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 $f$. $\square$

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