Proposition 7

(i) *Let V be a right vector K-space with a K'-structure V'. For every vector $x \in V$ (resp. every vector sub-K-space W of V), the set of subfields L of K containing $K'$ and such that $x$ (resp. W) is rational over L has a least element $K'(x)$ (resp. $K'(W)$).

(ii) Let $V_1, V_2$ be two right vector K-spaces with K'-structures $V_1', V_2'$ respectively. For every K-linear mapping f of $V_1$ into $V_2$, the set of subfields L of K containing $K'$ and such that f is rational over L has a least element $K'(f)$.

We first prove assertion (i) for a vector $x \in V$. Let B be a basis of V rational over $K'$; B is a basis of $V'$ over $K'$ and a basis of $V_L$ over L for every field L such that $K' \subset L \subset K$; for $x = \sum_{b \in B} b \xi_b$ to be rational over L, it is necessary and sufficient that the $\xi_b$ belong to L (no. 2) and hence the smallest field L with this property is the field generated by K', and the $\xi_b$ for $b \in B$.

We next prove (ii). Let $B_1, B_2$ be bases of $V_1, V_2$ respectively, which are rational over $K'$ and write, for all $b_1 \in B_1, f(b_1) = \sum_{b_2 \in B_2} b_2 \alpha_{b_2 b_1}$ (*the family $(\alpha_{b_2 b_1})$ is just the matrix of $f$ with respect to the bases $B_1$ and $B_2$; cf. § 10, no. 4*). As $B_1$ (resp. $B_2$) is a basis of $(V_1)_L$ (resp. $(V_2)_L$) over L for every field such that $K' \subset L \subset K$, for $f$ to be rational over L, it is necessary and sufficient that the $\alpha_{b_2 b_1}$ belong to L; the smallest field with this property is therefore the field generated by K' and the $\alpha_{b_2 b_1}$ for $b_1 \in B_1, b_2 \in B_2$.

Finally, to establish assertion (i) for a subspace W of V, we shall first prove the following lemma: