii) follows from Commutative Algebra, Chap. V, §2, no. 2, Cor. of Th. 2.