(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).

Clearly the bijection $\lambda^{-1}: V \to V' \otimes_{\mathbf{K'}} \mathbf{K}$, the inverse of the bijection $\lambda$ defined in no. 1, Proposition 1, maps every vector sub-K'-space W' of V' onto its image under the canonical injection $x' \mapsto x' \otimes 1$ and W'.K onto W' $\otimes_{\mathbf{K'}} \mathbf{K}$; the assertions of Proposition 2 are thus consequences of Definition 2 and § 7, no. 9, Proposition 19.

The assertion concerning sums is obvious. On the other hand, if $(W'_i)_{i \in I}$ is a family of vector sub-$\mathbf{K}'$-spaces of $V'$, then $\left( \bigcap_{i \in I} W'_i \right) \otimes_{\mathbf{K}'} \mathbf{K} = \bigcap_{i \in I} (W_i \otimes_{\mathbf{K}'} \mathbf{K})$ (§ 7, no. 7, Corollary to Proposition 14), which proves the corollary.

A basis $B$ of $V$ over $\mathbf{K}$ is said to be rational over $\mathbf{K}'$ if it consists of vectors which are rational over $\mathbf{K}'$.

If $W'$ is the vector sub-$\mathbf{K}'$-space of $V'$ generated by $B$, then $W'.\mathbf{K} = V = V'.\mathbf{K}$, whence $V' = W'$ by virtue of Proposition 2.

(i) *The mapping $f' \mapsto f' \otimes 1_{\mathbf{K}} = f'_{(\mathbf{K})}$ is a bijection of $\mathrm{Hom}_{\mathbf{K}'}(V'_1, V'_2)$ onto the set of $\mathbf{K}$-linear mappings of $V_1$ into $V_2$ which are rational over $\mathbf{K}'$; the inverse bijection associates with every $\mathbf{K}$-linear mapping $f : V_1 \to V_2$ rational over $\mathbf{K}'$ the $\mathbf{K}'$-linear mapping $f' : V'_1 \to V'_2$ with the same graph as the restriction of $f$ to $V'_1$.

(ii) *For every $\mathbf{K}$-linear mapping $f : V_1 \to V_2$ rational over $\mathbf{K}'$,

(i) Clearly with the above identifications, if $f' : V'_1 \to V'_2$ is a $\mathbf{K}'$-linear mapping, $f'_{(\mathbf{K})} = f' \otimes 1_{\mathbf{K}}$ is rational over $\mathbf{K}'$ and $f'$ is the mapping with the same graph as the restriction of $f'_{(\mathbf{K})}$ to $V'_1$. Conversely, if $f : V_1 \to V_2$ is a $\mathbf{K}$-linear mapping rational over $\mathbf{K}'$ and $f' : V'_1 \to V'_2$ has the same graph as the restriction of $f$ to $V'_1$, $f$ and $f'_{(\mathbf{K})}$ coincide on $V'_1$, which is a generating system of $V_1$ over $\mathbf{K}$, hence $f = f'_{(\mathbf{K})}$.

(ii) If $f = f' \otimes 1_{\mathbf{K}}$, then $f(V_1) = f(V'_1 \otimes_{\mathbf{K}'} \mathbf{K}) = f'(V'_1) \otimes_{\mathbf{K}'} \mathbf{K}$ and, as $f'(V'_1) \subset V'_2, f(V'_1) = f'(V'_1) \cap V'_2$ (§ 7, no. 9, Proposition 19); the formula $f^{-1}(V'_2) = V'_1 + \mathrm{Ker}(f)$ follows immediately.

This is a particular case of § 7, no. 9, Corollary to Proposition 18.

In the notation of Proposition 3, for every vector sub-$\mathbf{K}'$-space $W'_1$ of $V'_1$, $f'_{(\mathbf{K})}(W'_1 \otimes_{\mathbf{K}'} \mathbf{K}) = f'(W'_1) \otimes_{\mathbf{K}'} \mathbf{K}$; whence the assertion relating to $W_1$ (no. 2, Proposition 2). On the other hand, let $W'_2$ be a vector sub-$\mathbf{K}'$-space of $V'_2$ and let $g'$ be the canonical $\mathbf{K}'$-linear mapping $V'_2 \to V'_2/W'_2$; then $f'(W'_2) = \operatorname{Ker}(g' \circ f')$; hence, by Corollary 1,

whence the assertion relating to $W_2$.

Let $V_1, V_2$ be two right vector $\mathbf{K}$-spaces with $\mathbf{K}'$-structures $V'_1, V'_2$ respectively. It is immediate that $V'_1 \times V'_2$ is a $\mathbf{K}'$-structure on $V_1 \times V_2$, called the product of the $\mathbf{K}'$-structures $V'_1$ and $V'_2$.

Let $g$ be the mapping $x_1 \mapsto (x_1, f(x_1))$ from $V_1$ to $V_1 \times V_2$; this is a $\mathbf{K}$-linear mapping such that $\Gamma = g(V_1)$; if $f$ is rational over $\mathbf{K}'$, so is $\Gamma$ by virtue of Corollary 2 to Proposition 3. Conversely, suppose that $\Gamma$ is rational over $\mathbf{K}'$ and let it have the $\mathbf{K}'$-structure induced by that on $V_1 \times V_2$; it follows immediately from the definitions that the restrictions $p_1, p_2$ to $\Gamma$ of the projections $\mathrm{pr}_1, \mathrm{pr}_2$, are $\mathbf{K}$-linear mappings, rational over $\mathbf{K}'$ of $\Gamma$ into $V_1$ and $V_2$ respectively. As $p_1$ is bijective, its inverse mapping $q_1$ is rational over $\mathbf{K}'$ (Corollary 1 to Proposition 3) and hence so is $f = p_2 \circ q_1$.

Let H be a subset of V* whose elements are rational linear forms over K'. For all $x^* \in H$, the kernel of $x^*$ is a vector sub-K-space of V, rational over K' (no. 3, Corollary 2 to Proposition 3); the intersection of these kernels is therefore also a vector sub-K-space of V, rational over K' (no. 2, Corollary 1 to Proposition 2).

Conversely, let W be a vector sub-K-space of V rational over K', so that W is identified with $W' \otimes_{K'} K$, where $W' = W \cap V'$ (no. 2, Proposition 2). For a linear form ${x'}^* \in {V'}^*$ to be zero on W', it is necessary and sufficient that the linear form $x^* \in V^*$ which corresponds to it under (1) be zero on W, for by no. 3, Corollary I to Proposition 3,

Let H' be the orthogonal of W' in V'*; we know (§ 7, no. 5, Theorem 7) that W' is the orthogonal of H' in V'; if H is the image of H' in V* under the mapping (1), it follows from the above that W is the orthogonal of H in V, taking account of no. 7, Corollary to Proposition 14.

(2)

whose coefficients $\alpha_{\mu i}$ belong to K', every solution $(\xi_i)$ of this system consisting of elements of K is a linear combination with coefficients in K of solutions $(\xi'_i)$ of (2) consisting of elements of K'.

(ii) Given a system of linear equations

whose coefficients $\alpha_{\mu i}$ and right hand sides $\beta_\mu$ belong to $K'$, if there exists a solution to the system consisting of elements of $K$, there also exists a solution consisting of elements of $K'$.

(i) For every set S, let the right vector K-space $K_d^{(S)}$ be given the $K'$-structure $K_{d'}^{(S)}$. Let $f$ be the K-linear mapping of $K_d^{(I)}$ into $K_d^{(M)}$ mapping every vector $(\xi_i)_{i \in I}$ to the vector $(\zeta_\mu)_{\mu \in M}$ defined by $\zeta_\mu = \sum_{i \in I} \alpha_{\mu i} \xi_i$ for all $\mu \in M$. Clearly $f$ is rational over $K'$; its kernel V, which is the set of solutions in K of the system (2), is a subspace of $K_d^{(I)}$ rational over $K'$ (no. 3, Corollary 2 to Proposition 3) and hence generated by the solutions of (2) in $K'$.

(ii) We consider K as a left vector $K'$-space; there exists a $K'$-linear projector $p$ of K onto its vector subspace $K'_s$ (§ 7, no. 3, Proposition 4); if $(\xi_i)$ is a solution of (3) in K, then $\sum_{i \in I} \alpha_{\mu i} p(\xi_i) = p \left( \sum_{i \in I} \alpha_{\mu i} \xi_i \right) = p(\beta_\mu) = \beta_\mu$, which proves that $(p(\xi_i))$ is a solution of (3) in $K'$.

A ring K is called (left) faithfully flat over a subring $K'$ if Proposition 6 holds for K and $K'$; we shall study this notion in detail later (Commutative Algebra, I, § 3).

(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:

Let B be a basis of V rational over K'. Applying Theorem 2 of § 7, no. 1, to a basis of W over K, considered as a free subset of V, and the generating system the union of this free subset and B, it is seen that there exists a subset C of B such that V is the direct sum of W and the subspace $W_2$ of V generated by C. Also let $W_1$ be the subspace of V generated by $B - C$. As $B \subset V'$, clearly $W_1$ and $W_2$ are rational over K'. Moreover, for all $x \in W_1$, there exists one and only one vector $g(x)$ of $W_2$ such that $x + g(x) \in W$, since V is the direct sum of W and $W_2$; then W is the graph of g and g is K-linear since W is a vector sub-K-space of V.

Having proved this lemma, it is known that W is rational over a subfield L of K containing K' if and only if g is rational over L (no. 3, Proposition 4). The smallest field $K'(g)$ such that g is rational over $K'(g)$ is therefore also the smallest field over which W is rational.

(i) Let V be a right vector K-space with a K'-structure. For a vector $x \in V$ to be rational over L, it is necessary and sufficient that $\phi_V(x.\eta) = x.\phi(\eta)$ for all $\phi \in \mathscr{M}$ and all $\eta \in K$. For a vector sub-K-space W of V to be rational over L, it is necessary and sufficient that $\phi_V(W) \subset W$ for all $\phi \in \mathscr{M}$.

(ii) Let $V_1, V_2$ be two right vector K-spaces each with a K'-structure. For a K-linear mapping f of $V_1$ into $V_2$ to be rational over L, it is necessary and sufficient that $f(\phi_{V_1}(x_1)) = \phi_{V_2}(f(x_1))$ for all $x_1 \in V_1$ and all $\phi \in \mathscr{M}$.

We first prove assertion (i) for x. Let B be a basis of V rational over K' and write $x = \sum_{b \in B} b.\xi_b$; then, for $\phi \in \mathscr{M}$ and $\eta \in K$,

and therefore, the relations

"for all $\phi \in \mathscr{M}$ and all $\eta \in K$, $\phi_V(x.\eta) = x.\phi(\eta)$"

and

"for all $\phi \in \mathscr{M}$, all $b \in B$ and all $\eta \in K$, $\phi(\xi_b\eta) = \xi_b.\phi(\eta)$"

are equivalent. The second of these relations means that for all $b \in B$, $\xi_b \in \chi(\mathscr{M})$, which proves the first assertion of (i).

We next prove (ii). For f to be rational over L, it is necessary and sufficient that, for all $x'_1 \in V_1$ rational over K', $f(x'_1)$ is a vector in $V_2$ rational over L; this will imply that $f(x_1)$ is rational over L for every vector $x_1$ of $V_1$ rational over L, such a vector being a linear combination with coefficients in L of vectors rational over K'. The above condition is equivalent, by the first part of the argument, to the relation

which may also be written

As every element of $V_1$ is a linear combination with coefficients in K of elements of $V_1$ rational over K', condition (5) is equivalent to

for all $x_1 \in V_1$ and all $\phi \in \mathscr{M}$.

Finally, to prove the second assertion in (i), we use no. 6, Lemma 1: W is the graph of a K-linear mapping $g : W_1 \to W_2$ and W is rational over if and only if the mapping g is rational over L (no. 3, Proposition 4). By (ii), for g to be rational over L, it is necessary and sufficient that $g(\phi_{W_1}(x_1)) = \phi_{W_2}(g(x_1))$ for all $x_1 \in W_1$ and all $\phi \in \mathscr{M}$; as $\phi_v = \phi_{W_1} \times \phi_{W_2}$, the above condition means that the graph W of g is stable under $\phi_v$ for all $\phi \in \mathscr{M}$.