Proposition 1

Let V be a right vector space over K and V' a subset of V which is a vector subspace over K'. The following conditions are equivalent:

(a) The K-linear mapping $\lambda$ of $V'_{(K)} = V' \otimes_{K'} K$ into V such that $\lambda(x' \otimes \xi) = x'\xi$ for all $x' \in V', \xi \in K$, is bijective.

(b) Every K'-linear mapping $f'$ of V' into a vector K-space W can be extended uniquely to a K-linear mapping f of V into W.

(c) Every basis of V' over K' is a basis of V over K.

(d) There exists a basis of V' over K' which is also a basis of V over K.

(e) The vector K-space V is generated by V' and every subset of V' free over K' is free over K.

We know (§ 5, no. 1) that $V'_{(K)}$ has a right vector K-space structure under which $(x' \otimes \xi)\eta = x' \otimes (\xi\eta)$ ($\xi, \eta$ in K, $x' \in V'$) and that for every K'-linear mapping $f'$ of V' into a vector K-space W, there exists one and only one K-linear mapping $\bar{f}'$ of $V'_{(K)}$ into W such that $\bar{f}'(x' \otimes 1) = f'(x')$ for $x' \in V'$. If j is the canonical injection of V' into V, $\lambda$ is just the corresponding K-linear mapping $\bar{j}$. If $\lambda$ is bijective, then for every K'-linear mapping $f': V' \to W, \bar{f}' \circ \lambda^{-1}$ is the unique K-linear mapping of V into W extending $f'$; in other words, (a) implies (b). Conversely, if (b) holds, there exists in particular a K-linear mapping $\mu$ of V into $V'_{(K)}$ such that $\mu(x') = x' \otimes 1$ for all $x' \in V'$; it is immediate that $\mu \circ \lambda = 1_{V'_{(K)}}$; on the other hand, $\lambda(\mu(x')) = x'$ for all $x' \in V'$ and, as by hypothesis $j: V' \to V$ can be extended uniquely to an endomorphism of V, of necessity $\lambda \circ \mu = 1_V$, which completes the proof that (a) and (b) are equivalent.

For every basis B' of V' over K', the set B of elements $v' \otimes 1$ of $V'_{(K)}$, where $v'$ runs through B', is a basis of $V'_{(K)}$ over K (§ 5, no. 1, Proposition 4) and $\lambda(B) = B'$. For $\lambda$ to be bijective, it is necessary that the image under $\lambda$ of every basis of $V'_{(K)}$ be a basis of V over K and it is sufficient that it be so for a single basis of $V'_{(K)}$ (§ 1, no. 11, Corollary 2 to Proposition 17). This proves the equivalence of (a), (c) and (d).

As every subset of V' which is free over K' is contained in a basis of V' over K' (§ 7, no. 1, Theorem 2), (c) implies (e). Finally, suppose that (e) holds; if B' is a basis of V' over K', it is a free subset of V over K; on the other hand B' generates V' over K' and hence generates V over K by hypothesis; therefore B' is a basis of V over K, which proves that (e) implies (c).