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 $\varphi (M) = \overline{F}$.
$\square$

## 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Your email address will not be published.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: