Bourbaki

Lemma 1

Tag 05VF, permanent. Cite it as [Bourbaki, Tag 05VF].

Algebra, chapter II, § 8. Restriction of the field ofscalars in vector spaces

Let V be a right vector K-space with a K'-structure V' and W a vector sub-K-space of V. There exist two vector sub-K-spaces W1,W2W_1, W_2 of V, rational over K', such that V is the direct sum of W1W_1 and W2W_2 and, if V is identified with W1×W2W_1 \times W_2, W is the graph of a K-linear mapping g of W1W_1 into W2W_2.

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 W2W_2 of V generated by C. Also let W1W_1 be the subspace of V generated by B−CB - C. As B⊂V′B \subset V', clearly W1W_1 and W2W_2 are rational over K'. Moreover, for all x∈W1x \in W_1, there exists one and only one vector g(x)g(x) of W2W_2 such that x+g(x)∈Wx + g(x) \in W, since V is the direct sum of W and W2W_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)K'(g) such that g is rational over K′(g)K'(g) is therefore also the smallest field over which W is rational.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 1998, Springer. Not translated.