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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