Lemma 9.10.5. Let $F$ be a field. Let $\overline{F}$ be an algebraic closure of $F$. Let $M/F$ be an algebraic extension. Then there is a morphism of $F$-extensions $M \to \overline{F}$.

Proof. Consider the set $I$ of pairs $(E, \varphi )$ where $F \subset E \subset M$ is a subextension and $\varphi : E \to \overline{F}$ is a morphism of $F$-extensions. We partially order the set $I$ by declaring $(E, \varphi ) \leq (E', \varphi ')$ if and only if $E \subset E'$ and $\varphi '|_ E = \varphi$. If $T = \{ (E_ t, \varphi _ t)\} \subset I$ is a totally ordered subset, then $\bigcup \varphi _ t : \bigcup E_ t \to \overline{F}$ is an element of $I$. Thus every totally ordered subset of $I$ has an upper bound. By Zorn's lemma there exists a maximal element $(E, \varphi )$ in $I$. We claim that $E = M$, which will finish the proof. If not, then pick $\alpha \in M$, $\alpha \not\in E$. The $\alpha$ is algebraic over $E$, see Lemma 9.8.4. Let $P$ be the minimal polynomial of $\alpha$ over $E$. Let $P^\varphi$ be the image of $P$ by $\varphi$ in $\overline{F}[x]$. Since $\overline{F}$ is algebraically closed there is a root $\beta$ of $P^\varphi$ in $\overline{F}$. Then we can extend $\varphi$ to $\varphi ' : E(\alpha ) = E[x]/(P) \to \overline{F}$ by mapping $x$ to $\beta$. This contradicts the maximality of $(E, \varphi )$ as desired. $\square$

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