Lemma 3

(i) Let $m= (P(R) : Q(R))$. If $\varphi$ is the $m$th power of an element of $T_Q$, then $\varphi \in q(T_P)$.

(ii) If $k$ is algebraically closed, $q(T_P) = T_Q$.

There exist a basis $(\lambda_1, . . . , \lambda_l)$ of P(R) and integers $n_1\geq 1, . . . , n_l\geq 1$ such that $(n_1\lambda_1, . . . , n_l\lambda_l)$ is a basis of Q(R). We have $m=n_1. . . n_l$. Let $\psi \in T_Q$ and put $\psi (n_1\lambda_1) =t_1, . . . , \psi (n_l\lambda_l) =t_l$. For $i= 1, . . . , l$, put $m_i=$ $\prod_{j\not=i}n_j$. Let $\chi$ be the element of $T_P$ such that $\chi (\lambda_1) =t^{m_1}_1, . . . , \chi (\lambda_l) =t^{m_l}_l$. Then

so $\chi |Q(R) =\psi^m$. This proves (i). If $k$ is algebraically closed, every element of $k^*$ is the $m$th power of an element of $k^*$, so every element of $T_Q$ is the $m$th power of an element of $T_Q$; hence, (ii) follows from (i).