Lemma 9.8.11. Let $E/F$ an algebraic extension of fields. Any $F$-algebra map $f : E \to E$ is an automorphism.
Proof. If $E/F$ is finite, then $f : E \to E$ is an $F$-linear injective map (Lemma 9.6.1) of finite dimensional vector spaces, and hence bijective. In general we still see that $f$ is injective. Let $\alpha \in E$ and let $P \in F[x]$ be a polynomial such that $P(\alpha ) = 0$. Let $E' \subset E$ be the subfield of $E$ generated by the roots $\alpha = \alpha _1, \ldots , \alpha _ n$ of $P$ in $E$. Then $E'$ is finite over $F$ by Lemma 9.8.6. Since $f$ preserves the set of roots, we find that $f|_{E'} : E' \to E'$. Hence $f|_{E'}$ is an isomorphism by the first part of the proof and we conclude that $\alpha $ is in the image of $f$. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: