Bourbaki

Definition 6

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

Algebra, chapter V, § 14. Transcendental extensions

— Let LL be an extension of a field KK and (Ei)i∈I(E_i)_{i \in I} a family of subextensions of LL. The family (Ei)i∈I(E_i)_{i \in I} is said to be algebraically free if the following condition is satisfied:

(AF) For each i∈Ii \in I let AiA_i be a subset of EiE_i which is algebraically free over KK. Then Ai∩Aj=∅A_i \cap A_j = \varnothing for i≠ji \neq j and ⋃i∈IAi\bigcup_{i \in I} A_i is algebraically free over KK.

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.