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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: