Lemma 9.10.6. Any two algebraic closures of a field are isomorphic.
Proof. Let F be a field. If M and \overline{F} are algebraic closures of F, then there exists a morphism of F-extensions \varphi : M \to \overline{F} by Lemma 9.10.5. Now the image \varphi (M) is algebraically closed. On the other hand, the extension \varphi (M) \subset \overline{F} is algebraic by Lemma 9.8.4. Thus \varphi (M) = \overline{F}. \square
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