Bourbaki

Remark

Tag 06H2, permanent. Cite it as [Bourbaki, Tag 06H2].

Algebra, chapter V, § 2. Extensions

— Let P be the prime subfield of a field E (V, p. 2); for every subset A of E, P(A)P(A) is the least subfield of E containing A. In particular if K is a subfield of E, we have P(K∪A)=K(A)P(K \cup A) = K(A). If K and K' are two subfields of E, we thus have P(K∪K′)=K(K′)=K′(K)P(K \cup K') = K(K') = K'(K); this field is the least subfield of E containing K or K', or also the upper bound of K and K' in the set of subfields of E, ordered by inclusion; we sometimes say that this field is the field generated by K and K' in E.

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

Cited by

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Cites

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Transcribed from the printed edition, 2003, Springer. Not translated.