Proposition 1

Let $V_Q$ (resp. $V_Q^*$) be the Q-vector subspace of V (resp. $V^*$) generated by the $\alpha$ (resp. the $\alpha^{\sim}$). Then $V_Q$ (resp. $V_Q^*$) is a Q-structure on V (resp. $V^*$) (Algebra, Chap. II, § 8, no. 1). The restriction to $V_Q \times V_Q^*$ of the canonical bilinear form on $V \times V^*$ gives an identification of each of the spaces $V_Q, V_Q^*$ with the dual of the other. The set R is a root system in $V_Q$.

If $k = \mathbf{R}$, there exists a scalar product on V invariant under W(R) (Integration, Chap. VII, § 3, no. 1, Prop. 1); Lemma 2 now shows that the $\alpha^{\sim}$ generate $V^*$. By Remark 1, the $\alpha^{\sim}$ again generate $V^*$ if $k = \mathbf{Q}$. We now go to the general case. Put E = $V_Q$. By (RS_{\text{III}}), each $\alpha^{\sim}$ maps E to $\mathbf{Q}$, and so defines an element $\tilde{\alpha}$ of $E^*$. It is immediate that R is a root system in E, and that the element corresponding to $\alpha$ in $E^*$ is $\tilde{\alpha}$. By what we said above, the $\tilde{\alpha}$ generate the vector space $E^*$. Consider the canonical homomorphism $i : E \otimes_Q k \to V$, and its transpose $^t i : V^* \to E^* \otimes_Q k$. Since R generates V, $^t i$ is injective; but the image of $^t i$ contains the $\tilde{\alpha}$, so $^t i$ is surjective. From this we conclude finally that i and $^t i$ are isomorphisms. We can therefore identify V with $E \otimes k$, $V^*$ with $E^* \otimes k$, $\alpha^{\sim}$ with $\tilde{\alpha}$, and $V_Q^*$ with $E^*$. Thus, $V_Q$ (resp. $V_Q^*$) is a Q-structure on V (resp. $V^*$). The restriction to $V_Q \times V_Q^*$ of the canonical bilinear form on $V \times V^*$ can be identified with the canonical bilinear form on $E \times E^*$, hence the proposition.