Bourbaki

Proposition 7

Tag 06KI, permanent. Cite it as [Bourbaki, Tag 06KI].

Algebra, chapter V, § 7. Separable algebraic extensions

— Suppose that KK is infinite; let AA be a commutative KK-algebra possessing only a finite number of sub-algebras (for example an etale KK-algebra, VV, p. 30, Prop. 3) and let VV be a vector subspace of AA generating AA. Then there exists x∈Vx \in V such that A=K[x]A = K[x].

Let A1,…,AnA_1, \ldots, A_n be the subalgebras of AA distinct from AA. If x∉A1∪…∪Anx \notin A_1 \cup \ldots \cup A_n, then the sub-algebra K[x]K[x] cannot equal any of the AiA_i and so must coincide with AA. Further, since VV generates AA, it is not contained in any of the subspaces AiA_i. Prop. 7 is therefore a consequence of the following lemma:

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, 2003, Springer. Not translated.