Bourbaki

Theorem 1 (Steinitz)

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

Algebra, V, p. 20, § 4. Algebraically closed extensions

Let K be a field, E an algebraic extension of K and Ω\Omega an algebraically closed extension of K; then there exists a K-homomorphism of E into R.

By V, p. 13, Scholium, there exists an extension field Ω′\Omega' of Ω\Omega and a K-homomorphism u of E into Ω′\Omega'. Let x∈Ex \in E; since x is algebraic over K, u(x) is algebraic over u(K) and a fortiori over Ω\Omega (V, p. 17, Cor. 2); since Ω\Omega is algebraically closed, we thus have u(x)∈Ou(x) \in O. It follows that uu maps EE into OO.

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.